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Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

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Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Samacheer Kalvi 11th Chemistry Alkali and Alkaline Earth Metals Textual Evaluation Solved

I. Choose The Correct Answer:

Question 1.
For alkali metals, which one of the following trends are incorrect?
(a) Hydration energy : Li > Na > K > Rb
(b) Ionization energy : Li > Na > K > Rb
(c) Density : Li < Na < K < Rb
(d) Atomic size : Li < Na < K < Rb
Answer:
(c) Density : Li < Na < K < Rb
Potassium is lighter than sodium. The correct order of density is
Li < K< Na < Rb < Cs
0.54 < 0.86 < 0.97< 1.53< 1.90 (in g cm3).

Question 2.
Which of the following statements are incorrect?
(a) Li+ has minimum degree of hydration among alkali metal cations.
(b) The oxidation state of K in KO2 is +1.
(c) Sodium is used to make Na/Pb alloy.
(d) MgSO4 is readily soluble in water.
Answer:
(a) Li+ has minimum degree of hydration among alkali metal cations.
Li+ has maximum degree of hydration among alkali metal cations.
Li+ > Na+ > K+ > Rb+ > Cs+

Question 3.
Which of the following compounds will not evolve H2 gas on reaction with alkali metals?
(a) ethanoic acid
(b) ethanol
(c) phenol
(d) none of these
Answer:
(d) none of these
Hint:
All these compounds reacts with alkali metals to evolve hydrogen gas.

Question 4.
Which of the following has the highest tendency to give the reaction Aqueous –
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals
(a) Na
(b) Li
(c) Rb
(d) K
Answer:
(b) Li.
Hint:
Hydration energy of Li+ is more and hence Li+ is stabilized in aqueous medium.

Question 5.
Sodium is stored in ………..
(a) alcohol
(b) water
(c) kerosene
(d) none of these
Answer:
(c) kerosene

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 6.
RbO2 is ………….
(a) superoxide and paramagnetic
(b) peroxide and diamagnetic
(c) superoxide and diamagnetic
(d) peroxide and paramagnetic
Answer:
(a) superoxide and paramagnetic
Hint:
RbO2 is a super oxide which contains Rb+ and O2- ions. O2- contains one unpaired electron and hence it is paramagnetic.

Question 7.
Find the wrong statement …………
(a) sodium metal is used in organic qualitative analysis
(b) sodium carbonate is soluble in water and it is used in inorganic qualitative analysis
(c) potassium carbonate can be prepared by Solvay process
(d) potassium bicarbonate is acidic salt
Answer:
(c) Potassium carbonate can be prepared by Solvay process
Hint:
Potassium carbonate cannot be prepared by Solvay process. Potassium bicarbonate is fairly soluble in water and does not precipitate out.

Question 8.
Lithium shows diagonal relationship with
(a) sodium
(b) magnesium
(c) calcium
(d) aluminium
Answer:
(b) magnesium (diagram pending)

Question 9.
In case of alkali metal halides, the ionic character increases in the order
(a) MF < MCl < MBr < MI
(b) MI < MBr < MCl < MF
(c) MI < MBr < MF < MCl
(d) none of these
Answer:
(b) MI < MBr < MCl < MF
Hint:
Ionic character (difference in electronegativity) MI < MBr < MCl < MF

Question 10.
In which process, fused sodium hydroxide is electrolysed for extraction of sodium?
(a) Castner’s process
(b) cyanide process
(c) Down process
(d) All of these
Answer:
(a) Castners process Castner’s process
NaOH ⇌ Na+ + OH
Cathode : Na+ + e → Na
Anode : 2OH → H2O + 1/2 O2 + 2e

Question 11.
The product obtained as a result of a reaction of nitrogen with CaC2 is (NEET – Phase I)
(a) Ca(CN)3
(b) CaN2
(c) Ca(CN)2
(d) Ca3N2
Answer:
(c) Ca(CN)2
Hint:
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 12.
Which of the following has highest hydration energy?
(a) MgCl2
(b) CaCl2
(c) BaCl2
(d) SrCl2
Answer:
(a) MgCl2
Hint:
The order of hydration energy of alkaline earth metal is Be2+ > Mg2+ > Ca2+ > Sr2+ > Ba2+

Question 13.
Match the flame colours of the alkali and alkaline earth metal salts in the bunsen burner
(p) Sodium – (1) Brick red
(q) Calcium – (2) Yellow
(r) Barium – (3) Violet
(s) Strontium – (4) Apple green
(t) Cesium – (5) Crimson red
(u) Potassium –  (6) Blue
(a) p – 2, q – 1, r- 4, s – 5, t- 6, u – 3
(b) p – 1, q – 2, r – 4, s – 5, t – 6, u – 3
(c) p – 4, q – 1, r – 2, s – 3, t – 5, u – 6
(d) p – 6, q – 5, r – 4, s – 3, t – 1,u – 2
Answer:
(a) p – 2, q – 1, r – 4, s – 5, t – 6, u – 3
(p) sodium – yellow (2)
(p) calcium – brick red (1)
(r) barium – apple green (4)
(s) strontium – crimson red (5)
(t) cesium – blue (6)
(u) potassium – violet (3)

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 14.
Assertion : Generally alkali and alkaline earth metals form superoxides Reason : There is a single bond between O and O in superoxides.
(a) both assertion and reason are true and reason is the correct explanation of assertion
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(d) both assertion and reason are false
Hint:
Among alkali and alkaline earth metals, K, Rb and Cs alone forms superoxides. Superoxide O2- has 3 electron bond.

Question 15.
Assertion : BeSO4 is soluble in water while BaSO4 is not
Reason: Hydration energy decreases down the group from Be to Ba and lattice energy remains almost constant.
(a) both assertion and reason are true and reason is the correct explanation of assertion
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(a) both assertion and reason are true and reason is the correct explanation of assertion

Question 16.
Which is the correct sequence of solubility of carbonates of alkaline earth metals?
(a) BaCO3 > SrCO3 > CaCO3 > MgCO3
(b) MgCO3 > CaCO3 > SrCO3 > BaCO3
(c) CaCO3 > BaCO3 > SrCO3 > MgCO3
(d) BaCO3 > CaCO3 > SrCO3> MgCO3
Answer:
(b) MgCO3 > CaCO3 > SrCO3 > BaCO3
Hint:
Solubility of carbonates decreases down the group.

Question 17.
In context with beryllium, which one of the following statements is incorrect?
(a) It is rendered passive by nitric acid
(b) It forms Be2C
(c) Its salts are rarely hydrolyzed
(d) Its hydride is electron deficient and polymeric
Answer:
(c) Its salts are rarely hydrolyzed
Hint:
Correct statement is beryllium salts are easily hydrolyzed

Question 18.
The suspension of slaked lime in water is known as (NEET Phase – II)
(a) lime water
(b) quick lime
(c) milk of lime
(d) aqueous solution of slaked lime
Answer:
(c) milk of lime
Hint:
Slaked lime Ca(OH)2. The suspension is called milk of lime and the clear solution is called lime water

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 19.
A colourless solid substance (A) on heating evolved CO2 and also gave a white residue, soluble in water. Residue also gave CO2 when treated with dilute HCl.
(a) Na2CO3
(b) NaHCO3
(c) CaCO3
(d) Ca(HCO3)2
Answer:
(b) NaHCO3
Hint:
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 20.
The compound (X) on heating gives a colourless gas and a residue that is dissolved in water to obtain (B). Excess of CO2 is bubbled through aqueous solution of B, C is formed. Solid (C) on heating gives back X. (B) is ………..
(a) CaCO3
(b) Ca(OH)2
(c) Na2CO3
(d) NaHCO3
Answer:
(b) Ca(OH)2
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals
CaO + H2O → Ca(OH)2
Ca(OH)2 + CO2 → CaCO3 + H2O

Question 21.
Which of the following statement is false ? (NEET – Phase -1)
(a) Ca2+ ions are not important in maintaining the regular beating of the heart
(b) Mg2+ ions are important in the green parts of the plants
(c) Mg2+ ions form a complex with ATP
(d) Ca2+ ions are important in blood clotting
Answer:
(a) Ca2+ ions are not important in maintaining the regular beating of the heart
Hint:
Ca2+ ion plays an important role in maintaining regular heart beat.

Question 22.
The name ‘Blue John’ is given to which of the following compounds?
(a) CaH2
(b) CaF2
(c) Ca3(PO4)2
(d) CaO
Answer:
(b) CaF2
Hint:
‘Blue john’ – CaF2 (A variety of fluorite)

Question 23.
Formula of gypsum is ………….
(a) CaSO4.2H2O
(b) CaSO4. ½2H2O
(c) 3CaSO4.H2O
(d) 2CaSO4.2H2O
Answer:
(a) CaSO4.2H2O

Question 24.
When CaC2 is heated in atmospheric nitrogen in an electric furnace the compound formed is
(a) Ca(CN)2
(b) CaNCN
(c) CaC2N2
(d) CaNC2
Answer:
(b) CaNCN
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 25.
Among the following the least thermally stable is
(a) K2CO3
(b) Na2CO3
(c) BaCO3
(d) Li2CO3
Answer:
(d) Li2CO3
Hint:
Li2CO3 is the least stable.

II. Write a brief answer to the following questions.

Question 26.
Why sodium hydroxide is much more water-soluble than chloride?
Answer:

  1. Sodium hydroxide is a stronger base whereas sodium chloride is a salt.
  2. Sodium hydroxide dissolves freely in water with the evolution of much heat on account of intense hydration.
  3. In other words, when the Na+ and OH ions break up, the OH ions are much smaller than Cl ions and are able to form a hydrogen bond with water.
  4. Thus sodium hydroxide dissolves easily in water.

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 27.
Explain what to meant by efflorescence?
Answer:

  1. Efflorescence is the formation of powdery deposit on the surface of rock as a result of loss of moisture or water on exposure to air.
  2. Efflorescence is the formation of whitish powdery deposit on the surface of rocks like gypsum in dry regions. It is formed as mineral-rich water, rises to the surface through capillary action and then evaporates.
  3. Gypsum crystals are sometimes found to occur in the form that resembles the petals of flower. This happens mostly in arid areas or desert terrains, where there is rapid loss of water. This phenomenon is called as efflorescence.

Question 28.
Write the chemical equations for the reactions involved in Solvay process of preparation of sodium carbonate.
Answer:
Solvay process:
The Solvay process is represented by the below chemical equations:
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 29.
An alkali metal (x) forms a hydrated sulphate, X2SO2. 10H2O. Is the metal more likely to he sodium (or) potassium.
Answer:
Sodium: Because hydration is favoured by high charge density cations and of the two mono positive ions, sodium is smaller and will have higher charge density. Thus, Na2SO4.10H2O is more readily formed.

Question 30.
Write balanced chemical equation for each of the following chemical reactions.
(i) Lithium metal with nitrogen gas
(ii) Heating solid sodium bicarbonate
(iii) Rubidium with oxygen gas
(iv) Solid potassium hydroxide with CO2
(v) Heating calcium carbonate
(vi) Heating calcium with oxygen
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 31.
Discuss briefly the similarities between beryllium and aluminium.
Answer:

  1. Beryllium chloride forms a dimeric structure like aluminium chloride with chloride bridges. Beryllium chloride also forms a polymeric chain structure in addition to the dimer. Both are soluble in organic solvents and are strong Lewis acids.
  2. Beryllium hydroxide dissolves in excess of alkali and gives beryllate ion and [Be(OH)4]2- and hydrogen as aluminium hydroxide which gives aluminate ion, [Al(OH)4]2-.
  3.  Beryllium and aluminium ions have a strong tendency to form complexes, BeF42-, AlF63-.
  4. Both beryllium and aluminium hydroxides are amphoteric in nature.
  5. Carbides of beryllium (Be2C) like aluminium carbide (Al4C3) give methane on hydrolysis.
  6. Both beryllium and aluminium are rendered passive by nitric acid.

Question 32.
Give the systematic names for the following:

  1. milk of magnesia
  2. lye
  3. lime
  4. caustic potash
  5. washing soda
  6. soda ash and
  7. trona.

Answer:

  1. Milk of magnesia – Mg(OH)2 – Magnesium hydroxide
  2. Lye – NaOH – Sodium hydroxide
  3. Lime – Ca(OH)2 Calcium hydroxide
  4. Caustic potash – KOH – Potassium hydroxide
  5. Washing soda – Na2CO3. 10H2O – Sodium carbonate decahydrate
  6. Soda ash – Na2CO3 – Sodium carbonate (anhydrous)
  7. Trona – NaCO3.NaHCO3.2H2O – Sodium sesqui carbonate

Question 33.
Substantiate Lithium fluoride has the lowest solubility among group one metal fluorides.
Answer:
Lithium fluoride has the lowest solubility among alkali metal fluoride due to its small size of Li+ and F ions, lattice enthalpy is much higher than that of hydration enthalpy.

Question 34.
Mention the uses of Plaster of Paris.
Answer:

  • The largest use of Plaster of Paris is in the building industry as well as plasters.
  • It is used for immobilizing the affected part of organ, where there is a bone fracture or sprain.
  • It is also employed in dentistry, in ornamental work and for making casts of statues and busts.

Question 35.
Beryllium halides are covalent whereas magnesium halides are ionic why?
Answer:
Halogens are non-metals and beryllium is also non-metal. Since non-metals always form covalent bonds with each other due to almost similar ionization potential and electronegativity. And Beryllium is smaller in size and has high polarizing power therefore, beryllium halides are covalent.

Magnesium is a metal and metals mostly form ionic bonds with non-metals due to vast differences in their ionization potential and electronegativity, therefore magnesium halides are always ionic.

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 36.
Alkaline earth metal (A), belongs to 3rd period reacts with oxygen and nitrogen to form compound (B) and (C) respectively. It undergo metal displacement reaction with AgNO3 solution to form compound (D).
Answer:

  1. An alkaline earth (A) metal belongs to third period is magnesium (Mg).
  2. Magnesium reacts with oxygen to form magnesium oxide (MgO) (B).
    Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals
  3. Magnesium reacts with nitrogen to form magnesium nitride Mg3N2 (C).
    Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals
  4. Magnesium undergoes metal displacement reaction with AgNO3 solution to form magnesium nitrate Mg(NO3)3 (D).
    Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals
    Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 37.
Write balanced chemical equation for the following processes:
(a) heating calcium in oxygen
(b) heating calcium carbonate
(c) evaporating a solution of calcium hydrogen carbonate
(d) heating calcium oxide with carbon
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 38.
Explain the important common features of group 2 elements. Important common features of group 2 elements.
Answer:

  1. Group 2 is known as alkaline earth metals. It contains soft, silver metals that are less metallic in character than the Group 1 elements. Although many characteristics are common throughout the group, the heavier metals such as Ca, Sr, Ba, and Ra are almost as reactive as the Group 1 Alkali Metals.
  2. General electronic configuration can be represented as [Noble gas] ns2 where ‘n’ represents the valence shell.
  3. All the elements in Group 2 have two electrons in their valence shells, giving them an oxidation state of +2. This enables the metals to easily lose electrons, which increases their stability and allows them to form compounds via ionic bonds.
  4. The atomic and ionic radii of alkaline earth metals are smaller than the corresponding members of the alkali metals.
  5. On moving down the group, the radii increases due to gradual increase in the number of the shells and the screening effect.
  6. Down the group the ionisation enthalpy decreases as atomic size increases. They are less electropositive than alkali metals.
  7. Compounds of alkaline earth metals are more extensively hydrated than those of alkali metals because the hydration enthalpies of alkaline earth metal ions are larger than those of alkali metal ions.

Question 39.
Discuss the similarities between beryllium and aluminium.
Answer:
Similarities between Beryllium and Aluminium:

  1. Beryllium chloride forms a dimeric structure like aluminium chloride with chloride bridges. Beryllium chloride also forms a polymeric chain structure in addition to the dimer. Both are soluble in organic solvents and are strong Lewis acids.
  2. Beryllium hydroxide dissolves in excess of alkali and gives beryllate ion and [Be(OH)4]2-and hydrogen as aluminium hydroxide which gives aluminate ion, [Al(OH)4].
  3. Beryllium and aluminium ions have a strong tendency to form complexes, BeF42-, AlF63-.
  4. Both beryllium and aluminium hydroxides are amphoteric in nature.
  5. Carbides of beryllium (Be2C) like aluminium carbide (Al4C3) give methane on hydrolysis.
  6. Both beryllium and aluminum are rendered passive by nitric acid.

Question 40.
Why alkaline earth metals are harder than alkali metals?
Answer:
1. The strength of metallic bond in alkaline earth metals is higher than alkali metals due to the presence of 2 electrons in its outermost shell as compared to alkali metals, which have only 1 electron in valence shell. Therefore, alkaline earth metals are harder than alkali metals.

2. The alkaline earth metals have greater nuclear charge and more valence electrons, thus metallic bonding is more effective. Due to this they are harder than alkali metals.

Question 41.
How is plaster of paris prepared?
Answer:
Plaster of paris is a hemihydrate of calcium sulphate CaSO4. H2O. It is obtained by heating gypsum at 393 K.
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 42.
Give the uses of gypsum.
Answer:

  1. Gypsum is used in making drywalls or plasterboards. Plasterboards are used as the finish for walls and ceilings, and for partitions.
  2. Another important use of gypsum in the production of plaster of Paris. Gypsum is heated to about 300 degrees Fahrenheit to produce plaster of Paris, which is also known as gypsum plaster. It is mainly used as a sculpting material.
  3. Gypsum is used in making surgical and orthopedic casts, such as surgical splints and casting moulds.
  4. Gypsum plays an important role in agriculture as a soil additive, conditioner, and fertilizer. It helps loosen up compact or clay soil and provides calcium and sulphur, which are essential for the healthy growth of a plant. It can also be used for removing sodium from soils having excess salinity.
  5. Gypsum is used in toothpaste, shampoos, and hair products, mainly due to its binding and thickening properties.
  6. Gypsum is a component of Portland cement, where it acts as a hardening retarder to control the speed at which concrete sets.

Question 43.
Describe briefly the biological importance of calcium and magnesium.
Answer:

  1. An adult body contains about 25 g of Mg and 1200 g of Ca. The daily requirement in the human body has been estimated to be 200-300 mg.
  2. Magnesium is the co-factor of all enzymes that utilize ATP in phosphate transfer and energy release.
  3. The main pigment for the absorption of light in plants is chlorophyll which contains magnesium.
  4. About 99% of body calcium is present in bones and teeth.
  5. Calcium plays important roles in neuromuscular function, interneuronal transmission, cell membrane integrity and blood coagulation.
  6. The calcium concentration in plasma is regulated at about 100 mgL-1. It is maintained by two hormones: calcitonin and parathyroid hormone.
  7. Deficiency of magnesium results into convulsion and neuromuscular irritation.
  8. 2% of adult weight is made up of calcium. Calcium phosphate is present in teeth and Calcium carbonate is present in bones. They make the teeth and bone hard.
  9. Water in the human body such as inside the cell and in the blood contain dissolved calcium ions. These ions are involved in making muscles move and in sending electricity around the brain and along the nerves.
  10. Magnesium is an essential element in both plant and animal life.

Question 44.
Which would you expect to have a higher melting point, magnesium oxide or magnesium fluoride? Explain your reasoning.
Answer:

  • Magnesium fluoride – 1263°C
  • Magnesium oxide – 2852°C
  • The strength of ionic bonds usually depends on two factors – ionic radius and charge. Mg2+ and O2- have charges of +2 and -2 respectively. This is larger than the charge of other ions.
  • Magnesium ions and oxygen ions also have a small ionic radius.
  • Oxygen ion is smaller than fluoride
  • The smaller the ionic radii, the smaller the bond length and the stronger the bond. Therefore the ionic bond between magnesium and oxygen is very strong.

Samacheer Kalvi 11th Chemistry Alkali and Alkaline Earth Metal Additional Questions Solved

I. Choose the correct answer

Question 1.
The reducing property of alkali metals follows the order
(a) Na < K < Rb < Cs < Li
(b) K < Na < Rb < Cs < Li
(c) Li < Cs < Rb < K < Na
(d) Rb < Cs < K < Na < Li
Answer:
(a) Na < K < Rb < Cs < Li

Question 2.
The general electronic configuration of alkali metals is ………….
(a) [noble gas] ns2
(b) [noble gas] ns1
(c) ns2 np6
(d) ns2(n-1)d1-10
Answer:
(b) [noble gas] ns1

Question 3.
Li does not resemble other alkali metals in which of the following property?
(a) Li2CO3 decomposes into oxides while other alkali carbonates re thermally stable
(b) LiCl is predominantly covalent
(c) Li3N stable
(d) All of the above
Answer:
(d) All of the above

Question 4.
The half-life period of francium is ………….
(a) 21 days
(b) 21 years
(c) 2.1 minutes
(d) 21 minutes
Answer:
(d) 21 minutes

Question 5.
The alkali metal used in photoelectric cells is
(a) Na
(b) Cs
(c) Rb
(d) Fr
Answer:
(b) Cs

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 6.
The metal present in deposits of nitre is ……………
(a) lithium
(b) potassium
(c) rubidium
(d) francium
Answer:
(b) potassium

Question 7.
Be2C + 4H2O → 2X + CH4
X + 2HCl + 2H2O → Y
X and Y formed in the above two reactions is
(a) BeCO3 and Be(OH)2 respectively
(b) Be(OH)2 and BeCl2 respectively
(c) Be(OH)2 and [Be(OH)4]Cl2 respectively
(d) [Be(OH)4]2- and BeCl2 respectively
Answer:
(c) BQ(OH)2 and [Be(OH)4]Cl2 respectively

Question 8.
Which of the following are stored under oil?
(a) Alkali metals
(b) Coinage metals
(c) Noble metals
(d) Phosphorous
Answer:
(a) Alkali metals

Question 9.
Magnesium burns in the air to give
(a) MgO
(b) MgCO3
(c) MgCO3
(d) MgO and Mg3N2
Answer:
(d) MgO and Mg3N2

Question 10.
The most common oxidation state of alkali metals is ………….
(a) +1
(b) +2
(c) +3
(d) +5
Answer:
(a) +1

Question 11.
The word ‘alkali’ used for alkali metals indicates
(a) ashes of plants
(b) metallic luster
(c) soft metals
(d) reactive metals
Answer:
(a) ashes of plants

Question 12.
Which of the following salt is more soluble?
(a) NaClO4
(b) LiClO4
(c) CsBr
(d) KI
Answer:
(b) LiClO4

Question 13.
Select the correct statement?
LiOH > NaOH > KOH > RbOH
Li2CO3 > Na2CO3 > K2CO3 > Rb2CO3
(a) Solubility of alkali hydroxides is in order
(b) Solubility of alkali carbonates is in order
(c) both are correct
(d) None is correct
Answer:
(b) Solubility of alkali carbonates is in order

Question 14.
Which one of the following is less soluble in water?
(a) LiC
(b) NaCl
(c) KCl
(d) CsI
Answer:
(a) LlCl

Question 15.
Which of the following ions form a hydroxide highly soluble in water?
(a) Ni2+
(b) K2+
(c) Zn2+
(d) Al3+
Answer:
(b) K2+

Question 16.
Consider the following statements.
(1) Lithium does not have d-orbitais.
(ii) Lithium carbonate is more soluble than sodium carbonate in water.
(iii),The second ionization enthalpy of alkali metals are zero.
Which of the above statements is/are not correct?
(a) (i) only
(b) (ii) only
(c) (ii) and (iii)
(d) (i), (ii) and (iii)
Answer:
(c) (ii) and (iii)

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 17.
The carbide of which of the following metals on hydrolysis gives allylene or propyne?
(a) Be
(b) Ca
(c) Al
(d) Mg
Answer:
(d) Mg

Question 18.
Which colour is produced when alkali metals dissolved in liquid ammonia?
(a) Red
(b) Green
(c) Blue
(d) Violet
Answer:
(c) Blue

Question 19.
Consider the following statements.
(i) Superoxides of alkali metals are diamagnetic.
(ii) Superoxides of alkali metals are blue in colour.
(iii) Superoxides of alkali metals are paramagnetic.
Which of the above statements is/are not correct?
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (i) and (ii)
Answer:
(d) (i) and (ii)

Question 20.
Which of the following is known as a variety of gypsum?
(a) CaCO3
(b) CaSO4
(c) plaster of paris
(d) gypsum
Answer:
(d) gypsum

Question 21.
The colour produced by potassium when burnt in Bunsen flame is …………
(a) red
(b) blue
(c) green
(d) lilac
Answer:
(d) lilac

Question 22.
Which one of the following is a radioactive element of alkali metal?
(a) Cesium
(b) Francium
(c) Potassium
(d) Sodium
Answer:
(b) Francium

Question 23.
Which of the following ions perform important biological functions in maintenance of the ion balance and nerve impulse conduction?
(a) Li+, Rb+
(b) Na+, K+
(c) Cs+,Fr+
(d) Rb+, Cs+
Answer:
(b) Na+, K+

Question 24.
The colour of potassium salt in flame is
(a) Crimson red
(b) Lilac
(c) Blue
(d) Yellow
Answer:
(b) Lilac

Question 25.
Which of the following ions arc more responsible for transmission of nerve signal?
(a) Li+
(b) Rb+
(c) Cs+
(d) K+
Answer:
(d) K+

Question 26.
Which of the following fruits contain maximum of potassium?
(a) Grapes
(b) Potatoes
(c) Bananas
(d) Mangoes
Answer:
(c) Bananas

Question 27.
Which of the following element forms monoxide and peroxide?
(a) Lithium
(b) Potassium
(c) Rubedium
(d) Sodium
Answer:
(d) Sodium

Question 28.
Among the following, which is the fifth most abundant element?
(a) Beryllium
(b) Barium
(c) Radium
(d) Calcium
Answer:
(d) Calcium

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 29.
Celestite and strontianite are the ores of ………..
(a) cesium
(b) strontium
(c) magnesium
(d) barium
Answer:
(b) strontium

Question 30.
The products obtained on the reaction of Na2O2 with water are
(a) NaOH and H2O
(b) NaOH and H2O2
(c) Na2O and H2O2
(d) NaOH, Na2O
Answer:
(c) Na2O and H2O2

Question 31.
Which one of the following gives green spark in fire works?
(a) Magnesium chloride
(b) Sodium chloride
(c) Barium bromide
(d) Potassium iodide
Answer:
(a) Magnesium chloride

Question 32.
Statement – 1:
LiF has low solubility in water.
Statement – 2:
LiF has low lattice enthalpy.
In the above statements
(a) 1 alone is correct.
(b) Both 1 and 2 are correct
(c) 2 alone is correct
(d) Both 1 and 2 are incorrect
Answer:
(b) Both 1 and 2 are correct

Question 33.
Copper chloride produces colour in fire works.
(a) red
(b) green
(c) blue
(d) yellow
Answer:
(c) blue

Question 34.
The ammonia used in the Solvay proves- recovered by using
(a) calcium chloride
(b) Calcium hydroxide
(c) calcium carbonate
(d) calcium oxide
Answer:
(b) Calcium hydroxide

Question 35.
The most common oxidation state of alkaline earth metals is ……………
(a) +4
(b) +2
(c) + 1
(d) +3
Answer:
(b) +2

Question 36.
Sodium carbonate decahydrate on heating ab 373K gives
(a) Na2CO3 .3H2O
(b) N2CO3 .5H2O
(c) Na2CO3
(d) Na2CO3 .H2O
Answer:
(c) Na2CO3

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 37.
Hydroxides of beryllium are in nature.
(a) neutral
(b) basic
(c) acidic
(d) amphoteric
Answer:
(d) amphoteric

Question 38.
The product obtained on saturating a solution o sodium carbonate with carbon dioxide i,-.
(a) sodium bicarbonate
(b) sodium hydroxide
(c) sodium chloride
(d) sodium peroxide.
Answer:
(a) sodium bicarbonate

Question 39.
Which one of the following alkaline earth metal is not readily attacked by acids?
(a) Magnesium
(b) Calcium
(c) Beryllium
(d) Strontium
Answer:
(c) Beryllium

Question 40.
The used in baking cakes, pastries, etc., is
(a) sodium chloride
(b) sodium carbonate
(c) sodium bicarbonate
(d) sodium hydroxide
Answer:
(c) sodium bicarbonate

Question 41.
Which one of the following is used to build the beam pipe in accelerators?
(a) Be
(b) Ca
(c) Mg
(d) Sr
Answer:
(a) Be

Question 42.
Fluorapatite is the ore of
(a) magnesium
(b) beryllium
(c) potassium
(d) calcium
Answer:
(d) calcium

Question 43.
Which metal is used in photoengrave plates in printing industry?
(a) Co
(b) Pt
(c) Zn
(d) Mg
Answer:
(d) Mg

Question 44.
Strontium nitrate give _______ colour in fire works.
(a) violet
(b) bright red
(c) green
(d) orange
Answer:
(b) bright red

Question 45.
Which is used in dehydrating oils?
(a) Calcium
(b) Magnesium
(c) Beryllium
(d) Radium
Answer:
(a) Calcium

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 46.
Correctly match List-I and List-II using the code given below the list.
List-I
A. Beryllium
B. Calcium
C. Magnesium
D. Barium

List-II
1. Sacrificial anode
2. X-ray tube radiation window
3. Scavenger to remove oxygen in TV
4. Getter in vacuum tubes
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 47.
Correctly match List-I and List-II using the code given below the list.
List-I
A. Beryllium
B. Magnesium
C. Calcium
D. Strontium

List-II
1. Cement
2. Dating of rocks
3. X-ray detector
4. Missile construction
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 48.
Correctly match the list-I and List-II using the code given below the list.
List-I
A. Radium
B. Barium
C. Strontium
D. Calcium

List-II
1. Dehydration of oils
2. Aircraft and watches
3. Deoxidiser in copper refining
4. Radioactive tracer
https://samacheerkalvi.guru/samacheer-kalvi-11th-chemistry-solutions-chapter-4/
Answer:
https://samacheerkalvi.guru/samacheer-kalvi-11th-chemistry-solutions-chapter-4/

Question 49.
Consider the following statements.
(i) BeO is basic.
(ii) MgO is weakly basic.
(iii) BaO is strongly acidic.
Which of the above statements is/are not correct?
(a) (i) only
(b) (ii) only
(c) (ii) and (iii)
(d) (i) and (iii)
Answer:
(d) (i) and (iii)

Question 50.
______ is a major component of bones and teeth.
(a) Na
(b) Be
(c) Ca
(d) Mg
Answer:
(c) Ca

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 51.
Correctly match the list-I and List-II using the code given below the list.
List-I
A. Quick lime
B. Calcium hydroxide
C. Gypsum
D. Plaster of Paris

List-II
1. Casts of statues
2. Drying agent
3. White washing
4. Tooth paste
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 52.
Correctly match the list-I and List-II using the code given below the list.
List-I
A. CaO
B. Ca(OH)2
C. CaSO4 .2H2O
D. CaSO4 .½2H2O
List-II
1. Plaster of Paris
2. Quick lime
3. Slaked lime
4. Gypsum
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 53.
Which one of the following is the formula of limestone?
(a) CaO
(b) Ca(OH)2
(c) CaCO3
(d) CaCO3.MgCO3
Answer:
(c) CaCO3

Question 54.
Which one of the following is used in purification of sugar and as drying agent?
(a) Ca(OH)2
(b) MgSO4.7H2O
(c) CaSO4.2H2O
(d) CaO
Answer:
(d) CaO

Question 55.
Which one of the following is named as bleaching powder?
(a) CaCl2
(b) CaOCl
(c) Ca(OCl)2
(d) Ca(HCO3)2
Answer:
(c) Ca(OCl)2

Question 56.
Which one of the following is known as natural insulator?
(a) FeSO4.7H2O
(b) NaCO4.10H2O
(c) CaSO4.2H2O
(d) CaSO4.’/2H2O
Answer:
(c) CaSO4.2H2O

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 57.
Which one of the following is called ornamental stone?
(a) Alabaster
(b) Plaster of Paris
(c) Limestone
(d) Gypsum plaster
Answer:
(a) Alabaster

Question 58.
Which one of the following is used in toothpaste, shampoo and hair products?
(a) Plaster of Paris
(b) Limestone
(c) Quick lime
(d) Gypsum
Answer:
(d) Gypsum

Question 59.
Which one of the following plays an important role in agriculture as a soil additive, conditioner and fertilizer?
(a) Epsurn
(b) Gypsum
(c) Quick lime
(d) Salt petre
Answer:
(b) Gypsum

Question 60.
Which is used to treat upset stomach and eczema?
(a) MgSO4.7H2O
(b) FeSO4.7H2O
(c) CaSO4.2H2O
(d) 2CaSO4H2O
Answer:
(c) CaSO4.2H2O

Question 61.
Consider the following statements.
(i) Gypsum is used in making surgical and orthopedic casts.
(ii) Calcium nitrate acts a coagulator in making tofu.
(iii) Gypsum plays an important role in soap making.
Which of the above statements is/are not correct.
(a) (i) only
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (ii) only
Answer:
(b) (ii) and (iii)

Question 62.
About 393K, when Plaster of Paris is heated, it forms
(a) burnt alum
(b) dead burnt plast
(c) gypsum plaster
(d) alabaster
Answer:
(b) dead burnt plast

Question 63.
Which of the following is used in dentistry, ornamental works and making casts of statues?
(a) CaSO4.2H2O
(b) CaSO4.½H2O
(c) CaO
(d) Ca(OH)2
Answer:
(b) CaSO4.½H2O

Question 64.
Which one of the following metal act as co-factor in phosphate transfer of ATP by enzymes?
(a) Calcium
(b) Beryllium
(c) Magnesium
(d) Sodium
Answer:
(c) Magnesium

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 65.
The main pigment in plants is chlorophyll which contains ……………
(a) iron
(b) calcium
(c) barium
(d) magnesium
Answer:
(d) magnesium

Question 66.
Consider the following statements.
(i) 99% of body calcium is present in bones and teeth.
(ii) The calcium concentration in plasma is regulated at 10 mg L-1.
(iii) Calcium plays an important role in neuromuscular function, interneuronal transmission and in blood coagulation.
Which of the above statements is/are not correct?
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (i), (ii) and (iii)
Answer:
(b) (ii) only

Question 67.
Correctly match the list-I and list-Il using the code given below the list,
List-I
A. Chlorophyll
B. Bones
C. Dentistry
D. Cement
List-II
1. Plaster of paris
2. Gypsum
3. Magnesium
4. Calcium.
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 68.
Which one of the following is the most common alkaline earth metal found in the human body?
(a) Beryllium
(b) Magnesium
(c) Barium
(d) Calcium
Answer:
(d) Calcium

Question 69.
which alkaline earth metal do not import colour to a non-luminous flame?
(a) Beryllium
(b) Calcium
(c) Magnesium
(d) Barium
Answer:
(a) Beryllium

Question 70.
Statement-I: Alkali metals arc very soft metals.
Statement-II: Since the atoms of alkali metals have bigger kernels and smaller number ot valence electrons, the metallic bonds in them are very weak and hence they arc soft.
(a) Statements-I and II arc correct but statement-II is not the correct explanation of statement-I.
(b) Statements-I and II are correct and statement-II is the correct explanation of statement-I.
(c) Statement-I is correct but statement-lI is wrong.
(d) Statement-I is wrong but statement-II is correct.
Answer:
(b) Statements-I and II are correct and statement-II is the correct explanation of statement-I.

Question 71.
Statement-I: BeCl2 is soluble in organic solvent.
Statement-II: Since BeCl2 is a covalent compound, it is soluble in organic solvent.
(a) Statements-I and II arc correct and statement-II is the correct explanation of statement-I.
(b) Statements-I and II are correct but statement-II is not the correct explanation of statement-I.
(c) Statement-I is wrong but statement II is correct.
(d) Statement-I is correct but statement-II is wrong.
Answer:
(a) Statements-I and II are correct and statement-II is the correct explanation of statement-I.

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 72.
Which one of the following is more basic?
(a) Ca(OH)2
(b) Mg(OH)2
(c) NaOH
(d) Al(OH)3
Answer:
(c) NaOH

Question 73.
Statement-I: Cesium is considered as the most electropositive element.
Statement-II: Due to its lowest ionization energy, cesium is considered as the most electropositive element.
(a) Statements-I and II are correct and statement-II is the correct explanation of statemcnt-I.
(b) Statements-I and II are correct but statement-II is not the correct explanation of statement-I.
(c) Statement-I is correct but statement-II is wrong.
(d) Statement-I is wrong but statement-II is correct.
Answer:
(a) Statements-I and II are correct and statement-II is the correct explanation of statement-I.

Question 74.
The reducing property of alkali metals follows the order
(a) Na<K<Rb<Cs<Li
(b) K<Na<Rb<Cs<Li
(c) Li<Cs<Rb<K<Na
(a) Rb<Cs<K<Na<Li
Answer:
(a) Na<K<Rb<Cs<Li

Question 75.
Which of the following is the least thermally stable?
(a) MgCO3
(b) CaCO3
(c) SrCO3
(d) BeCO3
Answer:
(d) BeCO3

Question 76.
Which of the following is not a peroxide?
(a) KO2
(b) CrO5
(c) Na2O2
(d) BaO2
Answer:
(a) KO2

Question 77.
Which of the following is used in photoelectric cells?
(a)Na
(b) K
(e) Li
(d) Cs
Answer:
(d) Cs

Question 78.
When caesìum salt is subjected to flame test, the colour produced is
(a) lilac
(b) yellow
(c) blue
(d) crimson red
Answer:
(c) blue

Question 79.
Match the list-I and List-II using the correct code given below the list.
List-I
A. Lithium
B. Sodium
C. Potassium
D. Rubidium
List-II
1. Lilac
2. Reddish yiolet
3. Crimson red
4. Yellow
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 80.
Which of the following alloy is used in making white metal bearings for motor engines?
(a) Lithium + magnesium
(b) Lithium + lead
(c) Lithium + aluminium
(d) Lithium + copper
Answer:
(b) Lithium + lead

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 81.
Lithium aluminium alloy is used in making
(a) armour plates
(b) white metal bearings
(c) electrochemical cell
(d) aircraft parts
Answer:
(d) aircraft parts

Question 82.
Which of the following is used in making armour plates?
(a) Lithium + magnesium
(b) Lithium + aluminium
(c) Lithium + lead
(d) Sodium + lithium
Answer:
(a) Lithium + magnesium

Question 83.
Which metal is used in making electrochemical cells?
(a) Caesium
(b) Lithium
(c) Calcium
(d) Barium
Answer:
(b) Lithium

Question 84.
Which of the following is used as a coolant in fast breeder nuclear reactor?
(a) Liquid ammonia
(b) Liquid helium
(c) Liquid Na metal
(d) Solid CO2
Answer:
(c) Liquid Na metal

Question 85.
Which of the following is an excellent absorbent of carbon dioxide?
(a) K2 CO3
(b) CaCO3
(c) NaOH
(d) KOH
Answer:
(d) KOH

Question 86.
Which of the following is used in devising photoelectric cells?
(a) Li
(b) Cs
(c) Na
(d) K
Answer:
(b) Cs

Question 87.
The formula of washing soda is ……………..
(a) Na2 CO3
(b) NaHCO3
(c) Na,CO3 .10H2 O
(d) Ca(HCO3)2
Answer:
(c) Na2CO3 .10H2O

Question 88.
Match the list-I and List-II using the correct code given below the list.
List-I
A. Na2CO3
B. Na2CO3 .10H2O
C. NaHCO3
D. NaOH
List-II
1. Caustic soda
2. Baking soda
3. Soda ash
4. Washing soda
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Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 89.
Which one of the following is used for mercerising cotton fabrics?
(a) KOH
(b) NaOH
(c) Na2CO3
(d) NaHCO3
Answer:
(b) NaOH

Question 90.
Which one of the following is used in fire extinguishers?
(a) Washing soda
(b) Soda ash
(c) Baking soda
(d) Caustic soda
Answer:
(c) Baking soda

Question 91.
Assertion (A): Sodium hydrogen carbonate is used in baking cakes and pastries.
Reason (R): On heating sodium hydrogen carbonate, liberates bubbles of CO2  leaving holes in cakes and making them light and fluffy.
(a) both (A) and (R) are correct and (R) is the correct explanation of (A)
(b) both (A) and (R) are correct but (R) is not the correct explanation of(A)
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct
Answer:
(a) both (A) and (R) are correct and (R) is the correct explanation of (A)

Question 92.
Match the List-I and List-II using the correct code given below the list.
List-I
A. Manufacture of soap
B. Mild antiseptic
C. Softening of hard water
D. Coolant in nuclear reactor
List-II
1. Na2 CO3 . 10H2 O
2. Liquid Na metallic
3. NaOH
4. NaHCO3
https://samacheerkalvi.guru/samacheer-kalvi-11th-chemistry-solutions-chapter-4/
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Samacheer Kalvi 11th Chemistry Alkali and Alkaline Earth Metal 2-Mark Questions

Question 1.
What are s-block elements?
Answer:
The elements belonging to the group 1 and 2 in the modem periodic table are called s-block elements. The elements belonging to these two groups are commonly known as alkali and alkaline earth metals respectively.

Question 2.
Why group 1 elements are called alkali metals?
Answer:
Group I elements form strong hydroxides on reaction with water which are strong alkaline in nature. So, group 1 elements are called alkali metals.

Question 3.
Why does the ionization enthalpy of alkali metals decreases in a group?
Answer:
Alkali metals have the lowest ionisation enthalpy compared to other elements present in the respective period. As we go down the group, the ionisation enthalpy decreases due to the increase in atomic size. In addition, the number of inner shells also increases, which in turn increases the magnitude of screening effect and consequently, the ionisation enthalpy decreases down the group.

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 4.
Alkali metals are stored under oil. Give reason.
Answer:
Alkali metals are so reactive and they have to be stored under oil. Because when they are kept in air, they will bum immediately.

Question 5.
Name the list of elements present in alkali metal group. What is the general electronic configuration of them?
Answer:
Lithium, sodium, potassium, rubidium, caesium and francium are the elements present in alkali metal group. Their general electronic configuration is [noble gas] ns1 .

Question 6.
Why is lithium salts are more soluble than the salts of other metals of group-1?
Answer:
Lithium salts are more soluble than the salts of other metals of group 1. eg., LiClO4 is up to 12 times more soluble than NaClO4. KClO4, RbClO4, and CsClO4 have solubilities only 10-3 times that of LiClO4. The high solubility of Li salts is due to strong solvation of small size of Li+ ion.

Question 7.
The second ionization enthalpy of alkali metals are very high. Give reason.
Answer:
The removal of one electron from the alkali metals causes the formation of monovalent cations having very stable electronic configuration. Therefore it becomes very difficult to remove the second electron from the stable noble gas configuration, giving rise to very high second ionization energy values.

Question 8.
Why does the solubility of carbonates and bicarbonates decrease in a group?
Answer:
All the carbonates and bicarbonates are soluble in water and their solubilities increase rapidly on descending the group. This is due to the reason that lattice energies decrease more rapidly than their hydration energies on moving down the group.

Question 9.
Why lithium has anomalous behaviour than other elements ¡n the same group?
Answer:
The anomalous behaviour of lithium is due to the exceptionally small size of the atom and high polarizing power, which is a ratio of charge to radius and hydration energy.

Question 10.
What is washing soda?
Answer:
Sodium carbonate, commonly known as washing soda, crystallizes as decahydrate which is white in colour.

Question 11.
Lithium forms monoxide with oxygen whereas sodium forms peroxide with oxygen. Why?
Answer:

  • The fact that a small cation can stabilize a small anion and a large cation can stabilize a large anion explains the formation and stability of the oxides.
  • The size of Li+  ion is very small and it has a strong positive field around it. It can combine with only small anion, O2- ion, resulting in the formation of monoxide Li2O.
  • The Na ion is a larger cation and has a weak positive field around it and can stabilize a bigger peroxide ion, O22- or [-O-O-]2- resulting in the formation of peroxide Na2O2.

Question 12.
How does Lithium show similar properties with magnesium in its chemical behavior?
Answer:

  1. Both react with nitrogen to form nitrides.
  2. Both react with oxygen to give monoxides.
  3. Both the element have a tendency to form covalent compounds.
  4. Both can form complex compounds.

Question 13.
Alkali metal hydrides are strong reducing agents. Prove this statement.
Answer:
The decrease in ionization enthalpy down the group permits easy availability of electrons to forms H ions. So, the hydrides behave as reducing agent. Their reducing nature increases down the group.

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 14.
Write the chemical formula of the following compounds.
(a) Chile saltpeter
(b) marble
(c) Brine
Answer:
(a) Chile salt petre – NaNO3
(b) marble – CaCO3
(c) Brine – NaCl

Question 15.
Explain the action of sodium with water.
Answer:
Sodium reacts so rapidly with water with the evolution of heat. The metal whizzes around the surface of water. The hydrogen gas liberated may catch fire giving yellow coloured flame
because of sodium.
2Na + 2H2O → 2NaOH + H2↑ + heat

Question 16.
(a) Lithium Iodide is more covalent than Lithium fluoride
(b) Lattice enthalpy of LiF is maximum among all the alkali metals halides. Explain.
Answer:
(a) According to Fagan’s rule, Li+ ion can polarise I ion more than the F ion due to the bigger size of the anion. Thus, LiI has a more covalent character than LiF.

(b) Smaller the size (internuclear distance), more is the value of Lattice enthalpy since the internuclear distance is expected to be least in the LiF.

Question 17.
LiCl is soluble in water whereas LiBr and LiI are soluble In organic solvent. Give reason.
Answer:
Lithium chloride (LiCl) is ionic in nature and it is soluble in polar solvent water, whereas lithium bromide and lithium iodide are covalent and are soluble in non-polar organic solvents.

Question 18.
Give two uses of alkali metals.
Answer:

  • Lithium metal is used to make useful alloys. For example, lead is used to make ‘white metal’ bearings for motor engines, with aluminium to make aircraft parts, and with magnesium to make armour plates. It is used in thermonuclear reactions.
  • Lithium is also used to make electrochemical cells.

Question 19.
What are the elements present in group 2? Give their general electronic configuration.
Answer:
Group 2 contains beryllium, magnesium, calcium, strontium, barium and radium. They’re general electronic configuration is [noble gas] ns2.

Question 20.
What are alkaline earth metals?
Answer:
Group 2 in the modem periodic table contains the elements beryllium, magnesium, calcium, strontium, barium and radium are called alkaline earth metals.

Question 21.
Atomic radii of alkaline earth metals are smaller than the corresponding members of alkali metals. Why?
Answer:
The atomic radii of alkaline earth metals are smaller than alkali metals. This is due to the fact that group 2 elements have a higher nuclear charge, allowing the electrons to move towards the nucleus. This reduces the size of atomic and ionic radii.

Question 22.
Write the uses of Beryllium.
Answer:

  1. Because of its low atomic number and very low absorption for X-rays, it is used as radiation windows for X-ray tubes and X-ray detectors.
  2. The sample holder in X-ray emission studies usually made of beryllium
  3. Since beryllium is transparent to energetic particles it is used to build the ‘beam pipe’ in accelerators.
  4. Because of its low density and diamagnetic nature, it is used in various detectors.

Question 23.
Explain the action of halogen with alkaline earth metals.
Answer:
All the alkaline earth metals combine with halogen at elevated temperature to form their halides.
M + X2 → MX2, Where M = Be, Mg,Ca,Sr,Ba and Ra. X= F,Cl,Br and I.
For e.g., Be + Cl2 → BeCl2.

Question 24.
How beryllium chloride is prepared from beryllium oxide?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals
Beryllium oxide is heated with carbon and chloride to get BeCl2.

Question 25.
How does beryllium hydride can be prepared?
Answer:
All the elements except beryllium, combine with hydrogen on heating to form their hydrides with general formula MH2. BeH2 can be prepared by the reaction of BeCl2 with LiAlH4.
2BeCl2 + LiAlH4 → 2BeH2 + LiCl + AlCl3

Question 26.
Mention the uses of beryllium.
Answer:

  • Beryllium is used as radiation windows for X-ray tubes and X-ray detectors.
  • The sample holder in X-ray emission studies is made of beryllium.
  • Beryllium is used to build the beam pipe in accelerators.
  • Beryllium is used in detectors due to its low density and diamagnetic nature.

Question 27.
Write about the uses of strontium.
Answer:

  • 90Sr is used in cancer therapy.
  • 87Sr / 86Sr ratio is used in marine investigators as well as in teeth, tracking animal migrations or in criminal forensics.
  • Dating of rocks.
  • Strontium is used as a radioactive tracer in determining the source of archaeological materials such as timbers and coins.

Question 28.
Mention the uses of radium.
Answer:

  • In self-luminous paints for watches.
  • In nuclear panels.
  • In aircraft switches.
  • In clocks and instrument dials.

Question 29.
BeO is covalent where as MgO is ionic. Give reason.
Answer:
Beryllium oxide (BeO) is covalent due to the small size of Be2+ ion, while magnesium oxide (MgO) is ionic due to the bigger size of Mg2+ ion.

Question 30.
How is barium peroxide prepared?
Answer:
Barium peroxide is prepared by heating monoxides with oxygen at high temperature.
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 31.
How would you prepare quick lime?
Answer:
Quick lime is produced on commercial scale by heating limestone in a lime kiln at 1173K.
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals
This reaction being reversible, CO2 is removed as soon as it is produced to enable the reaction to proceed to completion.

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 32.
What is slaking of lime?
Answer:
The addition of limited amount of water breaks the lump of lime. This process is called slaking
of lime and the product is slaked lime.
CaO + H2O → Ca(OH)2

Question 33.
What happens when quick lime reacts with –
1. H2O
2. CO2?
Answer:
1. CaO + H2O → Ca(OH)2 (calcium hydroxide)
2. CaO + CO2 → CaCO3 (calcium carbonate)

Question 34.
Prove that calcium oxide is a basic oxide.
Answer:
Calcium oxide is a basic oxide. It combines with acidic oxides at high temperature.
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 35.
Mention the uses of quick lime.
Answer:
Calcium oxide (quick lime) is used

  1. to manufacture cement, mortar and glass.
  2. in the manufacture of sodium carbonate and slaked lime.
  3. in the purification of sugar.
  4. as drying agent.

Question 36.
What is milk of lime? How CO2 reacts with it?
Answer:
The aqueous solution of calcium hydroxide is known as lime water and a suspension of slaked lime in water is known as milk of lime. When carbon dioxide is passed through lime water, it turns milky due to the formation of calcium carbonate.
Ca(OH)2 + CO2 → CaCO2 + H2O

Question 37.
What happens when excess of CO2 reacts with calcium carbonate?
Answer:
CaCO3 + CO2 + H2O → Ca(HCO3)2 (Calcium bi-carbonate)

Question 38.
What is bleaching powder? How Is it prepared?
Answer:
Bleaching powder is Ca(OCl)2. it is prepared by treating chlorine with milk of lime.
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 39.
What are the uses of calcium hydroxide?
Answer:
Calcium hydroxide is used

  • in the preparation of mortar, a building material.
  • in white wash due to its disinfectant nature.
  • in glass making and tanning industry.
  • for the preparation of bleaching powder and for the purification of sugar.

Question 40.
How gypsum occurs in nature?
Answer:

  • Gypsum is CaSO4.2H2O. Gypsum beds were formed due to the evaporation of water from the massive prehistoric sea basins.
  • When water evaporates, the minerals present in it become concentrated and crystallized.
  • Gypsum is formed, due to evaporation, sulphur present in water bonds with oxygen to form a sulphate. The sulphate then combines with calcium and water to form gypsum.

Question 41.
How is gypsum synthesized?
Answer:
Gypsum can also be synthesized from coal-fired power plants, as a by-product of flue-gas de suiflirization. The process of scrubbing sulfur from flue gases produced when coal is burned results in the production of several by-products, including gypsum.

Question 42.
What is meant by retrograde solubility?
Answer:
Gypsum is a soft mineral and it is less soluble in water as the temperature increases. This is known as retrograde solubility, which is a distinguishing characteristic of gypsum.

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 43.
Write a note about physical appearance of gypsum.
Answer:

  • Gypsum is usually white, colorless or grey in colour.
  • It can also be found in the shades of pink, yellow, brown and light green, mainly due to the presence of impurities.
  • Gypsum crystals are found to occur in a form that resembles the petals of a flower. This type of formation is referred to as ‘desert rose’, as they mostly occur in arid areas or desert terrains.

Question 44.
Prove that gypsum is a natural insulator.
Answer:

  • Gypsum have low thermal conductivity.
  • It won’t allow the electric current to pass through it. So it is known as natural insulator.

Question 45.
Write a note about alabaster.
Answer:

  • Alabaster is a variety of gypsum.
  • It is highly valued as an ornamental stone.
  • it has been used by the sculptors for centuries.
  • Alabaster is granular and opaque.

Question 46.
What ¡s dead burnt plaster?
Answer:
When Plaster of Paris is heated above 393K, no water of crystallisation is left and anhydrous calcium sulphate (CaSO4) is formed. This is known as ‘dead burnt plaster’.

Question 47.
What ¡s meant by the setting of cement?
Answer:
When gypsum is added to cement by mixing with an adequate quantity of water, it forms a plastic mass that gets into a hard solid in 5 to 10 minutes.

Question 48.
Write notes on plaster of paris.
Answer:
It is a hemihydrate of calcium sulphate. It is obtained when gypsum, CaSO4.2H2O, is heated to 393 K.
2CaSO4 2H2O{s) → 2CaSO4 .H2)O + 3H2O
Above 393 K, no water of crystallisation is left and anhydrous calcium sulphate, CaSO4 is formed. This is known as ‘dead burnt plaster’. It has a remarkable property of setting with water. On mixing with an adequate quantity of water it forms a plastic mass that gets into a hard solid in 5 to 15 minutes.
Uses:
The largest use of Plaster of Paris is in the building industry as well as plasters. It is used for immobilizing the affected part of organ where there is a bone fracture or sprain. It is also employed in dentistry, in ornamental work and for making casts of statues and busts.

Question 49.
In what ways lithium shows similarities to magnesium in its chemical behaviour?
Answer:

  • Both react with nitrogen to form nitrides.
  • Both react with O2 to form monoxides.
  • Both the elements have the tendency to form covalent compounds.
  • Both can form complex compounds.

Question 50.
Explain why can alkali and alkaline earth metals not be obtained by the chemical reduction method.
Answer:
Alkali and alkaline earth metals, themselves acts as better recurring agents and reducing agents, better than alkali metals. That is why these metals are not obtained by chemical reduction methods.

Question 51.
Why are potassium and caesium. rather than lithium used ¡n photoelectric cells?
Answer:
Potassium and caesium have much lower ionization enthalpy than that of lithium. As a result, these metals easily emit electrons on exposure to light. Due to this, K and Cs are used in photoelectric cells rather than lithium.

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 52.
Berllium and magnesium do not give colour to flame whereas other alkaline earth metals do so. Why?
Answer:
Due to small size, the ionization enthalpies of Be and Mg are much higher than those of other alkaline earth metals. Therefore, a large amount of energy is needed to excite their valence electron and that’s why they do not impart colour to the flame.

Question 53.
Why are lithium salts commonly hydrated and those of the other alkali metal ions usually anhydrous?
Answer:
Due to its smallest size, Li+ can polarize water molecules easily than the other alkali metal ions.

Question 54.
Why are alkali metals always univalent? Which alkali metal ion forms largest hydrated ion in aqueous solution?
Answer:
They are always univalent because after losing one electron, they acquire nearest inert gas configuration. Li+ forms largest hydrated cations because it has the highest hydration energy.

Question 55.
What is the effect of heat on the following compounds (Give equations for the reactions)?
Answer:
1. CaCO3
2. CaSO4 .2H2O
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 56.
Explain the following:
Answer:
(a) Lithium Iodide is more covalent than lithium fluoride.
(b) Lattice enthalpy of LIF is maximuni among all the alkali metal halides.
Answer:
(a) According to Fazan’s rule, Li+ ion can polarise I ion more than the F ion due to bigger size of the anion. Thus Li+ has more covalent character than LiF.
(b) Smaller the size (internuclear distance), more is the value of lattice enthalpy since internuclear distance is expected to be least in the LiF.

Question 57.
Why alkali metals are soft and have low melting points?
Answer:
Alkali metals have only one valence electron per metal atom. As a result, the binding energy of alkali metal ions in the close-packed metal lattices are weak. Therefore, these are soft and have low melting point.

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 58.
Why is LiF almost insoluble In water whereas LiCl soluble not only in water but also in acetone?
Answer:
The low solubiLity of LiF in water is due to its very high Lattice enthalpy (F ion is very small in size). On the other hand, in lithium chloride (LiCl) the lattice enthaipy is comparatively very small. This means that the magnitude of hydration enthalpy is quite large. Therefore lithium chloride dissolves in water. It is also soluble in acetone due to dipolar attraction (Acetone is polar in nature).

Question 59.
The hydroxides and carbonates of sodium and potassium are easily soluble in water while the corresponding salts of magnesium and calcium are sparingly soluble in water. Explain.
Answer:
Since group 1 hydroxides and carbonates due to large size contain higher hydration energy than the lattice energy so, they are easily soluble in water. Where as, in magnesium and calcium due to small size their lattice energy dominates over hydration energy they are sparingly soluble in water.

Question 60.
Draw the tile structure of –
1. BeCl2 (vapour)
2. BeCl2 (solid).
Answer:
BeCl2 (vapour):
In the vapour state, it exists as a chlorobridged dimer.
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals
2. BeCl2 (solid):
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 61.
Why is Li2CO3 decomposed at a lower temperature whereas Na2CO3 at higher temperature?
Answer:
Li2CO3 is a covalent compound, whereas Na2CO3 is an ionic compound. Therefore, lattice energy of Na2CO3 is higher than that of Li2CO3. Thus, LiCO3 is decomposed at a lower temperature.

Question 62.
Alkali metals give colouration when heated in Bunsen flame. Give reason.
Answer:
1. When the alkali metals salts moistened with concentrated hydrochloric acid are heated on a platinum wire in a flame, they show characteristic coloured flame.

  • Lithium – Crimson red
  • Sodium – Yellow
  • Potassium – Lilac
  • Rubidium – Reddish violet
  • Caesium – Blue

2. The heat in the flame excites the valence electron to a higher energy level. When it drops back to its actual energy level, the excess energy is emitted as light whose wavelength is in the visible region produces colour.

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 63.
How sodium metal reacts with –
1. ethanol
2. acetylene.
Answer:
(i) Sodium metal reacts with ethanol to form sodium ethoxide and liberates H2 gas.
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 64.
Mention the uses of lithium and its compounds. .
Answer:

  • Lithium metal is used to make useful alloys. For e.g. with lead, lithium is used to make white metal bearings for motor engines, with aluminium to make aircraft parts and with magnesium to make armour plates. It is also used in thermonuclear reactions.
  • Lithium is used to make electrochemical cells.
  • Lithium carbonate is used in medicines.

Question 65.
What are the uses of sodium and its compounds?
Answer:

  • Sodium is used to make Na/Pb alloy needed to make Pb(Et)4 and Pb(Me)4. These organolead compounds were used as anti-knock additives to petrol in early days.
  • Liquid sodium metal is used as a coolant in fast breeder nuclear reactors.

Question 66.
What are the uses of potassiunt and its compounds?
Answer:

  • Potassium has a vital role in biological system.
  • Potassium chloride is used as a fertilizer.
  • Potassium hydroxide is used in the manufacture of soft soap.
  • Potassium hydroxide is also used as an excellent absorbent of carbon dioxide.

Question 67.
What is soda ash? How is it obtained?
Answer:
Sodium carbonate decahydrate commonly known as washing soda Na2CO3. 10H2O. Upon heating, it looses the water of crystallization to form monohydrate. Above 373K, the monohydrate becomes completely anhydrous and changes to a white powder called soda ash.
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 68.
Mention the uses of washing soda (or) sodium carbonate.
Answer:

  • Sodium carbonate is highly used in laundering.
  • It is an important laboratory reagent used in qualitative analysis and in volumetric analysis.
  •  It is also used in water treatment to convert hard water to soft water.
  • It is used in the manufacture of glass, paper, and paint.

Question 69.
How would you prepare pure sodium chloride from crude salt?
Answer:
(a) Crude salt contains sodium sulphate, calcium sulphate, calcium chloride and magnesium chloride as impurities along with sodium chloride.
(b) Pure NaCl is obtained from crude salt by removal of insoluble impurities through filtration from the crude salt solution with a minimum amount of water.
(c) Sodium chloride can be crystallized by passing HCl gas into this solution.
(d) Calcium and magnesium chloride being more soluble than NaCl, remain in the solution.

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 70.
Mention the uses of sodium chloride.
Answer:

  • It is used as a common salt (or) table salt for domestic purposes.
  • It is used for the preparation of many inorganic compounds such as NaOH and Na2CO3.

Question 71.
What are the uses of sodium hydroxides?
Answer:

  • Sodium hydroxide is used as a laboratory reagent.
  • It is used in the purification of bauxite and petroleum refining.
  • It is used in the textile industries for mercerizing cotton fabrics.
  • It is used in the manufacture of soap, paper, artificial silk, and a number of chemicals.

Question 72.
Give a reason why sodium bicarbonate is used in bakeries.
Answer:
Sodium bicarbonate is called baking soda. Because it decomposes on heating to generate bubbles of carbon dioxide, leaving holes in cakes or pastries and making them light and fluffy.

Question 73.
Write about the uses of sodium bicarbonate.
Answer:

  1. NaHCO3 is used as an ingredient in baking.
  2. It is used as a mild antiseptic for skin infections.
  3. It is also used in fire extinguishers.

Question 74.
Explain the action of soda-lime with
1. SiO3
2. P4O10.
Answer:
Quick lime mixed with soda gives solid soda lime. It combines with acidic oxides such as SiO2, ançi P4O10 to form calcium silicate and calcium phosphate, respectively.
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Samacheer Kalvi 11th Chemistry Alkali and Alkaline Earth Metal 3-Mark Questions

Question 1.
Explain the periodic nature of ionization enthalpy in the alkali group.
Answer:

  1. Alkali metals have the lowest ionization enthalpy in each period.
  2. Within the group, as we go down, the ionization enthalpies of alkali metals decrease due to the increase in atomic size.
  3. In large atoms, the valence electrons are loosely held by the nucleus and arc easily lost, leading them to have low ionization enthalpy and acquiring stable noble gas configuration.
  4. On moving down the group, the atomic size increases and the number of inner shells also increases, which in turn increases the magnitude of screening effect. So. the ionization enthalpies decreases down the group.

Question 2.
Discuss the reaction of alkali metals with liquid ammonia.
Answer:
Alkali metals dissolve in liquid ammonia to give deep blue solutions that are conducting in nature. The conductivity is similar to that of pure metals. This happens because the alkali metal atom readily loses its valence electron in ammonia solution. Both the cation and the electron are ammoniated to give ammoniated cation and ammoniated electron.
M + (x + y)NH3 → [M(NH3)x]+ [e(NH3)y)]

The blue colour of the solution is due to the ammoniated electron which absorbs energy in the visible region of light and thus imparts blue colour to the solution. The solutions are paramagnetic and on standing slowly liberate hydrogen resulting in the formation of an amide.
M+ + e + NH3 → MNH2 + \(\frac{1}{2}\)H2
In concentrated solution, the blue colour changes to bronze colour and become diamagnetic.

Question 3.
Explain the anomalous behaviour of lithium among the alkali metals.
Answer:

  • Lithium is extremely small.
  • It has great polarizing power.
  • It has least electropositive character.
  • In lithium, the non-availability of d-orbitals is observed.

Question 4.
Write the uses of sodium carbonate.
Answer:

  1. Sodium carbonate known as washing soda is used heavily for laundering
  2. It is an important laboratory reagent used in qualitative analysis and in volumetric analysis.
  3. It is also used in water treatment to convert hard water to soft water
  4. It is used in the manufacturing of glass, paper, paint, etc…

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 5.
How alkali metals react with liquid ammonia?
Answer:

  • Alkali metals dissolve in liquid ammonia to give deep blue solutions that are conducting in nature.
  • This happens because the alkali metal atom readily loses the valence electron in ammonia solution.
  • Both the cation and the electron combine with ammonia to form ammoniated cation and ammoniated electron. M + (x+v)NH3 → [M(NH3)x]+ + [e(NH3)y]

Question 6.
What is the reason behind the blue colouration of alkali metals with liquid ammonia?
Answer:
M + (x + v)NH3 → [M(NH3)x]+ + [e(NH3)y]
The blue colour of the solution is due to the ammoniated electron which absorbs energy in the visible region of light and thus imparts blue colour to the solution. The solutions are paramagnetic and on standing slowly liberate hydrogen resulting in the formation of amide. In concentrated solution, the blue colour changes to bronze colour and become diamagnetic.

Question 7.
Compare the ionization energy of alkali metals with alkaline earth metals.
Answer:
Members of group 2 have higher ionization enthalpy values than group 1 because of their smaller size, with electrons being more attracted towards the nucleus of the atoms. Correspondingly they are less electropositive than alkali metals.

Although IE1 values of alkaline earth metals are higher than that of alkali metals, the IE2 values of alkaline earth metals are much smaller than those of alkali metals. This occurs because in alkali metals the second electron is to be removed from a cation, which has already acquired a noble gas configuration.

In the case of alkaline earth metals, the second electron is to be removed from a monovalent cation, which still has one electron in the outermost shell. Thus, the second electron can be removed more easily in the case of group 2 elements than in group 1 elements.

Question 8.
Describe the fireworks of alkaline earth metals.
Answer:

  • Combined with the element of chlorine, barium sends up a green spark.
  • Strontium chloride flashes red.
  • Copper and chlorine compound makes a blue firework.
  • Magnalium – A mixture of the alkaline earth metal magnesium and aluminium boosts all fireworks colours, particularly makes the blue brighter.

Question 9.
Write the uses of calcium.
Answer:

  1. As a reducing agent in the metallurgy of uranium, zirconium and thorium.
  2. As a deoxidiser, desulphuriser or decarboniser for various ferrous and non-ferrous alloys.
  3. In making cement and mortar to be used in construction.
  4. As a getter in vacuum tubes.
  5. In dehydrating oils
  6. In fertilisers, concrete and plaster of paris.

Question 10.
IE1of alkaline earth metals are higher than that of alkali metals, but IE2 of alkaline earth metals are smaller than that of alkali metals. Give reason.
Answer:

  1. IE1 of alkaline earth metals > IE1 of alkali metals.
  2. IE2 of alkaline earth metals < IE2 of alkali metals.
  3. This occurs because in alkali metals the second electron is to be removed from a cation, which has already acquired a noble gas configuration.
  4. In the case of alkaline earth metals, the second electron is to be removed from a monovalent cation, which still has one electron in the outermost shell.
  5. Thus, the second electron can be removed more easily in the case of group 2 elements than in group I elements.

Question 11.
MgCl2 and CaCl2 are easily hydrated, while NaCl and KCl are not hydrated. Why?
Answer:
Compounds of alkaline earth metals are more extensively hydrated than those of alkali metals, because the hydration enthalpies of alkaline earth metal ions are larger than those of alkali metal ions.
e.g., MgCl2 and CaCl2 exist as MgCl2.6H2O and CaCl2.6H2O, while NaCl and KCl do not form such hydrates.

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 12.
What is the distinctive behaviour of beryllium?
Answer:

  • Beryllium is small in size.
  • It has high polarizing power.
  • Its electronegativity is relatively high.
  • It has high ionization enthalpy.
  • In valence shell, vacant d-orbitais are absent in beryllium.

Question 13.
Write about the important uses of calcium. Calcium is used
Answer:

  • As a reducing agent in the metallurgy of uranium, zirconium and thorium.
  • As a de oxidizer, desulfurizer or decabonizer for various ferrous and non-ferrous alloys.
  • in making of cements and mortars to be used in construction.
  • As a getter in vacuum tubes.
  • In dehydrating oils
  • In fertilizers, concrete and Plaster of Paris.

Question 14.
Mention about the uses of barium. Barium is used
Answer:

  • In metallurgy, its compounds are used in pyrotechnics, petroleum mining, and radiology.
  • De oxidizer in copper refining.
  • Its alloys with nickel readily emits electrons hence used in electron tubes and in spark plug electrodes.
  • As a scavenger to remove last traces of oxygen and other gases in television and other electronic tubes.
  • An isotope of barium ‘33Ba., used as a source in the calibration of gamma-ray detectors in nuclear chemistry.

Question 15.
Be(OH)2 is amphoteric in nature. Prove it.
Answer:
Be(OH)2 is amphoteric in nature as it reacts with both acid and alkali.
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 16.
Write a note about the structure of beryllium chloride.
Answer:

  • BeCl2 has a chain structure in the solid-state.
  • In the vapour phase BeCl2 tends to form a chloro-bridged dimer.
  • At high temperatures of the order of 1200K it gives linear monomer.

Question 17.
Draw the structure of BeCl2 in different physical states.
Answer:
1. In solid-state:
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals
2. in vapour state:
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals
3. in High temperature:
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 18.
Write about the suiphates of alkaline earth metals.
Answer:

  • The sulphates of the alkaline earth metals are all white solids and stable to heat.
  • BeSO4 and MgSO4 are readily soluble in water; the solubility decreases from CaSO4 to BaSO4.
  • The greater hydration enthalpies of Be2and Mg2 ions overcome the lattice enthalpy factor and therefore their suiphates are soluble in water.
    Solubility order : BeSO4 < CaSO4 < BaSO4

Question 19.
What are the common physical and chemical features of alkali metals?
Answer:
Physical properties of alkali metals :

  • Alkali metals have low ionization enthalpies.
  • Alkali metals are highly electropositive in nature.
  • Alkali metals exhibit +1 oxidation states in their compounds.
  • Alkali metals impart characteristic colours to the flame.

Chemical properties of alkali metals:

  • Alkali metals are highly reactive in nature.
  • Alkali metals hydroxides are highly basic in nature.
  • Alkali metals dissolve in liquid ammonia to form blue and conducting solution.

Question 20.
Compare the alkali metals and alkaline earth metals with respect to –

  1. ionization enthalpy
  2. basicity of oxides and
  3. solubility of hydroxides.

Answer:
1. Ionization enthalpy:
Because of high nuclear charge the ionization enthalpy of alkaline earth metals are higher than those of the corresponding alkali metals.

2. Basicity of oxides:
Basicity of oxides of alkali metals are higher than that of alkaline earth metals.

3. Solubility of hydroxides:
Solubility of hydroxides of alkali metals are higher than that of alkaline earth metals. Alkali metals due to lower ionization enthalpy are more electropositive than the corresponding group 2 elements.

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 21.
Why is Li2CO3 decomposed at a lower temperature, whereas Na2CO3 at a higher temperature?
Answer:
Li2CO3 is a covalent compound, whereas Na4CO3 is an ionic compound. Therefore, the lattice energy of Na2CO3 is higher than that of Li2CO4. Thus, Li2CO3 is decomposed at a lower temperature.

Question 22.
What happens when

  • Sodium metal is dropped in water?
  • Sodium metal is heated in free supply of air?
  • Does sodium peroxide dissolve in water?

Answer:

  • 2Na + 2H2O —‘ 2NaOH + H2
  • 2Na + O2 —+ Na2O2
  • Na2O2+ 2H2O —‘ 2NaOH + H202

Question 23.
Write balanced equations for reactions between

  1. Na2O2 and water
  2. KO2and water
  3. Na2O and CO2

Answer:

  1. Na2O2 + 2H2O → 2NaOH + H2O2
  2. 2KO2 + 2H2O → 2KOH + O2 + H2O2
  3. Na2O + CO2 → Na2CO3

Question 24.
How would you explain the following observations?

  1. BeO is almost insoluble but BeSO4 is soluble ¡n water.
  2. BaO is soluble but BaSO4 is insõluble in water.
  3. LiI is more soluble than KI in ethanol.

Answer:

  1. Lattice energy of BeO is comparatively higher than the hydration energy. Therefore, it is almost insoluble in water. Whereas BeSO4 is ionic in nature and its hydration energy dominates the lattice energy.
  2. Both BaO and BaSO4 are ionic compounds but the hydration energy of BaO is higher than the lattice energy, therefore it is soluble in water.
  3. Since the size of Li+ ion is very small in comparison to K+ ion, it polarises the electronb cloud of I ion to a great extent. Thus LiI dissolves in ethanol more easily than the KI.

Question 25.
Explain the following:

  1. Why Cs is considered as the most electropositive element?
  2. Lithium cannot be used in making photoelectric cells.
  3. Lithium does not form alums.

Answer:

  1. Due to its lowest ionization energy, Cs is considered as the most electropositive element.
  2. Lithium cannot be used in making photoelectric cells because out of all the alkali metals, it has highest ionization energy and thus cannot emit electrons when exposed to light.
  3. Due to small size, lithium does not form alums.

Question 26.
Give the important uses of the following compounds.
1. NaHCO3
2. NaOH
Answer:
1. Uses of NaHCO3

  • It is used in fire extinguisher.
  • It is used as mild antiseptic for skin infections.
  • It is used as antacid.

2. Uses of NaOH

  • It is used in soap industry
  • It is used as reagent in laboratory
  • It is used in absorbing poisonous gases.

Question 27.
The hydroxides and carbonates of sodium and potassium are easily soluble in water, while the corresponding salts of magnesium and calcium are sparingly soluble in water. Explain.
Answer:
All the compounds are crystalline solids and their solubility in water is guided by both lattice enthalpy and hydration enthalpy. The magnitude of lattice enthalpy is quite small in case of sodium and potassium compounds, hence they are readily dissolved in water, when compared to magnesium and calcium compounds.

However, in case of corresponding magnesium and calcium compounds, the cations have smaller sizes and more magnitude of positive charge. This means that their lattice enthalpies are more, when compared to the sodium and potassium compounds. Therefore, the hydroxi des and carbonates of these metals are only sparingly soluble in water.

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 28.
Why is LiF almost insoluble in water, whereas LiCl soluble not only in water but also In acetone?
Answer:
The low solubility of LiF in water is due to its very high lattice enthalpy (F ion is very small in size). On the other hand, in lithium chloride (LiCl) the lattice enthalpy is comparatively very small. This means that the magnitude of hydration enthalpy is quite large. Therefore lithium chloride dissolves in water. It is also soluble in acetone due to dipolar attraction (Acetone is polar in nature).

Question 29.
Which out of the following can be used to store an alkali metal?
1. H2O
2. C2H5 OH
3. benzene
Answer:
3. Benzene can be used to store an alkali metal, because other substances react with alkali metal as below:
Na + H2O → NaOH + ½H2
Na + C2H5COH → C2H5ONa + ½H2

Samacheer Kalvi 11th Chemistry Alkali and Alkaline Earth Metal 5 – Mark Questions

Question 1.
Explain in what respects lithium ¡s different from other metals of the same group.
Answer:
lithium:

  • Very hard.
  • High melting and boiling point.
  • Least reactive.
  • Reacts with nitrogen to get Li3N.
  • Reacts with bromine slowly.
  • Burnt in air gives monoxide only.
  • Compounds are partially soluble in water.
  • Lithium nitrate decomposes to fòrm an oxide.
  • Extremely small in size.
  • Li+ has greater polarizing power.

Other elements of the family:

  • Very Soft.
    Low melting and boiling point.
  • More reactive.
  • No reaction.
  • Reacts violently.
  • Burnt in air gives peroxides also, apart from monoxides. K, Rb and Cs gave super oxides.
  • Highly soluble in water.
  • Other metals on heating gives nitrite.
  • Comparatively large in size.
  • Other M+ ions have comparatively larger polarizing power.

Question 2.
Discuss the diagonal relationship between Lithium and Magnesium.
Answer:
Similarity between the first member of group 1 (Li) and the diagonally placed second clement of group 2 (Mg) is called diagonal relationship. It is due to similar size (rLi+ = 0.766 Å and Mg2+ = 0.72 Å) and comparable electronegativity values (Li = 1.0; Mg = 1.2). Similarities between Lithium and Magnesium are

  1. Both lithium and magnesium are harder than other elements in the respective groups
  2. Lithium and magnesium react slowly with water. Their oxides and hydroxides are much less soluble and their hydroxides decompose on heating.
  3. Both form a nitride, Li3N, and Mg3N2, by direct combination with nitrogen
  4. They do not give any superoxides and form only oxides, Li2O and MgO
  5. The carbonates of lithium and magnesium decompose upon heating to form their respective oxides and CO2.
  6. Lithium and magnesium do not form bicarbonates.
  7. Both LiCl and MgCl2 are soluble in ethanol and are deliquescent. They crystallize from aqueous solution as hydrates, LiCl.2H2O and MgCl28H2O

Question 3.
Compare the properties of beryllium with the other elements in the same group.
Answer:

Beryllium:

  • Forms covalent compounds.
  • High melting and boiling point.
  • Does not react with water even at elevated temperature.
  • Does not combine directly with hydrogen.
  • Halides are covalent.
  • Hydroxides and oxides of beryllium are amphoteric in nature.
  • It is not readily attacked by acids because of the presence of an oxide film.
  • Beryllium carbide evolves methane withwater.
  • Salts of Be are extensively hydrolyzed.
  • It has no vacant ‘d’ orbitals in the outermost shell.

Other elements of the family:

  • Forms ionic compounds.
  • Lower melting and boiling point.
  • React with water.
  • Combine directly with hydrogen.
  • Halides are ionic or electrovalent.
  • Hydroxides and oxides are basic in nature.
  • Readily attacked by acids.
  • Other carbides evolves acetylene with water.
  • Hydrolyzed.
  • They have vacant ‘d’ orbitais in the outermost shell.

Question 4.
Write the uses of alkali metals.
Answer:

  1. Lithium metal is used to make useful alloys. For example, lead is used to make ‘white metal’ bearings for motor engines, with aluminium to make aircraft parts, and with magnesium to make armour plates. It is used in thermonuclear reactions.
  2. Lithium is also used to make electrochemical cells.
  3. Lithium carbonate is used in medicines
  4. Sodium is used to make Na/Pb alloy needed to make Pb(Et)4 and Pb(Me)4. These organolead compounds were earlier used as anti-knock additives to petrol, but nowadays lead-free petrol in use.
  5. Liquid sodium metal is used as a coolant in fast breeder nuclear reactors. Potassium has a vital role in biological systems.
  6. Potassium chloride is used as a fertilizer. Potassium hydroxide is used in the manufacture of soft soap. It is also used as an excellent absorbent of carbon dioxide.
  7. Caesium is used in devising photoelectric cells.

Question 5.
Distinguish between alkali metals and alkaline earth metals.
Answer:
Alkali Metals:

  • Alkali metals are soft.
  • They have a single electron in the valence shell and their electronic configuration is [noble gas] ns1.
  • They have low melting points.
  • Hydroxides are strongly basic.
  • Carbonates do not decompose.
  • Carbonates do not decompose.
  • Nitrates give corresponding nitrites and oxygen as products.
  • They show +1 oxidation states.
  • Their carbonates are soluble in water except for Li2CO3.
  • Except for Li, alkali metals do not form complex compounds.

Alkaline earth metals:

  • Alkaline earth metals are hard.
  • They have two electrons in the valence shell and their electronic configuration is [noble gas] ns2.
  • They have relatively high melting points.
  • Hydroxides are less basic.
  • Carbonates decompose to form oxide when heated to high temperatures.
  • Carbonates decompose to form oxide when heated to high temperatures.
  • Nitrates give corresponding oxides, nitrogen dioxide, and oxygen as products.
  • They show +2 oxidation states.
  • Their carbonates are insoluble in water.
  • They can form complex compounds.

Question 6.
Explain the diagonal relationship of Beryllium with Aluminium.
Answer:
Beryllium (the first member of group 2) shows a diagonal relationship with aluminium. In this case, the size of these ions (rBe2+ = 0.45 Å and rAl3+ = 0.54 Å) is not as close. However, their charge per unit area is closer. (Be2+ = 2.36 and Al3+ = 2.50) They also have same electronegativity values (Be = 1.5; Al = 1.5).
Properties:

  1. Beryllium chloride forms a dimeric structure like aluminium chloride with chloride bridges. Beryllium chloride also forms polymeric chain structure in addition to dimer. Both are soluble in organic solvents and are strong Lewis acids.
  2. Beryllium hydroxide dissolves in excess of alkali and gives beryllate ion and [Be(OH)4]2- and hydrogen as aluminium hydroxide which gives aluminate ion. [Al( OH4)
  3. Beryllium and aluminum ions have strong tendency to form complexes, BeF42-, AlF63-.
  4. Both beryllium and aluminium hydroxides are amphoteric in nature.
  5. Carbides of beryllium (Be2C) like aluminum carbide (Al4C3) give methane on hydrolysis
  6. Both beryllium and aluminium are rendered passive by nitric acid.

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 7.
An alkali metal (A) belongs to period number II and group number I react with oxygen to form (B). (A) reacts with water to form (C) with the liberation of hydrogen compound (D). Identify A, B, C, and D.
Answer:
1. An alkali metal (A) belongs to period number II and group number I is lithium.
2. Lithium reacts with oxygen to form simple oxide lithium oxide (B).
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals
3. Lithium reacts with water to form lithium hydroxide with the liberation of hydrogen.
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals
4. Lithium directly react with carbon to form an ionic compound lithium carbide.
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 8.
Describe the Solvay process (or) how is washing soda (or) sodium carbonate prepared in industries?
Answer:
(i) Solvay process – in this process ammonia is converted to ammonium carbonate, which is then converted to ammonium bicarbonate bypassing excess carbon dioxide in sodium chloride solution saturated with ammonia.
(ii) The ammonium bicarbonate formed reacts with sodium chloride to give sodium bicarbonate.
As sodium bicarbonate has poor solubility, it gets precipitated.
(iii) The sodium bicarbonate is isolated and is heated to give sodium carbonate.
(iv) The equations involved in this process is as below:

Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Question 9.
Describe the method of electrolysis of brine solution? (or) how is sodium hydroxide prepared commercially?
Answer:
(a) Sodium hydroxide is prepared commercially by the electrolysis of brine solution in Castner-Kellner cell using a mercury cathode and a carbon anode.
(b) Sodium metal is discharged at the cathode and combines with mercury to form sodium amalgam.
(c) Chlorine gas is evolved at the cathode.
(d) The sodium amalgam thus obtained is treated with water to give sodium hydroxide.
Samacheer Kalvi 11th Chemistry Solutions Chapter 5 Alkali and Alkaline Earth Metals

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Students can Download Physics Chapter 1 Electrostatics Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Physics Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Samacheer Kalvi 12th Physics Electrostatics Textual Evaluation Solved

Samacheer Kalvi 12th Physics Electrostatics Multiple Choice Questions

Question 1.
Two identical point charges of magnitude -q are fixed as shown in the figure below. A third charge +q is placed midway between the two charges at the point P. Suppose this charge +q is displaced a small distance from the point P in the directions indicated by the arrows, in which direction(s) will +q be stable with respect to the displacement?
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-1
(a) A1 and A2
(b) B1 and B2
(c) both directions
(d) No stable
Answer:
(b) B1 and B2

Question 2.
Which charge configuration produces a uniform electric field?
(a) point charge
(b) the infinite uniform line charge
(c) uniformly charged infinite plane
(d) uniformly charged spherical shell
Answer:
(c) uniformly charged infinite plane

Question 3.
What is the ratio of the charges \(\left|\frac{q_{1}}{q_{2}}\right|\) for the following electric field line pattern?
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-2
(a) \(\frac { 1 }{ 5 }\)
(b) \(\frac { 25 }{ 11 }\)
(c) 5
(d) \(\frac { 12 }{ 25}\)
Answer:
(d) \(\frac { 12 }{ 25}\)

Question 4.
An electric dipole is placed at an alignment angle of 30° with an electric field of 2 x 105 N C-1. It experiences a torque equal to 8 N m. The charge on the dipole if the dipole length is 1 cm is-
(a) 4 mC
(b) 8 mC
(c) 5 mC
(d) 1 mC
Answer:
(b) 8 mC

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 5.
Four Gaussian surfaces are given below with charges inside each Gaussian surface. Rank the electric flux through each Gaussian surface in increasing order-
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-3

(a) D < C < B < A
(b) A < B = C < D
(c) C < A = B < D
(d)D > C > B > A
Answer:
(a) D < C < B < A

Question 6.
The total electric flux for the following closed surface which is kept inside water-
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-4
(a) \(\frac { 80q }{{ ε }_{0}}\)
(b) \(\frac { q }{{ 40ε }_{0}}\)
(c) \(\frac { q }{{ 80ε }_{0}}\)
(d) \(\frac { q }{{ 40ε }_{0}}\)
Answer:
(b) \(\frac { q }{{ 40ε }_{0}}\)

Question 7.
Two identical conducting balls having positive charges q1 and q2 are separated by a center to center distance r. If they are made to touch each other and then separated to the same distance, the force between them will be- (NSEP 04-05)
(a) less than before
(b) same as before
(c) more than before
(d) zero
Answer:
(c) more than before

Question 8.
Rank the electrostatic potential energies for the given system of charges in increasing order
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-5
(a) 1 = 4 < 2 < 3
(b) 2 = 4 < 3 < 1
(c) 2 = 3 < 1 < 4
(d) 3 < 1 < 2 < 4
Answer:
(a) 1 = 4 < 2 < 3

Question 9.
An electric field \(\vec { E } \) = 10x\(\hat{i} \) exists in a certain region of space. Then the potential difference V = V0 – VA, Where V0 is the potential at the origin and VA is the potential at x = 2 m is-
(a) 10 J
(b) -20 J
(c) + 20 J
(d) – 10 J
Answer:
(a) 10 J

Question 10.
A thin conducting spherical shell of radius R has a charge Q which is uniformly distributed on its surface. The correct plot for electrostatic potential due to this spherical shell is-
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-6
Answer:
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-7

Question 11.
Two points A and B are maintained at a potential of 7 V and -4 V respectively. The work done in moving 50 electrons from A to B is-
(a) 8.80 x 10-17 J
(b) -8.80 x 10-17 J
(c) 4.40 x 10-17 J
(d) 5.80 x 10-17 J
Answer:
(a) 8.80 x 10-17 J

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 12.
If voltage applied on a capacitor is increased from V to 2V, choose the correct conclusion.
(a) Q remains the same, C is doubled
(b) Q is doubled, C doubled
(c) C remains same, Q doubled
(d) Both Q and C remain same
Answer:
(c) C remains same, Q doubled

Question 13.
A parallel plate capacitor stores a charge Q at a voltage V. Suppose the area of the parallel plate capacitor and the distance between the plates are each doubled then which is the quantity that will change?
(a) Capacitance
(b) Charge
(c) Voltage
(d) Energy density
Answer:
(d) Energy density

Question 14.
Three capacitors are connected in a triangle as shown in the figure. The equivalent capacitance between points A and C is
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-8
(a) 1 μF
(b) 2 μF
(c) 3 μF
(d) \(\frac { 1 }{ 4 }\) μF
Answer:
(b) 2 μF

Question 15.
Two metallic spheres of radii 1 cm and 3 cm are given charges of -1 x 10-2 C and 5 x 10-2 C respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is (AIIPMT 2012)
(a) 3 x 10-2 C
(b) 4 x 10-2 C
(c) 1 x 10-2 C
(d) 2 x 10-2 C
Answer:
(a) 3 x 10-2 C

Samacheer Kalvi 12th Physics Electrostatics Short Answer Questions

Question 1.
What is meant by quantisation of charges?
Answer:
The charge q on any object is equal to an integral multiple of the fundamental unit of charge ‘e’.
q = ne
Where ‘n’ is an integer e
e = charge of an electron =1.6 × 10-19 C.

Question 2.
Write down Coulomb’s law in vector form and mention what each term represents.
Answer:
The force on a charge q1 exerted by a point charge q1 is given by
\(\vec { F } \)12 = \(\frac { 1 }{{ 4πε }{0}}\) \(\frac {{ q }_{1}{ q }_{2}}{{ r }^{2}}\) \(\hat{r} \)21
Here \(\hat{r} \)21 is the unit vector from charge q1 to q1.
But \(\hat{r} \)21 = –\(\hat{r} \)12,
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-9
Therefore, the electrostatic force obeys Newton’s third law.

Question 3.
What are the differences between the Coulomb force and the gravitational force?
Answer:

Coulomb force Gravitational force
1. It can be attractive or repulsive depends on the nature of the charge 1. It is always attractive
2. The value of Proportionality constant K = 9 x 109 Nm2 C-2 2. The value of Gravitational constant G = 6.626 x 1011 Nm2 Kg-2
3. It depends on the medium which it exists. 3. It is independent of the medium which it exists.

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 4.
Write a short note on the superposition principle.
Answer:
According to this superposition principle, the total force acting on a given charge is equal to the vector sum of forces exerted on it by all the other charges.
\({ \vec { F } }_{ 1 }^{ tot }\) = \(\vec { F } \)12 + \(\vec { F } \)13 + \(\vec { F } \)14 + \(\vec { F } \)1n

Question 5.
Define ‘Electric field’.
Answer:
It is defined as the force experience by a unit positive charge, kept at that point
It is a Vector quantity.
Unit: NC-1

Question 6.
What is mean by ‘Electric field lines’?
Answer:
Electric field vectors are visualized by the concept of electric field lines. They form a set of continuous lines which are the visual representation of the electric field in some region of space.

Question 7.
The electric field lines never intersect. Justify.
Answer:
If two lines cross at a point, then there will be two different electric field vectors at the same point which is not possible. hence, they do not intersect.

Question 8.
Define ‘Electric dipole’
Answer:
Two equal and opposite charges separated by a small distance constitute an electric dipole.

Question 9.
What is the general definition of electric dipole moment?
Answer:
The electric dipole moment for a collection of ‘n point charges is given by
\(\overrightarrow{\mathrm{P}}=\sum_{i=1}^{\mathrm{n}} \mathrm{q}_{\mathrm{i}} \mathrm{r}_{\mathrm{i}}\)
where r̂i is the position ofvector of change qi from origin.

Question 10.
Define “electrostatic potential”.
Answer:
The electric potential at a point P is equal to the work done by an external force to bring a unit positive charge with constant velocity from infinity to the point P in the region of the external
electric field \(\vec { E } \).

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 11.
What is an equipotential surface?
Answer:
An equipotential surface is a surface on which all the points are at the same electric potential.

Question 12.
What are the properties of an equipotential surface?
Answer:
Properties of equipotential surfaces
(i) The work is done to move a charge q between any two points A and B,
W = q (VB – VA). If the points A and B lie on the same equipotential surface, work done is zero because of VA = VB.

(ii) The electric field is normal to an equipotential surface. If it is not normal, then there is a component of the field parallel to the surface. Then work must be done to move a charge between two points on the same surface. This is a contradiction. Therefore the electric field must always be normal to the equipotential surface.

Question 13.
Give the relation between electric field and electric potential.
Answer:
The electric field is the negative gradient of the electric potential.
\(\mathrm{E}=-\frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}}\)

Question 14.
Define electrostatic potential energy?
Answer:
The potential energy of a system of point charges may be defined as the amount of work done in assembling the charges at their locations by bringing them in from infinity.

Question 15.
Define ‘electric flux’.
Answer:

  1. The number of electric field lines crossing a given area kept normal to the electric field lines is called electric flux.
  2. Scalar quantity.
  3. Unit: Nm2C-1

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 16.
What is meant by electrostatic energy density?
Answer:
The energy stored per unit volume of space is defined as energy density uE = \(\frac { U }{ Volume }\)
From equation uE = \(\frac { 1 }{ 2 }\) \(\frac{\left(\varepsilon_{0} A\right)}{d}\) (Ed)2 = \(\frac { 1 }{ 2 }\) ε0 (Ad) E2 or uE = \(\frac { 1 }{ 2 }\) ε0E2

Question 17.
Write a short note on ‘electrostatic shielding’.
Answer:

  1. The process of isolating a certain region of space from the external field. It is based on the fact that the electric field inside a conductor is zero.
  2. Whatever the charges at the surfaces and whatever the electrical disturbance outside, the electric field inside the cavity are zero.

Question 18.
What is Polarisation?
Answer:
Polarisation \(\vec { P } \) is defined as the total dipole moment per unit volume of the dielectric.
\(\vec { P } \) = Xe \(\vec { P } \)ext

Question 19.
What is dielectric strength?
Answer:

  1. The maximum electric field the dielectric can withstand before it breakdowns is called dielectric strength.
  2. The dielectric strength of air 3 × 106 Vm-1

Question 20.
Define ‘capacitance’. Give its unit.
Answer:
The capacitance C of a capacitor is defined as the ratio of the magnitude of charge on either of the conductor plates to the potential difference existing between the conductors.
C = \(\frac { q }{ V }\) or Q ∝ V.
The SI unit of capacitance is coulomb per volt or farad (F).

Question 21.
What is corona discharge?
Answer:
The total charge of the conductor near the sharp edge gets reduces due to ionization of surrounding air. It is called corona discharge.

Samacheer Kalvi 12th Physics Electrostatics Long Answer Questions

Question 1.
Discuss the basic properties of electric charges.
Answer:
The electric charge is an inherent property of particles.
Conservation of electric charge:

  1. Total electric charge in the universe is constant.
  2. Charge can be neither created nor destroyed.
  3. In any physical process, the net change in charge will always be zero.
  4. The charge ‘q’ of any object is equal to an integral multiple of the fundamental unit of charge ‘e’.
    q = ne
  5. n is any integer
  6. e is charge of an electron = 1.6 × 10-19C.

Question 2.
Explain in detail Coulomb’s law and its various aspects.
Answer:
Consider two point charges q1 and q2 at rest in vacuum, and separated by a distance of r. According to Coulomb, the force on the point charge q2 exerted by another point charge q1 is
\(\vec { F } \) 21 = K\(\frac{q_{1} q_{2}}{r_{2}}\) \(\hat{r} \)12,
where [/latex] \(\hat{r} \)12 is the unit vector directed from charge q1 to charge q2 and k is the proportionality constant.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-10

Important aspects of Coulomb’s law:
(i) Coulomb’s law states that the electrostatic force is directly proportional to the product of the magnitude of the two point charges and is inversely proportional to the square of the distance between the two point charges.

(ii) The force on the charge q2exerted by the charge q1 always lies along the line joining the two charges. \(\hat{r} \)21is the unit vector pointing from charge q1 to q2 Likewise, the force on the charge q1 exerted by q2 is along – (i.e., in the direction opposite to \(\hat{r} \)21).

(iii) In SI units, k = \(\frac { 1 }{{ 4πε }_{0}}\) and its value is 9 x 109 Nm2C-2. Here e0 is the permittivity of free space or vacuum and the value of ε0 = \(\frac { 1 }{{ 4πε }_{0}}\) = 8.85 x 10-12 C2 N-1 m-2

(iv) The magnitude of the electrostatic force between two charges each of one coulomb and separated by a distance of 1 m is calculated as follows:
[F] = \(\frac{9 \times 10^{9} \times 1 \times 1}{1^{2}}\) = 9 x 109N. This is a huge quantity, almost equivalent to the weight of one million ton. We never come across 1 coulomb of charge in practice. Most of the electrical phenomena in day-to-day life involve electrical charges of the order of pC (micro coulomb) or nC (nano coulomb).

(v) In SI units, Coulomb’s law in vacuum takes the form \(\vec { F } \) 21 = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{r_{2}}\) \(\hat{r} \)12. sin Since ε > ε0, the force between two point charges in a medium other than vacuum is always less than that in vacuum. We define the relative permittivity for a given medium as ε = \(\frac { ε }{{ ε }_{0}}\) .For vacuum or air, εr = 1 and for all other media εr > 1

(vi) Coulomb’s law has same structure as Newton’s law of gravitation. Both are inversely proportional to the square of the distance between the particles. The electrostatic force is directly proportional to the product of the magnitude of two point charges and gravitational force is directly proportional to the product of two masses.

(vii) The force on a charge q1 exerted by a point charge q2 is given by \(\vec { F } \)12 = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{r_{2}}\) \(\hat{r} \)12 Here \(\hat{r} \)21 is sthe unit vector from charge q2 to q1.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-11
Therefore, the electrostatic force obeys Newton’s third law.

(viii) The expression for Coulomb force is true only for point charges. But the point charge is an ideal concept. However we can apply Coulomb’s law for two charged objects whose sizes are very much smaller than the distance between them. In fact, Coulomb discovered his law by considering the charged spheres in the torsion balance as point charges. The distance between the two charged spheres is much greater than the radii of the spheres.

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 3.
Define ‘Electric field’ and discuss its various aspects.
Answer:
The electric field at the point P at a distance r from the point charge q is the force experienced by a unit charge and is given by
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-12
Here \(\hat{r} \) is the unit vector pointing from q to the point of interest P. The electric field is a vector quantity and its SI unit is Newton per Coulomb (NC-1).

Important aspects of the Electric field:
(i) If the charge q is positive then the electric field points away from the source charge and if q is negative, the electric field points towards the source charge q.

(ii) If the electric field at a point P is \(\vec { E } \), then the force experienced by the test charge qo placed at the point P is \(\vec { F } \) = q0 \(\vec { E } \). This is Coulomb’s law in terms of the electric field. This is shown below Figure.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-13

(iii) The equation implies that the electric field is independent of the test charge q0 and it depends only on the source charge q.

(iv) Since the electric field is a vector quantity, at every point in space, this field has a unique direction and magnitude as shown in Figures (a) and (b). From the equation, we can infer that as distance increases, the electric field decreases in magnitude. Note that in Figures (a) and (b) the length of the electric field vector is shown for three different points. The strength or magnitude of the electric field at point P is stronger than at the point Q and R because the point P is closer to the source charge.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-14

(v) In the definition of the electric field, it is assumed that the test charge q0 is taken sufficiently small, so that bringing this test charge will not move the source charge. In other words, the test charge is made sufficiently small such that it will not modify the electric field of the source charge.

(vi) The expression is valid only for point charges. For continuous and finite-size charge distributions, integration techniques must be used. However, this expression can be used as an approximation for a finite-sized charge if the test point is very far away from the finite-sized source charge.

(vii) There are two kinds of electric field: uniform (constant) electric field and non-uniform electric field. A Uniform electric field will have the same direction and constant magnitude at all points in space. The non-uniform electric field will have different directions or different magnitudes or both at different points in space. The electric field created by a point charge is basically a non-uniform electric field. This non-uniformity arises, both in direction and magnitude, with the direction being radially outward (or inward), and the magnitude changes as distance increases.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-15

Question 4.
How do we determine the electric field due to a continuous charge distribution? Explain. Electric field due to continuous charge distribution.
Answer:
The electric charge is quantized microscopically. The expressions of Coulomb’s Law, superposition principle force and electric field are applicable to only point charges. While dealing with the electric field due to a charged sphere or a charged wire etc., it is very difficult to look at individual charges in these charged bodies. Therefore, it is assumed that charge is distributed continuously on the charged bodies, and the discrete nature of charges is not considered here. The electric field due to such continuous charge distributions is found by invoking the method of calculus.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-16
Consider the following charged object of irregular shape. The entire charged object is divided into a large number of charge elements ∆q1, ∆q2, ∆q3 ……..∆qn,…… and each charge element Δq is taken as a point charge.
The electric field at a point P due to a charged object is approximately given by the sum of the fields at P due to all such charge elements.
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Here ∆ qi is the ith charge element, rip is the distance of the point P frome the ith charge element, rip is the unit vector from ith charge element to the pont P.
However the equation is only an approximation. To incorporate the continuous distribution of charge, we take the limit ∆q → 0(= dq). In this limit, the summation in the equation becomes an integration and takes the following form
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-17-1
Here r is the distance of the point P from the infinitesimal charge dq and \(\hat{r} \) is the unit vector from dq to point P. Even though the electric field for a continuous charge distribution is difficult to evaluate, the force experienced by some test charge q in this electric field is still given by \(\vec { F } \) = q\(\vec { E } \).

(a) Line charge distribution: If the charge Q is uniformly distributed along the wire of length L, then linear charge density (charge per unit length) is λ = \(\frac { Q }{ L }\). Its unit is colomb per meter (Cm-1). The charge present in the infinitestimal length dl is dq = λdl.
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The electric field due to the line of total charge Q is given by
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-19

(b) Surface charge distribution: If the charge Q is uniformly distributed on a surface of area A, then surface charge density (charge per unit area) is σ = \(\frac { Q }{ A }\). Its unit is coulomb per square meter (C m-2). The charge present in the infinitesimal area dA is dq = σdA. The electric field due to a of total charge Q is given by
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-20

(c) Volume charge distribution: If the charge Q is uniformly distributed in a volume V, then volume charge density (charge per unit volume) is given by ρ = \(\frac { Q }{ V }\). Its unit is coulomb per cubic meter (Cm-3) The charge present in the infinitesimal volume element dV is dq = ρdV. The electric field due to a volume of total charge Q is given by
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-21

Question 5.
Calculate the electric field due to a dipole on its axial line and the equatorial plane.
Answer:
Case (I) :
Electric field due to an electric dipole at points on the axial line. Consider an electric dipole placed on the x-ax is as shown in the figure. A point C is located at a distance of r from the midpoint O of the dipole along the axial line. Axial line
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-22
The electric field at a point C due to +q is
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-23
Since the electric dipole moment vector \(\vec { P } \) is from -q to +q and is directed along BC, the above equation is rewritten as
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-24
where \(\hat{p} \) is the electric dipole moment unit vector from -q to +q. The electric field at a point C due to -q is
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-25
Since +q is located closer to the point C than -q, \(\vec { E } \) _. \(\vec { E } \) + us stronger than \(\vec { E } \). Therefore, the length of the E + vector is drawn large than that of \(\vec { E } \) _vector.
The total electric field at point C is calculated using the superposition principle of the electric field.
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Note that the total electric field is along \(\vec { E } \)+, since +q is closer to C than -q.
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The direction of \(\vec { E } \)tot is shown in Figure
If the point C is very far away from the dipole then (r >> a). Under this limit the term(r2 – a2)2 ≈ r4 Substituting this into equation, we get
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-28
If point C is chosen on the left side of the dipole, the total electric field is still in the

Case (II) :
Electric field due to an electric dipole at a point on the equatorial plane
Consider a point C at a distance r from the midpoint O of the dipole on the equatorial plane as shown in Figure. Since point C is equidistant from +q and -q, the magnitude of the electric fields of +q and -q are the same. The direction of E+ is along with BC and the direction of E is along with CA. E+ and E_ are resolved into two components; one component parallel to the dipole axis and the other perpendicular to it.

The perpendicular components \(\left|\vec{E}_{+}\right|\) sin θ and \(\left|\vec{E}_{-}\right|\) sin θ are oppositely directed and cancel each other. The magnitude of the total electric field at point C is the sum of the paralle component of \(\vec { E } \)+ and \(\vec { E } \) and its direction is along \(\hat{-p} \).
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-29
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The magnitudes \(\vec { E } \)+ and \(\vec { E } \) are the same and are given by
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By substituting equation (1) into equation (2), we get
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At very large distances (r >> a), the equation becomes
\(\vec { E } \)tot \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { p }{{ r }^{3}}\) (r >>) …… (4)

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 6.
Derive an expression for the torque experienced by a dipole due to a uniform electric field.
Answer:
Torque experienced by an electric dipole in the uniform electric field:
Consider an electric dipole of dipole moment \(\vec { p } \) placed in a uniform electric field E whose field lines are equally spaced and point in the same direction. The charge +q will experience a force q\(\vec { E } \) in the direction of the field and charge -q will experience a force -q\(\vec { E } \) in a direction opposite to the field.

Since the external field \(\vec { E } \) is uniform, the total force acting on the dipole is zero. These two forces acting at different points will constitute a couple and the dipole experience a torque. This torque tends to rotate the dipole. (Note that electric field lines of a uniform field are equally spaced and point in the same direction). The total torque on the dipole about the point O
\(\vec { τ } \) = \(\overrightarrow{\mathrm{OA}}\) × (-q\(\vec { E } \)) + \(\overrightarrow{\mathrm{OB}}\) × q\(\vec { E } \)
Using the right-hand corkscrew rule, it is found that total torque is perpendicular to the plane of the paper and is directed into it.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-33
The magnitude of the total torque
\(\vec { τ } \) = \(|\overrightarrow{\mathrm{OA}}|\)(-q\(\vec { E } \)) sin θ + \(|\overrightarrow{\mathrm{OB}}|\) \(|q \overrightarrow{\mathrm{E}}|\) sin θ
where θ is the angle made by \(\vec { P } \) with \(\vec { E } \). Since p = 2aq, the torque is written in terms of the vector product as
\(\vec { τ } \) = \(\vec { p } \) x \(\vec { E } \)
The magnitude of this torque is τ = pE sin θ and is maximum Torque on dipole
when θ =90°.
This torque tends to rotate the dipole and align it with the electric field \(\vec { E } \). Once \(\vec { E } \) is aligned with \(\vec { E } \), the total torque on the dipole becomes zero.

Question 7.
Derive an expression for electrostatic potential due to a point charge.
Answer:
Electric potential due to a point charge:
Consider a positive charge q kept fixed at the origin. Let P be a point at distance r from the charge q.
The electric potential at point P is
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Electric field due to positive point charge q is
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-36
The infinitesimal displacement vector, d\(\vec { r } \) = dr\(\hat{r} \) and using \(\hat{r} \) . \(\hat{r} \) = 1, we have
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-37
After the integration,
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-38
Hence the electric potential due to a point charge q at a distance r is
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{ r }\) …… (2)
Important points (If asked in the exam)
(i) If the source charge q is positive, V > 0. If q is negative, then V is negative and equal to
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{ r }\)
(ii) The description of the motion of objects using the concept of potential or potential energy is simpler than using the concept of field.
(iii) From expression (2), it is clear that the potential due to positive charge decreases as the distance increases, but for a negative charge, the potential increases as the distance are increased. At infinity (r = ∞) electrostatic potential is zero (V = 0).
(iv) The electric potential at a point P due to a collection of charges q1,q2,q3… qn is equal to the sum of the electric potentials due to individual charges.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-39
Where r1, r2,r3,…..rn are the distances of q1,q2,q3… qn
respectively from P
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-40

Question 8.
Derive an expression for electrostatic potential due to an electric dipole.
Answer:
Electrostatic potential at a point due to an electric dipole:
Consider two equal and opposite charges separated by a small distance 2a. The point P is located at a distance r from the midpoint of the dipole. Let 0 be the angle between the line OP and dipole axis AB.
Let r1 be the distance of point P from +q and r1 be the distance of point P from -q.
Potential at P due to charge +q = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{{ r }_{1}}\)
Potential at P due to charge -q = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{{ r }_{2}}\)
Total Potential at the point P,
V = \(\frac { 1 }{{ 4πε }_{0}}\)q \(\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right)\) ….. (1)
Suppose if the point P is far away from the dipole, such that r >> a, then equation can be expressed in terms of r. By the cosine law for triangle BOP,
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-41
\(r_{1}^{2}\) = r2 + a2 – 2ra cos θ = r2 \(\left(1+\frac{a^{2}}{r^{2}}-\frac{2 a}{r} \cos \theta\right)\)
Since the point P is very far from dipole, then r >> a. As a result the term \(\frac {{ a }^{ 2 }}{{ r }^{ 2 }}\) is very small and can be neglected. Therefore
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-42
since \(\frac { a }{ r }\) << 1, we can use binominal theorem and retain the terms up to first order
\(\frac { 1 }{{ r_{1}} }\) = \(\left(1+\frac{a}{r} \cos \theta\right)\) ……. (2)
Similarly applying the cosine law for triangle AOP,
\(r_{2}^{2}\) = r2 + a2 – 2ra cos (180 – θ)
Since cos (180 – θ) = cos θ we get
\(r_{2}^{2}\) = r2 + a2 + 2ra cos θ
Neglecting the term \(\frac {{ a }^{ 2 }}{{ r }^{ 2 }}\) (because r >> a)
\(r_{2}^{2}\) = r2 \(\left(1+\frac{2 a \cos \theta}{r}\right)\) (or) r2 = r \(\left(1+\frac{2 a \cos \theta}{r}\right)^{\frac{1}{2}}\)
Using Binomial theorem, we get
\(\frac { 1 }{{ r_{2}} }\) = \(\frac { 1 }{ r }\) \(\left(1-a \frac{\cos \theta}{r}\right)\)
Substituting equations (3) and (2) in equation (1)
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-43
But the electric dipole moment p = 2qa and we get,
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\left(\frac{p \cos \theta}{r^{2}}\right)\)
Now we can write p cos θ = \(\vec { P } \), \(\hat{r} \) where \(\hat{r} \) is the unit vector from the point O to point P. Hence the electric potential at a point P due to an electric dipole is given by
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{\vec{p} \cdot \hat{r}}{r^{2}}\) (r >> a) ….. (4)
Equation (4) is valid for distances very large compared to the size of the dipole. But for a point dipole, the equation (4) is valid for any distance.
Special cases:

Case (I):
If the point P lies on the axial line of the dipole on the side of +q, then θ = 0. Then the electric potential becomes
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { p }{{r}^{ 2 }}\)

Case (II):
If the point P lies on the axial line of the dipole on the side of -q, then θ = 180°, then
V = – \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { p }{{r}^{ 2 }}\)

Case (III):
If the point P lies on the equatorial line of the dipole, then θ = 90°. Hence, V = 0.

Question 9.
Obtain an expression for potential energy due to a collection of three-point charges which are separated by finite distances.
Answer:
Electrostatic potential energy for a collection of point charges:
The electric potential at a point at a distance r from point charge ql is given by
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ q }_{ 1 }}{r}\) …… (1)
This potential V is the work done to bring a unit positive charge from infinity to the point. Now if the charge q2 is brought from infinity to that point at a distance r from qp the work done is the product of q2 and the electric potential at that point. Thus we have W = q2V …… (2)
This work done is stored as the electrostatic potential energy U of a system of charges q1 and q2 separated by a distance r. Thus we have
U = q2 V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{r}\) …… (3)
The electrostatic potential energy depends only on the distance between the two point charges. In fact, the expression (3) is derived by assuming that q1 is fixed and q2 is brought from infinity. The equation (3) holds true when q2 is fixed and q1 is brought from infinity or both q2and q2 are simultaneously brought from infinity to a distance r between them.
Three charges are arranged in the following configuration as shown in Figure.
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To calculate the total electrostatic potential energy, we use the following procedure. We bring all the charges one by one and arrange them according to the configuration.
(i) Bringing a charge q1 from infinity to point A requires no work, because there are no other charges already present in the vicinity of charge q1

(ii) To bring the second charge q2 to point B, work must be done against the electric field created by the charge q1 So the work done on the charge q1 is W = q2V1B. Here V1B is the electrostatic potential due to the charge q1 at point B.
U = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{{r}_{12}}\) ….. (4)
Note that the expression is the same when q2 is brought first and then q1 later.

(iii) Similarly to bring the charge q3 to point C, work has to be done against the total electric field due to both charges q1 and q2. So the work done to bring the charge q3 is = q3 (V1C + V2C). Here V1C is the electrostatic potential due to charge q1 at point C and V2C is the electrostatic potential due to charge q2 at point C. The electrostatic potential is
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-45

(iv) Adding equations (4) and (5), the total electrostatic potential energy for the system of three charges q1,q2, and q3 is
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Note that this stored potential energy U is equal to the total external work done to assemble the three charges at the given locations. The expression (6) is the same if the charges are brought to their positions in any other order. Since the Coulomb force is a conservative force, the electrostatic potential energy is independent of the manner in which the configuration of charges is arrived at.

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 10.
Derive an expression for the electrostatic potential energy of the dipole in a uniform electric field.
Answer:
The electrostatic potential energy of a dipole in a uniform electric field:
Consider a dipole placed a torque when kept in an uniform electric field \(\vec { E } \). A dipole experiences a torque when kept in an uniform electric field \(\vec { E } \). This torque rotates the dipole to align it with the direction of the electric field. To rotate the dipole (at constant angular velocity) from its initial angle θ’ to another angle θ against the torque exerted by the electric field, an equal and opposite external torque must be applied on the dipole.
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The work done by the external torque to rotate the dipole from angle θ’ to θ at constant angular velocity is
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-47-1
Since τext is equal and opposite to τE = \(\vec { P } \) x \(\vec { E } \), we have
\(\left|\overrightarrow{\mathrm{r}}_{\mathrm{ext}}\right|\) = \(\left|\overrightarrow{\mathrm{r}}_{\mathrm{E}}\right|\)= \(|\overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{E}}|\) …. (2)
Substituting equation (2) in equation (1) We get,
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-48
This work done is equal to the potential energy difference between the angular positions θ and θ’.
U(θ) – (Uθ’) = AU = -pE cos θ +PE cos θ’.
If the initial angle is = θ’ = 90° and is taken as reference point, then U(θ’) + pE cos θ’ = θ.
The potential energy stored in the system of dipole kept in the uniform electric field is given by El = -pE cos θ = –\(\vec { P } \) . \(\vec { E } \) ….. (3)
In addition to p and E, the potential energy also depends on the orientation θ of the electric dipole with respect to the external electric field.
The potential energy is maximum when the dipole is aligned anti-parallel (θ = π) to the external electric field and minimum when the dipole is aligned parallel (θ = 0) to the external electric field.

Question 11.
Obtain Gauss law from Coulomb’s law.
Answer:
Gauss law: Gauss’s law states that if a charge Q is enclosed by an arbitrary closed surface, then the total electric flux ΦE through the closed surface is
ΦE = \(\oint { \vec { E } } \) .d \(\vec { A } \) = \(\frac{\mathrm{Q}_{\mathrm{end}}}{\varepsilon_{0}}\)
A positive point charge Q is surrounded by an imaginary sphere of radius r as shown in the figure. We can calculate the total electric flux through the closed surface of the sphere using the equation.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-49
ΦE = \(\oint { \vec { E } } \) .d \(\vec { A } \) = \(\oint { EdA } \) cos θ …… (1)
The electric field of the point charge is directed radially outward at all points on the surface of the sphere. Therefore, the direction of the area element d \(\vec { A } \) is along the electric field \(\vec { E } \) and θ = 0°.
ΦE = \(\oint { EdA } \) since cos 0° = 1 ….. (2)
E is uniform on the surface of the sphere,
ΦE = \(\oint { EdA } \) ….. (3)
Substituting for
\(\oint { dA } \) = 4π2 and E = \(\frac { 1 }{{ 4πε }_{0}}\) Q in equation 3, we get
ΦE = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{{ r }^{2}}\) × 4π2 = 4π \(\frac { 1 }{{ 4πε }_{0}}\) = \(\frac { q }{{ ε }_{0}}\) ……. (4)
The equation (4) is called as Gauss’s law. The remarkable point about this result is that the equation (4) is equally true for any arbitrary shaped surface which encloses the charge Q.

Question 12.
Obtain the expression for electric field due to an infinitely long charged wire.
Answer:
Electric field due to an infinitely long charged wire:
Consider an infinitely long straight wire having uniform linear charge density λ. Let P be a point located at a perpendicular distance r from the wire. The electric field at the point P can be found using Gauss law. We choose two small charge elements A1 and A1 on the wire which are at equal distances from the point P.

The resultant electric field due to these two charge elements points radially away from the charged wire and the magnitude of electric field is same at all points on the circle of radius r. From this property, we can infer that the charged wire possesses a cylindrical symmetry.
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Let us choose a cylindrical Gaussian surface of radius r and length L. The total electric flux in this closed surface is
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It is seen that for the curved surface, \(\vec { E } \) is parallel to \(\vec { A } \) and \(\vec { E } \).d \(\vec { A } \) = EdA. For the top and bottom surface, \(\vec { E } \) is perpendicular to \(\vec { A } \) and \(\vec { E } \).d\(\vec { A } \) = 0
Substituting these values in equation (2) and applying Gauss law
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Since the magnitude of the electric field for the entire curved surface is constant, E is taken out of the integration, and Qencl is given by Qencl = λL.
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Here,
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dA = total area of the curved surface = 2πrL. Substituting this in
equation (4), We get
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The electric field due to the infinite charged wire depends on \(\frac { 1 }{ r }\) rather than \(\frac { 1 }{{r}^{ 2 }}\) for a point charge.
Equation (6) indicates that the electric field is always along the perpendicular direction (\(\hat{r} \) ) to wire. In fact, if λ > 0 then E points perpendicular outward (\(\hat{r} \) ) from the wire and if λ < 0, then E points perpendicular inward (- \(\hat{r} \) ).

Question 13.
Obtain the expression for the electric field due to a charged infinite plane sheet.
Answer:
Electric field due to the charged infinite plane sheet: Consider an infinite plane sheet of charges with uniform surface charge density o. Let P be a point at a distance of r from the sheet. Since the plane is infinitely large, the electric field should be the same at all points equidistant from the plane and radially directed at all points. A cylindrical shaped Gaussian surface of length 2r and area A of the flat surfaces is chosen such that the infinite plane sheet passes perpendicularly through the middle part of the Gaussian surface.
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Applying Gauss law for this cylindrical surface,
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The electric field is perpendicular to the are element at all points on the curved surface and is parallel to the surface areas at P and P’. Then,
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Since the magnitude of the electric field at these two equal surfaces is uniform, E is taken out of the integration, and Qencl is given by Qencl = σA, we get
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The total area of surface either at P or P’
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Hence 2EA = \(\frac { σA }{{ ε }_{0}}\) or E = \(\frac { σ }{{ 2ε }_{0}}\) …… (3)
In vector from, E = \(\frac { σ }{{ 2ε }_{0}}\) \(\hat{n} \) ….. (4)
Hence \(\hat{n} \) is the outward unit vector normal to the plane. Note that the electric field due to an infinite plane sheet of the charge depends on the surface charge density and is independent of the distance r.

The electric field will be the same at any point farther away from the charged plane. Equation (4) implies that if o > 0 the electric field at any point P is outward perpendicular n to the plane and if σ < 0 the electric field points inward perpendicularly (\(\hat{n} \) ) to the plane. For a finite charged plane sheet, equation (4) is approximately true only in the middle region of the plane and at points far away from both ends.

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 14.
Obtain the expression for the electric field due to a uniformly charged spherical shell.
Answer:
Electric field due to a uniformly charged spherical shell:
Consider a uniformly charged spherical shell of radius R and total charge Q. The electric field at points outside and inside the sphere is found using Gauss law.
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Case (a):
At a point outside the shell (r > R): Let us choose a point P outside the shell at a distance r from the center as shown in figure (a). The charge is uniformly distributed on the surface of the sphere (spherical symmetry). Hence the electric field must point radially outward if Q > 0 and point radially inward if Q < 0. So we choose a spherical Gaussian surface of radius r and the total charge enclosed by this Gaussian surface is Q. Applying Gauss law,
\(\oint { \vec { E } } .d\vec { A } \) = \(\frac { Q }{{ ε }_{0}}\) …….(1)
The electric field \(\vec { E } \) and d\(\vec { A } \) point in the same direction (outward normal) at all the points on the Gaussian surface. The magnitude of \(\vec { E } \) is also the same at all points due to the spherical symmetry of the charge distribution.
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But
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dA = total area of Gaussian surface = 4πr2. Substituting this value in equation (2).
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The electric field is radially outward if Q > 0 and radially inward if Q < 0. From equation (3), we infer that the electric field at a point outside the shell will be same as if the entire charge Q is concentrated at the center of the spherical shell. (A similar result is observed in gravitation, for gravitational force due to a spherical shell with mass M)

Case (b):
At a point on the surface of the spherical shell (r = R): The electrical field at points on the spherical shell (r = R) is given by
\(\vec { E } \) = \(\frac{\mathrm{Q}}{4 \pi \varepsilon_{0} \mathrm{R}^{2}}\) \(\hat{r} \) …… (4)

Case (c):
At a point inside the spherical shell (r < R): Consider a point P inside the shell at a distance r from the center. A Gaussian sphere of radius r is constructed as shown in the figure (b). Applying Gauss law
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Since Gaussian surface encloses no charge, So Q = 0. The equation (5) becomes E = 0 (r < R) …(6)
The electric field due to the uniformly charged spherical shell is zero at all points inside the shell.

Question 15.
Discuss the various properties of conductors in electrostatic equilibrium.
Answer:
Properties of conductors in electrostatic equilibrium:
(i) The electric field is zero everywhere inside the conductor. This is true regardless of whether the conductor is solid or hollow. This is an experimental fact. Suppose the electric field is not zero inside the metal, then there will be a force on the mobile charge carriers due to this electric field.

As a result, there will be a net motion of the mobile charges, which contradicts the conductors being in electrostatic equilibrium. Thus the electric field is zero everywhere inside – the conductor. We can also understand this fact by applying an external uniform electric field on the conductor.
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Before applying the external electric field, the free electrons in the conductor are uniformly distributed in the conductor. When an electric field is applied, the free electrons accelerate to the left causing the left plate to be negatively charged and the right plate to be positively charged.

Due to this realignment of free electrons, there will be an internal electric field created inside the conductor which increases until it nullifies the external electric field. Once the external electric field is nullified the conductor is said to be in electrostatic equilibrium. The time taken by a conductor to reach electrostatic equilibrium is in the order of 10-6s, which can be taken as almost instantaneous.

(ii) There is no net charge inside the conductors. The charges must reside only on the surface of the conductors. We can prove this property using Gauss law. Consider an arbitrarily shaped conductor. A Gaussian surface is drawn inside the conductor such that it is very close to the surface of the conductor.
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Since the electric field is zero everywhere inside the conductor, the net electric flux is also zero over this Gaussian surface. From Gauss’s law, this implies that there is no net charge inside the conductor. Even if some charge is introduced inside the conductor, it immediately reaches the surface of the conductor.

(iii) The electric field outside the conductor is perpendicular to the surface of the conductor and has a magnitude of \(\frac { σ }{{ ε }{0}}\) where a is the surface charge density at that point. If the electric field has components parallel to the surface of the conductor, then free electrons on the surface of the conductor would experience acceleration. This means that the conductor is not in equilibrium. Therefore at electrostatic equilibrium, the electric field must be perpendicular to the surface of the conductor.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-68
We now prove that the electric field has magnitude \(\frac { σ }{{ ε }{0}}\) just outside the conductor’s surface. Consider a small cylindrical Gaussian surface. One half of this cylinder is embedded inside the conductor. Since electric field is normal to the surface of the conductor, the curved part of the cylinder has zero electric flux. Also inside the conductor, the electric field is zero. Hence the bottom flat part of the Gaussian surface has no electric flux. Therefore the top flat surface alone contributes to the electric flux. The electric field is parallel to the area vector and the total charge inside the surface is σA. By applying Gauss’s law,
EA = \(\frac { σA }{{ ε }{0}}\)
In vector from, \(\vec { E } \) = \(\frac { σ }{{ ε }{0}}\) \(\hat{n} \)
Here n represents the unit vector outward normal to the surface of the conductor. Suppose σ < 0, then electric field points inward perpendicular to the surface.

(iv) The electrostatic potential has the same value on the surface and inside of the conductor. We know that the conductor has no parallel electric component on the surface which means that charges can be moved on the surface without doing any work. This is possible only if the electrostatic potential is constant at all points on the surface and there is no potential difference between any two points on the surface. Since the electric field is zero inside the conductor, the potential is the same as the surface of the conductor. Thus at electrostatic equilibrium, the conductor is always at equipotential.

Question 16.
Explain the process of electrostatic induction.
Answer:
Whenever a charged rod is touched by another conductor, charges start to flow from the charged rod to the conductor. This type of charging without actual contact is called electrostatic induction:

(i) Consider an uncharged (neutral) conducting sphere at rest on an insulating stand. Suppose a negatively charged rod is brought near the conductor without touching it, as shown in figure (a). The negative charge of the rod repels the electrons in the conductor to the opposite side.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-69
Various steps in electrostatic induction
As a result, positive charges are induced near the region of the charged rod while negative charges on the farther side. Before introducing the charged rod, the free electrons were distributed uniformly on the surface of the conductor and the net charge is zero. Once the charged rod is brought near the conductor, the distribution is no longer uniform with more electrons located on the farther side of the rod and positive charges are located closer to the rod. But the total charge is zero.

(ii) Now the conducting sphere is connected to the ground through a conducting wire. This is called grounding. Since the ground can always receive any amount of electrons, grounding removes the electron from the conducting sphere. Note that positive charges will not flow to the ground because they are attracted by the negative charges of the rod (figure (b)).

(iii) When the grounding wire is removed from the conductor, the positive charges remain near the charged rod (figure (c)).

(iv) Now the charged rod is taken away from the conductor. As soon as the charged rod is removed, the positive charge gets distributed uniformly on the surface of the conductor (figure (d)). By this process, the neutral conducting sphere becomes positively charged.

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 17.
Explain dielectrics in detail and how an electric field is induced inside a dielectric.
Answer:
Induced Electric field inside the dielectric:
When an external electric field is applied to a conductor, the charges are aligned in such a way that an internal electric field is created which cancels the external electric field. But in the case of a dielectric, which has no free electrons, the external electric field only realigns the charges so that an internal electric field is produced.

The magnitude of the internal electric field is smaller than that of the external electric field. Therefore the net electric field inside the dielectric is not zero but is parallel to an external electric field with a magnitude less than that of the external electric field. For example, let us consider a rectangular dielectric slab placed between two oppositely charged plates (capacitor) as shown in the figure.
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The uniform electric field between the plates Induced electric field lines inside the dielectric acts as an external electric field \(\vec { E } \)ext which polarizes the dielectric placed between plates. The positive charges are induced on one side surface and negative charges are induced on the other side of the surface But inside the dielectric, the net charge is zero even in a small volume. So the dielectric in the external field is equivalent to two oppositely charged sheets with the surface charge densities +σb and -σb. These charges are called bound charges. They are not free to move like free electrons in conductors. This is shown in the figure.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-71
(a) Balloon sticks to the wall
(b) Polarisation of the wall due to the electric field created by the balloon
For example, the charged balloon after rubbing sticks onto a wall. The reason is that the negatively charged balloon is brought near the wall, it polarizes opposite charges on the surface of the wall, which attracts the balloon.

Question 18.
Obtain the expression for capacitance for a parallel plate capacitor.
Answer:
The capacitance of a parallel plate capacitor:
Consider a capacitor with two parallel plates each of cross-sectional area A and separated by a distance d. The electric field between two infinite parallel plates is uniform and is given by E = \(\frac { σ }{{ ε }{0}}\) where σ is the surface charge density on the plates σ = \(\frac { Q }{ A }\) .If the separation distance d is very much smaller than the size of the plate (d2 << A), then the above result is used even for finite-sized
parallel plate capacitor.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-72
The capacitance of a parallel plate capacitor
The electric field between the plates is
E = \(\frac { Q }{{ Aε }{0}}\) ….. (1)
Since the electric field is uniform, the electric potential between the plates having separation d is given by
V = Ed = \(\frac { Qd }{{ Aε }{0}}\) ….. (2)
Therefore the capacitance of the capacitor is given by
C = \(\frac { Q }{ V }\) = \(\frac{\mathrm{Q}}{\left(\frac{\mathrm{Q} d}{\mathrm{A} \varepsilon_{0}}\right)}\) = \(\frac{\varepsilon_{0} \mathrm{A}}{d}\) ….. (3)
From equation (3), it is evident that capacitance is directly proportional to the area of cross-section and is inversely proportional to the distance between the plates. This can be understood from the following.

  • If the area of cross-section of the capacitor plates is increased, more charges can be distributed for the same potential difference. As a result, the capacitance is increased.
  • If the distance d between the two plates is reduced, the potential difference between the plates (V = Ed) decreases with the E constant.

Question 19.
Obtain the expression for energy stored in the parallel plate capacitor.
Answer:
Energy stored in the capacitor:
Capacitor not only stores the charge but also it stores energy. When a battery is connected to the capacitor, electrons of total charge -Q are transferred from one plate to the other plate. To transfer the charge, work is done by the battery. This work done is stored as electrostatic potential energy in the capacitor. To transfer an infinitesimal charge dQ for a potential difference V, the work done is given by
dW = VdQ ….. (1)
Where V = \(\frac { Q }{ C }\)
The total work done to charge a capacitor is
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-73
This work done is stored as electrostatic potential energy (UE) in the capacitor.
UE = \(\frac {{ Q }^{2}}{ 2C }\) = \(\frac { 1 }{ 2 }\) CV2 (∴ Q = CV) ….. (3)
where Q = CV is used. This stored energy is thus directly proportional to the capacitance of the capacitor and the square of the voltage between the plates of the capacitor. But where is this energy stored in the capacitor? To understand this question, the equation (3) is rewritten as follows using the results
C = \(\frac{\varepsilon_{0} \mathrm{A}}{d}\) and V = Ed
UE = \(\frac { 1 }{ 2 }\) \(\left(\frac{\varepsilon_{0} \mathrm{A}}{d}\right)\) (Ed)2 = \(\frac { 1 }{ 2 }\) ε0(Ad)2 …… (4)
where Ad = volume of the space between the capacitor plates. The energy stored per unit volume of space is defined as energy density \(\overline { Volume } \). Frome equation (4) we get
uE = \(\frac { 1 }{ 2 }\) ε0E2
From equation (5), we infer that the energy is stored in the electric field existing between the plates of the capacitor. Once the capacitor is allowed to discharge, the energy is retrieved.

Question 20.
Explain in detail the effect of a dielectric placed in a parallel plate capacitor.
Answer:
(i) When the capacitor is disconnected from the battery:
Consider a capacitor with two parallel plates each of cross-sectional area A and are separated by a distance d. The capacitor is charged by a battery of voltage V0 and the charge stored is Q0. The capacitance of the capacitor without the dielectric is
C0 = \(\frac {{ Q }_{0}}{{ V }_{0}}\) ….. (1)
The battery is then disconnected from the capacitor and the dielectric is inserted between the plates. The introduction of dielectric between the plates will decrease the electric field. Experimentally it is found that the modified electric field is given by
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-74
(a) Capacitor is charged with a battery
(b) Dielectric is inserted after the battery is disconnected
E = \(\frac {{ E }_{0}}{{ ε }_{r}}\) …… (2)
Here E0 is the electric field inside the capacitors when there is no dielectric and εr is the relative permeability of the dielectric or simply known as the dielectric constant. Since εr > 1, the electric field E < E0. As a result, the electrostatic potential difference between the plates (V = Ed) is also reduced. But at the same time, the charge Q0 will remain constant once the battery is disconnected. Hence the new potential difference is
V = Ed = \(\frac {{ E }_{0}}{{ ε }_{r}}\)d = \(\frac {{ V }_{0}}{{ ε }_{r}}\) ….. (3)
We know that capacitance is inversely proportional to the potential difference. Therefore as V decreases, C increases. Thus new capacitance in the presence of a dielectric is
C = \(\frac {{ Q }_{0}}{ V }\) = εr \(\frac {{ Q }_{0}}{{ V }_{0}}\) = εr C0 …… (4)
Since εr > 1, we have C > C0. Thus insertion of the dielectric constant εr increases the capacitance. Using equation,
C = \(\frac { { \varepsilon }_{ 0 }A }{ d } \)
C = \(\frac{\varepsilon_{r} \varepsilon_{o} A}{d}\) = \(\frac { εA }{ d }\) …… (5)
where ε = εrε0 is the permittivity of the dielectric medium. The energy stored in the capacitor before the insertion of a dielectric is given by U0 = \(\frac { 1 }{ 2 }\) \(\frac{\mathrm{Q}_{0}^{2}}{\mathrm{C}_{0}}\) ….. (6)
After the dielectric is inserted, the charge Q0 remains constant but the capacitance is increased. As a result, the stored energy is decreased.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-75
Since εr> 1 we get U < U0. There is a decrease in energy because, when the dielectric is inserted, the capacitor spends some energy in pulling the dielectric inside.

(ii) When the battery remains connected to the capacitor: Let us now consider what happens when the battery of voltage V0 remains connected to the capacitor when the dielectric is inserted into the capacitor.
The potential difference V0 across the plates remains constant. But it is found experimentally (first shown by Faraday) that when the dielectric is inserted, the charge stored in the capacitor is increased by a factor εr.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-76
(a) Capacitor is charged through a battery
(b) Dielectric is inserted when the battery is connected.
Q = εrQ0 ….. (1)
Due to this increased charge, the capacitance is also increased. The new capacitance is
C = \(\frac {{ Q }_{0}}{ V }\) = εr \(\frac {{ Q }_{0}}{{ V }_{0}}\) = εr C0 …… (2)
However, the reason for the increase in capacitance in this case when the battery remains connected is different from the case when the battery is disconnected before introducing the dielectric.
Now, C0 = \(\frac {{ ε }{_0}A}{ d }\) and, C = \(\frac { εA }{ d }\) …… (3)
U0 = \(\frac { 1 }{ 2 }\) C0 \({ V }_{ 0 }^{ 2 }\) ….. (4)
Note that here we have not used the expression
U0 = \(\frac { 1 }{ 2 }\)\({{ V }_{ 0 }^{ 2 }}{{C}_{0}}\)
because here, both charge and capacitance are changed, whereas in equation 4, V0 remains constant. After the dielectric is inserted, the capacitance is increased; hence the stored energy is also increased.
U = \(\frac { 1 }{ 2 }\) \({ CV }_{ 0 }^{ 2 }\) = \(\frac { 1 }{ 2 }\) εr \({ CV }_{ 0 }^{ 2 }\) = εr U0
Since er > 1 we have U > U0
It may be noted here that since voltage between the capacitor V0 is constant, the electric field between the plates also remains constant.

Question 21.
Derive the expression for resultant capacitance, when capacitors are connected in series and in parallel.
Answer:
capacitors in series and parallel:
(i) Capacitors in series:
Consider three capacitors of capacitance C1, C2 and C3 connected in series with a battery of voltage V as shown in figure (a).
As soon as the battery is connected to the capacitors in series, the electrons of charge -Q are transferred from the negative terminal to the right plate of C3which pushes the electrons of the same amount -Q from left plate of C3 to the right plate of C2 due to electrostatic induction. Similarly, the left plate of C2 pushes the charges of Q to the right plate of which induces the positive charge +Q on the left plate of C1 At the same time, electrons of charge -Q are transferred from the left plate of C1 to the positive terminal of the battery.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-77
By these processes, each capacitor stores the same amount of charge Q. The capacitances of the capacitors are in general different so that the voltage across each capacitor is also different and are denoted as V1, V2 and V3 respectively.
The total voltage across each capacitor must be equal to the voltage of the battery.
V = V1 + V2 + V3 ….. (1)
Since Q = CV, we have V = \(\frac { Q }{{ C }_{1}}\) + \(\frac { Q }{{ C }_{2}}\) + \(\frac { Q }{{ C }_{3}}\)
Q = \(\left( \frac { 1 }{ { C }_{ 1 } } +\frac { 1 }{ { C }_{ 2 } } +\frac { 1 }{ { C }_{ 3 } } \right) \) ….. (2)
If three capacitors in series are considered to form an equivalent single capacitor Cs shown in figure (b), then we have V = \(\frac { Q }{{ C }_{s}}\)
Substituting this expression into equation (2) we get
V = \(\frac { Q }{{ C }_{s}}\) = Q\(\left( \frac { 1 }{ { C }_{ 1 } } +\frac { 1 }{ { C }_{ 2 } } +\frac { 1 }{ { C }_{ 3 } } \right) \)
\(\frac { 1 }{{ C }_{s}}\) = \(\frac { 1 }{{ C }_{1}}\) + \(\frac { 1 }{{ C }_{2}}\) + \(\frac { 1 }{{ C }_{3}}\) ….. (3)
Thus, the inverse of the equivalent capacitance Cs of three capacitors connected in series is equal to the sum of the inverses of each capacitance. This equivalent capacitance Cs is always less than the smallest individual capacitance in the series.

(ii) Capacitance in parallel:
Consider three capacitors of capacitance C1,C2 and C3 connected in parallel with a battery of voltage V as shown in figure (a).
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-78
Since corresponding sides of the capacitors are connected to the same positive and negative terminals of the battery, the voltage across each capacitor is equal to the battery’s voltage. Since the capacitance of the capacitors is different, the charge stored in each capacitor is not the same. Let the charge stored in the three capacitors be Q1,Q2, and Q2 respectively. According to the law of conservation of total charge, the sum of these three charges is equal to the charge Q transferred by the battery,
Q = Q1 + Q2 + Q3 ….. (1)
Now, since Q = CV, we have
Q = C1V + C2 V + C3 V ….. (2)
If these three capacitors are considered to form a single capacitance CP which stores the total charge Q as shown in figure (b), then we can write Q = CPV. Substituting this in equation (2), we get
Cp V = C1 V + C2 V + C3 V
Cp = C1 + C2 + C3
Thus, the equivalent capacitance of capacitors connected in parallel is equal to the sum of the individual capacitance. The equivalent capacitance Cp in a parallel connection is always greater than the largest individual capacitance. In a parallel connection, it is equivalent as an area of each capacitance adds to give a more effective area such that total capacitance increases.

Question 22.
Explain in detail how charges are distributed in a conductor, and the principle behind the lightning conductor.
Answer:
Distribution of charges in a conductor: Consider two conducting spheres A and B of radii r1 and r2 respectively connected to each other by a thin conducting wire as shown in the figure. The distance between the spheres is much greater than the radii of either sphere.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-79
If a charge Q is introduced into any one of the spheres, this charge Q is redistributed into both the spheres such that the electrostatic potential is same in both the spheres. They are now uniformly charged and attain electrostatic equilibrium. Let q1 be the charge residing on the surface of sphere A and q2 is the charge residing on the surface of sphere B such that Q = q1 + q2 The charges are distributed only on the surface and there is no net charge inside the conductor. The electrostatic potential at the surface of sphere A is given by
VA = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ q }_{ 2 }}{{ r }_{ 2 }}\) …. (1)
The electrostatic potential at the surface of sphere B is given by
VB = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ q }_{ 2 }}{{ r }_{ 2 }}\) ….. 2)
The surface of the conductor is an equipotential. Since the spheres are connected by the conducting wire, the surfaces of both the spheres together form an equipotential surface. This implies that
VA = VB or \(\frac {{ q }_{ 1 }}{{ r }_{ 1 }}\) = \(\frac {{ q }_{ 2 }}{{ r }_{ 2 }}\) ….. (3)
Let us take the charge density on the surface of sphere A is σ1 and charge density on the surface of sphere B is σ1. This implies that q1 = \({ 4\pi r }_{ 1 }^{ 2 }\)σ1 and q1 = \({ 4\pi r }_{ 1 }^{ 2 }\)σ2. Substituting these values into equation (3), we get
σ1 r1 = σ2r2 ….. (4)
from which we conclude that
σr = constant …. (5)
Thus the surface charge density o is inversely proportional to the radius of the sphere. For a smaller radius, the charge density will be larger and vice versa.

Lightning arrester or lightning conductor:
This is a device used to protect tall buildings from lightning strikes. It works on the principle of action at points or corona discharge. The device consists of a long thick copper rod passing from top of the building to the ground. The upper end of the rod has a sharp spike or a sharp needle.

The lower end of the rod is connected to the copper plate which is buried deep into the ground. When a negatively charged cloud is passing above the building, it induces a positive charge on the spike. Since the induced charge density on thin sharp spike is large, it results in a corona discharge.

This positive charge ionizes the surrounding air which in turn neutralizes the negative charge in the cloud. The negative charge pushed to the spikes passes through the copper rod and is safely diverted to the Earth. The lightning arrester does not stop the lightning; rather it divers the lightning to the ground safely.

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 23.
Explain in detail the construction and working of a Van de Graaff generator.
Answer:
Principle: Electrostatic induction and action at points.
Construction:
A large hollow spherical conductor is fixed on the insulating stand. A pulley B is mounted at the center of the hollow sphere and another pulley C is fixed at the bottom. A belt made up of insulating materials as silk or rubber runs over both pulleys. The pulley C is driven continuously by the electric motor.

Two comb-shaped metallic conductors E and D are fixed near the pulleys. The comb D is maintained at a positive potential of 104 V by a power supply. The upper comb E is connected to the inner side of the hollow metal sphere.

Working:
Due to the high electric field near comb D, the air between the belt and comb D gets ionized. The positive charges are pushed towards the belt and negative charges are attracted towards the comb D. The positive charges stick to the belt and move up. When the positive charges reach the comb E, a large amount of negative and positive charges are induced on either side of comb E due to electrostatic induction. As a result, the positive charges are pushed away from the comb E and they reach the outer surface of the sphere. Since the sphere is a conductor, the positive charges are distributed uniformly on the outer surface of the hollow sphere. At the same time, the negative charges nullify the positive charges in the belt due to corona discharge before it passes over the pulley.
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When the belt descends, it has almost no net charge. At the bottom, it again gains a large positive charge. The belt goes up and delivers positive charges to the outer surface of the sphere. This process continues until the outer surface produces the potential difference of the order of 107 which is the limiting value. We cannot store charges beyond this limit since the extra charge starts leaking to the surroundings due to the ionization of air. The leakage of charges can be reduced by enclosing the machine in a gas-filled steel chamber at very high pressure. Uses: The high voltage produced in this Van de Graaff generator is used to accelerate positive ions (protons and deuterons) for nuclear disintegrations and other applications.

Samacheer Kalvi 12th Physics Electrostatics Numerical Problems

Question 1.
When two objects are rubbed with each other, approximately a charge of 50 nC can be produced in each object. Calculate the number of electrons that must be transferred to produce this charge.
Solution:
Charge produced in each object q = 50 nC
q = 50 x 10-9 C
Charge of electron (e) = 1.6 x 10-9 C
Number of electron transferred, n = \(\frac { q }{ e }\) = \(\frac {{ 50 × 10 }^{-9}}{{ 1.6 × 10 }^{-19}}\)
=31. 25 × 10-9 × 1019
n = 31.25 x 1010 electrons
Ans. n = 31.25 x 1010 electrons

Question 2.
The total number of electrons in the human body is typically in the order of 1028. Suppose, due to some reason, you and your friend lost 1% of this number of electrons. Calculate the electrostatic force between you and your friend separated at a distance of 1 m. Compare this with your weight. Assume the mass of each person is 60 kg and use point charge approximation.
Solution:
Number of electrons in the human body = 1028
Number of electrons in me and my friend after lost of 1% = 1028 x 1%
= 1028 x \(\frac { 1 }{ 100 }\)
n = 1026 electrons
Separation distance d = 1 m
Charge of each person q = 1026 x 1.6 x 10-19
q = 1.6 x 107 C
Electrostatic force, F = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{r^{2}}\) = \(\frac{9 \times 10^{9} \times 1.6 \times 10^{7} \times 1.6 \times 10^{7}}{1^{2}}\)
F = 2.304 x 1024N
Mass of the person, M = 60 kg
Acceleration due to gravity, g = 9.8 ms-2
Weight (W) = mg
= 60 x 9.8
W = 588 N
Comparison: Electrostatic force is equal to 3.92 x 1021 times of weight of the person.

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 3.
Five identical charges Q are placed equidistant on a semicircle as shown in the figure. Another point charge q is kept at the center of the circle of radius R. Calculate the electrostatic force experienced by the charge q.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-81
Solution:
Force acting on q due to Q1 and Q5 are opposite direction, so cancel to each other.
Force acting on q due to Q3 is F3 = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ qQ }_{3}}{{ R }^{2}}\)
Force acting on q due to Q2 and Q4

Resolving in two-component method:
(i) Vertical Component:
Q2 Sin θ and Q4 Sinθ are equal and opposite directions, so they cancel to each other.

(ii) Horizontal Component:
Q2 Sin θ and Q4 cos θ are equal and same direction, so they can get added.
F24 = F2q + F4q = F2 cos 55° + F4 cos 45°
F24 = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ qQ }_{2}}{{ R }^{2}}\) cos 45° + \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { qQ 4}{{ R }^{2}}\) cos 45°
Resultant net force F
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-82
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-83

Question 4.
Suppose a charge +q on Earth’s surface and another +q charge is placed on the surface of the Moon, (a) Calculate the value of q required to balance the gravitational attraction between Earth and Moon (b) Suppose the distance between the Moon and Earth is halved, would the charge q change? (Take mE = 5.9 x 1024 kg, mM = 7.348 x 1022 kg)
Solution:
Mass of the Earth, ME = 5.9 x 1024 kg
Mass of the Moon, MM = 7.348 x 1022 kg
Charge placed on the surface of Earth and Moon = q
(a) Required charge to balance the FG between Earth and Moon
FC = FG (or) \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ q }^{2}}{{ r }^{2}}\) = \(\frac{\mathrm{G} \mathrm{M}_{\mathrm{E}} \times \mathrm{M}_{\mathrm{M}}}{r^{2}}\)
q2 = G × ME × MM × 4πε0 = 320.97 × 1025
q = \(\sqrt { 320.97\times { 10 }^{ 25 } } \) = 5.665 x 1013 = 5.67 x 1013 C

(b) The distance between the Moon and Earth is
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-84
so q = 5.67 x 1013 C
There is no change.

Question 5.
Draw the free body diagram for the following charges as shown in figure (a), (b) and (c).
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-85
Solution:

Question 6.
Consider an electron travelling with a speed VΦ and entering into a uniform electric field \(\vec { E } \) which is perpendicular to \(\overrightarrow{\mathrm{V}_{0}}\) as shown in the Figure. Ignoring gravity, obtain the electron’s acceleration, velocity and position as functions of time.
Solution:
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-86
Speed of an electron = V0
Uniform electric field = \(\vec { E } \)
(а) Electron’s acceleration:
Force on electron due to uniform electric field, F = Ee
Downward acceleration of electron due to electric field, a = \(\frac { F }{ m }\) = – \(\frac { eE }{ M }\)
Vector from, \(\vec { a } \) = – \(\frac { eE }{ M }\) \(\hat{j} \)

(b) Electron’s velocity:
Speed of electron in horizontal direction, u = V0 From the equation of motion, V = u + at
V = V0 \(\frac { eE }{ M }\) t
Vector from \(\vec { V } \) = V0 \(\hat{j} \) – \(\frac { eE }{ M }\) t \(\hat{j} \)

(c) Electron’s position:
Position of electron, s = r
From equation of motion, r = V0 t + \(\frac { 1 }{ 2 }\) \(\left(-\frac{e \mathrm{E}}{\mathrm{M}}\right)\) t2
r = V0 t + \(\frac { 1 }{ 2 }\) \(\frac { eE }{ M }\) t2 \(\hat{j} \)
Vector from,
\(\vec { r } \) = V0 t \(\hat{j} \) \(\frac { 1 }{ 2 }\) \(\frac { eE }{ M }\) t2 \(\hat{j} \)

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 7.
A closed triangular box is kept in an electric field of magnitude E = 2 × 103 N C-1 as shown in the figure.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-87
Calculate the electric flux through the
(a) vertical rectangular surface
(b) slanted surface and
(c) entire surface.
Answer:
Electric field of magnitude E = 2 × 103 NC-1
(a) Vertical rectangular surface:
Rectangular area A= 5 × 10-2 × 15 × 10-2
A = 75 × 10-24 m2
θ = 180°
⇒ cos 180° = -1
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-88
Electric flux, Φv.s = EA cos θ
= 2 × 103 × 75 × 10-4 × cos 180°
= -150 × 10-1
Φv.s = -15 Nm2 C-1

(b) Slanted surface:
cos θ = cos 60° = 0.5
sin θ = sin 30° = \(\frac { Opposite }{ hyp }\)
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-89
hyp = \(\frac {{ 5 × 10 }^{2}}{ 0.5 }\)
hyp = 0.1m
Area of slanted surface A2 = (0.1 × 15 × 10-2)
A2 = 0.015 M2
Electric flux, Φv.s = EA = cos θ
= 2 × 103 × 0.015 × cos 60°
= 2 × 103 × 0.015 × 103
= 0.015 × 103
Φv.s = 15 Nm2 C-1
Horizontal surface
θ = 90° ; cos 90° = 0
Electric flux, ΦH.S = E. A3 Cos 90° = 0

(c) Entire surface:
ΦTotal = ΦV.S + ΦS.S + ΦH.S = -15 + 15 + 0
ΦTotal = 0

Question 8.
The electrostatic potential is given as a function of x in figure (a) and (b). Calculate the corresponding electric fields in regions A, B, C and D. Plot the electric field as a function of x for the figure (b).
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-90
Answer:
The relation between electric field and potential
E = – \(\frac { dv }{ dx }\)

(a) Region A :
dv = -3V ; dx = 0.2 m
Electric field, EA = \(\frac { (-3) }{ 0.2 }\) = 15 V m-1
Region B:
dv = 0V ; dx = 0.2 m
Electric field, EB = \(\frac { 0 }{ 0.2 }\) = 0
Region C:
dv = 2V ; dx = 0.2 m
Electric field, EC = \(\frac { -2 }{ 0.2 }\) = 10 V m-1
Region D:
dv = -6V ; dx = 0.2 m
Electric field, ED = \(-\left(\frac{-6}{0.2}\right)\) = 10 V m-1 = 30 V m-1

Electric field, EA = 15 V m-1
Electric field, EB = 0
Electric field, EC = \(\frac { (-3) }{ 0.2 }\) = 10 V m-1
Electric field, ED = 30 V m-1

(b)
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-91

Question 9.
A spark plug in a bike or a car is used to ignite the air-fuel mixture in the engine. It consists of two electrodes separated by a gap of around 0.6 mm gap as shown in the figure.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-92
To create the spark, an electric field of magnitude 3 x 106Vm-1 is required, (a) What potential difference must be applied to produce the spark? (b) If the gap is increased, does the potential difference increase, decrease or remains the same? (c) find the potential difference if the gap is 1 mm.
Answer:
Separation gap between two electrodes, d = 0.6 mm
d = 0.6 × 10-3 m
Magnetude of electric field Electric field = E = 3 × 106 V m-1
Electric field E = \(\frac { V }{ d }\)
(a) Applied potential difference, V = E . d
= 3 × 106 × 0.6 10-13 = 1.8 × 103
V = 1800 V

(b) From equation, V = E . d
If the gap (distance) between the electrodes increased, the potential difference also increases.

(c) Gap between the electrodes, d = 1mm = 1 x 10-3 m
Potential difference, V = E.d
= 3 × 106 × 1 × 10-3 = 3 × 103
V = 3000 V

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 10.
A point charge of +10 μC is placed at a distance of 20 cm from another identical point charge of +10 μC. A point charge of -2 μC is moved from point a to b as shown in the figure. Calculate the
change in potential energy of the system? Interpret your result.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-93
q1 = 10μC = 10 x 10-6 C
q2 = 2μC = -2 x 10-6 C
distance, r = 5cm = 5 x 10-2 m
Answer:
Change in potential energy,
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-94
= -36 × 1 × 109 × 10-12 × 102 = -36 × 10-1
∆ U = -3.6 J

Negative sign implies that to move the charge -2pC no external work is required. System spends its stored energy to move the charge from point a to point b.
Ans:
∆ U = -3.6 J, negative sign implies that to move the charge -2μC no external work is required. System spends its stored energy to move the charge from point a to point b.

Question 11.
Calculate the resultant capacitances for each of the following combinations of capacitors.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-95
Answer:
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-96
Parallel combination of capacitor 1 and 2
Cp = C0 + C0 = 2C0
Series combination of capacitor Cp and 3
\(\frac { 1 }{{ C }_{S}}\) = \(\frac { 1 }{{ C }_{p}}\) + \(\frac { 1 }{{ C }_{3}}\) = \(\frac { 1 }{{ 2C }_{0}}\) + \(\frac { 1 }{{ C }_{0}}\) = (or) \(\frac { 1 }{{ C }_{S}}\) = \(\frac { 3 }{ 2 }\) C(or)CS = \(\frac { 2 }{ 3 }\) C

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-97
\(\frac { 1 }{ { C }_{ { S }_{ 1 } } } \) = \(\frac { 1 }{{ C }_{1}}\) + \(\frac { 1 }{{ C }_{2}}\) = \(\frac { 1 }{{ C }_{0}}\) + \(\frac { 1 }{{ C }_{0}}\) = \(\frac { 1 }{{ C }_{0}}\) (or)
\(\frac { 1 }{ { C }_{ { S }_{ 1 } } } \) = \(\frac { 2 }{{ C }_{0}}\) (or) \({ C }_{ { S }_{ 1 } }\) = \(\frac {{ C }_{0}}{ 2 }\)
Similarly 3 and 4 are series combination
\(\frac { 1 }{ { C }_{ { S }_{ 2 } } } \) = \(\frac { 1 }{{ C }_{3}}\) + \(\frac { 1 }{{ C }_{4}}\) = \(\frac { 1 }{{ C }_{0}}\) + \(\frac { 1 }{{ C }_{0}}\) = \(\frac { 2 }{{ C }_{0}}\) (or) \({ C }_{ { S }_{ 2 } }\) = \(\frac {{ C }{0}}{ 2 }\)
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-98
\({ C }_{ { S }_{ 1 } }\) and \({ C }_{ { S }_{ 2 } }\) are in parallel combination
Cp = \({ C }_{ { S }_{ 1 } }\) + \({ C }_{ { S }_{ 2 } }\) = \(\frac {{ C }_{0}}{ 2 }\) + \(\frac {{ C }_{0}}{ 2 }\) (or) Cp = \(\frac {{ 2C }_{0}}{ 2 }\) Cp = C0

(c) Capacitor 1, 2 and 3 are in parallel combination
Cp = C0 + C0 + C0 = 3C0
Cp = 3C0
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-99

(d) Capacitar C1 and C2 are in combination
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-100

Similarly C3 and C4 are in series combination
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-101
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-102
\({ C }_{ { S }_{ 1 } }\) and \({ C }_{ { S }_{ 2 } }\) are in parallel combination across RS:
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-103

(e) Capacitor 1 and 2 are series combination
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-104
Similarly 3 and 4 are series combination
\(\frac { 1 }{ { C }_{ { S }_{ 2 } } } \) = \(\frac { 2 }{{ C }_{0}}\) (or) \({ C }_{ { S }_{ 2 } }\) = \(\frac {{ C }_{0}}{ 2 }\)
Three capacitors are in parallel combination
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-105

Question 12.
An electron and a proton are allowed to fall through the separation between the plates of a parallel plate capacitor of voltage 5 V and separation distance h = 1 mm as shown in the figure.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-106
(a) Calculate the time of flight for both electron and proton
(b) Suppose if a neutron is allowed to fall, what is the time of flight?
(c) Among the three, which one will reach the bottom first?
(Take mp = 1.6 x 10-27 kg, me= 9.1 x 10-31 kg and g = 10 m s-2)
Answer:
Potential difference between the parallel plates V = 5 V
Separation distance, h = 1 mm =1 x 10-3 m
Mass of proton, mp = 1.6 x 10-27 kg
Mass of proton, m =9.1 x 10-31 kg
Charge of an a proton (or) electron, e— 1.6 x 10-19 C
[u = 0; s = h]
From equation of motion, S = ut + \(\frac { 1 }{ 2 }\) at2
From equation of motion, h = \(\frac { 1 }{ 2 }\) at2
t = \(\sqrt { \frac { 2h }{ a } } \)
Acceleration of an electron due to electric field, a = \(\frac { F }{ m }\) = \(\frac { eE }{ m }\)
[E = \(\frac { V }{ d }\)]

(a) Time of flight for both electron and proton,
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-107
tp = 63 ns……. (2)

(b) time of flight of neutron tn = \(\sqrt { \frac { 2h }{ g } } \) = \(\sqrt{\frac{2 \times 1 \times 10^{3}}{10}}\) = \(\sqrt{0.2 \times 10^{-3}}\)
tn = 0.0141 s = 14.1 x 10-3 s
tn = 14.1 x 10-3 ms ……. (3)
(c) Compairision of values 1,2 and 3. The electron will reach the bottom first.

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 13.
During a thunderstorm, the movement of water molecules within the clouds creates friction, partially causing the bottom part of the clouds to become negatively charged. This implies that the bottom of the cloud and the ground act as a parallel plate capacitor. If the electric field between the cloud and ground exceeds the dielectric breakdown of the air (3 x 106 Vm-1 ), lightning will occur.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-108
(a) If the bottom part of the cloud is 1000 m above the ground, determine the electric potential difference that exists between the cloud and ground.
(b) In a typical lightning phenomenon, around 25C of electrons are transferred from cloud to ground. How much electrostatic potential energy is transferred to the ground?
Answer:
(a) Electric field between the cloud and ground,
V = E.d
V= 3 x 106 x 1000 = 3 x 109V
(a) Electrons transfered from cloud to ground,
q = 25 C
Electron static potential energy,
U = \(\frac { 1 }{ 2 }\) CV2
[C = \(\frac { q }{ V }\)]
= \(\frac { 1 }{ 2 }\) qV = \(\frac { 1 }{ 2 }\) x 25 x 3 x 109
U = 37.5 x 109 J

Question 14.
For the given capacitor configuration
(a) Find the charges on each capacitor
(b) potential difference across them
(c) energy stored in each capacitor.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-109
Answer:
Capacitor b and c in parallel combination
Cp = Cb + Cc = (6 + 2) μF = 8 μF
Capacitor a, cp and d are in series combination, so the resulatant copacitance
\(\frac { 1 }{{ C }_{s}}\) = \(\frac { 1 }{{ C }_{a}}\) + \(\frac { 1 }{{ C }_{cp}}\) + \(\frac { 1 }{{ C }_{d}}\) = \(\frac { 1 }{ 8 }\) + \(\frac { 1 }{ 8 }\) + \(\frac { 1 }{ 8 }\) = \(\frac { 3 }{ 8 }\)
Cs  = \(\frac { 8 }{ 3 }\) μF
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-110

(a) Charge on each capacitor,
Charge on capacitor a, Qa = Cs V = \(\frac { 8 }{ 3 }\) x 9
Qa = 24 μC
Charge on capacitor, d, Qd = Cs V = \(\frac { 8 }{ 3 }\) x 9
Qd = 24 μC
Capacitor b and c in parallel
Charge on capacitor, b, Qb = \(\frac { 6 }{ 3 }\) x 9 = 18
Qb = 18 μC
Charge on capacitor, c, Qc = \(\frac { 2 }{ 3 }\) x 9 = 6
Qc = 6 μC

(b) Potential difference across each capacitor, V = \(\frac { q }{ C }\)
Capacitor Ca, Va = \(\frac{ { q }_{a}}{{ C }_{a}}\) = \(\frac {{ 24 × 10 }^{6}}{{ 8 × 10 }^{6}}\) = 3 V
Capacitor Cb, Vb = \(\frac{ { q }_{b}}{{ C }_{b}}\) = \(\frac {{ 18 × 10 }^{6}}{{ 6 × 10 }^{6}}\) = 3 V
Capacitor Cc, Vc = \(\frac{ { q }_{c}}{{ C }_{c}}\) = \(\frac {{ 6 × 10 }^{6}}{{ 2 × 10 }^{6}}\) = 3 V
Capacitor Cd, Vd = \(\frac{ { q }_{d}}{{ C }_{d}}\) = \(\frac {{ 24 × 10 }^{6}}{{ 8 × 10 }^{6}}\) = 3 V

(c) Energy stores in a capacitor, U = \(\frac { 1 }{ 2 }\) CV2
Energy in capacitor Ca, Ua = \(\frac { 1 }{ 2 }\) Ca \({ V }_{ a }^{ 2 }\) = \(\frac { 1 }{ 2 }\) x 8 x 10-6 x (3)2
Ua = 36 μJ
Capacitor Cb, Ub = \(\frac { 1 }{ 2 }\) Cb \({ V }_{ b }^{ 2 }\) = \(\frac { 1 }{ 2 }\) x 6 x 10-6 x (3)2
Ua = 27 μJ
Cc, Uc = \(\frac { 1 }{ 2 }\) Cc \({ V }_{ c }^{ 2 }\) = \(\frac { 1 }{ 2 }\) x 2 x 10-6 x (3)2
Ua = 9 μJ
Cd, Ud = \(\frac { 1 }{ 2 }\) Cd \({ V }_{ d }^{ 2 }\) = \(\frac { 1 }{ 2 }\) x 8 x 10-6 x (3)2
Ua = 36 μJ

Question 15.
Capacitors P and Q have identical cross-sectional areas A and separation d. The space between the capacitors is filled with a dielectric of dielectric constant as shown in the figure. Calculate the capacitance of capacitors P and Q.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-111
Answer:
Cross-sectional area of parallel plate capacitor = A
Each area of different medium between parallel plate capacitor = \(\frac { A }{ 2 }\)
Separation distance = d
Capacitance of parallel plate capacitor, C = \(\frac { εA }{ d }\)
Air medium of dielectric constant, εr = 1
dielectric medium of dielectric constant = εr

Case 1:
Capacitance of air filled capacitor
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-112
Capacitance of dielectric-filled capacitor
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-113
Capacitance of parallel plate capacitor
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-114

Case 2:
Each distance of different medium between the parallel plate capacitor = \(\frac { d }{ 2 }\)
Capacitance of dielectric-filled capacitor
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-115
Capacitance of air filled capacitor,
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-116
Capacitance of parallel plate capacitor,
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-117

Samacheer Kalvi 12th Physics Electrostatics Additional Questions Solved

I. Multiple Choice Questions

Question 1.
When a solid body is negatively charged by friction, it means that the body has
(a) acquired excess of electrons
(b) lost some, problems
(c) acquired some electrons and lost a lesser number of protons
(d) lost some positive ions
Answer:
(a) acquired excess of electrons

Question 2.
A force of 0.01 N is exerted on a charge of 1.2 x 10-5 G at a certain point. The electric field at that point is
(a) 5.3 x 104 NC-1
(b) 8.3 x 10-4 NC-1
(c) 5.3 x 102 NC-1
(d) 8.3 x 104 NC-1
Answer:
(d) 8.3 x 104 NC-1
Hint:
E = \(\frac { F }{ q }\) = \(\frac { 0.01 }{{ 1.2 × 10 }^{-5}}\) = 8.3 x 102 NC-1

Question 3.
The electric field intensity at a point 20 cm away from a charge of 2 x 10 5 C is
(a) 4.5 x 106 NC-1
(b) 3.5 x 105 NC-1
(c) 3.5 x 106 NC-1
(d) 4.5 x 105 NC-1
Answer:
(a) 4.5 x 106 NC-1
Hint:
E = \(\frac{q}{4 \pi \varepsilon_{0} r^{2}}\) = \(\frac{9 \times 10^{9} \times 2 \times 10^{-5}}{(0.2)^{2}}\) = 4.5 x 106 NC-1

Question 4.
How many electrons will have a charge of one coulomb?
(a) 6.25 x 1018
(b) 6.25 x 1019
(c) 1.6 x 1018
(d) 1.6 x 1019
Answer:
(a) 6.25 x 1018
Hint:
Number of electron, n = \(\frac { q }{ e }\) = \(\frac { 1 }{{ 1.6 × 10 }^{-19}}\) = 6.25 × 1018

Question 5.
The ratio of the force between two charges in air and that in a medium of dielectric constant K is
(a) K : 1
(b) 1 : K
(c) K2 : 1
(d) 1 : K2
Answer:
(a) K : 1

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 6.
The work done in moving a positive charge on an equipotential surface is
(a) finite and positive
(b) infinite
(c) finite and negative
(d) zero
Answer:
(d) zero

Question 7.
If a charge is moved against the Coulomb force of an electric field.
(a) work is done by the electric field
(b) energy is used from some outside source
(c) the strength of the field is decreased
(d) the energy of the system is decreased
Answer:
(b) energy is used from some outside source

Question 8.
No current flows between two charged bodies when connected
(a) if they have the same capacitance
(b) if they have the same quantity of charge
(c) if they have the same potential
(d) if they have the same charge density
Answer:
(c) if they have the same potential

Question 9.
Electric field lines about a negative point charge are
(a) circular, anticlockwise
(b) circular, clockwise
(c) radial, inwards
(d) radial, outwards
Answer:
(c) radial, inwards

Question 10.
Two plates are 1 cm apart and the potential difference between them is 10 V. The electric field between the plates is
(a) 10 NC-1
(b) 250 NC-1
(c) 500 N-1
(d) 1000 NC-1
Answer:
(d) 1000 NC-1
Hint:
E = \(\frac { V }{ d }\) = \(\frac { 10 }{{ 1 × 10 }^{-2}}\) = 8.3 x 102 NC-1

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 11.
At a large distance (r), the electric field due to a dipole varies as
(a) \(\frac { 1 }{ r }\)
(b) \(\frac { 1 }{{ r }^{2}}\)
(c) \(\frac { 1 }{{ r }^{3}}\)
(d) \(\frac { 1 }{{ r }^{4}}\)
Answer:
(c) \(\frac { 1 }{{ r }^{3}}\)

Question 12.
Two thin infinite parallel plates have uniform charge densities +c and -σ. The electric field in the space between then is
(a) \(\frac { σ }{{ 2ε }_{0}}\)
(b) \(\frac { σ }{{ ε }_{0}}\)
(c) \(\frac { 2σ }{{ 2ε }_{0}}\)
(d) Zero
Answer:
(b) \(\frac { σ }{{ ε }_{0}}\)

Question 13.
Two isolated, charged conducting spheres of radii R1, and R2 produce the same electric field near their surfaces. The ratio of electric potentials on their surfaces is-
(a) \(\frac {{ R }_{1}}{{ R }_{2}}\)
(b) \(\frac {{ R }_{2}}{{ R }_{1}}\)
(c) \(\frac { { R }_{ 1 }^{ 2 } }{ { R }_{ 2 }^{ 2 } } \)
(d) \(\frac { { R }_{ 2 }^{ 2 } }{ { R }_{ 1 }^{ 2 } } \)
Answer:
(b) \(\frac {{ R }_{2}}{{ R }_{1}}\)

Question 14.
A 100 μF capacitor is to have an energy content of 50 J in order to operator a flash lamp. The voltage required to charge the capacitor is
(a) 500 V
(b) 1000 V
(c) 1500 V
(d) 2000 V
Answer:
(b) 1000 V
Hint:
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-118

Question 15.
A 1 μF capacitor is placed in parallel with a 2 μF capacitor across a 100 V supply. The total charge on the system is
(a) \(\frac { 100 }{ 3 }\) μC
(b) 100 μC
(c) 150 μC
(d) 300 μC
Answer:
(d) 300 μC
Hint:
Equivalent capacitor = 1 + 2 = 3 μF
Total charge, q = CV = 3 x 100 = 300 μF

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 16.
A parallel plate capacitor of capacitance 100 μF is charged to 500 V. The plate separation is then reduced to half its original value. Then the potential on the capacitor becomes
(a) 250 V
(b) 500 V
(c) 1000V
(d) 2000 V
Answer:
(a) 250 V
Hint:
Here, C’ = 2C, since the charge remains the same.
q = C’V’ = CV ⇒ V = \(\frac { CV }{ 2C }\) = \(\frac { 500 }{ 2 }\) = 250 V

Question 17.
A point charge q is placed at the midpoint of a cube of side L. The electric flux emerging from the cube is ‘
(a) \(\frac { q }{{ ε }_{0}}\)
(b) \(\frac { q }{{ 6Lε }_{0}}\)
(c) \(\frac { 6Lq }{{ ε }_{0}}\)
(d) zero
Answer:
(a) \(\frac { q }{{ ε }_{0}}\)

Question 18.
The capacitor C of a spherical conductor of radius R is proportional to
(a) R2
(b) R
(c) R-1
(d) R0
Answer:
(b) R

Question 19.
Energy of a capacitor of capacitance C, when subjected to a potential V, is given by
(a) \(\frac { 1 }{ 2 }\) CV2
(b) \(\frac { 1 }{ 2 }\) C2V
(c) \(\frac { 1 }{ 2 }\) CV
(d) \(\frac { 1 }{ 2 }\) \(\frac { C }{ V }\)
Answer:
(a) \(\frac { 1 }{ 2 }\) CV2

Question 20.
The electric field due to a dipole at a distance r from its centre is proportional to
(a) \(\frac { 1 }{{ r }^{3/2}}\)
(b) \(\frac { 1 }{{ r }^{3}}\)
(c) \(\frac { 1 }{ r }\)
(d) \(\frac { 1 }{{ r }^{3}}\)
Answer:
(b) \(\frac { 1 }{{ r }^{3}}\)

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 21.
A point charge q is rotating around a charge Q in a circle of radius r. The work done on it by the Coulomb force is
(a) 2πrq
(b) 2πQq
(c) \(\frac { Q }{{ 2ε }^{0}r}\)
(d) zero
Answer:
(d) zero

Question 22.
The workdone in rotating an electric dipole of moment P in an electric field E through an angle 0 from the direction of the field is
(a) pE (1 – cos θ)
(b) 2pE
(c) zero
(d) -pE cos θ
Answer:
(a) pE (1 – cos θ)
Hint:
W = pE(cos θ0 – cos θ)
0 = cos 0, cos 0 = 1]
W = pE(1 – cos θ)

Question 23.
The capacitance of a parallel plate capacitor can be increased by
(a) increasing the distance between the plates
(b) increasing the thickness of the plates
(c) decreasing the thickness of the plates
(d) decreasing the distance between the plates
Answer:
(d) decreasing the distance between the plates

Question 24.
Two charges are placed in vacuum at a distance d apart. The force between them is F. If a medium of dielectric constant 2 is introduced between them, the force will now be
(a) 4F
(b) 2F
(c) F/2
(d) F/4
Answer:
(d) F/4

Question 25.
An electric charge is placed at the centre of a cube of side a. The electric flux through one of its faces will be
(a) \(\frac { q }{{ 6ε }^{0}}\)
(b) \(\frac { q }{ { ε }_{ 0 }{ a }^{ 2 } } \)
(c) \(\frac { q }{ { 4πε }_{ 0 }{ a }^{ 2 } } \)
(a) \(\frac { q }{{ ε }^{0}}\)
Answer:
(a) \(\frac { q }{{ 6ε }^{0}}\)
Hint:
According to Gauss’s law, the electric flux through the cube is \(\frac { q }{{ ε }^{0}}\). Since there are six faces, the flux through one face is \(\frac { q }{{ 6ε }^{0}}\).

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 26.
The electric field in the region between two concentric charged spherical shells-
(a) is zero
(b) increases with distance from centre
(c) is constant
(d) decreases with distance from centre
Answer:
(d) decreases with distance from centre

Question 27.
A hollow metal sphere of radius 10 cm is charged such that the potential on its surface is 80V. The potential at the centre of the sphere is-
(a) 800 V
(b) zero
(c) 8 V
(d) 80 V
Answer:
(d) 80 V

Question 28.
A 4 μF capacitor is charged to 400 V and then its plates are joined through a resistance of 1 K Ω. The heat produced in the resistance is-
(a) 0.16 J
(b) 0.32 J
(c) 0.64 J
(d) 1.28 J
Answer:
(b) 0.32 J
Hint:
The energy stored in capacitor is converted into heat
U = H = \(\frac { 1 }{ 2 }\) CV2 = \(\frac { 1 }{ 2 }\) x 4 x 10-6 x (400)2 = 0.32 J

Question 29.
The work done in carrying a charge Q, once round a circle of radius R with a charge Q2 at the centre is-
(a) \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}^{2}}\)
(b) zero
(c) \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}}\)
(d) infinite
Answer:
(b) zero
Hint:
The electric field is conservative. Therefore, no work is done in moving a charge around a closed path in an electric field.

Question 30.
Two plates are 2 cm apart. If a potential difference of 10 V is applied between them. The electric field between the plates will be
(a) 20 NC-1
(b) 500 NC-1
(c) 5 NC-1
(d) 250 NC-1
Answer:
(b) 500 NC-1
Hint:
\(\frac { V }{ d }\) = \(\frac { 10 }{{ 2 ×10 }^{-2}}\) 500 NC-1

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 31.
The capacitance of a parallel plate capacitor does not depend on
(a) area of the plates
(b) metal of the plates
(c) medium between the plates
(d) distance between the plates
Answer:
(b) metal of the plates

Question 32.
A capacitor of 50 μF is charged to 10 volts. Its energy in joules is
(a) 2.5 x 10-3
(b) 5 x 10-3
(c) 10 x 10-4
(d) 2.5 x 10-4
Answer:
(a) 2.5 x 10-3
Hint:
U = \(\frac { 1 }{ 2 }\) CV2 = \(\frac { 1 }{ 2 }\) x 50 x 10-6 x (10)2 = 2.5 x 10-3 J

Question 33.
A cube of side b has a charge q at each of its vertices. The electric field due to this charge distribution at the centre of the cube is
(a) \(\frac { q }{{b}^{ 2 }}\)
(b) \(\frac { q }{{2b}^{ 2 }}\)
(c) \(\frac { 32q }{{b}^{ 2 }}\)
(d) zero
Answer:(d) zero
Hint:
There
is an equal charge at diagonally opposite comer. The fields due the these at the centre cancel out. Therefore, the net field at the centre is zero.

Question 34.
Total electric fulx coming out of a unit positive charge put in air is
(a) ε0
(b) \({ \varepsilon }_{ 0 }^{ -1 }\)
(c) (4πε0)-1
(d) 4πε0
Answer:
(b) \({ \varepsilon }_{ 0 }^{ -1 }\)

Question 35.
Electron volt (eV) is a unit of
(a) energy
(b) potential
(c) current
(d) charge
Answer:
(a) energy

Question 36.
A point Q lies on the perpendicular bisector of an electric dipole of dipole moment P. If the distance of Q from the dipole is r, then the electric field at Q is proportional to-
(a) p-1 and r-2
(b) p and r-2
(c) p and r-3
(d) p2 and r-3
Answer:
(c) p and r-3

Question 37.
A hollow insulated conducting sphere is given a positive charge of 10 μC. What will be the electric field at the centre of the sphere is its radius is 2 metres?
(a) zero
(b) 8 μCm-2
(c) 20 μCm-2
(d) 5 μCm-2
Answer:
(d) zero

Question 38.
A particle of charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is-
(a) qE2y
(b) q2Ey
(c) qEy2
(d) qEy
Answer:
(d) qEy
Hint:
Force on the particle = qE
KE = Work done by the force = F.y = qEy

Question 39.
Dielectric constant of metals is-
(a) 1
(b) greater then 1
(c) zero
(d) infinite
Answer:
(d) infinite

Question 40.
When a positively charged conductor is earth connected
(a) protons flow from the conductor to the earth
(b) electrons flow from the earth to the conductor
(c) electrons flow from the conductor to the earth
(d) no charge flow occurs
Answer:
(b) electrons flow from the earth to the conductor

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 41.
The SI unit of electric flux is
(a) volt metre2
(b) newton per coulomb
(c) volt metre
(d) joule per coulomb
Answer:
(c) volt metre

Question 42.
Twenty seven water drops of the same size are charged to the same potential. If they are combined to form a big drop, the ratio of the potential of the big drop to that of a small drop is-
(a) 3
(b) 6
(c) 9
(d) 27
Answer:
(c) 9
Hint:
V’ = n2/3 V
⇒ \(\frac { V’ }{ V }\) = (27)2/3 = 9

Question 43.
A point charge +q is placed at the midpoint of a cube of side l. The electric flux emerging ’ from the cube is-
(a) \(\frac { q }{{ ε }^{0}}\)
(b) \(\frac {{ 6ql }^{2}}{{ ε }^{0}}\)
(c) \(\frac { q }{ { 6l }^{ 2 }{ { ε }^{ 0 } } } \)
(d) \(\frac { { C }^{ 2 }{ V }^{ 2 } }{ 2 } \)
Answer:
(a) \(\frac { q }{{ ε }^{0}}\)

Question 44.
The energy stored in a capacitor of capacitance C, having a potential difference V between the plates, is-
Answer:
(c)

Question 45.
The electric potential at the centre of a charged conductor is-
(a) zero
(b) twice that on the surface
(c) half that on the surface
(d) same as that on the surface
Answer:
(d) same as that on the surface

Question 46.
The energy stored in a capacitor is given by
(a) qV
(b) \(\frac { 1 }{ 2 }\)qV
(c) \(\frac { 1 }{ 2 }\) CV
(d) \(\frac { q }{ 2C }\)
Answer:
(b) \(\frac { 1 }{ 2 }\)qV

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 47.
The unit of permitivity of free space so is
(a) coulomb/newton-metre
(b) newton-metre2/coulomb2
(c) coulomb2/newton-metre2
(d) coulomb/(newton-metre)2
Answer:
(c) coulomb2/newton-metre2

Question 48.
An electric dipole has the magnitude of its charge as q and its dipole moment is p. It is placed in a uniform electric field E. If its dipole moment is along the direction of the field, the force on it and its potential energy are, respectively.
(a) 2qE and minimum
(b) qE and pE
(c) zero and minimum
(d) qE and maximum
Answer:
(c) zero and minimum
Hint:
Potential energy, U = -pE cos θ
For q = 0°; U = -pE, which is minimum.

Question 49.
An electric dipole of moment \(\vec { P } \) is lying along a uniform electric field \(\vec { E } \) . The workdone in rotating the dipole by 90° is
(a) \(\frac { pE }{ 2 }\)
(b) 2pE
(c) pE
(d) √2pE
Answer:
(c) pE

Question 50.
A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates
(a) does not charge
(b) becomes zero
(c) increases
(d) decreases
Answer:
(c) increases

Question 51.
When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two charges separated by a distance
(a) increases K times
(b) increases K-1 times
(c) decreases K times
(d) remains constant
Answer:
(c) decreases K times

Question 52.
A comb runs through one’s dry hair attracts small bits of paper. This is due to the fact that
(a) comb is a good conductor
(b) paper is a good conductor
(c) the atoms in the paper get polarised by the charged comb
(d) the comb posseses magnetic properties
Answer:
(c) the atoms in the paper get polarised by the charged comb

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 53.
Which of the following is not a property of equipotential surfaces?
(a) they do not cross each other
(b) they are concentric spheres for uniform electric field
(c) the rate of change of potential with distance on them is zero
(d) they can be imaginary spheres.
Answer:
(b) they are concentric spheres for uniform electric field

Question 54.
A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will be
(a) reduced to half
(b) doubled
(c) becomes 4 times
(d) remains the same
Answer:
(d) remains the same

Question 55.
If the electric field in a region is given by \(\vec { E } \) = 5\(\hat{j} \) + 4\(\hat{j} \) + 9\(\hat{k} \) , then the electric flux through a surface of area 20 units lying in the y-z plane will be-
(a) 20 units
(b) 80 units
(c) 100 units
(d) 180 units
Answer:
(c) 100 units
Hints:
The area vector \(\vec { A } \) = 20\(\hat{j} \); \(\vec { E } \) = (5\(\hat{j} \) + 4\(\hat{j} \) + 9\(\hat{k} \))
Flux (Φ) = \(\vec { E } \) – \(\vec { A } \) = 5 x 20 =100 units

Question 56.
A, B and C are three points in a uniform electric field. The electric potential is-
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-119
(a) maximum at A
(b) maximum at B
(c) maximum at B
(d) same at all the three points A, B, and C
Answer:
(b) maximum at B
Hint:
The potential decreases in the direction of the field. Therefore VB > VC>CA.

Question 57.
A conducting sphere of radius R is give a charge Q. The electric potential and the electric field at the centre of the sphere are, respectively-
(a) zero, \(\frac { Q }{ { 4\pi ε }_{ 0 }{ R }^{ 2 } } \)
(b) \(\frac { Q }{ { 4\pi ε }_{ 0 }{ R } } \)
(c) \(\frac { Q }{ { 4\pi ε }_{ 0 }{ R } } \), zero
(d) zero,zero
Answer:
(c) \(\frac { Q }{ { 4\pi ε }_{ 0 }{ R } } \), zero.

II. Fill in the blanks

Question 1.
A dipole is placed in a uniform electric field with its axis parallel to the field. It experiences …………………
Answer:
neither a net force nor a torque

Question 2.
The unit of permittivity is…………………
Answer:
C2N-1m-2

Question 3.
The branch of physics which deals with static electric charges or charges at rest is …………………
Answer:
electrostatics

Question 4.
The charges in an electrostatic field are analogous to ………………… in a gravitational field.
Answer:
mass

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 5.
The substances which acquire charges on rubbing are said to be …………………
Answer:
electrified

Question 6.
Electron means …………………
Answer:
amber

Question 7.
A glass rod rubbed with a silk cloth. Glass rod and silk cloth acquire…………………
Answer:
positive and negative charge respectively

Question 8.
When ebonite rod is rubbed with fur, ebonite rod and fur acquires …………………
Answer:
negative and positive charge respectively

Question 9.
………………… termed the classification of positive and negative charges.
Answer:
Franklin

Question 10.
Applications such as electrostatic point spraying and powder coating, are based on the property of ………………… between charged bodies.
Answer:
attraction and repulsion

Question 11.
Bodies which allow the charge to pass through them are called …………………
Answer:
conductor

Question 12.
Bodies which do not allow the charge to pass through them are called …………………
Answer:
insulators

Question 13.
The unit of electric charge is …………………
Answer:
coulomb

Question 14.
Total charge in an isolated system …………………
Answer:
remains a constant

Question 15.
The force between two charged bodies was studied by …………………

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics
Answer:
coulomb

Question 16.
The unit of permittivity in free space (s0) is …………………
Answer:
C2N-1m-2

Question 17.
The value of s, for air or vacuum is …………………
Answer:1

Question 18.
Charges can neither be created nor be destroyed is the statement of the law of conservation of …………………
Answer:
charge

Question 19.
The space around the test charge, in which it experiences a force is known as field …………………
Answer:
electric

Question 20.
Electric field at a point is measured in terms of …………………
Answer:
electric field intensity

Question 21.
The unit of electric field intensity is …………………
Answer:
NC-1.

Question 22.
The lines of force are far apart, when electric field E is …………………
Answer:
small

Question 23.
The lines of force are close together when electric field E is …………………
Answer:
large

Question 24.
Electric dipole moment …………………
Answer:
P = 2qd

Question 25.
Torque experienced by electric dipole is …………………
Answer:
x = PE sin θ

Question 26.
An electric dipole placed in a non-uniform electric field at an angle θ experiences …………………
Answer:
both torque and force

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 27.
When thee dipole is aligned parallel to the field, its electric potential energy is …………………
Answer:
u = -PE

Question 28.
Change of potential with distance is known as …………………
Answer:
potential distance

Question 29.
The number of electric lines of force crossing through the given area is …………………
Answer:
electric flux

Question 30.
The process of isolating a certain region of space from the external field is called …………………
Answer:
electrostatic shielding

Question 31.
A capacitor is a device to store …………………
Answer:
charge

Question 32.
The charge density in maximum at …………………
Answer:
pointed

Question 33.
The principle made use of lightning arrestor is …………………
Answer:
action of points

Question 34.
Van de Graaff generator producers large electrostatic potential difference of the order of …………………
Answer:
107 V

III. Match the following

Question 1.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-120
Answer:
(i) → (d)
(ii) → (a)
(iii) → (b)
(iv) → (c)

Question 2.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-121
Answer:
(i) → (c)
(ii) → (d)
(iii) → (a)
(iv) → (b)

Question 3.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-122
Answer:
(i) → (b)
(ii) → (d)
(iii) → (a)
(iv) → (c)

Question 4.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-123
Answer:
(i) → (b)
(ii) → (d)
(iii) → (a)
(iv) → (c)

IV. Assertion and reason type

(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.
(b) If both assertion and reason are true but the reason is not correct explanation of the assertion.
(c) If the assertion is true but the reason is false.
(d) If the assertion and reason both are false.
(e) If the assertion is false but the reason is true.

Question 1.
Assertion: Electric lines of force cross each other.
Reason: Electric field at a point superimposed to give one resultant electric field.
Answer:
(e) Both assertion and reason are true but the reason is not the correct explanation of the assertion.
Explanation: If electric lines of forces cross each other, then the electric field at the point of intersection will have two directions simultaneously which is not possible physically.

Question 2.
Assertion: Charge is quantized.
Reason: Charge, which is less than 1 C is not possible.
Answer:
(c) If assertion is true but reason is false.
Explanation: Q = ±ne and charge lesser than 1 C is possible.

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 3.
Assertion:
A point charge is brought in an electric field. The field at a nearby point will increase, whatever be the nature of the charge.
Reason: The electric field is independent of the nature of the charge.
(d) If the assertion and reason both are false.
Explanation: Electric field at the nearby-point will be resultant of the existing field and field due to the charge brought. It may increase or decrease if the charge is positive or negative depending on the position of the point with respect to the charge brought.

Question 4.
Assertion: The tyre’s of aircraft are slightly conducting.
Reason: If a conductor is connected to the ground, the extra charge induced on the conductor will flow to the ground.
Answer:
(b) Both assertion and reason are true but the reason is not the correct explanation of the assertion.
Explanation: During take-off and landing, the friction between treys and the runway may cause electrification of treys. Due to conducting to a ground and election sparking is avoided.

Question 5.
Assertion: The lightning conductor at the top of a high building has sharp ends.
Reason: The surface density of charge at sharp points is very high, resulting in the setting up of electric wind.
Answer:
(a) Both assertion and reason are true and the reason is the correct explanation of the assertion.

Samacheer Kalvi 12th Physics Electrostatics Short Answer Questions

Question 1.
What is meant by triboelectric charging?
Answer:
Charging the objects through rubbing is called triboelectric charging.

Question 2.
What is meant by the conservation of total charges?
Answer:
The total electric charge in the universe is constant and the charge can neither be created nor be destroyed. In any physical process, the net change in charge will always be zero.

Question 3.
State Gauss’s Law?
Answer:
Definition:
Gauss’s law states that if a charge Q is enclosed by an arbitrary closed surface, then the total electric flux OE through the closed surface is
ΦE = \(\oint { \vec { E } } \) .d\(\vec { A } \) = \(\frac {{ q }_{encl}}{{ ε }_{0}}\)

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 4.
What is meant by electrostatic shielding?
During lightning accompanied by a thunderstorm, it is always safer to sit inside a bus than in open ground or under a tree. The metal body of the bus provides electrostatic shielding, since the electric field inside is zero. During lightning, the charges flow through the body of the conductor to the ground with no effect on the person inside that bus.

Question 5.
What is meant by dielectric?
Answer:
A dielectric is a non-conducting material and has no free electrons. The electrons in a dielectric are bound within the atoms. Ebonite, glass and mica are some examples of dielectrics.

Question 6.
What are non-polar molecules? Give examples.
A non-polar molecule is one in which centers of positive and negative charges coincide. As a result, it has no permanent dipole moment. Examples of non-polar molecules are hydrogen (H2), oxygen (O2), and carbon dioxide (CO2) etc.

Question 7.
What are polar molecules? Give examples.
Answer:
In polar molecules, the centers of the positive and negative charges are separated even in the absence of an external electric field. They have a permanent dipole moment.
The net dipole moment is zero in the absence of an external electric field. Examples of polar molecules are H2O, N2O, HCl, NH3.

Question 8.
What is a capacitor?
Answer:
A capacitor is a device used to store electric charge and electrical energy. Capacitors are widely used in many electronic circuits and have applications in many areas of science and technology.

Samacheer Kalvi 12th Physics Electrostatics Long Answer Questions

Question 1.
Derive an expression for the electric field due to the system of point charges?
Answer:
Electric field due to the system of point charges:
Suppose a number of point charges are distributed in space. To find the electric field at some point P due to this collection of point charges, the superposition principle is used. The electric field at an arbitrary point due to a collection of point charges is simply equal to the vector sum of the electric fields created by the individual point charges. This is called the superposition of electric fields.
Consider a collection of point charges q1, q2, q3,…., qn located at various points in space. The ‘ total electric field at some point P due to all these n charges is given by
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-124
Here r1p, r2p, r3p,…., rnp, are the distance of the charges 1, q2, q3,…., qn from the point respectively. Also \(\hat{r} \)1p + \(\hat{r} \)2p + \(\hat{r} \)3p,…., \(\hat{r} \)np are the corresponding unit vectors directed from q1, q2, q3,…., qn tpo P.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-125
Equation (2) can be re-written as,
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-126
For example in figure, the resultant electric field due to three point charges q1, q2, q3 at point P is shown. Note that the relative lengths of the electric field vectors for the charges depend on relative distantes of the charges to the point P.

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 2.
Derive an expression for the electric flux of rectangular area placed in a uniform electric field.
Answer:
(i) Electric flux for uniform Electric field:
Consider a uniform electric field in a region of space. Let us choose an area A normal to the electric field lines as shown in figure (a). The electric flux for this case is
ΦE = EA ….. (1)
Suppose the same area A is kept parallel to the uniform electric field, then no electric field lines pierce through the area A, as shown in figure (b). The electric flux for this case is zero.
ΦE = 0 ….. (2)
If the area is inclined at an angle θ with the field, then the component of the electric field perpendicular to the area alone contributes to the electric flux. The electric field component parallel to the surface area will not contribute to the electric flux. This is shown in figure (c). For this case, the electric flux
ΦE = (E cosθ) A …(3)
Further, θ is also the angle between the electric field and the direction normal to the area. Hence in general, for uniform electric field, the electric flux is defined as
ΦE= \(\vec { E } \).\(\vec { A } \) = EA cos θ …(4)
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-127
Here, note that \(\vec { A } \) is the area vector \(\vec { A } \) = A\(\hat{n} \). Its magnitude is simply the area A and the direction is along the unit vector h perpendicular to the area. Using this definition for flux, ΦE= \(\vec { E } \).\(\vec { A } \), equations (2) and (3) can be obtained as special cases.
In figure (a), θ = 0° so ΦE= \(\vec { E } \).\(\vec { A } \) = EA
In figure (b), θ = 90° so ΦE= \(\vec { E } \).\(\vec { A } \) = 0

(ii) Electric flux in a non uniform electric field and an arbitrarily shaped area: Suppose the electric field’is not uniform and the area A is not flat, then the entire area is divided
into n small area segments ∆\(\vec { A } \)1 ∆\(\vec { A } \)2, ∆\(\vec { A } \)3,…..∆\(\vec { A } \)n, such that each area element is almost flat and the electric field over each area element is considered to be uniform.
The electric flux for the entire area A is approximately written as
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-128
By taking the limit ∆\(\vec { A } \)1 → 0 (for all i) the summation in equation (5) becomes integration. The total electric flux for the entire area is given by
ΦE = ∫\(\vec { E } \).d\(\vec { A } \) ….. (6)
From Equation (6), it is clear that the electric flux for a given surface depends on both the electric field pattern on the surface area and the orientation of the surface with respect to the electric field.

(iii) Electric flux for closed surfaces: In the previous section, the electric flux for any arbitrary curved surface is discussed. Suppose a closed surface is present in the region of the non-uniform electric field as shown in figure (a).
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-129
The total electric flux over this closed surface is written as
ΦE = \(\oint { \vec { E } } \).d\(\vec { A } \) …… (7)
Note the difference between equations (6) and (7). The integration in equation (7) is a closed surface integration and for each areal element, the outward normal is the direction of d\(\vec { A } \) as shown in figure (b).
The total electric flux over a closed surface can be negative,
positive or zero. In figure (b), it is shown that in one area element, the angle between d\(\vec { A } \) and \(\vec { E } \) is less than 90°, then the electric flux is positive and in another areal element, the angle between dA and E is greater than 90°, then the electric flux is negative. In general, the electric flux is negative if the electric field lines enter the closed surface and positive if the electric field lines leave the closed surface.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-130

Samacheer Kalvi 12th Physics Electrostatics Numerical Problems

Question 1.
Electrons are caused to fall through a potential difference of 1500 volts. If they were initially at rest. Then calculate their final speed.
Solution:
The electrical potential energy is converted into kinetic energy. If v is the final speed then
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-131

Question 2.
Small mercury drops of the same size are charged to the same potential V. If n such drops coalesce to form a single large drop, then calculate its potential.
Solution:
Let r be the radius of a small drop and R that of the large drop. Then, since the volume remains conserved,
\(\frac { 1 }{ 2 }\) πR2 = \(\frac { 4 }{ 3 }\) πR3n
⇒ R3 = r3n
R = r3(n)1/3
Further, since the total charge remains conserved, we have, using Q = CV
Clarge V = n Csmall v
Where V is the potential of the large drop.
4πε0 RV = n (4πε0r)v
V = \(\frac { nrv }{ R }\) = \(\frac { nrv }{{ r(n) }^{1/3}}\)
V = vn2/3

Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

Question 3.
Two particles having charges Q1 and Q2 when kept at a certain distance, exert a force F on each other. If the distance between the two particles is reduced to half and the charge on each particle is doubled. Find the force between the particles.
Solution:
F = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{r^{2}}\)
If the distance is educed by half and two particles of charges are doubled.
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-132

Question 4.
Two charged spheres, separated by a distance d, exert a force F on each other. If they are immersed in a liquid of dielectric constant 2, then what is the force.
Solution:
Force between the charges (vacuum)
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-133
Force between the charges (medium)
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-134

Question 5.
Find the force of attraction between the plates of a parallel plate capacitor.
Solution:
Let d be the distance between the plates. Then the capacitor is
C = \(\frac { { \varepsilon }_{ 0 }A }{ d } \)
Energy stored in a capacitor,
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-135
Energy magnitude of the force is,
Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics-136

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Students can Download Physics Chapter 3 Magnetism and Magnetic Effects of Electric Current Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Physics Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Samacheer Kalvi 12th Physics Magnetism and Magnetic Effects of Electric Current Textual Evaluation Solved

Samacheer Kalvi 12th Physics Magnetism and Magnetic Effects of Electric Current Multiple Choice Questions   

Question 1.
The magnetic field at the center O of the following current loop is-
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-1
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-1-1
Answer:
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-2

Question 2.
An electron moves straight inside a charged parallel plate capacitor of uniform charge density σ The time taken by the electron to cross the parallel plate capacitor when the plates of the capacitor are kept under constant magnetic field of induction \(\vec { B } \) is-
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-3
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-4
Answer:
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-5

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 3.
The force experienced by a particle having mass m and charge q accelerated through a potential difference V when it is kept under perpendicular magnetic field \(\vec { B } \) is-
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-6
Answer:
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-7

Question 4.
A circular coil of radius 5 cm and 50 turns carries a current of 3 ampere. The magnetic dipole moment of the coil is-
(a) 1.0 amp – m2
(b) 1.2 amp – m2
(c) 0.5 amp – m2
(d) 0.8 amp – m2
Answer:
(b) 1.2 amp – m2

Question 5.
A thin insulated wire forms a plane spiral of N = 100 tight turns carrying a current 1 = 8 m A (milli ampere). The radii of inside and outside turns are a = 50 mm and b = 100 mm respectively. The magnetic induction at the center of the spiral is
(a) 5 μT
(b) 7 μT
(c) 8 μT
(d) 10 μT
Answer:
(b) 7 μT

Question 6.
Three wires of equal lengths are bent in the form of loops. One of the loops is circle, another is a semi-circle and the third one is a square. They are placed in a uniform magnetic field and same electric current is passed through them. Which of the following loop configuration will experience greater torque ?
(a) circle
(b) semi-circle
(c) square
(d) all of them
Answer:
(a) circle

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 7.
Two identical coils, each with N turns and radius R are placed coaxially at a distance R as shown in the figure. If I is the current passing through the loops in the same direction, then the magnetic field at a point P which is at exactly at \(\frac { R }{ 2 }\) distance between two coils is-
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-8
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-8-1
Answer:
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-9

Question 8.
A wire of length l carries a current I along the Y direction and magnetic field is given by \(\vec { B } \) = \(\frac { β }{ √3 }\) \((\hat{i}+\hat{j}+\hat{k})\). The magnitude of Lorentz force acting on the wire is-
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-10
Answer:
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-11

Question 9.
A bar magnet of length l and magnetic moment M is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be- (NEET 2014)
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-12
(a) M
(b) \(\frac { π }{ 3 }\) M
(c) \(\frac { 2 }{ π }\) M
(d) \(\frac { 1 }{ 2 }\) M
Answer:
(b) \(\frac { π }{ 3 }\) M

Question 10.
A non-conducting charged ring of charge q, mass m and radius r is rotated with constant angular speed ω. Find the ratio of its magnetic moment with angular momentum is
(a) \(\frac { q }{ m }\) M
(b) \(\frac { 2q }{ 3 }\) M
(c) \(\frac { q }{ 2m }\) M
(d) \(\frac { q }{ 4m }\) M
Answer:
(c) \(\frac { q }{ 2m }\) M

Question 11.
The BH curve for a ferromagnetic material is shown in the figure. The material is placed inside a long solenoid which contains 1000 turns/ cm. The current that should be passed in the solenoid to demagnetize the ferromagnet completely is-
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-13
(a) 1.00 m A(milli ampere)
(b) 1.25 mA
(c) 1.50 mA
(d) 1.75 mA
Answer:
(b) 1.25 mA

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 12.
Two short bar magnets have magnetic moments 1.20 Am2 and 1.00 Am2respectively. They are kept on a horizontal table parallel to each other with their north poles pointing towards the south. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the mid-point O of the line joining their centers is (Horizontal components of Earth’s magnetic induction is 3.6 x 10-5 Wb m-2) (NSEP 2000-2001)
(a) 3.60 x 10-5 Wb m-1
(b) 3.5 x 10-5 Wb m-1
(c) 2.56 x 10-4 Wb m-1
(d) 2.2 x 10-4 Wb m-1
Answer:
(c) 2.56 x 10-4 Wb m-1

Question 13.
The vertical component of Earth’s magnetic field at a place is equal to the horizontal component. What is the value of angle of dip at this place?
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer:
(b) 45°

Question 14.
A flat dielectric disc of radius R carries an excess charge on its surface. The surface charge density is σ. The disc rotates about an a × is perpendicular to its plane passing through the center with angular velocity ω. Find the magnitude of the torque on the disc if it is placed in a uniform magnetic field whose strength is B which is directed perpendicular to the a × is of rotation
(a) \(\frac { 1 }{ 4 }\) σωπ BR
(b) \(\frac { 1 }{ 4 }\) σωπ BR2
(c) \(\frac { 1 }{ 4 }\) σωπ BR3
(d) \(\frac { 1 }{ 4 }\) σωπ BR4
Answer:
(d) \(\frac { 1 }{ 4 }\) σωπ BR4

Question 15.
A simple pendulum with charged bob is oscillating with time period T and let θ be the angular displacement. If the uniform magnetic field is switched ON in a direction perpendicular to the plane of oscillation then-
(a) time period will decrease but θ will remain constant
(b) time period remain constant but θ will decrease
(c) both T and θ will remain the same
(d) both T and θ will decrease
Answer:
(c) both T and θ will remain the same

Magnetism and Magnetic Effects of Electric Current Short Answer Questions

Question 1.
What is meant by magnetic induction?
Answer:

  1. The process by which an object or material is magnetized by an external magnetic field.
  2. S.I unit of magnetic induction is the tesla (T) or Wbm-2
  3. Dimensional Formula is MT2A-1

Question 2.
Define magnetic flux.
Answer:
The number of magnetic field lines crossing per unit area is called magnetic flux B.
ΦB = \(\vec { B } \) .\(\vec { A } \) = BA cos θ = B ⊥ A

Question 3.
Define magnetic dipole moment.
Answer:

  1. The product of the pole strength and magnetic length.
  2. It is a vector quantity, denoted by \(\overrightarrow{\mathrm{p}}_{\mathrm{m}}\) ;
    \(\overrightarrow{\mathrm{p}}_{\mathrm{m}}=\overrightarrow{\mathrm{q}}_{\mathrm{m}} \overrightarrow{\mathrm{d}}\)
  3. The magnitude of magnetic dipole moment Pm = 2qml

Question 4.
State Coulomb’s inverse law.
Answer:
The force of attraction or repulsion between two magnetic poles is directly proportional to the product of their pole strengths and inversely proportional to the square of the distance between them.
\(\overrightarrow{\mathrm{F}} \propto \frac{q_{m_{\mathrm{A}}} q_{m_{b}}}{r^{2}} \hat{r}\)

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 5.
What is magnetic susceptibility?
Answer:

  1. The ratio of the intensity of magnetisation \((\overrightarrow{\mathrm{M}})\) induced in the material due to the magnetising field \((\overrightarrow{\mathrm{H}})\)
  2. \(\mathrm{X}_{\mathrm{m}}=\frac{|\overrightarrow{\mathrm{M}}|}{|\overrightarrow{\mathrm{H}}|}\)
  3. It is a dimensionless quantity.
  4. Isotropic medium – Scalar, Non-isotropic medium – vector.

Question 6.
State Biot-Savart’s law.
Answer:
The magnitude of magnetic field d\(\vec { B } \) at a point P at a distance r from the small elemental length taken on a conductor carrying current varies-

  • directly as the strength of the current I
  • directly as the magnitude of the length element \(\vec { dl } \)
  • directly as the sine of the angle (say θ) between \(\vec { dl } \) and \(\hat{r}\) .
  • inversely as the square of the distance between the point P and length element \(\vec { dl } \).
  • This is expressed as \(d \mathrm{B} \propto \frac{\mathrm{I} d l}{r^{2}} \sin \theta\)

Question 7.
What is magnetic permeability?
Answer:

  1. The measure of the ability of the material to allow the passage of magnetic field lines through it. (or)
  2. The measure of the capacity of the substance to take magnetisation. (or)
  3. The degree of penetration of the magnetic field through the substance.

Question 8.
State Ampere’s circuital law.
Answer:
The line integral of magnetic field over a closed-loop is µ0 times net current enclosed by the loop.
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-14

Question 9.
Compare dia, para and ferromagnetism.
Answer:
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-15

Question 10.
What is meant by hysteresis?
Answer:

  1. Lagging of magnetic induction behind the magnetising field.
  2. Hysteresis means lagging behind.

Magnetism and Magnetic Effects of Electric Current Long Answer Questions

Question 1.
Discuss Earth’s magnetic field in detail.
Answer:
Gover suggested that the Earth’s magnetic field is due to hot rays coming out from the Sun. These rays will heat up the air near-equatorial region. Once air becomes hotter, it rises above and will move towards northern and southern hemispheres and get electrified. This may be responsible to magnetize the ferromagnetic materials near the Earth’s surface.
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-16
The north pole of magnetic compass needle is attracted towards the magnetic south pole of the Earth which is near the geographic north pole. Similarly, the south pole of magnetic compass needle is attracted towards the geographic north pole of the Earth which is near magnetic north pole. There are three quantities required to specify the magnetic field of the Earth on its surface, which are often called as the elements of the Earth’s magnetic field. They are:

(a) Magnetic declination (D): The angle between magnetic meridian at a point and geographical meridian is called the declination or magnetic declination (D).

(b) Magnetic dip or inclination (I): The angle subtended by the Earth’s total magnetic field \(\vec { B } \) with the horizontal direction in the magnetic meridian is called dip or magnetic inclination (I) at that point.

(c) The horizontal component of the Earth’s magnetic field (BH): The component of Earth’s magnetic field along the horizontal direction in the magnetic meridian is called horizontal component of Earth’s magnetic field, denoted by BH.

Let BE be the net Earth’s magnetic field at a point P on the surface of the Earth. BE can be resolved into two perpendicular components.
Horizontal component BH = BE cos I …… (1)
Vertical component BV = BE sin I …… (2)
Dividing the equation, we get tan I = \(\frac {{ B }_{V}}{{ B }_{H}}\) …….(3)

(i) At magnetic equator The Earth’s magnetic field is parallel to the surface of the Earth (i.e., horizontal) which implies that the needle of magnetic compass rests horizontally at an angle of dip, I = 0°.
B BE = BE
Bv = 0
This implies that the horizontal component is maximum at equator and vertical component is zero at equator.

(ii) At magnetic poles. The Earth’s magnetic field is perpendicular to the surface of the Earth (i.e., vertical) which implies that the needle of magnetic compass rests vertically at an angle of dip, I = 90°
Hence, BH = 0
Bv = BE
This implies that the vertical component is maximum at poles and horizontal component is zero at poles.

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 2.
Deduce the relation for the magnetic induction at a point due to an infinitely long straight conductor carrying current.
Answer:
Magnetic field due to a long straight conductor carrying current:
Consider a long straight wire NM with current I flowing from N to M. Let P be the point at a distance a from point O. Consider an element of length dl of the wire at a distance l from point O and \(\vec { r } \) be the vector joining the element dl with the point P. Let θ be the angle between \(\vec { dl } \) and \(\vec { r } \). Then, the magnetic field at P due to the element is d\(\vec { B } \) = \(\frac { { \mu }_{ 0 }I\vec { dl } }{ 4{ \pi r }^{ 2 } } \) sinθ (unit vector perpendicular to \(\vec { dl } \) and \(\vec { r } \)) …… (1)

The direction of the field is perpendicular to the plane of the paper and going into it. This can be determined by
taking the cross product between two vectors \(\vec { dl } \) and \(\vec { r } \) (let it be \(\hat{n}\) ). The net magnetic field can be determined by integrating equation with proper limits.
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-17
\(\vec { B } \) = \(\int { d\vec { B } } \)
From the figure, in a right angle triangle PAO,
tan (π – θ) = \(\frac { a }{ l }\)
l = \(\frac { a }{ tan θ }\) (since tan (π – θ) = -tan θ) ⇒ \(\frac { 1 }{ tan θ }\)
l = a cot θ and r = a cosec θ
differentiating,
dl = a cosec2 θdθ
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-18
This is the magnetic field at a point P due to the current in small elemental length. Note that we have expressed the magnetic field OP in terms of angular coordinate i.e. θ. Therefore, the net magnetic field at the point P which can be obtained by integrating d\(\vec { B } \) by varying the angle from θ = (φ1 to θ = φ1 is
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-19
For a an infinitely long straight wire, 1 = 0 and 2 = , the magnetic field is
\(\vec { B } \) = \(\frac {{ µ }_{0}}{ 2πa }\) \(\hat{n}\) …… (3)
Note that here \(\hat{n}\) represents the unit vector from the point O to P.

Question 3.
Obtain a relation for the magnetic induction at a point along the axis of a circular coil carrying current.
Answer:
Magnetic field produced along the axis of the current-carrying circular coil:
Consider a current-carrying circular loop of radius R and let I be the current flowing through the wire in the direction. The magnetic field at a point P on the axis of the circular coil at a distance z from its center of the coil O. It is computed by taking two diametrically opposite line elements of the coil each of length \(\vec { dl } \) at C and D. Let \(\vec { r } \) be the vector joining the current element (1 \(\vec { dl } \)) at C to the point P.
PC = PD = T = \(\sqrt { { R }^{ 2 }+{ Z }^{ 2 } } \) and angle ∠CPO = ∠DPO = θ

According to Biot-Savart’s law, the magnetic field at P due to the current element I \(\vec { dl } \) is
d\(\vec {B} \) = \(\frac {{ µ }_{0}}{ 4π }\) \(\frac{\mathrm{I} d \vec{l} \times \hat{r}}{r^{2}}\) ….. (1)
The magnitude of magnetic field due to current element l dl at C and D are equal because of equal distance from the coil. The magnetic field dB due to each current element I \(\vec { dl } \) is resolved into two components; dB sin θ along y-direction and dB cos θ along z-direction. Horizontal components of each current element cancels out while the vertical components (dB cos θ \(\hat{k}\)) alone contribute to total magnetic field at the point P.
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-20
If we integrate \(\vec { dl } \) around the loop, d\(\vec { B } \) sweeps out a cone, then the net magnetic field \(\vec { B } \) at point P is
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-21
Using Pythagorous theorem r2 = R2 + Z2 and integrating line element from 0 to 2πR, we get
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-22
Note that the magnetic field \(\vec { B } \) points along the direction from the point O to P. Suppose if the current flows in clockwise direction, then magnetic field points in the direction from the point P to O.

Question 4.
Compute the torque experienced by a magnetic needle in a uniform magnetic field.
Answer:
Torque Acting on a Bar Magnet in Uniform Magnetic Field:
Consider a magnet of length 21 of pole strength qm kept in a uniform magnetic field \(\vec { B } \) Each pole experiences a force of magnitude qmB but acts in opposite direction.

Therefore, the net force exerted on the magnet is zero, so that there is no translatory motion. These two forces constitute a couple (about midpoint of bar magnet) which will rotate and try to align in the direction of the magnetic field \(\vec { B } \).

The force experienced by north pole, \(\vec { F } \)N = qm \(\vec { B } \) ……. (1)
The force experienced by south pole, \(\vec { F } \)S = qm \(\vec { B } \) …….. (2)
Adding equations (1) and (2), we get the net force acting on the dipole as
\(\vec { F } \) = \(\vec { F } \)N + \(\vec { F } \)S = \(\vec { O } \)
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-23
This implies, that the net force acting on the dipole is zero, but forms a couple which tends to rotate the bar magnet clockwise (here) in order to align it along \(\vec { B } \).
The moment of force or torque experienced by the north and south pole about point O is
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-24
By using right hand cork screw rule, we conclude that the total torque is pointing into the paper. Since the magnitudes
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-25
The magnitude of total torque about point O
τ = l × qm B sin θ +l × qm B sin θ
τ = 2l xqm B sin θ
τ = Pm B sin θ
(∴ qm × 2l = Pm )
In vector notation, τ = pm × \(\vec { B } \)

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 5.
Calculate the magnetic induction at a point on the a×ial line of a bar magnet.
Answer:
The magnetic field at a point along the axial line of the magnetic dipole (bar magnet):
Consider a bar magnet NS. Let N be the North Pole and S be the south pole of the bar magnet, each of pole strength qm and separated by a distance of 2l. The magnetic field at a point C (lies along the axis of the magnet) at a distance from the geometrical center O of the bar magnet can be computed by keeping unit north pole (qMC = 1 A m) at C. The force experienced by the unit north pole at C due to pole strength can be computed using Coulomb’s law of magnetism as follows:
The force of repulsion between the north pole of the bar magnet and the unit north pole at point C (in free space) is
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-26
where r – l is the distance between north pole of the bar magnet and unit north pole at C. The force of attraction between the South Pole of the bar magnet and unit North Pole at point C (in free space) is
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-27
where r + 1 is the distance between south pole of the bar magnet and unit north pole at C.
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-28
From equation (1) and (2), the net force at point C is \(\vec { F } \) = \(\vec { F } \)N + \(\vec { F } \)S.
From definition, this net force is the magnetic field due to magnetic dipole at a point C(\(\vec { F } \) = \(\vec { B } \))
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-29
Since, magnitude of magnetic dipole moment is \(\left| { \vec { P } }_{ m } \right| \) pm = qm. 2l the magnetic field point C equation (3) can be written as
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-30
If the distance between two poles in a bar magnet is small (looks like a short magnet) compared to the distance between geometrical centre O of a bar magnet and the location of point C i.e.,
r >>1 then, (r2 – l2)2 ≈ r4 ….. (5)
Therefore, using equation (5) in equation (4), we get
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-31

Question 6.
Obtain the magnetic induction at a point on the equatorial line of a bar magnet. The magnetic field at a point along the equatorial line due to a magnetic dipole (bar magnet)
Answer:
Consider a bar magnet NS. Let N be the north pole and S be the south pole of the bar magnet, each with pole strength qm and separated by a distance of 2l. The magnetic field at a point C (lies along the equatorial line) at a distance r from the geometrical center O of the bar magnet can be computed by keeping unit north pole (qmC = 1 A m) at C. The force experienced by the unit north pole at C due to pole strength N-S can be computed using Coulomb’s law of magnetism as follow’s:
The force of repulsion between the North Pole of the bar magnet and the unit north pole at point C (in free space) is
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-32
\(\vec { F } \)N = -FN cos θ \(\hat{i}\) + FN sin θ \(\hat{j}\) …… (1)
Where FN = \(\frac { { µ }_{ 0 } }{ 4\pi } \) \(\frac { { q }_{ m } }{ { r’ }^{ 2 } } \)
The force of attraction (in free space) between south pole of the bar magnet and unit north pole at point C is
\(\vec { F } \)S = -FS cos θ \(\hat{i}\) + FS sin θ j …… (2)
Where \(\vec { F } \)S = \(\frac { { µ }_{ 0 } }{ 4\pi } \) \(\frac { { q }_{ m } }{ { r’ }^{ 2 } } \)
From equation (1) and equation (2), the net force at point C is \(\vec { F } \) = FN + FS This net force is equal to the magnetic field at the point C.
\(\vec { B } \) = -(FN + FS) cos θ \(\hat{i}\)
Since, FN = FS
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-33
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-34
In a right angle triangle NOC as shown in the Figure 1
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-35
Substituting equation 4 in equation 3 We get
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-36
Since, magnitude of magnetic dipole moment is \(\left| { \vec { P } }_{ m } \right| \) = Pm = qm. 2l and substituting in equation (5), the magnetic field at a point C is
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-37
If the distance between two poles in a bar magnet are small (looks like a short magnet) when compared to the distance between geometrical center O of a bar magnet and the location of point C i.e., r>> l, then,
(r2 + l2)\(\frac { 3 }{ 2 }\)
Therefore, using equation (7) in equation (6), we get ≈ r3 ……… (7)
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-38
Since Pm \(\hat{i}\) = \(\left| { \vec { P } }_{ m } \right| \)m, in general, the magnetic field at equatorial point is given by
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-39
Note that magnitude of Baxial is twice that of magnitude of Bequatorial and the direction of are opposite.

Question 7.
Find the magnetic induction due to a long straight conductor using Ampere’s circuital law. Magnetic field due to the current-carrying wire of infinite
Answer:
length using Ampere’s law:
Consider a straight conductor of infinite length carrying current I and the direction of magnetic field lines. Since the wire is geometrically cylindrical in shape C and symmetrical about its axis, we construct an Amperian loop in the form of a circular shape at a distance r from the centre of the conductor. From the Ampere’s law, we get
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-40
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-41
Where dl is the line element along the amperian loop (tangent to the circular loop). Hence, the angle between the magnetic field vector and the line element is zero. Therefore,
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-42
where I is the current enclosed by the Amperian loop. Due to the symmetry, the magnitude of the magnetic field is uniform over the Amperian loop, we can take B out of the integration.
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-43
For a circular loop, the circumference is 2πr, which implies,
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-44
In vector form, the magnetic field is
\(\vec { B } \) = \(\frac { { μ }_{ 0 }I }{ 2\pi r } \)\(\hat{n}\)
Where \(\hat{n}\) is the unit vector along the tangent to the Amperian loop. This perfectly agrees with the result obtained from Biot-Savarf s law as given in equation
\(\vec { B } \) = \(\frac { { μ }_{ 0 }I }{ 2\pi a } \)\(\hat{n}\)

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 8.
Discuss the working of the cyclotron in detail.
Answer:
Cyclotron:
A cyclotron is a device used to accelerate the charged particles to gain large kinetic energy. It is also called as high energy accelerator. It was invented by Lawrence and Livingston in 1934.

Principle:
When a charged particle moves normal to the magnetic field, it experiences magnetic Lorentz force.

Construction:
The particles are allowed to move in between two semicircular metal containers called Dees (hollow D – shaped objects). Dees are enclosed in an evacuated chamber and it is kept in a region with uniform magnetic field controlled by an electromagnet. The direction of the magnetic field is normal to the plane of the Dees. The two Dees are kept separated with a gap and the source S (which ejects the particle to be accelerated) is placed at the center in the gap between the Dees. Dees are connected to high frequency alternating potential difference.
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-q8

Working:
Let us assume that the ion ejected from source S is positively charged. As soon as ion is ejected, it is accelerated towards a Dee (say, Dee – 1) which has negative potential at that time. Since the magnetic field is normal to the plane of the Dees, the ion undergoes circular path. After one semi-circular path in Dee-1, the ion reaches the gap between Dees. At this time, the polarities of the Dees are reversed so that the ion is now accelerated towards Dee-2 with a greater velocity. For this circular motion, the centripetal force of the charged particle q is provided by Lorentz force.
\(\frac {{ mv }^{2}}{ r }\) qvB ⇒ r = \(\frac { m }{ qb }\) v ⇒ r ∝ v
From the equation, the increase in velocity increases the radius of circular path. This process continues and hence the particle undergoes spiral path of increasing radius. Once it reaches near the edge, it is taken out with the help of deflector plate and allowed to hit the target T. Very important condition in cyclotron operation is the resonance condition. It happens when the frequency ƒ at which the positive ion circulates in the magnetic field must be equal to the constant frequency of the electrical oscillator fosc From equation
fosc = \(\frac { qB }{2πm}\) T = \(\frac { 1 }{{ f }_{osc}}\)
The time period of oscillation is
T = \(\frac { 2πm }{qB}\)
The kinetic energy of the charged paricle is
K E = \(\frac { 1 }{ 2 }\) mv2 = \(\frac {{ q }^{2}}{{ B }^{2}}{{ r }^{2}}{ 2m }\)

Limitations of cyclotron:
(a) the speed of the ion is limited
(b) electron cannot be accelerated
(c) uncharged particles cannot be accelerated

Question 9.
What is tangent law? Discuss in detail.
Answer:
Tangent law:
Statement:
When a magnetic needle or magnet is freely suspended in two mutually perpendicular uniform magnetic fields, it will come to rest in the direction of the resultant of the two fields.

Explanation:
Let B be the magnetic field produced by passing current through the coil of the tangent Galvanometer and BH be the horizontal component of earth’s magnetic field. Under the action of two magnetic fields, the needle comes to rest making angle θ with BH, such that
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-45
tan θ = \(\frac { B }{{ B }_{H}}\)
B = BH tan θ ……. (1)
When no current is passed through the coil, the small magnetic needle lies along the horizontal component of Earth’s magnetic field. When the circuit is switched ON, the electric current will pass through the circular coil and produce magnetic field. Now there are two fields which are acting mutually perpendicular to each other. They are:

  • the magnetic field (B) due to the electric current in the coil acting normal to the plane of the coil.
  • the horizontal component of Earth’s magnetic field (BH)

Because of these crossed fields, the pivoted magnetic needle deflects through an angle θ. From tangent law, B = BH tan θ When an electric current is passed through a circular coil of radius R having N turns, the magnitude of the magnetic field at the center is
B = µ0 \(\frac { NI }{ 2R }\) …….. (2)
From equation (1) and equation (2), we get
µ0 \(\frac { NI }{ 2R }\) = BH tan θ
The horizontal component of Earth’s magnetic field can be determined as
B = µ0 \(\frac { NI }{ 2R tan θ }\) in tesla ……. (3)

Question 10.
Explain the principle and working of a moving coil galvanometer.
Answer:
Moving coil galvanometer:
Moving coil galvanometer is a device which is used to indicate the flow of current in an electrical circuit.

Principle:
When a current-carrying loop is placed in a uniform magnetic field it experiences a torque.

Construction:
A moving coil galvanometer consists of a rectangular coil PQRS of insulated thin copper wire. The coil contains a large number of turns wound over a light metallic frame. A cylindrical soft-iron core is placed symmetrically inside the coil. The rectangular coil is suspended freely between two pole pieces of a horse-shoe magnet.

The upper end of the rectangular coil is attached to one end of fine strip of phosphor bronze and the lower end of the coil is connected to a hair spring which is also made up of phosphor bronze. deflection of the coil with the help of lamp and scale arrangement. The other end of the mirror is connected to a torsion head T. In order to pass electric current through the galvanometer, the suspension strip W and the spring S are connected to terminals.

Working:
Consider a single turn of the rectangular coil PQRS whose length be l and breadth b. PQ = RS = l and QR = SP = b.
Let I be the electric current flowing through the rectangular coil PQRS. The horse-shoe magnet has hemispherical magnetic poles which produce a radial magnetic field.
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-46
Due to this radial field, the sides QR and SP are always parallel to to the B-field (magnetic field) and experience no force. The sides PQ and RS are always parallel to the B-field and experience force and due to this, torque is produced.
For a single turn, the deflection couple.
τ = bF = bBIl= (lb) BI = ABI since, area of the coil
A = lb For a coil with N turns, we get r = NABI …… (1)
Due to this deflecting torque, the coil gets twisted and restoring torque (also known as restoring couple) is developed. Hence the magnitude of restoring the couple is proportional to the amount of twist θ. Thus τ = K θ ……. (2)
where K is the restoring couple per unit twist or torsional constant of the spring. At equilibrium, the deflection couple is equal to the restoring couple. Therefore by comparing equation (1) and (2), we get
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-47

NABI = Kθ ⇒ I = \(\frac { K }{ NAB }\)θ …….. (3)
(or) I = G θ
Where, G = \(\frac { K }{ NAB }\) is called is called galvanometer constant or current reduction factor of the galvanometer. Since, suspended moving coil galvanometer is very sensitive, we have to handle with high care while doing experiments. For most of the galvanometer, we use arc pointer type moving coil galvanometer.

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 11.
Discuss the conversion of a galvanometer into an ammeter and also a voltmeter.
Answer:
Conversion of galvanometer into ammeter and voltmeter:
A galvanometer is a very sensitive instrument to detect the current. It can be easily converted into an ammeter and voltmeter.
(i) Galvanometer to an Ammeter:
An ammeter is an instrument used to measure the current flowing in the electrical circuit. The ammeter must offer low resistance such that it will not change the current passing through it. So ammeter is connected in series to measure the circuit current.
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-48
A galvanometer is converted into an ammeter by connecting a low resistance in parallel with the galvanometer. This low resistance is called shunt resistance S. The scale is now calibrated in ampere and the range of ammeter depends on the values of the shunt resistance.

Let I be the current passing through the circuit. When current I reach junction A, it divides into two components. Let Ig be the current passing through the galvanometer of resistance Rg through a path AGE and the remaining current (I – Ig) passes along the path ACDE through shunt resistance S.

The value of shunt resistance is so adjusted that current I produce full-scale deflection in the galvanometer. The potential difference across galvanometer is the same as the potential difference across shunt resistance.
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Since, the deflection in the galvanometer is proportional to the current passing through it.
θ = \(\frac { 1 }{ g }\) Ig ⇒ θ Ig ⇒ θ ∝ I
So, the deflection in the galvanometer measures the current I passing through the circuit (ammeter). Shunt resistance is connected in parallel to a galvanometer. Therefore, the resistance of the ammeter can be determined by computing the effective resistance, which is
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-50
Since, the shunt resistance is a very low resistance and the ratio \(\frac { S }{{R}_{ g }}\) is also small. This means, Rg is also small, i.e., the resistance offered by the ammeter is small. So, when we connect ammeter in series, the ammeter will not change the resistance appreciably and also the current in the circuit. For an ideal ammeter, the resistance must be equal to zero. Hence, the reading in the ammeter is always lesser than the actual current in the circuit. Let Iideal be current measured from ideal ammeter and Iactual be the actual current measured in the circuit by the ammeter. Then, the percentage error in measuring a current through an ammeter is
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-51

(ii) Galvanometer to a voltmeter: A voltmeter is an instrument used to measure potential difference across any two points in the electrical circuits. It should not draw any current from the circuit otherwise the value of potential difference to be measured will change. Voltmeter must have high resistance and when it is connected in parallel, it will not draw appreciable current so that it will indicate the true potential difference.
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-52

A galvanometer is converted into a voltmeter by connecting high resistance Rh in series with a galvanometer. The scale is now calibrated in volt and the range of voltmeter depends on the values of the resistance connected in series i.e. the value of resistance is so adjusted that only current Ig produces full scale deflection in the galvanometer.

Let Rg be the resistance of galvanometer and ‘g be the current with which the galvanometer produces full scale deflection. Since the galvanometer is connected in series with high resistance, the current in the electrical circuit is same as the current passing through the galvanometer.
Potential difference
I = Ig
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-53

Since the galvanometer and high resistance are connected in series, the total resistance or effective resistance gives the resistance of the voltmeter. The voltmeter resistance is Rv Rg + Rh
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-54
Note that Ig ∝ V
The deflection in the galvanometer is proportional to the current I. But current is proportional to the potential difference. Hence the deflection in the galvanometer is proportional to a potential difference. Since the resistance of voltmeter is very large, a voltmeter connected in an electrical circuit will draw least current in the circuit. An ideal voltmeter is one which has infinite resistance.

Question 12.
Calculate the magnetic field inside and outside of the long solenoid using Ampere’s circuital law.
Answer:
Magnetic field due to a long current-carrying solenoid:
Consider a solenoid of length L having N turns. The diameter of the solenoid is assumed to be much smaller when compared to its length and the coil is wound very closely.

In order to calculate the magnetic field at any point inside the solenoid, we use Ampere’s circuital law. Consider a rectangular loop abed. Then from Ampere’s circuital law,
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-55
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-56
= µ0 x (total current enclosed by Amperian loop)
The left-hand side of the equation is
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-57
Since the elemental lengths along bc and da are perpendicular to the magnetic field which is along the axis of the solenoid, the integrals
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-58
Since the magnetic field outside the solenoid is zero, the integral
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-59
For the path along ab, the integral is
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-60
where the length of the loop ab is h. But the choice of length of the loop ab is arbitrary. We can take a very large loop such that it is equal to the length of the solenoid L. Therefore the integral is
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-61
let N I be the current passing through the solenoid of N turns, then
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-62
The number of turns per unit length is given by \(\frac { NI }{ L }\) = n, then
B = µ0 \(\frac { nLI }{ L }\) = µ0nI
Since n is a constant for a given solenoid and p0 is also constant. For a fixed current I, the magnetic field inside the solenoid is also a constant.

Magnetism and Magnetic Effects of Electric Current Numerical problems

Question 1.
A bar magnet having a magnetic moment \(\overrightarrow { M } \) is cut into four pieces i.e., first cut in two pieces along the axis of the magnet and each piece is further cut into two pieces. Compute the magnetic moment of each piece.
Solution:
Consider a bar magnet of magnetic moment \(\overrightarrow { M } \) . When a bar magnet first cut in two pieces
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-63
Their magnetic moment of each pieces \(\frac { \overrightarrow { M } }{ 4 } \)
Their magnetic moment of each pieces \(\overrightarrow { M } \)new \(\frac { 1 }{4} \) \(\overrightarrow { M } \)

Question 2.
A conductor of linear mass density 0.2 g m-1 suspended by two flexible wires as shown in the figure. Suppose the tension in the supporting wires ¡s zero when it is kept inside the magnetic field of I T whose direction is into the page. Compute the current inside the conductor and also the direction of the current. Assume g = 10 m s-2.
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-64
Solution:
Downward force, F = mg
Linear mass density, \(\frac { m }{ l }\) = 0.2 gm-1
\(\frac { m }{ l }\) = 0.2 x 10-3 kg m-1
m = (0.2 x 10-3 x l) kg m-1
F = (0.2 x 10-3 x l x 10) N
The upward magnetic force acting on the wire
F = BIl
0.2 x 10-3 x l x 10 = 1 x I x l
I = 2 x 10-3
I = 2 m A

Question 3.
A circular coil with cross-sectional area 0.1 cm2 is kept in a uniform magnetic field of strength 0.2 T. If the current passing in the coil is 3 A and plane of the loop is perpendicular to the direction of magnetic field. Calculate
(a) total torque on the coil
(b) total force on the coil
(c) average force on each electron in the coil due to the magnetic field of the free electron density for the material of the wire is 1028 m-3
Solution:
Cross-sectional area of coil, A = 0.1 cm2
A = 0.1 x 10-4m2
Uniform magnetic field of strength, B = 0.2T
Current passing in the coil, I = 3A
The angle between the magnetic field and normal to the coil, θ = 0°
(a) Total torque on the coil,
τ = ABI sin θ = 0.1 x 10-4 x 0.2 x 3 sin0° sin0° = 0
τ = 0

(b) Total force on the coil
F = BIl sin θ = 0.2 x 3 x l x sin 0°
F = 0

(c) Average force:
F = qVdB
drift velocity, Vd = \(\frac { 1 }{ ne A }\)
[∵ q = e]
F = e \(\left( \frac { 1 }{ neA } \right) \)B
[∵n = 1028 m-3
\(\frac { IB }{ n A }\) = \(\frac{3 \times 0.2}{10^{28} \times 0.1 \times 10^{-4}}\) = 6 x 10-24
Fav = 0.6 x 10-23 N

Question 4.
A bar magnet is placed in a uniform magnetic field whose strength is 0.8 T. Suppose the bar magnet orient at an angle 30° with the external field experiences a torque of 0.2 N m. Calculate:
(i) the magnetic moment of the magnet
(ii) the work done by an applied force in moving it from most stable configuration to the most unstable configuration and also compute the work done by the applied magnetic field in this case.
Solution:
Uniform magnetic field strength B = 0.8T
Bar magnet orient an angle with magnetic field θ = 30°
Torque τ = 0.2 Nm
(i) Magnetic moment of the magnet,
Torque τ = Pm B Sin θ
∴ Magnetic moment, Pm = \(\frac { τ }{ B sin θ }\) = \(\frac { 0.2 }{ 0.8 × Sin 30° }\) = \(\frac { 0.2 }{ 0.4 }\)
Pm = 0.5 Am2

(ii) Work done by external torque is stored in the magnet as potential energy.
W = U = – Pm B Sin θ
Here, the applied force acting on the magnet is moving from most stable θ’ to most unstable θ.
θ’= 0° and θ = 180°
So, workdone W = U = – Pm B (Cosθ – Cosθ’)
= – Pm B (Cos 180° – Cos 0°) = – 0.5 x 0.8 ((-1) – 1) = – 0.4 (-2)
W = U = 0.8
W= 0.8 J

Question 5.
A non – conducting sphere has a mass of 100 g and a radius 20 cm. A flat compact coil of wire with turns 5 is wrapped tightly around it with each turn concentric with the sphere. This sphere is placed on an inclined plane such that plane of coil is parallel to the inclined plane. A uniform magnetic field of 0.5 T exists in the region in the vertically upward direction. Compute the current 1 required to rest the sphere in equilibrium.
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-65
Solution:
At equilibrium
fs R-pm B sin θ = 0
mgR = NBAI
I = \(\frac { mg R }{ N B A }\) = \(\frac { mg R }{{ N B π R }^{2}}\)
I = \(\frac { mg }{ π R N B }\)
Mass of the sphere, m= 100 g = 100 x 10-3 kg
Radius of the sphere R = 20 cm = 20 x 10-2 m
Number of turns n = 5
Uniform magnetic field B = 0.5 T
I = \(\frac{100 \times 10^{-3} \times 10}{\pi \times 20 \times 10^{-2} \times 5 \times 0.5}\) = \(\frac{1000 \times 10^{-3}}{\pi \times 50 \times 10^{-2}}\) = \(\frac{20 \times 10^{-1}}{\pi}\)
I = \(\frac { 2 }{ π }\) A.

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 6.
Calculate the magnetic field at the center of a square loop which carries a current of 1.5 A, length of each loop is 50 cm.
Solution:
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-66
Current through the square loop, I = 1.5A
Length of each loop, l = 50cm 50 x 10-2 m
According to Biot-Savart Law.
Magnetic field due to a current-carrying straight wire
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-67
B =0.084866 x 10-5 T
Magnetic field at a point p? of centre of current carrying square ioop
B’ =4 sides x B
= 4 x 0.08487 x 10-5 = 0.33948 x 10-5
B’ =3.4 x 10-6 T

Question 7.
Show that the magnetic field at any point on the axis of the solenoid having n turns per unit length is B = \(\frac { 1 }{ 2 }\) µ0 nI (cos θ1 – cosθ2)
Solution:
A solenoid is a cylindrical coil having a number of circular turns. Consider a solenoid having radius R consists of n number of turns per unit length.
Let ‘P’ be the point at a distance ‘x’ from the origin of the solenoid. The current carrying element dx at a distance x from origin and the distance r from point ‘P’.
r = \(\sqrt{\mathrm{R}^{2}+\left(x_{0}-x\right)^{2}}\)
The magnetic field due to current carrying circular coil at any axis is
dB = \(\frac{\mu_{0} \mathrm{IR}^{2}}{2 r^{3}}\) × N
Where N = ndx, then
dB = \(\frac {{ µ }_{0}}{ 2 }\) \(\frac{n \mathrm{IR}^{2} d x}{r^{3}}\) ……. (1)
sin θ = \(\frac { R }{ r }\)
r = R × \(\frac { 1 }{ sin θ }\) = R cosec θ ……. (2)
tan θ = \(\frac { R }{{ x }_{0}-x}\)
x0 – x = R × \(\frac { 1 }{ tan θ }\) = R cot θ
\(\frac { dx }{ dθ }\) = R cosec2 θ
⇒ dx = R cosec2 θ dθ ……. (3)
From above three equations, wer get
dB = \(\frac {{ µ }_{0}}{ 2 }\) \(\frac{n \mathrm{IR}^{2}\left(\mathrm{R} \csc ^{2} \theta d \theta\right)}{\mathrm{R}^{3} \csc ^{3} \theta}\)
dB = \(\frac {{ µ }_{0}}{ 2 }\) n I sin θ dθ
Now total magnetic field can be obtained by integrating from θ1 to θ2, we get
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-68

Question 8.
Let I1 and I2 be the steady currents passing through a long horizontal wire XY and PQ respectively. The wire PQ is fixed in the horizontal plane and the wire XY be is allowed to move freely in a vertical plane. Let the wire XY is in equilibrium at a height d over the parallel wire PQ as shown in the figure.
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-69
Solution:
If the wire XY is slightly displaced and released, it executes simple harmonic motion due to the force of repulsion produced between the current-carrying wire.
Acceleration of the wire, a = -ω2y
Time period of oscillation of the wire,
T = 2π \(\sqrt { \frac { d }{ g } } \)

Samacheer Kalvi 12th Physics Magnetism and Magnetic Effects of Electric Current Additional Questions Solved

I. Choose the Correct Answer

Question 1.
Source (s) of a magnetic field is (are)-
(a) an isolated magnetic pole
(b) a static electric charge
(c) a moving electric charge
(d) all of these
Answer:
(c) a moving electric charge

Question 2.
Magnetic field lines-
(a) cannot intersect at all
(b) intersect at infinity
(c) intersect within the magnet
(d) intersect at the neutral points
Answer:
(a) cannot intersect at all

Question 3.
A magnetic needle is kept in a non-uniform magnetic field. It experiences-
(a) a force and a torque
(b) a force but not a torque
(c) a torque but not a force
(d) neither a force nor a torque
Answer:
(a) a force and a torque

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 4.
The SI unit of pole strength is-
(a) Am2
(b) Am-1
(c) Am-2
(d) Am
Answer:
(d) Am

Question 5.
Earth’s magnetic field always has a horizontal component except for at-
(a) equator
(b) magnetic pole
(c) a latitude of 60°
(d) a latitude of 50°
Answer:
(b) magnetic pole

Question 6.
The angle of dip at the magnetic equator is-
(a) 0°
(b) 30°
(c) 60°
(d) 90°
Answer:
(a) 0°

Question 7.
At a certain place the horizontal component of earth’s magnetic field is √3 times vertical component. The angle of dip at that place is-
(a) 75°
(b) 60°
(c) 45°
(d) 30°
Answer:
(d) 30°
Hint:
tan δ = \(\frac {{ B }_{V}}{{ B }_{H}}\) = \(\frac { 1 }{ √3 }\) ⇒ δ = tan-1 \(\left( \frac { 1 }{ \sqrt { 3 } } \right) \) = 30°

Question 8.
At magnetic poles the angle of dip is-
(a) 45°
(b) 30°
(c) 0°
(d) 90°
Answer:
(d) 90°

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 9.
The horizontal component of earth’s magnetic field at a place is 3.6 x 10-5T. If the angle of dip at this place is 60°. the vertical components of earth’s field at this place is-
(a) 1.2 x 10-5T
(b) 2.4 x 10-5T
(c) 4 x 10-5T
(d) 6.2 x 10-5T
Answer:
(d) 6.2 x 10-5T
Hint:
Bv = BH tan δ = 3.6 x 10-5 x tan 60°
Bv = 6.2 x 10-5T

Question 10.
A bar magnet of magnetic moment M is cut into two parts of equal length. The magnetic moment of either part is-
(a) M
(b) 2M
(c) \(\frac { M }{ 2 }\)
(d) Zero
Answer:
(c) \(\frac { M }{ 2 }\)

Question 11.
A magnetic needle suspended by a silk thread is vibrating in the earths magnetic field, If the temperature of the needle is increased by 500°C, then-
(a) time period decreases
(b) time period remains unchanged
(c) time period increases
(d) the needle stops vibrating
Answer:
(c) time period increases
Hint:
Magnet moment decreases with temperature. Therefore the time period will increase.

Question 12.
Demagnetisation of a magnet can be done by-
(a) rough handling
(b) magnetising in the opposite direction
(c) heating
(d) all the above
Answer:
(d) all the above

Question 13.
All the magnetic materials lose their magnetic properties when-
(a) dipped in water
(b) dipped in oil
(c) brought near a piece of iron
(d) strongly heated
Answer:
(d) strongly heated

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 14.
The relative permeability of a paramagnetic material is-
(a) greater than unity
(b) less than unity
(c) equal to unity
(d) negative
Answer:
(a) greater than unity

Question 15.
The relative permeability of a diamagnetic material is-
(a) greater than unity
(b) less than unity
(c) equal to unity
(d) negative
Answer:
(b) less than unity

Question 16.
Which of the following is most suitable for the core of an electromagnet?
(a) air
(b) soft iron
(c) steel
(d) Co-Ni alloy
Answer:
(b) soft iron

Question 17.
Soft Iron is used in many parts of electrical machines for-
(a) low hysteresis loss and low permeability
(b) low hysteresis less and high permeability
(c) high hysteresis loss and low permeability
(d) high hysteresis loss and high permeability
Answer:
(b) low hysteresis loss and high permeability

Question 18.
When a diamagnetic substance is brought near the north or the south pole of a bar magnet, it is-
(a) attracted by the poles
(b)repel led by the poles
(c) attracted by the north pole and repelled by the south pole
(d) attracted by the south pole and repelled by the north pole
Answer:
(b) repelled by the poles

Question 19.
At Curie point, a ferromagnetic material becomes-
(a) non-magnetic
(b) diamagnetic
(c) paramagnetic
(d) antiferromagnetic
Answer:
(c) paramagnetic

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 20.
Magnetic permeability is maximum for-
(a) diamagnetic substances
(b) paramagnetic substances
(c) ferromagnetic substances
(d) all of these
Answer:
(c) ferromagnetic substances

Question 21.
The material of a permanent magnet has-
(a) high retentivity, low coercivity
(b) low retentivity, low coercivity
(c) low retentivity, low coercivity
(d) high retentivity, high coercivity
Answer:
(d) high retentivity, high coercivity

Question 22.
Which one of the following is not made of soft iron?
(a) electromagnet
(b) core of transformer
(c) core of dynamo
(d) magnet of loudspeaker
Answer:
(d) magnet of loudspeaker

Question 23.
A dip circle k at right angles to the magnetic meridian. The apparent dip is-
(a) 0°
(b) 30°
(c) 600
(d) 90°
Answer:
(d) 90°

Question 24.
A magnetic needle is placed in a uniform magnetic field. It experience-
(a) a force and a torque
(b) a force but not a torque
(c) a torque but not a force
(d) neither a force nor a torque
Answer:
(c) a torque but not a force

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 25.
The relative permeability of iron is 5500. Its magnetic susceptibility is-
(a) 5501
(b) 5500 x 10-7
(c) 5500 x 107
(d) 5499
Answer:
(d) 5499
Hint:
µr = 1 + x
⇒ x = µr -1 = 5500 – 1 = 5499

Question 26.
The inherent property of all matter is/are-
(a) paramagnetism
(b) diamagnetism
(c) ferromagnetism
(d) all the above
Answer:
(b) diamagnetism

Question 27.
A magnetic needle suspended freely-
(a) orients itself in a definite direction
(b) remains in any direction
(c) become vertical with N-pole up
(d) become vertical with N-pole down
Answer:
(a) orients itself in a definite direction

Question 28.
The earth’s magnetic field at a given point is 0.5 x 10-5 Wb/m2. This field is to be annulled by magnetic induction at the center of circular conducting loop of radius 5 cm. The current required to be flown in the loop is nearly-
(a) 0.2 A
(b) 0.4 A
(c) 4 A
(d) 40 A
Answer:
(b) 0.4 A
Hint:
B = \(\frac {{ µ }_{0}I}{ 2r }\) ⇒ I = \(\frac {2r B}{{ µ }_{0}}\) = \(\frac{2 \times 0.05 \times 0.5 \times 10^{-5}}{4 \pi \times 10^{-7}}\) = 04 A

Question 29.
A frog can be levitated in a magnetic field produced by a current in a vertical solenoid placed below the frog. This is possible because the body of the frog is-
(a) paramagnetic
(b) diamagnetic
(c) ferromagnetic
(d) antiferromagnetic
Answer:
(a) paramagnetic

Question 30.
The magnetic moment of a current-carrying circular coil of radius r varies as-
(a) \(\frac { 1 }{{ r }^{2}}\)
(b) \(\frac { 1 }{ r }\)
(c) r
(d) r2
Answer:
(d) r2
Hint:
Magnetic moment, M = 1A = I(πr2) = M αr2

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 31.
For a paramagnetic material, the dependence of the magnetic susceptibility % on the absolute temperature T is given as-
(a) x ∝ r
(b) x ∝ \(\frac { 1 }{{T}^{ 2 }}\)
(c) x ∝ \(\frac { 1 }{ T }\)
(d) x ∝ T2
Answer:
(c) x ∝ \(\frac { 1 }{ T }\)

Question 32.
A charged particle (charge q) is moving in a circle of radius R with uniform speed V. The associated magnetic moment is given by-
(a) qVR2
(b) \(\frac {{qVR}^{ 2 }}{2}\)
(c) qVR
(d) \(\frac { qVR }{ 2 }\)
Answer:
(d) \(\frac { qVR }{ 2 }\)
Hint:
Magnetic moment, M = IA = \(\frac { q }{ T }\) (π R2)
M = \(\frac { qv }{ 2πR }\) (π R2) = \(\frac { qvR }{ 2 }\)

Question 33.
Nickel shows ferromagnetic property at room temperature. If the temperature is increased . beyond Curie temperature, then it will show-
(a) antiferromagnetism
(b) no magnetic property
(c) diamagnetism
(d) para magnetism
Answer:
(d) paramagnetism

Question 34.
A proton enters a magnetic field of flux density 1.5 Wb/m2 with a speed of 2x 107 m/s at an angle of 30° with the field. The force on the proton will be-
(a) 0.24 x 10-12 N
(b) 2.4 x 10-12 N
(c) 24 x 10-12 N
(d) 0.024 x 10-12 N
Answer:
(b) 2.4 x 10-12 N
Hint:
F = Bqv sin θ =1.5 x 1.6 x 10-19 x 2 x 107 x sin 30° = 2.4 x 10-12 N

Question 35.
A moving charge produces-
(a) an electric field only
(b)a magnetic field only
(c) both electric and magnetic fields
(d) neither an electric nor a magnetic field
Answer:
(c) both electric and magnetic fields

Question 36.
A straight conductor carrying a current I, is split into a circular loop of radius r as shown in the figure. The magnetic field at the centre O of the circle, in tesla, is-
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-70
(a) \(\frac {{ µ }_{0}I}{ 2r }\)
(b) \(\frac {{ µ }_{0}I}{ 2πr }\)
(c) \(\frac {{ µ }_{0}I}{ πr }\)
(d) zero
Answer:
(d) zero
Hint:
Field due to the upper and lower semicircles will cancel out.

Question 37.
In a moving coil galvanometer the current ‘i’ is related to the deflection θ as-
(a) i α θ
(b) i α tan θ
(c) i α θ2
(d) i α √θ
Answer:
(a) i α θ

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 38.
The magnetic field due to a current carrying circular coil on the axis, at a large distance r from the centre of the coil, varies approximately as-
(a) \(\frac { 1 }{ r }\)
(b) \(\frac { 1 }{ { r }^{ \frac { 3 }{ 2 } } } \)
(c) \(\frac { 1 }{{ r }^{3}}\)
(d) \(\frac { 1 }{{ r }^{2}}\)
Answer:
(c) \(\frac { 1 }{{ r }^{3}}\)

Question 39.
A moving charge is subjected to an external magnetic field. The change in the kinetic energy of the particle-
(а) increases with the increase in the field strength
(b) decreases with the increase in the field strength
(c) is always zero
(d) depends upon whether the field is uniform or non-uniform
Answer:
(c) is always zero

Question 40.
Lorentz force is given by-
(a) q (E + V × B)
(b)q (E -V × B)
(c) q (E + V.B)
(d) q (E × B × V)
Answer:
(a) q (E + V × B)

Question 41.
A circular loop has radius R and a current I flows through it. Another circular loop has a radius of 2R and a current 21 flows through it. The ratio of the magnetic fields at their centres is-
(a) \(\frac { 1 }{ 4 }\)
(b) 1
(c) 2
(d) 4
Answer:
(b) 1
Hint:
B1 = \(\frac {{ µ }_{0}I}{ 2r }\) and B2 = \(\frac { µ(2I) }{ 2(2R) }\); \(\frac {{ B }_{1}}{ { B }_{2}}\) = 1

Question 42.
The magnetic field inside a solenoid is-
(a) directly proportional to current
(b) inversely proportional to current
(c) directly proportional to its length
(d) inversely proportional to the total number of turns
Answer:
(a) directly proportional to current

Question 43.
A circular loop of area 0.01 m2 and carrying a current of 10A is placed parallel to a magnetic field of intensity 0.1T. The torque acting on the loop, in Nm is-
(a) 1.1
(b) 0.8
(c) 0.001
(d) 0.01
Answer:
(d) 0.01
Hint:
τ = BIA = 0.1 x 10 x 0.01 = 0.01 Nm

Question 44.
In a current-carrying long solenoid, the field produced does not depend upon-
(a) number of turns per unit length
(b) current flowing
(c) the radius of the solenoid
(d) all of the above
Answer:
(c) radius of the solenoid

Question 45.
When a charged particle enters a uniform magnetic field its kinetic energy-
(a) remains constant
(b) increases
(c) decreases
(d) becomes zero
Answer:
(a) remains constant

II. Fill in the blanks.

Question 1.
At Curie point, a ferromagnetic material becomes ………………
Answer:
paramagnetic

Question 2.
Electromagnets are made of soft iron because soft iron has ………………
Answer:
high susceptibility and low retentivity

Question 3.
The word magnetism is derived from iron ore ………………
Answer:
haematite

Question 4.
The chemical formula of magnetises is ………………
Answer:
Fe3 O4

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 5.
The iron ore magnetite was found on the island of ………………
Answer:
Magnesia

Question 6.
……………… suggested that earth behaves as a giant bar magnet
Answer:
Gilbert

Question 7.
The field at the surface of the earth is approximately equal to
Answer:
10-4T

Question 8.
The natural magnets have ………………
Answer:
the irregular shape and they are weak

Question 9.
Pieces of iron or steel that acquires magnetic properties when it is rubbed with a magnet are called ………………
Answer:
artificial magnet

Question 10.
The artificial magnet in the form of a rectangular or cylindrical bar is called ………………
Answer:
bar magnet

Question 11.
In a magnet, the magnetic attraction is maximum at the ……………… of the magnet.
Answer:
poles

Question 12.
The unit of pole strength is ………………
Answer:
ampere meter

Question 13.
The unit of magnetic flux density is ………………
Answer:
Weber metre2 (or) tesla

Question 14.
The value of po is equal to ………………
Answer:
4π x 10-7 Hm-1

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 15.
A unit pole experiences a force of ………………
Answer:
10-7N

Question 16.
At neutral points the resultant magnetic field due to the magnet and earth is ………………
Answer:
the magnetic field of the earth

Question 17.
The circular scale of deflection magneto meter is divided into ……………… quadrants?
Answer:
4

Question 18.
Each quadrant of the circular scale of deflection magnetometer is graduated from ………………
Answer:
0 – 90°

Question 19.
The sensitivity of the deflection magneto meter is more at ………………
Answer:
45°

Question 20.
The magnetic field used to magnetise a material is called the ………………
Answer:
magnetic intensity

Question 21.
The unit of magnetising field or magnetic intensity is ………………
Answer:
ampere metre 1

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 22.
The pole strength per unit area of the cross-section of the material is termed as
Answer:
intensity of magnetisation ………………

Question 23.
The magnetic moment per unit volume of the material is termed as ………………
Answer:
intensity of magnetization

Question 24.
In diamagnetic materials the net magnetic moment of atoms is ………………
Answer:
zero

Question 25.
The susceptibility of diamagnetic materials has a ……………… value.
Answer:
low negative

Question 26.
The susceptibility (xm) of bismuth is ………………
Answer:
0.00017

Question 27.
The relative permeability of diamagnetic material is ………………
Answer:
less than one

Question 28.
Ferromagnetic substances have ………………magnetic moment.
Answer:
spontaneous net

Question 29.
As the temperature increase, the value of susceptibility of the ferromagnetic substance ………………
Answer:
decrease

Question 30.
The phenomenon of lagging of magnetic induction behind the magnetising field is called ………………
Answer:
hysteresis

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 31.
The direction of the magnetic field in a current-carrying conductor is given by ………………
Answer:
Maxwell’s right cork screw rule

Question 32.
The relative permeability (µr) for air is ………………
Answer:
one

Question 33.
The instrument used for measuring current is ………………
Answer:
Tangent galvanometer

Question 34.
Tangent galvanometer works on the principle of ………………
Answer:
Tangent Law

Question 35.
The tangent law is applied in ………………
Answer:
Tangent

III Match the following

Question 1.
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-71
Answer:
(i) → (c)
(ii) → (d)
(iii) → (a)
(iv) → (b)

Question 2.
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-72
Answer:
(i) → (c)
(ii) → (d)
(iii) → (b)
(iv) → (a)

Question 3.
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-73
Answer:
(i) → (c)
(ii) → (a)
(iii) → (d)
(iv) → (b)

Question 4.
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-74
(i) → (d)
(ii) → (c)
(in) →(a)
(iv) → (b)

Question 5.
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-75
Answer:
(i) → (d)
(ii) → (a)
(iii) → (b)
(iv) → (c)

IV Assertion and Reason Questions

(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) If assertion is true but reason is false.
(d) If assertion is false but reason is true.
(e) If both assertion and reason are false.

Question 1.
Assertion: The poles of magnet cannot be separated by breaking into two pieces.
Reason: The magnetic moment will be reduced to half when a magnet is broken into two equal pieces.
Answer:
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
Explanation: As we know every atom of a magnet acts as a dipole, so poles cannot be separated. When magnet is broken into two equal pieces, magnetic moment of each part will be half of the original magnet.

Question 2.
Assertion: When radius of the circular loop carrying current is doubled, its magnetic moment becomes four times.
Reason: Magnetic moment depends on area of the loop.
Answer:
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
Explanation: Magnetic moment M = IA = I (πr2)
New magnetic moment Ml = I π (2r)2 = 4πIr2 = 4M

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 3.
Assertion: The ferromagnetic substance do not obey Curie’s Law.
Reason: At Curie point, a ferromagnetic substance starts behaving as a paramagnetic substance.
Answer:
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
Explanation: The susceptibility of ferromagnetic substance decreases with the rise of temperature in a complicated manner. After Curie point, the susceptibility ferromagnetic substance varies inversely with its absolute temperature.

Question 4.
Assertion: Soft iron is used as transformer core.
Reason: Soft iron has narrow hysteresis loop.
Answer:
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
Explanation: For high efficiency of transformer, the energy loss will be lesser if the hysteresis loop is lesser area, i.e narrow.

Question 5.
Assertion: Cyclotron does not accelerate electrons.
Reason: Mass of the electron is very small
Answer:
(a) If both assertion and reason are true and the reason is the correct explanation of assertion.
Explanation: Cyclotron is suitable for accelerating heavy particles like protons, x-particles. etc. and not for electrons because of low mass.

Question 6.
Assertion: Cyclotron is a device which is used to accelerate the positive ion.
Reason: Cyclotron frequency depends upon the velocity.
Answer:
(c) If assertion is true but reason is false.
Explanation: Cyclotron is utilised to accelerate the positive ion. And cyclotron frequency is given by v =\(\frac { Be }{ 2πm }\). It means cyclotron frequency doesn’t depend upon velocity.

Magnetism and Magnetic Effects of Electric Current 2-mark Questions

Question 1.
Define magnetic declination (D).
Answer:
The angle between magnetic meridian at a point and geographical meridian is called the declination or magnetic declination (D).

Question 2.
Define magnetic inclination (I).
Answer:
The angle subtended by the Earth’s total magnetic field B with the horizontal direction in the magnetic meridian is called dip or magnetic inclination (I) at the point.

Question 3.
Define magnetic field.
Answer:
The magnetic field \(\vec { B } \) at a point is defined as a force experienced by the bar magnet of unit pole strength.
B = \(\frac { 1 }{{ q }_{m}}\) \(\vec { F } \) Its unit is N A-1 m-1

Question 4.
Define magnetic flux density.
Answer:
The magnetic flux density is defined as the number of magnetic field lines crossing unit area kept normal to the direction of the line of force. Its unit is Wb m-2 or tesla.

Question 5.
State ‘Tangent Law’.
Answer:
When a magnetic needle or magnet is freely suspended in two mutually perpendicular uniform magnetic fields, it will come to rest in the direction of the resultant of the two fields.

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 6.
Define in terms of magnetising field.
Answer:
The magnetic field which is used to magnetize a sample or specimen is called the magnetising field. Magnetising field is a vector quantity and it denoted by \(\vec { H } \) and its unit is A m-1 .

Question 7.
State Right hand thumb rule.
Answer:
Answer:
If we curl the fingers of right hand in the direction of current in the loop, then the stretched thumb gives the direction of the mag netic moment associated with the loop.

Question 8.
State ‘One ampere’.
Answer:
One ampere is defined as that current when it is passed through each of the two infinitely long parallel straight conductors kept at a distance of one meter apart in a vacuum causes each conductor to experience a force of 2 x 10-7 newton per meter length of the conductor.

Magnetism and Magnetic Effects of Electric Current 3-Mark Questions

Question 1.
Explain Curie’s Law of magnetism.
Answer:
When the temperature is increased, thermal vibration will upset the alignment of magnetic dipole moments. Therefore, the magnetic susceptibility decreases with an increase in temperature. In many cases, the susceptibility of the materials is This relation is called Curie’s law.
xm ∝\(\frac { 1 }{ T }\) or xm = \(\frac { C }{ T }\)
This relation is called Curie’s law.

Question 2.
Write down the concept of Maxwell’s right-hand cork screw rule.
Answer:
This rule is used to determine the direction of the magnetic field. If we rotate a right-handed screw using a screw driver, then the direction of current is same as the direction in which screw advances, and the direction of rotation of the screw gives the direction of the magnetic field.

Question 3.
Define in terms of Voltage Sensitivity of a galvanometer.
Answer:
It is defined as the deflection produced per unit voltage applied across it.
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-76

Question 4.
Define in terms of the Current Sensitivity of a galvanometer.
Answer:
It is defined as the deflection produced per unit current flowing through it.
Is = \(\frac { θ }{ I }\) = \(\frac { N A B }{ K }\) ⇒ Is = \(\frac { 1 }{ G }\)

Magnetism and Magnetic Effects of Electric Current 5-Marks Questions

Question 1.
Derive an expression for the potential energy of a bar magnet in a uniform magnetic field.
Answer:
When a bar magnet (magnetic dipole) of dipole moment \(\vec { B } \)m is held at an angle 0 with the direction of a uniform magnetic field B , the magnitude of the torque acting on the dipole is
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-77
\(\left| { \vec { \tau } }_{ B } \right| \) = \(\left| { \vec { P } }_{ B } \right| \) \(\left| \vec { B } \right| \) sin θ ….. (1)
If the dipole is rotated through a very small angular displacement dθ against the torque τB at constant angular velocity, then
the work done by external torque (\({ \vec { \tau } }_{ ext }\)) for this small angular displacement is given by
dW = (\({ \vec { \tau } }_{ ext }\)) dθ …… (2)
The bar magnet has to be moved at constant angular velocity, which implies that
\(\left| { \vec { \tau } }_{ B } \right| \) = \(\left| { \vec { \tau } }_{ ext } \right| \)
dW = Pm B sin θ d θ
Total work done in rotating the dipole from θ’ to θ is e e
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-78
W = -Pm B (cosθ – cosθ’) …… (3)
This work done is stored as potential energy in bar magnet at an angle θ when it is rotated from θ’ to θ and it can be written as
U = -Pm B (cosθ – cosθ’) …(4)
In fact, equation (4) gives the difference in potential energy between the angular positions θ’ and θ. We can choose the reference point θ’ = 90°, so that the second term in the equation becomes zero and the equation 4 can be written as
U = -Pm B(cosθ) …(5)
U = -Pm • \(\vec { B } \)
Case 1:
(i) If θ = 0°, then
U= —Pm B (cos 0° ) = Pm B
(ii) If θ = 180°, then
U=-pm B(cos 180°) = pm B
The potential energy stored in a bar magnet in a uniform magnetic field is given by

Question 2.
Write down the application of Hysteresis loop.
Answer:
Applications of hysteresis loop:
The significance of hysteresis loop is that it provides information such as retentivity, coercivity, permeability, susceptibility and energy loss during one cycle of magnetisation for each ferromagnetic material. Therefore, the study of hysteresis loop will help us in selecting proper and suitable material for a given purpose. Some examples:

(i) Permanent magnets:
The materials with high retentivity, high coercivity and high permeability are suitable for making permanent magnets.
Examples:
Steel and Alnico

(ii) Electromagnets:
The materials with high initial permeability, low retentivity, low coercivity and thin hysteresis loop with smaller area are preferred to make electromagnets.
Examples:
Soft iron and Mumetal (Nickel Iron alloy).

(iii) Core of the transformer:
The materials with high initial permeability, large magnetic induction and thin hysteresis loop with smaller area are needed to design transformer cores.
Examples:
Soft iron

Question 3.
Write down the difference between soft and hard ferromagnetic materials.
Answer:
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-79

Question 4.
State and explain the Biot – Savart Law.
Answer:
Biot and Savart experimentally observed that the magnitude of magnetic field d\(\vec { B } \) at a point P at a distance r from the small elemental length taken on a conductor carrying current varies
(i) directly as the strength of the current I
(ii) directly as the magnitude of the length element \(\vec { dl } \)
(iii) directly as the sine of the angle (say,θ) between \(\vec { dl } \) and \(\hat{r}\).
(iv) inversely as the square of the distance between the point P and length element \(\vec { dl } \). This is expressed as
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-80
dB ∝ \(\frac { I dl }{{r}^{ 2 }}\) sin θ
dB = k \(\frac { I dl }{{r}^{ 2 }}\) sin θ
Where K = \(\frac {{μ }_{0}}{ 4π }\) in SI units and k = 1 in CGS units.
In vector notation,
d\(\vec { B } \) = \(\frac {{μ }_{0}}{ 4π }\) \(\frac{\mathrm{I} d \vec{l} \times \hat{r}}{r^{2}}\) ….. (1)
Here vector d\(\vec { B } \) is perpendicular to both I \(\vec { dl } \) (pointing current carrying conductor the direction of current flow) and the unit vector and \(\hat{r}\) directed from \(\vec { dl } \) toward point P The equation 1 is used to compute the magnetic field only due to a small elemental length \(\vec { dl } \) of the conductor. The net magnetic field at P due to the conductor is obtained from principle of superposition by considering the contribution from all current elements I \(\vec { dl } \). Hence integrating equation (1), we get
\(\vec { B } \) = ∫ d\(\vec { B } \) = \(\frac {{ µ }_{0}I}{ 4π }\) ∫ \(\frac { \vec { dl } \times \quad \hat { r } }{ { r }^{ 2 } } \) ……. (2)
where the integral is taken over the entire current distribution.
Case:
1. If the pont P lies on the ckonductor, then θ = 0°. Therefore, d\(\vec { B } \) is zero.
2. If the point lies perpendicular to the conductor, then θ = 90°. Therefore, d\(\vec { B } \) is maximum and is given by d\(\vec { B } \) = \(\frac { I dl }{{ r }{2}}\) \(\hat{n}\).
where \(\hat{n}\) is the unit vector perpendicular to both I \(\vec { dl } \) and \(\hat{r \).

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 5.
Write down the difference between Coulomb’s Law and Biot-Savart Law
Answer:
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-81

Question 6.
Obtain an expression for the magnetic dipole moment of a revolving electron.
Answer:
Magnetic dipole moment of revolving electron:
Suppose an electron undergoes circular motion around the nucleus. The circulating electron in a loop is like current in a circular loop (since flow of charge is current). The magnetic dipole moment due to current carrying circular loop is
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-82
\(\vec { µ } \)L = \(\overrightarrow { A } \) ….. (1)
In magnitude, µ = I A
If T is the time period of an electron, the current due to circular motion of th electron is
I = \(\frac { -e }{ T }\) ……. (2)
where -e is the charge of an electron. If R is the radius of the circular orbit and v is the velocity of the electron in the circular orbit, then
T = \(\frac { 2πR }{ v }\) …… (3)
Using equation (2) and equation (3) in equation (1), we get
µL = –\(\frac{\frac{e}{2 \pi \mathrm{R}}}{v}\) πR2 = \(\frac { evR }{ 2 }\) ……. (4)
Where A = πR2 is the area of the circular loop. By definetion, angular moment of the electron about O is
\(\overrightarrow { L } \) = \(\overrightarrow { R } \) x \(\overrightarrow { P } \)
In magnitude,
L = PR = mvR …….. (5)
Using equation (4) and equation (5), we get
\(\frac { { \mu }_{ L } }{ L } \) = –\(\frac { evR }{ \frac { 2 }{ mvR } } \) = \(\frac { e }{ 2m }\) \(\overrightarrow { L } \) ……. (6)
The negative sign indicates that the magnetic moment and angular momentum are in opposite direction.
In magnitude,
\(\frac { { \mu }_{ L } }{ L } \) = \(\frac { e }{ 2m }\) = \(\frac{1.60 \times 10^{-19}}{2 \times 9.11 \times 10^{-31}}\) = 0. 0878 x 1012
\(\frac { { \mu }_{ L } }{ L } \) = 8.78 x 1010 C kg-1 = content
The ration \(\frac { { \mu }_{ L } }{ L } \) is a constant and also known as gyro-magnetic ration \(\frac { { \mu }_{ L } }{ L } \). It must be noted that the gyro-magnetic ratio is a constant of proportionality which connects angular momentum of the electron and the magnetic moment of the electron. According to Neil’s Bohr quantization rule, the angular momentum of an electron moving in a stationary orbit is quantized, which means,
l = nh = n \(\frac { h }{ 2π }\)
where, h is the planck’s constant (h = 6.63 x 10-34 J s) and number n takes natural numbers (i.e., n = 1,2,3,…..). Hence,
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-83
µL = n x 9.27 x 10-24 A m2
The minimum magnetic moment can be obtained by subsituting by sustituting n = 1,
µL = n x 9.27 x 10-24 A m2 = 9.27 x 10-24 J T-1 = (µL)min = µB
Where, µB = \(\frac { eh }{ 4πm }\) = 9.27 x 10-24 A m2
is called Bohr magneton. This is a convenient unit with which one can measure atomic magnetic moments.

Question 7.
Apply Ampere’s Circuital Law to find the magnetic field both inside and outside of a toroidal solenoid.
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-84
Answer:
A solenoid is bent in such a way its ends are joined together to form a closed ring shape, is called a toroid. The magnetic field has constant magnitude inside the toroid whereas in the interior region (say, at point P) and exterior region (say, at point Q), the magnetic field is zero.

(a) Open space interior to the toroid:
Let us calculate the magnetic field Bp at point P. We construct an Amperian loop 1 of radius r1 around the point P. For simplicity, we take circular loop so that the length of the loop is its circumference.
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-85
L1 = 2π1
Ampere’s circuital law for the loop 1 is
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-86
Scince, the loop 1 encloses no current, Ienclosed = 0
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-87
This is possible only if the magnetic field at point P vanishes i.e.
\(\vec { B } \)p = 0

(b) Open space exterior to the toroid:
Let us calculate the magnetic field BQ at point Q. We construct an Amperian loop 3 of radius r3 around the point Q. The length of the loop is L3= 2πr3 Ampere’s circuital law for the loop 3 is
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-88
Since, in each turn of the toroid loop, current coming out of the plane of paper is cancelled by the current going into the plane of paper. Thus, Ienclosed = 0
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-89
This is possible only if the magnetic field at point Q vanishes i.e. \(\vec { B } \)Q = 0

(c) Inside the toroid:
Let us calculate the magnetic field BS at point S by constructing an Amperian loop 2 of radius r2 around the point S. The length of the loop is L2= 2πr2 Ampere’s circuital law for the loop 2 is
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-90
Let I be the current passing through the toroid and N be the number of turns of the toroid, then Ienclosed = NI
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-91
The number of turns per unit length is n = \(\frac { n }{{ 2πr }_{2}}\) then the magnetic field at point S is BS = μ0nI

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 8.
Obtain an expression for force on a moving charge in a magnetic field.
Answer:
When an electric charge q is moving with velocity \(\vec { v } \) in the magnetic field \(\vec { B } \), it experiences a force, called magnetic force \(\vec { F } \)m. After careful experiments, Lorentz deduced the force experienced by a moving charge in the magnetic field \(\vec { F } \)m
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-92
\(\vec { F } \)m = q(\(\vec { v } \) x \(\vec { B } \)) ……. (1)
In magnitude, Fm=qvB sin θ ……. (2)
The equations (1) and (2) imply

  1. \(\vec { F } \)m is directly proportional to the magnetic field \(\vec { B } \)
  2. \(\vec { F } \)m is directly proportional to the velocity \(\vec { v } \)
  3. \(\vec { F } \)m is directly proportional to sine of the angle between the velocity and magnetic field
  4. \(\vec { F } \)m is directly proportional to the magnitude of the charge q
  5. The direction of \(\vec { F } \)m is always perpendicular to \(\vec { v } \) and g as \(\vec { F } \)m is ti’e cross product of \(\vec { v } \) and \(\vec { B } \)
  6. The direction of jprn is on negative charge is opposite to the direction of F charge provided other factors are identical.
  7. If velocity v of the charge q is along magnetic field \(\vec { B } \) then, \(\vec { F } \)m is zero.

Magnetism and Magnetic Effects of Electric Current Numerical Problems

Question 1.
The radius of the first orbit of hydrogen atom is 0.5 Å. The electron moves in an orbit with a uniform speed of 2.2 x 106 ms-1. What is the magnetic field produced at the centre of the nucleus due to the motion of this electron? Use µ0 = 4π x 10-7 Hm-1 and electric charge = 1.6 x 10-19C
Solution:
Here, r = 0.5 Å = 0.5 x 10-10m
v = 2.2 x 106 ms-1
Period of revolution of electron
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-93
= 1.12 x 10-3 A
Magnetic field produced at the centre of the nucleus,
Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current-94
B = 14. 07 T

Question 2.
A positive charge of 1.5 µC is moving with a speed of 2 x 10-6 ms-1 along the positive X-axis. A magnetic field, \(\overrightarrow { B } \) = (o.2 \(\hat{j}\) + 0.4 \(\hat{k}\)) tesla acts in space. Find the magnetic force acting on the charge.
Solution:
Here q = 1.5 µC = 1.5 x 10-6C
\(\vec { v } \) = 2 x 10-6 \(\hat{i}\) ms-1 ; \(\overrightarrow { B } \) = (o.2 \(\hat{j}\) + 0.4 \(\hat{k}\)) T
Magnetic force on the positive charge is
\(\vec { F } \) = (\(\vec { v } \) x \(\overrightarrow { B } \))
= 1.5 x 10-6 [2 x 106\(\hat{i}\) x (o.2 \(\hat{j}\) + 0.4 \(\hat{k}\))]
3.0 [o.2 \(\hat{j}\) x \(\hat{j}\) + 0.4 \(\hat{i}\) x \(\hat{k}\)]
= (0.6 \(\hat{k}\) – 1.2 \(\hat{j}\)) N
[∴ i x \(\hat{j}\) = \(\hat{k}\), \(\hat{i}\) x \(\hat{k}\) = –\(\hat{j}\)]

Question 3.
Copper has 8.0 x 1028 electrons per cubic meter carrying a current and lying at right angle to a magnetic field of strength 5 x 10-3 T. experiences a force of 8.0 x 10-2 N. Calculate the drift velocity of free electrons in the wire.
Solution:
n = 8 x 1028 m-3 ; l = lm
A= 8 x 1028 m-2 ; e = 1.6 x 10-9C
Total charge contained in the wire
q = volume of wire x ne = Alne = 8 x 10-6 x 1 x 8 x1028 x 1.6 x 10-19C = 102.4 x 103 C
If vd is the drift speed of electrons, then
F = qvd B sin 90° = qvd B
∴ vd = \(\frac { F }{ qB }\) = \(\frac{8.0 \times 10^{-2}}{102.4 \times 10^{3} \times 5 \times 10^{-3}}\) ms-1
vd = 1.56 x 10-4 ms-1

Samacheer Kalvi 12th Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Question 4.
An electron is moving at 106 ms-1 in a direction parallel to a current of 5A flowing through an infinitely long straight wire, separated by a perpendicular distance of 10cm in air. Calculate the magnitude of the force experienced by the electron.
Solution:
Magnetic field of the straight wire carrying a current of 2 A, at a distance of 10cm or 0. lm from it is
B = \(\frac {{ µ }_{0} I}{ 2πr }\) = \(\frac{4 \pi \times 10^{-7} \times 5}{2 \pi \times 0.1}\) = 10-5T
This field acts perpendicular to the direction of the electron. So magnetic force on the electron is
F = qv B sin 90° = 1.6 x 10-19 x 106 10-5 x 1
F = 1.6 x 10-18 N

Common Errors and Its Rectifications

Common Errors:

  1. Students may not know about arrow mark in the figures, (diagram)
  2. Students wrongly mention the unit of magnetic induction. Eg. 0.5 x 10-5 Tesla. Most of the children makes the mistakes in this area.
  3. Student may confuse the direction of current versus produced magnetic induction.

Rectifications:

  1. Arrow mark is importance to show where the electric and magnetic fields are directed.
  2. The correct unit of magnetic induction of T (or) tesla.
  3. If the current moves in upward the magnetic induction is anti-clock. If the current in downward, the magnetic induction is clockwise.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Students can Download Chemistry Chapter 3 Periodic Classification of Elements Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Chemistry Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Samacheer Kalvi 11th Chemistry Chapter 3 Periodic Classification of Elements Textual Evaluation Solved

Choose The Correct Answer from The Following

Question 1.
What would be the IUPAC name for an element with atomic number 222?
(a) bibibiium
(b) bididium
(c) didibium
(d) bibibium
Answer:
(d) bibibium

Question 2.
The electronic configuration of the elements A and B are 1s2, 2s2, 2p6, 3s2 and 1s2, 2s2, 2p5, respectively. The formula of the ionic compound that can be formed between these elements is ……….
(a) AB
(b) AB2
(c) A2B
(d) none of the above.
Answer:
(a) AB2

Question 3.
The group of elements in which the differentiating electron enters the anti-penultimate shell of atoms are called –
(a) p-block elements
(b) d-block elements
(c) s-block elements
(d) f-block elements
Answer:
(d) f-block elements

Question 4.
In which of the following options the order of arrangement does not agree with the variation of property indicated against it? (NEET 2016 Phase 1)
(a) I < Br < Cl < F (increasing electron gain enthalpy)
(b) Li < Na < K < Rb (increasing metallic radius)
(c) Al3+< Mg2+< Na+ < F(increasing ionic size)
(d)  B < C < O < N (increasing first ionization enthalpy)
Answer:
(a) I < Br < Cl < F (increasing electron gain enthalpy)

Question 5.
Which of the following elements will have the highest electro negativity?
(a) Chlorine
(b) Nitrogen
(c) Cesium
(d) Fluorine
Answer:
(d) Fluorine

Question 6.
Various successive ionization enthalpies (in kJ mol-1) of an element are given below. The element is ………….
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
(a) phosphorus
(b) sodium
(c) aluminium
(d) silicon table
Answer:
(c) aluminium

Question 7.
In the third period, the first ionization potential is of the order …………..
(a) Na > Al > Mg > Si > P
(b) Na < Al < Mg < Si < P
(c) Mg > Na > Si > P > Al
(d) Na< Al < Mg < Si < P
Answer:
(b) Na < Al < Mg < Si < P

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 8.
Identify the wrong statement ……………..
(a) Among st the iso electronic species, smaller the positive charge on cation, smaller is the ionic radius
(b) Among-st iso electric species greater the negative charge on the anion, larger is the ionic radius
(c) Atomic radius of the elements increases as one moves down the first group of the periodic table
(d) Atomic radius of the elements decreases as one moves across from left to right in the 2nd period of the periodic table.
Answer:
(a) Among-st the iso electronic species, smaller the positive charge on cation, smaller is the ionic radius

Question 9.
Which one of the following arrangements represent the correct order of least negative to most negative electron gain enthalpy?
(a) Al< O<C< Ca< F
(b) Al < Ca<O< C< F
(c) C < F < O < Al < Ca
(d) Ca < Al < C < O < F
Answer:
(d) Ca < Al < C < O < F

Question 10.
The correct order of electron gain enthalpy with negative sign of F, Cl, Br and I having atomic number 9, 17, 35 and 53, respectively is ………..
(a) J > Br > Cl >F
(b) F > Cl > Br >I
(c) Cl > F > Br >I
(d) Br > I > Cl > F
Answer:
(c) Cl > F > Br > I

Question 11.
Which one of the following is the least electro negative element?
(a) Bromine
(b) Chlorine
(c) Iodine
(d) Hydrogen
Answer:
(d) Hydrogen.
Solution:
Hydrogen is the least electro negative element. Since electro negativity increases across the period from left to right. Hydrogen is the first element and it has less electro negativity and down the group electro negativity decreases.

Question 12.
The element with positive electron gain enthalpy is ……….
(a) hydrogen
(b) sodium
(c) argon
(d) fluorine
Answer:
(c) argon
Solution:
Argon has completely filled configuration. So addition of the electron is not possible and has positive electron gain enthalpy.

Question 13.
The correct order of decreasing electro negativity values among the elements X, Y, Z and A with atomic numbers 4, 8, 7 and 12 respectively –
(a) Y > Z > X > A
(b) Z > A > Y > X
(c) X > Y > Z > A
(d) X > Y >A >Z
Answer:
(a) Y > Z > X > A

Question 14.
Assertion: Helium has the highest value of ionization energy among all the elements known.
Reason: Helium has the highest value of electron affinity among all the elements known –
(a) Both assertion and reason are true and reason is correct explanation for the assertion
(b) Both assertion and reason are true but the reason is not the correct explanation for the assertion
(c) Assertion is true and the reason is false
(d) Both assertion and the reason are false
Answer:
(c) Assertion is true and the reason is false

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 15.
The electronic configuration of the atom having maximum difference in first and second ionization energies is ……….
(a) 1s2, 2s2, 2p6, 3s1
(b) 1s2, 2s2, 2p6, 3s2
(c) 1s2, 2s2, 2p6, 3s2, 3s2, 3p6, 4s1
(d) 1s2, 2s2, 2p6, 3s2, 3p1
Answer:
(a) 1s2, 2s2, 2p6, 3s1

Question 16.
Which of the following is second most electro negative element?
(a) Chlorine
(b) Fluorine
(c) Oxygen
(d) Sulphur
Answer:
(a) Chlorine

Question 17.
IE1  and IE2 of Mg are 179 and 348 k cal mol-1 respectively. The energy required for the reaction
Mg → Mg2+ + 2e is ……..
(a) +169 kcal mol-1
(b) -169 kcal mol-1
(c) +527 kcal mol-1
(d) -527 kcal mol-1
Answer:
(c) +527 kcal mol-1

Question 18.
In a given shell the order of screening effect is …………..
(a) s > p > d > f
(b) s > p > f > d
(c) f > d > p > s
(d) f > p > s > d
Answer:
(a) s > p > d > f

Question 19.
Which of the following orders of ionic radii is correct?
(a) H > H+ > H
(b) Na+ > F“ > O
(c) F > O2- > Na+
(d) None of these
Answer:
(d) None of these

Question 20.
The first ionization potential of Na, Mg and Si are 496, 737 and 786 kJ mol-1 respectively. The ionization potential of Al will be closer to
(a) 760 kJ mol-1
(b) 575 kJ mol-1
(c) 801 kJ mol-1
(d) 419 kJ mol-1
Answer:
(b) 575 kJ mol-1

Question 21.
Which one of the following is true about metallic character when we move from left to right in a period and top to bottom in a group?
(a) Decreases in a period and increases along the group
(b) Increases in a period and decreases in a group
(c) Increases both in the period and the group
(d) Decreases both in the period and in the group
Answer:
(a) Decreases in a period and increases along the group

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 22.
How does electron affinity change when we move from left to right in a period in the periodic table?
(a) Generally increases
(b) Generally decreases
(c) Remains unchanged
(d) First increases and then decreases
Answer:
(a) Generally increases.

Question 23.
Which of the following pairs of elements exhibit diagonal relationship?
(a) Be and Mg
(b) Li and Mg
(c) Be and B
(d) Be and Al
Answer:
(d) Be and Al

 II. Write a brief answer to the following questions.

Question 24.
Define modern periodic law.
Answer:
The physical and chemical properties of the elements are periodic functions of their atomic numbers.

Question 25.
What are isoelectronic ions? Give examples.
Answer:
There are some ions of different elements having the same number of electrons are called isoelectronic ions.
Example:
Na+, Mg2+, Al3+, F , O2- and N3-

Question 26.
What is effective nuclear charge?
Answer:
The net charge experienced by valence electrons in the outermost shell is called the effective nuclear charge. It is approximated by the equation Zeff = Z – S, where Z is the atomic number and S is the screening constant which can be calculated using Slater’s rules.

Question 27.
Is the definition given below for ionization enthalpy is correct?
“Ionization enthalpy is defined as the energy required to remove the most loosely bound electron from the valence shell of an atom”
Answer:
No. It is not correct. The accurate and absolute definition is as follows:
Ionization energy is defined as the minimum amount of energy required to remove the most loosely bound electron from the valence shell of the isolated neutral gaseous atom in its ground state.

Question 28.
Magnesium loses electrons successively to form Mg+, Mg2+ and Mg3+ ions. Which step will have the highest ionization energy and why?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

The third step will have the highest ionization energy. I.E3>I.E2>I.E1
Because from a neutral gaseous atom, the electron removal is easy and less amount of energy is required. But from a di positive cation, there will be more number of protons than the electrons and there is more forces of attraction between the nucleus and electron. So the removal of electron in a di positive cation, becomes highly difficult and more energy is required.

Question 29.
Define electro negativity.
Answer:
Electro negativity is the relative tendency of an element present in a covalently bonded molecule, to attract the shared pair of electrons towards itself.

Question 30.
How would you explain the fact that the second ionization potential is always higher than first ionization potential?
Answer:
The minimum amount of energy required to remove a unipositive cation is called second ionization energy. It is represented by the following equation,
M+(g) + IE2 – M2+(g) + 1e,
The total number of electrons is less in the cation than the neutral atom while the nuclear charge remains the same. Therefore, the effective nuclear charge of the cation is higher than the corresponding neutral atom. Thus, the successive ionization energies, always increase in the following order I.E1 < I.E2. Hence, the second ionization potential is always higher than the first ionization potential.

Question 31.
Energy of an electron in the ground state of the hydrogen atom is -2.18 x 10-18 J. Calculate the ionization enthalpy of atomic hydrogen in terms of kJ mol-1.
Answer:
Energy of an electron in the ground state of the hydrogen atom = -2.18 x 10-18 J
H → H+ + e
Energy required to ionize 1 mole of hydrogen atoms, we multiply by the Avogadro constant.
E = 2.18 x 10-18 x 6.023 x 1023
= 13.123 x 105 J mol-1
I.E = +1312 K J mol-1

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 32.
The electronic configuration of an atom is one of the important factor which affects the value of ionization potential and electron gain enthalpy. Explain.
Answer:
The plot of atomic number vs ionization energy shows that there are two deviations in the trends of ionization energy. It is expected that boron has higher ionization energy than beryllium since it has a higher nuclear charge. However, the actual ionization energies of beryllium and boron are 899 and 800 kJ mol-1 respectively contrary to the expectation. It is due to the fact that beryllium with completely filled 2s orbital, is more stable than partially filled valence shell electronic configuration of boron. (2s2 2p1).

Similarly, nitrogen with a 1s2 2s2 2p3 electronic configuration has higher ionization energy (1402 kJ mol-1) than oxygen (1314 kJ mol-1). Since the half-filled electronic configuration is more stable, it requires higher energy to remove an electron from 2p orbital of nitrogen. Whereas the removal of one 2p electron from oxygen leads to a stable half-filled configuration. This makes it comparatively easier to remove 2p electron from oxygen.

As we move from alkali metals to halogens in a period, generally electron affinity increases. This is due to an increase in the nuclear charge and a decrease in the size of the atoms. However, in the case of elements such as beryllium (1s2 2s2), nitrogen (1s2 2s2 2p3 ) the addition of extra electrons will disturb their stable electronic configuration and they have almost zero electron affinity.

Noble gases have a stable ns2 np6 configuration, and the addition of further electrons is unfavourable and requires energy. Halogens having the general electronic configuration of ns2 np5 readily accept an electron to get the stable noble gas electronic configuration {ns2 np6) and therefore, in each period the halogen has a high electron affinity.

Question 33.
In what period and group will an element with Z = 118 will be present?
Answer:
The element with atomic number Z = 118 is present in 7th period and 18th group.

Question 34.
Justify that the fifth period of the periodic table should have 18 elements on the basis of quantum numbers.
Answer:
In the fifth period, the filling of valence electrons starts with 5s orbital followed by 4d and 6p orbitals. The filling of 4d orbitals starts with Yttribium and ends with cadmium. There are 10 elements present in the second transition series. The period starts with Rubidium (Rb – 5s1 ) and ends with Xenon (Xe – 5s2 5p6).

Question 35.
Elements a, b, c and d have the following electronic configurations:
a : 1s2, 2s2, 2p6
b : 1s2, 2s2, 2p6, 3s2, 3p1
c : 1s2, 2s2, 2p6 3s2,3p6
d : 1s2, 2s2, 2p1
Which elements among these will belong to the same group of periodic table?
Answer:

  1. Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
  2. In the above elements, Ne and Ar belong to same group (Noble gases – 18th group).
  3. Al and B belong to the same group (13th group).

Question 36.
Give the general electronic configuration of lanthanides and actinides?
Answer:
The general electronic configuration of lanthanides is 4f1-14 5d0-1 6s2 and for Actinides is 5f1-14 6d0-1 7s2.

Question 37.
Why halogens act as oxidizing agents?
Answer:
Halogens act as oxidizing agents. Their electronic configuration is ns2 np5. So all the halogens are ready to gain one electron to attain the nearest inert gas configuration. An oxidizing agent is the one which is ready to gain an electron. So all the halogens act as oxidizing agents. Also halogens are highly electro negative with low dissociation energy and high negative electron gain enthalpies. Therefore, the halogens have a high tendency to gain an electron. Hence they act as oxidizing agents.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 38.
Mention any two anomalous properties of second period elements.
Answer:
The anomalous properties of second-period elements are as follows. The ionization energy of Boron is greater than that of Beryllium due to the fact that Be has completely filled 2s orbital which is more stable than the partially filled valence shell electronic configuration of Boron. Similarly, nitrogen with a half-filled electronic configuration has higher ionization energy than oxygen because the half-filled electronic configuration is more stable.

Question 39.
Explain the Pauling’s method for the determination of ionic radius.
Answer:
1. Ionic radius is defined as the distance from the center of the nucleus of the ion up-to which it exerts its influence on the electron cloud of the ion.
2. Ionic radius of uni-univalent crystal can be calculated from the inter-ionic distance between the nuclei of the cation and anion.
3. Pauling assumed that ions present in a crystal lattice are perfect spheres and they are in contact with each other, therefore
d = rC+ + rA ………(1)
Where, d = distance between the center of the nucleus of cation C+ and the anion A-
rC+ = radius of cation
rA = radius of anion.
4. Pauling assumed that the radius of the ion having noble gas configuration (Na+ and F having 1s2, 252, 2p6 configuration) is inversely proportional to the effective nuclear charge felt at the periphery of the ion.
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
Where Zeff is the effective nuclear charge
Zeff = Z – S
5. Dividing the equation (2) by (3)
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
On solving equation (1) and (4), the values of rC+ and rA can be obtained.

Question 40.
Explain the periodic trend of ionization potential.
Answer:
Variation along the period:
The ionization energy usually increases along a period with few exceptions. When we move from left to right along a period, the valence electrons are added to the same shell, at the same time protons arc added to the nucleus. This successive increase of nuclear charge increases the electrostatic attractive force on the valence electron and more energy is required to remove the valence electron resulting in high ionization energy.

Consider the variation in ionization energy of second-period elements. The plot of atomic number vs ionization energy is given below. The plot of atomic number vs ionization energy shows that there are two deviations in the trends of ionization energy. It is expected that boron has higher ionization energy than beryllium since it has a higher nuclear charge.

However, the actual ionization energies of beryllium and boron are 899 and 800 kJ mol-1 respectively contrary to the expectation. It is due to the fact that beryllium with completely filled 2s orbital, is more stable than partially filled valence shell electronic configuration of boron. (2s2, 2p1).

Similarly, nitrogen with 1s2, 2s2, 2p3 electronic configuration has higher ionization energy (1402 kJ mol-1) than oxygen (1314 kJ mol-1). Since the half-filled electronic configuration is more stable, it requires higher energy to remove an electron from the 2p orbital of nitrogen. Whereas the removal of one 2p electron from oxygen leads to a stable half-filled configuration. This makes it comparatively easier to remove 2p electron from oxygen.

Variation along with the group:
The ionization energy decreases down a group. As we move down a group, the valence electron occupies new shells, the distance between the nucleus and the valence electron increases. So, the nuclear forces of attraction on valence electron decreases, and hence, ionization energy also decreases down a group.

As we move down a group, the number of inner-shell electrons increases which in turn increases the repulsive force exerted by them on the valence electrons, i.e., the increased shielding effect caused by the inner electrons decreases the attractive force acting on the valence electron by the nucleus. Therefore, the ionization energy decreases.

Question 41.
Explain the diagonal relationship.
Answer:

  • On moving diagonally across the periodic table, the second and the third period elements show certain similarities.
  • Even though the similarity is not same as we see in a group, it is quite pronounced in the following pair of elements.
  • Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
  • The similarity in properties existing between the diagonally placed elements is called “diagonal relationship”.

Question 42.
Why the first ionization enthalpy of sodium is lower than that of magnesium while its second ionization enthalpy is higher than that of magnesium?
Answer:
The first ionization enthalpy of sodium is lower than that of magnesium.
Na(1s2, 2s2, 2p6, 3s1)+ IE1 → Na+ (1s2, 2s2, 2p6) + e

Mg(1s2, 2s2, 2p6, 3s2) + IE1 → Mg+(1s2, 2s2, 2p6, 3s1) + e

Magnesium has completely filled 3s orbital (1s2, 2s2, 2p6, 3s2) , is more stable than partially filled valence shell electronic configuration of sodium (1s2, 2s2, 2p6, 3s1).

Na+ (1s2, 2s2, 2p6) + IE2 → Na2+ (1s2, 2s2, 2p5) + e
Mg+ (1s2, 2s2, 2p6, 3s1) + IE2 → Mg2+ (1s2, 2s2, 2p6, 3s2) + e

Na+ has completely filled 2p orbital (1s2, 2s2, 2p6), is more stable than partially filled valence shell electronic configuration of Mg+ (1s2, 2s2, 2p6). Hence, the second ionization energy of sodium is higher than that of magnesium.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 43.
By using Pauling’s method calculate the ionic radii of K+ and Cl ions in the potassium chloride crystal. Given that \(\mathrm{d}_{\mathrm{K}}+_{-} \mathrm{cl}^{-}\) = 3.14 Å
Answer:
Given
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
We know that,
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
(Zeff)Cl = Z – S
= 17 – [(0.35 x 7) + (0.85 x 8) + (1 x 2)]
= 17 – 11.25 = 5.75
(Zeff)K+ = Z – S
= 19 – [(0.35 x 7) + (0.85 x 8) + (1 x 2)]
= 19 – 11.25 = 7.75
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
r(K+) = 0.74 r(Cl)
Substitute the value of r(K+) in equation (1)
0.74 r(Cl) + r(Cl) = 3.14 Å
1.74 r(Cl) = \(\frac {3.14 Å}{1.74}\) = 1.81 Å.

Question 44.
Explain the following, give appropriate reasons.

  1. Ionization potential of N is greater than that of O.
  2. First ionization potential of C-atom is greater than that of B-atom, where as the reverse is true for second ionization potential.
  3. The electron affinity values of Be, Mg and noble gases are zero and those of N (0.02 eV) and P (0.80 eV) are very low
  4. The formation of F (g) from F(g) is exothermic while that of O2-(g) from O (g) is endothermic.

Answer:
1. N (Z = 7) 1s2 2s2 2px1 12py1 2pz1. It has exactly half-filled electronic configuration and it is more stable. Due to stability, ionization energy of nitrogen is high.
O (Z = 8) 1s2 2s2 2px1 2py1 2pz1. It has incomplete electronic configuration and it requires less ionization energy.
I.E1 N > I.E1O

2. C (Z = 6) 1s2 2s2 2px1 2py1. The electron removal from p orbital is very difficult. So carbon has highest first ionization potential. B (Z = 5) 1s2 2s2 2p1. In boron nuclear charge is less than that of carbon, so boron has lowest first ionization potential.
I.E1 C > I.E1 B
But it is reverse in the case of second ionization energy. Because in case of B+ the electronic configuration is 1s2 2s2, which is completely filled and it has high ionization energy. But in C+ the electronic configuration is 1s2 2s2 2p1, one electron removal is easy so it has low ionization energy.
I.E2 B > I.E2 C

3. Be (Z = 4) 1s2 2s2
Mg (Z = 12) 1s2 2s2 2p6 3s2
Noble gases has the electronic configuration of ns2 np6. All these are completely filled and are more stable. For all these elements Be, Mg and noble gases, addition of electron is unfavorable and so they have zero electron affinity.

Nitrogen (Z = 7) 1s2 2s2 2px1 2py1 2pz1. It has half filled electronic configuration. So addition of electron is unfavorable and it has very low electron affinity value of 0.02 eV. Phosphorus (Z = 15) 1s2 2s2 2p6 3s2 3px1 3py1 3pz1. It also has half filled electronic configuration. Due to the symmetry and more stability, it has very low electron affinity value of 0.80 eV.

4. F(g) + e → F(g) exothermic
F (Z = 9) 1s2 2s2 2p5. It is ready to gain one electron to attain the nearest inert gas configuration. By gaining one electron, energy is released, so it is an exothermic reaction.
O(g) + 2e → O2-(g) endothermic
O (Z = 8) 1s2 2s2 2px1 2py1 2pz1. It is the small atom with high electron density. The first electron affinity is negative because energy is released in the process of adding one electron to the neutral oxygen atom. Second electron affinity is always endothermic (positive) because the electron is added to an ion which is already negative, therefore it must overcome the repulsion.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 45.
What is screening effect? Briefly give the basis for Pauling’s scale of electro negativity. Screening effect:
Answer:
In addition to the electrostatic forces of attraction between the nucleus and the electrons, there exists repulsive forces among the electrons. The repulsive force between the inner shell electrons and the valence electrons leads to a decrease in the electrostatic attractive forces acting on the valence electrons by the nucleus. Thus, the inner shell electrons act as a shield between the nucleus and the valence electrons. This effect is called the shielding effect.

Electronegativity is defined as the relative tendency of an element present in a covalently bonded molecule, to attract the shared pair of electrons towards itself. Electronegativity is not a measurable quantity. In the Pauling scale, an arbitrary values of electronegativities for hydrogen and fluorine are assigned as 2.2 and 4.0 respectively. Based on this, the electronegativity values for other elements can be calculated using the following expression,
A – χB) = 0.182 \(\sqrt{E_{A B}}\) – (EAA × EBB)1/2
where EAB, EAA, and EBB are the bond dissociation energies of AB, A2, and B2molecules respectively.

The electronegativity of any given element is not a constant and its value depends on the element to which it is covalently bound. The electronegativity values play an important role in predicting the nature of the bond.

Question 46.
State the trends in the variation of electro negativity in period and group.
Answer:
Variation of electron negativity in a period:
The electro negativity increases across a period from left to right. Since the atomic radius decreases in a period, the attraction between the valence electron and the nucleus increases. Hence the tendency to attract shared pair of electrons increases. Therefore, electro negativity increases in a period.
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Variation of electro negativity in a group:
The electro negativity decreases down a group. As we move down a group, the atomic radius increases and the nuclear attractive force on the valence electron decreases. Hence electro negativity decreases in a group.
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Samacheer Kalvi 11th Chemistry Periodic Classification of Elements  In-Text Question – Evaluate your self

Question 1.
What is the basic difference in approach between Mendeleev’s periodic table and modern periodic table?
Answer:
The main basic difference between Mendeleev’s periodic table and modem periodic table is that first one is constructed on the basis of atomic weight and the later is constructed on the basis of atomic number.

Question 2.
The element with atomic number 120 has not been discovered so far. What would be the IUPAC name and the symbol for this element? Predict the possible electronic configuration of this element.
Answer:
Atomic number : 120
IUPAC temporary symbol : Unbinilium
IUPAC temporary symbol : Ubn
Possible electronic configuration : [Og] 8s2

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 3.
Predict the position of the element in periodic table satisfying the electronic configuration (n – 1 )d2 ns2 where n = 5?
Answer:
Electronic Configuration : (n – 1 )d2 ns2
for n = 5, the electronic configuration is,
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d2 5s2
Atomic number : 40
4th group 5th period (d block element) = Zirconium

Question 4.
Using Slater’s rule calculate the effective nuclear charge on a 3p electron in aluminium and chlorine. Explain how these results relate to the atomic radii of the two atoms.
Answer:
Electronic Configuration of Aluminium
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
Effective nuclear charge = Z – S = 13 – 9.5
(Zeff)Al = 3.5
Electronic Configuration of chlorine
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
Effective nuclear charge = Z- S = 17 – 10.9
(Zeff)Cl = 6.1
(Zeff)Cl > (Zeff)Cl and hence rCl< rAl

Question 5.
A student reported the ionic radii of iso electronic species X3+ , Y2+ and Z as 136 pm, 64 pm and 49 pm respectively. Is that order correct? Comment.
Answer:
X3+, Y2+, Z are iso electronic.
∴ Effective nuclear charge is in the order
(Zeff)Cl < (Zeff)YY2+ < (Zeff)X3+ and hcnce, ionic radii should be in the order rZ > rY2+ > rX3+
∴ The correct values are:
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 6.
The first ionisation energy (IE1) and second ionisation energy (IE2) of elements X, Y and Z are given below.
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
Which one of the above elements is the most reactive metal, the least reactive metal and a noble gas?
Answer:
Noble gases:
Ionisation energy ranging from 2372 KJmol-1 to 1037 kJ mol-1. For element X, the IE1 value is in the range of noble gas, moreover for this element both IE1 and IE2 are higher and hence X is the noble gas. For Y, the first ionisation energy is low and second ionisation energy is very high and hence Y is most reactive metal.
For Z, both IE1 and IE2 are higher and hence it is least reactive.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 7.
The electron gain enthalpy of chlorine is 348 kJ mol-1. How much energy in kJ is released when 17.5 g of chlorine is completely converted into Cl ions in the gaseous state?
Cl(g)+ e → Cl(g)
∆H = 348 kJ mol-1
For one mole (35.5g) 348 kJ is released.
∴ For 17.5g chlorine, Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements  energy leased.
∴ The amount of energy released = \(\frac {348}{2}\) = 174 kJ.

Samacheer Kalvi 11th Chemistry Periodic Classification of Elements  Additional Questions

Question 1.
The chemical symbol of carbon and cobalt are
(a) Ca and CO
(b) Ca and Cl
(c) C and CO
(d) Cr and Cb
Answer:
(c) C and CO

Question 2.
Consider the following statements.
(i) The chemical symbol of nickel is N.
(ii) An element is a material made up of different kind of atoms.
(iii) The physical state of bromine is liquid.
Which of the above statement is/are not correct?
(a) (i) and (iiii)
(b) (iii) only
(c) (ii) and (iii)
(d) (i) and (ii)
Answer:
(d) (i) and (ii)

Question 3.
Match the list-I and List-II using the correct code given below the list.
List – I
A. Jewels
B. Bolts and cot
C. Table salt
D. Utensils

List – II
1. Sodium chloride
2. Copper
3. Gold
4. Iron
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 4.
The law of triads is not obeyed by
(a) Ca, Sr, Ba
(b) Cl, Br, I
(c) Li, Na, K
(d) Be, B, C
Answer:
(d) Be, B, C

Question 5.
The law of triads is obeyed by
(a) Fe, CO, Ni
(b) C, N, O
(c) He, Ne, Ar
(d) Al, Si, P
Answer:
(a) Fe, CO, Ni

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 6.
Match the List-I and List-II using the code given below the list.
List-I
A. Law of triads
B. Law of octaves
C. First periodic law
D. Modem periodic law

List-II
1. Chancourtois
2. Henry Moseley
3. Newland
4. Johann Dobereiner
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 7.
Consider the following statements.
(i) In Chancourtois classification, elements differed from each other in atomic weight by 16 or multiples of 16 fell very nearly on the same vertical line.
(ii) Mendeleev’s periodic law is based on atomic weight.
(iii) Mendeleev listed the 117 elements known at that time and are arranged in the order of atomic numbers.
Which of the following statement is/are not correct?
(a) (i) only
(b) (ii) and (iii)
(c) (iii) only
(d) (i),(ii), (iii)
Answer:
(c) (iii) only

Question 8.
Which of the following elements were unknown at that time of Mendeleev?
(a) Na, Mg
(b) Fe, CO
(c) K, Cu
(d) Ga, Ge
Answer:
(d) Ga, Ge

Question 9.
Consider the following statements.
(i) Position of hydrogen could not be made clear.
(ii) Isotopes find correct place in Mendeleev’s periodic table.
(iii) Mendeleev’s periodic table could not explain the variable valencies of elements.
Which of the above statement is/are not correct?
(a) (i) only
(b) (i) and (iii)
(c) (ii) only
(d) (i), (ii), (iii)
Answer:
(c) (ii) only

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 10.
According to modem periodic law, the physical and chemical properties of the elements are periodic functions of their
(a) atomic volume
(b) atomic numbers
(c) atomic weights
(d) valency
Answer:
(B) atomic numbers

Question 11.
Which period contain 32 elements?
(a) Period 1
(b) Period 4
(c) Period 5
(d) Period 6
Answer:
(d) Period 6

Question 12.
There are horizontal rows of the periodic table known as
(a) groups
(b) periods
(c) families
(d) chalcogens
Answer:
(b) periods

Question 13.
The shortest period contains elements.
(a) H, He
(b) Li, Be
(c) B, C
Answer:
(a) H, He

Question 14.
The longest form of periodic table was constructed by
(a) Dmitri Mendeleev
(b) Henry Moseley
(c) Lothar Meyer
(d) New lands
Answer:
(b) Henry Moseley

Question 15.
Match the List-I and List-II using the correct code given below the list.
List – I
Z = 100
Z = 101
Z = 102
Z = 103

List – II
1. Mendelevium
2. Lawrencium
3. Fermium
4. Nobelium
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 16.
Which one of the following is the first transition series?
(a) Sc
(b) Zn
(c) Ti
(d) Cu
Answer:
(a) Sc

Question 17.
Which period mostly includes man-made radioactive elements?
(a) 4th period
(b) 7th period
(c) 6th period
(d) 31 period
Answer:
(b) 7th period

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 18.
Which one of the following is called halogen family?
(a) Group 17
(b) Group 16
(c) Group 1
(d) Group 2
Answer:
(a) Group 17

Question 19.
Group 16 constitutes family.
(a) halogen
(b) Nobel gas
(c) chalcogen
(d) alkali metals
Answer:
(c) chalcogen

Question 20.
Consider the following statements.
(i) The valency of the elements increases from left to right in a period.
(ii) Valency decreases from 7 to I with respect to oxygen.
(iii) The metallic character of the elements decreases across a period.
Which of the above statement is/are not correct?
(a) (i) only
(b) (ii) only
(c) (i) and (iii)
(d) (i) (ii) and (iii)
Answer:
(b) (ii) only

Question 21
Match the list – I and list -II using the correct code given below the list.
List – I
A. Li
B. Na
C. K
D. Cs

List – II
1. 2, 8, 8, 1
2. 2, 1
3. 2, 8, 18, 18, 8, 1
4. 2, 8, 1
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 22.
What will be the change in valency down the group in the periodic table?
(a) increases
(b) decreases
(c) remains same
(d) zero
Answer:
(c) remains same

Question 23.
Which one of the following is a metalloid?
(a) N
(b) P
(c) Bi
(d) Sb
Answer:
(d) Sb

Question 24.
Which one of the following is a metal?
(a) N
(b) Br
(c) Bi
(d) As
Answer:
(c) Bi

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 25.
Match the list – I and list – II using the correct code given below the list.
List – I
A. Alkali metal
B. Alkaline earth metals
C. d-block elements
D. p-block elements

List – II
1. ns2 np1-6
2. ns1
3. ns2
4. (n – 1)d1-10 ns0-2
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 26.
Consider the following statements.
(i) Oxidation character increases from left to right in a period.
(ii) Reducing character increases from left to right in a period.
(iii) Metallic character increases from left to right in a period.
Which of the above statement is/are not correct?
(a) (i) only
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (i), (ii), (iii)
Answer:
(c) (ii) and (iii)

Question 27.
The general electronic configuration of d-block elements is
(a) ns2 nd1-10 10
(b) (n-1)d1-10 ns0-2
(c) (n-2)d1-10 (n – 1)0-2
(d) ns2nd5
Answer:
(b) (n-1)d1-10 ns0-2

Question 28.
Consider the flowing statements.
(i) d-block elements show variable oxidate states.
(ii) Mostly d-block elements form colourless compounds.
(iii) Mostly d-block elements are diamagnetic due to paired electrons.
Which of the above statement is/are not correct?
(a) (i) only
(b) (ii) only
(c) (i) and (ii)
(d) (ii) and (iii)
Answer:
(d) (ii) and (iii)

Question 29.
All the s – block and p-block elements excluding l8 group are called elements.
(a) representative
(b) transition
(c) inner – transition
(d) trans uranium
Answer:
(a) representative

Question 30.
Which of the following is the correct electronic configuration of noble gases?
(a) ns2 np6 nd10
(b) ns2 np5
(c) ns2 np6
(d) ns2 np3
Answer:
(c) ns2 np6

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 31.
Group numbers 13 to 12 in the periodic table are called …………..
(a) inner transition elements
(b) Representative elements
(c) synthetic elements
(d) transition elements
Answer:
(d) transition elements

Question 32.
Which one of the following is in solid state at room temperature?
(a) Bromine
(b) Mercury
(c) Bismuth
(d) Gallium
Answer:
(c) Bismuth

Question 33.
Which of the following is not a metalloid (or) semi-metal?
(a) Silicon
(b) Arsenic
(c) Germanium
(d) Sodium
Answer:
(d) Sodium

Question 34.
Which of the following metal is not in liquid state?
(a) Gallium
(b) Aluminium
(c) Mercury
(d) Calcium
Answer:
(b) Aluminium

Question 35.
Which of the following is not a periodic property?
(a) Atomic radius
(b) Ionization enthalpy
(c) Electron affinity
(d) Oxidation number
Answer:
(d)Oxidation number

Question 36.
Which of the following property increases as we go down the group in the periodic property?
(a) ionization energy
(b) Electro negativity
(c) Atomic radius
(d) Electron affinity
Answer:
(c) Atomic radius

Question 37.
The metallic radius of copper is ………………
(a) 0.99 Å
(b) 1.28 Å
(c) 1.98 Å
(d) 2.56 Å
Answer:
(5) 1.28 Å

Question 38.
Consider the following statements………………
(i) Atomic radius of elements increases with increase in atomic number as we go down the group.
(ii) Atomic radius of elements increases with increase in atomic number as we go across the period.
(iii) Atomic radius of elements decreases as we go from left to right in a period.
Which of the above statement is/are not correct?
(a) (i) only
(b) (i) and (ii)
(c) (ii) only
(d) (i) and (iii)
Answer:
(c) (ii) only

Question 39.
Which one of the following is not an iso electronic ion?
(a) Na+
(b) Mg2+
(c) Cl
(d) O2-
Answer:
(c) Cl

Question 40.
Which one of the following is not an isoelectronic ion?
(a) Al3+
(b) N3-
(c) Mg2+
(d) K+
Answer:
(d) K+

Question 41.
Which of the following possess almost same properties due to lanthanide contraction?
(a) Zr, HF
(b) Na, K
(c) Zn, Cd
(d) Ag. Au
Answer:
(a) Zr, HF

Question 42.
Consider the following statements.
(i) Ionization is always an exothermic process.
(ii) Ionization energies always increase in the order I.E1> IE2>I.E3.
(iii) Ionization energy measurements are carried out with atoms in the solid-state.
Which of the above statement is/are not correct?
(a) (i) only
(b) (i) and (ii)
(c) (iii) only
(d) (i), (ii) and (iii)
Answer:
(d) (i), (ii) and (iii)

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 43.
Statement-I : Ionization enthalpy of Be is greater than that of 13.
Statement-II : The nuclear charge of B is greater than that of Be.
(a) Statement-I and II are correct and statement-II is the correct explanation of statement-I.
(b) Statement-I and II are correct but statement-II is not the correct explanation of statement-I.
(c) Statement-I is correct but statement-II is wrong.
(d) Statement-I is wrong but statement-II is correct.
Answer:
(b) Statement-I and II are correct but statement-II is not the correct explanation of statement-I.

Question 44.
Statement-I: Ionization enthalpy of nitrogen is greater than that of oxygen.
Statement-II: Nitrogen has exactly a half-filled electronic configuration which is more stable than electronic configuration of oxygen.
(a) Statement-I is wrong but statement-II is correct.
(b) Statement-I is correct but statement-II is wrong.
(c) Statement-I and II are correct and statement-TI is the correct explanation of statement-I.
(d) Statement-I and II are correct but statement-II is not the correct explanation of statement-I.
Answer:
(c) Statement-I and II are correct and statement-II is the correct explanation of statement-I.

Question 45.
Which of the following does not have zero electron gain enthalpy?
(a) Be
(b) Cl
(c) Mg
(d) N
Answer:
(b) Cl

Question 46.
Which of the following have zero electron gain enthalpy?
(a) Halogens
(b) Noble gases
(c) Chalcogens
(d) Gold
Answer:
(b) Noble gases

Question 47.
Which of the following have the highest value of electronegativity?
(a) Halogens
(b) Alkali metals
(c) Alkaline earth metals
(d) Transition metals
Answer:
(a) Halogens

Question 48.
Among all the elements which one has the highest value of electronegativity?
(a) Chlorine
(b) Bromine
(c) Fluorine
(d) Iodine
Answer:
(c) Fluorine

Question 49.
Among the alkali metals which one form compounds with more covalent character?
(a) Sodium
(b) Potassium
(c) Rubidium
(d) Lithium
Answer:
(d) Lithium

Question 50.
Which of the following pair is not diagonally related?
(a) Li, Mg
(b) Li, Na
(c) Be, Al
(d) B, Si
Answer:
(b) Li, Na

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 51.
In the modern periodic table, the period indicates the value of
(a) atomic number
(b) mass number
(c) principal quantum number
(d) azimuthal quantum number
Answer:
(c) principal quantum number
Hint:
In the modern periodic table, each period begins with the filling of a new shell. Therefore, the period indicates the value of principal quantum number. Thus, option (c) is correct.

Question 52.
Which of the following statements related to the modern periodic table is incorrect?
(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitais in a p-subshell.
(b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitais in a d-subshell.
(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
(d) The block indicates value of azimuthal quantum number (1) for the last subshell that received electrons in building up the electronic configuration.
Answer:
(b)The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitais in a d-subshell.

Question 53.
The size of isoelectronic species- F, Ne and Na+ is affected by
(a) nuclear charge (Z)
(b) valence principal quantum number (n)
(c) electron-electron interaction in the outer orbitais
(d) none of the factors because their size is the same
Answer:
(a) nuclear charge (Z).

Question 54.
Which of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electrons from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitais bearing lower n value is easier than from orbital having high n value.
Answer:
(d)Removal of electron from orbitais bearing lower n value Is easier than from orbital having high n value.

Question 55.
Considering the elements B, Al, Mg and K, the correct order of their metallic character is:
(a) B > Al >Mg > K
(b) Al > Mg > B > K
(c) Mg > Al > K > B
(d) K > Mg > Al > B
Answer:
(d) K > Mg > Al> B
Hint:
In a period, metallic character decreases as we move from left to right. Therefore, metallic character of I< Mg and Al decreases in the order: K> Mg > Al. However, within a group, the metallic character, increases from top to bottom. Thus, Al is more metallic than B.Therefore, the correct sequence of decreasing metallic character is K> Mg >Al > B, i.e,
option (d) is correct.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 56.
Considering the elements B, C, N, F and Si, the correct order of their non-metallic character is
(a) B>C>Si>N>F
(b) Si>C>B>N>F
(c) F>N>C>B>Si
(d) F>N>C>Si>B
Answer:
(c) F>N>C>B> Si
Hint:
In a period, the non-metallic character increases from left to right. Thus, among B, C, N and F, non-metallic character decreases in the order: F > N > C> B. However, within a group, non-metallic character decreases from top to bottom. Thus, C is more non-metallic than Si. Therefore, the correct sequence of decreasing non-metallic character is: F> N > C > B > Si, i.e., option (c) is correct.

Question 57.
Considering the elements F, Cl, O, and N, the correct order of their chemical reactivity in terms of oxidizing property is
(a) F>Cl>O>N
(b) F>O>Cl>N
(c) Cl>F>O>N
(d) O>F>N>Cl
Answer:
(b) F>O>Cl>N.
Hint:
Within a period, the oxidizing character increases from left to right. Therefore, among F, O and N,oxidizing power decreases in the order: F> O> N. However, within a group, oxidizing power decreases from top to bottom. Thus, F is a stronger oxidizing agent than Cl. Further because O is more electronegative than Cl, therefore, O is a stronger oxidizing agent than Cl. Thus, overall decreasing order of oxidizing power is: F > O > Cl > N, i.e., option (b) is correct.

Question 58.
The highest ionization energy is exhibited by ………………
(a) halogens
(b) alkaline earth metals
(c) transition metals
(d) noble gases
Answer:
(b) alkaline earth metals

Question 59.
Which of the following is arranged in order of increasing radius?
(a) K+(aq) < Na+(aq) <Li+(aq)
(b) K+(aq)> Na+(aq)> Zn2+(aq)
(c) K+(aq)> Li+(aq) > Na+(aq)
(d) Li+(aq)< Na+(aq) < K+(aq)
Answer:
(d) Li+(aq)< Na+(aq) < K+(aq)

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 60.
Among the following elements, which has the least electron affinity?
(a) Phosphorous
(b) Oxygen
(c) Sulphur
(d) Nitrogen
Answer:
(d) Nitrogen

Question 61.
Which one of the following is isoelectronic with Ne?
(a) N3-
(b) Mg2+
(c) Al3+
(d) All the above
Answer:
(d) All the above

Question 62.
Which clement has smallest size?
(a) B
(b) N
(c) Al
(d) P
Answer:
(b) N

Question 63.
In halogens, which of the following decreases from iodine to fluorine?
(a) Bond length
(b) Electronegativity
(c) ionization energy
(d) Oxidizing power
Answer:
(a) Bond length

Question 64.
What is the electronic configuration of the elements of group 14?
(a) ns2 np4
(b) ns2 np6
(c) ns2 np2
(d) ns2
Answer:
(c) ns2 np2

Samacheer Kalvi 11th Chemistry Periodic Classification of Elements  2-Mark Questions

Write brief answer to the following questions:

Question 1.
State Johann Dobereiner’s law of triads.
Answer:
Johann Dobereiner noted that elements with similar properties occur in groups of three which he called triads. It was seen that invariably, the atomic weight of the middle number of the triad was nearly equal to the arithmetic mean of the weights of the other two numbers of the triad.
For e.g.,
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 2.
What is Dobereiner law of triads?
Answer:
In triads, the atomic weight of the middle element nearly equal to the arithmetic mean of the atomic weights of the remaining two elements.

Question 3.
State the New land’s law of octaves.
The Law of octaves states that, “when elements are arranged in the order of increasing atomic weights, the properties of the eighth element are a repetition of the properties of the first element.

Question 4.
What is periodicity?
Answer:
Elements with similar properties recur after regular intervals in the periodic table. The repetition of physical and chemical properties at regular intervals is called periodicity.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 5.
Explain about the relationship between the atomic number of an element and frequency of the X-ray emitted from the elements.
Answer:
Henry Moseley studied the X-ray spectra of several elements and determined their atomic numbers (Z). He noticed that the frequencies of X-ray emitted from the elements concerned could be correlated by the equation
\(\sqrt{υ}\) = a(Z – b)
Where, υ Frequency of the X-ray emitted by the element.
a and b = Constants and have same values for all the elements.
Z = Atomic number of the element.

Question 6.
What are s-block elements?
Answer:
The elements of group-1 and group-2 are called s-block elements since the last valence electron enters the ns orbital. The group-1 elements are called alkali metals while the group-2 elements are called alkaline earth metals.

Question 7.
What are the anomalies of the long form of periodic table?
Answer:
The long form of periodic table need clarification about the following:

  • The position of hydrogen is not defined till now.
  • Lanthanides and actinides still find place in the bottom of the table.

Question 8.
What are d-block elements? Give their general electronic configuration.
Answer:
The elements of groups 3 to 12 are called d-block elements or transition elements with general valence
shell electronic configuration ns1 – 2, (n- 1)d1 – 10.

Question 9.
Write a note about the electronic configuration of elements in groups.
Answer:
A vertical column of the periodic table is called a group. A group consists of a series of elements having a similar configuration to the outermost shell. There are 18 groups in the periodic table. it may be noted that the elements belonging to the same group are said to constitute a family. For example, elements of group 17 are called halogen family.

Question 10.
Write the general valence shell electronic configuration of lanthanides and actinides.
Answer:
The general valence shell electronic configuration of lanthanides is 4f1 – 14, 5d0 – 1, 6s2 and for the actinides is 5f1 – 14, 6d0 – 1, 7s2.

Question 11.
Write any two characteristic properties of alkali metals.
Answer:

  • Alkali metals readily lose their outermost electron to form +1 ion.
  • Alkali metals are soft metals with low melting and boiling points.

Question 12.
What is a covalent radius?
Answer:
Covalent radius is one-half of the internuclear distance between two identical atoms linked together by a single covalent bond.

Question 13.
Groups from 13 to 1 in the periodic table are called p-block elements. Give reason.
Answer:

  • The elements whose last electron enters into the p-orbital of the outermost shell are having similar properties and thus form a group. The ‘np’ orbital of these elements is being progressively tilled. Hence, these elements are named as p-block elements.
  • The groups of 3rd to 18th in the periodic table belong to the p-block.

Question 14.
Why noble gases do not show much of chemical reactivity?
Answer:
Noble gases having closed valence shell configuration as ns2 np6. The valence shell orbitais of noble gases are completely filled by electrons and it is very difficult to alter this stable arrangement by the addition or removal of electrons. Because of these reasons, noble gases do not show much chemical reactivity.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 15.
What is the shielding effect?
Answer:
The inner shell electrons in an atom can act as a shield between the nucleus and the valence electrons. This effect is called the shielding effect.

Question 16.
What are d-biock elements? Why are they called so?
Answer:

  • The elements whose last electron enters into the d-orbital of the penultimate shell (n-1) are having similar properties and called as d-block elements.
  • The groups of 3 to 12 in the center of the periodic table belongs to d-block.

Question 17.
Define ionic radius.
Answer:
Ionic radius is defined as the distance from the centre of the nucleus of the ion upto which it exerts its influence on the electron cloud of the ion.

Question 18.
Why Zn, Cd and Hg are considered as soft metals?
Answer:

  • Zinc, cadmium and mercury are metals with low melting points. This is because they have an especially stable electronic configuration.
  • Mercury is so poor at forming metallic bonds that it is liquid at room temperature.
  • Zinc and cadmium arc soft metals that oxidize to the +2 oxidation states.

Question 19.
What is second ionization energy?
Answer:
The minimum amount of energy required to remove an electron from a unipositive cation is called second ionization energy.

Question 20.
What are f-block elements? how many series are there? Why they are called f-block elements?
Answer:

  • The elements, whose last electron enters into the f-orbital of the ante-penultimate shell (n-2) are having similar properties are called f-block elements. In these elements (n-2)f orbitais are being filled progressively.
  • The two rows of elements placed at the bottom of the periodic table. They are lanthanides and actinides.

Question 21.
Ionization energy decreases down a group. Give reason.
Answer:
As we move down a group, the valence electron occupies new shells, the distance between the nucleus and the valence electron increases. So, the nuclear forces of attraction on valence electron decreases, and hence, ionization energy also decreases down a group.

Question 22.
What are lanthanides and actinides? ,
Answer:

  • In 4f senes, 4f’orbitals arc being progressively filled with electrons, 4f1-14 5d0-1 6s2. These elements lie in 6th period and are called rare earths or lanthanides or lanthanones.
  • In the 5f series, 5f orbitais are being progressively filled with electrons, 5f16d0-1 7s2. These elements lie in the 7th period and are called actinides or actonones.

Question 23.
What are semi-metals? Give example.
Answer:

  • Some elements in the periodic table show properties that are characteristic of both metals and non-metals. These elements are called semi-metals or metalloids.
  • Example: Silicon. germanium, arsenic, antimony, and tellurium.

Question 24.
Why is the electron affinity of Beryllium almost zero?
Answer:
Beryllium has completely filled stable (1s2, 2s2, 2p3) electronic configuration. The addition of extra electrons will disturb their stable electronic configuration and it has almost zero electron affinity.

Question 25.
Define ionic radius.
Answer:
The ionic radius of an ion is the distance between the center of the ion and the outermost point of its electron cloud.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 26.
Why halogens have high electron affinity values?
Answer:
Halogens having the general electronic configuration of ns2, np5 readily accept an electron to get the stable noble gas electronic configuration (ns2, np6), and therefore, in each period the halogen has high electron affinity.

Question 27.
The anionic radius is higher than the corresponding neutral atom. Give reason.
Answer:
When an atom gains one or more electrons it forms anion. During the formation of anion, the number of orbital electrons becomes greater than the nuclear charge. Hence, the electrons are not strongly attracted by the lesser number of nuclear charges. Hence anionic radius is higher than the corresponding neutral atom.

Question 28.
What is the variation of electronegativity in a period?
Answer:
The electronegativity generally increases across a period from left to right. When the atomic radius decreases in a period, the attraction between the valence electron and the nucleus increases. Hence, the tendency to attract shared pair of electrons increases. Therefore, electronegativity also increases in a period.

Question 29.
Define ionization energy. Give its unit.
Answer:
The energy required to remove the most loosely held electron from an isolated gaseous atom is called ioniiation energy.
M(g) + energy M+(g) + electron
The unit of ionization energy is KJ mole-1.

Question 30.
What is mean by valence?
Answer:
The valence of an atom is the combining capacity relative to a hydrogen atom. It is usually equal to the total number of electrons in the valence shell or equal to eight minus the number of valence electrons.

Question 31.
The ionization energy of nitrogen is greater than the ionization energy of oxygen. Give reason.
Answer:
7N 1s2 2s1 2p3 (or) Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements . Nitrogen has exactly half filled valence electrons, which requires high I.E1.
8O 1s2 2s2 2p4 (or)Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements . Oxygen has incomplete valence shell electrons, which requires comparatively less I.E1.
I.E1N>I.E1O

Question 32.
Write the anomalies of the Mendeleev periodic table.
Answer:
In Mendeleev periodic table, some elements with similar properties were placed in different groups and those with dissimilar properties were placed in the same group.
Example: Tellurium was placed in the VI group but Iodine was placed in VII group.
Similarly, elements with higher atomic weights were placed before lower atomic weights based on their properties in contradiction to his periodic laws.
Example: 59Co27 was placed before 58M28.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 33.
Electron gain enthalpy of F ¡s less negative than Cl. Why?
Answer:
When an electron is added to F, the added electron goes to the L shell (n = 2). As the ‘L’ shell possesses a smaller region of space, the added electron feels significant repulsion from the other electrons present at this level.
E.A of F = -328 KJ mole-1
E.A of Cl = -349 KJ mole-1

Question 34.
Electron affinity of oxygen is less negative than sulphur. Justify this statement.
Answer:
When an electron is added to oxygen, the added electron goes to the ‘L’ shell (n 2). As the ‘L’ shell possess smaller region of space, the added electron feels significant repulsion from the other electrons present in this level.
E.A of O = – 141 KJ mole.
E.A of S = – 200 KJ mole.

Question 35.
Explain about the factors that affect electro negativity.
Answer:

  • Effective nuclear charge: As the nuclear charge increases, electro negativity also increases along the periods.
  • Atomic radius: The atoms in smaller size will have larger electronegativity.

Question 36.
Explain about periodic variation of electro negativity across a period.
Answer:
As we move from left to right in a period, electro negativity increases. This is due to the following reasons:

  • Nuclear charge increases in a period
  • Atomic size decrease in a period

Halogens have the highest value of electro negativity in their respective periods.

Question 37.
Explain about the period variation of electro negativity along a group.
Answer:
As we move down from top to bottom in a group, electro negativity decreases due to increased atomic radius. Fluorine has the highest value of clectro negativity among all the elements.

Question 38.
Define valency. How is it determined?
Answer:
The valency of an element may be defined as the combining capacities of elements. The electrons present in the outermost shell are called valence electrons and these electrons determine the valency of the atom.

Question 39.
What is the basic difference in approach between Mendeleev’s periodic law and the modern periodic law?
Answer:
The basic difference in approach between Mendeleev’s periodic law and modern periodic law is the change in basis of classification of elements from atomic weight to atomic number.

Question 40.
On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
Answer:
The sixth period corresponds to sixth shell. The orbitais present in this shell are 6s, 4f. 5p and 6d. The maximum number of electrons which can be present in these sub-shell is 2 + 14 + 6 + 10 = 32. Since the number of elements in a period corresponds to the number of electrons in the shells, the sixth period should have a maximum of 32 elements.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 41.
Why do elements in the same group have similar physical and chemical properties?
Answer:
The elements in a group have same valence shell electronic configuration and hence have similar physical and chemical properties.

Question 42.
How do atomic radius vary in a period and in a group? How do you explain the variation.
Answer:
Within a group atomic radius increases down the group Reason :
This is due to continuous increases in the number of electronic shells or orbit numbers in the structure of atoms of the elements down a group.

Variation across period:
Atomic radii:
From left to right across a period atomic radii generally decreases due to increase in effective nuclear charge from left to right across a period.

Question 43.
Explain why cation are smaller and anions are larger ¡n radii than their parent atoms?
Answer:
A cation is smaller than the parent atom because it has fewer electrons while its nuclear charge remains the same. The size of anion will be larger than that of parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge.

Question 44.
What is basic difference between the terms electron gain enthalpy and electronegativity?
Answer:
Electron gain enthalpy refers to tendency of an isolated gaseous atom to accept an additional electron to form a negative ion. Whereas electronegativity refers to tendency of the atom of an clement to attract shared pair of electrons towards it in a covalent bond.

Question 45.
Would you expect the first ionization enthalpies of two isotopes of the same element to be the same or different? Justify your answer.
Answer:
Ionization enthalpy, among other things, depends upon the electronic configuration (number of electrons) and nuclear charge (number of protons). Since isotopes of an element have the same electronic configuration and same nuclear charge, they have same ionization enthalpy.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 46.
Write the general electronic configuration of s-, p-, d-, and f-block elements?
Answer:

  • s-block elements : ns1-2 where n 2 – 7.
  • p-block elements : ns2 np1-6 where n = 2 – 6.
  • d-block elements : (n – 1) d1-0 ns0-2 where n = 4 – 7.
  • f-block elements : (n – 2) f0-14 (n – 1) d0-1 ns2 where n = 6 – 7.

Samacheer Kalvi 11th Chemistry Periodic Classification of Elements 3-Mark Questions

Question 1.
Why there is a need for classification of elements?
Answer:

  • Classification is a fundamental and essential process in our day-to-day life for the effective utilization of resources, daily events and materials.
  • In such a way, for the effective utilization of discovered elements becomes fundamentally essential process.
  • The periodic classification of the elements is one of the outstanding contributions to the progress of chemistry.

Question 2.
Prove that the halogens, chlorine, bromine and iodine follow the law of triads.
Answer:
When the halogens, chlorine, bromine and iodine are placed on below the others, they had similar properties. The atomic weight of bromine was close to the average of the atomic weights of chlorine and iodine.
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 3.
What are the salient features of New land’s law of octaves?
Answer:

  1. This law is quite well for lighter elements but not supported to heavier elements.
  2. Elements were arranged in increasing atomic masses without taking an account on the properties of elements.
  3. This law was seemed to be applicable only for elements upto calcium.

Question 4.
How the properties of Eka – silicon was related to germanium?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 5.
Compare the properties of Eka – aluminium and gallium.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 6.
Write notes on Dobereiner’s classification of elements.
Answer:
Dobereiner classified some elements such as chlorine, bromine, and iodine with similar chemical properties into the group of three elements called triads. In triads, the atomic weight of the middle element nearly equal to the arithmetic mean of the atomic weights of the remaining two elements. However, only a limited number of elements can be grouped as triads.
Example: Elements in the Triad: Li, Na, and K.
Atomic weight of middle element (Na) = 23.
Average atomic weight of the remaining elements = \(\frac{(7+39)}{2}\) = 23.
This concept cannot be extended to some triads which have nearly the same atomic masses.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 7.
Write any three significance of modern periodic table.
Answer:

  1. The physical and chemical properties of the elements are correlated to the arrangement of electrons in their outermost shell.
  2. All the elements are arranged in the modem periodic table which contains 18 vertical columns and 7 horizontal rows.
  3. Each period starts with the element having general outer electronic configuration ns1 and ends with np6.

Question 8.
Draw a simplified form of periods and elements present in the modern periodic table.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 9.
Write the electronic configuration of alkali metals 2Li,11Na, 19K, 37Rb, 55Cs and 87Fr.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 10.
What are p-block elements? Give their properties.
Answer:
The elements of groups 13 to 18 are called p-block elements or representative elements and have a general electronic configuration ns2, np1 – 6.

  • The elements of group 16 and 17 are called chalcogens and halogens respectively.
  • The elements of the 18th group contain a completely filled valence shell electronic configuration and are called inert gases.
  • The elements of the p-block have high negative electron gain enthalpies.
  • The ionization energies are higher than that of s-block elements.
  • They form mostly covalent compounds and show more than one oxidation state in their compounds.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 11.
Define ionization energy. Discuss the order of successive ionization energies.
Answer:
Ionization energy is defined as the minimum amount of energy required to remove the most loosely bound electron from the valence shell of the isolated neutral gaseous atom in its ground state.
M(g) + IE1 → M+(g) + 1e
The minimum amount of energy required to remove an electron from a unipositive cation is called second ionization energy. It is represented by the following equation.
M+(g) + IE2 → M2+(g) + 1e
The total number of electrons is less in the cation than the neutral atom while the nuclear charge remains the same. Therefore, the effective nuclear charge of the cation is higher than the corresponding neutral atom. Thus, the successive ionization energies, always increase in the following order
IE1 < IE2 < IE3 < …

Question 12.
How many elements are there in the 4th period? Prove it.
Answer:
In the fourth period, 18 elements are present. In this period electrons are entering into the fourth energy level, i.e., n = 4. it starts with the filling of 4s-orbitals. However, after the 4s, but before the 4p orbitais, there are five 3d-orbitais also to be filled. Thus, nine orbitais (one 4s, five 3d, and three 4p) have to be filled. These nine orbitais can accommodate (9 x 2 = 18)18 electrons. Hence, the period contains 18 elements in it.

Question 13.
How many elements are there in the 6th period? Prove it.
Answer:
In the sixth period, 32 elements are present. This period starts with the filling of the 6th energy shell, n = 6. There are sixteen orbitais (one 6s, seven 4f, five 4d, and three 6p) to be filled. These sixteen orbitais can accommodate 32 (16 × 2 = 32) electrons. Hence, 32 elements are present in the sixth period.

Question 14.
What is meant by valence? Discuss its periodicity.
Answer:
The valence of an atom is the combining capacity relative to a hydrogen atom. It is usually equal to the total number of electrons in the valence shell or equal to eight minus the number of valence electrons.

The valence of an atom primarily depends on the number of electrons in the valence shell. As the number of valence electrons remains the same for the elements in the same group, the maximum valence also remains the same. However, in a period the number of valence electrons increases, hence the valence also increases.

In addition to that some elements have a variable valence, For example, most of the elements of group 15 which have 5 valence electrons show two valences 3 and 5. Similarly, transition metals and inner transition metals also show variable oxidation states.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 15.
Explain the salient features of metals.
Answer:

  • Metals comprise more than 78% of all known elements. They are present on the left side of the periodic table.
  • They are usually solids at room temperature. [Mercury is an exception (Hg-liquid), gallium (303 K) and cesium (302 K) also have very low melting points].
  • Metals usually have high melting and boiling points.
  • They are good conductors of heat and electricity.
  • They are malleable and ductile, and also can be flattened into thin sheets by hammering and drawn into thin wires.

Question 16.
Explain the characteristics of non-metals.
Answer:

  • Non-metals are located at the top right-hand side of the periodic table.
  • In a period, as we move from left to right the non-metallic character increases while the metallic character increases as we go down a group.
  • Non-metals are usually solids or liquids or gases at room temperature with low melting and low boiling points (Exceptions: boron and carbon).
  • They are poor conductors of heat and electricity.
  • Most of the non-metallic solids are brittle and are neither malleable nor ductile.

Question 17.
Periodic change in electronic configuration is responsible for the physical and chemical properties of elements. Justify this statement.
Answer:

  • The electronic configuration of the elements changes periodically in a period and group as well.
  • We could find a pattern in the physical and chemical properties as we go down in a group or move across a period.
  • For example, The chemical reactivity is high at the beginning, lower at the middle, and increases to a maximum at group 17 in a period.
  • The reactivity increases on moving down the group of alkali metals. But the reactivity decreases on moving down the group of halogens.
  • Atomic radii increase down the group and decrease across the period.

Question 18.
What is a covalent radius? How would you determine the covalent radius of a chlorine atom?
Answer:
The distance between the nuclei of two covalent bonded atoms is known as covaLent distance or inter-nuclear distance. The one – half of this inter-nuclear distance is called covalent radius. The covalent distance (Cl – Cl) of Cl2 molecule is experimentally found as 198 pm (1.98 A). Its covalent radius is 99 pm (0.99 Å).
Cl – Cl Inter nuclear distance = 1.98 Å
∴ rCl = 1.98 / 2=0.99 Å.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 19.
Write a note about metallic radius.
Answer:

  • It is defined as one half of the distance between the centers of nuclei of the two adjacent atoms in the metallic crystal.
  • The metallic radius is always larger than its covalent radius.
  • The distance between two adjacent copper atoms in solid copper is 2.56 A. Hence, the metallic radius of copper is 1.28 A.

Question 20.
Arrange Na+, Mg2+ and Al3+ in the increasing order of ionic radii. Give reason.
Answer:
Na+, Mg2+ and Al3+ are iso electronic cations.
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
The cation with the greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus. Hence the increasing order of ionic radii is,
rNa+ > rMg2+ > rAl3+

Question 21.
Arrange the ions F, O2- and N3- ¡n the increasing order of their ionic radii. Give reason.
Answer:
F, O2- and N3- are isoelectronic species.
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
The anion with the greater negative charge will have a larger radius because of the lesser attraction of the electrons to the nucleus. Hence the increasing order of ionic radii is,
rNa3 > rO2 > rF

Question 22.
Mention some characteristics of ionization energy.
Answer:

  • Ionization is always an endothermic process. It absorbs energy.
  • Ionization energies always increase in the order, I.E1< I.E2<IE3.
  • Ionization energy measurements are carried out with atoms in the gaseous state.

Question 23.
Why ionization energy and electron affinity are calculated in gaseous state?
Answer:

  • Inter molecular force can affect the value of ionization energy and electron affinity.
  • In gaseous state, there is little inter molecular force in a substance and it can be considered negligible In some cases. So the value of I.E and E.A are almost unaffected if they are calculated for gaseous atoms.
  • When we arc talking about ionization energy and electron affinity, we need to consider
    atoms and we can find free atoms only when the substance is in gaseous state.

Question 24.
How is screening effect of inner electrons affect ionization energy?
Answer:

  • The electrons of inner shells form a cloud of negative charge. This cloud shields the outer electron from the nucleus and also acts as a screen between the nucleus and the outer electrons.
  • This screen reduces the coulombic attraction between the positive nucleus and the negative outer electrons. Therefore, the outer electrons do not feel the full effect of the positive charge of the nucleus.
  • If screening effect increases, ionization energy decreases.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 24.
How is screening effect of inner electrons affect ionization energy?
Answer:

  • The electrons of inner shells form a cloud of negative charge. This cloud shields the outer electron from the nucleus and also acts as a screen between the nucleus and the outer electrons.
  • This screen reduces the coulombic attraction between the positive nucleus and the negative outer electrons. Therefore, the outer electrons do not feel the full effect of the positive charge of the nucleus.
  • If screening effect increases, ionization energy decreases.

Question 25.
Ionization energy of Mg is greater than that of Al. Why?
Answer:
Mg (Z = 12) 1s2 2s2 2p6 3s2.
Al (Z = 13) 1s2 2s2 2p6 3s2 3p1.
Although the nuclear charge of aluminium is greater than that of magnesium, I.E of Mg is greater than that of Al. It is because Mg atom has more stable configuration than Al atom. IE1 of Mg > IE1 of Al.

Question 26.
What are all the factors that influences electron gain enthalpy?
Answer:
1. Size of the atom:
The new electron which was added experiences stronger attraction to its nucleus if the atoms are smaller in size.
Atomic size α \(\frac {1}{ Electron gain enthalpy }\)

2. Nuclear charge:
The new electron which was added experiences stronger attraction to its nucleus if the atom possess greater nuclear charge. Nuclear charge α Electron gain enthalpy

3. Electronic configuration:
An atom with stable electronic configuration has no tendency to gain an electron. Such atoms have zero or almost zero electron gain enthalpy.

Question 27.
Explain about the periodic variation of electron gain enthalpy in a period and in a group.
Answer:
1. The electron gain enthalpy increases as we move from left to right in a period due to the increase of nuclear charge. However, Be, Mg, N and noble gases have almost zero value of electron gain enthalpy due to extra stability of completely and half filled orbitais.

2. When we move in a group of periodic table, the size and nuclear charge increase. But the effect of increase in atomic size is much more pronounced than that of nuclear charge and thus the additional electron feels less attraction by the large atom. Consequently, electron gain enthalpy decreases.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 28.
Explain the electronegativity and non-metallic character across the period and down the group.
Answer:
Electronegativity α Non-metallic character:

  • As the electronegativity is directly proportional to the non-metallic character, thus across the period, with an increase in electronegativity. the non-metallic character also increases.
  • As we move down the group. the decrease in electronegativity is accompanied by a decrease in non-metallic character.

Question 29.
Prove that valency is a periodic property.
Answer:
Variation in period:
The number of valence electrons increases from I to 8 on moving across a period. The valency of the elements with respect to hydrogen and chlorine increases from 1 to 4 and then decreases from 4 to zero.

Variation in group:
On moving down a group, the number of valence electrons remains same. All the elements in a group exhibit same valency. For example, all the elements of group I have valency equal to 1. Hence. valency is a periodic property.

Question 30.
Write a note about periodic trends and chemical reactivity.
Answer:

  • The group 1 elements are extremely reactive because these elements can lose one electron to form cation. Their ionization enthalpy is also least.
  • The high reactivity of halogens is due to the ease with which these elements can gain an electron to form anion. Their electron gain enthalpy is most negative.
  • The elements at the extreme left (alkali) exhibit strong reducing behavior, whereas the elements at the extreme right (halogens) exhibit strong oxidizing behaviour.
  • The reactivity of elements at the center of the periodic table becomes low when compared with extreme right and left.

Question 31.
How would you explain the fact that the first ionization enthalpy of sodium ¡s lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
Answer:
Electronic configuration of Na and Mg are
Na = 1s2 2s2 2p6 3s1
Mg = 1s2 2s2 2p6 3s2
First electron in both cases has to be removed from 3s-orbital but the nuclear charge of Na (+11) is lower than that of Mg (+12) therefore first ionization energy of sodium is lower than that of magnesium.

After the loss of first electron, the electronic configuration of
Na = 1s2 2s2 2p6
Mg = 1s22s22p63s1
Here electron is to be removed from inert (neon) gas configuration which is very stable and hence removal of second electron requires more energy when compared to Mg. Therefore, second ionization enthalpy of sodium is higher than that of magnesium.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 32.
What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down the group?
Answer:
Atomic size:
With the increase in atomic size, the number of electron shells increase. Therefore, the force that binds the electrons with the nucleus decreases. The ionization enthalpy thus decreases with the increase in atomic size.

Screening or shielding effect of inner shell electron:
With the addition of new shells, the number of inner electron shells which shield the valence electrons increases. As a result, the force of attraction of the nucleus for the valence electrons further decreases and hence the ionization enthalpy decreases.

Question 33.
Which of the following pairs of elements would have more negative electron gain enthalpy?

  1. O or F
  2. F or Cl.

Answer:
1. O or F. Both O and F lie in 2e” period. As we move from O to F the atomic size decreases. Due to smaller size off nuclear charge increases. Further, gain of one electron by
F → F
F ion has inert gas configuration, While the gain of one electron by
O → O
gives O ion which does not have stable inert gas configuration. Consequently, the energy released is much higher in going from
F → F
than going from O → O. In other words electron gain enthalpy off is much more negative than that of oxygen. –

2. The negative electron gain enthalpy of Cl (e.g. ∆H = – 349 mol-1) is more than that of F (e.g. ∆H = -328 U mol-1).

The reason for the deviation is due to the smaller size off. Due to its small size, the electron repulsion in the relatively compact 2p-subshell are comparatively large and hence the attraction for incoming electron is less as in the case of Cl.

Question 34.
Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.
For oxygen atom:
O(g) + e → O-1(g) (e.g. ∆H = -141 Id mor-1)
O-1(g) + e → O2-(g) (e.g. ∆H = + 780 kJ mol-1)
The first electron gain enthalpy of oxygen is negative because energy is released when a gaseous atom accepts an electron to form monovalent anion. The second electron gain enthalpy is positive because energy is needed to overcome the force of repulsion between monovalent anion and second incoming electron.

Question 35.
What are major differences between metals and non-metals?
Metals:

  • Have a strong tendency to lose electrons to form cations.
  • Metals are strong reducing agents.
  • Metals have low ionization enthalpies.
  • Metals form basic oxides and ionic compounds.

Non-Metals:

  • Non-metals have a strong tendency to accept electrons to form anions.
  • Non-metals are strong oxidizing agents.
  • Non-metals have high ionization enthalpies.
  • Non-metals form acidic oxides and covalent compounds.

Question 36.
The increasing order of reactivity among group 1 elements is Li < Na < K < Rb Cl> Br>. Explain.
Answer:
The elements of group 1 have only one electron in their respective valence shells and thus have a strong tendency to lose this electron. The tendency to lose electrons in turn, depends upon the ionization enthalpy. Since the ionization enthalpy decreases down the group therefore, the reactivity of group 1 elements increases in the same order Li

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 37.
Arrange the following as stated:

  1. N2 O2, F2, Cl2 (Increasing order of bond dissociation energy)
  2. F, Cl, Br, I (Increasing order of electron gain enthalpy)
  3. F2, N2, Cl2, O2 (Increasing order of bond length).

Answer:

  1. F2, N2, Cl2, O2
  2. I< Br < F < Cl
  3. N2 O2, F2, Cl2

Question 38.
Write the salient features of the modern periodic table.
Answer:
The physical and chemical properties of the elements are correlated to the arrangement of electrons in their outermost shell. Different elements having similar outer shell electronic configuration possess similar properties, For example, elements having one electron in their valence shell s-orbital possess similar physical and chemical properties, These elements are grouped together in the modem periodic table as first group elements.

Similarly, all the elements are arranged in the modem periodic table which contains 18 vertical columns and 7 horizontal rows. The vertical columns are called groups and the horizontal rows are called periods. Groups are numbered 1 to 18 in accordance with the IUPAC recommendation.

Each period starts with the element having general outer electronic configuration ns1 and ends with np6. Here,’ n’ corresponds to the period number. The Aufbau principle and the electronic configuration of atoms provide a theoretical foundation for the modem periodic table.

Question 39.
Give reasons:

  1. lE1 of sodium is lower than that of magnesium whereas IE2 of sodium is higher than that of magnesium.
  2. Noble gases have a positive value of electron gain enthalpy.

Answer:
1. The effective nuclear charge of magnesium is higher than that of sodium. For these reasons, the energy required to remove an electron from magnesium is more than the energy required in sodium. Hence, the first ionization enthalpy of sodium is lower than that of magnesium.

2. Noble gases have completely filled electronic configuration and they are more stable. So in Noble gases addition of electron is not possible. Electron gain enthalpy is always the amount of energy released (-ve sign) when an electron is added to an atom. – Butin noble gases, if an electron is added, they have positive value of electron gain enthalpy.

Samacheer Kalvi 11th Chemistry Periodic Classification of Elements 5-Mark Questions

Question 1.
(a) State Mendeleev’s periodic law.
(b) Describe about the merits of Mendeleev’s periodic table.
Answer:
(a) Mendeleev’s periodic law:
Mendeleev’s periodic law states that the physical and chemical properties of elements are a periodic function of their atomic weights.

(b) Merits of Mendeleev’s periodic table:

  • The comparative studies of elements were made easier.
  • The table shóws the relationship in properties of elements in a group.
  • The table helped to correct the atomic masses of some elements, later on, At the time of Mendeleev, the atomic weight of Au and Pt were known as 196.2 and 196.7 respectively. However, Mendeleev placed Au (196.2) after Pt (196.7) saying that the atomic weight of Au is incorrect, which was later on found to be 197.
  • At the time of Mendeleev, about 70 elements were known and thus blank spaces were left for unknown elements which helped further discoveries.
  • Both Gallium (Ga) in the III group and Germanium (Ge) in the IV group, were unknown at that time by Mendeleev predicted their existence and properties.
  • He referred to the predicted elements as eka-aluminum and eka-silicon. After the discovery of the actual elements, their properties were found to match closely to those predicted by Mendeleev.

Question 2.
Explain the anomalies of Mendeleev’s periodic table. Anomalies of Mendeleev’s periodic table
Answer:

  1. Some elements with similar properties were placed in different groups whereas some elements having dissimilar properties were placed in the same group, but iodine (127) was placed in the VII group.
    Example: Tellurium (127.6) was placed in VI group.
  2. Some elements with higher atomic weights were placed before lower atomic masses in order to maintain the similar chemical nature of elements. This concept was called the inverted pair of elements concept.
    Example: 5927CO and 58.728Ni
  3. Isotopes did not find any place in Mendeleev’s periodic table.
  4. The position of hydrogen could not be made clear.
  5. He did not leave any space for lanthanides and actinides which were discovered later on.
  6. Elements with different nature were placed in one group,
    Example: Alkali metals and coinage metals were placed together.
  7. Diagonal and horizontal relationships were not explained.

Question 3.
Explain the structural features of Moseley’s long form of the periodic table.
Answer:

  • The long form of the periodic table of the elements is constructed on the basis of modem periodic law.
  • The arrangement resulted in repeating electronic configurations of atoms at regular intervals.
  • The elements placed in horizontal rows are called periods and in vertical columns are called groups.
  • According to IUPAC, the groups are numbered from I to 18.
  • There are 18 vertical columns which constitute 18 groups or families. All the members of a particular group have similar outer shell electronic configurations.
  • There are 7 horizontal rows called periods.
    Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
    The elements are shown in the above table along with their atomic number.
  • The atomic number also indicates the number of electrons in the atoms of an element.
  • This periodic table is important and useful because we can predict the properties of any element using periodic trends, even though that element may be unfamiliar to us.

Question 4.
Explain the variation of electronic configuration in the groups.
Answer:
Elements of a group have similar electronic configurations in the outer shell. The groups can be combined as s, p, d, and f block elements on the basis of the orbital in which the last valence electron enters. The elements of group 1 and group 2 are called s-block elements since the last valence electron enters the ns orbital. The group 1 elements are called alkali metals while the group 2 elements are called alkaline earth metals.

These are soft metals and possess low melting and boiling points with low ionization enthalpies. They are highly reactive and form ionic compounds. They are highly electropositive in nature and most of the elements impart colour to the flame.

The elements of groups 13 to 18 are called p-block elements or representative elements and have a general electronic configuration ns2, np1 – 6. The elements of group 16 and 17 are called chalcogens and halogens respectively. The elements of the 18th group contain completely filled valence shell electronic configuration (ns2, np6) and are called inert gases or nobles gases.

The elements of the p-block have high negative electron gain enthalpies. The ionisation energies are higher than that of s-block elements. They form mostly covalent compounds and show more than one oxidation state in their compounds.

The elements of groups 3 to 12 are called d-block elements or transition elements with general valence shell electronic configuration ns1 – 2, (n – l)1 – 10. These elements also show more than one oxidation state and form ionic, covalent, and coordination compounds.

They can form interstitial compounds and alloys which can also act as catalysts. These elements have high melting points and are good conductors of heat and electricity.

The lanthanides (4f1 – 14, 5d0 – 1, 6s2) and the actinides (5f0 – 14, 6d0 – 2, 7s2) are called f-block elements. These elements are metallic in nature and have high melting points. Their compounds are mostly coloured. These elements also show variable oxidation states.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 5.
Explain about the general characteristics of periods.
Answer:
1. Number of electrons in outermost shell:
The number of electrons present in the outermost shell increases from 1 to 8 as we proceed in a period.

2. Number of shells:
As we move from left to right in a period the shells remains the same. The number of shells present in the elements corresponds to the period number. For example : all the elements of 2 period have on 2 shells (K, L)

3. Valency:
The valency of the elements increases from left to right in a period. With respect to hydrogen. the valency of period elements increases from 1 to 4 and then falls to one. With respect to oxygen, the valency increases from1 to 7.

4. Metallic character:
The metallic character of the elements decreases across a period.
For example:
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 6.
What is electronegativity? Discuss its variation along with the period and group.
Answer:
Electronegativity is defined as the relative tendency of an element present in a covalently bonded molecule, to attract the shared pair of electrons towards itself. Electronegativity is not a measurable quantity. However, a number of scales are available to calculate its value. One such method was developed by Pauling, he assigned an arbitrary value of electronegativities for hydrogen and fluorine as 2.2 and 4.0 respectively. Based on this the electronegativity values for other elements can be calculated using the following expression
A – χB) = 0.182 \(\sqrt{E_{A B}}\) – (EAA – EBB)\(\frac{1}{2}\)

Where EAB, EAA and EBB are the bond dissociation energies of AB, A2 and B2 molecules respectively.

The electronegativity of any given element is not a constant and its value depends on the element to which it is covalently bound. The electronegativity values play an important role in predicting the nature of the bond.

Variation of Electronegativity in a period:
The electronegativity generally increases across a period from left to right. As discussed earlier, the atomic radius decreases in a period, as the attraction between the valence electron and the nucleus increases. Hence the tendency to attract shared pair of electrons increases. Therefore, electronegativity also increases in a period.

Variation of Electronegativity in a group:
The electronegativity generally decreases down a group. As we move down a group the atomic radius increases and the nuclear attractive force on the valence electron ‘ decreases. Hence, the electronegativity decreases. Noble gases are assigned zero electronegativity. The electronegativity values of the elements of 5-block show the expected decreasing order in a group. Except 13th and 14th group all other p-block elements follow the expected decreasing trend in electronegativity.

Question 7.
Explain the classification of elements based on chemical behaviour and on physical properties.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Based of chemical behavior:

  1. Main group elements:
    All s-block and p-block elements excluding group elements are called representative elements.
  2. Noble gases:
    The 18th group elements are exclusively called noble gases. They have completely filled electronic configuration as ns2 np6. These elements are highly stable.
  3. Transition elements:
    The elements of d-block are called transition elements. These include elements of groups from 3 to 12 lying between s-block and p-block elements.
  4. Inner transition elements:
    The elements of the f-block are called inner-transition elements. These consist of lanthanides and actinides, with 14 elements in each.

Based on physical properties:

  1. Metals:
    Metals comprise more than 78% of all known elements. They are usually solids at room temperature (except Hg, Ga and Cs). They have high melting and boiling points. They are good conductors of heat and electricity.
  2. Non-metals:
    Non-metals are usually solids or liquids or gases at room temperature with low melting and low boiling points (except boron and carbon). They are poor conductors of heat and electricity. Most of the non-metallic solids are brittle and are neither malleable nor ductile.
  3.  Metalloids or Semi-metals:
    Some elements in the periodic tables show properties that are characteristic of both metals and non-metals. They are called metalloids. Example: Silicon, germanium, arsenic, antimony, and tellurium.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 8.
(a) Define atomic radius.
(b) What are the difficulties in determining atomic radius?
Answer:
(a) Atomic radius is the distance between the center of its nucleus and the outermost shell
containing the electron.
(b) Difficulties in determining the atomic radius

  • The size of an atom is very small (∼ 1.2Å i.e 1.2 × 1010)
  • The atom is not a rigid sphere; it is more like a spherical cotton ball rather than like a cricket ball.
  •  It is not possible to isolate an atom and measure its radius.
  • The size of an atom depends upon the type of atoms in its neighborhood and also the nature of bonding between them.

Question 9.
Prove that the atomic radii is a periodic property.
Answer:
Atomic radius is the distance between the center of its nucleus and the outermost shell containing the electron.

Atomic radius is a periodic property.
1. Variation in periods:
The atomic radius decreases while going from left to right in a period. As we move from left to right in a period, the nuclear charge increases by one unit in each succeeding element. But the number of the shell remains the same. Hence, the electrons are attracted strongly by the nucleus. Hence the atomic radius decreases along the period. In 2nd period rLi>rBe>rB>rC>rN>rO>rF

2. Variation in a group:
The atomic radius of elements increases with an increase in atomic number as we move from top to bottom in a group. The attraction of the nucleus for the electrons decreases as the shell number increases. Hence atomic radius increases along with the group. In 1 group rLi < rNa < rK <rRb < rCs
Hence, atomic radii is a periodic property.

Question 10.
Explain the factors that influence the ionization enthalpy. Factors influencing ionization enthalpy:
Answer:
1. Size of the atom:
If the size of an atom is larger, the outermost electron shell from the nucleus is also larger and hence the outermost electrons experience the lesser force of attraction. Hence it would be easier to remove an electron from the outermost shell. Thus, ionization energy decreases with increasing atomic sizes.
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

2. Magnitude of nuclear charge:
As the nuclear charge increases, the force of attraction between the nucleus and valence electrons also increases. So, more energy is required to remove a valence electron. Hence I.E increases with an increase in nuclear charge.
Ionization enthalpy α nuclear charge

3. Screening or shielding effect of the inner electrons:
The electrons of inner shells form a cloud of negative charge and this shields the outer electron from the nucleus. This screen reduces the coulombic attraction between the positive nucleus and the negative outer electrons. 1f screening effect increases, ionization energy decreases.
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

4. Penetrating power of subshells s, p, d and f:
The s-orbital penetrates more closely to the nucleus as compared to p-orbitais. Thus, electrons in s-orbitals are more tightly held by the nucleus than electrons in p-orbitais. Due to this, more energy is required to remove an electron from an s-orbital as compared to a p-orbital. For the same value of ‘n’, the penetration power decreases in a given shell in the order.
s > p > d > f.

5. Electronic configuration:
If the atoms of elements have either completely filled or exactly half-filled electronic configuration, then the ionization energy increases.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

Question 11.
Define ionization energy.
Prove that ionization energy is a periodic property.
Answer:
(a) The energy required to remove the most loosely held electron from an isolated gaseous atom is called as ionization energy.
(b) (i) Variation in a period:
On moving across a period from left to right, the ionization enthalpy value increases. This is due to the following reasons:

  • Increase of nuclear charge in a period
  • Decrease of atomic size in a period

Because of these reasons, the valence electrons are held more tightly by the nucleus, thus ionization enthalpy increases. Hence, ionization energy is a periodic property.

(ii) Variation in a group:
As we move from top to bottom along a group, the ionization enthalpy decreases. This is due to the following reasons:

  • A gradual increase in atomic size
  • Increase of screening effect on the outermost electrons due to the increase of number of inner electrons.
    Hence, ionization enthalpy is a periodic property.

Question 12.
Distinguish between electron affinity and electron negativity.
Answer:
Electron affinity:

  • It is the tendency of an isolated gaseous atom to gain an electron.
  • Ills the property of an isolated atom.
  • It does not change regularly in a period or a group.
  • It is measured in electron volts/atom or kcal/mole or kJ/mole.

Electron negativity:

  • It is the tendency of an atom in a molecule to attract the shared pair of electrons.
  • It is the property of bonded atom.
  • It changes regularly in a period or a group.
  • It is a number and has no units.

Question 13.
What are the anomalous properties of second-period elements?
Answer:

  1. In the Pt group, lithium differs in many aspects from its own family elements. Similarly, in the 2rd group, beryllium differs in many aspects from its own family.
  2. For example. lithium forms compounds with more covalent character. But other alkali metals of this group form only ionic compounds.
  3. Similarly, beryllium forms compounds with more covalent character while the other elements of this family form only ionic compounds.
  4. Lithium and beryllium resemble more with the elements lying at their right hand side in the 3t1 period than with the other members of their own family.
  5. These kinds of anomalies are also observed from the 13th to 17th groups.
  6. This sort of similarity is commonly referred to the as a diagonal relationship in the periodic properties.
  7. The anomalous behaviors are attributed to the following factors:
    • Smaller atomic size
    • Higher ionization enthalpy
    • High electronegativity

Activity 3.1

Covalent radii (in Å) for sonic elements of different groups and periods are listed below. Plot these values against the atomic numbers. From the plot, explain the variation along a period and a group.
2nd group elements : Be (0.89), Mg (1.36), Ca (1.74), Sr (1.91) Ba( 1.98)
17th group elements : F (0.72), Cl (0.99), Br (l.14),I (1.33)
3nd Period elements : Na (1.57), Mg(1.36),AI (1.25), Si (1.17), P(1.10), S (1.04), Cl (0.99)
4thperiod elements : K (2.03), Ca (1.74), Sc (l.44), Ti(1.32), V (1.22), Cr (1.17), Mn (1.17), Fe( 1.17), CO (1.16), Ni (1.15), Cu (1.17), Zn(1.25), Ga(1.25), Ge(1.22), As(1.21), Se(1.14), Br( 1. 14)
Solution:
2nd group elements:
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
17th group elements:
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

As we move down the group, atomic radii increases with the increase in atomic number . As we move down the group, the number energy levels increases, as the number of electrons Periodic.

Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

17th group elements: F (4.i), Cl (3.0), Br (2.8), I (2.5)
3rd Period elements : Na(0.9), Mg(l.2), Al (1.5), Si(l.8), P(2.1), S(2.5), Cl(3.0)
4th period elements : K(O.8), Ca (1.0), Sc (1.3), Ti (l.5), V(1.6), Cr(1.6), Mn(1.5), Fe(l.8), CO(1.9), Ni(1.9), Cu(1.9), Zn(1.6), Ga(1.6), Ge(1.8), As(2.0), Se(2.4), Br(2.8)
Solution:
2nd group elements:
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
17th group elements:
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
As we go down the group, the electronegativity value decreases. Moving down the group, the electronegativity decreases due to the long distance between the nucleus and the valence electron shell thereby decreasing the attraction making the atom have less attraction for electrons or protons.
3rd period:
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
4th period:
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements
The positively charged protons in the nucleus attract the negatively charged electrons. As the number of protons in the nucleus increases, the electronegativity or attraction will increase. Therefore electronegativity increases from left to right across the period. This occurs due to the greater charge on the nucleus, causing the electron bonding pairs to be very attracted to atoms placed further right on the periodic table.

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

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Samacheer Kalvi 11th Chemistry Chapter 2 Quantum Mechanical Model of Atom Textual Evaluation Solved

I. Choose the correct answer
Question 1.
Electronic configuration of species M2+ is 1s2 2s2 2p63s2 3p6 3d6 and its atomic weight is 56. The number of neutrons in the nucleus of species M is ………..
(a) 26
(b) 22
(c) 30
(d) 24
Answer:
(c) 30
Solution:
M2+ : 1s2 2s2 2p63s2 3p6 3d6
M : 1s2 2s2 2p63s2 3p6 3d8
Atomic number = 26
Mass number = 56
No. of neutrons = 56 – 26 = 30.

Question 2.
The energy of light of wavelength 45 nm is
(a) 6.67 x 1015 J
(b) 6.67 x 1011 J
(c) 4.42 .x 1018 J
(d) 4.42 x 10-15 J
Answer:
(c) 4.42 x 1018 J
Solution:
E = hv = hc / λ
\(\frac{6.626 \times 10^{-34} \mathrm{J} \mathrm{s} \times 3 \times 10^{8} \mathrm{ms}^{-1}}{45 \times 10^{-9} \mathrm{m}}\) = 4.42 .x 1018 J.

Question 3.
The energies E1 and E2 of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths i.e. λ1 and λ2 will be …………
(a) \(\frac{\lambda_{1}}{\lambda_{2}}=1\)
(b) λ1 = 2 λ2
(c) λ1 = \(\sqrt{25 \times 50} \lambda_{2}\)
(d) 2 λ1 = λ2
Answer:
(b) λ1 = 2 λ2
Solution:
\(\frac{E l}{E 2}\) = \(\frac{25eV}{50eV}\) = \(\frac{1}{2}\)
\(\frac{\mathrm{hc}}{\lambda_{1}} \times \frac{\lambda_{2}}{\mathrm{hc}}\) = \(\frac{1}{2}\)
2 = λ1.

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 4.
Splitting of spectral lines in an electric field is called …………….
(a) Zeeman effect
(b) Shielding effect
(c) Compton effect
(d) Stark effect
Answer:
(d) Stark effect
Solution:
The splitting of spectral lines in a magnetic field is called the Zeeman effect and the splitting of spectral lines in an electric field is called the Stark effect.

Question 5.
Based on equation E = -2.178 x 1018 J \(\left(\frac{z^{2}}{n^{2}}\right)\) certain conclusions are written. Which of them is not correct? (NEET)
(a) Equation can be used to calculate the change in energy when the electron changes orbit
(b) For n = 1, the electron has more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit
(c) The negative sign in the equation simply means that the energy of an electron bound to the nucleus is lower than it would be if the electrons were at an infinite distance from the nucleus.
(d) Larger the value of n, the larger is the orbit radius.
Answer:
(b) For n = 1, the electron has more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit
Solution:
Correct statement:
For n = 6, the electron has more negative energy than it does for n = 6 which means that the electron is strongly bound in the smallest allowed orbit.

Question 6.
According to the Bohr Theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon?
(a) n = 6 to n = 1
(b) n = 5 to n = 4
(c) n = 5 to n = 3
(d) n = 6 to n = 5
Answer:
(d) n = 6 to n = 5
Solution:
n = 6 to n = 5
E6 = -13.6 / 62 ; E5 = – 13.6 / 52
E6 – E5 = (-13.6 / 62) – (-13.6 / 52)
= 0.166 eV atom-1
E5 – E4 = (-13.6 / 52) – (-13.6 / 42)
= 0.306 eV atom-1

Question 7.
Assertion: The spectrum of He+ is expected to be similar to that of hydrogen.
Reason: He+ is also a one-electron system.
(a) If both assertion and reason are true and the reason is the correct explanation of assertion.
(b) If both assertion and reason are true but the reason is not the correct explanation of assertion.
(c) If the assertion is true but the reason is false
(d) If both assertion and reason are false
Answer:
(a) If both assertion and reason are true and the reason is the correct explanation of assertion.

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 8.
Which of the following pairs of d-orbitals will have electron density along the axes ? (NEET Phase – II)
(a) dz2, dxz
(b) dxz, dyz
(c) dz2, \(d_{x^{2}-y^{2}}\)
(d) dxy, \(d_{x^{2}-y^{2}}\)
Answer:
(c) dz2, \(d_{x^{2}-y^{2}}\)

Question 9.
Two electrons occupying the same orbital are distinguished by …………
(a) azimuthal quantum number
(b) spin quantum number
(c) magnetic quantum number
(d) orbital quantum number
Answer:
(b) spin quantum number
Solution:
Spin quantum number For the first electron ms = +\(\frac {1}{2}\)
For the second electron ms = –\(\frac {1}{2}\).

Question 10.
The electronic configuration of Eu (atomic no. 63) Gd (atomic no. 64) and Tb (atomic no. 65) are (NEET – Phase II)
(a) [Xe] 4f7 5d1 6s2, [Xe] 4f7 5d1 6s2 and [Xe] 4f8 5d1 6s2
(b) [Xe] 4f7, 6s2, [Xe] 4f7 5d1 6s2 and [Xe] 4f9 6s2
(c) [Xe] 4f7 , 6s2, [Xe] 4f8 6s2 and [Xe] 4f8 5d1 6s2
(d) [Xe] 4f8 5d1 6s2[Xe] 4f7 5d1 6s2 and [Xe] 4f9 6s2
Answer:
(b) [Xe] 4f7, 6s2, [Xe] 4f7 5d1 6s2 and [Xe] 4f9 6s2
Solution:
Eu : [Xe] 4f7, 5d0, 6s2
Gd : [Xe] 4f7, 5d1, 6s2
Tb : [Xe] 4f9, 5d0,6s2

Question 11.
The maximum number of electrons in a subshell is given by the expression …………..
(a) 2n2
(b) 21 + 1
(c) 41 + 2
(d) none of these
Answer:
(c) 41 + 2
Solution:
2 (21 + 1) = 41 + 2.

Question 12.
For d-electron, the orbital angular momentum is ………….
(a) \(\frac{\sqrt{2} h}{2 \pi}\)
(b) \(\frac{\sqrt{2 \mathrm{h}}}{2 \pi}\)
(c) \(\sqrt{2 \times 4}\)
(d) \(\frac{\sqrt{6} h}{2 \pi}\)
Answer:
(d) \(\frac{\sqrt{6} h}{2 \pi}\)
Solution:
Orbital angular momentum
= \(\sqrt{(1(1+1)}) \mathrm{h} / 2 \pi\)
For d orbital = \(\sqrt{(2 × 3)} \mathrm{h} / 2 \pi\) = \(\sqrt{6} \mathrm{h} / 2 \pi\).

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 13.
What is the maximum numbers of electrons that can be associated with the following set of quantum numbers ? n = 3,l = 1 and m = -1
(a) 4
(b) 6
(c) 2
(d) 10
Answer:
(c) 2
Solution:
n = 3; l = 1; m = -1 either 3px or 3py

Question 14.
Assertion: The number of radials and angular nodes for 3p orbital are l, l respectively.
Reason: The number of radials and angular nodes depends only on the principal quantum number.
(a) both assertion and reason are true and the reason is the correct explanation of assertion.
(b) both assertion and reason are true but the reason is not the correct explanation of assertion.
(c) the assertion is true but the reason is false
(d) both assertion and reason are false
Answer:
(c) assertion is true but reason is false
Solution:
No. of radial node = n- l – 1
No. of angular node = l for 3p orbital
No. of angular node = l = 1
No. of radial node = n – l – 1 = 3 – 1 – 1 = 1.

Question 15.
The total number of orbitals associated with the principal quantum number n = 3 is ………..
(a) 9
(b) 8
(c) 5
(d) 7
Answer:
(a) 9
Solution:
n = 3; l = 0; m1 = 0 – one s orbital n = 3; l = 1; m1 = -1, 0, 1 – three p orbitals n = 3; l = 2; m1 = -2, -1, 0, 1, 2 – five d orbitals, overall nine orbitals are possible.

Question 16.
If n = 6, the correct sequence for filling of electrons will be, …………
(a) ns → (n – 2) f → (n – 1)d → np
(b) ns → (n – 1) d → (n – 2) f → np
(c) ns → (n – 2) f → np → (n – 1) d
(d) none of these are correct
Answer:
(a) ns → (n – 2)f → (n – l)d → np
Solution:
n = 6 According Aufbau principle,
6s → 4f → 5d → 6p
ns → (n – 1)f → (n – 2)d → np.

Question 17.
Consider the following sets of quantum numbers:
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
Which of the following sets of quantum numbers is not possible?
(a) (i), (ii), (iii) and (iv)
(b) (ii), (iv) and (v)
(c) (z) and (iii)
(d) (ii), (iii) and (iv)
Answer:
(b) (ii), (iv) and (v)
Solution:
(ii) l can have the values from 0 to n – 1 n = 2; possible l values are 0, 1 hence l = 2 is not possible.
(iv) for l = 0; m = -1 not possible
(v) for n = 3 l = 4 and m = 3 not possible.

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 18.
How many electrons in an atom with atomic number 105 can have (n + 1) = 8?
(a) 30
(6) 17
(c) 15
(d) unpredictable
Answer:
(b) 17
Solution:
n + 1 = 8
Electronic configuration of atom with atomic number 105 is [Rn] 5f14 6d3 7s2

Orbital (n+1) No. of electrons
5f 5 + 3 = 8 14
6d 6 + 2 = 8 3
7s 7 + 0 = 0 2
                 No. of electrons = 14 + 3 = 17

Question 19.
Electron density in the yz plane of 3 dx2-y2 orbitals is …………….
(a) zero
(b) 0.50
(c) 0.75
(d) 0.90
Answer:
(a) zero
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 20.
If uncertainty in position and momentum are equal, then minimum uncertainty in velocity is ……….
(a) \(\frac{1}{m} \sqrt{\frac{h}{\pi}}\)
(b) \(\sqrt{\frac{\mathrm{h}}{\pi}}\)
(c) \(\frac{1}{2 m} \sqrt{\frac{h}{\pi}}\)
(d) \(\frac{\mathrm{h}}{4 \pi}\)
Answer:
(c) \(\frac{1}{2 m} \sqrt{\frac{h}{\pi}}\)
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 21.
A macroscopic particle of mass 100 g and moving at a velocity of 100 cm s-1d will have a de Broglie wavelength of ………….
(a) 6.6 x 10-29 cm
(b) 6.6 x 10-30 cm
(c) 6.6 x 10-31 cm
(d) 6.6 x 10-32 cm
Answer:
(c) 6.6 x 10-31 cm
Solution:
m = 100 g = 100 x 10-3 kg
v = 100 cm s-1 = 100 x 10-2 m s-1
λ = \(\frac{h}{mv}\) =Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
= 6.626 x 10-31 ms-1
= 6.626 x 10-31 cm s-1

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 22.
The ratio of de Broglie wavelengths of a deuterium atom to that of an α-particle, when the velocity of the former is five times greater than that of later, is ……………
(a) 4
(b) 0.2
(c) 2.5
(d) 0.4
Answer:
(d) 0.4

Question 23.
The energy of an electron in the 3rd orbit of a hydrogen atom is -E. The energy of an electron in the first orbit will be ……………..
(a) – 3E
(b) – E /3
(c) – E / 9
(d) – 9E
Answer:
(c) – E / 9
Solution:
En  = \(\frac{-13.6}{n^{2}}\) eV atom-1
E1 = \(\frac{-13.6}{1^{2}}\)13.6 = \(\frac{-13.6}{9}\)
Given that,
E3 = – E
\(\frac{-13.6}{9}\) = -E
13.6 = – 9E = E1 = – 9E
E1 = – 9E

Question 24.
Time independent Schrodinger wave equation is ………….
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
Answer:
(a) \(\widehat{\mathrm{H}} \psi=\mathrm{E} \psi\).

Question 25.
Which of the following does not represent the mathematical expression for the Heisenberg uncertainty principle?
(a) ∆E.∆p ≥ h/4π
(b) ∆E.∆v ≥ h/4πm
(c) ∆E.∆t ≥ h/4π
(d) ∆E.∆x ≥ h/4π
Answer:
(d) ∆E.∆x ≥ h/4π.

 II. Write brief answer to the following questions

Question 26.
Which quantum number reveals information about the shape, energy, orientation, and size of orbitals?
Answer:
The information about the shape, energy, orientation, and size of the orbitals are respectively given by Azimuthal, spin, magnetic, and principal quantum numbers.

Question 27.
How many orbitals are possible for n = 4?
Answer:
If n = 4, the possible number of orbitals are calculated as follows –
n = 4, main shell = N
If n = 4, l values are 0, 1, 2, 3
If l = 0,  4s orbital = 1 orbital
If l = 1,  m = -1,0, +1 = 3 orbitals
If l = 2,  m = -2,-1,0, +1,+2 = 5 orbitals
If l = 3,  m = -3,-2,-1,0, +1,+2,+3 = 7 orbitals
∴ Total number of orbitals = 16 orbitals

Question 28.
How many radial nodes for 2s, 4p, 5d, and 4f orbitals exhibit? How many angular nodes?
Answer:
The formula for the total number of nodes = n – 1

1. For 2s orbital: Number of radial nodes =1.

2. For 4p orbital: Number of radial nodes = n – l – 1. = 4 – 1 – 1 = 2
Number of angular nodes = l
∴ Number of angular nodes = 1
So, 4p orbital has 2 radial nodes and 1 angular node.

3. For 5d orbital:
Total number of nodes = n – 1 = 5 – 1 = 4 nodes
Number of radial nodes = n – l – 1 = 5 – 2 – 1 = 2 radial nodes.
Number of angular nodes = l = 2
∴ 5d orbital have 2 radial nodes and 2 angular nodes.

4. For 4f orbital:
Total number of nodes = n – 1 = 4 – 1 = 3 nodes
Number of radial nodes = n – 7 – 1 = 4 – 3 – 1 = 0 node.
Number of angular nodes = l = 3 nodes
∴ 4f orbital have 0 radial node and 3 angular nodes.

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 29.
The stabilization of a half-filled d-orbital is more pronounced than that of the p-orbital why?
Answer:
The half-filled orbitals are more stable due to symmetry and exchange energy. In the case of half-filled d-orbitals, there are ten possible exchanges whereas in p-orbitals three possible exchanges only, If more exchanges are possible, more exchange energy is released and more stable. Hence, the stabilization of a half-filled d-orbital is more pronounced than that of the p-orbital.

Question 30.
Consider the following electronic arrangements for the d5 configuration.
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
(1) Which of these represents the ground state
(2) Which configuration has the maximum exchange energy.
Answer:
(1) Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom – This represents the ground state.
(2)  Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom-65 – This represents the maximum exchange energy.

Question 31.
State and explain Pauli’s exclusion principle.
Answer:
Pauli’s exclusion principle states that “ No two electrons in an atom can have the same set of values of all four quantum numbers”. It means that each electron must have unique values for the four quantum numbers.

For the lone electron present in hydrogen atom, the four quantum numbers are: n = 1, l = 0, m = 0 and s = +1/2. For the two electrons present in helium, one electron has the quantum numbers same as the electron of the hydrogen atom,
n = 1, l = 0, m = 0, and s = +1/2

For other electron, the fourth quantum number is different, i.e., n = 1, l = 0, m = 0 and s = -1/2.

Question 32.
Define orbital? What are the n and l values for 3px and 4 dx2-y2 electron?
Answer:
(i) Orbital is a three-dimensional space in which the probability of finding the electron is maximum.
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 33.
Explain briefly the time-independent Schrodinger wave equation?
Answer:
The time independent Schrodinger equation can be expressed as
\(\widehat{\mathrm{H}}\) Ψ = EΨ ……………(1)
Where \(\widehat{\mathrm{H}}\) is called Hamiltonian operator.
Ψ is the wave function.
E is the energy of the system.
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
Since Ψ is a function of position coordinates of the particle and is denoted by Ψ (x, y, z)
∴ Equation (1) can be written as,
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
Multiply the equation (3) by \(\widehat{\mathrm{H}}\) and rearranging
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
The above equation (4) Schrodinger wave equation does not contain time as a variable and is referred to as time-independent Schrodinger wave equation.

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 34.
Calculate the uncertainty in position of an electron, if ∆v = 0.1% and n = 2.2 x 106 ms-1.
Answer:
Mass of an electron = m = 9.1 x 10-31 kg.
∆v = Uncertainty in velocity = \(\frac {0.1}{100}\) x 2.2 x 103 ms-1 .
∆v = 0.22 x 104 = 2.2 x 103 ms-1
∆x . ∆v . m = \(\frac {h}{4π}\)
∆x = \(\frac {h}{∆v . m x 4π}\)
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
= 0.02635 x 10-6
∆x = 2.635 x 10-8
Uncertainty in position = 2.635 x 10-8.

Question 35.
Determine the values of all the four quantum numbers of the 8th electron in O – atom and 15th electron in Cl atom and the last electron in chromium.
Answer:
Electronic configuration of Oxygen (Atomic Number = 8) is 1s2 2s2 2py2 2pz1 2px1
The eighth electron is present in 2px orbital and the quantum numbers are
n = 2, l = 1, m = +1 or -1 and s = +1/2.

Electronic configuration of Chlorine (Atomic Number = 17) is 1s2 1s2 2p6 3s2 3p5. Therefore, 15th (last) electron is present in 3pz orbital and the quantum numbers are n = 32, l = 1, m = +1 or -1 and s = +1/2

Electronic configuration of Chromium (Atomic Number = 24) is 1s2 2s2 2p6 3s2 3p6 3d5 4s1. Therefore, last electron is present in 4s orbital and the quantum numbers are n = 4, l = 0, m = 0 and s = +1/2.

Question 36.
The quantum mechanical treatment of the hydrogen atom gives the energy value:
En = \(\frac{-13.6}{n^{2}}\) eV atom-1

  1. use this expression to find ∆E between n = 3 and n = 4
  2. Calculate the wavelength corresponding to the above transition.

Answer:
(1) When n = 3
E3 = \(\frac{-13.6}{3^{2}}\) = \(\frac {-13.6}{9}\) = – 1.511 eV atom-1
When n = 4 E4 = \(\frac{-13.6}{4^{2}}\) = – 0.85 eV atom-1
∆E = E4 – E3 = – 0.85 – (-1.511) = + 0.661 eV atom
∆E = E3 – E4
= – 1.511 – (-0.85)
= – 0.661 eV atom-1

(2) Wave length = λ
∆E = \(\frac {hc}{ λ}\)
λ = \(\frac {hc}{∆E}\)
h = Planck’s constant = 6.626 x 10-34 Js-1
c = 3 x 108 m/s
λ = \(\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{0.661}\)
= 10.02 x 10-34 x 3 x 108
= 30 x 10-26
λ = 3 x 10-25 m.

Question 37.
How fast must a 54 g tennis ball travel in order to have a de Broglie wavelength that is equal to that of a photon of green light 5400 Å?
Answer:
m = mass of tennis ball = 54 g = 5.4 x 10-2 kg.
λ = de Broglie wavelength = 5400 Å. = 5400 x 10-10 m.
V = velocity of the ball = ?
λ = \(\frac {h}{mV}\)
V = \(\frac {h}{λ.m}\)
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
= 0.2238 x 10-24
= 2.238 x 10-25 m.

Question 38.
For each of the following, give the sub level designation, the allowable m values and the number of orbitals.

  1. n = 4, l = 2,
  2. n = 5, l = 3
  3. n = 7, l = 0

Answer:
1. n = 4, l = 2
If l = 2, ‘m’ values are -2, -1, 0, +1, +2
So, 5 orbitals such as dxy,dyz,dxz,\(d_{x^{2}-y^{2}}\) and dz

2. n = 5 , l = 3
If l = 3, ‘m’ values are -3, -2, -1, 0, +1, +2, +3
So, 7 orbitals such as f z, fxz, fyz, fxyz, fz(x2 y2)’ ^x(x2-3y2)’ ^y(3×2 ??y

3. n = 7 , l = 0
If l = 0, ‘m’ values are 0. Only one value.
So, 1 orbital such as 7s orbital.

Question 39.
Give the electronic configuration of Mn2+ and Cr3+
Answer:
Mn (z = 25). Electronic configuration of
Mn2+ = 1s2 2s2 2p6 3s2 3p6 4s0 3d5.

Cr (z =24) Electronic configuration of
Cr3+ = 1s2 2s2 2p6 3s2 3p6 4s0 3d3.

Question 40.
Describe the Aufbau principle.
Answer:
In the ground state of the atoms, the orbitals are filled in the order of their increasing energies. That is the electrons first occupy the lowest energy orbital available to them. Once the lower energy orbitals are completely filled, then the electrons enter the next higher energy orbitals.
The order of filling of various orbitals as per the Aufbau principle is –
1 s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d ………..
For e.g., K (Z =19)
The electronic configuration is 1s2 2s2 2p6 3s2 3p6 4s1.
After filling 4s orbital only we have to fill up 3d orbital.

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 41.
An n atom of an element contains 35 electrons and 45 neutrons. Deduce

  1. the number of protons
  2. the electronic configuration for the element
  3. All the four quantum numbers for the last electron

Answer:
An element X contains 35 electrons, 45 neutrons

  1. The number of protons must be equal to the number of electrons. So the number of protons = 35.
  2. Number of electrons = 35. So the electronic configuration is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5.
  3. The last electron i.e. 5th electron in 4p orbital has the following quantum numbers. n = 4, l = 1, m =+1, s = + \(\frac {1}{2}\)

Question 42.
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the nucleus.
Answer:
In order for the electron wave to exist in phase, the circumference of the orbit should be an integral multiple of the wavelength of the electron wave. Otherwise, the electron wave is out of phase.
mvr = nh / 2π, 2πr = nλ,
where mvr = angular momentum
where 2πr = circumference of the orbit
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
n = 3, n = 4

Question 43.
Calculate the energy required for the process.
He+(g) → He2+(g) + e
The ionization energy for the H atom in its ground state is – 13.6 eV atom-1.
Answer:
The ionization energy for the H atom in its ground state =-13.6 eV atom-1.
Ionization energy = \(\frac{13.6 z^{2}}{n^{2}}\) eV
Z = atomic number
n = principal quantum number or shell number
For He, n = 1, z = 2
IE = \(\frac{-13.6 \times 2^{2}}{1^{2}}\)eV.

Question 44.
An ion with mass number 37 possesses a unit negative charge. If the ion contains 11.1% more neutrons than electrons. Find the symbol of the ion.
Answer:
Mass number (A) = Number of Protons + Nnumber of neutrons = 37.
Number of neutrons = Number of electrons (x) + 11.1 % of x.
Given:
(x – 1) ion + 1.111x = 37.
x = \(\frac{38}{2}\) . 11 = 18 = 18.
Atomic Number(z) = 18 – 1 = 17.
Symbol of the ion is \({ }_{17}^{37} X\).

Question 45.
The Li2+ ion is a hydrogen-like ion that can be described by the Bohr model. Calculate the Bohr radius of the third orbit and calculate the energy of an electron in 4th orbit.
Answer:
Li2+ hydrogen-like ion.
Bohr radius of the third orbit = r3 = ?
r3 = \(\frac{(0.529) n^{2}}{Z}\) A
Where n = shell number, Z = atomic number.
r3 = \(\frac{(0.529) 3^{2}}{3}\) A [∴for lithium Z = 3, n = 3]
= \(\frac{0.529 x 9}{3}\)
r3 = l.587Å
En = \(\frac{(-13.6) Z^{2}}{n^{2}}\) eV atom-1.
E4 = Energy of the fourth orbit = ?
E4 = \(\frac{(-13.6) \times 3^{2}}{4^{2}}\) = \(\frac{-13.6 \times 9}{16}\) = -7.65 eV atom-1
E4 = – 7.65 eV atom-1

Question 46.
Protons can be accelerated in particle accelerators. Calculate the wavelength (in Å)of such accelerated proton moving at 2.85 × 108 ms-1 (the mass of proton is 1.673 x 10-27 Kg).
Answer:
m = mass of the proton = 1.673 x 10-27 Kg
v = velocity of the proton = 2.85 x 108 ms-1
λ = \(\frac{h}{mv}\)
h = Planck’s constant = 6.626 x 1034 Kg m2 s-1
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
Wavelength of proton = λ = 1.389 x 10-15 m.

Question 47.
What is the de Broglie wavelength (in cm) of a 160g cricket ball travelling at 140 Km hr-1.
Answer:
m = mass of the cricket ball = 160g = 0.16 kg.
v = velocity of the cricket ball =140 Km h-1
= \(\frac {140 x 5}{18}\) = 38.88 ms-1
de Broglie equation = λ = \(\frac{h}{mv}\)
h = Planck’s constant = 6.626 x 10-34 kg m2 s-1
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
λ = 1.065 x 10-34m
Wave length in cm = 1.065 x 10-34 x 100
= 1.065 x 10-32 cm.

Question 48.
Suppose that the uncertainty in determining the position of an electron in an orbit is 0.6 A. What is the uncertainty in its momentum?.
Answer:
∆x = uncertainty in position of an electron = 0.6 Å = 0.6 x 10-10 m.
∆p = uncertainty in momentum = ?
Heisenberg’s uncertainty principle states that,
∆x.∆p ≥ \(\frac{h}{4π}\)
∆p = \(\frac{h}{4π.∆x}\)
h = Planck’s constant = 6.626 x 10-34 kg m2 s-1
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
Uncertainty in momentum = 0.8792 x 10-24 kg ms-1 (or) = 8.792 x 10-25 kg ms-1

Question 49.
Show that if the measurement of the uncertainty in .the location of the particle is equal to its de Broglie wavelength, the minimum uncertainty in its velocity is equal to its velocity /4π
Answer:
If, uncertainty in position = ∆x = λ , the value of uncertainty in velocity = \(\frac{v}{4π}\)
Heisenberg’s principle states that
∆x.∆v. m = \(\frac{h}{4π}\) …………(1)
de Broglie equation states that
λ = \(\frac{h}{mv}\) ………….(2)
∴ h = λ .m.v …………(3)
∆x = \(\frac{h}{∆v.4π}\) ………….(4)
Substituting the value of h in equation (4)
∆x = \(\frac{λ x m. v}{∆v.4π.m}\)
if ∆x = λ
∆v = Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom = \(\frac{v}{4π}\)

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 50.
What is the de Broglie wave length of an electron, which is accelerated from the rest, through a potential difference of 100V?
Answer:
Potential difference = V = 100 V
Potential energy = eV = 1.609 x 10-19c x 100V
\(\frac{v}{4π}\) m v2 = 1.609 x 10-19 x 100
\(\frac{v}{4π}\) m v2 = 1.609 x 10-19 J
v2 = \(\frac{2 \times 1.609 \times 10^{-17}}{m}\)
m = mass of electron = 9.1 x 10-31 Kg
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
v = 5.93 x 106 m/s
λ = \(\frac{h}{mv}\) where h = 6.62 x 10-34 JS
= \(\frac{6.62 \times 10^{-34}}{9.1 \times 10^{-31} \times 5.93 \times 10^{6}}\)
= 1.2x 10-10m
A= 1.2 Å.

Question 51.
Identify the missing quantum numbers and the sub energy level
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Samacheer Kalvi 11th Chemistry Quantum Mechanical Model of Atom In-Text Questions – Evaluate Yourself

Question 1.
Calculate the de Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1 k eV.
Answer:
λ = \(\frac{h}{mv}\)
Potential difference of an electron = V = 1 keV.
Potential energy = \(\frac{1}{2}\) mv2 = eV
e = charge of an electron = 1.609 x 10-19c
l k V = 1000 V
:. Potential energy = 1.609 x 10-19 x 1000 = 1.609 x 10-19
\(\frac{1}{2}\) mv2 = 1.609 x 10-16V
m = 9.1 x 10-31 kg
λ = \(\frac{h}{mv}\)
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
= 1.2 x 10-11 m
λ = 1.2 x 10-11 m.

Question 2.
Calculate the uncertainty in the position of an electron, if the uncertainty in its velocity is 5.7 x 10 s ms-1.
Answer:
Uncertainty in velocity = Av = 5.7 x 105 ms-1
Mass of an electron = m = 9.1 x 10-31 kg.
Uncertainty in position = ∆x = ?
∆x.m.∆v = \(\frac{h}{4π}\)
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
= 1 x 10-10m.
Uncertainty in position = 1 x 10-10 m.

Question 3.
How many orbitals are possible in the 4th energy level? (n = 4)
Answer:
n = 4
Number of orbitals in 4th energy level = ?
When n = 4, l = 0,1,2,3
If l = 0 orbital = 4s =1
If l = 1 orbital = 4px, 4py, 4pz = 2
If l = 2 orbital = \(4 \mathrm{d}_{\mathrm{xy}}, 4 \mathrm{d}_{\mathrm{yz}}, 4 \mathrm{d}_{\mathrm{zx}}, 4 \mathrm{d}_{\mathrm{x}} 2_{\mathrm{y}}, 2,4 \mathrm{d}_{\mathrm{z}^{2}}\) = 5
If l = 3 orbital = -3,-2, -1, 0, +1, +2, +3 = 7
Number of orbitals in 4th energy level = 16.

Question 4.
Calculate the total number of angular nodes and radial nodes present in 3d and 4f orbitals.
Answer:
The number of angular nodes in 3d orbital =?
A number of radial nodes in 3d orbital =?
Number of angular nodes = l
Number of radial nodes = n – l – 1

1. For 3d orbital:
Number of angular nodes = 2 because l = 2
Number of radial nodes = 3 – 2 -1 = 0
Total number of nodes in 3d orbital = 2

2. For 4f orbital:
Number of angular nodes = 3 because l = 3
Number of radial nodes = n – l – l =4 – 3 – 1 = 0
Total number of nodes in 4f orbital = 3.

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 5.
The energy of an electron in a hydrogen atom in the ground state is -13.6 eV. What is the energy of the electron in the second excited state?
Answer:
Energy of an electron in ground state = -13.6 eV.
∴ The energy of an electron in the second excited state = E2.
n = 2
E2 = \(\frac{-13.6 \mathrm{eV}}{\mathrm{n}^{2}}\) = \(\frac{-13.6}{2^{2}}\) = \(\frac{-13.6}{4}\) = -3.4 eV.

Question 6.
How many unpaired electrons are present in the ground state of Fe3+ (z = 26), Mn2+ (z = 25) and argon (z=18)?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

1s2 2s2 2p6 3s2 3p6 3d6 4s2 for Fe atom.
1s2 2s2 2p6 3s2 3p6 3d6 3d5 for Fe3+ ion.
So, it contains 5 unpaired electrons.
Mn (Z = 25). Electronic configuration is 1s2 2s2 2p6 3s2 3p6 3d6 3d5
Mn → Mn2+ + 2e
Number of unpaired electrons in Mn2+ = 5
Ar (Z = 18). Electronic configuration is 1s2 2s2 2p6 3s2 3p6.
All orbitals are completely filled. So, no unpaired electrons in it.

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 7.
Explain the meaning of the symbol 4f2. Write all the four quantum numbers for these electrons.
Answer:
4f2 : It means that the element has 2 electrons in outermost 4f shell.
Quantum number values are, Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
n = principal quantum number = 4
l = azimuthal quantum number = 3
m = magnetic quantum number = – 3, -2
s = spin quantum number = +\(\frac {1}{2}\) – \(\frac {1}{2}\).

Question 8.
Which has the stable electronic configuration? Ni2+ or Fe3+
Answer:
Ni (Z = 28). 1s2 2s2 2p6 3s2 3p64s23d8
Ni2+ electronic configuration = Is2 2s22p6 3s2 3p6 3d8
Fe (Z = 26). 1s2 2s2 2p6 3s2 3p6 4s2 3d6
Fe3+ Is2 2s2 2p6 3s23p6 3d5
If d orbital is half filled, according to Aufbau principle, it is more stable. So Fe3+ is more stable than Ni2+.

Samacheer Kalvi 11th Chemistry Solutions Quantum Mechanical Model of Atom Additional Questions Solved

I. Choose the correct answer

Question 1.
Which of the following experiment proves the presence of an electron in an atom?
(a) Rutherford’s α-ray scattering experiment
(b) Davisson and Germer experiment
(c) J.J. Thomson cathode ray experiment
(d) G.R Thomson gold foil experiment
Answer:
(c) J.J. Thomson cathode ray experiment.

Question 2.
Consider the following statements regarding Rutherford’s α-ray scattering experiment.
i. Most of the α-particles were deflected through a small angle.
ii. Some of α-particles passed through the foil.
iii. Very few α-particles were reflected back by 180°.
Which of the above statements is/are not correct.
(a) i and ii
(b) ii and iii
(c) i and iii
(d) i ii and iii
Answer:
(a) i and ii.

Question 3.
Considering Bohr’s model which of the following statements is correct?
(a) The energies of electrons are continuously reduced in the form of radiation.
(b) The electron is revolving around the nucleus in a dynamic orbital.
(c) Electrons can revolve only in those orbits in which the angular momentum (mvr) of the electron must be equal to an integral multiple of h/2 π.
(d) In an atom, electrons are embedded like seeds in watermelon.
Answer:
(c) Electrons can revolve only in those orbits in which the angular momentum (mvr) of the electron must be equal to an integral multiple of h/2π.

Question 4.
The energy of an electron of hydrogen atom in 2nd main shell is equal to
(a) – 13.6 eV atom-1
(b) – 6.8 eV atom-1
(c) – 0.34 eV atom-1
(d) – 3.4 eV atom-1
Answer:
(d) -3.4 eV atom-1
Hints:
Energy of an electron in 2nd main shell = \(\frac{(-13.6) Z^{2}}{n^{2}}\); Z = 1, n = 2
E = \(\frac{-13.6}{2^{2}}\) = \(\frac{-13.6}{4}\) = -3.4 atom-1.

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 5.
The energy of an electron of Li2+ in the 3rd main shell is …………..
(a) – 1.51 eV atom-1
(b) – 6.8 eV atom-1
(c) + 1.51 eV atom-1
(d) – 3.4 eV atom-1
Answer:
(a) -1.51 eV atom-1
Hints:
E = \(\frac{(-13.6) Z^{2}}{n^{2}}\) eV atom-1
Li2+= H atom. So Z = 1, n = 3.
E = \(\frac{(-13.6) 1^{2}}{3^{2}}\) = \(\frac{-13.6}{9}\) = -1.51 eV atom-1

Question 6.
The energy of an electron of a hydrogen atom in the main shell in terms of U mold is
(a) – 1312.8 k J mol-1
(b) – 82.05 k J mol-1
(c) – 328.2 kJ mol-1
(d) – 656.4 k J mol-1
Answer:
(b) – 82.05 k J mol-1
Hints:
E = \(\frac{(-13.6) Z^{2}}{n^{2}}\) kJ mol-1 , Z = 1, n = 4
∴ E = \(\frac{-1312.8}{16}\) = -82.50 kJ mol-1

Question 7.
The Bohr’s radius of Li2 0f21d orbit is
(a) 0.529 Å
(b) 0.0753 Å
(c) 0.7053 Å
(d) 0.0529 Å
Answer:
(c) 0.7053 Å
Hints:
rn = \(\frac{(0.529) Z^{2}}{n^{2}}\)Å, n = 2, Z = 3(for Li2+)
r = \(\frac{(0.529) 3^{2}}{2^{2}}\) = \(\frac{0.529 x 4}{3}\) = 0.7053 Å.

Question 8.
The formula used to calculate the Boh’s radius is ………..
(a) rn = \(\frac{(-13.6) Z^{2}}{n^{2}}\) eV atom-1
(b) rn = \(\frac{(0.529) Z^{2}}{n^{2}}\) A
(c) rn= \(\frac{(-1312.8) Z^{2}}{n^{2}}\) kJ mol-1
(d) rn = \(\frac{(+1312.8) Z^{2}}{n^{2}}\) kJ mol-1
Answer:
(b) rn = \(\frac{(0.529) Z^{2}}{n^{2}}\) A.

Question 9.
Who proposed the dual nature of light to all forms of matter?
(a) John Dalton
(b) Neils Bohr
(c) Albert Einstein
(d) J.J. Thomson
Answer:
(c) Albert Einstein

Question 10.
dc Brogue equation is ………..
(a) E = h γ
(b) E = mc2
(c) γ = \(\frac{\mathrm{E}_{2}-\mathrm{E}_{1}}{\mathrm{h}}\)
(d) λ = \(\frac{h}{mv}\)
Answer:
(d) λ = \(\frac{h}{mv}\).

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 11.
The crystal used in Davison and Germer experiment is …………….
(a) nickel
(b) zinc suiphide
(c) gold foil
(d) NaCl
Answer:
(a) nickel.

Question 12.
Which one of the following is the time independent Schrodinger wave equation?
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 13.
Match the list-I and list-II correctly using the code given below the list.
List – I
A. Principal quantum number
B. Azimuthal quantum number
C. Magnetic quantum number
D. Spin quantum number

List – II
1. represents the directional orientation of orbital
2. represents the spin of the electron
3. represents the main shell
4. represents the subshell
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 14.
The maximum number of electrons that can be accommodated in N shell is …………..
(a) 8
(b) 18
(c) 32
(d) 36
Answer:
(c) 32
Hints:
Number of electrons in the main shell = 2n2 n = 4, for N shell.
∴ Maximum number of electrons in N shell = 2(4)2 = 32.

Question 15.
The maximum number of electrons that can be accommodated in f orbital is ………….
(a) 10
(b) 14
(c) 16
(d) 6
Answer:
(6) 14
Hints:
f orbital – l = 3.
Maximum number of electrons in sub shell = 2(2l + 1)
∴ For ‘f’ orbital, the maximum number of electrons = 2(2 x 3 + l) = 14.

Question 16.
When l = 0, the number of electrons that can be accommodated in the subshell is ……………..
(a) 0
(b) 2
(c) 6
(d) 8
Answer:
(b) 2
Hints:
If l = 0, number of electrons = (2l + 1)
= 2 (2 x 0 + 1) = 2.

Question 17.
Which one of the quantum numbers is used to calculate the angular momentum of an atom?
(a) n
(b) m
(c) l
(d) s
Answer:
(c) l

Question 18.
What is the formula used to calculate the angular momentum?
(a) \(\sqrt{l^{(l+1)}} \frac{h}{2 \pi}\)
(b) \(\frac{\mathrm{mvr}}{2 \pi}\)
(c) \(\frac{mvr}{2}\)
(d) m . ∆v
Answer:
(a) \(\sqrt{l^{(l+1)}} \frac{h}{2 \pi}\).

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 19.
Which of the following provides the experimental justification of magnetic quantum number?
(a) Zeeman effect
(b) Stark effect
(c) Uncertainty principle
(d) Quantum condition
Answer:
(a) Zeeman effect.

Question 20.
What are the values of n, l, m, and s for 3px electron?
(a) 3, 2, 1, 0
(b) 3, 1,-l, +½
(c) 3, 2, +1, -½
(d) 3, 0, 0, +½]
Answer:
(b) 3, 1, -1, +½
Hint:
3px electron ; n = 3 (main shell)
for px orbitaI, l = 1, m = -1, s = \(\frac {1}{2}\).

Question 21.
Identify the quantum number for \(4 d_{x^{2}-y^{2}}\) electron.
(a) 4, 2, -2, +½
(b) 4, 0, 0, +½
(c) 4, 3, 2, +½
(d) 4, 3, 2, -½
Answer:
(a) 4, 2, -2, +½.

Question 22.
How many orbitals are possible in 3rd energy level?
(a) 16
(6) 9
(c) 3
(d) 27
Answer:
(b) 9
Hints:
3rd energy level Number of orbitals = ?
n = 3 main shell = m
l = 0, 1,2 m = 0, -1,0, +1
Total = 9 orbitals.

Question 23.
The region where the probability density function of electron reduces to zero is called
(a) orbit
(b) orbital
(c) nodal surface
(d) subshell
Answer:
(c) nodal surface.

Question 24.
Consider the following statements.
(i) The region where the probability density of electron is zero, called nodal surface.
(ii) The probability of finding the electron is independent of the direction of the nucleus.
(iii) The number of radial nodes is equal to n + l + 1 Which of the above statements is/are correct?
(a) (i) and (iii)
(b) (i) and (ii)
(c) (iii) only
(d) (ii) and (iii)
Answer:
(b) (i) and (ii).

Question 25.
Match the list-I and List-II correctly using the code given below the list.
List-I
A. s – orbital
B. p – orbital
C. d – orbital
D. f – orbital

List-II
1. complex three-dimensional shape
2. symmetrical sphere
3. dumb-bell shape
4. cloverleaf shape
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 26.
Which one of the following is the correct increasing order of effective nuclear charge felt by an electron?
(a) s>p>d>f
(b) s<p<d<f
(c) s>p>f>d
(d) f<p<d<s
Answer:
(a) s>p>d>f.

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 27.
The value of n, l, m and s of 8th electron in an oxygen atom are respectively
(a) 1, 0, 0, + ½
(b) 2, 1, +1, – ½
(c) 2, 1, -1, – ½
(d) 2, 1, 0, +½
Answer:
(a) 2, 1, +1, – ½.

Question 28.
The number of impaired electrons in carbon atom in the gaseous state is
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(c) 2

Question 29.
Which of the following is not used in writing electronic configuration of an atom?
(a) Aufbau principle
(b) Hund’s rule
(c) Pauli’s exclusion principle
(d) Heisenberg’s uncertainty principle
Answer:
(d) Heisenberg’s uncertainty principle.

Question 30.
Which of the following is the expected configuration of Cr (Z = 24)?
(a) 1s2 2s2 2p6 3s2 3p6 3d4 4s2
(b) 1s2 2s2 2p6 3s2 3p6 3d5 4s1
(c) 1s2 2s2 2p6 3s2 3p6 3d6
(d) 1s2 2s2 2p6 3s2 3p6 3d5 4s3
Answer:
(a) 1s2 2s2 2p6 3s2 3p6 3d4 4s2

Question 31.
Which of the following is the actual configuration of Cr (Z = 24)?
(a) 1s2 2s2 2p6 3s2 3p6 3d4 4s2
(b) 1s2 2s2 2p6 3s2 3p6 3d5 4s1
(c) 1s2 2s2 2p6 3s2 3p6 3d6
(d) 1s2 2s2 2p6 3s2 3p6 3d5 4s3
Answer:
(b) 1s2 2s2 2p6 3s2 3p6 3d5 4s1

Question 32.
Assertion (A) : Cr with electronic configuration [Ar] 3d5 4s1 is more stable than [Ar] 3d4 4s1.
Reason(R ): Half-filled orbitals have been found to have extra stability than partially filled orbitals.
(a) A and R are correct and R is the correct explanation of A.
(b) A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) A and R are correct and R is the correct explanation of A.

Question 33.
Assertion (A): Copper (Z = 29) with electronic configuration [Ar] 4s1 3d10 is more stable than [Ar] 4s1 3d10.
Reason(R): Copper with [Ar] 4s2 3d9 is more stable due to symmetrical distribution and exchange energies of d electrons.
(a) A and R are correct and R is the correct explanation of A.
(b) A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) A and R are correct and R is the correct explanation of A.

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 34.
In a sodium atom (atomic number = 11 and mass number = 23) and the number of neutrons is …………..
(a) equal to the number of protons
(b) less than the number of protons
(c) greater than the number of protons
(d) none of these
Answer:
(c) greater than the number of protons.

Question 35.
The idea of stationary orbits was first given by …………
(a) Rutherford
(b) J.J. Thomson
(c) Nails Bohr
(d) Max Planck
Answer:
(c) Niels Bohr.

Question 36.
de Broglie equation is ……………
(a) λ = \(\frac {h}{mv}\)
(b) λ = \(\frac {hv}{m}\)
(c) λ = \(\frac {mv}{h}\)
(d) λ = hmv
Answer:
(a) λ = \(\frac {h}{mv}\).

Question 37.
The orbital with n = 3 and l = 2 is …………..
(a) 3s
(b) 3p
(c) 3d
(d) 3J
Answer:
(c) 3d

Question 38.
The outermost electronic configuration of manganese (at. no. = 25) is …………
(a) 3d5 4s2
(b) 3d6 4s1
(c) 3d7 4s0
(d) 3d6 4s2
Answer:
(a) 3d5 4s2

Question 39.
The maximum number of electrons in a sub-shell is given by the equation
(a) n2
(b) 2n2
(c) 2l – l
(d) 2l + 1
Answer:
(d) 2l + 1

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 40.
Which of the following statements is correct for an electron that has the quantum numbers n = 4 and m = -2.
(a) The electron may be in 2 p orbital
(b) The electron may be in 4 d orbital
(c) The electron is in the second main shell
(d) The electron must have spin quantum number as +\(\frac {1}{2}\).
Answer:
(b) The electron may be in 4d orbital.

Samacheer Kalvi 11th Chemistry Quantum Mechanical Model of Atom 2 – Marks Questions

Question 1.
What is a stationary orbit?
Answer:
The electron is revolving around the nucleus in a certain fixed circular path called the stationary orbit.

Question 2.
Explain the theory of electromagnetic radiation.
Answer:

  • The theory of electromagnetic radiation states that a moving charged particle should continuously lose its energy in the form of radiation.
  • Therefore, the moving electron in an atom should continuously lose its energy and finally collide with the nucleus resulting in the collapse of the atom.

Question 3.
Explain how the matter has dual character?
Answer:

  • Albert Einstein proposed that light has dual nature, i.e. like photons behave both like a particle and as a wave.
  • Louis de Broglie extended this concept and proposed that all forms of the matter showed dual character.
  • He combined the following two equations of the energy of which one represents wave character (hυ) and the other represents the particle nature (mc2).

Question 4.
Explain the significance of the de Broglie equation.
Answer:

  • X = \(\frac {h}{mv}\). This equation implies that a moving particle can be considered as a wave and a wave can exhibit the properties of a particle.
  • For a particle with high linear momentum (mv), the wavelength will be so small and cannot be observed.
  • For a microscopic particle such as an electron, the mass is of the order of 10-31 kg, hence the wavelength is much larger than the size of an atom and it becomes significant.
  • For the electron, the de Broglie wavelength is significant and measurable while for the iron ball it is too small to measure, hence it becomes insignificant.

Question 5.
How many electrons can be accommodated in the main shell l, m, and n?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 6.
How many electrons can be accommodated in the sub-shell s, p, d, f?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 7.
Why is de Broglie concept is insignificant for macroscopic particles?
Answer:
The de Broglie wavelength is too small to measure for macroscopic particles and hence, it becomes insignificant.

Question 8.
How many orbitals are possible in the 3rd energy level?
Answer:
n = 3, main shell is m.
Total number of orbitals in 3rd energy level = ?
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
Total number of orbitals = 9.

Question 9.
What are Ψ and Ψ2?
Answer:

  • Ψ itself has no physical meaning but it represents an atomic orbital.
  • Ψ2 is related to the probability of finding the electrons within a given volume of space.

Question 10.
What is meant by nodal surface?
Answer:

  • The region, where there is a probability density function, reduces to zero is called nodal surface or a radial node.
  • For ns orbital, (n-1) nodes are found in it.

Question 11.
Mention the shape of s, p, d orbitals.
Answer:

  • The shape of s – orbital – sphere
  • The shape of p – orbital – dumbbell
  • The shape of d – orbital – cloverleaf

Question 12.
What is the physical significance of ψ and ψ2?
Answer:
The wave function ψ itself has no physical meaning. However, the probability of finding the electron in a small volume dxdydz around a point (x, y, z) is proportional to |ψ(x, y, z)|2 dxdydz. |ψ(x, y, z)|2 is known as probability density and is always positive.

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 13.
Write the equation to calculate the energy of nth orbit.
Answer:
En = \(\frac{(-1312.8) Z^{2}}{n^{2}}\) kJ mol-1
Where Z = atomic number, n = principal quantum number.

Question 14.
what are degenerate orbitals?
Answer:

  • Three different orientations in space that are possible for a p-orbital. All the three p-orbitals, namely px, py and pz have same energies and are called degenerate orbitals.
  • In the presence of magnetic or electric field, the degeneracy is lost.

Question 15.
Energy of an electron in hydrogen atom in ground state is -13.6 eV. What is the energy of the electron in the third excited state?
Answer:
E1 = – 13.6 eV
E3 = \(\frac{-13.6}{n^{2}}\) Where n = 3
E3 = \(\frac{-13.6}{9}\) = 1.511 eV
Energy of the electron in the third excited state = 1.511 eV.

Question 16.
What are quantum numbers?
Answer:
The electron in an atom can be characterized by a set of four quantum numbers, namely principal quantum number(n), azimuthal quantum number (l), magnetic quantum number (m), and spin quantum number (s).

Question 17.
State Hund’s rule of maximum multiplicity.
Answer:
It states that electron pairing in the degenerate orbitals does not take place until all the available orbitals contain one electron each.

Question 18.
How many unpaired electrons are present in the ground state of –
1. Cr3+ (Z = 24)
2. Ne (Z = 10)
Answer:
1. Cr3+ (Z = 24) 1s2 2s2 2p6 3s2 3p6 3d5 4s1
Cr3+ – 1s2 2s2 2p6 3s2 3p6 3d4.
It contains 4 unpaired electrons.

2. Ne (Z = 10) 1s22s22p6. No unpaired electrons in it.

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 19.
What is meant by electronic configuration? Write the electronic configuration of N (Z = 7).
Answer:
The distribution of electrons into various orbitals of an atom is called its electronic configuration.
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 20.
Which is the actual configuration of Cr (Z = 24) Why?
Answer:
Cr (Z = 24) 1s22s22p6.
The reason for this is, Cr with 3d5 configuration is half filled and it will be more stable. Chromium has [Ar] 3d5 4s1 and not [Ar] 3d4 4s2 due to the symmetrical distribution and exchange energies of d electrons.

Question 21.
What is the actual configuration of copper (Z = 29)? Explain about its stability.
Answer:
Copper (Z = 29)
Expected configuration : 1s2 2s2 2p6 3s2 3p6 3d9 4s2
Actual configuration : 1s2 2s2 2p6 3s2 3p6 3d10 4s1
The reason is that fully filled orbitals have been found to have extra stability. Copper has the electronic configuration [Ar] 3d10 4s1 and not [Ar] 3d9 4s2 due the symmetrical distribution and exchange energies of d electrons. Symmetry leads to stability. The full filled configuration have symmetrical distribution of electrons and hence they are more stable than unsymmetrical configuration.

Samacheer Kalvi 11th Chemistry Quantum Mechanical Model of Atom 3 – Mark Questions

Question 1.
Write the observations of Rutherford’s α-ray scattering experiment.
Answer:
The observations of the α-ray scattering experiment are
(i) most of the α-particles passed through the foil
(ii) some of them were deflected through a small angle and
(iii) very few a -particles were reflected back by 180°.

Question 2.
What are the limitations of Bohr’s atom model?
Answer:

  • The Bohr’s atom model is applicable only to species having one electron such as hydrogen, Li2+ etc, and not applicable to multi-electron atoms.
  • It was unable to explain the splitting of spectral lines in the presence of a magnetic field (Zeeman effect) or an electric field (Stark effect).
  • Bohr’s theory was unable to explain why the electron is restricted to revolve around the nucleus in a fixed orbit in which the angular momentum of the electron is equal to nh / 2π.

Question 3.
Illustrate the significance of de Broglie equation with an iron ball and an electron.

  1. 6.626 kg iron ball moving with 10 ms-1.
  2. An electron moving at 72.73 ms-1.

Answer:
1. λiron ball = \(\frac{h}{mv}\)
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
= 1 x 10-35m

2. λiron ball = \(\frac{h}{mv}\)
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
= \(\frac{6.626}{662.6}\) x 10-3m = 1 x 105m
For an electron, the de Broglie wavelength is significant and measurable while for iron ball it is too small to measure, hence it becomes insignificant.

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 4.
What are the limitations of Bohr’s atom model?
Answer:
The Bohr’s atom model is applicable only to species having one electron such as hydrogen, Li2+, etc., and not applicable to multi-electron atoms. It was unable to explain the splitting of spectral lines in the presence of a magnetic field (Zeeman effect) or an electric field (Stark effect). Bohr’s theory was unable to explain why the electron is restricted to revolve around the nucleus in a fixed orbit in which the angular momentum of the electron is equal to \(\frac{n h}{2 \pi}\)

Question 5.
Bohr radius of 1st orbit of a hydrogen atom is 0.529 Å. Assuming that the position of an electron in this orbit is determined with the accuracy of 0.5% of the radius, calculate the uncertainty in the velocity of the electron in the hydrogen atom.
Answer:
Uncertainty in position = ∆x
= \(\frac{0.5}{100}\) x 0.529 Å
= \(\frac{0.5}{100}\) x 10-10 x 0.529 m
∆x = 2.645 x 10-13 m
From Heisenberg’s uncertainty principle,
∆x.∆p ≥ \(\frac{h}{4π}\)
∆x.m.∆p ≥ \(\frac{h}{4π}\)
∆v ≥ \(\frac{h}{∆x.m.4π}\)
∆v = Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
∆v = 2.189 x 108m.

Question 6.
State and explain Heisenberg’s principle.
Answer:
The dual nature of matter imposes a limitation on the simultaneous determination of position and velocity of a microscopic particle. Based on this, Heisenberg arrived at his uncertainty principle, which states that ‘It is impossible to accurately determine both the position as well as the momentum of a microscopic particle simultaneously’. The product of uncertainty in the measurement is expressed as follows.
∆x . ∆p ≥ \(\frac{h}{4 \pi}\)
where, ∆x and ∆p are uncertainties in determining the position and momentum, respectively. The uncertainty principle has negligible effect for macroscopic objects and becomes significant only for microscopic particles such as electrons.

Question 7.
Explain the azimuthal quantum number.
Answer:

  • It is represented by the letter 7′ and can take integral values from zero to n – 1, where n is the principal quantum number.
  • Each l value represents a subshell (orbital). l = 0, 1, 2, 3 and 4 represents the s, p, d, f and g orbitals respectively.
  • The maximum number of electrons that can be accommodated in a given subshell (orbital) is 2(2l + 1).
    It is used to calculate the orbital angular momentum using the expression Angular momentum = \(\sqrt{l(l+1)} \frac{h}{2 \pi}\).

Question 8.
Draw the shapes of 1s, 2s and 3s orbitals
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 9.
Explain how effective nuclear charge is related with stability of the orbital.
Answer:

  • In a multi-electron atom, in addition to the electrostatic attractive force between the electron and nucleus, there exists a repulsive force among the electrons.
  • These two forces are operating in the opposite direction. This results in the decrease in the nuclear force of attraction on electron.
  • The net charge experienced by the electron is called effective nuclear charge.
  • The effective nuclear charge depends on the shape of the orbitals and it decreases with increase in azimuthal quantum number l.
  • The order of the effective nuclear charge felt by a electron in an orbital within the given shell is s > p > d > f.
  • Greater the effective nuclear charge, greater is the stability of the orbital. Hence, within a given energy level, the energy of the orbitals are in the following order s < p < d < f.

Question 10.
Calculate the wavelength of an electron moving with a velocity of 2.05 x 107 ms-1.
Answer:
According to de Broglie’s equation, λ = \(\frac {h}{mv}\)
Mass of electron (m) = 9.1 x 10-31 kg
Velocity of electron (υ) = 2.05 x 107 ms-1
Planck’s constant (h) = 6.626 x 10-34 kg m2 s-1
λ = Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom =355 x 10-4m.

Question 11.
The mass of an electron is 9.1 x 10-31 kg. If its kinetic energy is 3.0 x 10-25 J, calculate its wavelength.
Answer:
Step I.
Calculation of the velocity of electron
Kinetic energy = 1 / 2 mυ2 = 3.0 x 10-25 kg m2 s-2
υ2Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom = 65.9 x 104 m2 s-2
υ = (65.9 x 104 m s-2) = 8.12 x 102 ms-1

Step II.
Calculation of wavelength of the electron
According to de Broglie’s equation,
λ = \(\frac {h}{mv}\) = Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
=  0.08967 x l0-5 m = 8967 x 10-10 m = 8967 Å (∴1Å = 10-10m).

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 12.
From the following sets of quantum numbers, state which are possible. Explain why the others are not possible.

  1. n = 0, l = 0, ml = 0, ms = + \(\frac {1}{2}\)
  2. n = 1, l = 0, ml = 0, ms = – \(\frac {1}{2}\)
  3. n = 1, l = 1, ml = 0, ms = + \(\frac {1}{2}\)
  4. n = 1, l = 0, ml = +1, ms= +\(\frac {1}{2}\)
  5. n = 3, l = 3, ml = -3, ms = + \(\frac {1}{2}\)
  6. n = 3, l = 1, ml = 0, ms = +\(\frac {1}{2}\)

Answer:

  1. The set of quantum numbers is not possible because the minimum value of n can be 1 and not zero.
  2. The set of quantum numbers is possible.
  3. The set of quantum numbers is not possible because, for n = 1,1 cannot be equal to 1. It can have 0 value.
  4. The set of quantum numbers is not possible because for l = 0, ml; cannot be +1. It must be zero.
  5. The set of quantum numbers is not possible because, for n = 3, l = 3.
  6. The set of quantum numbers is possible.

Question 13.
How many electrons in an atom may have the following quantum numbers?
(a) n = 4; ms = – ½
(b) n = 3, l = 0.
Answer:
(a) For n = 4
1 Total number of electrons = 2n2 = 2 x 16 = 32
Half out of these will have ms = – \(\frac {1}{2}\)
Total electrons with ms (-½) = 16.

(b) For n = 3
l = 0; m1 = 0, ms = + ½ – ½ (two e).

Question 14.
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.
Answer:
According to Bohr’s theory,
mυr = \(\frac {nh}{2π}\) (n = 1,2,3, …… so on)
or 2πr = \(\frac {nh}{mυ}\) or mυ = \(\frac {nh}{2πr}\) ………..(i)
According to de Brogue equation,
λ = \(\frac {h}{mυ}\) or mυ = \(\frac {h}{λ}\) ……….(ii)
Comparing (i) and (ii),
\(\frac {nh}{2πr}\) = \(\frac {h}{λ}\) or 2πr = nλ
Thus, the circumference (2πr) of the Bohr orbit for hydrogen atom is an into the de Broglie wave length.

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 15.
An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign symbol to the ion.
Answer:
Let the no. of electrons in the ion = x
the no. of the protons = x + 3 (as the ion has three units positive charge)
and the no. of neutrons = x + \(\frac {30.4x }{100}\) = x + 0.304 x
Now, mass number of ion = Number of protons + Number of neutrons
= (x + 3) + (x + 0.304 x)
∴ 56 = (x + 3) + (x + 0.304 x) or 2.304 x = 56 – 3 = 53
x = \(\frac {53}{2.304}\) = 23
Atomic number of the ion (or element) = 23 + 3 = 26
The element with atomic number 26 is iron (Fe) and the corresponding ion is Fe3+.

Question 16.
The uncertainty in the position of a moving bullet of mass 10 g is 10 s m. Calculate the uncertainty in its velocity?
Answer:
According to uncertainty principle,
∆x.m∆υ = \(\frac {h}{4π}\) or ∆υ = \(\frac {h}{4πm∆x}\);
h = 6.626 x 1o-34 kg m2 s-1; m = 10 g = 10-2 kg
∆x = 10-5m; ∆v = Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom = 5.27 x 10-28mv
= 1.6 x 10-15 kg m2 s-15
Or
\(\frac {1}{2}\) mv2 = 1.6 x 10-15kg m2s-2
v = Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom = 5.93 x 107m-1

Question 17.
The uncertainty in the position and velocity of a particle are 10-10 m and 5.27 × 10-24 ms-1 respectively. Calculate the mass of the particle.
Answer:
According to uncertainty principle.
∆x. m∆υ = \(\frac {h}{4π}\)
or
m = \(\frac {h}{4π∆x.∆υ}\);
h = 6.626 x 10-34 kg m2 s-1
∆x = 10-10 m; ∆x = 5.27 x 10-24ms-1
m Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom = 0.1 kg.

Question 18.
With what velocity must an electron travel so that its momentum Is equal to that of a photon of wave length = 5200 A?
Answer:
According to de Brogue equation, λ = \(\frac {h}{mv}\)
Momentum of electron, mv = \(\frac {h}{λ}\) Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
= 1.274 x 10-27 kg ms-1 ………(i)
The momentum of electron can also be calculated as = mv = (9.1 x 10-31kg) x v ………(ii)
Comparing (i) and (ii)
(9.1 X 10-31kg) v = (1.274 x 10-27 kg ms-1)
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom = 1.4 x 103 ms-1

Question 19.
Using Aufbau principle, write the ground state electronic configuration of following atoms.

  1. Boron (Z = 5)
  2. Neon (Z = 10)
  3. Aluminium (Z = 13)
  4. Chlorine (Z = 17)
  5. Calcium (Z = 20)
  6. Rubidium (Z = 37)

Answer:

  1. Boron (Z = 5) ; 1s2 2s2 2p1
  2. Neon (Z = 10) ; 1s2 2s2 2p6
  3. Aluminium (Z = 13) ; 1s2 2s2 2p6 3s2 3p1
  4. Chlorine(Z = 17) ; 1s2 2s2 2p6 3s2 3p5
  5. Calcium (Z = 20) ; 1s2 2s2 2p6 3s2 3p6 4s2
  6. Rubidium (Z = 37) ; 1s2 2s22p6 3s2 3p63d10 4s2 4p6 5s1

Question 20.
Calculate the de Broglie wavelength of an electron moving with 1 % of the speed of light?
Answer:
According to de Brogue equation, A = \(\frac {h}{mv}\)
Mass of electron = 9.1 x 10-31 kg; Planck’s constant 6.626 x 10-34 kg m2 s-1
Velocity of electron = 1% of speed of light = 3.0 x 108 x 0.01 = 3 106 ms-1
Wavelength of electron (λ) = \(\frac {h}{mv}\) Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
= 2.43 x 10-10m.

Question 21.
What is the wavelength for the electron accelerated by 1.0 X i0 volts?
Answer:
Step I.
Calculation of the velocity of electron
Energy (kinetic energy) of electron = 1.0 x 104 volts.
= 1.0 x 104 x 1.6 x 10-19 J = 1.6 x 10-15J.

Step II.
Calculation of wavelength of the electron
According to de Broglie equation,
λ = \(\frac {h}{mυ}\); λ  Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
= 1.22 x 10-11m .

Question 22.
In a hydrogen atom, the energy of an electron in first Bohr’s orbit is 13.12 x 105 J mol-1. What is the energy required for its excitation to Bohr’s second orbit?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
The energy required for the excitation is:
∆E = E2 – E1 = (-3.28 x l05) – (- 13.12 x 105) = 9.84 x 105 J mol-1

Question 23.
Dual behavior of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 x 106 ms-1, calculate de Broglie wavelength associated with this electron.
Answer:
λ = \(\frac {h}{mυ}\); λ = Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
= 0.455 x 10-34 + 25 m = 0.455 nm = 455 pm.

Question 24.
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the symbol to the element.
Answer:
An element can be identified by its atomic number only. Let us find the atomic number.
Let the number of protons = x
∴ Number of neutrons = x + \(\frac {x × 31.7}{100}\) = (x + 0.317 x)
Now, Mass no. of element = No. of protons + No. of neutrons
81 = x + x + 0.317 x = 2.317 x
Or
x = \(\frac {81}{2.317}\) = 35
∴ No. of protons = 35, No. of neutrons = 81 – 35 = 46
Atomic number of element (Z) = Number of protons = 35
The element with atomic number (Z) 35 is bromine 8135Br.

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 25.
The electron energy in hydrogen atom is given by En = (- 2.18 × 10-18) / n2 J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?
Answer:
Step I.
Calculation of energy required
The energy required is the difference in the energy when the electron jumps from orbit with
n = ∞ to orbit with n = 2.
The energy required (∆E) = E – E2
= 0 – \(\left(-\frac{2.18 \times 10^{-18}}{4} \mathrm{J}\right)\) = 5.45 x 10-19 J.

Step II.
Calculation of the longest wavelength of light in cm used to cause the transition
∆E = hv = hc / λ.
λ = \(\frac {hc}{∆E}\) = Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atomn= 3.644 x 10-7
m = 3.644 x 10-7 x 102 = 3.645 x 10-5 cm.

Samacheer Kalvi 11th Chemistry Quantum Mechanical Model of Atom 5-Mark Questions

Question 1.
Describe about Bohr atom model.
Answer:
Assumptions of Bohr atom model.
1. The energies of electrons are quarantined

2. The electron is revolving around the nucleus in a certain fixed circular path called the stationary orbit.

3. Electron can revolve only in those orbits in which the angular momentum (mvr) of the electron must be equal to an integral multiple of h/2π
mvr = \(\frac {nh}{2π}\) where n = 1,2,3,…etc.,

4. As long as an electron revolves in a fixed stationary orbit, it doesn’t lose its energy. But if an electron jumps from a higher energy state (E2) to a lower energy state (E1), the excess energy is emitted as radiation. The frequency of the emitted radiation is E2 – E1= hv.
∴ v = \(\frac{\mathrm{E}_{2}-\mathrm{E}_{1}}{\mathrm{h}}\)
Conversely, when suitable energy is supplied to an electron, it will jump from lower energy orbit to a higher energy orbit.

5. Bohr’s postulates are applied to a hydrogen like atom (H, He+ and Li2+ etc..) the radius of the nth orbit and the energy of the electron revolving in the th orbit were derived.
rn = \(\frac{(0.529) n^{2}}{Z}\) A(0.529) n2
En = \(\frac{(-1 3.6) Z^{2}}{n}\) eV atom-1
En = \(\frac{(1312.8) Z^{2}}{n}\) kJ mol-1

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 2.
Derive de Brogue equation and give its significance.
Answer:
1. Louis de Brogue extended the concept of dual nature of light to all forms of matter. To quantify this relation, he derived an equation for the wavelength of a matter-wave.

2. He combined the following two equations of the energy of which one represents wave character (hu) and the other represents the particle nature (mc2).
Planck’s quantum hypothesis:
E = hv ……….(1)
Einsteins mass-energy relationship:
E = mc2 ………(2)
From (1) and (2)
hv = mc2
hc/λ = mc2
∴ λ = \(\frac{h}{mc}\) ………(3)
The equation (3) represents the wavelength of photons whose momentum is given by mc. (Photons have zero rest mass).

3. For a particle of matter with mass m and moving with a velocity y, the equation (3) can be written as λ = \(\frac{h}{mc}\) ………(4)

4. This is valid only when the particle travels at speed much less than the speed of Light.

5. This equation implies that a moving particle can be considered as a wave and a wave can exhibit the properties of a particle (i.e momentum).

6. Significance of de Brogue equation:
For a particle with high linear momentum, the wavelength will be too small and cannot be observed. For a microscopic particle such as an electron, the mass is 9.1 x 10-31 kg. Hence the wavelength is much larger than the size of atom and it becomes significant.

Question 3.
What are the main features of the quantum mechanical model of an atom.
Answer:
1. The energy of electrons in an atom is quarantined.

2. The existence of quarantined electronic energy levels is a direct result of the wave-like properties of electrons. The solutions of the Schrodinger wave equation gives the allowed energy levels (orbits).

3. According to Heisenberg’s uncertainty principle, the exact position and momentum of an electron cannot be determined with absolute accuracy. As a consequence, quantum mechanics introduced the concept of orbital. Orbital is a three-dimensional space in which the probability of finding the electron is maximum.

4. The solution of the Schrodinger wave equation for the allowed energies of an atom gives the wave function Ψ, which represents an atomic orbital. The wave nature of the electron present in an orbital can be well defined by the wave function Ψ.

5. The wave function Ψ itself has no physical meaning. However, the probability of finding the electron in a small volume dx, dy, dz around a point (x,y,z) is proportional to |Ψ (x,y,z)|2 dx dy dz |Ψ (x,y,z)|2 is known as probability density and is always positive.

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 4.
Explain about –
(1) Magnetic quantum number
(2) Spin quantum number
Answer:
(1) Magnetic quantum number:

  • It is denoted by the letter ml. It takes integral values ranging from – l to +l through 0.
    i.e. if l = 1; m = -1, 0 and +1.
  • The Zeeman Effect (the splitting of spectral lines in a magnetic field) provides the experimental justification for this quantum number.
  • The magnitude of the angular momentum is determined by the quantum number l while its direction is given by magnetic quantum number.

(2) Spin quantum number:

  • The spin quantum number represents the spin of the electron and is denoted by the letter ‘ms‘.
  • The electron in an atom revolves not only around the nucleus but also spins. It is usual to write this as electron spins about its own axis either in a clockwise direction or in anti-clockwise direction.
  • Corresponding to the clockwise and anti-clockwise spinning of the electron, maximum two values are possible for this quantum number.
  • The values of ‘ms‘ is equal to –\(\frac {1}{2}\) and +\(\frac {1}{2}\).

Question 5.
Explain about the shape of orbitals.
Answer:
Orbital: The solution to Schrodinger equation gives the permitted energy values called eigen values and the wave functions corresponding to the eigen values are called atomic orbitals.

Shape of orbital:
s – orbital:
For Is orbital, l = 0, m = 0, f(θ) = 1√2 and g(φ) = 1/√2π. Therefore, the angular distribution function is equal to 1/√2π. i.e. it is independent of the angle θ and φ. Hence, the probability of finding the electron is independent of the direction from the nucleus. So, the shape of the s orbital is spherical.
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

p – orbital:
For p orbitals l = 1 and the corresponding m values are -1, 0 and +1. The three different m values indicates that there are three different orientations possible for p orbitals. These orbitals are designated as px, py and pz. The shape of p orbitals are dumb bell shape.
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

d – orbital:
For ‘d’ orbital 1 = 2 and the corresponding m values are -2, -1, 0, +l,+2. The shape of the d orbital looks like a clover leaf. The five m values give rise to five d orbitals namely dxy, dyz, dzx, dx2-y2 and dz2 The 3d orbitals contain two nodal planes.
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

f – orbital
For f orbital, 1 = 3 and the m values are -3, -2,-1, 0, +1, +2, +3 corresponding to seven f orbitals, \(\mathrm{f}_{\mathrm{z}^{3}}, \mathrm{f}_{\mathrm{xz}^{2}}, \mathrm{f}_{\mathrm{yz}^{2}}, \mathrm{f}_{\mathrm{xyz}}, \mathrm{f}_{\mathrm{z}\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right)}, \mathrm{f}_{\mathrm{x}\left(\mathrm{x}^{2}-3 \mathrm{y}^{2}\right)}, \mathrm{f}_{\mathrm{y}\left(3 \mathrm{x}^{2}-\mathrm{y}^{2}\right)}\). They contain 3 nodal planes.
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Question 6.
What is exchange energy? How it is related with stability of atoms? Explain with suitable examples.
Answer:
1. If two or more electrons with the same spin are present in degenerate orbitals, there is a possibility for exchanging their positions. During exchange process, the energy is released and the released energy is called exchange energy.

2. If more number of exchanges are possible, more exchange energy is released. More number of exchanges are possible only in the case of half filled and fully filled configurations.

3. For example, in chromium, the electronic configuration is [Ar]3d5 4s1. The 3d orbital is half filled and there are ten possible exchanges.
Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

4. On the other hand only six exchanges are possible for [Ar] 3d4 4s2 configuration.

5. Hence, exchange energy for the half filled configuration is more This increases the stability of half filled 3d orbitals.

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Students can Download Chemistry Chapter 4 Hydrogen Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Chemistry Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Samacheer Kalvi 11th Chemistry Hydrogen Textual Evaluation Solved

I. Choose The Correct Answer:
Question 1.
Which of the following statements about hydrogen is incorrect ? (NEET – 2016)
(a) Hydrogen ion, H3O+ exists freely in solution.
(b) Dihydrogen acts a,s a reducing agent.
(c) Hydrogen has three isotopes of which tritium is the most common.
(d) Hydrogen never acts as a cation in ionic salts.
Answer:
(c) Hydrogen has three isotopes of which tritium is the most common.
Hint:
Correct statement:
Hydrogen has three isotopes of which protium is the most common.

Question 2.
Water-gas is ………..
(a) H2 O(g)
(b) CO + H2O
(C) CO + H2
(d) CO + N2
Answer:
(c) CO + H2

Question 3.
Which one of the following statements is incorrect with regard to ortho and para dihydrogen?
(a) They are nuclear spin isomers
(b) Ortho isomer has zero nuclear spins whereas the para isomer has one nuclear spin
(c) The para isomer is favoured at low temperatures
(d) The thermal conductivity of the para isomer is 50% greater than that of the ortho isomer.
Answer:
(b) Ortho isomer has zero nuclear spins whereas the para isomer has one nuclear spin
Hints:
Correct statement:
Ortho isomer – one nuclear spin Para isomer – zero nuclear spin

Question 4.
Ionic hydrides are formed by …………….
(a) halogens
(b) chalcogens
(c) inert gases
(d) group one elements
Answer:
(d) group one elements
e.g., Sodium hydride (Na+ H )

Question 5.
Tritium nucleus is contains ……………..
(a) 1p + 0n
(b) 2p + 1n
(c) 1p + 2n
(d) none of these
Answer:
(c) lp + 2n
1T3 (le, lp, 2n)

Question 6.
Non-stoichiometric hydrides are formed by……………..
(a) palladium, vanadium
(b) carbon, nickel
(c) manganese, lithium
(d) nitrogen, chlorine
Answer:
(a) palladium, vanadium

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 7.
Assertion: Permanent hardness of water is removed by treatment with washing soda.
Reason: Washing soda reacts with soluble calcium and magnesium chlorides and sulphates in hard water to form insoluble carbonates
(a) Both assertion and reason are true and the reason is the correct explanation of assertion.
(b) Both assertion and reason are true but the reason is not the correct explanation of assertion.
(c) Assertion is true but the reason is false
(d) Both assertion and reason are false
Answer:
(a) Both assertion and reason are true and the reason is the correct explanation of assertion.
Ca2+ + Na2CO3 → CaCO3↓ + 2Na+

Question 8.
If a body of a fish contains 1.2 g hydrogen in its total body mass, if all the hydrogen is replaced with deuterium then the increase in body weight of the fish will be ……………
(a) 1.2 g
(b) 2.4 g
(c) 3.6 g
(d) \(\sqrt{4.8}\) g
Answer:
(a) 1.2 g
Hints:
Mass of deuterium = 2 × mass of protium
If all the 1.2 g hydrogen is replaced with deuterium, the weight will become 2.4g. Hence the increase in body weight is (2.4 – 1.2 = 1.2 g).

Question 9.
The hardness of water can be determined by volumetrically using the reagent …………..
(a) sodium thiosulphate
(b) potassium permanganate
(c) hydrogen peroxide
(d) EDTA
Answer:
(d) EDTA

Question 10.
The cause of permanent hardness of water is due to ………….
(a) Ca(HCO3)2
(b) Mg(HCO3k)3
(c) CaCl2
(d) MgCO3
Answer:
(c) CaCl2
Hints:
The permanent hardness of water is due to the presence of the chlorides, nitrates and sulphates of Ca2+ and Mg2+ ions.

Question 11.
Zeolite used to soften hardness of water is, hydrated ………….
(a) Sodium aluminium silicate
(b) Calcium aluminium silicate
(c) Zinc aluminium borate
(d) Lithium aluminium hydride
Answer:
(a) Sodium aluminium silicate
Zeolite is sodium aluminium silicate.
(NaAlSi2O6 .H2O)

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 12.
A commercial sample of hydrogen peroxide marked as 100 volume H2O2, it means that ……………
(a) 1 ml of H2O2 will give 100 ml O2 at STP
(b) 1 L of H2O2 will give 100 ml O2 at STP
(c) 1 L of H2O2 will give 22.4 L O2
(d) 1 ml of H2O2 will give 1 mole of O2 at STP
Answer:
(a) 1 ml of H2O2 will give 100 ml O2 at STP

Question 13.
When hydrogen peroxide is shaken with an acidified solution of potassium dichromate in presence of ether, the ethereal layer turns blue due to the formation of
(a) Cr2 O3
(b) CrO42-
(c) CrO(O2)2
(d) none of these
Answer:
(c) CrO(O2)2 CrO(O2)2
Hints:
Cr2O72- + 2H+ + 4H2O2 → 2CrO(O2)2 + 5H2O

Question 14.
For decolorization of 1 mole of acidified KMnO4, the moles of H2O2 required is …………….
(a) \(\frac {1}{2}\)
(b) \(\frac {3}{2}\)
(c) \(\frac {5}{2}\)
(d) \(\frac {7}{2}\)
Answer:
(c) \(\frac {5}{2}\)
Hints:
2MnO4 + 5H2O2(aq) + 6H+ → 2Mn2++ 5O2 + 8H2O

Question 15.
Volume strength of 1.5 N H2O2 is ……………..
(a) 1.5
(b) 4.5
(c) 16.8
(d) 8.4
Answer:
(d) 8.4
Hints:
Volume strength of hydrogen peroxide = Normality of hydrogen peroxide × 5.6 = 1.5 x 5.6 = 8.4
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
Volume strength of hydrogen peroxide Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
= Normality × \(\frac {17 × 22.4}{68}\)
Volume strength of hydrogen peroxide = Normality x 5.6

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 16.
The hybridization of oxygen atom is H2O and H2O2 are respectively
(a) sp and sp3
(b) sp and sp
(c) sp and sp2
(d) sp3 and sp3
Answer:
(d) sp3 and sp3

Question 17.
The reaction H3PO2 + D2O → H2DPO2 + HDO indicates that hypo-phosphorus acid is ……………
(a) tri basic acid
(b) di basic acid
(c) mono basic acid
(d) none of these
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
Answer:
(c) monobasic acid
Hints:
Hypophosphorous acid on reaction with D2O, only one hydrogen is replaced P by deuterium and hence it is monobasic.

Question 18.
In solid ice, the oxygen atom is surrounded
(a) tetrahedrally by 4 hydrogen atoms
(b) octahedrally by 2 oxygen and 4 hydrogen atoms
(c) tetrahedrally by 2 hydrogen and 2 oxygen atoms
(d) octahedrally by 6 hydrogen atoms
Answer:
(a) tetrahedrally by 4 hydrogen atoms

Question 19.
The type of H-bonding present in ortho nitrophenol and p-nitrophenol is respectively ……………
(a) intermolecular H-bonding and intramolecular H-bonding
(b) intramolecular H-bonding and intermolecular H-bonding
(c) intramolecular H – bonding and no H – bonding
(d) intramolecular H – bonding and intramolecular H – bonding
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
Answer:
(A) intramolecular H-bonding and intermolecular H-bonding

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 20.
Heavy water is used as ……………
(a) modulator in nuclear reactions
(b) coolant in nuclear reactions
(c) both (a) and (b)
(d) none of these
Answer:
(c) both (a) and (A)
Hints:
Heavy water is used as a moderator as well as a coolant in nuclear reactions.

Question 21.
Water is a ……………
(a) basic oxide
(b) acidic oxide
(c) amphoteric oxide
(d) none of these
Answer:
(c) amphoteric oxide

II. Write brief answer to the following questions

Question 22.
Explain why hydrogen is not placed with the halogen in the periodic table.
Answer:
The electron affinity of hydrogen is much less than that of halogen atoms. Hence, the tendency of hydrogen to form hydride ions is low compared to that of halide ions. In most of its compounds hydrogen exists in a +1 oxidation state. Therefore, it is reasonable to place the hydrogen in group -1 along with alkali metals and not placed with halogens.

Question 23.
the cube at 0°C is placed in some liquid water at 0°C, the ice cube sinks – Why?
Answer:

  • In an ice cube, each atom is surrounded tetrahedrally by four water molecules through a hydrogen bond and its density is low.
  • Liquid water at 0°C has a density of 999.82 kg/cm3. Maximum density is attained by water only at 4°C as 1000 kg/cm3.
  • When the temperature changed from 4°C to 0°C, the density of water decreases rather than increases. This is called the anomalous expansion of water.
  • The reason for this phenomenon lies in the structure of ice lattice and hydrogen bonding in water.
  • At 0°C, the ice cube sinks in liquid water at 0°C because of the lesser density and greater volume of water.

Question 24.
Discuss the three types of Covalent hydrides.
Answer:
Covalent hydrides are the compound in which hydrogen is attached to another element by sharing electrons. The most common examples of covalent hydrides of non-metals are methane, ammonia, water, and hydrogen chloride. Covalent hydrides are further divided into three categories, viz., electron precise (CH4 C2H6), electron-deficient (B2H6), and electron-rich hydrides (NH3H2 O). Since most of the covalent hydrides consist of discrete, small molecules that have relatively weak intermolecular forces, they are generally gases or volatile liquids.

Question 25.
Predict which of the following hydrides is gas on a solid (a) HCl (b) NaH. Give your reason.
Answer:

  • At room temperature, HCl is a colourless gas, and the solution of HCl in water is called hydrochloric acid and it is in a liquid state.
  • Sodium hydride NaH is an ionic compound and it is made of sodium cations (Na+) and hydride (H) anions. It has an octahedral crystal structure. It is an alkali metal hydride.

Question 26.
Write the expected formulas for the hydrides of 4th-period elements. What is the trend in the formulas? In what way the first two numbers of the series different from the others?
Answer:
The expected formulas of the hydrides of 4th-period elements are MH or MH2. However, except for the first two members, many of the elements form non-stoichiometric interstitial hydrides with variable composition. The first two members of the period alkali metal (Potassium) and alkali earth metal (Calcium) forms ionic hydrides.

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 27.
Write the chemical equation for the following reactions.

  1. the reaction of hydrogen with tungsten (VI) oxide NO3 on heating.
  2. hydrogen gas and chlorine gas.

Answer:

  1. 3H2 + WO2 → W + 3H2O
    Hydrogen reduces tungsten (VI) oxide. WO3 to tungsten at high temperature.
  2. H2 + Cl2 → 2HCl (Hydrogen Chloride)
    Hydrogen reacts with chlorine at room temperature under light to give hydrogen chloride.

Question 28.
Complete the following chemical reactions and classify them into (a) hydrolysis (b) redox (c) hydration reactions.
(i) KMnO4 + H2O2
(ii) CrCl3+ H4O →
(iii) CaO + H2O →
Answer:
(i) KMnO4 + H2O2 → 2KMnO2 + 2KOH + 2H2O + 3O2
The reaction of potassium permanganate with hydrogen peroxide is a redox reaction.

(ii) CrCl3 + H2O → [Cr(H2O)6]Cl3
It is a hydration reaction. Many salts crystallized from aqueous solutions form hydrated crystals. The water in the hydrated salt may form co-ordinate bonds.

(iii) CaO + H2O → Ca(OH)2
It is a hydrolysis reaction. Calcium oxide hydrolyses to calcium hydroxide.

Question 29.
Hydrogen peroxide can function as an oxidizing agent as well as a reducing agent. Substantiate this statement with suitable examples.
Answer:
Hydrogen peroxide can function as an oxidizing agent as well as a reducing agent.

  • H2O2 acts as an oxidizing agent in an acidic medium. For example,
    Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
  • H2O2 acts as a reducing agent in a basic medium. For example,
    Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen>

Question 30.
Do you think that heavy water can be used for drinking purposes?
Answer:

  • Heavy water (D2O) contains a proton and a neutron. This makes deuterium about twice as heavy as protium, but it is not radioactive. So heavy water is not radioactive.
  • If you drink heavy water, you don’t need to worry about radiation poisoning. But it is not completely safe to drink, because the biochemical reaction in our cells is affected by the difference in the mass of hydrogen atoms.
  • If you drink an appreciable volume of heavy water, you might feel dizzy because of the density difference. It would change the density of the fluid in your inner ear. So it is unlikely to drink heavy water.

Question 31.
What is the water-gas shift reaction?
Answer:
The carbon monoxide of water gas can be converted to carbon dioxide by mixing the gas mixture with more steam at 400°C and passed over a shift converter containing iron/copper catalysts. This reaction is called the water-gas shift reaction.
CO + H2O → CO2 + H2

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 32.
Justify the position of hydrogen in the periodic table?
Answer:
The hydrogen has the electronic configuration of 1s1 which resembles with ns1 general valence shell configuration of alkali metals and shows similarity with them as follows:
1. It forms unipositive ion (H+) like alkali metals (Na+, K+, Cs+)
2. It forms halides (HX), oxides, (H2O), peroxides (H2O2), and sulphides (H2S) like alkali metals (NaX, Na2O, NaH2OH2, NaH2S)
3. It also acts as a reducing agent.

However, unlike alkali metals which have ionization energy ranging from 377 to 520 kJ mol-1, hydrogen has 1,314 kJ mol-1 which is much higher than alkali metals.

Like the formation of halides (X) from halogens, hydrogen also has a tendency to gain one electron to form a hydride ion (H) whose electronic configuration is similar to the noble gas, helium. However, the electron affinity of hydrogen is much less than that of halogen atoms. Hence, the tendency of hydrogen to form hydride ion is low compared to that of halogens to form the halide ions as evident from the following reactions:
\(\frac{1}{2}\) H2 + e → H                         ∆H = +36 kcalmol-1
\(\frac{1}{2}\) Br2 + e → Br                       ∆H = -55 kcalmol-1

Since hydrogen has similarities with alkali metals as well as halogens; it is difficult to find the right position in the periodic table. However, in most of its compounds hydrogen exists in a +1 oxidation state. Therefore, it is reasonable to place the hydrogen in group 1 along with alkali metals as shown in the latest periodic table published by IUPAC.

Question 33.
What are isotopes? Write the names of isotopes of hydrogen.
Answer:

  1. Isotopes are atoms of the same element that have the same atomic number but having different mass numbers (or) Isotopes are atoms with the same number of protons and electrons but differ in a number of neutrons.
  2. Hydrogen has three naturally occurring isotopes namely Protium (1H1), Deuterium (1H2), and Tritium (1H3).
    Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 34.
Give the uses of heavy water.
Answer:
The uses of heavy water are as follows:

  1. Heavy water is widely used as a moderator in nuclear reactors as it can lower the energies of fast neutrons
  2. It is commonly used as a tracer to study organic reaction mechanisms and mechanism of metabolic reactions
  3. It is also used as a coolant in nuclear reactors as it absorbs the heat generated.

Question 35.
Explain the exchange reactions of deuterium.
Answer:
Deuterium can replace reversibly hydrogen in compounds either partially or completely depending upon the reaction conditions. These reactions occur in the presence of deuterium.
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 36.
How do you convert para-hydrogen into ortho hydrogen?
Answer:
At room temperature, normal hydrogen consists of about 75% ortho-form and 25% para-form. As the ortho-form is more stable than para-form, the conversion of one isomer into the other is a slow process. However, the equilibrium shifts in favour of para hydrogen when the temperature is lowered.

The para-form can be catalytically transformed into an I ortho-form using platinum or iron. Alternatively, it can also be converted by passing an electric j discharge, heating above 800°C, and mixing with paramagnetic molecules such as O2, NO, NO2, or with nascent/atomic hydrogen.

Question 37.
Mention the uses of deuterium.
Answer:

  • Deuterium is used as a tracer element.
  • Deuterium is used to study the movement of groundwater by isotopic effect.

Question 38.
Explain the preparation of hydrogen using electrolysis.
Answer:
High purity hydrogen (> 99.9 %) is obtained by the electrolysis of water containing traces of acid or alkali or the electrolysis of an aqueous solution of sodium hydroxide or potassium hydroxide using a nickel anode and iron cathode. However, this process is not economical for large-scale production.
At anode:
2OH → H2O + \(\frac{1}{2}\)O2 + 2e
At cathode:
2H2O + 2e → 2OH + H2
Overall reaction:
H2O → H2 + \(\frac{1}{2}\)O2

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 39.
A group metal (A) which is present in common salt reacts with (B) to give compound (C) in which hydrogen is present in a -1 oxidation state. (B) on reaction with a gas (C) to give universal solvent (D). The compound (D) reacts with (A) to give (B), a strong base. Identify A, B, C, D, and E. Explain the reactions.
Answer:
1.Group (1) metal (A) is present in common salt NaCl. So, (A) is sodium – Na.
2. Sodium reacts with hydrogen (B) to give sodium hydride – NaH (C) in which hydrogen is in a -1 oxidation state.
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
3. Hydrogen on reaction with oxygen (O2) gas which is (C) to give universal solvent water (D).
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
4. Water (D) reacts with sodium metal (A) to give a strong base sodium hydroxide NaOH which is (E).
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 40.
An isotope of hydrogen (A) reacts with a diatomic molecule of the element which occupies group number 16 and period number 2 to give compound (B) is used as a modulator in a nuclear reaction. (A) adds on to a compound (C), which has the molecular formula C3H6 to give (D). Identify A, B, C, and D.
Answer:
1. An isotope of hydrogen Deuterium (A) reacts with a diatomic molecule of element belongs to group number 16 and period number 2 oxygen O2 to give a compound (B) which is heavy water D2O. D2O is used as a moderator in a nuclear reaction.
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
2. Deuterium reacts with C3H6 propane (C) to give Deutero propane C2D6 (D).
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 41.
NH3 has an exceptionally high melting point and boiling point as compared to those of the hydrides of the remaining element of group 15- Explain.
Answer:

  1. NH3 has an exceptionally high melting point and boiling point due to hydrogen bonding between NH3 molecules.
  2. Each molecule can form a maximum of 4 hydrogen bonds but on average 1 hydrogen bond per molecule as there is only one lone pair on NH3 available for hydrogen bonding.
  3. Hydrogen bonding is a strong intermolecular attraction as H on NH3 acts as a proton due to partial positive on it the whole N has the partial negative charge. Thus when the very polarized H comes close to an N atom in another NH3 molecule, a very strong hydrogen bond is formed.
  4. Due to many strong intermolecular interactions compared to weaker permanent dipole-dipole interactions between other XH3 molecules in group 15, a large amount of energy are required to overcome the forces, giving it the highest boiling point and highest melting point.

Question 42.
Why interstitial hydrides have a lower density than the parent metal.
Answer:
In interstitial hydrides, hydrogen occupies the interstitial sites. These hydrides show properties similar to parent metals. Most of these hydrides are non-stoichiometric with variable composition. Hence, interstitial hydrides have a lower density than the parent metal.

Question 43.
How do you expect the metallic hydrides to be useful for hydrogen storage?
Answer:
In metallic hydrides, hydrogen is adsorbed as H-atoms. Due to the adsorption of H atoms, the metal lattice expands and becomes unstable. Thus, when metallic hydride is heated, it decomposes to form hydrogen and finely divided metal. The hydrogen evolved can be used as fuel.

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 44.
Arrange NH3, H2O, and HF in the order of increasing magnitude of hydrogen bonding and explain the basis for your arrangement.
Answer:
When a hydrogen atom is covalently bonded to a highly electronegative atom such as nitrogen, the bond is polarized. Due to this effect, the polarized hydrogen atom is able to form a weak electrostatic interaction with another electronegative atom present in the vicinity. This interaction is called hydrogen bonding. The magnitude of hydrogen bonding increases with the increase in the electronegativity of the atom. Hence, the increasing magnitude of hydrogen bonding of NH3, H2O, and HF follows the order NH3 < H2O < HF.

Question 45.
Compare the structures of H2O and H2O2.
Answer:
In water, O is sp3 hybridized. Due to stronger lone pair-lone pair repulsions than bond pair-bond pair repulsions, the HOH bond angle decreases from 109.5° to 104.5°. Thus water molecule has a bent structure.
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
H2O2 has a non-planar structure. The 0 – H bonds are in different planes. Thus, the structure of H2O2 is like an open book.
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Samacheer Kalvi 11th Chemistry Hydrogen Additional Questions Solved

I. Choose the correct answer:

Question 1.
The simplest atom which contains one electron and one proton is
(a) Helium
(b) Deuterium
(c) Hydrogen
(d) Tritium
Answer:
(c) Hydrogen

Question 2.
is the most abundant 90% of all atoms…………
(a) Lithium
(b) Hydrogen
(c) Oxygen
(d) Silicon
Answer:
(A) Hydrogen

Question 3.
Which of the following properties of hydrogen similar to alkali metals?
1. It forms a uni negative ion.
2. It forms halides, oxides, and sulphides similar to alkali metals.
3. It also acts as a reducing agent.
(a) 1 and 2
(b) 2 and 3
(c) 1 and 3
(d) 1, 2 and 3
Answer:
(b) 2 and 3

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 4.
Which one of the metal is used to convert para-hydrogen into ortho hydrogen?
(a) Copper
(b) Aluminium
(c) Sodium
(d) Platinum
Answer:
(d) Platinum

Question 5.
Hydrogen is placed in the ________ of the periodic table.
(a) group – 1
(b) group – 17
(c) group – 18
(d) group – 2
Answer:
(a) group – 1

Question 6.
Which of the following is not used in the conversion of para hydrogen into ortho hydrogen?
(a) by heating more than 800°C
(b) bypassing an electric discharge
(c) by mixing with atomic hydrogen
(d) by mixing with diamagnetic molecules
Answer:
(d) by mixing with diamagnetic molecules

Question 7.
The radioactive isotope of hydrogen is
(a) protium
(b) deuterium
(c) tritium
(d) heavy hydrogen
Answer:
(c) tritium

Question 8.
Which one of the following does not contain neutron?
(a) ordinary hydrogen
(b) Heavy hydrogen
(c) Radioactive hydrogen
(d) Deuterium
Answer:
(a) ordinary hydrogen

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 9.
The half-life period of tritium is ______(in years).
(a) 13.2
(b) 10.5
(c) 12.3
(d) 15.8
Answer:
(c) 12.3

Question 10.
Which of the following is used in illumination of wristwatches?
(a) Phosphorous
(b) Radon
(c) Tritium
(d) Deuterium
Answer:
(c) Tritium

Question 11.
The percentage of ortho and para forms of normal hydrogen at room temperature is
(a) 25% and 75%
(b) 40% and 60%
(c) 60% and 40%
(d) 75% and 25%
Answer:
(d) 75% and 25%

Question 12.
By which rays nuclear reactions are induced in upper atmosphere to produce tritium?
(a) α-rays
(b) β-rays
(c) γ-rays
(d) cosmic rays
Answer:
(d) cosmic rays

Question 13.
The composition of syngas is
(a) CO + N2
(b) CO + H2O
(c) CO + H2
(d) CO2 + H2
Answer:
(c) CO + H2

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 14.
Consider the following statements ……………..
(i) Tritium is a beta-emitting radioactive isotope of hydrogen.
(ii) Deuterium is known as heavy hydrogen.
(iii) Deuterium is used in emergency exit signs.
Which of the following statement is/are not correct?
(a) (i) only
(b) (iii) only
(c) (i) and (ii)
(d) (i) (ii) and (iii)
Answer:
(b) (iii) only

Question 15.
The conversion of carbon monoxide of the water gas into carbon dioxide is called water gas _______ reaction.
(a) displacement
(b) decomposition
(c) shift
(d) conversion
Answer:
(c) shift

Question 16.
Consider the following statements.
(i)Hydrogen is a colourless, odourless, tasteless heavy and highly inflammable gas.
(ii) Hydrogen is a good reducing agent.
(ii) Hydrogen can be liquefied under low pressure and high temperature.
Which of the above statements is/are not correct?
(a) (i) only
(b) (ii) only (c)
(c) (i) and (iii)
(d) (ii) and (Hi)
Answer:
(c) (i) and (iii)

Question 17.
The CO2 formed in the water gas shift reaction is absorbed in a solution of
(a) Potassium bicarbonate
(b) Sodium chloride
(c) Potassium sulphate
(d) Potassium carbonate
Answer:
(d) Potassium carbonate

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 18.
Which one of the following is manufactured in Haber’s process?
(a) SO3
(b) NH3
(C) N2
(d) H2
Answer:
(b) NH3

Question 19.
When water is completely electrolyzed, the gas liberated is/are
(a) H2
(b) D2
(c) H2 and D2
(d) H2 and T2
Answer:
(c) H2 and D2

Question 20.
Match the List-I and List-II using the correct code given below the list.
List-I
A. Hydrogenation of unsaturated vegetable oils
B. Calcium hydride
C. Liquid hydrogen
D. Atomic hydrogen

List-II
1. Rocket fuel
2. Welding of metals
3. Desiccant
4. Margarine
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 21.
Statement-I: Hydrogen is placed at the top of the group which is in-line with the latest periodic table.
Statement-II: Hydrogen has a tendency to lose its electron to form H+, thus showing electropositive character like alkali metals. On the other hand, hydrogen has a tendency to gain an electron to yield H, thus showing electronegative character like halogens.
(a) Statements-I and II are correct and Statement-II is the correct explanation of statement-I.
(b) Statements-I and II are correct but Statement-II is not the correct explanation of statement-I.
(c) Statement-I is correct but Statement-II is wrong.
(d) Statement-I is wrong but Statement-II is correct.
Answer:
(a) Statements-I and II are correct and Statement-II is the correct explanation of statement-I.

Question 22.
Statement-I: The magnetic moment of para-hydrogen is zero.
Statement-II: The spins of two hydrogen atoms in para H2 molecule neutralise each other.
(a) Statements-I and II are correct and Statement-II is the correct explanation of statement-I.
(b) Statements-I and II are correct but Statement-II is not the correct explanation of statement-I.
(c) Statement-I is correct but Statement-II is wrong.
(d) Statement-I is wrong but Statement-II is correct.
Answer:
(a) Statements-I and II are correct and Statement-II is the correct explanation of statement-I.

Question 23.
Statement – I: Tritium is a β-emitter.
Statement – II: Radioactive decay of tritium gives \({ }_{2}^{3} \mathrm{He}\) and \({ }_{1}^{0} e\).
The correct statement/s is/are
(a) I alone
(b) II alone
(c) both I and II
(d) both are incorrect.
Answer:
(c) both I and II

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 24.
Which of the following is used as desiccants to remove moisture from organic solvents?
(a) Calcium hydride
(b) LiAlH4
(c) Sodium boro hydride
(d) Sodium hydride
Answer:
(a) Calcium hydride

Question 25.
Chlorine reacts with water and forms _____ and _____ respectively.
(a) HCl and HOCl
(b) H2 and HCl
(c) HOCl and H2
(d) HCl and ClO2
Answer:
(c) HOCl and H2

Question 26.
Liquid hydrogen is used as
(a) a rocket fuel as well as in space research
(b) fuel cells for generating electrical energy
(c) cutting and welding torch
(d) desiccant to remove moisture from organic solvent
Answer:
(a) a rocket fuel as well as in space research

Question 27.
Hydrolysis of P4O10 gives
(a) HPO2
(b) H4P2O7
(c) H3PO3
(d) H2PO4
Answer:
(d) H2PO4

Question 28.
At the temperature conditions of the earth (300K) the OPR of H2O is ……………
(a) 2.5
(b) 3
(c) 1
(d) zero
Answer:
(b) 3

Question 29.
Fluorine reacts with water and liberates
(a) hydrogen
(b) oxygen
(c) Fluorine dioxide
(d) HOF
Answer:
(b) oxygen

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 30.
Which of the following does not have any effect on water?
(a) Sodium
(b) Iron
(c) Lead
(d) Mercury
Answer:
(d) Mercury

Question 31.
The most common metal ions present in hard water are
(a) Magnesium and Iron
(b) Calcium and Aluminium
(c) Magnesium and Calcium
(d) Manganese and Calcium
Answer:
(c) Magnesium and Calcium

Question 32.
Which of the following non-metal reacts with ordinary water?
(a) Carbon
(b) Sulphur
(c) Chlorine
(d) Phosphorous
Answer:
(c) Chlorine

Question 33.
The permanent hardness of water is due to the presence of soluble salts of _____ and ______ of magnesium and calcium.
(a) carbonates and bicarbonates
(b) chlorides and carbonates
(c) bicarbonates and sulphates
(d) chlorides and sulphates
Answer:
(d) chlorides and sulphates

Question 34.
Which one of the following is used as a bleach?
(a) Cl2 water
(b) Br, water
(c) Water gas
(d) Liquid hydrogen
Answer:
(a) Cl2 water

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 35.
The general formula of zeolites is
(a) NaOAl2O3. xSiO2. yH2O
(b) Na2O.Al2O3. ySiO2. xH2O
(c) NaOH.Al2O3. xSiO2. yH2O
(d) NaO.Al (OH)3 .xSiO2. yH2O
Answer:
(a) NaOAl2O3. xSiO2. yH2O

Question 36.
Permanent hardness of water is removed by
(a) boiling
(b) lime
(c) washing soda
(d) chlorine
Answer:
(c) washing soda

Question 37.
_______ is used as a moderator and coolant in nuclear reactors.
(a) Heavy hydrogen
(b) Ortho hydrogen
(c) Hydrogen peroxide
(d) Heavy water
Answer:
(d) Heavy water

Question 38.
In chelating method of softening of hard water is used.
(a) magnesia
(b) lime
(c) EDTA
(d) washing soda
Answer:
(c) EDTA

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 39.
The percentage of hydrogen peroxide in ‘100 volume’ is
(a) 40
(b) 30
(c) 50
(d) 20
Answer:
(b) 30

Question 40.
Match the List-I and List-Il using the code given below the list.
List-I
A. Heavy water
B. Hydrogen peroxide
C. Heavy hydrogen
D. Lithium Aluminium hydride

List-Il
1. Antiseptic
2. Moderator
3. Reducing agent
4. Tracer
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 41.
______ present in the glass catalyses the disproportionation reaction of hydrogen peroxide.
(a) Silica
(b) Alkali metals
(c) fluorine
(d) Oxygen
Answer:
(b) Alkali metals

Question 42.
Statement-I: Heavy water has been widely used as moderator in nuclear reactors.
Statement-Il: Heavy water can lower the energies of fast moving neutrons.
(a) Statements-I and II are correct and Statement-Il is the correct explanation of statement-I.
(b) Statements-I and II are correct but Statement-IT is not the correct explanation of statement-I;
(c) Statement-I is correct but Statement-Il is wrong.
(d) Statement-I is wrong but Statement-Il is correct.
Answer:
(a) Statements-I and II are correct and Statement-Ills the correct explanation of statement-I.

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 43.
Which of the following statement/s are true about hydrogen peroxide?
1. It can act both as an oxidizing agent and a reducing agent.
2. It is used in water treatment to oxidize pollutants.
3. It is used as a mild analgesic.
4. It restores the white colour of the old paintings,
(a) 1, 2 and 3
(b) 1, 3 and 4
(c) 1, 2 and 4
(d) 2, 3 and 4
Answer:
(c) 1, 2 and 4

Question 44.
Which one of the following is an ionic or saline hydride?
(a) SiH4
(b) GeH4
(c) B2H6
(cl) LiH
Answer:
(d) LiH

Question 45.
The smallest molecule which shows hindered rotation about a single bond is
(a) Hydrogen peroxide
(b) Water
(c) Deuterium oxide
(d) hydrogen
Answer:
(a) Hydrogen peroxide

Question 46.
Which of the following pair is an electron rich hydride?
(a) NH3, H2O
(b) CH4, C2H6
(c) B2H6, GeH4
(d) CH4, SiH4
Answer:
(a) NH3, H2O

Question 47.
Which of the following molecule shows an intramolecular hydrogen bond?
(a) Water
(b) Ammonia
(c) Salicylaldehyde
(d) Para-nitrophenol
Answer:
(c) Salicylaldehyde

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 48.
Metallic hydrides are otherwise called …………….
(a) Salt hydrides
(b) Saline hydrides
(c) molecular hydrides
(d) Interstitial hydrides
Answer:
(d) Interstitial hydrides

Question 49.
Which one of the following is a covalent hydride?
(a) NH3
(b) BeH2
(c) NaH
(d) ZrH2
Answer:
(a) NH3

Question 50.
Which one of the following is known as a Hydrogen sponge?
(a) Lithium hydride
(b) Diborane
(c) Palladium hydride
(d) Ammonia
Answer:
(c) Palladium hydride

Question 51.
Which of the following is the correct order of stability of bonds?
(a) Hydrogen bond < CovaLent bond < Vanderwaals bond
(b) Vanderwaals bond < Hydrogen bond <Covalent, bond
(c) Vanderwaals bond > Hydrogen bond > Covalent bond
(d) Covalent bond < Hydrogen bond <Vanderwaals bond
Answer:
(b) Vanderwaals bond < Hydrogen bond <Covalent, bond

Question 52.
Which one of the following does not have intramolecular hydrogen bonding?
(a) water
(b) o-nitrophenol
(c) Salicylaldehyde
(d) Salicylic acid
Answer:
(a) water

Question 53.
Which of the following contains intramolecular hydrogen bonding?
(a) Acetic acid
(b) o-nitrophenol
(c) Hydrogen fluoride
(d) water
Answer:
(b) o-nitrophenol

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 54.
Consider the following statements.
(i) In ice, each oxygen atom is surrounded by hydrogen atoms tetrahedrally to four water molecules.
(ii) Acetic acid exists as a dimer due to intramolecular hydrogen bonding.
(iii) Strong hydrogen bonds lead to an increase in the melting and boiling points.
Which of the above statements is/are not correct? ,
(a) (ii) only
(b) (i) and (iii)
(c) (i) (ii) and (iii)
(d) (i) only
Answer:
(a) (ii) only

Question 55.
Which one of the following is an example for Clathrate hydrate?
(a) CuSO4.5H2O
(b) Na2CO3. 10H2O
(c) CH4. 20 H2O
(d) FeSO3.7H2O
Answer:
(c) CH4. 20 H2O

Question 56.
Which one of the following is not a crystalline hydrate?
(a) CH4. 20H2O
(b) Na2,CO3. 10H2O
(c) CuSO4.5H2O
(d) FeSO4.7H20
Answer:
(a) CH4. 20H2O

Question 57.
Statement-I: Hydrogen can be used as a clean-burning fuel.
Statement-II: Hydrogen on combustion give only water as end product and it is free from pollutants.
(a) Statements-I and II are correct and Statement-lI is the correct explanation of statement-I.
(b) Statements-I and II are correct but Statement-Il is not the correct explanation of statement-I.
(e) Statement-I is correct but Statement-Il is wrong.
(d) Statement-I is wrong but Statement-lI is correct.
Answer:
(a) Statements-I and II are correct and Statement-Il is the correct explanation of statement-I.

Question 58.
Which isotope of hydrogen is radioactive?
(a) Protium
(b) Deuterium
(c) Tritium
(d) H
Answer:
(c) Tritium

Question 59.
Which type of elements form interstitial hydrides ………..
(a) s-block and p-block
(b) p-block only
(c) d-block and f-block
(d) s-block only
Answer:
(c) d-block and f-block

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 60.
Which of the following is named perhydrol and used as an antiseptic?
(a) D2O
(b) H2O2
(c) NaH
(d) B2H6
Answer:
(b) H2O2

Question 61.
Which of the following causes temporary hardness of water?
(a) MgCl2
(b) Na2SO4
(c) Mg(HCO3)2
(d) NaCl
Answer:
(c) Mg(HCO3)2

Question 62.
Which of the following can oxidise Hydrogen peroxide?
(a) acidified KMnO4
(b) Cu
(c) dil. HNO4
(d) CrO2Cl2
Answer:
(a) acidified KMnO4

Question 63.
Which type of hydrides are generally non-stoichiometric in nature?
(a) Metallic hydride
(b) Covalent hydrides
(c) Ionic hydride
(d) Saline hydride
Answer:
(a) Metallic hydride

Question 64.
Hydrogen gas is generally prepared by the ………….
(a) reaction of granulated zinc with dilute H2SO4
(b) reaction of zinc with cone, H2SO4
(c) reaction of pure zinc with dil. H2SO4
(d) action of stream on red hot coke
Answer:
(a) reaction of granulated zinc with dilute H2SO4

Question 65.
The higher density of water than that of ice is due to ……………
(a) dipole-dipole interaction
(b) dipole-induced dipole interaction
(c) hydrogen bonding
(d) all of these
Answer:
(c) hydrogen bonding

Question 66.
Hydrogen resembles halogens in many respects for which several factors are responsible. Of the following factors which one is most important in this respect?
(a) Its tendency to lose an electron to form a cation.
(b) Its tendency to gain an electron to attain stable electronic configuration.
(c) Its low negative electron gain enthalpy value.
(d) It’s small in size.
Answer:
(b) Its tendency to gain an electron to attain stable electronic configuration.

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 67.
Metal hydrides are ionic, covalent, or molecular in nature. Among LiH, NaH, KH, RbH, CsH, the correct order of increasing ionic character is ………….
(a) LiH>NaH>CsH>KH>RbH
(b) LiH<NaH<KH<RbHCsH>NaH>Kil>LIH
(d) NaH>CsH>RbH>LiH>KH
Answer:
(b) LiH<NaH<KH<RbH<CsH

Question 68.
Statement-I: Permanent hardness of water is removed by treatment with washing soda ………..
Statement-II: Washing soda reacts with soluble magnesium and calcium sulphate to form insoluble carbonate.
(a) Statements-I and II are correct and Statement-Il is the correct explanation of statement-I.
(b) Statements-I and II are correct but Statement-Il is not the correct explanation of statement-I.
(c) Statement-I is correct but Statement-II is wrong.
(d) Statement-I is wrong but Statement-II is correct.
Answer:
(a) Statements-I and II are correct and Statement-Ills the correct explanation of statement-I.

Question 69.
Which of the following is a saline hydride?
(a) HCl
(b) NH3
(c) NaH
(d) PbH
Answer:
(e) NaH

Question 70.
Which metal does not liberate H2 gas from dilute aqueous hydrochloric acid at 298 K?
(a) Mg
(b) Zn
(c) Al
(d) Cu
Answer:
(d) Cu

Question 71.
When zeolite is treated with hard water, the sodium ions are exchanged with …………
(a) H+ ions and Cl ions
(b) Ca2+ ions
(c) Cl ions
(d) Ca2+ ions and Mg2+ ions
Answer:
(d) Ca2+ ions and Mg2+ ions

Question 72.
The most abundant element in the universe is …………..
(a) Carbon
(b) Nitrogen
(c) Silicon
(d) Hydrogen
Answer:
(d) Hydrogen

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 73.
Which of the following can effectively remove all types of the hardness of water?
(a) Soap
(b) Slaked lime
(c) Washing soda
(d) Zeolite
Answer:
(a) Soap

Question 74.
A commercial sample of H2O2 is labelled as 100 volume. Its percentage strength is nearly
(a) 10%
(b) 30%
(c) 100%
(d) 90%
Answer:
(b) 30%

Question 75.
Which of the following will not produce dihydrogen gas?
(a) Cu + dil (HCl)
(b) CH2(g) + H2O(g)
(c) Zn + dil. HCl
(cl) C(s) + H2O(g)
Answer:
(a) Cu + dil. (HCl)

Samacheer Kalvi 11th Chemistry Hydrogen 2-Mark Questions
Question 1.
Draw and define ortho and para hydrogen molecule.
Answer:
Molecular hydrogen have ortho and para form in which the nuclear spins are aligned or opposed, respectively.
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 2.
Write the physical properties of Hydrogen?
Answer:
Hydrogen is a colorless, odorless, tasteless, lightest, and highly flammable gas. It is a non-polar diatomic molecule. It can be liquefied under low temperature and high pressure. Hydrogen is a good reducing agent.

Question 3.
Mention the uses of tritium.
Answer:

  • Tritium has replaced radium in applications such as emergency exit signs.
  • Tritium is used in illumination of wristwatches.

Question 4.
Draw the structures of three isotopes of hydrogen, Hydrogen Deuterium
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 5.
What is the half-life period of tritium? How is it undergoes radioactive disintegration?
Answer:

  • The half-life of tritium = 12.33 years.
  • Tritium is a beta-emitting radioactive isotope of hydrogen.
    Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 6.
How will you convert para-hydrogen into ortho hydrogen?
Answer:
The para-form can be catalytically transformed into ortho-form using platinum or iron. Alternatively, it can also be converted by passing an electric discharge, heating above 800°C, and mixing with paramagnetic molecules such as O2, NO, NO2, or within ascent/atomic hydrogen.

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 7.
How ammonia is manufactured from Hydrogen? Give the uses of ammonia.
Answer:

  • Ammonia is manufactured by Habe’s prôcess in which the largest consumer of hydrogen is used.
    N2(g) + 3H2(g) ⇌ 2 NH3(g)
  • Ammonia is employed for the manufacture of nitric acid, fertilizers, and explosives.

Question 8.

Question 9.
What is hydrogenation? Give one example.
Answer:
Hydrogenation is a reaction in which the addition of hydrogen to an alkene /alkyne. Compounds containing multiple bonds.
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 10.
What is syngas? Why is it called so?
Answer:
Water gas is also called syngas as it is used in the synthesis of organic compounds such as methanol and simple hydrocarbons.

Question 11.
How alkali metals react with water? Give an equation?
Answer:
The most reactive alkali metals decompose water in the cold with the evolution of hydrogen and leaving an alkali solution.
2Na(s) +2H2O(l) → 2NaOH(aq) +H2(g)

Question 12.
How is Tritium prepared?
Answer:
Tritium is present only in trace amounts. So it can be artificially prepared by bombarding lithium with slow neutrons in a nuclear fission reactor. The nuclear transmutation reaction for this process is as follows.
\({ }_{3}^{6} L i\) + \({ }_{0}^{1} n\) → \({ }_{2}^{4} \mathrm{He}\) + \({ }_{1}^{3} T\)

Question 13.
Explain the action of chlorine with water.
Answer:
Chlorine reacts with the water to form hydrochloric acid and hypochlorous acid.
Cl2(g) + H2O(l) → HCl(aq) + HOCl(aq)

Question 14.
What is the deuterium exchange reaction?
Answer:
Deuterium can replace reversibly hydrogen in compounds either partially or completely depending upon the reaction conditions. These reactions occur in the presence of deuterium or heavy water.
CH4 + 2D2 → CD4 + 2H2
2NH3 + 3D2 → 2ND3 + 3H2

Question 15.
What ¡s permanent hardness of water? How ¡t will be removed?
Answer:
The permanent hardness of water is due to the presence of soluble salts of magnesium and calcium in the form of chlorides and suiphates in water. It can be removed by adding washing soda which reacts with these metal chlorides and suiphates in hard water to form insoluble carbonates.
MCl2(aq) + Na2CO3(aq) → MCO3(s) + 2NaCl(aq)
MSO4(aq) + Ni2 CO3(aq) → MCO3(s) + Na2SO4(s)
Where M = Ca or Mg.

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 16.
Water is an amphoteric oxide. Give reason.
Answer:
Water is an amphoteric oxide. It has the ability to accept as well as donate protons and hence it can act as an acid or a base. For example, in the reaction with HCl, it accepts proton whereas in the reaction with weak base ammonia it donates a proton.
NH3 + H2O → NH4+ + OH
HCl + H2O → H3O+ + Cl

Question 17.
H2O2 is always stored in plastic bottles. Why?
Answer:
The aqueous solution of hydrogen peroxide is spontaneously disproportionate to give oxygen. The reaction is slow but it is explosive when it is catalyzed by metal or alkali dissolved from glass. For this reason, its solution are stored in plastic bottles.
H2O2(aq) → H2O(l) + ½ O2(g)

Question 18.
Why H2O2 ¡s used as mild antiseptic?
Answer:
The oxidising property of hydrogen peroxide and harmless nature of its products such as water and oxygen, leads to oxidation of pollutants in water and act as a mild antiseptic.

Question 19.
What is the permanent hardness of water?
Answer:
The permanent hardness of water is due to the presence of soluble salts of magnesium and calcium in the form of chlorides and sulphates.

Question 20.
What is meant by 100 – volume hydrogen peroxide?
Answer:
A 30% solution is marketed as 100 – volume hydrogen peroxide indicating that at STP, 100 volumes of oxygen are liberated per millimeter of the solution.

Question 21.
What is heavy water? How is it obtained?
Answer:
Heavy water (D2O) is the oxide of heavy hydrogen. One part of heavy water is present in 5000 parts of ordinary water. It is mainly obtained as the product of the electrolysis of water.

Question 22.
What is meant by binary hydride? Give example.
Answer:
A binary hydride is a compound formed by hydrogen with other electropositive elements including metals and non-metals, e.g., LiH or MgH2.

Question 23.
How does hard water produces less foam with detergents?
Answer:
The cleaning capacity of soap is reduced when used in hard water. Soaps are sodium or potassium salts of long chain fatty acids (e.g., coconut oil). When soap is added to hard water, the divalent magnesium and calcium ions present in hard water react with soap. The sodium salts present in soaps are converted to their corresponding magnesium and calcium salts which are precipitated as scum/precipitate.
M2+ + 2RCOONa → (RCOO)2M + 2Na+
M = Ca or Mg, R = C17H35

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 24.
What are the different types of hydrides?
Answer:
The hydrides are classified as Ionic, Covalent and Metallic Hydrides.
Ionic hydride – LiH
Covalent hydride – CH4
Metallic hydride -TiH

Question 25.
Why metallic hydrides are called interstitial hydrides? Give one example.
Answer:
Metallic hydrides are usually obtained by hydrogenation of metals and alloys in which hydrogen occupies the interstitial sites (voids). Hence, they are called interstitial hydrides. e.g., PdH.

Question 26.
What is hydrogen bonding?
Answer:
When a hydrogen atom (H) is covalently bonded to a highly electronegative atom such as fluorine (F) or oxygen (O) or nitrogen (N), the bond is polarized in such a way that the hydrogen atom is able to form a weak bond between the hydrogen atom of a molecule and the electronegative atom of a second molecule. The bond thus formed is a hydrogen bond and it is denoted by dotted lines (……)

Question 27.
What are the types of hydrogen bonding? Give example.
Answer:
There are two types of hydrogen bonding.

  • Intramolecular hydrogen bonding: It can occur with a molecule. e.g., o – nitrophenol.
  • Intermolecular hydrogen bonding: It is formed between two molécules of the same type or different type. e.g., H2O.

Question 28.
Explain the type of bonding present in hydrogen fluoride?
Answer:
In hydrogen fluoride (HF), for example, one molecule is strongly attracted to the fluorine on its neighboring hydrogen. In both liquid and solid, hydrogen fluoride forms long hydrogen-bonded zig-zag chains as a consequence of the orientation of the lone pairs on the fluorine atoms.
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 29.
Ice is less dense than water at 0°C. Justify this statement.
Answer:
In ice, each oxygen atom is surrounded tetrahedrally by through hydrogen bonds to four water molecules. That is, the presence of two hydrogen atoms and two lone electron pairs (on oxygen) in each water molecule results in a three – dimensional structure. The arrangements creates an open structure, which in turn accounts for the fact that ice is less dense than water at 0°C.

Question 30.
Draw the structure of –

  1. Acetic acid
  2. Water.

Answer:
1. Acetic acid:
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
2. Water:
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 31.
What is a hydrogen bond?
Answer:
When a hydrogen atom (H) is covalently bonded to a highly electronegative atom such as fluorine (F) or oxygen (O) or nitrogen (N), the bond is polarized. Due to this effect, the polarized hydrogen atom is j able to form a weak electrostatic interaction with another electronegative atom present in the vicinity, f This interaction is called a hydrogen bond.

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 32.
Give the advantages of future Fuel – Hydrogen.
Answer:
Hydrogen is considered as a potential candidate for this purpose as it is a clean-burning fuel. Hence, hydrogen can directly be used as a fuel and can replace existing gasoline (petrol) diesel kerosene-powered engines, and indirectly be used with oxygen in fuel cells to generate electricity. One major advantage of using hydrogen is that the combustion product is essentially free from pollutants; the end-product formed in both cases is water.

Question 33.
What do you understand by the term non-stoichiometric hydrides’? Do you expect this type of hydrides to be formed by alkali metals? Justify your answer.
Answer:
Those hydrides which do not have fixed composition are called non-stoichiometric hydrides, and the composition varies with temperature and pressure. This type of hydrides is formed by d and f-block elements. They cannot be formed by alkali metals because alkali metal hydrides form ionic hydrides.

Question 34.
How does the atomic hydrogen or oxy – hydrogen torch function for cutting and welding purposes? Explain.
Answer:
When hydrogen is burnt in oxygen the reaction is highly exothermic, it produces very high temperature nearly 4000°C which is used for cutting and welding purposes.

Question 35.
How does H2O2 behave as bleaching agent?
Answer:
Bleaching action of H2O2 is due to the oxidation of colouring matter by nascent oxygen.
H2 O2 → H2O(l) + O(g)

Question 36.
Write the properties of hydrogen similar to alkali metals.
Answer:
The hydrogen has the electronic configuration of 1s1 which resembles with ns1 general valence shell configuration of alkali metals and shows similarity with them as follows:

  • It forms unipositive ion (H+) like alkali metals (Na+, K+, Cs+)
  • It forms halides (HX), oxides (H2O), peroxides (H2O2), and sulphides (H2S) like alkali metals (NaX, Na2O, Na2O2, Na2S)
  • It also acts as a reducing agent.

Question 37.
Write the chemical properties of deuterium.
Answer:
Like hydrogen, deuterium also reacts with oxygen to form deuterium oxide called heavy water. It also reacts with halogen to give corresponding halides.
2D2 + O2 → 2D2O
D2 + X2 → 2DX (X = F, Cl, Br & I)

Deuterium can replace reversibly hydrogen in compounds either partially or completely depending upon the reaction conditions. These reactions occur in the presence of deuterium or heavy water
CH4 + 2D2 → CD4 + 2H2
2NH3 + 3D2 → 2ND3 + 3H2

Question 38.
Why is hydrogen peroxide stored ¡n wax-lined plastic coloured bottles?
Answer:
The decomposition of H2O2 occurs readily in the presence of rough surface (acting as catalyst). It is also decomposed by exposure of light. Therefore, wax-lined smooth surface and coloured bottles retard the decomposition of H2O2.

Samacheer Kalvi 11th Chemistry Hydrogen 3 – Mark Questions

Question 1.
Compare the properties of ortho and para hydrogen.
Answer:

s.no Properties Ortho hydrogen Para hydrogen
1. Melting point 13.95 K 13.83 K
2. Boiling point 20.39 K 20.26 K
3. Vapour pressurec Normal higher
4. Magnetic moment Twice zero

Question 2.
Compare the properties of isotopes of hydrogen.
Answer:

s.no Property Protium Deuterium Tritium
1. Atomic nature H D T
2. Atomic mass 1.008 2.014 3.016
3. Nuclear stability Stable Stable Radioactive
4. Molecular hydrogen H2 D2 T2
5. Abundance(%) 99.985 0.015 ᷉10-16
6. Molecular mass 2.016 4.028 6.032

Question 3.
Draw the structure of the isotopes of hydrogen and distinguish them.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 4.
Explain the different methods of preparation of Tritium with the equation.
Answer:
It occurs naturally as a result of nuclear reactions induced by cosmic rays in the upper atmosphere.
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 5.
How would you prepare hydrogen in the laboratory?
Answer:
Small amounts of hydrogen are conveniently prepared in laboratory by the reaction of metals, such as zinc, iron and tin, with dilute acid.
Zn(s)  + 2HCl(aq) → ZnCl2(s) + H2(g)
In principle, any metal with a negative standard reduction potential will react with an acid to generate hydrogen.
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 6.
What happens when hydrogen reacts with –

  1. O2
  2. Cl2
  3. Na ?

Answer:

  1. 2 H2(g) + O2(g) → 2 H2O(l) – Water
  2. H2(g) + Cl2(g) → 2 HCl(g) – Hydrogen Chloride
  3. 2 Na(s) + H2(g) → 2 NaH(s) – Sodium hydride

Question 7.
Write a note about ortho water and para water.
Answer:

  1. Water exists in space in the interstellar clouds, ¡n proto-planetary disks, in the comets and icy satellites of the solar system, and on the Earth.
  2. In particular, the ortho-to-para ratio (OPR) of water in space has recently received attention. Like hydrogen, water can also be classified into ortho-H2O and para-H2O, in which the directions are antiparallel.
    Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
  3. At the temperature conditions of the Earth (300 K), the OPR of H2O is 3.
  4. At low temperatures below (< 50 K) the amount of para-H2O increases. It is known that the OPR of water in interstellar clouds and comets has more para-H2O (OPR = 2.5) than on Earth.

Question 8.
Water ¡s an amphoteric oxide. Justify this statement.
Answer:

  1. Water is an amphoteric oxide. It has the ability to act an acid as well as a base. That is, water shows this behavior when it reacts with hydrogen chloride and ammonia.
  2. When water reacts with ammonia, it behaves as an acid.
    Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
  3. When water reacts with hydrogen chloride, behaves as a base.
    Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
    So water is an amphoteric oxide.

Question 9.
How is the temporary hardness of water removed by boiling?
Answer:
Temporary hardness is primarily due to the presence of soluble bicarbonates of magnesium and calcium. This can be removed by boiling the hard water followed by filtration. Upon boiling, these salts decompose into insoluble carbonate which leads to their precipitation. The magnesium carbonate thus formed further hydrosol used to give insoluble magnesium hydroxide.
Ca(HCO3)2 → CaCO3 + H2O + CO2Mg(HCO3) → MgCO3 + H2O
CO2MgCO3 + H2O → Mg(OH)2 + CO2
The resulting precipitates can be removed by filtration.

Question 10.
Explain the action of soap with hard water.
Answer:

  • The cleaning capacity of soap is reduced when used in hard water.
  • Soaps are sodium or potassium salts of long-chain fatty acids.
  • When soap is added to hard water, the divalent magnesium and calcium cations present in hard water react with soap.
  • The sodium salts present in soaps are converted to their corresponding magnesium and calcium salts which are precipitated as scum or precipitate.
    M2+ + 2RCOONa (RCOO)2M(s) + 2Na+(aq);
    Where, M Ca or Mg;
    R = C17H35.

Question 11.
Write the chemical properties of heavy water.
Answer:
When compounds containing hydrogen are treated with D2O, hydrogen undergoes an exchange for deuterium 2NaOH + D2O → 2NaOD + HOD
HCl + D2O → DCl + HOD
NH4Cl + 4D2O → ND4Cl + 4HOD

These exchange reactions are useful in determining the number of ionic hydrogens present in a given compound. For example, when D2O is treated with hypo-phosphorus acid only one hydrogen atom is exchanged with deuterium. It indicates that it is a monobasic acid.
H3PO2 + D2O → H2DPO2 + HDO
It is also used to prepare some deuterium compounds:
Al4C3 + 12D2O → 4Al(OD)3 + 3CD4
CaC2 + 2D2O → Ca(OD)2 + C2D2
MgN2 + 6D2O → 3Mg(OD)2 + 2ND3
Ca3P2 + 6D2O → 3Ca(OD)2 + 2PD3

Question 12.
Explain the exchange reactions of deuterium oxide.
Answer:
When compounds containing hydrogen are treated with D2O, hydrogen undergoes an exchange for Deuterium.
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 13.
Complete the following reactions.
A14C3 + D2O → ?
CaC2 + D2O →?
Mg3N2 + D2O →?
Ca3P2 + D2O →?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 14.
What are the uses of hydrogen peroxide?
Answer:

  • H2O2 is used in water treatment to oxidize pollutants.
  • H2O2 is used as a mild antiseptic.
  • H2O2 is used as a bleach in textile, paper and hair-care industry.

Question 15.
Prove that H2O2 act as reducing agent ¡n alkaline medium.
Answer:
In alkaline conditions, H2O2 act as a reducing agent.
2KMnO4(aq)(Potassium permanganate) + 3 H2SO4(aq) + 5H2O2(aq) → K2 SO4 + 2MnSO4 + 8H2 O(l) + 5O2(g)

Question 16.
Write a note about saline (or) ionic hydride.
Answer:
Ionic hydrides composed of an electropositive metal, generally, an alkali or alkaline-earth metal (except beryllium and magnesium) formed by transferring of electrons from metal to hydrogen atoms. They can be prepared by the reaction of elements at about 400°C. These are salt-like, high-melting, white, crystalline solids having hydride ions (H) and metal cations (Mn+).
2Li(s) + H2(g) → 2LiH(s) (Lithiuinhydride)
2 Ca(s) + H2(g) → 2 CaH2(s) (Calcium hydride)

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 17.
Write the uses of hydrogen peroxide.
Answer:
The oxidizing ability of hydrogen peroxide and the harmless nature of its products, i.e., water and oxygen, lead to its many applications. It is used in water treatment to oxidize pollutants, as a mild antiseptic, and as bleach in the textile, paper and hair – care industry.

Hydrogen peroxide is used to restore the white colour of the old paintings which was lost due to the reaction of hydrogen sulphide in the air with the white pigment Pb3(OH)2 (CO3)2 to form black colored lead sulphide. Hydrogen peroxide oxidises black coloured lead sulphide to white coloured lead sulphate, thereby restoring the colour.
PbS + 4H2O2 → PbSO4 + 4H2O

Question 18.
What is intramolecular hydrogen bonding? Explain with an example.
Answer:

  1. Intramolecular hydrogen bonds are those which occur within one single molecule. This occurs between two functional groups within a molecule.
    Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
  2. An intramolecular hydrogen bond (dashed lines) joins the OH group to the doubly bonded oxygen atom of the carboxyl group on the same molecule. e.g., Salicylic acid.
    Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
  3. Salicylic acid act as an analgesic and antipyretic.

Question 19.
What are intermolecular hydrogen bonds? Explain with example.
Answer:

  1. Intermolecular hydrogen bonds occur between two separate molecules. They can occur between any numbers of like or unlike molecules as long as hydrogen donors and acceptors are present in positions in which they can interact.
  2. For e.g., Intermolecular hydrogen bonds can occur between ammonia molecules alone, between water molecules alone or between ammonia and water.
    Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
    Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 20.
Explain about the importance of hydrogen bonding ¡n proteins.
Answer:

  • Hydrogen bonds occur in complex biomolecules such as proteins and in biological systems.
  • For example, hydrogen bonds play an important role in the structure of deoxyribonucleic acid (DNA), since it holds together the two helical nucleic acid chains.
  • In these systems, hydrogen bonds are formed between specific pairs, for example. with a thymine unit in one chain bonding to an adenine unit in another; similarly, a cytosine unit in one chain bonds to a guanine unit in another.
  • Intramolecular hydrogen bonding also plays an important role in the structure of polymers, both synthetic and natural.

Question 21.
What are Clatharate hydrate? Explain it with suitable example.
Answer:

  • Gas hydrates ¡n which the guest molecules are not bonded chemically but retained by the structure of host is called Clatharates.
  • Water forms clatharate hydrates, e.g., methane hydrate (CH4 20H2O) which arc a type of ice that will bum when a lit match is held to it.
  • The structure of methane hydrate is made of linked polyhedra that contains 20 hydrogen-bonded water molecules forming a cage in which methane molecule is trapped.
  • Deposits of methane clatharates occur naturally in deep sea bed.
  • Hydrates are commonly obtained when water is frozen in presence of a gas such as argon, methane, etc.
  • Most gases form hydrates under high pressure.

Question 22.
What are crystalline hydrates? Explain it with example.
Answer:

  1. In these, hydrogen bonding is very important. Often the water molecules serve to fill in the interstices and bind together structure.
  2. A specific example is CuSO4  5H2O.
  3. Although there are five water molecules for every divalent copper ion, only four are coordinated to the cation, it’s six-coordination being completed from sulphate anions. The fifth water molecule is held in place of hydrogen bonds, O – H – O, between it and two coordinated water molecules and then coordinate sulphate anion.
  4. Water forms hydrated salts during crystallization. Examples, Na2 CO3 . 10H2O, FeSO4 .7H2O.
  5. The water present in the hydrates is called as water of hydration.

Question 23.
Write the chemical properties and uses of heavy water.
Answer:
When compounds containing hydrogen are treated with D2O, hydrogen undergoes an exchange for deuterium 2NaOH + D2O → 2NaOD + HOD
HCl + D2O → DCl + HOD
NH4Cl + 4D2O → ND4Cl + 4HOD

These exchange reactions are useful in determining the number of ionic hydrogens present in a given compound. For example, when D2O is treated with hypo-phosphoric acid only one hydrogen atom is exchanged with deuterium. It indicates that it is a monobasic acid.
H3PO2 +D2O → H2DPO2 + HDO

It is also used to prepare some deuterium compounds:
Al4Cl3 + 12D2O → 4Al(OD)3 + 3CD4
CaC2 + 2D2O → Ca(OD)2 + C2D2
Mg3N2 + 6D2O → 3Mg(OD)2 + 2ND3
Ca3P2 + 6D2O → 3Ca(OD)2 + 2PD3

Uses of Heavy Water:

  1. Heavy water is widely used as a moderator in nuclear reactors as it can lower the energies of fast neutrons
  2. It is commonly used as a tracer to study organic reaction mechanisms and mechanism of metabolic reactions
  3. It is also used as a coolant in nuclear reactors as it absorbs the heat generated.

Question 24.
What causes the temporary and permanent hardness of water?
Answer:
Temporary hardness of water is due to the presence of bicarbonates of calcium and magnesium in water i.e., Ca(HCO3)2 and Mg(HCO3) in water. The permanent hardness of water is due to the presence of soluble chlorides and sulphates of calcium and magnesium i.e., CaCl2, CaSO4, MgCl2, and MgSO4.

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 25.
Write chemical reaction to show the amphoteric nature of water.
Answer:
Water is amphoteric in nature because it acts as an acid.
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
Question 26.
What is the difference between the terms ‘hydrolysis’ and ‘hydration’?
Answer:
Hydrolysis is a chemical reaction in which a substance reacts with water under neutral, acidic or alkaline conditions.
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
Hydration on the other hand is the property of a chemical compound to take up molecules of water of crystallization and get hydrated.
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Samacheer Kalvi 11th Chemistry Hydrogen 5-Mark Questions

Question 1.
Explain about the different industrial preparation of hydrogen.
Answer:
In the large scale, hydrogen is produced currently by steam reforming of hydrocarbons. Steam and methane reacts with each other in the presence of nickel catalyst at 35 atm and at a temperature of 800°C gives hydrogen.
CH4(g) + H2O(g) → CO(g) +3H2(g)
Steam is passed over a red hot coke to produce carbon monoxide and hydrogen.
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
Water is reduced to hydrogen with carbon monoxide by passing over iron oxide catalyst at 400°C.
CO(g) + H2O(g) → CO2(g) + H2(g)
Hydrogen is produced as a by-product in oil refining industry during the cracking of long chain hydrocarbons.
C6H12(g) → CH66(g) + 3H2(g)
Hydrogen is also obtained in the manufacture of chlonne and sodium hydroxide via electrolysis of a concentrated solution of sodium chloride.

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 2.
What are hydrides? How are they classified?
Answer:
Hydrogen forms binary hydrides with many electropositive elements including metals and non-metals. It also forms ternary hydrides with two metals. E.g., LiBH4 and LiAlH4. The hydrides are classified as ionic, covalent, and metallic hydrides according to the nature of bonding.

Hydrides formed with elements having lower electronegativity than hydrogen are often ionic, whereas with elements having higher electronegativity than hydrogen form covalent hydrides.

Ionic (Saline) hydrides:
These are hydrides composed of an electropositive metal, generally, an alkali or alkaline-earth metal, except beryllium and magnesium, formed by the transfer of electrons from the metal to hydrogen atoms. They can be prepared by the reaction of elements at about 400° These are salt-like, high-melting, white crystalline solids having hydride ions (H) and metal cations (Mn+).
2 Li + H2 → 2LiH

Covalent (Molecular) hydrides:
They are compounds in which hydrogen is attached to another element by sharing of electrons. The most common examples of covalent hydrides of non-metals are methane, ammonia, water and hydrogen chloride. Covalent hydrides are further divided into three categories, viz., electron precise (CH4, C2H6, SiH4, GeH4), electron-deficient(B2H6) and electron-rich hydrides (NH3, H2O). Since most of the covalent hydrides consist of discrete, small molecules that have relatively weak intermolecular forces, they are generally gases or volatile liquids.

Metallic (Interstitial) hydrides:
Metallic hydrides are usually obtained by hydrogenation of metals and alloys in which hydrogen occupies the interstitial sites (voids). Hence, they are called interstitial hydrides; the hydrides show properties similar to parent metals and hence they are also known as metallic hydrides.

Most of the hydrides are non-stoichiometric with variable composition (TiH1.5 – 1.8 and PdH0.6 – 0.8), some are relatively light, inexpensive and thermally unstable which make them useful for hydrogen storage applications. Electropositive metals and some other metals form hydrides with the stoichiometry MH or sometimes MH2 (M = Ti, Zr, Hf, V, Zn).

Question 3.
Describe the process of water softening and purification.
Answer:

  1. An idealised image of water softening process involves replacement of cations such as Mg, Ca and Fe in water with sodium ions.by a cation exchange zeolite. The ion exchange zeolites or resins are used to replace the Mg and Ca ions found in hard water with sodium ions.
  2. They can be recharged by washing it with a solution containing a high concentration of sodium ions.
  3. The calcium and magnesium ions migrate from the zeolite or resin being replaced by sodium ions from the solution until a new equilibrium is reached. That is, the salt is used to recharge an ion exchange medium, which itself is used to soften the water.
  4. A couple of other methods, namely chelating method, and reverse osmosis are also used to soften hard water. The chelating method employs a polydentate ligand such as EDTA, while reserve osmosis uses high pressure to force the water through a semi-permeable membrane.
  5. In the case of water purification application, ion-exchange zeolites or resins are used to remove toxic (eg., copper) and heavy metal (e.g., cadmium or lead) ions from solution, replacing them with harmless sodium or potassium ions.

Question 4.
(a) How is H2O2 prepared?
(b) Explain about the structure of H2O2.
Answer:
(a) Hydrogen peroxide can be made by adding a metal peroxide to dilute acid.
BaO2(s) + H(s)SO4 → BaSO4(s) + H2O2(aq)
(b) Structure of H2O2
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

1. H2O2 has a non-polar structure. The molecular dimensions in the gas phase and solid phase differ as shown in the figure.

2. Both in gas phase and solid phase, the H2O2 molecule adopt a skew configuration due to repulsive interaction of the – OH bonds with lone pairs of electrons on each oxygen atom.

3. Indeed, it is the smallest molecule known to show hindrance rotation about a single bond. in solid phase, the dihedral angle is sensitive and hydrogen bonding decreasing from 111.50 in the gas phase to 90.2°, in the solid phase.

4. Structurally, H2O2 is represented by the dihydroxyl formula in which the two O-H groups do not lie in the same plane. In the solid phase of molecule, the dihedral angle reduces to 90.2° due to hydrogen bonding and the O-O-H angle expands from 94.8° to 101.9°.

5. One way of explaining the shape of hydrogen peroxide is that the hydrogen atoms would lie on the pages of a partly opened book, and the oxygen atoms along the spin.

Question 5.
Explain about Hydrogen sponge.
Answer:
1. Hydrogen sponge (or) Metal hydride e.g., the palladium-hydrogen system is a binary hydride (PdH).

2. Upon heating, H atoms diffuse through the metal to the surface and recombine to form molecular hydrogen. Since no other gas behaves this way with palladium, this process has been used to separate hydrogen gas from other gases:
2Pd(s) + H2(g) ⇌ 2PdH(s)

3. The hydrogen molecule readily adsorbs on the palladium surface, where it dissociates into atomic hydrogen. The dissociated atoms dissolve into the interstices or voids (octahedral or tetrahedral) of the crystal lattice.
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

4. Technically, the formation of metal hydride is by chemical reaction but it behaves like a physical storage method, i.e., it is absorbed and released like a water sponge. Such a reversible uptake of hydrogen in metals and alloys is also attractive for hydrogen Storage and for rechargeable metal hydride battery applications.

Question 6.
How are reducing agents in synthetic organic chemistry prepared?
Answer:
Hydrogen has a tendency to react with reactive metals like lithium, sodium to give corresponding hydrides.
2Li+H2 → 2LiH
2Na+H2 → 2NaH
These hydrides are used as reducing agents in synthetic organic chemistry. It is also used to prepare important hydrides such as lithium aluminium hydride and sodium borohydride (organic reducing agents).
4 LiH + AlCl3 → Li[AIH4] + 3 LiCl
4 NaH + B(OCH3)3 → Na[BH4] + 3 CH3ONa

Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 7.
How does water react with –
1. SiCl4
2. P4O10
Answer:
1. Water reacts with SiCl4 to give silica.
SiCl4 + 4H2O → Si(OH)4 + 4HCl
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen
2. Water reacts with P4O10 to give orthophosphoric acid.
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 8.
Explain the structure of CuSO4.5H2O
Answer:
Copper sulphate pentahydrate CuSO4.5H2O. In this compound, 4 water molecules form coordinate bonds while the fifth water molecule present outside the coordination can form an intermolecular hydrogen bond with another molecule as [Cu(H2O)2] SO2.H2O.
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 9.
How is hydrogen peroxide prepared on industrial scale?
Answer:
On an industrial scale, hydrogen peroxide is prepared by auto oxidation of of 2-alkyl antliraquinol.
Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

Question 10.
How is hydrogen peroxide is used to restore the white colour of old paintings.
Answer:
Hydrogen peroxide is used to restore the white colour which was lost due to the reaction of hydrogen suiphide in air with the white pigment Pb3(OH)2(CO3)2 to form black colored lead sulphide (PbS) Hydrogen peroxide oxidises black coloured lead sulphide to white coloured lead sulphate, thereby restoring the colour.
PbS + 4H2O2 → PbSO2 + 4H2O

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