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## Samacheer Kalvi 10th Science Model Question Paper 1 English Medium

Students can Download Samacheer Kalvi 10th Science Model Question Paper 1 English Medium Pdf, Samacheer Kalvi 10th Science Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## Tamil Nadu Samacheer Kalvi 10th Science Model Question Paper 1 English Medium

General Instructions:

1. The question paper comprises of four parts
2. You are to attempt all the questions in each part. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 12 in Part I are Multiple Choice Questions of one mark each.
These are to be answered by writing the correct answer along with the corresponding option code.
5. Question numbers 13 to 22 in Part II are of two marks each. Any one question should be answered compulsorily.
6. Question numbers 23 to 32 in Part III are of four marks each. Any one question should be answered compulsorily.
7. Question numbers 33 to 35 in Part IV are of seven marks each. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 75

Part – I

(i) Answer all the questions. [12 × 1 = 12]
(ii) Choose the most suitable answer and write the code with the corresponding answer.

Question 1.
The unit of ‘g’ is ms-2. It can be also expressed as _______.
(a) cm s-1
(b) N kg-1
(c) N m2 kg-1
(d) cm2 s-2
N kg-1

Question 2.
A convex lens forms a real, diminished point sized image at focus. Then the position of the object is at _____.
(a) focus
(b) infinity
(c) at 2f
(d) between f and 2 f
(b) infinity

Question 3.
Velocity of sound is maximum in ________.
(a) Solid
(b) Liquid
(c) Gases
(d) All the above
(a) Solid

Question 4.
The volume occupied by 4.4 g of CO2 at S.T.P _______.
(a) 22.4 litre
(b) 2.24 litre
(c) 0.24 litre
(d) 0.1 litre
(b) 2.24 litre

Question 5.
Which of the following have inert gases 2 electrons in the outermost shell?
(a) He
(b) Ne
(c) Ar
(d) Kr
(a) He

Question 6.
Deliquescence is due to ________.
(a) Strong affinity to water
(b) Less affinity of water
(c) Strong hatred to water
(d) Inertness to water
(a) Strong affinity to water

Question 7.
Decrease in leucocytes is _______.
(a) Leucopenia
(b) Leukemia
(c) Anaemia
(d) None of the above
(a) Leucopenia

Question 8.
Cancer cells are more easily damaged by radiations than normal cells because they are ______.
(a) Different in structure
(b) Non-dividing
(c) Mutated Cells
(d) Undergoing rapid division
(d) Undergoing rapid division

Question 9.
The xylem and phloem arranged side by side on same radius is called ________.
(b) amphivasan
(c) conjoint
(d) amphicribral
(c) conjoint

Question 10.
______ is the green coloured plastids.
(a) Chloroplast
(b) Chromoplast
(c) Leucoplast
(d) Stele
(a) Chloroplast

Question 11.
The best way of direct dating fossils of recent origin is by ________.
(c) Potassium-argon method
(d) Both (a) and (c)

Question 12.
All files are stored in the _______.
(a) Box
(b) Pai
(c) Scanner
(d) Folder
(d) Folder

Part – II

Answer any seven questions. (Q.No: 22 is compulsory) [7 × 2 = 14]

Question 13.
Define moment of a couple.
When two equal and unlike parallel forces applied simultaneously at two distinct points constitute couple. A couple results in causes the rotation of the body. This rotating effect of a couple is known as moment of a couple.

Question 14.
Distinguish between the resistivity and conductivity of a conductor?

 Resistivity Conductivity 1. It is the resistance of a conductor of unit length and unit area of cross section. 1. It is the reciprocal of electrical resistivity of material. 2. It’s unit is ohm metre 2. Its unit is ohm–1 metre-1 or ohm metre-1

Question 15.
Why does sound travel faster on a rainy day than on a dry day?
When humidity increases, the speed of sound increases. That is why you can hear sound from long distances clearly during rainy seasons.

Question 16.
What is rust? Give the equation for formation of rust?
When iron is exposed to moist air, it forms a layer of brown hydrated ferric oxide on its surface. This compound is known as rust and the phenomenon of formation of rust is known as rusting.

Question 17.
Give an example each
(i) gas in liquid
(ii) solid in liquid
(iii) solid in solid
(iv) gas in gas.
(i) Carbon-di-oxide dissolved in water (Soda water)
(ii) Sodium chloride dissolved in water.
(iii) Copper dissolved in gold (Alloy)
(iv) Mixture of Helium – Oxygen gases

Question 18.
How does leech respire?
Respiration takes place through the skin in leech. The exchange of respiratory gases takes place by diffusion. Oxygen dissolved in water diffuses through the skin into haemocoelic fluid, while carbon dioxide diffuses out.

Question 19.
What is bolting? How can it be induced artificially?
Treatment of rosette plants with gibberellin induces sudden shoot elongation followed by flowering is called bolting.

Question 20.
Match the following:

(a) (iii)
(b) (i)
(c) (iv)
(d) (ii)

Question 21.
What is Atavism?
The reappearance of ancestral characters in some individuals is called atavism. E.g. Presence of rudimentary tail in new born babies.

Question 22.
Calculate the velocity of a moving body of mass 7 kg whose linear momentum is 5 kg ms-1
Given, p = 5 kg ms-1
m = 7 kg
p =m × v
v = $$\frac{p}{m}=\frac{5}{7}$$
∴ v = 0.71 ms-1

Part – III

Answer any seven questions (Q.No: 32 is compulsory) [7 × 4 = 28]

Question 23.
State and prove the law of conservation of linear momentum.
(i) There is no change in the linear momentum of a system of bodies as long as no net external force acts on them.
(ii) Let us prove the law of conservation of linear momentum with the following illustration:

(iii) Let two bodies A and B having masses m1 and m2 move with initial velocity u1 and u2 in a straight line.
(iv) Let the velocity of the first body be higher than that of the second body. i.e., U1 > u2.
(v) During an interval of time t second, they tend to have a collision. After the impact, both of them move along the same straight line with a velocity v, and v2 respectively.

By Newton’s III law of motion
Action force = Reaction force
FA = -FB
$$\frac{m_{1}\left[v_{1}-u_{1}\right]}{t}=-\frac{m_{2}\left[v_{2}-u_{2}\right]}{t}$$
m1v1 – m1u1 = m2u2 + m2u2
m1v1 + m2v2 = m1u1 + m2u2

The above equation confirms in the absence of an external force, the algebraic sum of the momentum after collision is numerically equal to the algebraic sum of the momentum before collision. Hence the law of conservation linear momentum is proved.

Question 24.
(i) Explain about domestic electric circuits.

• Electricity is distributed through the domestic electric circuits wired by the electricians.
• The first stage of the domestic circuit is to bring the power supply to the main-box from a distribution panel, such as a transformer.
• The important components of the main-box are: (i) a fuse box and (ii) a meter. The meter is used to record the consumption of electrical energy.
• The fuse box contains either a fuse wire or a miniature circuit breaker (MCB).
• The function of the fuse wire or a MCB is to protect the house hold electrical appliances from overloading due to excess current.

(ii) What connection is used in domestic appliances and why?

• In domestic appliance is used parallel connection to avoid short circuit and breakage.
• One more advantage of the parallel connection of circuits is that each electric appliance gets an equal voltage.

Question 25.
(i) What is an echo?
An echo is the sound reproduced due to the reflection of the original sound from various rigid surfaces such as walls, ceilings, surfaces of mountains, etc.

(ii) State two conditions necessary for hearing an echo.
(i) The persistence of hearing for human ears is 0.1 second. This means that we can hear two sound waves clearly, if the time interval between the two sounds is atleast 0.1 s. Thus, the minimum time gap between the original sound and an echo must be 0.1 s.

(ii) The above criterion can be satisfied only when the distance between the source of sound and the reflecting surface would satisfy the following equation:

Question 26.
(i) What are the advantages of detergents over soaps?
Detergents are better than soaps because they:

• can be used in both hard and soft water and can clean more effectively in hard water than soap.
• can also be used in saline and acidic water.
• do not leave any soap scum on the tub or clothes.
• dissolve freely even in cool water and rinse freely in hard water.
• can be used for washing woollen garments, where as soap cannot be used.
• have a linear hydrocarbon chain, which is biodegradable.
• are active emulsifiers of motor grease.
• do an effective and safe cleansing, keeping even synthetic fabrics brighter and whiter.

(b) Mention the disadvantages of detergents.

• Some detergents having a branched hydrocarbon chain are not fully biodegradable by micro-organisms present in water. So, they cause water pollution.
• They are relatively more expensive than soap.

Question 27.
(a) Why does the reaction rate of a reaction increase on raising the temperature?
On increasing temperature heat is supplied to the reactant. This energy breaks more bonds and thus speed up the chemical reaction. Foods kept at room temperature spoils – faster than that kept in the refrigerator.

(b) Explain the types of double displacement reactions with examples.
There are two major classes of double displacement reactions. They are,
(i) Precipitation Reactions:
When aqueous solutions of two compounds are mixed, if they react to form an insoluble compound and a soluble compound, then it is called precipitation reaction.
Pb(NO3)2(aq) + 2KI(aq) → PbI2(s)↓+ 2KNO3(aq)

(ii) Neutralisation Reactions:
Another type of displacement reaction in which the acid reacts with the base to form a salt and water. It is called ‘neutralisation reaction’ as both acid and base neutralize each other.
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(i)

Question 28.
How nerve impulses are transferred from one neuron to next neuron?
All information from the environment are detected by the receptors, located in the sense organs such as the eyes, nose, skin, etc.

Information from the receptors is transmitted as electrical impulse and is received by the dendritic tips of the neuron. This impulse travels from the dendrite to the cell body and then along the axon to its terminal end.
On reaching the axonal end, it causes the nerve endings to release a chemical (neurotransmitter) which diffuses across a synapse and starts a similar electrical impulse in the dendrites of the next neuron, then to their cell body to be carried along the axon.

In this way, the electrical signal reaches the brain or spinal cord. The response from brain (or spinal cord) is similarly passed on to the effector organs such as the muscle or gland cell, that undergoes the desired response. The flow of nerve impulses from axon end of one neuron to dendrite of another neuron through a synapse is called synapatic transmission.

Question 29.
(a) How is dihybrid cross differ from monohybrid cross?
Monohybrid cross:

• It is a genetic cross, that involves a single pair of genes.
• In this cross, Parents differ by single trait (Eg : height)
• Monohybrid ratio in F2 generation is 3 : 1

Dihybrid cross:

• It is a genetic cross, that involves two pairs of genes, which are responsible for two traits.
• In this cross, the parents have two different independent traits. Eg: Flower colour, stem length
• The dihybrid ratio in F2 generation is 9 : 3 : 3 : 1

(b) What is cohesion?
The force of attraction between molecules of water is called cohesion.

Question 30.
(a) Give an account on vascular bundle of dicot stem.
The vascular bundles of dicot stem (Eg. sunflower) are conjoint, collateral, endarch and open. They are arranged in the form of a ring around the pith.

(b) Name the three basic tissue system in flowering plants.
The three basic tissue system in flowering plants are

1. Dermal or Epidermal tissue system
2. Ground tissue system
3. Vascular tissue system

Question 31.
(a) What are psychotropic drugs?
The drugs which act on the brain and alter the behaviour, consciousness, power of thinking and perception, are called Psychotropic drugs. They are also called Mood altering drugs.

(b) State the applications of DNA fingerprinting technique.

• DNA fingerprinting technique is widely used in forensic applications like crime investigation such as identifying the culprit.
• It is also used for paternity testing in case of disputes.
• It also helps in the study of genetic diversity of population, evolution and speciation.

Question 32.
(a) What are the applications of Doppler effect.
(i) To measure the speed of an automobile:

• A source emitting electromagnetic wave is attached to a police car
• The wave is reflected by a moving source.
• From the frequency shift, the over speeding vehicles can be determined.

(ii) Tracking a satellite:

• The frequency of radio waves decreases as the satellite passes away from the Earth.
• By this change in frequency, the location of the satellites is determined.

• The frequency change tracks the speed and location of the aircraft’s.

(iv) SONAR:

• There is a frequency change between the sent signal and received signal.
• This helps to determine the speed of submariners and marine animals.

(ii) A solution is made from 25 mi of ethanol and 75 ml of water. Calculate the volume percentage.
Volume of ethenol = 25 ml
Volume of water = 75 ml

Part – IV

(1) Answer all the questions. [3 x 7 = 21]
(2) Each question carries seven marks.
(3) Draw diagram wherever necessary.

Question 33.
(a) (i) What are the characteristic features of heat energy transfer?

• Heat always flows from a system of higher temperature to a system at lower temperature.
• The mass of a system is not altered when it is heated or cooled.
• Heat gained by the cold system is equal to heat lost by the hot system.

(ii) Explain linear expansion with diagram?

• When a body is heated or cooled, the length of the body changes due to change in its temperature.
• Then the expansion is said to be linear or longitudinal expansion.
• The ratio of increase in length of the body per degree rise in temperature to its unit length is called as the coefficient of linear expansion.
• It is different for different materials. Its SI unit is K-1.
• The equation is $$\frac{\Delta L}{L_{0}}$$ = αLΔT
ΔL – Change in length (Final length – Original length)
L0 – Original length (Final temperature – Initial temperature)
αL – Coefficient of linear expansion

[OR]

(b) (i) In the circuit diagram given below, three resistors R1, R2 and R3 of 5 Ω, 10Ω and 20Ω respectively are connected as shown.
Calculate:
(A) Current through each resistor
(B) Total current in the circuit
(C) Total resistance in the circuit

(A) Since the resistors are connected in parallel, the potential difference across each resistor
is same (i.e. V = 10V)
Therefore, the current through R1 is,

(B) Total current in the circuit,
I = I1 + I2 + I3
= 2 + 1 + 0.5 = 3.5 A

(C) Total resistance in the circuit

(ii) A charge of 12 coulomb flows through a bulb in 6 seconds. What is the current through the bulb?
Charge Q = 12 C, Time t = 6 s. Therefore,
Current I = $$\frac{Q}{t}=\frac{12}{6}$$ = 2A

Question 34.
(a) (i) Calculate the number of moles in 27 g of Al.

(ii) Find the percentages of nitrogen in ammonia.
Molar mass of NH3 = 1 × atomic mass of nitrogen + 3 × atomic mass of hydrogen
= (1 × 14) + (3 × 1)
= 14 + 3 = 17g

The percentage of nitrogen in ammonia = 82.35 %

(iii) Calculate the molecular mass of C2H6.
Molecular mass of C2H6 = (2 × atomic mass of C) + (6 × atomic mass of H)
= (2 × 12) + (6 × 1)
= 24 + 6 = 30 g

[OR]

(b) Write notes on various factors affecting solubility.
Factors affecting solubility : There are three main factors which govern the solubility of a solute. They are Nature of the solute and solvent, Temperature, and Pressure.

Nature of the solute and solvent:

• The nature of the solute and solvent plays an important role in solubility.
) Solubility is based on the phrase “like dissolves like.” The expression means that dissolving occurs when similarities exist between the solvent and the solute.
For example: Common salt is a polar compound and dissolves readily in polar solvent like water.
(iii) Non-polar compounds are soluble in non-polar solvents. For example, Sulphur dissolves in carbon di sulphide.

Effect of Temperature:
Solubility of Solids in Liquid:

• Generally, solubility of a solid solute in a liquid solvent increases with increase in temperature.
• In endothermic process, solubility increases with increase in temperature.
• In exothermic process, solubility decreases with increase in temperature.

Solubility of Gases in liquid:
Solubility of gases in liquid decrease with increase in temperature.

Effect of Pressure:
When the pressure is increased, the solubility of a gas in liquid increase. When pressure is increased, more gas molecules strike the surface of the liquid and mix into the solution.

Question 35.
(i) What is Cretinism?
Cretinism: It is caused due to decreased secretion of thyroid hormone (thyroxine) in children. The conditions are stunted growth, mental defect, lack of skeletal development and deformed bones.

(ii) Write the characteristics of insect pollinated flowers?

• Brightly coloured flowers, with smell and nectar.
• Pollen grain are larger in size, the exine is pitted, spiny so they can be adhered firmly on the sticky stigma.

(iii) What are the advantages of using bio-gas?
The advantages of using biogas are as follows:

• It bums without smoke and therefore causing less pollution.
• An excellent way to get rid of organic wastes like bio-waste and sewage material.
• Left over slurry is a good manure rich in nitrogen and phosphorus.
• It is safe and convenient to use.
• It can reduce the amount of green house gases emitted.

[OR]

(b) (i) What is hermaphrodite?
Hermaphrodite: If the male and female reproductive organs are present in the same individual, it is called hermaphrodite animal. Eg. Leech.

(ii) What are the effects of hybrid vigour?
The superiority of the hybrid obtained by cross breeding is called as heterosis or hybrid vigour. Its effects in animal breeding are:

• Increased production of milk by cattle
• Increased production of egg by poultry
• High quality of meat is produced
• Increased growth rate in domesticated animals.

## Tamil Nadu 12th Tamil Model Question Papers 2020-2021

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12th Tamil Model Question Papers 2019-2020 Tamil Nadu

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## Tamil Nadu 12th English Model Question Papers 2019-2020

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## Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium

Students can Download Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium Pdf, Tamil Nadu 12th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## TN State Board 12th Chemistry Model Question Paper 1 English Medium

Instructions:

1. The question paper comprises of four parts
2. You are to attempt all the parts. An internal choice of questions is provided wherever: applicable
3. All questions of Part I, II, III and IV are to be attempted separately
4. Question numbers 1 to 15 in Part I are Multiple choice Questions of one mark each.  These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are lo be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are lo be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are lo be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 70

Part-I

Answer all the questions. Choose the correct answer. [15 x 1 = 15]

Question 1.
The change in Gibbs free energy for a reaction is expressed by ……………… .
(a) ∆G = ∆H + T∆S
(b) ∆G = ∆H – TS
(c)G = H-TS
(d) ∆G = ∆H – T∆S
(d) ∆G = ∆H – T∆S

Question 2.
Consider the following statements
(i) In interhalogen compounds, the central atom will be the smaller one.
(ii) It can be formed only between two halogen and not more than two halogens.
(iii) They are strong reducing agents.
Which of the above statement(s) is / are not correct?
(a) (i) only
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (iii) only
(c) (i) and (iii)

Question 3.
Thermodynamically the most stable form of carbon is ………….
(a) Diamond
(b) graphite
(c) Fullerene
(d) none of these
(b) graphite

Question 4.
Which one of the following transition element has maximum oxidation states?
(a) Manganese
(b) Copper
(c) Scandium
(d) Titanium
(a) Manganese

Question 5.
An excess of silver nitrate is added to 100ml of a 0.01M solution of penta aquachlorido chromium (III) chloride. The number of moles of AgCl precipitated would be
(a) 0.02
(b) 0.002
(c) 0.01
(d) 0.2
(b) 0.002

Question 6.
Match the following.

Question 7.
If the initial concentration of the reactant is doubled, the time for half reaction is also doubled. Then the order of the reaction is ………..
(a) Zero
(b) one
(c) Fraction
(d) none
For a first order reaction t1/2 is independent of initial concentration.
i.e n ≠ 1; for such cases.

Question 8.
Which of the following is not a buffer solution?
(a) CH3COOH + CH3COONa
(b) NH4OH + NH4Cl
(c) H2CO3 + NaHCO3
(d) NaOH + NaCl
(d) NaOH + NaCl

Question 9.
In the electrochemical cell: ZnlZnSO4 (0.01M)||CuSO4 (1.OM)|Cu, the emf of this Daniel cell is E1 When the concentration of ZnSO4 is changed to 1.0M and that CuSO4 changed to 0.01M, the emf changes to E2 From the followings, which one is the relationship between E1 and E2?
(a) E1 < E2
(b) E1 > E2
(c) E2 = 0↑E1
(d) E1 = E2

(b) E1 > E2

Question 10.
Which method is used to prepare metal sols?
(a) ultrasonic dispersion
(b) mechanical dispersion
(c) Bredig’s arc method
(d) peptisation
(c) Bredig’s arc method

Question 11.
What are the products formed when methoxy ethane is treated with hydroiodic acid?
(a) Phenol + iodomethane
(b) Iodomethane + Ethanol
(c) Iodoethane + Methanol
(d) Iodobenzene + Methane
(b) Iodomethane + Ethanol

Question 12.
Predict the product Z in the following series of reactions

(a) (CH3)2C(OH)C6H5

Question 13.
Assertion(A): 2-nitro propane is more acidic than nitro methane.
Reason (R): When the number of alkyl group attached to a carbon increases, acidity decreases, due to +I effect of alkyl groups.
(a) Both A and R are correct but R is not the correct explanation of of A
(b) Both A and R are correct and R is the correct explanation of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
(d) A is wrong but R is correct

Question 14.
Glucose(HCN) Product (hydrolysis) Product (HI + Heat) A, the compound A is
(a) Heptanoic acid
(b) 2-Iodohexane
(c) Heptane
(d) Heptanol
(a) Heptanoic acid

Question 15.
Which one of the following is used to provide relief from the allergic effects?
(a) cetrizine
(b) ampicillin
(c) erythromycin
(d) milk of magnesia
(a) cetrizine

Part – II

Answer any six questions. Question No. 21 is compulsory. [6 x 2 = 12]

Question 16.
Magnesite (Magnesium carbonate) is calcined to obtain magnesia, which is used to make refractory bricks. Write the decomposition reaction.
Magnesite is a carbonate of magnesium. Magnesite when heated at 800°C to 1000°C at the CO2 content in it is driven off. The residue so obtained is known as calcined magnesite.

Question 17.
Mention the application of Xenon?

• Xenon is used in fluorescent bulbs, flash bulbs and lasers.
• Xenon emits an intense light in discharge tubes instantly. Due to this it is used in high speed electronic flash bulbs used by photographers.

Question 18.
Compare the stability of Ni4+ and Pt4+ from their ionisation enthalpy values.

 lE Ni Pt I 737 864 II 1753 1791 III 3395 2800 IV 5297 4150

(i) Ni4+I.E. = 737+ 1753 + 3395 + 5297= 11182 kJmol-1
(ii) Pt4+ I.E. = 864 + 1791 + 2800 + 4150 = 9605 kJ mol-1
Pt4+ compounds are stable than Ni4+ compounds because the energy needed to remove 4 electrons in Pt is less than that of Ni.

Question 19.
Why is glass considered as super cooled liquid?
Glass is an amorphous solid. Like liquids it has a tendency to flow, though very slowly. The proof of this fact is that glass panes in the windows or doors of old buildings are invariably found to be slightly thicker at the bottom than at the top.

Question 20.
Write Arrhenius equation and explains the terms involved.
Arrhenius equation : $$\mathrm{k}=\mathrm{Ae}^{-}\left(\frac{\mathrm{E}_{a}}{\mathrm{RT}}\right)$$
where, k = Rate constant
A = Arrhenius factor (frequency factor)
Ea = Activation energy
R = Gas constant
T = Absolute temperature (in K)

Question 21.
Can Fe3+ oxidises Bromide to bromine under standard conditions?

Required half cell reaction

If E°cell is -ve; ∆G is +ve and the cell reaction is non-spontaneous.
Hence, Fe3+ cannot oxidise Bromide to Bromine.

Question 22.
What are used of Urotropine? Give its structure.
This is the structure of Urotropine (Hexamethylene tetramine)

• Urotropine is used as a medicine to treat urinary infection.
• Nitration of Urotropine under controlled condition gives an explosive RDX (Research and development explosive). It is also called cyclonite or cyclotri methylene trinitramine.

Question 23.
Ethylamine is soluble in water whereas aniline is not. Give reason.
Ethylamine when added to water forms intermolecular H-bonds with water. And therefore it is soluble in water. But aniline does not form H-bond with water to a very large extent due to the presence of a large hydrophobic – C H group. Hence, aniline is insoluble in water.

Question 24.
What are antihistamines? Give example and mention its use.

• Antihistamines block histamine release from histamine – 1 receptors.
• It is used to provide relief from the allergic effects.

Part – III

Answer any six questions. Question No. 31 is compulsory. [6 x 3 = 18]

Question 25.
Explain the action of heat on boric acid.
Boric acid when heated at 373 K gives metaboric acid and at 413 K, it gives tetraboric acid. When heated at red hot, it gives boric anhydride which is a glassy mass.

Question 26.
Draw the structure of (a) white phosphorous (b) red phosphorous
(a) White phosphorous

(b) Red phosphorous

Question 27.
Why do zirconium and Hafnium exhibit similar properties?

1. The element of second and third transition series resemble each other more closely than the elements of first and second transition series due to lanthanoid contraction.
2. e.g., Zr- 4d series -Atomic radius 145 pm
Hf — 5d series – Atomic radius 144 pm
3. The radii are very similar even though the number of electrons increases.
4. Zr and Hf have very similar chemical behaviour, having closely similar radii and electronic configuration.
5. radius dependent properties such as lattice energy, solvation energy are similar.
6. Thus lanthanides contraction leads to formation of pair of elements and those known as chemical twins, e.g., Zr – Hf

Question 28.
How will derive the formula of density of a unit cell?
Using the edge length of a unit cell, we can calculate the density (p) of the crystal by considering a cubic unit cell as follows.

Substitute the value (3) in (2)
Mass of the unit cell = $$n \times \frac{\mathbf{M}}{N_{\mathrm{A}}}$$
For a cubic unit cell, all the edge lengths are equal, i.e., a = b = c
Volume of the unit cell = a x a x a = a3
∴ Density of the unit cell = $$\rho=\frac{n M}{a^{3} N_{A}}$$

Question 29.
The salt of strong acid and strong base does not undergo hydrolysis. Explain.
(i) In this case, neither the cations nor the anions undergo hydrolysis. Therefore the solution remains neutral.
(ii) For example, in the aqueous solution of NaCl, its ions Na+ and Cl ions have no tendency to react with H+ or OH ions of water. This is because the possible products of such reaction are NaOH and HC1 which are completely dissociated. As a result, there is no change in the concentration of H+ and OH ions and hence the solution continues to remain neutral.

Question 30.
Write a note about medicinal applications of colloids.

• Antibodies such as penicillin and streptomycin are produced in colloidal form for suitable injections. They cure pneumonia.
• Colloidal gold and colloidal calcium are used as tonics.
• Milk of magnesia is used for stomach troubles.
• Silver sol protected by gelatine known as Argyrol is used as eye lotion.

Question 31.
Predict the major product, when 2-methyl but -2-ene is converted into an alcohol in each of the following methods.
(i) Acid catalysed hydration (ii) Hydroboration (iii) Hydroxylation using bayers reagent

Question 32.
Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.
Aromatic primary amines react with HNO2 at 273-278 K to form aromatic diazonium salts.

Aliphatic primary amines also react with HNO2 at 273-278 K to form aliphatic diazonium salts. But these are unstable even at low temperature and thus decomposes readily to form a mixture of compounds consisting of alkyl chlorides, alkenes and alcohols, out of which ‘ alcohols generally predominates.

Question 33.

• Uses of preservatives reduce the product spoilage and extend the shelf-life of food.
• Addition of vitamins and minerals reduces the mall nutrient.
• Flavouring agents enhance the aroma of the food.
• Antidxidants prevent the formation of potentially toxic oxidation products of lipids and other food constituents.

Part – IV

Answer all the questions. [5 x 5 = 25]

Question 34.
(a) (i) Out of coke and CO, which is better reducing agent for the reduction of ZnO? Why? (2)
(ii) An element (A) extracted from kernite. A reacts with nitrogen at high temperature gives B. A reacts with alkali to form C. Find out A, B and C. Give the chemical equations. (3)
[OR]
(b) (i) Deduce the oxidation number of oxygen in hypofluorous acid – HOF. (2)
(ii) Which catalyst is used in the conversion of acetaldehyde to acetic acid? Give equation. (3)
(a) (i) Coke (C) is a better reducing agent for the reduction of ZnO.
Because, when we use coke, the reduction can be easily carried out at 673 K. Thus Carbon (Coke) reduces zinc oxide more easily than carbon monoxide (CO). From the Ellingham diagrams, it is quite clear that the reduction of zinc oxide is more favourable using coke (∆G for the formation of carbon monoxide from carbon is more negative).

(ii)

• An element (A) extracted from kernite is boron.
• Boron reacts with nitrogen at high temperature to give. Boron nitride (B)
• Boron reacts with alkali (NaOH) to give sodium borate (C).

 A Boron B B Boron nitride BN C Sodium borate Na3BO3

(b) (i)

[OR]

In case of O – F bond is HOF, fluorine is most electronegative element. So its oxidation number is -1. Thereby oxidation number of O is +1. Similarly in case of O – H bond is HOF. O is highly electronegative than H. So its oxidation number is -1 and oxidation number of H is +1. So, Net oxidation of oxygen is – 1 + 1 = 0.
(ii)

Question 35.
(a) (i) Prove that H2SO4 is a strong dibasic acid. (2)
(ii) Distinguish tetrahedral and octahedral voids. (3)
[OR]
(b) In an octahedral crystal field, draw the figure to show splitting of d orbitals. (5)
(a) (i) Sulphuric acid forms two types of salts namely sulphates and bisulphates.

(ii)

 Tetrahedral void Octahedral void 1. A single triangular void in a crystal is surrounded by four (4) spheres and is called a tetrahedral void A double triangular void like c is  surrounded by six(6) spheres and is called an octahedral void 2. A sphere of second layer is above the void of the first layer, a tetrahedral void is formed The voids ¡n the first layer are partially covered by the spheres of layer now such a void is called a octahedral void 3. This constitutes four spheres, three in the lower and one in upper layer. When the centres of these four spheres are joined a tetrahedron is formed This constitutes six spheres, three in the lower layer and three in the upper layer. When the centers of these six spheres are joined an octahedron is formed 4. The radius of the sphere which can be accommodated in an octahedral hole without disturbing the structure should not exceed 0.414 times that of the structure forming sphere The sphere which can be placed in  a tetrahedral hole without disturbing the close packed structure should not have a radius larger than 0.225 times the radius of the sphere.forming the structure 5. Radius of an tetrahedral void r/R = 0.225 Radius of a octahedral void r/R = 0.414

[OR]

(b) Step 1: In an isolated gaseous state, all the five d orbitals of the central metal ion are degenerate.
Initially, the ligands form a spherical field of negative charge around the metal. In this filed, the energies of all the five d orbitals will increase due to the repulsion between the electrons of the metal and the ligand.

Step 2: The ligands are approaching the metal atom in actual bond directions. To illustrate this let us consider an octahedral field, in which the central metal ion is located at the origin and the six ligands are coming from the +x, -x, +y, -y,
+z and -z directions as shown below.

As shown in the figure, the orbitals lying along the axes dx2 – y2 and dz2 orbitals will experience strong repulsion and raise in energy to a greater extent than the orbitals with lobes directed between the axes (dxy, dyz and dzx). Thus the degenerate d orbitals now split into two sets and the process is called crystal field splitting.

Step 3: Up to this point the complex formation would not be favoured. However, when the ligands approach further, there will be an attraction between the negatively charged electron and the positively charged metal ion, that results in a net decrease in energy. This decrease in energy is the driving force for the complex formation.
During crystal field splitting in octahedral field, in order to maintain the average energy of the orbitals (barycentre) constant, the energy of the orbitals dx2 y2 and d z2 (represented as e2 orbitals) = will increase by 3/5 ∆0 while that of the- other three orbitais dxy , dyz and dzx (represented t2gas t orbitais) decrease by 2/5 ∆0 . Here, ∆0 represents the crystal field splitting energy in the octahedral field.

Question 36.
(a) Explain briefly the collision theory of bimolecular reactions. (5)
[OR]
(b) Discuss about the hydrolysis of salt of weak acid and weak base and derive pH value for the solution. (5)
(a) Collision theory is based on the kinetic theory of gases. According to this theory, chemical reactions occur as a result of collisions between the reacting molecules. Let its understand this theory by considering the following reaction.
A2(g) + B2(g) → 2AB(g)

If we consider that, the reaction between A2 and B2 molecules proceeds through collisions between them, then the rate would be proportional to the number of collisions per second.
Rate ∝ Number of molecules colliding per litre per second (or) Rate ∝ Collision rate.
The number of collisions is directly proportional to the concentration of both A2 and B2.
Collision rate ∝ [A2][B2] ; Collision rate = Z [A2][B2] where, Z is a constant.

The collision rate in gases can be calculated from kinetic theory of gases. For a gas at room temperature (298K) and 1 atm pressure, each molecule undergoes approximately 109 collisions per second, i.e., 1 collision in 10-9 second. Thus, if every collision resulted in reaction, the reaction would be complete in 10-9 second. In actual practice this does not happen. It implies that all collisions are not effective to lead to the reaction.

In order to react, the colliding molecules must possess a minimum energy called activation energy.
The molecules that collide with less energy than activation energy will remain intact and no reaction occurs.

Fraction of effective collisions (f) is given by the following expression, $$\mathbf{f}=e^{\frac{-\mathbf{E}_{\mathrm{a}}}{\mathrm{RT}}}$$

Fraction of collisions is further reduced due to orientation factor i.e., even if the reactant collide with sufficient energy, they will not react unless the orientation of the reactant molecules is suitable for the formation of the transition state.

The fraction of effective collisions (f) having proper orientation is given by the steric factor R ’ Rate = P x f x collision rate

Where k is the rate constant.
On comparing equation (1) and (2), the rate constant k is,
k = PZ$$\frac{-E_{a}}{R T}$$

[OR]

(b) i. Consider the hydrolysis of ammonium acetate
CH3COONH4(aq) CH3COO(aq) + NH+4(aq)

ii. In this case both the cation (NH4+ ) and (CH3COO) anion have the tendency to react with water.
CH3COO+H2 ⇌ CH3COOH + OH
NH4+ + H2O ⇌ NH4OH + H+

iii. The nature of the solution depends on the strength of acid (or) base i.e., if Ka > Kb, then the solution is acidic and pH < 7, if Ka < Kb then the solution is basic and pH > 7.
If Ka= Kb, then the solution is neutral.
iv. The relation between the dissociation constant Ka, Kb and hydrolysis constant is given
Ka. Kb .Kh = Kw
v. pH of the solution pH = 7 + 1/2 PKa – 1/2 pKb

Question 37.
(a) (i) Ionic conductance at infinite dilution of Al3+and SO42- are 189 and 160 mho cm2 equiv-1 . Calculate the equivalent and molar conductance of the electrolyte Al2(SO4)3 at infinite dilution. (3)
(ii) Suggest a way to determine λ0m value of water (2)

(OR)

(b) (i) Why does bleeding stop by rubbing moist alum. (2)
(ii) How is glycerol reacts with fuming nitric acid? (or) How would you convert glycerol into nitroglycerine? (3)

(b) (i) Blood is a colloidal sol. When we rub the injured part with moist alum then coagulation of blood takes place. Hence main reason is coagulation, which stops the bleeding. Therefore bleeding stop by rubbing moist alum.
(ii)

Question 38.
(a) (i) What is crossed cannizaro reaction? Explain it. (2)
(ii) Complete the following reactions.(3)

(b) (i) Is the following sugar, D – sugar or L – sugar? (2)

(ii) Write a note on vulcanization of rubber. (3)
(a) (i) When Cannizaro reaction (Auto redox reaction) takes place between two different aldehyde, the reaction is called as crossed cannizaro reaction.

In crossed cannizaro reaction more reactive aldehyde is oxidized and less reactive aldehyde is reduced.

[OR]

(b) (i) L-Sugar
(ii)

• Natural rubber is very soft and brisky. it has high water absorption capacity and low tensile strength. Its properties can be improved by a process called vulcanization.
• Natural rubber is mixed with 3-5% sulphur and heated at 100-150°C causes cross linking of the cis – 1, 4 polyisoprene chains through disulphide (-S-S-) bonds.
• The physical properties of rubber can be altered by controlling the amount of sulphur that is used for vulcanization. When 3 to 10% sulphur is used the resultant rubber is somewhat harder but flexible.

Following properties of rubber improved by vulcanization:

• Tensile strength
• Elasticity
• Hardness
• Tear strength
• Resistance to solvants.

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