Class 11

Samacheer Kalvi 11th English Solutions Prose Chapter 6 The Accidental Tourist

Students can Download English Lesson 6 The Accidental Tourist Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 11th English Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations. Learn the Samacheer Kalvi 11th English Grammar to enhance your grammar skills like reading comprehension, passage writing, parts of speech, tenses, passive and active voice, and many other concepts in no time.

Tamilnadu Samacheer Kalvi 11th English Solutions Prose Chapter 6 The Accidental Tourist

Warm up

Question 1.
Often on formal occasions, we admire friends and strangers who appear elegant, who are pleasant to converse with and who conduct themselves gracefully. At times, we also see people who are awkward, nervous and doubtful about their next move.
Recall a few examples of awkward actions that can cause discomfort or disturbance to others like spilling a cup of hot drink on someone nearby.
Answer:
Mr. X has travelling sickness. Once he was in a city bus, his wife advised him to take a polythene bag so that he can vomit inside it if he felt like it. But he forgot her advice. As the bus moved on, he had a feeling that he wasn’t fine. He craned his neck out of the window of the bus and vomited his undigested breakfast.

As the bus was moving fast, people seated behind him had their shirts and sarees sprinkled with the undigested breakfast and dinner. The conductor and driver got upset. The whole day, passengers refused to sit on those three seats.

Samacheer Kalvi 11th English The Accidental Tourist Textual Questions

A. Based on your understanding of the lesson, answer the following questions in one or two sentences each:

Question 1.
Give a few instances of Bryson’s confusing acts.
Answer:
He would end up standing in an alley on the wrong side of a self-locking door trying to locate a restroom in a cinema theatre. He would often go to the hotel desk, atleast two to three times a day asking what was his room number. The author had a serious problem of orientation in any new place:

Question 2.
What were the contents of the bag?
Answer:
The contents of the bag were newspaper cuttings, loose papers, 14-ounce tin of pipe tobacco, magazines, passport, English money, and film.

Samacheer Kalvi 11th English Solutions Prose Chapter 6 The Accidental Tourist

Question 3.
Describe the Buttery cascade of things tumbling from the bag.
Answer:
The side of the bag flew open and everything within was extravagantly ejected over an area about the size of a tennis court. The carefully stored documents came raining down in a fluttery cascade. Coins bounced to a variety of noisy oblivions. Lidless tobacco tin rolled crazily across the concourse disgorging its content as it went.

Question 4.
Why did the author’s concern over tobacco shift to his finger?
Answer:
The author s concern over tobacco shifted to his finger as he gashed his finger on the zip and blood started shedding in a lavish manner.

Question 5.
What happened to Bryson when he leaned to tie his shoelace?
Answer:
When Bryson leaned to tie a shoe lace inside the air-craft, someone in the seat ahead of him threw his seat back into full recline. The author found himself pinned helplessly in a crash position.

Question 6.
How did Bryson free himself from the crash position?
Answer:
By clawing (hurting) the leg of the man sitting next to him, Bryson managed to free himself.

Question 7.
Give a brief account of the embarrassing situation of Bryson when he knocked down the drink.
Answer:
Bryson rose from the dinner table looking as if he had just experienced a localized seismic event. While opening the lid, he used to spill the contents on his family and fellow passengers and thus embarrassing them. To avoid such anxious moments, his wife used to give directions to the children. “Take the lids off the food for daddy”.

Question 8.
What was Bryson’s worst accident on a plane?
Answer:
His worst accident as he fell into a conversation with a young lady in the next seat for 20 minutes sucking his pen. Later he discovered that his pen had leaked and that his mouth, chin, tongue, teeth, and gums remained navy blue for several days.

Question 9.
What did Bryson wish to avoid in his life?
Answer:
Bryson wished to avoid air travel especially with his family members.

Question 10.
How would staying away from liquid mischief benefit Bryson?
Answer:
Staying away from liquid mischief benefitted Bryson by cutting down his laundry bills.

B. Answer the following questions in about three to four sentences each:

Question 1.
Why doesn’t Bryson seem to be able to do easily what others seem to? Give a few reasons.
Answer:
Bryson lacks orientation. He has. great capacity in forgetting hotel rooms, the location of rest rooms in a cinema hall and the number of his room in a hotel. He goes atleast three times a day to the reception desk to ask which room is allotted to him.

Question 2.
What was the reaction of Bryson’s wife to his antics?
Answer:
Bryson’s wife was not angry or irritated by looking at his behaviour. She was rather struck with wonder thinking how a normal human being could behave in that manner.

Question 3.
Briefly describe the ‘accidents’ encountered on the flight by Bryson.
Answer:
Once the author was trying to tie a shoe lace. He bent to tie it. Just that moment, the passenger in the front seat reclined in full swing. The author was pinned down in a crash position. On another occasion, he engaged an attractive woman in a witty conversation. While talking to her, he sucked his pen. The pen left a scrub-resistant navy blue on his gum, mouth, teeth and his chin for several days.

C. Based on your understanding of the text, answer the following questions in a paragraph of about 100 – 150 words:

Question 1.
‘To this day, I don’t know how I did it’ – What does ‘it’ refer to?
Answer:
‘It’ refers to the incident which Bill Bryson experienced while knocking down a soft drink to the lap of a sweet little lady sitting beside him. It was really the most embarrassing moment for him. When he knocked a soft drink onto the lap of the lady, the flight attendant came and cleaned her up. He was then provided with a replacement drink which he knocked it onto the woman again.

He was just thinking about it and could not understand how it happened repeatedly. He felt that it so happened in a strange manner as what usually happened in 1950s horror movies with the title ‘The undead Limb’. The lady looked at him with the stupefied expression which was quite natural and uttered an oath that was not heard by Bryson before. Thus ‘It ‘was really an embarrassing moment for Bryson.

Question 2.
But, when it’s my own – well, I think hysterics are fully justified’ – How?
Answer:
The author had planned to go to England with all his family members. He arrived at the Logan airport at Boston. When they were checking in, he suddenly remembered that he forgot to use his frequent flier card (British Airways). He also remembered how he had left it in a bag. He tried to open the bag. The zip was jammed. He tried to open it by force. After several attempts, it gave away spilling all the contents in a sprawling corridor in the airport. He ignored the flying documents, silver coins and even passport.

He worried about the tobacco-box which was rolling away crazily disgorging its content on the way. He cried “My Tobacco” remembering how expensive it would be to buy tobacco for his pipe in England. Just then he realized that he was bleeding profusely. He had made a gash on his finger while trying to open the zip of his bag by force. He cried hysterically on seeing his own blood, “My finger” My finger”. In general, he was not comfortable flowing other’s blood. But when it came to spilling his own blood “hysterics” was really justified.

“Relived stress through hysterical screaming.”

Question 3.
Bring out the pun in the title ‘The Accidental Tourist’ (one who happens to travel by accident or one who meets with accidents often on his or her trips!).
Answer:
A pun is the humorous use of words that have two meanings. The title ‘The Accidental tourist’ is a pun which brings out two meanings. ‘One who travels by accident’ and the other meaning is the one who meets with an accident often in his/her trip. The second meaning is what aptly suits this story. We come to know that Bill Bryson, a globetrotter often meets with catastrophes when he is there on travel.

Once on an aeroplane when he tried to tie his shoelace leaning down he got pinned helplessly in crash position just as the person ahead of him threw his seat back to rest. The next accident has he knocked down a soft drink onto the lap of a young lady who was sitting beside him. The worst accident was sucking his pen thoughtfully without knowing that it was leaking and found his mouth remained navy blue for several days. All the above incidents clearly bring out the pun in the title.

Question 4.
Can a clumsy person train himself/herself to overcome short comings? How could this be done?
Answer:
Yes, a clumsy person can train himself to overcome his short comings. People with severe Parkinson’s disease to learn to hold a spoon and eat with great difficulties. Children with multiple disabilities and nervous problems learn to button up their shirts, tie the shoe laces and even assist friends with similar ailments. There is no difficulty in the world that is insurmountable. Practice makes one perfect. The author’s wife, without rebuking him publicly for his callous clumsiness, trains her children to be supportive of their father Mr. Bryson who always spills drinks or bumps on something or even sits on chewing gum or spilled oil.

As clumsiness is not a welcome behavior pattern among adults, one must learn how to mend oneself and try to overcome one eccentricity per day with deliberate effort. The author does admit to restrain himself to reduce the laundry bill. But if he extends his efforts even during the presence of his family on long distance air-travels, it would be nice. Cultivating acceptable behaviour in public places is not an impossible task for any sensible man.

‘‘I am not clumsy. It s just the floor that hates me.
The tables and chairs are bullies
And the wall gets in the way.”

Question 5.
As a fellow passenger of Bill Bryson on the flight, make a diary entry describing his clumsy behaviour during the trip and the inconveniences caused to others as a result of his nervousness.
Answer:
I had a trip on a plane where I happened to meet a clumsy person named Bryson. There arose many inconveniences to fellow passengers because of his clumsy behaviour. He leaned over to tie his shoelace and was pinned in the crash position as the person ahead threw his seat back to relax.

He might have informed the person in his front row to pull the seat front. Instead, he hurt the leg of the man sitting next to him and freed himself. Later he knocked down a soft drink onto the lap of a young lady who was sitting next to him. The lady got drenched and was helped by the flight assistant to clean herself. To her surprise, Bryson knocked down yet another cup of drink on her lap. She was really shocked by his behaviour and didn’t know what to do. I understood that everything happened because of his nervousness.

‘‘I have always had a reputation as a Bufoon.”

Additional Questions

Question 6.
Bill Bryson “ached to be suave”. Was he successful in his mission? List his “unsuave ways.
Answer:
Bill Bryson expresses his genuine desire to be “suave”. He would love just once in his life time to rise from the dinner table as if he had experienced an “extremely localized seismic event, get into a car without leaving 14 inch coat outride, wear light-coloured trousers without ever discovering at the end of the day that he had at various times and places sat on chewing gum, ice-cream cough syrup and motor oil. No, Bill Bryson was not successful in his mission. Twice he spilled his drinks on a sweet nun who happened to sit next to him. He tried to show off his wisdom to another attractive lady. As usual, he was sucking his pen.

His shirt, teeth and gum carried the unscrubbable navy blue stain for many days. He always did “liquid mischief’. His clumsy behaviour in the aeroplane made the saintly mm use abusive language. To avoid unsuave ways, he gave up air-travel with his family members. His wife and children supported him yet failed to be refined in manners.
‘‘Heroes, well, they don’t live so long.
But they ’re too suave, and we all admire them.”

Vocabulary

A. Foreign words and phrases .
You have come across the French phrases ‘en famille’ an <famille> and ‘bons mots’ ,ban ma:ts in the lesson. Now look at the following phrases and their meanings.

  • viva voce – /vaivo vausi /- a spoken examination
  • sine die -/sina’dAii:/- without a date being fixed
  • resume -/rezju:mei/- a brief summary
  • rapport -/rae’pa:(r)/- close relationship with good understanding
  • bonafide – /bauna faidi/- genuine

B. Refer to the dictionary and find out the meanings of the following foreign words /phrases. Use them in sentences of your own:

  1. bon voyage
  2. in toto
  3. liaison
  4. ex gratia
  5. en masse
  6. en route
  7. ad hoc
  8. faux pas

1. Bon voyage – Express good wishes to some on leaving for a long journey. Hemalatha went to the airport and said “Bon voyage” to Keerthi who was about to leave for UK.

2. In toto- as a whole they accepted the business plan of Murali in toto.

3. Liaison – a close working relationship between people and organization. .The bank clerk regretted his liaison with the watchman who robbed the bank and vanished.

4. Ex gratia-compensation paid by the Government to the victim of an accident.
The Hon’ble Chief Minister gave each of the eight survivors of the Road accident an ex gratia of . two lakh rupees.

5. En masse – in a group all together
The striking workers ran en masse to the gate when the boss arrived.

6. Enroute – on the way
He stopped in Mumbai en route to Kolkatta.

Samacheer Kalvi 11th English Solutions Prose Chapter 6 The Accidental Tourist

7. Adhoc – created for a particular
An Adhoc committee was set up to probe the scam in the universities.

8. Faux pas – an embarrassing or tactless act, blunder
I did not disclose his fauxpas till he joined a college. .

Here is a list of some words borrowed from Indian languages and have been included in the Dictionary of English. Add more words to the table.

WORD ORIGIN MEANING
veranda Hindi a roofed platform along the outside of a house
bungalow Hindi a house in the Bengal style
chutney Hindi a ground or mashed relish
cheetah Sanskrit uniquely marked
coir Malayalam rope
bamboo Kannada wood
bandicoot Telugu kind of rat
catamaran Tamil multi-hulled watercraft
guru Sanskrit master

C. Idioms

Look at the list of idioms given below. Find their meanings from a dictionary. Read the sentences that follow and replace the words in italics with the appropriate idioms, making suitable changes wherever necessary.

(a) right up one’s alley –
(b) drive one up the wall –
(c) hit the road-
(d) take (one) for a ride –
(e) in panic mode-

(a) The old man got irritated at the loud noise outside.
(b) We were driving, when it started raining heavily. After stopping for an hour, we began the journey again.
(c) Ramesh gave false excuses for not attending the meeting and deceived me.
(d) At the interview when questions were fired at me rapidly, I forgot every thing and grew irritated.
(e) I love thrillers and this book appeals to me strongly.
Answers
(a) driven up the wall
(b) hit the road
(c) he took me for a ride
(d) got into a panic mode
(e) is right up to my alley

listening Activity

Bala : Hello, Mahesh; where did you spend your holidays?
Mahesh : Well Bala, I had been to a place of ethereal beauty, Kerala.
Bala : Did you spend the three day holiday at Kerala?
Mahesh : Yes Bala. I visited Trivandrum, Quilon and Cochin.
Bala : What are the important tourist spots you visited there?
Mahesh : I shall just mention a few – The Art Museum called Chitralayam at Trivandrum, the King’s palace and the Kovalam Beach.
Bala : What did you see in Quilon?
Mahesh : The lovely scenery at Varkala, the oldest port of Quilon and the Periyar Lake Wild Life Sanctuary.
Bala : Tell me something about Cochin.
Mahesh : Cochin has earned the title ‘Venice of the East’. It is famous for coir and there are many historical monuments there. .
Bala : Oh, I see! I think you should take me to these spots next time you visit Kerala.
Mahesh : Of course! We shall explore its beauties and learn more about our motherland.

Listen to the dialogue read out by the teacher or to the recorded version and answer the questions that follow:

Question (i)
______ was one of the places visited by Mahesh.
(a) Srilanka
(b) Goa
(c) Kasi
(d) Cochin
Answer:
(d) Cochin

Question (ii)
The Art Museum at Trivandrum is called ________
(a) Swamalayam
(b) Gitalayam
(c) Chitralayam
(d) Saranalayam
Answer:
(c) Chitralayam

Question (iii)
Varkala is the oldest port of ________
(a) Quilon
(b) Andhra
(c) Puducherry
(d) the Andamans
Answer:
(a) Quilon

Question (iv)
Mahesh had been to the ________ Lake Wild Life Sanctuary.
(a) Chidambaram
(b) Pulicat
(c) Kovalam
(d) Periyar
Answer:
(d) Periyar

Question (v)
Cochin is called the ________ of the East.
(a) Granary
(b) Cuba
(c) Venice
(d) Pearl
Answer:
(c) Venice

Speaking Activity

(a) Build a dialogue of 8-10 exchanges between your friend and yourself, on the following situation:

You were to board a train to Delhi. By mistake you got into the wrong train and fought for your seat there. On realising your mistake, you left the train shamefaced, after creating a commotion there. Role-play this situation before the class.
James : Last Friday I was caught in an embarrassing situation.
Rahim: Where? How?
James : All the passengers were waiting at the Chennai central station.
Rahim : Where were you heading to? .
James : Well, I was heading to New Delhi.
Rahim: What was the cause of your embarrassment?
James : I did not know about the last-minute change of platform. I had got an open ticket to Delhi. Along with many other passengers, I rushed to the unreserved compartment and I boarded it in platform 6. But the platform was changed to 11. I got a comer seat and had a sense of pride.
Rahim : Then when did you realize your mistake?
James : A well-dressed boy asked me if I was going to Mysore. I was surprised and said that I was going to Delhi. He laughed aloud and said to the fellow passengers look, this young man is going to Delhi by Cauvery express. The train had started moving. I rushed out with my luggage and jumped out.
Rahim : Did you catch Tamil Nadu express or not?
James : Of course I did, but I had to travel standing for one night.
Rahim : I’m really sorry to hear it. Be careful in future.

(b) Speak to the class for a minute, as to how one should conduct oneself on formal occasions. (You could talk about table-manners especially while eating, general appearance, manner of speaking, etc.)

Manners maketh a man. While talking, one must look into the eyes of the person spoken to.
When someone else is talking, one should have the decency to pay attention and restate what was gathered to confirm one has understood the essence of the talk. When someone is asking something or enquiring something, one should not be meddling with other things. Such a behaviour amounts to an insult to the speaker. While eating along with friends, when you finish, never get up and leave the table to wash the hands. While eating, one must eat without making much noise. Care must be taken not to spill curry, drinks on the fellow diners.

While attending an interview or going to the college, one must wear neat dress and greet others with a cheerful face. While talking to elders one must always use “Sir” or Madam. While requesting use “please”. When receiving a help, remember to say ‘thank you’. When someone is hurt by an un wise remark, be quick to apologise or say “sorry”.

Reading

Caesar, the Hero of Mumbai on 26/11

1. Mumbai Caesar, the last surviving hero of his kind, died after the attack on one Thursday. Caesar, a Labrador retriever, was covered with tri-colour and given an emotional farewell from the city Police Force. The Mumbai Police Commissioner too marked the passing of the hero with a tweet.

2. Caesar, who was 11 years old was the sole survivor among the dogs of Mumbai Police who took part in bomb detection operations during the terrorist attack on Mumbai that began on November 26, .2008. He died of heart attack at a farm in Virar where he and his three canine buddies had been sent after retirement. During the terror attack in Mumbai, Caesar saved several lives when he sniffed out the hand grenades left by the terrorists at the busy CST railway station.

Samacheer Kalvi 11th English Solutions Prose Chapter 6 The Accidental Tourist

3. Caesar was also a part of the search team at Nariman house, where terrorists were held up for three days. Earlier he was also pressed into service for bomb search operation after the 2006 serial train blasts and July 2017 blast in Mumbai. The Mumbai police officials also tweeted their grief saying, “Services of retired members of Dog Squad during 26/11 will be unforgettable. We will remember our heroes forever.”

Answer the following questions.

Question 1.
Labrador retriever was covered with tri-colour. What does this signify?
The use of tri colour flag on the body of Labrador retriever signifies that it has served the nation like a soldier and deserves our homage.

Question 2.
How did Caesar save several lives at the CST railway station?
Answer:
During the terror attack at Mumbai, Caesar saved several lives when he sniffed out the hand grenades left by the terrorists at the busy CST railway station.

Question 3.
Which word in the passage (paragraph 3) means the same as ‘forced’.
Answer:
Pressed into means “forced”.

Question 4.
“Services of retired members of Dog Squad during 26/11 will be unforgettable”. Mention three services rendered by Caesar.
Answer:
Caesar had taken part in the bomb detection operations during the terrorist attack on Mumbai that began on November 26, 2008. Caeser saved several lives when he sniffed out the hand grenades left by the terrorists at the busy CST railway station. Caesar was also a part of bomb search operation after 2006 serial train blasts and July 2017 blast in Mumbai.

Question 5.
Caesar is a Labrador breed of dogs. Name a few other native breeds that are used by the Police force.
Answer:
The Blood hound, The German short haired pointer, The Boxer, The Doberman, The Dutch German Shepherd and The Giant Schnauzer are some of the famous breeds used by police force.

Question 6.
Try to rewrite the news item in your mother tongue without losing the spirit and flavour of the text. Give a suitable title to your translated version.
Answer:
வீரமரணம்
மும்பை 26\11 தாக்குதலில் குண்டு கண்டுபிடிக்கும் வீரநாய் சீசர் மரணம்.
1. மும்பை சீசர், அவரது சகாக்களில் கடைசியாக எஞ்சியிந்த மும்பை தாக்குதலுக்குப் பின் ஓர் வியாழக்கிழமை மரணமடைந்தது மீட்புப் பணியில் ஈடுபட்ட அந்த லப்ராடர் இனநாய் மூவ்வண்ணக் கொடியால் போர்த்தப்பட்டு ஓர் உணர்ச்சி மயமான பிரிவு உபச்சாரம் நகர காவல்துறையால் வழங்கப்பட்டது. சீசரின் வீரமரணத்தைப் பற்றி மும்பை காவல் ஆணையர் ட்வீட் செய்துள்ளார்.

2. பதினோறு வயது நிரம்பிய சீசர் மும்பை காவல்துறையில் பணியில் இருந்தது. தீவிரவாதத் தாக்குதலின் போது குண்டு கண்டுபிடிக்கும் பணியில் நவம்பர் 26, 2008 முதல் ஈடுபடுத்தப்பட்டு பயிற்சி பெற்ற பல நாய்களில் உயிர்தப்பிய ஒரே ஒரு நாயாகும். விரார் அருகே உள்ள பண்ணை வீட்டில் அது மாரடைப்பால் மரணமடைந்தது. அங்குதான் அதுவும் அதன் மூன்று தோழர்களும் பணி ஓய்வுக்குப் பின் அனுப்பப் பட்டிருந்தனர். மும்பை தீவிரவாதத் தாக்குதலின் போது, சீசர் தீவிரவாதிகள் விட்டுச் சென்ற பல கையெறிகுண்டுகளை மோப்பம் பிடித்து கண்டுபிடித்து மிகவும் பரப்பரப்பாக இயங்கிக் கொண்டிருந்த சத்ரபதிசிவாஜி இரயில் நிறுத்தத்தில் (CST) பல உயிர்களைக் காப்பாற்றியது.

3. மூன்று நாட்களாக தீவிரவாதிகள் பதுங்கியிருந்த நரிமன் பாயின்டில் தேடும் குழுவின் ஓர் அங்கமாக சீசரும் இருந்தது. 2006 ஆம் ஆண்டு தொடர்வண்டி குண்டுவெடிப்புச் சம்பவத்தின் போதும், ஜுலை 2017 மும்பை குண்டுவெடிப்புக்கு முன்னரும் பல முறை குண்டு தேடும் பணியில் அது ஈடுபடுத்தப்பட்டது. மும்பை காவல் அதிகாரிகளும் தமது துயரத்தை ட்விட்டர் வலை தளத்தில் “26/11 நாய் படையிலிருந்து ஓய்வுபெற்றவர்களது சேவை மறக்க இயலாதது. எங்களது வீரர்களை நாங்கள் என்றும் நினைவில் வைத்துப் போற்றுவோம்” எனப் பதிவு செய்தனர்.

Grammar

Now complete the following.

(a) Do as directed.

Question 1.
Dinesh and Prabhu wanted to meet Varsha at the bus stop. They went to the bus stop. (Change into a compound sentence)
Answer:
Dinesh and Prabhu wanted to meet Varsha at the bus stop and so they went to the bus stop.

Question 2.
Varsha reached the railw ay station. She was waiting for them there.(Change into a compound sentence)
Answer:
Varsha reached the railway station and she was waiting for them there.

Samacheer Kalvi 11th English Solutions Prose Chapter 6 The Accidental Tourist

Question 3.
While she waited at the train station, Varsha realized that the train was late. (Change into a simple sentence)
Answer:
Waiting at the railway station, Varsha realized that her train was late.

Question 4.
Dinesh and Prabhu left the bus stop. Varsha rang them. (Change into a complex sentence)
Answer:
After /When Dinesh and Prabhu left the bus stop, Varsha rang them.

Question 5.
The trio met at the station. Varsha left for Madurai. (Change into a complex sentence)
Answer:
After the trio met at the railway station, Varsha left for Madurai.

Question (b)
This paragraph has only simple sentences. Combine them into compound and complex sentences. The first one is done for you.
Answer:
One day Ajay and Tijo went to the canal. They wanted to catch some fish. Some people were playing nearby. They chose a better place. They took out the fishing rods. Suddenly there was a loud splash. They also heard a loud scream. Both Ajay and Tijo looked up. They saw something moving in the water. Then they saw a hand waving. Someone had fallen in the water. It was Yusuf. He had jumped into the water. He wanted to swim.
One day Ajay and Tijo went to the canal to catch some fish

One day Ajay and Tijo wanted to catch some fish and so they went to the canal. As some people were playing nearby, they chose a better place. When they took out the fishing rods, there was a loud splash and scream. As they looked up, they saw something like a hand waving, in the water. It was obvious that someone had fallen into the water. It was Yusuf who had jumped into the water to swim.

Question (c)
Here is one long sentence. Split them into smaller sentences.
Like all living things, human beings also need food in order to live as every part of the body must get a steady supply of food so that it can work properly, but first the food eaten has to be broken down through a process called digestion so that it can dissolve in the blood and carried to all parts of the body.
Answer:
Like all living things, human beings need food to live. Every part of the body must get a steady ‘ supply of food. Only then it can work properly. First the eaten food has to be broken down through digestion. The digested food dissolves into the blood. It is then carried to all parts of the body.

Writing

Now write a short story to explain these proverbs.

Question 1.
Actions speak louder than words.
Answer:
Vivek Pradhan was not a happy man.. Even the plush comfort of the air-conditioned compartment of the Shatabdi express could not cool his frayed nerves. He was the Project Manager and still not entitled to air travel. It was not the prestige he sought, he had tried to reason with the admin person, it was the savings in time. As Project Manager, he had so many things to do!!
He opened his case and took out the laptop, determined to put the time to some good use.

‘Are you from the software industry sir?’ the man beside him was staring appreciatively at the laptop. Vivek glanced briefly and mumbled in affirmation, handling the laptop now with exaggerated care and importance as if it were an expensive car.

‘You people have brought so much advancement to the country, Sir. Today everything is getting computerized.’ ‘Thanks,’ smiled Vivek, turning around to give the man a look. He always found it difficult to resist appreciation. The man was young and stockily built like a sportsman. He looked simple and strangely out of place in that little lap of luxury like a small-town boy in a prep school. He probably was a railway sportsman making the most of his free travelling pass.

‘You people always amaze me,’ the man continued, ‘You sit in an office and write something on a computer and it does so many big things outside.’ Vivek smiled deprecatingly. Naiveness demanded reasoning not anger. ‘It is not as simple as that my friend. It is not just a question of writing a few lines. There is a lot of process that goes behind it.’ For a moment, he was tempted to explain the entire Software Development Lifecycle but restrained himself to a single statement.

‘It is complex, very complex.’ ‘It has to be. No wonder you people are so highly paid,’ came the reply. This was not turning out as Vivek had thought. A hint of belligerence crept into his so far affable, persuasive tone. ‘Everyone just sees the money. No one sees the amount of hard work we have to put in. Indians have such a narrow concept pf hard work. Just because we sit in an air-conditioned office, does not mean our brows do not sweat. You exercise the muscle;
‘we exercise the mind and believe me that is no less taxing.’

He could see, he had the man where he wanted, and it was time to drive home the point.

‘Let me give you an example. Take this train. The entire railway reservation system is computerized. You can book a train ticket between any two stations from any of the hundreds of computerized booking centers across the country. Thousands of transactions accessing a single database, at a time concurrently; data integrity, locking, data security. Do you understand the complexity in designing and coding such a system?’ The man was awestruck; quite like a child at a planetarium. This was something big and beyond his imagination.

‘You design and code such things?’ ‘I used to,’ Vivek paused for effect, ‘but now I am the Project Manager.’ ‘Oh! ’ sighed the man, as if the storm had passed over, ‘so your life is easy now.’ This was like the last straw for Vivek. He retorted, ‘Oh come on, does life ever get easy as you go up the ladder. Responsibility only brings more work. Design and coding! That is i the easier part. Now I do not do it, but I am responsible for it and believe me, that is far more

stressful. My job is to get the work done in time and with the highest quality. To tell you about the pressures, there is the customer at one end, .always changing his requirements, the user at the other, wanting something else, and your boss, always expecting you to have finished it i yesterday.’ Vivek paused in his diatribe, his belligerence fading with self-realization. What he had said, was not merely the outburst of a wronged man, it was the truth. And one need not get angry while defending the truth. ‘My friend,’ he concluded triumphantly, ‘you don’t know what it is to be in the Line of Fire’. The man sat back in his chair, his eyes closed as if in realization. When he spoke after sometime, it was with a calm certainty that surprised Vivek. ‘I know sir, I know what it is to be in the Line of Fire ’He was staring blankly, as if no passenger,
no train existed, just a vast expanse of time.

‘There were 30 of us when we were ordered to capture Point 4875 in the cover of the night. The enemy was firing from the top. There was no knowing where the next bullet was going to come from and for whom. In the morning when we finally hoisted the tri-colour at the top only four of us were alive.’ ‘You are a…?’ ‘I am Subedar Sushant from the 13 J&K Rifles on duty at Peak 4875 in Kargil. They tell me 1 have completed my term and can opt for a soft assignment. But, tell me sir, can one give up duty just because it makes life easier? On the dawn of that capture, one of my colleagues lay injured in the snow, open to enemy fire while we were hiding behind a bunker.

It was my job to go and fetch that soldier to safety. But my captain sahib refused me permission and went ahead himself. He said that the first pledge he had taken as a Gentleman Cadet was to put the safety and welfare of the nation foremost followed by the safety and welfare of the men he commanded his own personal safety came last, always and every time.’

‘He was killed as he shielded and brought that injured soldier into the bunker. Every morning thereafter, as we stood guard, I could see him taking all those bullets, which were actually meant for me . I know sir….I know, what it is to be in the Line of Fire.’ Vivek looked at him in disbelief not sure of how to respond. Abruptly, he switched off the laptop. It seemed trivial, even insulting to edit a Word document in the presence of a man for whom valour and duty was a daily part of life; valour and sense of duty which he had so far attributed only to epical heroes. The train slowed down as it pulled into the station, and Subedar Sushant picked up his bags to alight.

‘It was nice meeting you sir.’ Vivek fumbled with the handshake.

This hand… had climbed mountains, pressed the trigger, and hoisted the tri-colour. Suddenly, as if by impulse, he stood up in attention and his right hand went up in an impromptu salute,… It was the least he felt he could do for the country.

PS: The incident he narrated during the capture of Peak 4875 is a true-life incident during the Kargil war. Capt. Batra sacrificed his life while trying to save one of the men he commanded, as victory was within sight. For this and various other acts of bravery, he was awarded the Param Vir Chakra, the nation’s highest military award. Live humbly, there are great people around us, let us learn! Action speaks louder than words

Samacheer Kalvi 11th English Solutions Prose Chapter 6 The Accidental Tourist

Question 2.
Despair gives courage to a coward.
Answer:
Most of the time, we do not realize the talent that is within us. Certain situations help to bring out these special qualities to the fore. Just as a drowning man tries to hold on to anything that comes his way to save his life, we too come up with remedies beyond our imagination in desperate situations. We find the courage to do extraordinary things when we are in despair. Even a coward can be turned into a brave warrior in a life-threatening situation. Once there was an ant that lived in a tree by the side of a pond. He was very timid and was scared of even the slightest disturbance in the tree. One day a heavy wind blew across the tree and the ant fell into the pond and was struggling to swim and stay afloat. A pigeon that lived in the same tree broke a twig of the tree and threw it to the ant. The ant got on to the twig and the pigeon swooped down picked up the twig and placed it back on the tree. The ant thanked the pigeon profusely for saving his life.

As the days went by the ant and the pigeon became good friends. The pigeon always told the ant that he should learn to be a bit braver and face life boldly. But the ant remained the same. One day a hunter came to the pond to quench his thirst. As he was drinking water from the pond he noticed the pigeon sitting on the tree. The hunter wanted to shoot the pigeon and got his bow and arrow ready. The ant noticed this and wanted to desperately save his friend. The pigeon was fast asleep and there was no way to warn him since he was sitting a long way the ant jumped down from the tree and crawled as fast as he could and bit the toe of the hunter hard as he could. The hunter cried out in pain and missed his aim.

The pigeon woke up hearing the noise and noticed the hunter with his bow and arrow. It flew away to safety. The ant was very happy that it could save his friend and felt elated for having acted bravely for once in his life.

The story clearly illustrates that when placed in a desperate situation the ant could rise above its limitations and act bravely in order to save the pigeon’s life. So when the situation demands even a coward can turn into a courageous person.

Develop the following hints into a paragraph

As is the King, So are the Subjects

Once upon a time two kings ruled neighbouring kingdoms. King Arya was a great warrior. He looked after his subjects very well. People loved him. He was always looking for ways to increase their safety and welfare. All his subjects were happy. On the other hand, king Vaishali was a very lazy man. He spent his time entertaining himself. There was always singing, dancing, and merry-making. His subjects were very angry with him as he never came out of the palace to listen to their woes.

A powerful Sultan attacked both the kingdoms. King Arya’s army, being well-prepared was

Very alert. The enemy forces were really powerful. But men, women and even children i joined the hands of army to protect their king. King Vaishal’s subject soon after sensing the

impending war, started fleeing the kingdom. King was left all alone. People were not interested in protecting him as he was not interested in their welfare and safety.

King Vaishali realized his foolishness but it was too late. He was defeated in war and fled his country in disgrace to save his own life. King Arya defeated Sultan. This happened because King Arya always kept the welfare of his people in his heart all the time. His subjects reciprocated his love. They were loyal and supportive during testing times.

Writing A Curriculum Vitae

Vijayraj Joseph
Task: Write a CV for the post of a DTP operator at ABC Publishing House and send it to P.O. Box No. 2345 or E-mail it to [email protected]

Curriculum Vitae
Name : Vijay
Mobile No : 8765412385
email : [email protected]
Address : 17/2 Beach road,
Neelangarai Chennai

Career objective:
Looking for a challenging job in the field of Desk top publishing-which requires the optimum use of my skills in typing and designing and provides me opportunities for vertical growth.

Synopsis:
A graduate with a degree in computer science from Loyola College, Madras

Profile

  • Good working knowledge of computers.
  • Good at typing .
  • Excellent at wrapper designing and editing
  • Profound knowledge in Corel draw and photoshop
  • Expertise in MS Word and Excel file handling.

Educational Qualification

  • B.Sc Computer Science with a second class
  • 12th with aggregate of 67% from GHSS, Koovathur
  • 10th with aggregate of 55% from GHSS, Pudhupattinam, Kalpakkam

Previous Experience

  • Two years experience in student Xerox as a DTP operator

Projects done

  • Typed about 50 Ph.D. dissertation and 20 M. Phil thesis

Extra-curricular Activities

  • NSS volunteer in college
  • Volleyball District player Strength:
  • A team player
  • Devoted and smart in work
  • Optimistic

Personal Details:
Date of birth: 06.09.1993
Sex: Male
Marital status : Single
Languages known : Tamil and English

Declaration
I hereby declare that the above furnished information is true to the best of my knowledge. If I am offered an opportunity, I shall prove my mettle and be worthy of your choice
S/d
Vijay

Task 1:
You see an advertisement in the newspaper. A publishing house in Chennai has brought out a paperback edition of the complete works of Khushwant Singh. You want to buy it. You are asked to send a Demand Draft for Rs.1000/- Fill in the following DD challan in favour of ‘X publishing house, New Delhi’, payable at Chennai. The surcharge for Rs.1000/- is Rs.25/-

Samacheer Kalvi 11th English Solutions Prose Chapter 6 The Accidental Tourist 4
Answer:
Samacheer Kalvi 11th English Solutions Prose Chapter 6 The Accidental Tourist 1
Task 2:
Fill in the following forms with imaginary details.
Question (a)
Samacheer Kalvi 11th English Solutions Prose Chapter 6 The Accidental Tourist 2

Answer:

Samacheer Kalvi 11th English Solutions Prose Chapter 6 The Accidental Tourist 5

Question (b)

Samacheer Kalvi 11th English Solutions Prose Chapter 6 The Accidental Tourist 3

Answer:

Samacheer Kalvi 11th English Solutions Prose Chapter 6 The Accidental Tourist 6

Question (c)

Samacheer Kalvi 11th English Solutions Prose Chapter 6 The Accidental Tourist 8

Answer:

Samacheer Kalvi 11th English Solutions Prose Chapter 6 The Accidental Tourist 9

The Accidental Tourist About the Author

Samacheer Kalvi 11th English Solutions Prose Chapter 6 The Accidental Tourist 7

Bill Bryson is an Anglo-American author of books on travel, English language, science and non-fiction topics. He stayed in Great Britain during his adult years. He served as the Chancellor of Durham University from 2005 to 2011. ’Neither here not there’, ‘Notes from a small island’, ‘A Walk in the Woods’
‘A Short History of nearly everything’, ‘The Life and Times of the Thunderbolt Kid’, and Icons of England are some his famous

The Accidental Tourist Summary

The whole story revolves around the author’s inability to do ordinary things that other people do in public places like a cinema theatre or a plane. He has problems of orientation. He would try to reach a rest room but find himself behind a self-locking door. The author recounts how his anxiety ruined the happiness of people at the airport and how he himself suffered. For fear of such life-time incompatibility, the author refrained from making frequent trips and could never use his frequent flier miles for a trip to Bali island. The author had a flash of memory that he had joined British Airways frequent flier programme. His entire family arrived at Logan Airport in Boston. Unfortunately, he had placed the card in a carry-on-bag hanging around his neck.

The zip on’ the bag was jammed. He tried pulling it hard. He used force instead of commonsense, the zip gave way spilling the contents over an area as large as tennis court.

To the great embarrassment of all fellow passengers waiting at the airport, 14 ounce tin pipe, coins and other important documents flew helter-skelter. He was annoyed to see the snuff box rolling and emptying its content. Instead of trying to collect coins, document,s and important papers, he went behind the tobacco-box crazily shouting, “My tobacco”. When he sensed oozing of blood from his finger that he had gashed while opening the bag’s jammed zip he shifted his attention immediately to his finger. The author’s wife, wondering at his hysterics said, “I can’t believe you do this for a living”.

Once in an aeroplane, the author leaned to tie his shoe lace. Just then some one in the front seat leaned back into full recline. The author found himself pinned helplessly in a crash position. Once the author spilled drinks on a fellow passenger too. He did it twice. He did not know how he did it.

The author was recording his important thoughts in a note book. They were as silly as he was “Buy socks, clutch drinks carefully”. He had fallen into a deep conversation with an attractive young lady in the next seat. Having lost his head, he amused her for about twenty minutes exchanging witty remarks to impress her. When he went to a rest room, he realized that the pen head leaked. His mouth, teeth and gums appeared in striking, scrub-resistent navy blue. To his disgust, it remained blue for many days.

The author wanted to be polished and sophisticated in his manners; But as serendipity would have it, he never left a dinner without making fellow diners realize that he had just experienced a local seismic event. Whenever he entered a car, he closed the door when his coat was still 14 inches outside the door. His trousers always brought evidence of his having sat on chewing gum, ice-cream, cough syrup or motor oil.

The author had a clear prediction of the catastrophe he intended to venture into unconsciously. So, his wife would give directions to children like,

“Take the lid off the food for daddy” or put your hoods up children, Daddy’s going to cut his meat”. The author does all such cranky things only during his flight with his family.

While staying alone the author sat on his own down turned palms to prevent himself from doing some “liquid mischief ”. When alone, he never does things which would land him in some sure catastrophe.

The author admits his peculiar inability of earning flyer miles. He has a habitual forgetting of his “Frequent flyer mile card” while collecting his Boarding pass. Though he deserves to have earned 100,000 miles a year, and visited Bali on a free air ticket, he had earned only 212 air miles. While on a flight to Australia, the airline agency refused to grant air miles because the flyer mile card read as W. Bryson where as the name in the ticket read as B. Bryson.

Samacheer Kalvi 11th English Solutions Prose Chapter 6 The Accidental Tourist

He admits that he would never go to Bali because he has a peculiar habit of spilling the contents of food on board. People who have clumsy habits do overcome their awkwardness in public places with careful practice. As we live in a society, we should leam to conform to the norms of the society including table manners.

Textual:

  • alley – a narrow passage-way between or behind buildings
  • Bill – William (the letter W is changed to B and William is called Bill)
  • bons mots – (French) witty remarks
  • cascade – waterfall
  • catastrophe – a terrible disaster
  • concourse – the open central area in a
  • large public building (here ‘airport’) /hall
  • consternation – worry
  • disgorging -discharging
  • en famille – (French) as a family
  • exasperation – irritation
  • extravagantly – excessively
  • gashed – cut deeply
  • hysterics – a fit of uncontrollable laughing or crying
  • suave – polite and sophisticated
  • venerable – valued
  • yanked – pulled with a jerk

Additional:

  • abruptly – suddenly
  • accumulated – collected
  • annoyed – angry and irritated
  • anxiety – worry
  • confused – disoriented
  • constantly – all the time
  • dumbstruck – shocked and speechless
  • entitled – deserving special treatment
  • evident – obvious
  • frequent – often
  • frustration – vexation
  • hysterics – uncontrolled emotion
  • jammed – packed tightly
  • oblivions – forgetfulness
  • panic – intense fear
  • specialty – particular skill

The Accidental Tourist Synonyms

Choose the most appropriate synonyms for the underlined words,

Question 1.
I am constantly filled with wonder.
(a) hardly
(b) regularly
(c) rarely
(d) sparsely
Answer:
(b) regularly

Question 2.
I yanked at the zip of the bag.
(a) pushed
(b) dragged
(c) jerked
(d) closed
Answer:
(c) jerked

Question 3.
He had always ended up standing in an alley.
(a) maze
(b) road
(c) pathway
(d) hospital
Answer:
(c) pathway

Question 4.
He tried pulling it with great consternation.
(a) anger
(b) hatred
(c) pity
(d) worry
Answer:
(d) worry

Samacheer Kalvi 11th English Solutions Prose Chapter 6 The Accidental Tourist

Question 5.
All the contents of the bag were extravagantly ejected.
(a) frugally
(b) judiciously
(c) vainly
(d) lavishly
Answer:
(d) lavishly

Question 6.
The tin of tobacco rolled crazily across the concourse,
(a) closure
(b) locked up
(c) hall
(d) strait
Answer:
(c) hall

Question 7.
I had gashed my finger.
(a) bandaged
(b) plastered
(c) cut
(d) healed
Answer:
(c) cut

Question 8.
The tobacco tin went disgorging its content.
(a) attacking
(b) discharging
(c) collecting
(d) dusting
Answer:
(b) discharging

Question 9.
I always have a catastrophic when I travel.
(a) condy
(b) disasters
(c) jokes
(d) joy
Answer:
(b) disasters

Question 10.
The author’s wife did not show anger or exasperation.
(a) pleasure
(b) irritation
(c) pain
(d) hatred
Answer:
(b) irritation

Question 11.
He amused her with urbane bons mots.
(a) news
(b) puzzles
(c) witticism/repartee jokes
(d) joked
Answer:
(c) witticism/repartee jokes

Question 12.
He explained the venerable relationship between Bill and William but in vain.
(a) loathsome
(b) valued
(c) hurt
(d) howled
Answer:
(b) valued

Question 13.
I ache to be suave.
(a) rude
(b) indecent
(c) dishonest
(d) polite / sophisticated
Answer:
(d) polite / sophisticated

Samacheer Kalvi 11th English Solutions Prose Chapter 6 The Accidental Tourist

Question 14.
When it’s my own blood, I think hysterics is justified.
(a) frenzied
(b) whisper
(c) rustle
(d) hustle
Answer:
(a) frenzied

Question 15.
My hair went into panic mode.
(a) courage
(b) fear
(c) joy
(d) pleasure
Answer:
(b) fear

The Accidental Tourist Antonyms

Choose the most appropriate antonyms for the underlined words.

Question 1.
I was shedding blood in a lavish manner.
(a) Extravagant
(b) frugal
(c) easy
(d) tough
Answer:
(b) frugal

Question 2.
I am constantly filled with wonder.
(a) inconstantly/rarely
(b) incorrectly
(c) infrequently
(d) incessantly
Answer:
(a) inconstantly/rarely

Question 3.
I pulled it with a frown.
(a) Scowl
(b) grimace
(c) mockery
(d) smile
Answer:
(d) smile

Question 4.
I managed to get myself freed.
(a) cleared
(b) disentangled
(c) imprisoned trapped
(d) discharged
Answer:
(c) imprisoned trapped

Question 5.
I watched dumbstruck as the carefully sorted documents came down in a cascade.
(a) ruffled
(b) excited
(c) expected
(d) petrified
Answer:
(c) expected

Question 6.
I was talking to an attractive lady.
(a) charming
(b) captivating
(c) unattractive / ugly
(d) ravishing
Answer:
(c) unattractive / ugly

Question 7.
I cried in horror.
(a) alarm
(b) antipathy
(c) disgust
(d) pleasure
Answer:
(d) pleasure

Question 8.
The author amused the lady with some witty remarks.
(a) bored
(b) doted
(c) delighted
(d) entertained
Answer:
(a) bored

Question 9.
Coins bounced to a variety of noisy oblivions.
(a) Nuisance
(b) awareness
(c) forgetfulness
(d) unconscious
Answer:
(b) awareness

Samacheer Kalvi 11th English Solutions Prose Chapter 6 The Accidental Tourist

Question 10.
This had become a real frustration.
(a) Disappointment
(b) cramp
(c) discontentment
(d) fulfilment
Answer:
(d) fulfillment

Samacheer Kalvi 11th English Solutions Poem Chapter 5 Everest is not the Only Peak

Students can Download English Poem 5 Everest is not the Only Peak Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 11th English Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations. Learn the Samacheer Kalvi 11th English Grammar to enhance your grammar skills like reading comprehension, passage writing, parts of speech, tenses, passive and active voice, and many other concepts in no time.

Tamilnadu Samacheer Kalvi 11th English Solutions Poem Chapter 5 Everest is not the Only Peak

Warm up

Identify the following personalities and their fields of achievement.

Samacheer Kalvi 11th English Solutions Poem Chapter 5 Everest is not the Only Peak

Answer:

Samacheer Kalvi 11th English Solutions Poem Chapter 5 Everest is not the Only Peak

Question 1.
Mention a remarkable achievement of any of these personalities
Answer:
Kailash Satyarthi has rescued thousands of child labourers and gave them new life.

Question 2.
What quality do you admire the most in each of these achievers?
Answer:
They were persistent and never accepted failure.

Samacheer Kalvi 11th English Solutions Poem Chapter 5 Everest is not the Only Peak

Question 3.
What are the qualities that you may share with them?
Answer:
Each one has a concern for the country.

Question 4.
Name a few more popular personalities who have made our nation proud.
Answer:

  1. Rabindranath Tagore
  2. Mother Teresa
  3. Sarojini Naidu
  4. Sania Mirza
  5. Mary Kom
  6. Virat Kohli
  7. Saina Nehwal
  8. Abhinav Bindra
  9. MS Dhoni
  10. AR Rahman

Samacheer Kalvi 11th English Everest is not the Only Peak Textual Questions

A. Based on your understanding of the poem, answer the following questions in a sentence or two each.

We are proud and feel so tall,
Our virtues though are few and small
Our nature is that whatever we try
We do with devotion deep and true.

Defeat we repel, courage our fort;
Cringing from others we haven’t done,
To seek a gain we adore none:
We are proud and feel so tall.

We deem it our duty and mission in life,
To bless and praise the deserving ones;
Never shall we fail in what we commit,
Shall nourish the ones that nourish the world.

Samacheer Kalvi 11th English Solutions Poem Chapter 5 Everest is not the Only Peak

We are proud of the position we
Hold; humble as we are,
Our pride springs from the way we live.
Ours is a path of dignity and honour,
A life that knows no kneeling and bending.
We are proud and feel so tall.

Everest is not the only peak,
Every hillock has a summit to boast!
The height you reach is not that we care;
He, who does not stop, is a king we adore.
We bow before competence and merit;
The ones that are true and stand on their own
Are really the ladder for the rise of Man.
Honour is a property, common to all:
In dignity and pride, no one needs to be poor.
We are proud and feel so tall.

Question 1.
Which line is repeated in the poem? What is the effect created by this repetition?
Answer:
‘We are proud and feel so tall” This line is repeated in the poem. We should have pride and courage to reach the highest level of the position.

Question 2.
Who are the ‘deserving ones’?
Answer:
Those who have merit and competence are the deserving ones.

Question 3.
Which quality does the speaker wish to nourish? What is his mission?
Answer:
The speaker wishes to nourish the quality of doing work with determination. His mission is doing his work with full dedication and never give it up.

Question 4.
Which path should we follow in life?
Answer:
We should follow the path of dignity and honour. We should never stoop before others for any ‘gains’.

Question 5.
What does ‘Everest’ in the title stand for?
Answer:
‘Everest’ in the title stands for the highest position.

Question 6.
What does ‘hillock’ refer to in the line ‘Every hillock has a summit to boast!’?
Answer:
The hillock is a small hill. Everyone need not become Tenzing. Each one can achieve some ordinary pursuit and be proud of his achievements.

Question 7.
Why does the speaker say ‘Everest is not the only peak’?
Answer:
Every hillock which one reaches is considered as a peak or height of achievement of him. So the speaker says “Everest is not the only Peak”.

Question 8.
What does the ladder symbolize?
Ladder symbolizes help given to enable others to climb up to a higher position in life.

B. Read the given lines and answer the questions that follow.

1. Our nature is that whatever we try
We do with devotion deep and true.

Question (a)
Who does ‘we’ refer to?
Answer:
‘We’ refers to ordinary people.

Samacheer Kalvi 11th English Solutions Poem Chapter 5 Everest is not the Only Peak

Question (b)
How should we carry out our duties?
Answer:
We should carry out our duties with devotion deep and true.

2. Defeat we repel, courage our fort;

Question (a)
How do we react to defeat?
Answer:
We defy defeat.

(b) Which is considered as our stronghold?
Answer:
Courage is our stronghold.

3. We are proud of the position we hold;
humble as we are,

Question (a)
What is the speaker proud of?
Answer:
The speaker is proud of the position people hold on to.

Question (b)
How is the speaker both humble and proud?
Answer:
The ordinary position they hold keeps them humble. But the path of self-dignity and honour they tread, makes them feel proud.

Question (c)
Pick out the alliteration in these lines.
Answer:
proud, position, hold, humble are the words which alliterate.

4. He, who does not stop, is a king we adore.
We bow before competence and merit;

Question (a)
Who is adored as a king?
Answer:
An upright or straight forward person is adored as a king.

Question (b)
What is the figure of speech used in the first line?
Answer:
Metaphor

5. Honour is a property, common to all:
In dignity and pride, no one needs to be poor.

Question (a)
Who is considered rich?
Answer:
Those who possess dignity and pride are considered rich.

Question (b)
What is their asset?
Answer:
Honour is their asset.

C. Answer the following questions in a paragraph of 100-150 words each.

Question 1.
In what way is every hillock similar to Everest?
Answer:
The poet does not compare rare feats of athletes, mountaineers or horsemen. He does not attach great value to positions or possessions. He scoffs at those who pull strings to achieve their ends. The means must justify the ends. One should not stoop to underhand dealings to achieve their desired goals in life. Those who reach great heights in lifelike Everest due to their hardwork, perseverance, and competence are adorable. At the same time, those who trek any small hillocks can’t be underestimated. The efforts made in reaching even the smallest positions in life, if done with sincerity of purpose and deep devotion, is worthy of hearty appreciation.

One who holds a humble position, but upright and serves as a ladder for fellow humans to reach great heights deserve our respect. The poet admits that he is proud of people’s humble positions because their pride springs not from positions or possessions but the way they live. Their life knows no bending. The poet just doesn’t bother the height of the peak one reaches. It could even be a hillock. What matters is how one reaches that spot. If merit and competence have paved the way for their success and positions, however humble they are, the poet admires them.

“Take on risks and ride the journey called life with no regrets. ”

Samacheer Kalvi 11th English Solutions Poem Chapter 5 Everest is not the Only Peak

Question 2.
The poem does not focus on the destination but the journey towards it. Discuss.
Answer:
The poet discusses the merits of efforts, duty and devotion, and values of honesty, uprightness and service-mindedness. He does not have any special appreciation to those who reach great peaks like the Himalayas. He appreciates the process, the journey, and not the destination. When the whole world has a perspective of seeking glory using any foul method or underhand dealing, the poet differs from it. For him, the means is more important than the end. However modest may be one’s position is, it is adorable if attained by competence and merit. Pride is not in heights one reaches but in a life that knows no bending or kneeling. The poet respects one who does not stoop as a king. Thus the poet pays importance to the journey of life, not the destination.

“The journey of life is not meant to be feared and planned; It is meant to be travelled and enjoyed. ”

Creative Activity

D. Write eight words you associate with success.

  • Use the words to write eight lines that mean success to you or how success makes you feel.
  • Arrange your lines into a poem.
  • Share your poem with the class and post a copy on the notice board.
    1. strive
    2. flaunt
    3. determination
    4. destination
    5. wise
    6. want
    7. success
    8. kind

Success
Poem:

Real success is when you strive
for what you want
when you have that guts to flaunt
Not thinking about the world and wise
when you can alone suffice
With upright will and determination
where finally you reach your destination.
Real success is hard to find. But it’s one of a kind.

(OR)

  1. brave
  2. hunt
  3. learn
  4. mistakes
  5. encounter
  6. failure
  7. never
  8. If

Poem

You’ll never be brave.
If you don’t get hurt
you’ll never learn.

If you don’t make mistakes
you’ll never be successful.
If you don’t encounter failure.

Speaking Activity

E. Discuss the following topics in groups of five and choose a representative to sum up the views and share them with the class.

Question (a)
To succeed in life, one must have single-minded devotion to duty.
Answer:
Singleminded determination is necessary to achieve success in life. Legend says Dronacharya was training Pandavas to shoot arrows in the jungle. Once guru saw a bird at the top branch of a tall tree. He assigned the task of shooting the bird on its right eye.
Bhim, Nakulan, Sakadevan and Dharman were denied the chance to aim at it because they all told Dronacharya that they saw the whole jungle, tree, sky, and bird respectively. But it was Arjun who said that he saw only the right eye of the bird and nothing else. Indo-Pak war was in progress. Four bombers were prowling into Indian airspace.

Indian fighter bomber pilot realized all the bombs were exhausted. He had a single-minded determination to prevent the four bombers from bombing India harm and force them to surrender. He tried a trick. He called them and spoke to them. He informed them that he was a thousand feet above them with his finger on a missile. If they just surrender, their lives would be spared. He radioed to his chief that they should welcome four prisoners of war along with their fighter bombers.

Napoleon Boneparte was once watching a battle from a hillock. One of his soldiers came with an urgent message. Napoleon looking at the badly wounded soldier thought that the war was lost. He called one of his aides and gave instructions for the pullout. But the soldier said, “Sire, we‘ve won”. Then he gave a slip from another pocket. Being a practical leader, he had alternate plans always at hand. Such leaders never accept failure as permanent. People who pursue their goals in life with single-minded determination always win.

Samacheer Kalvi 11th English Solutions Poem Chapter 5 Everest is not the Only Peak

Question (b)
‘Success is not final, failure is not fatal.’ It is courage and perseverance that Counts.
Answer:
India’s cricket team was beaten in Test series in Africa. In fact, it was routed. The team’s morale was a little down. The team captain Virat Kohli told his team to focus on what they were good at (i.e) sterling performance in One Day Internationals. He believed in the youngest bowlers and told them to play the game, the way they loved to play. He still believed in them. They would have to prove who are the masters of the game. Gathering their broken hopes and courage, the Indian team players, snatched the ODI series from the overconfident South Africans. They went on to win the T20 series too against the hosts. Thus they proved that success is not final and failure never fatal. One may bounce back from failure if one persists long enough.

A king had lost a battle. All his soldiers had been scattered across the country. Heartbroken king Bruce was hiding in a cave. He saw a spider failing a number of times to spin a web. But it made it after about 20 attempts. This bolstered the confidence of the king. He refused to be controlled by failure. He defied defeat. He gathered his soldiers again and won the battle. These incidents throw much light on the truth that perseverance and courage count for success in life.

Question (c)
Successful people neither brood over the past nor worry about the future.
Answer:
Mahatma Gandhi and his followers were arrested and jailed many times. Gandhi’s followers were brutally lathi-charged. Gandhi had decided to silence the guns of the British with Ahimsha passive non-violent resistance. The brutal suppression of the struggle for freedom did not dishearten Gandhi. He did not brood about the strength of the British army and the weaknesses of unarmed peasants who believed in his leadership. He was a Karmayogi. Whatever the duty to be done it must be done with steadfast devotion and sincerity. Other leaders got worried. Some angry young men resorted to violence.

They burnt down a police station at Chauri chaura too. But Gandhi declared a fast unto death. He plunged into action. If he had worried about the unpleasant developments, he wouldn’t have launched Quit India Movement or Salt Satyagraha effectively.

Thomas Alva Edison was not able to find the element that would glow if electricity was passed. He had failed 1000 times to invent the bulb. But he said, “The light bulb was an invention with 1000 steps”. Each step taught him what did not work. He lost his hearing capacity. He had many failures. His teachers believed him to be mad and unteachable.

His entire schooling was only a few years. His mom taught him and made him believe in himself. This man who had been ill-treated in school and faced many challenges had no time to brood. He went on to make 1093 inventions and got them patented. Those who are busy building facilities for transforming the world have neither the time nor the inclination to brood about failures or about possibilities of success in the future.

Everest is not the Only Peak About the poet:

Samacheer Kalvi 11th English Solutions Poem Chapter 5 Everest is not the Only Peak

Prof. V.C. Kulandtiisamy ( 14th July 1929 10th Dec 2016) is known as Kulotliimgan. He is a prolific Tamil writer with six volumes of poems and seven of prose essays. He has 1 been conferred Thiruvalluvar Award by the Government of Tamil Nadu (1999) and Sahitva j Academy’s Award in 1988 for his outstanding masterpiece “Vaazhum Valluvam”. Ills i poems, as evident in the poem “Everest is not the only Peak” deal with the gamut of human [ progress and all-pervasive human effort to better the world.

Samacheer Kalvi 11th English Solutions Poem Chapter 5 Everest is not the Only Peak

Everest is not the Only Peak Summary

This poem discusses in depth the virtues that make people greater than those who scale Himalayas. We have a few virtues which make us feel proud and tall. By nature we do every assigned duty with deep and true devotion.

We repel defeats because courage is our fort. We have never cringed from others. We don’t lick others’ shoes to achieve personal gains. We are proud and tall.

We deem it a duty and a mission to appreciate those who deserve appreciation. We shall carry on doing unfailingly what we are committed to do. We shall nourish those who nourish the world (i.e) those who better the world with their services, inventions and discoveries.

Despite living a simple and humble life, we have pride over our path of dignity and honour. We feel proud and tall for we never bend before the mighty.

We value individual differences. Every small effort to succeed in life matters. Peaks alone don’t matter. Even a hillock is a symbol of human achievement. We respect those who don’t stoop to conquer. We are proud to be rich in honour, dignity, and pride. Those who stand on their own merit and serve as a ladder for the progress of fellowmen are worthy of our appreciation. We are proud and feel so tall.

Not the worldly riches, power, and positions impress us. We derive our happiness in serving those who serve the nation. We respect all who are upright in their dealings. So, as honest citizens, we are proud.

Everest is not the Only Peak Glossary

Textual:

  • adore – worship someone
  • competence – the ability to do something efficiently
  • cringing – behaving in an excessively humble or servile way
  • devotion – loyal commitment towards a particular activity
  • merit -the quality of being particularly good or worthy
  • nourish – to help the growth and development of someone
  • repel – hate or detest
  • stoop – yield or submit, to descend from dignity
  • summit – the highest point of a hill or mountain peak
  • virtues – good qualities

Samacheer Kalvi 11th English Solutions Poem Chapter 5 Everest is not the Only Peak

Additional:

  • dignity – deserves respect
  • fort – huge building for protection
  • hillock – a small hill
  • humble -modest
  • mission – vocation
  • praise – appreciate
  • property – possession
  • proud – feeling important
  • springs – arises

Samacheer Kalvi 11th English Solutions Prose Chapter 5 Convocation Address

Students can Download English Lesson 5 Convocation Address Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 11th English Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations. Learn the Samacheer Kalvi 11th English Grammar to enhance your grammar skills like reading comprehension, passage writing, parts of speech, tenses, passive and active voice, and many other concepts in no time.

Tamilnadu Samacheer Kalvi 11th English Solutions Prose Chapter 5 Convocation Address

Warm up

A. Work in pairs. Look at the mind map below and add a few other features of democracy you can think of. Share your answers with the class.

  • e.g. sharing and caring
  • Protecting nature
  • Cleaning environment
  • Doing one’s duties
  • Enjoying freedom of speech

Answer:

Samacheer Kalvi 11th English Solutions Prose Chapter 5 Convocation Address

  • People
  • Rights
  • Majority
  • Voting
  • Individualism
  • Politics

B. Given below is a list of items or features you find in a good school. Study the list and classify the items as individual and common.

(textbooks – uniform – smart class room – laboratory – competent teachers – library – mid-day meals – blackboard – stationery – play ground – sports equipment-washrooms)

Individual Common
e.g. textbooks play ground

Answer:

Individual Common
e.g. textbooks play ground
uniform smart classroom
stationery library
teachers laboratory
mid-day meals washroom
sports equipment blackboard

C. Who provides these common facilities? Tick the appropriate source.

  • Parents
  • Educational institutions
  • The Department of Education
  • Social Service Organizations
  • Other agencies

Answer:
The Department of Education provides these common facilities.

D. Discuss this aspect with your partner and share your views on how students can give back to society.

Students can do a lot in generating ideas which could be helpful to society. Many illiterate people and young children succumb to dengue, malaria, and other diseases as they don’t keep their surroundings clean. NSS, Red Cross, National Green Corps cadets can run successful campaigns to present the spread of such diseases. Water is the most precious wealth of the nation. People tend to waste it without realizing the sordid and potential threat of “zero water day” in Tamil Nadu if we don’t prevent the sand mafia’s from looting riverbed sand.

Besides, farmers do not use advanced scientific methods to make judicious use of water or water harvesting. As a result, their crops fail along with rainfall. More than double, the quantity of water, we beg from Karnataka goes and ends up in the Bay of Bengal. So, students could cite Kajender Singh, a water specialist who made parts of Rajasthan bloom due to water conservation techniques. To popularize water harvesting, judicious use of water and reduce pollution and enhance recycling practices, the student community can interact with a community, conduct rallies or campaigns, and be proud to be a part of the development of the nation.

Samacheer Kalvi 11th English Convocation Address Textual Questions

A. Based on your reading of the speech, answer the following questions briefly in a sentence or two.

Question 1.
Who does the speaker claim to represent?
Answer:
The speaker wishes to represent some of the cardinal principles enunciated by those who offered to graduates in the past. He remarks that his speech would bear the layman’s point of view to the findings of experts in various fields of education. He promises to annotate the speeches of great educationists for the benefit of the graduates.

Samacheer Kalvi 11th English Solutions Prose Chapter 5 Convocation Address

Question 2.
Why are universities necessary for society?
Answer:
Universities, as the repositories of knowledge and nursing ground for the emissaries of thought wisdom and service, have got a prominent role to play (i.e) to equip graduates to better the society. Universities are necessary to trim, train, guide and lead the youth to the development task. It kindles sweet hopes but demands patience, perseverance, faith and confidence. He should have faith in himself and others. The faith should also be in his inherent ability to shoulder the responsibilities. The universities today have the responsibility to fashion out of the graduate an individual fitted and equipped for the task of making democracy fruitful and effective. So, universities are essential.

Question 3.
What was the role of scholars and poets in the olden days?
Answer:
Scholars and poets adorned the chambers of royalty or the gilded mansions of lords and nobles. Their wisdom was meant for the mansion, not for the market place. Poets and scholars were content to work in secluded spheres. Their work was to weave the costly fabric of philosophy or poetry which, in turn, was to be converted into dazzling garments for the select and the privileged royalty.

Question 4.
In what ways have universities improved society?
Answer:
Universities have improved society by training and guiding an individual before he is asked to do his duty as a citizen of a democracy.

Question 5.
Universities develop broad-mindedness. How does Dr. Radhakrishnan drive home this idea?
Answer:
The true spirit of democracy can be developed only in universities. A broad-minded person alone can appreciate other’s points of view and adjust ideological differences through discussions. In universities, a lot of opportunities are made available for debates. Friendly discussions provide ample scope for nurturing the broad-mindedness of the youth.

Question 6.
What should the youngsters aim in life after graduation?
Answer:
The youngsters should aim in life after graduation for an aristocracy of achievements arising out of a democracy of opportunities. As youngsters they should aim at toning up the society, bringing light into dark alleys sunshine into dingy places, solace into the afflicted, hope unto the despondent, and a new life into everyone.

Question 7.
How can a graduate give back to his/her society?
Answer:
The graduates are indebted to society for higher education. They need to replenish the social chest on which they have drawn largely as students. They can contribute to society, not in terms of material goods. But in terms of quality services, through services the graduates, can confer a rich benefit to humanity.

B. Based on your understanding, answer the following questions in three or four sentences each.

Question 1.
‘Wisdom was meant for the mansion, not for the market place’ What does this statement signify?
Answer:
During the monarchical or feudal days, universities had to train scholars and poets to adorn the chambers of royalty or the gilded mansions of the lords and nobles. Those were the days when eminent scholars were not asked to confront the problems of the masses. They were content to work in secluded places. So, the speaker says that their wisdom was meant for the mansion and not for the market place.

Question 2.
According to the speaker, how should Universities mould the students of the present day?
Answer:
Before asking graduates to do the duties as citizens of a democracy, the university must trim, train, guide and lead them to a task which kindles sweet hopes. It should also demand patience and perseverance from them. Universities should inculcate faith and confidence. Faith should be in the graduates and others to shoulder the responsibilities. The universities today should fashion out of them individuals fitted and equipped for the task of making democracy fruitful and effective.

Question 3.
How does Arignar Anna highlight the duties and responsibilities of graduates to society?
Answer:
The graduates must acquire the means of a decent living. But it should be the only objective. As their education is funded by the tax from poor people, they have an obligation to pay back to society if not in cash in terms of service. They must bring light into the dark alleys, sunshine into dingy places, solace into the affiliated hope unto the despondent, and a new life into everyone.

Question 4.
Students are instilled with some of the essential values and skills by the universities. Enumerate them.
Answer:
Universities instill the values of robust optimism, respect for democracy, and appreciation of others’ points of view. It also develops adjustment of differences through discussion; develop patience, perseverance, confidence, faith in themselves and others. They also instill confidence in their ability to shoulder responsibilities.

Question 5.
What are the hindrances a graduate faces in his/her way?
Answer:
The world sometimes dims the hopes of graduates and disturbs their determination. They come face to face with unpleasant kinds of practices contrary to their principles. Tyranny of all sorts stares at them. Self-seekers are enthroned and patient workers are decried. People with robust optimism are discouraged. These hindrances are faced by graduates on their way.

C. Answer the following questions, based on your understanding of the speech of Dr. Arignar Annadurai, adding your own ideas, in a paragraph of about 100-150 words.

Question 1.
How do Universities mould students apart from imparting academic education to them?
Answer:
Universities mould students by providing various opportunities to develop their soft skills and to develop values that would contribute to the process of nation-building. They enable graduates to develop patience and perseverance. They help them develop faith in their own inherent ability to shoulder responsibilities. They are oriented to become citizens of a democracy and repay to the society quality services which would reform the lives of the poor people.

They develop the true spirit of democracy among young graduates. They enable appreciation of others’ points of view. The graduates are also provided opportunities to adjust to differences through amicable discussions. The universities, apart from imparting education mould the students’ character and personality too.

“The quality of a university is measured more by the kind of student it turns out than the kind it takes in.”

Samacheer Kalvi 11th English Solutions Prose Chapter 5 Convocation Address

Question 2.
The common men contribute to the maintenance of institutions of higher education. Explain this statement.
Answer:
Higher institutes of learning like universities require a lot of funds for their maintenance. These funds often come from the community’s taxes, through the Government students should remember the fact a good proportion of these funds come from farmers, peasants and coolies who did not enjoy even schooling facilities. These illiterates but cultured men willingly face discomforts to ensure better life for the younger generation. These common men contribute to the development and maintenance of institutes of higher education.

“Every child should have the opportunity to receive a quality education.”

Question 3.
How does the speaker highlight the importance of giving back to society?
Answer:
The graduates have drawn largely from the social chest. The largest taken from society needs to be replenished. If graduates fail to pay back, ordinary people’s coffer will be empty. The supervisor’s education enjoins greater responsibility to society. Apart from their own individual advancement, society has got a right to expect an adequate return from the graduates.

Society does not expect them to pay back in cash. But they must payback in terms of service. They should tone up society by bringing light into the dark alleys. They should herald sunshine into dingy places. They must give solace to the affiliated people. They should also give hope unto the despondent and thus ensure a new life unto everyone.

“We Must stop thinking of the individual and start thinking about what’s best for society.”

Question 4.
You were one of the fresh graduates at the convocation function of the University. You had the rare privilege of listening to the enlightening speech of Dr. Arignar Anna. Write a letter to your friend describing the core ideas of his speech and the impact of the speech
on you.
Answer:

3rd March, 20xx
Dear Tim,

I was lucky to be present when Dr. Annadurai gave the Convocation Address. Very rarely one gets such a lifetime opportunity. His genius was explicit in his talk. I could see a great leader and his concern for both the educated and unsung heroes fighting to eke out a living. At the outset, he greeted the graduates and thanked Annamalai University for the unique opportunity. His modesty was evident when he admitted that he was not adequately prepared to offer them appropriate guidelines to succeed but, as a layman, he could quote and annotate the ideas of experts connected to the field of education. He showered rich encomiums on the universities for being the repositories of thought, wisdom, and service. He admired the universities for their role in equipping the youth to better the society.

He compared the universities of feudal days which turned scholars into adornments of royalty but modem universities which are surely for the market and not for mansions. He explained that universities have a role in preparing the graduates to serve society apart from earning a decent living for themselves. He advises the graduates not to be daunted by the adverse and demotivating environment but to draw inspiration from Purananooru, our ancient classic, and overcome societal obstacles to serve the society in an exemplary manner. He completed his convocation address with the wish ‘May your life be a bright one and may its luster brighten the entire land’. The words of the great leader are still ringing in my memory.

With love
Your sincere friend
Joe

To
Tim Smith,
12, Sterling Road,
Chennai

Vocabulary

A. Based on your understanding of paragraphs 6 and 7, complete the mind map.

Samacheer Kalvi 11th English Solutions Prose Chapter 5 Convocation Address

Answer:

Samacheer Kalvi 11th English Solutions Prose Chapter 5 Convocation Address

B. Words belonging to the different semantic fields.

Question 1.
Now, read the following statements taken from Arignar Anna’s address:
Answer:
‘The role of the universities today is not cloistered and confined as in the past.
The highlighted word ‘universities’ is used in the field of education. Find other words pertaining to the field of education from the speech of Anna.
Graduate, convocation, theories, principles, annotations, point of view, experts, finding, knowledge, thought, wisdom, service, higher education, guide, lead.

C. Look at the following words and classify them according to their fields.

Clinical, orthopedic, dividend, operations, fertile, Carnatic, diagnostics, industries, keyboard, hacker, desktop, vocal, cultivation, organic, unplugged, disease, harvest, livestock, motherboard, investment, internet, proxy, recycle bin, orchestra, trade, hip-hop, uprooting, guitar, cure, contracts

Music
Agriculture
Computer
Commerce
Medicine

Answer:

Music Carnatic Orchestra Keyboard Hip hop Vocal Guitar
Agriculture Fertile Livestock Cultivation Harvesting Organic Uprooting
Computer Internet Motherboard Recycle bin Hacker Desktop Proxy
Commerce Trade contracts Investment Divided Industries
Medicine Cure Diagnostics Orthopaedic Clinical Operations Disease

D. Words with different functions

Now, choose the appropriate words to complete the sentences. The first one is done for you.

(frame, guide, book, play, print, plan)

Question 1.
a. We usually _____ tickets for movies in advance.
b. Thirukkural is my favourite _____
Answer:
(a) book
(b) book

Question 2.
a. The _____ of the photo is broken.
b.We _____ questions on all topics.
Answer:
(a) frame
(b) frame

Question 3.
a. My teachers _____ me towards the path of success.
b. The tourist _____ explained the historical importance of the site.
Answer:
(a) guide
(b) guide

Question 4.
a. We enacted a humorous _____ in our school function.
b. The children _____ in the ground every afternoon.
Answer:
(a) play
(b) play

Question 5.
a. My _____ worked out well.
b. We should _____ our work well in advance.
Answer:
(a) plan
(b) plan

Question 6.
a. The _____ is not clear; we cannot read the sentences.
b.We _____ wedding cards here.
Answer:
(a) print
(b) print

E. A belief, attitude, theory, etc. that is referred to by a word ends with the suffix ‘-ism’. Here is a sentence from the speech of Dr. Annadurai: ‘I admit that the environment is such that even people with robust optimism will be discouraged and forced to take to the path of ease and comfort’.

The meaning of the word ‘optimism’ is the hopeful feeling that all is going to turn out well’. Match the ‘ism’ words with the appropriate meanings.

S.No. Meanings Words
I. e.g. love of country and willingness to sacrifice for it patriotism
2. a brutal barbarous, savage act nationalism
3. the doctrine that your country’s interests are superior egocentrism
4. participating in sports as a hobby rather than for money feminism
5. belief that the best possible concepts should be pursued criticism
6. a serious examination and judgment of something amateurism
7. habitual failure to be present at work barbarism
8. a doctrine that advocates equal rights for women idealism
9. concern for your own interests and welfare heroism
10. exceptional courage when facing danger absenteeism

Answer:

  1. patriotism
  2. barbarism
  3. nationalism
  4. amateurism
  5. idealism
  6. criticism
  7. absenteeism
  8. feminism
  9. egocentrism
  10. heroism

Listening Activity

Listen to the information about Vishalini and complete her profile with suitable words/ phrases.

(For listening to the passage refer to our website www.fullcircleeducation.in)

Born in Tirunelveli, a small city of Tamil Nadu, in a middle-class family, Vishalini is no different from other children of her age who love watching cartoons, riding bicycles, and playing games; however, her IQ and intelligence is not of an average 11 -year-old girl. She is a child prodigy blessed with exceptional computer and analytical skills. According to reports, her IQ stands at 225, which is considerably higher than the previous Guinness World Record holder, Kim Ung-Yong, whose IQ is approximately 210. Vishalini’s wonderful accomplishments include the Microsoft Certified Professional and Cisco Certified Network Associate.

Vishalini holds the unique pride of addressing various International conferences as a Chief Guest and Keynote Speaker in the presence of delegates from various countries when she was an 11-year-old child. At the Google India Summit held in New Delhi, the international delegates from about 75 countries were astonished and astounded at Vishalini’s keynote address on ‘Cloud Computing in Google Apps for Education’. There she was also honoured as “The Youngest Google Speaker”. She is the recipient of five international awards. Vishalini, the pride of Tamil Nadu, is undoubtedly a wonderful girl.

Vishalini, hailing from (1) ________ in Tamilnadu is (2) ________ years old. She is endowed with an outstanding (3) ________ and (4) ________ skills. Her IQ is (5) ________ higher than the previous score of (6) ________ She has been the (7) ________ speaker in International conferences. At the Google India Summit, she was honoured as the youngest (8) ________ She is the recipient of (9) ________ international awards. This child prodigy is considered a (10) ________ girl.
Answer:

  1. Tirunelveli
  2. 11
  3. computer
  4. analytical
  5. 225
  6. Kim-ungyoug
  7. keynote
  8. Google speaker
  9. five
  10. wonder

Speaking Activity

Prepare a formal five-minute speech on the topic ‘The Importance of Education’ and deliver it at your School Assembly.

Samacheer Kalvi 11th English Solutions Prose Chapter 5 Convocation Address

The importance of Education:

Good morning to one and all. Respected Principal, teachers, and my dear friends, I stand before you to speak my views on “The importance of education”. Education gives people the capacity to reason out and choose the best course of action for themselves. It gives them the necessary skills to earn a living. They often can invest the capital gained in schools and colleges to start industries or farms which can offer employment opportunities to their less fortunate kinsmen. Only educated folks can generate awareness among the masses against population explosion, pollution, and promote water harvesting practices to sustain life on earth. Those who are educated in the real sense have a passion for the upliftment of society. They always look for ways to repay the society which invested in their education.

They employ various strategies to enhance the quality or standard of living of laymen. Education teaches young men and women to accept different points of view without coming to blows. It provides them the capacity to appreciate the points of view of others and tolerate differences. Education develops the values of democracy, patience, and perseverance and also the soft skills required to face challenges in life. Immortal poet Thiruvalluvar said, “Those who are educated have a vision and the rest have two sores in the place of eyes. So, I conclude, education is important for all of us. I thank you all for the unique opportunity given to me to share my ideas on the importance of education.

Reading

Have you heard of Open Letters?
An open letter is a letter that is intended to be read by a wide audience, or a letter intended for ‘ an individual, but that is nonetheless widely distributed intentionally through newspapers and other media, such as a letter to the editor or a blog.

Sociologists, scholars, anthropologists, and historians have written open letters. Letters patent are another form of an open letter in which a legal document is both mailed to a person by the government and publicized so that all are made aware of it. Open letters can also be addressed directly to a group rather than any individual.

Reading such letters will also help to improve your vocabulary. Here are some easy ways to improve and expand your vocabulary.

  • Read as much as you can. As you read, try to understand the meaning of new words from the context or by referring to a dictionary.
  • When you refer to a dictionary, leam the pronunciation, meaning, synonyms, antonyms, and derivatives of the word.
  • Note down the new words you leam for future reference.
  • Make it a habit to leam one new word a day.
  • Play word games such as word search, crossword, scrabble etc. to expand your vocabulary.
  • Listen to conversations, leam and discover new words, and then start using them.

The following is a letter by Nobel Laureate Rabindranath Tagore to Mahatma Gandhi. Follow the diction, fluency, and style of the great Indian writer who has contributed excellent writings to Indian Literature. You can improve your vocabulary by familiarising some of the words used in the letter using a dictionary.

Gandhi Letter 23A: From Rabindranath Tagore
Shanti Niketan,

April 12, 1919.

Dear Mahatmaji,
Power in all its forms is irrational – it is like the horse that drags the carriage blindfolded. The moral element in it is only represented in the man who drives the horse. Passive resistance, a force which is not necessarily moral in itself, can be used against truth as well as for it. The danger inherent in all force grows stronger when it is likely to gain success, for then it becomes a temptation. I know your teaching is to fight against evil by the help of the good. But such a fight is for heroes and not for men led by impulses of the moment. Evil on one side naturally begets evil on the other, injustice leading to violence and insult to vengefulness.

Unfortunately, such a force has already been started, and either through panic or through wrath our authorities have shown us the claws, whose sure effect is to drive some of us into the secret path of resentment and others into utter demoralization. In this crisis you, as a great leader of men, have stood among us to proclaim your faith in the ideal which you know to be that of India, the ideal which is both against the cowardliness of hidden revenge and the cowed submissiveness of the terror-stricken…

I have always felt, and said accordingly, that the great gift of freedom can never come to a people through charity. We must win it before we can own it. And India’s opportunity for winning it will come to her when she can prove that she is morally superior to the people who rule her by their right of conquest. Armed with her utter faith in goodness she must stand unabashed before the arrogance that scoffs at the power of the spirit. And you have come to your motherland in the time of her need to remind her of her mission, to lead her in the true path of conquest, to purge her present-day politics of its feebleness which imagines that it has gained its purpose when it struts in the borrowed feathers of diplomatic dishonesty.

This is why I pray most fervently that nothing that tends to weaken our spiritual freedom may intrude into your marching line, that martyrdom for the cause of truth may never degenerate into fanaticism for mere verbal forms, descending into the self-deception that hides behind sacred names. With these few words for an introduction allow me to offer the following as a poet’s contribution to your noble work:

Give me the supreme courage of love,
this is my prayer,
the courage to speak,
to do, to suffer at thy will,
to leave all things or be left alone.

Give me the supreme faith of love,
this is my prayer,
the faith of life in death,
of the victory in defeat,
of the power hidden in the frailness of beauty,
of the dignity of pain that accepts hurt,
but disdains to return it.

Very sincerely yours,
Rabindranath Tagore

A. Answer the following questions.

Question 1.
Who according to Gandhi, can light against evil and how?
Answer:
One can fight against evil with the help of good. Such a fight is led by heroes like Gandhi himself.

Samacheer Kalvi 11th English Solutions Prose Chapter 5 Convocation Address

Question 2.
What is Gandhiji’s ideal?
Answer:
Gandhi’s ideal is to observe passive resistance against the cowardliness of hidden revenge and the cowed submissiveness of terror.

Question 3.
According to Tagore, when will India get the opportunity to win the gift of freedom?
Answer:
According to Tagore, India’s opportunity of winning freedom will come to her when she can prove that she is morally superior to the people who rule by the right of conquest.

Question 4.
How does Tagore acknowledge Gandhi’s noble work?
Answer:
Tagore acknowledges the noble work of Gandhi in his poem. He appreciates the moral courage he had instilled in Indians to suffer at his will. He appreciates the faith of love, faith of life in death, of the victory in defeat, of the power hidden in the frailness of beauty, and of the dignity of pain that accepts hurt but disdains to return it.

Question 5.
Find words from the passage which mean the same as the following.
(a) a malevolent desire for revenge (para 1)
(b) tactful (para 2)
(c) despise (para 3)
Answer:
(a) vengefulness
(b) diplomatic
(c) disdain

Question 6.
Find words from the passage which are antonyms of the following.
(a) artificially (para 1)
(b) strength (para 2)
Answer:
(a) naturally
(b) feebleness

Grammar

A. Report on the following dialogue.

Prabhu: What are you doing here, Kiran? I haven’t seen you for a few months.
Kiran: I have just come back from my native town Virudhunagar.
Prabhu: Did you enjoy your vacation?
Kiran: Yes. I love the place. It is a clean and busy town.
Prabhu: Where did you go and what did you see?
Kiran: I went to Courtallam falls in Tenkasi.
Prabhu: Share some pictures of your trip.
Kiran: Sure. See you later.

Answer:

Prabhu asked Kiran what she was doing there and said that he hadn’t seen her for a few months. Kiran replied that she had just come back from her native town Virudhunagar. Prabhu asked her if she had enjoyed her vacation. Kiran replied affirmatively and added that she loved the place and it was a clean and busy town. Prabhu further required her where she had gone and what she had seen. Kiran replied that she had gone to Courtallam falls in Tenkasi. Prabhu requested her to share some of the pictures of her trip. Kiran agreed and bade farewell to him.

B. Virat Kohli, the Man of the Match and Man of the Series in the one-day international series between India and South Africa in February 2018 had this to say during the post-match presentation. India won the match by 8 wickets and won the series by 5 – 1, a historic win against South Africa on their home soil.

Rewrite his words in Reported Speech.

Samacheer Kalvi 11th English Solutions Prose Chapter 5 Convocation Address

Question 1.
“It was a day I felt really good. Last game, I was not in the right kind of mindset. This is a beautiful place to bat under lights. That’s the idea behind bowling first. I like setting up for the short ball. It was a blessing in disguise, and they kept bowling short. I think the pitch got better to bat on under lights! It has been a roller coaster till now. People who are close to me deserve a lot of credit. Obviously, you want to lead from the front, and that’s a wonderful feeling.

I have got eight or nine years left in my career and I want to make the most of every day. It’s a blessing that I am healthy and getting to captain my country. They have shown great character – especially the two young spinners. The way the series went augurs well for us. We’re looking forward to the T20s. The tour is not over yet. After losing the Test series, I am talking to you. I am here talking to you after winning the ODI series.”

Answer:

lt was a day Kohli had felt really good. The previous game he was not in the right mindset. It was a beautiful place to bat under lights. That was the idea behind bowling first. He liked setting up for a short ball. It had been a blessing in disguise and they had kept bowling short. He thought that the pitch got better to bat on under lights. It had been a roller coaster till then. People who were close to him deserved a lot of credit. Obviously one would want to lead from the front and that was a wonderful feeling.

He had got eight or nine years left in his career and he wanted to make the most of it. It was a blessing that he was healthy and got to captain his country. They had shown great character, especially the two young spinners. The way the series had gone augured well for them. They were looking forward to the T20s. The tour was, not over yet. After having lost the test series, he had been talking to them. He was there talking to them after winning the ODI series.

C. Read the given passages and rewrite them in direct speech.

Question 1.
The cyclist warned the driver not to move his car till the police arrived. The driver pleaded that it was not his fault; he was ready to pay a hundred rupees to repair the damaged cycle. The cyclist refused the money and insisted that the police be called.
Answer:
The cyclist : Don’t move your car till the police arrive.
Car driver : Sir, it is not my fault. I’m ready to pay hundred rupees for repairing the damaged cycle.
The cyclist: I don’t want the money. I want to call the police

Question 2.
The striking workers demanded an increase in salary and asked for the withdrawal of all cases. They threatened to continue the strike if the demands were not met. The manager insisted on them calling off the strike and invited them for a discussion. He agreed to listen to their demands
Answer:
Workers: We demanded an increase in salary to withdraw all the cases.
Manager: Call off the strike
Workers: If our demands are not met, we will continue the strike.
Manager: I am willing to look into your demands, please come for a discussion.

Writing

Samacheer Kalvi 11th English Solutions Prose Chapter 5 Convocation Address

Question 1.

Question (a)
Write an e-mail to your uncle thanking him for the gift that he had sent from abroad.
Answer:

From: [email protected]
To: [email protected]
Sub: Thanks a lot for the gift!
Dear uncle,

I’m fine and I hope my e-mail finds you in robust health. Your gift exactly on my birthday was a bolt from the blue. You had remembered the promise you had made three months ago and sent a marvelous Lap Top. It is light but very cool. I am really proud of you. I pray to God that you get more success in your business venture abroad and help other helpless ’

children too.
Yours lovingly
S. Philip

Question (b)
Write an e-mail to a charitable trust requesting for a scholarship.
Answer:

From: [email protected]
To: [email protected]
Sub: Seeking scholarship

Dear Sir,
I am Ritika about to complete my Std XI in Government Higher Secondary School, Nandhivaran. My father, the only breadwinner, got hurt in an accident. He was working in a private firm. My mom is a homemaker. I am at crossroads now. My mom has taken my dad to Nagercoil. I need to stay in a ladies’ hostel to continue my studies. Could you please consider my request favorably and release a scholarship till I complete my XII public exam – successfully. I hope to avail a Bank loan after I get my results.
Awaiting your favourable reply,

Yours sincerely,
S. Ritika
XI-BGGHSS,
Nandhivaran
Kanchipuram Dt.

Formal Letters

1. Write a letter to the editor of a newspaper about the need to wear seat belts while driving. 18.03.2019 Virudhu Nagar

From
M. Kamaraj,
18, Anna Street,
Kabilan avenue,
Virudhu Nagar

To
The Editor in Chief,
The Hindu,
Mount Road Chennai Dear sir,

Sub: Importance of seat belts
I’m M. Kamaraj living in Virudhunagar. I have been reading every day about road accidents. Though the law-enforcing officers are making appeals to all car drivers to wear a seat-belt, very few do it. It is because cheap car engines start even when the driver has not attached the seat belt. Technically, advanced cars which are above 10 lakhs are better designed and wouldn’t budge if the driver doesn’t wear the seat-belt. Such cars have additional safety mechanisms such as air-pillows that prevent serious injuries in case of accidents. But the cheap cars crash and kill not only the drivers but also their loved ones.

Thrill kills! The government of India should insist on air-bags and technical viability of connecting seat belts to ignition points in a car to ensure the safety of all drivers of cars irrespective of the cost of the cars. I request you to publish this in all your city editions including that of Delhi. This may influence the Transport Minister of Government of India to take positive action on this.

Thank you,
Yours sincerely,
M. Kamaraj

To
The Editor in Chief
The Hindu
Mount Road, Chennai

2. Write a letter to the commissioner of the corporation complaining about the sanitary conditions of the street in your locality.

18.03.2019
Madurai

From
K. Lavanya MIG 14, Anna St,
Amma Colony, Shanthi Nagar Madurai

To
The Commissioner
Corporation of Madurai
Madurai -2
Sir,

Sub: Deteriorating sanitary conditions – a complaint I wish to submit the following painful facts for your consideration and immediate action. There are four big garbage bins kept in my street. For the past three weeks, your corporation vehicles have not come to clear the garbage. Now dust bins are overflowing. People are throwing garbage around the garbage bins. But the stench arising out of the accumulated garbage is nauseating. I’m really worried that some serious diseases may break out. Dengue is still causing anxiety among the residents of Madurai as it has already taken a toll on seven babies, two young boys, and three old men. Please instruct your officers to send vehicles immediately to collect the garbage and send your sanitary inspectors to sprinkle/spray disinfectants to ensure the prevention of infectious diseases like cholera too.

Thanking you in anticipation
Yours truly,
K. Lavanya

To
The Commissioner,
Corporation of Madurai
Madurai -2

3. Write a letter to a sports company ordering sports items for the Physical Education Department of your school.

15.7.2018
Alanganallur
From

M. Jothi,
Sports Secretary,
Government Girl HSS,
Alanganallur-625501

To
The Manager,
M.J. Sports Company,
Good shed Street,
Trichy-1
Sir,
Sub: Ordering sports goods
We propose to organize an Inter-school sports contest in November. We like to buy the following sports materials from you. Please send us the price of the items with the discount, you usually give for school. On delivery, we shall send the money by RTGS. Kindly send your bank account details and reply by e.mail to [email protected] Materials required

Samacheer Kalvi 11th English Solutions Prose Chapter 5 Convocation Address

Kindly send the mail with your estimated price inclusive of transportation and GST. Looking forward to your prompt response.
With warm regards Yours sincerely
M. Jothi

To
The Manager,
M.J. Sports Company,
Good shed Street,
Trichy-1

4. Write a letter to the Chief Reservation Supervisor of Railways requesting him to grant a concession for your educational tour.

10.07.2018
Nanguneri
From

M. Prabu,
SPL.GHSS.
Nanguneri.

To
The Chief Reservation Supervisor,
Southern Railways,
Thirunelveli.

Sir,
Sub: GHSS- Educational tour to Bangalore requesting to grant concession – Reg.
I am M. Prabu studying in the XI-D section. I am the school pupil leader. I’ve been asked by my school Headmaster to work out the cost for our educational tour to Bangalore. A team of eighty students intends to go the Bangalore in the month of November. Could I request you to consider extending us concession? Please let us know either through a reply letter or by mail what would be the cost of all 80 tickets (to and fro) and the dates ideal for us to travel. We intend to stay in Bangalore for two days. Kindly check the availability in the 3rd week. If possible could you please request authorities to allot two bogies for us exclusively both ways? I have enclosed the name list of the students for your kind consideration. My e.mail id is [email protected]. An early reply will help me collect the money and meet you to book our onward and return journey.

Thank you,
Yours sincerely,
M. Prabu

To
The Chief Reservation Supervisor,
Southern Railways,
Thirunelveli.

5. Write a letter to the curator of the museum seeking permission for a school visit.

17.06.2018
Thindivanam

From
M. Padmavathy Class leader,
XI-C section,
GGHSS,
Thindivanam

To
The curator,
Museum
Egmore
Chennai

Sir,
Sub: GGHSS – Thindivanam seeking permission to visit the museum – Reg.
A team of SO students wishes to visit your museum on 17.08.2018. We will be coming on two buses. Kindly arrange a guide who can detail each exhibit with its relevance to Indian culture and Tamil History. I leam that curators can help children understand the importance of every exhibit in the museum and effortlessly relate it to the age it belongs to and its relevance to modem times. Could I request you to grant us permission to visit the museum on 17.08.2018 and get the benefit of your expert guidance and a well-experienced guide who can explain things well in Tamil?

Thank you,
Yours sincerely,
M. Padmavathy.

To
The curator,
Museum, Egmore
Chennai

Task:

(a) Respond to the following advertisements.

Classified Advertisement

WANTED 100 part-time Graphic Artists
Experienced in Photoshop and Indesign-
Salary negotiable. Apply to: MM Graphics,
Triplicane, Chennai – 5 or Mail your Resume to [email protected]
Answer:
From: [email protected]
To: [email protected]
Sub: Applying for a part-time graphic artist
Dear Sir,

With reference to your advertisement in The Hindu for the post of 100 – part-time Graphic Artist with experience in Photoshop and Indesign. I hereby wish to apply for the same.

I am doing my second-year B.Sc. (Computer science) in SIEVT college as a part-time student from 4 to 9 p.m. Thereby I am free in the morning from 9 a.m. to 3 p.m. I have a working knowledge too of Photoshop and Indesign. I have also completed a crash course CSC. Please find enclosed my resume for your kind perusal.

Resume

Name : A. Parveen
Father’s Name : K. Akbar
Date of birth : 02.04.1988
Language known : English, Hindi, Urdu & Tamil
Qualification : BSC II year, CSC (Doing) SSLC : 450/500 in Nirmala Girls Hr. Sec School, Triplicane in 2014 HSC : 1100/1200 in Nirmala Girls Hr. Sec school Triplicane in 2016
Technical Qualification : 3 months crash course Indesign in 2015-16 3 months crash course in Photoshop in 2016-17
Salary Expected: Rs. 12000/- P.M
Address For Communication: A. Parveen D/o. Akbar No. 18, Pallivasal Street, Triplicane Chennai. Email Id: [email protected]

Awaiting your e-mail
Yours sincerely,
A. Parveen

Samacheer Kalvi 11th English Solutions Prose Chapter 5 Convocation Address

b. Block Advertisement

Female IELTS Teachers Required
Postgraduates in English with a minimum
of 3 years’ experience.
Must be bold and confident
Good salary, transport, and food allowance
provided, Apply to:
Arv Institute of Languages,
149B, Bose Road, Coimbatore
Ph: 98400 XXXXX
Email: [email protected]
Samacheer Kalvi 11th English Solutions Prose Chapter 5 Convocation Address

From: [email protected]
To: [email protected]
Sub: Applying for the post of IELTS teachers- Reg.

Sir

I am glad to learn that you have a position to handle IELTS Class. I got through IELT courses way back in 2015. I’ve been working as a language teacher in Millennium school near Airport. Though I am getting a decent salary, I wish to switch over to training. Training youngsters whose ambitions are sky-high is thrilling. Teaching English to adults is my forte. I’ve already been a trainer in the British council for a year. I’ve enclosed photocopies of my credentials for your reference. Send you a call letter to [email protected]. Looking forward to meeting you.

Yours sincerely,
K. Suguna.

Encl:
1. Resume
2. Photocopies of certificates

Resume

Name: K. Suguna
Date of birth : 03.03.1990
Address : No. 18, Kamban Street,
Ambattur,
Chennai.
Qualification: B.Ed -Teacher Education University 2015-2016 – 55%
MA English – University of Madras 2013-2014 – 60%
B.A. English – Women’s Christian College
Nungambakkam 2012-2013-58%
HSC – Presidency girls HSS Egmore 2008-09 – 80%
SSLC – Presidency girls HSS Egmore 2005-06 — 88%
Experience: Millennium school Alandur since June 2016
The salary drawn: Rs.20,000/- PM
Salary Expected: Rs. 30,000/-PM
Address For communication: [email protected]

Declaration
The details furnished above is a true letter to my knowledge. Looking forward to receiving your call letter
Date: 14.04.20xx
Location: XYZ

x x x x
Signature of the Candidate

Convocation Address About the Author

Samacheer Kalvi 11th English Solutions Prose Chapter 5 Convocation Address

Dr. Conjccvaram Natarajan Annadurai (15 September 1909 – 3 February 1969), popularly called Arignar Anna was an Indian politician who served as the Chief Minister of Tamil Nadu (1967 – 69). He was a great orator and an acclaimed writer in Tamil. Jawaharlal Nehru hailed him as one of the greatest parliamentarians. He has published several novels, short stories and plays which incorporate political and social themes. C.N. Annadurai was awarded the Clubb Fellowship at Yale University, the first non-American to receive this honour. The same year he was awarded an honorary doctorate from Annamalai University.

Convocation Address Summary

Societal development happens only through the hard work of many unsung heroes. Those who develop their potential by drawing on the resources of society need to pay them back after they graduate. Dr. C. M. Annadurai conveyed his gratitude for inviting him to deliver the Convocation Address at Annamalai University in 1967. He wished the fresh graduates a prosperous future. He accepted that it was difficult to provide them with appropriate guidelines for success in life. However, he shared cardinal principles enunciated by those eminent scholars who had done so earlier. He chose to replay those ideas as a common man as he interpreted them without any sophistication of language. Universities in the past gave access to scholars and poets to adorn the chambers of royalty.

Samacheer Kalvi 11th English Solutions Prose Chapter 5 Convocation Address

They were for the royal mansion and not for the market place. Today universities have the responsibility to fashion out of every individual fitted and equipped for the task of making democracy fruitful and effective. According to Dr. Radhakrishnan, “Universities ensure the democratic way of life for the future generations.” Though earning a decent living is an important objective, it shouldn’t be the only objective. The university graduate should feel indebted to the community. Universities are sustained by the revenue paid by peasants and poor people. Unless graduates replenish the social chest, it will become an empty offer.

The speaker was quite aware of the social challenges the graduates might face. He advised them to emulate men and women endowed with the spirit of service and carry on the crusade against poverty and other social ills. He was hopeful that Tamils, being the inheritors of a great culture will have the strength to overcome social challenges and serve society to the best of their abilities. He wished that they were bound to win and brighten the entire land.

Convocation Address Glossary

Textual:

  • autocracy – government by one ruler
  • cloistered – restricted
  • conferred – granted a title, degree, benefit or right
  • confronting – aggressively resisting
  • crusade – campaign for a good cause/holy war
  • despondent – depressed, frustrated
  • emissaries – deputies
  • endowed – gifted
  • enunciated – spoke clearly
  • eschewed – have nothing to do with
  • feudal – having to do with the Middle Ages, old
  • indebted – obliged to repay
  • inherent – inborn
  • inheritors – successors
  • lustre – glow of reflected light
  • perils – dangers and risks
  • perseverance – continued effort, steadfastness
  • reiterate – say or do again
  • replenish – refill
  • repositories – storehouses
  • ruggedness – strength, toughness
  • secluded spheres – isolated areas
  • sermons – speeches on moral subjects
  • tillers – persons who produce crops / raise animals, cultivators
  • toilers – workers, people who perform

Samacheer Kalvi 11th English Solutions Prose Chapter 5 Convocation Address

Additional:

  • appropriate – suitable
  • cardinal (adj) – most important
  • concern -worry
  • eminent – famous
  • emulate – follow
  • gratitude – thanks
  • indebted – bound
  • peasant – poor farmers
  • prosperous -rich
  • responsibility – obligation
  • sophistication – refinement

Convocation Address Synonyms

Choose the correct synonyms from the options given below. (Exam model)

Question 1.
A student embarks on a career after graduation.
(a) rupture
(b) begins
(c) completes
(d) summons
Answer:
(b) begins

Question 2.
Society has cradled and nurtured every graduate.
(a) sowed
(b) protected
(c) irrigated
(d) annoyed
Answer:
(b) protected

Question 3.
Graduation is the apt time to give back to society.
(a) misfit
(b) appropriate
(c) irrelevant
(d) ineligible
Answer:
(b) appropriate

Question 4.
I am thankful for the unique honour.
(a) common
(b) special/unusual
(c) drab
(d) conventional
Answer:
(b) special/unusual

Question 5.
This Institution has conferred on my honour.
(a) fielded
(b) granted
(c) defied
(d) withdraw
Answer:
(b) granted

Question 6.
I shall only reiterate some of the cardinal principles.
(a) repeat/say or do again
(b) suspend
(c) reexamine
(d) reciprocate
Answer:
(a) repeat/say or do again

Samacheer Kalvi 11th English Solutions Prose Chapter 5 Convocation Address

Question 7.
Wise men have enunciated those principles in the past.
(a) beat about the bush
(b) confused
(c) asserted
(d) spoke harshly
Answer:
(c) asserted

Question 8.
I do claim to represent him in all his ruggedness.
(a) meekness
(b) weakness
(c) toughness
(d) feebleness
Answer:
(c) toughness

Question 9.
Universities are repositories of knowledge.
(a) rented houses
(b) own houses
(c) storehouses
(d) leased houses
Answer:
(c) storehouses

Question 10.
Universities are the nursing ground for the emissaries of thought.
(a) admirably
(b) adversary
(c) foe
(d) ambassador/deputies
Answer:
(d) ambassador/deputies

Question 11.
We have eschewed monarchy.
(a) embraced
(b) accepted
(c) avoided
(d) relinquished
Answer:
(c) avoided

Question 12.
We have stayed away from autocracy.
(a) dictatorship
(b) militarism
(c) mobs
(d) royalist
Answer:
(a) dictatorship

Question 13.
We have inaugurated the era of democracy.
(a) Self-government
(b) Monarchy
(c) Bureaucracy
(d) Autocracy
Answer:
(a) Self-government

Question 14.
During the feudal days, universities were different.
(a) old/middle ages
(b) recent
(c) future
(d) modem
Answer:
(a) old/middle ages

Question 15.
Eminent scholars were not asked to confront the problems of the masses.
(a) face/resist
(b) accept
(c) cowed down
(d) brow beaten
Answer:
(a) face/resist

Question 16.
They were content to work in secluded spheres.
(a) isolated
(b) swarmed
(c) crowded
(d) mobbed/congested
Answer:
(a) isolated

Question 17.
University today is not cloistered.
(a) liberated
(b) restricted
(c) limitless
(d) infinite
Answer:
(b) restricted

Question 18.
The task demands patience and perseverance.
(a) low effort
(b) persistence
(c) high loss
(d) heavy loss
Answer:
(b) persistence

Samacheer Kalvi 11th English Solutions Prose Chapter 5 Convocation Address

Question 19.
Graduates should have faith in their inherent ability to shoulder responsibility.
(a) cultivated
(b) artificial
(c) inborn
(d) nurtured
Answer:
(c) inborn

Question 20.
Their philosophy was turned into the dazzling garments for the privileged.
(a) lackluster
(b) gloomy
(c) glaring
(d) dim
Answer:
(c) glaring

Question 21.
The immediate concern of every individual is to acquire the means for a decent living.
(a) kindness
(b) worry
(c) indifference
(d) love
Answer:
(b) worry

Question 22.
You are deeply indebted to the community.
(a) thankless
(b) grateful
(c) loud
(d) quiet
Answer:
(b) grateful

Question 23.
That revenue comes from tillers of the soil.
(a) manufactures
(b) farmers
(c) pollutants
(d) wasters
Answer:
(b) farmers

Question 24.
The toilers did not enjoy this privilege.
(a) bosses
(b) workers
(c) kings
(d) leaders
Answer:
(b) workers

Question 25.
You should replenish it.
(a) exhaust
(b) refill
(c) consume
(d) deplete
Answer:
(b) refill

Question 26.
You must return hope unto the despondent.
(a) empowered
(b) rejoiced
(c) depressed/frustrated
(d) energized
Answer:
(c) depressed/frustrated

Question 27.
The world will disturb your determination.
(a) instability
(b) laziness
(c) uncertainty
(d) firmness
Answer:
(d) firmness

Question 28.
The sermons would become sweet nothings. .
(a) doctrines
(b) poems
(c) dramas
(d) debates
Answer:
(a) doctrines

Question 29.
Tyranny of all sorts may stare at you.
(a) kindness
(b) autocracy
(c) love
(d) happiness
Answer:
(b) autocracy

Samacheer Kalvi 11th English Solutions Prose Chapter 5 Convocation Address

Question 30.
A continuous stream of men and women are endowed with the spirit of service.
(a) narrated
(b) gifted
(c) withdrew
(d) submitted
Answer:
(b) gifted

Question 31.
These service-minded people have been carrying on the crusade successfully.
(a) campaign fight
(b) killing
(c) campaign
(d) strike
Answer:
(c) campaign

Question 32.
Tamil are the inheritors of the rich legacy of culture.
(a) proprietors
(b) owners
(c) successors/heirs
(d) legacy
Answer:
(c) successors/heirs

Question 33.
The legacy of culture enables us to overcome even the adverse environment.
(a) bequest
(b) curse
(c) bane
(d) perdition
Answer:
(a) bequest

Question 34.
You are adequately equipped.
(a) legitimately
(b) aptly
(c) sufficiently
(d) legibility
Answer:
(c) sufficiently

Question 35.
I am confident that you are being sent into the wide world.
(a) despondent
(b) hopeful
(c) desperate
(d) annoyed
Answer:
(b) hopeful

Convocation Address Antonyms

Choose the most appropriate antonyms for the underlined words. (Exam model)

Question 1.
A qualified student embarks on his career.
(a) boards
(b) lands/disembarks
(c) climbs
(d) takes off
Answer:
(b) lands/disembarks

Question 2.
Society has nurtured graduates.
(a) caned
(b) protected
(c) contemplated
(d) neglected
Answer:
(d) neglected

Question 3.
I intend to reiterate some of the cardinal principles enunciated by educational experts.
(a) spoken vividly
(b) faltered
(c) delivered
(d) articulated
Answer:
(b) faltered

Question 4.
I do represent him in all ruggedness.
(a) toughness
(b) tenderness
(c) strength
(d) might
Answer:
(b) tenderness

Question 5.
We have eschewed monarchy.
(a) avoided
(b) disturbed
(c) embraced
(d) dodge
Answer:
(c) embraced

Samacheer Kalvi 11th English Solutions Prose Chapter 5 Convocation Address

Question 6.
It requires individual responsibility.
(a) personal
(b) modem
(c) classical
(d) empty
Answer:
(d) empty

Question 7.
It is the immediate concern of every individual.
(a) worry
(b) apprehension
(c) indifference
(d) anxiety
Answer:
(c) indifference

Question 8.
You are deeply indebted to society.
(a) grateful
(b) thankful
(c) obliged
(d) unaligned
Answer:
(d) unaligned

Question 9.
You need to replenish the social chest.
(a) fill
(b) refill
(c) require
(d) empty
Answer:
(d) empty

Question 10.
Turn despondent into optimistic citizens.
(a) desperate
(b) frustrated
(c) hopeless
(d) hopeless
Answer:
(d) hopeless

Question 11.
You are adequately equipped.
(a) sufficiently
(b) inadequately / insufficiently
(c) enough
(d) required
Answer:
(b) inadequately / insufficiently

Question 12.
They were content to work in secluded spheres.
(a) isolated
(b) congested
(c) interior
(d) inaccessible
Answer:
(b) congested

Question 13.
It is not just material advancement.
(a) progress
(b) backwardness
(c) development
(d) refinement
Answer:
(b) backwardness

Question 14.
In feudal days universities were different.
(a) old
(b) middle
(c) ancient
(d) modem
Answer:
(d) modem

Question 15.
Their wisdom meant for the selected was privileged.
(a) knowledge
(b) information
(c) intelligence
(d) foolishness
Answer:
(d) foolishness

Samacheer Kalvi 11th English Solutions Supplementary Chapter 6 The Never – Never Nest

Students can Download English Lesson 6 The Never – Never Nest Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 11th English Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations. Learn the Samacheer Kalvi 11th English Grammar to enhance your grammar skills like reading comprehension, passage writing, parts of speech, tenses, passive and active voice, and many other concepts in no time.

Tamilnadu Samacheer Kalvi 11th English Solutions Supplementary Chapter 6 The Never – Never Nest

Warm up

A. What are the essentials one needs to lead a comfortable life? Fill in the empty bubbles with some of them.

Samacheer Kalvi 11th English Solutions Supplementary Chapter 6 The Never - Never Nest
Answer:

Samacheer Kalvi 11th English Solutions Supplementary Chapter 6 The Never - Never Nest

B. List six gadgets that you want to purchase. Write them according to your priorities and state the reasons.

S.No. Gadgets Reasons
1.
2.
3.
4.
5.
6.

Answer:

S.No. Gadgets Reasons
1. Refrigerator to preserve vegetables in shape
2. Gas stove to cook food quickly
3. Micro oven to bake things quickly
4. Washing Machine to wash clothes
5. Television to entertain self
6. Mobile for outside contact

C. Answer the following questions.

Question (a)
Do you think you can afford to buy all of these at once?
Answer:
No, we can’t afford to buy all gadgets at a time

Question (b)
We may not have money to buy all our wants at the same time. In such a situation, what are the options available?
Answer:
We can buy them on loan by paying equal monthly instalment schemes.

Samacheer Kalvi 11th English Solutions Supplementary Chapter 6 The Never - Never Nest

Question (c)
Expand EMI
Answer:
EMI – Equal Monthly Instalment

Samacheer Kalvi 11th English The Never – Never Nest Textual Questions

A. Reading Comprehension Questions:

Question 1.
What did Aunt Jane like about Jack’s ‘little nest?
Answer:
Aunt Jane liked the furniture, the car, the piano, the refrigerator and the radio in the house.

Question 2.
Aunt Jane seemed to think that there was a mistake in the wedding present she had given Jack. Why?
Answer:
Aunt Jane had given Jack and Jill just 200 pounds as wedding gift. But Jack claimed to have bought the house. He had the car and all gadgets at home. So, she wondered if she had given them 2000 pounds instead.

Question 3.
What would make Jack the owner instead of being the tenant?
Answer:
Paying ten pounds and a few quarterly payments have made Jack Mr. Owner of the house

Question 4.
What sounded absurd to Aunt Jane?
Answer:
Aunt Jane realized that the house and all the gadgets in Jacks’s home were bought on loan. He was paying EMI more than his salary. He was steadily borrowing to pay his EMI. So, living beyond the means sounded absurd for her.

Question 5.
How did Jack manage to pay seven pounds eighty and eighty pence out of six pounds? Jack managed to pay seven pounds eight and eight pence out of his salary of six pounds by borrowing the rest of the money from “Thrift and providence trust corporation”

Question 6.
What advice did Aunt Jane offer the couple?
Answer:
Aunt Jane advised Jack not to continue their borrowing spree. She gave them ten pounds to settle atleast one of their bills. –

Question 7.
For what purpose did Aunt Jill wish to use the cheque given by Aunt Jane?
Answer:
Jill wished to pay Dr. Martin who had helped her deliver the baby. She owed still 20 dollars more to him. She used ten dollars to pay the doctor.

Question 8.
‘Just one more instalment and BABY’S REALLY OURS!’ This tells us that the couple
Answer:
The couple did not even have sufficient money to pay the doctor’s bill when Jill delivered the baby. Jane had a guilty feeling that she should pay ten more pounds to claim the baby as “theirs”.

B. Answer the following questions in about a paragraph of 100 -150 words each:

Question 1.
Why is there a double negative in the title: The Never – Never Nest? Elucidate with reasons from the play.
Answer:
The title of the play ‘The Never- Never Nest’ has two never in it, ensuring that the nest would never be built. The double negative is emphasizing the impossibility of home The ‘nest in the title, literally refers to the home of birds. Birds make their home by collecting straws and twigs of various trees.

The nest acts as their temporary home as they do migration with respect to the changing weather conditions. Also, they are not safe, as different animals might attack their nest anytime. The same is in the case with Jack and Jill. The couple can be attacked by the money – lenders anytime if the instalments are not paid on the designated time.

Question 2.
Bring out the humorous elements in the play.
Answer:
Jane, Aunt of Jack, gives a wedding gift of two hundred pounds to him and Jill to start a contented life. But Jane buys a home, a car, piano, dining table, cot, sofa and even a radio on loan. Every month he pays more than his salary towards the payment of EMIs. Initially Aunt Jane is pleased with Jack’s comfortable life. The dramatic irony of the whole play starts when Jack tells Jane that he owes his comfortable life to her. She asks if she had given to them two thousand pounds instead of two hundred by an oversight.

Aunt Jane infers that the house would cost a great deal of rent. It was only then she has a rollercoaster ride on the shocking truth about all Jack’s worldly possessions. He says that they thought it uneconomic to go on paying rent and be Mr. Tenant. To be wise, he decided to become Mr. Owner by just paying ten pounds and a few quarterly payments. Jill supports her husband’s wisdom saying that he had a pay hike of five shillings. Then it dawns on Jane that all the possessions may be bought on hire purchase of long term EMI based loans. Such loans collect a lot of interest concealed under EMIs. She asks if they own the car. Jack says the steering wheel, one of the tyres , about two of the cylinders belong to them . Jack and Jill reveal the wonderful strategy of living on future earnings.

They explain that radio, piano and the furniture have been purchased on EMI from Mr. Sage. Mrs. Jane is dumbstruck. She abhors sitting on the furniture and the cot which do not belong to her relatives but to some creditors like Mr. Sage or Spencer. She gives 10 pounds as gift and leaves. She turns down the offer of a car ride as one tyre and two cylinders only belong to Jack. As soon as they leave, Jill sends the giftcheque to Dr. Martin. The black comedy apparent is that with one more installment, she could call the baby their own.

Question 3.
How does the play ‘The Never – Never Nest’ expose the harsh reality of modern living?
Answer:
Jack and Jill bought each and every luxury of life cheerfully Jack and Jill believe in buying furniture to house in easy installments. They have recently purchased a house, a car and furniture in installments are higher than his income Sometimes Jack borrows from money lenders to pay the installments. Even they have got their baby in installment.

The couple can be attacked by the moneylender anytime if the installments are not paid on the designated time. Such a couple would make a nest but they will never settle happily. Thus the play ‘The Never-Never Nest’ exposes the harsh reality of modern living.

Question 4.
Jill said that they owned the steering wheel of a car, one of the tyres, two of the cylinders and leg of the sofa. What does this convey?
Answer:
Jills claim that they owned the steering wheel of the car, one of the tyres, two of the cylinders and leg of the sofa. This implies that none of the things that are in their home and even their home did not belong to them. His salary is six pounds but he has to pay seven pounds and eight pence. The Equated monthly installments ate away all his salary.

Jack was borrowing for the excess money from “ Thrift and providence Trust corporation”. The pathetic things that melts one’s heart is seen towards the end of the play. The gift cheque of 10 pounds is released to Mr. Martin. When Jack says doctors don’t expect to be paid so soon, Jill says that with one more installment the baby will be absolutely “theirs”. This reveal the fact, Jack and Jill do not have anything left even for medical emergencies like child – delivery or sickness. Such life is not to be flaunted but pitied.

listening Activity

C. Listen to the passage read out aloud by the teacher or played on a recorder and answer the questions that follow.

Listen to the views of a leading economist R. Azhagarasan about EMI in an interview.
We live in a world where everyone likes to lead a comfortable and luxurious life, though our salaries are not enough to meet our needs. In such a situation, the EMI seems to be the only option. It helps people pay for the expensive things, they have purchased, in parts as installments. It gives monetary power to buy expensive things beyond the capacity of a common man. Otherwise, many people will never be able to buy such things. This proves beneficial for people in purchasing a house and other essential appliances and accessories.

It also has some negative unavoidable results. If people are not able to pay EMI on time, they have to keep on paying the loan amount for years together with interest. If people are not able pay EMI on time, they may face serious problems such as the penalty for default, repossession of the purchased items and might even be subjected to legal actions. Moreover, people may have the tendency to buy things that are not at all necessary, paying EMIs over a longer period of time. This may lead them into never-ending debts. Hence, people should learn to spend within their means.

Questions:

Question (i)
The speaker says that our income is sufficient to meet our needs. Is it true or false?
Answer:
False

Question (ii)
EMI is the only ______ for people who buy veiy expensive things.
(a) consolation
(b) setback
(c) option
(d) debt
Answer:
(b) setback

Samacheer Kalvi 11th English Solutions Supplementary Chapter 6 The Never - Never Nest

Question (iii)
One is able to buy costly things with ______ power offered by EMI.
(a) physical
(b) withstanding
(c) honorary
(d) monetary
Answer:
(d) monetary

Question (iv)
If people fail to pay EMIs, they may be subjected to ______
(a) expulsion
(b) high pressure
(c) legal action
(c) dejection
Answer:
(c) legal action

Question (v)
People should learn to spend ______ their means.
(a) within
(b) beyond
(c) above
(d) beneath
Answer:
(a) within

Speaking Activity

D. Based on your understanding of the play ‘The Never-Never Nest’, organise a classroom debate on the topic: ‘Is Equated Monthly Instalment Scheme – a boon or bane to middle- class families?’

A : Equated Monthly Installment scheme is a boon. Middle-class people who can’t dream of owning a home or even buying a two-wheeler get the benefit.
B : How do you prove it?
A : Well, a person who earns 15,000 rupees a month is able to take a house loan. Shri Ram, Finance Investment company of Bajaj Fin Serve offers scooter, car loans on as low as 7% interest. So a man can avail the benefits first but pay later.
B : My dear friend you present the matter as if it is just easy. What happens if a medical emergency happens and if the person is unable to pay back EMI for a housing loan or vehicle loan. Don’t you know a farmer was beaten by “Collection agents” for not . being able to pay back EMI arrears of just 50,000 rupees? The farmer, unable to face
the disgrace committed suicide. Didn’t you read it in the newspapers? EMI scheme is a bane.
A : My dear friends, such a rare occurrence should not be blown out of proportion. For health emergencies, one can join Prime Minister’s health insurance scheme by just paying 12 rupees a month. Besides, there are also Health insurance schemes by Tamil Nadu state Government which covers the cost of critical Health care upto 60%. I assert the EMI scheme is a boon for all middle class people who keep EMI to less than 50% of their total earnings and have 10% for other unforeseen expenses. If one leads a planned life, EMI scheme, I reiterate is a bane.
Judge : Now both the teams have argued their case very well. Now team A has almost justified that EMI scheme is a boon.
B : Sir, I would like you to take a look at the notice released by all leading bank containing the list of housing property confiscated due to non – payment of EMIs.
A : Every middle-class Indian has dreamed of owning a house. If one save money say for twenty years to purchase a land or a house the cost would definitely go up by 10 times. Save now and buy later doesn’t suit the modern lifestyle. It is true that we should not speculate much. At the same time one can calculate earnings, future pay hike and keep the EMIs within payable limits – such judicious spending will really make EMI schemes a great boon.
J : Listening to the arguments of both the team, I conclude EMI scheme is both a bane and a boon. For those who are always on a spending spree irrespective of their limited income, EMI scheme is a bane. For those who think twice before making a hasty purchase and narrow down their choice only to vital needs, definitely, EMI scheme is a boon. Use EMI scheme wisely. Good luck.

Writing

Writing Task

E. Aunt Jane was shocked to see the life of Jack and Jill during her visit to their place. Put yourself in the place of Aunt Jane and write a letter to the couple advising them not to spend beyond their means. Make more suggestions to enable them lead a debt-free life.

West Avenue,
New York, .
18.10.20XX
Dear Jack & Jill,

I reached back home safely. After returning from your home, I could not sleep properly for three days, I have always purchased things with cash. I feel it is a disgrace to buy things on long term hire purchase schemes. I gave you two hundred pounds to help you start your married life. But you have squandered all that seed money in making advance payment on expensive purchases like piano, furniture, car and even a home on loan. You pay equated.

monthly installments that exceed your monthly salary. You are borrowing every month to pay your dues I’m really upset. I don’t know what you did with the ten pounds, I gave you. I wish to give you the following recommendation. You can print them and display them in your drawing-room. So, that you‘11 try to follow them seriously in your life.

  1. Let your first expenditure be saving 10% of salary for the future.
  2. Never buy things you don’t need.
  3. Cut your cloth according to your size.
  4. If you spend on future income you will become a pauper.
  5. A penny saved, is a penny earned.
  6. Contentment is the greatest wealth in the world.
  7. Lead a simple life befitting your income.
  8. Don’t ever try to complete with others.
  9. Those who are thrifty invariably become rich.
  10. Those who spend all their money in haste will regret in leisure.

Your loving aunt,
Jane

To
Jack,
15, Gandhi Road,
New Jersey,
United States of America

Additional Questions

I. Fill in the blanks with the right options:

Question 1.
Jack insisted that one must have ______ these days.
(a) TV
(b) Tab
(c) Phone
(d) Radiogram
Answer:
(d) Radiogram

Samacheer Kalvi 11th English Solutions Supplementary Chapter 6 The Never - Never Nest

Question 2.
When Jack is away ______ gets all her entertainment from her radiogram at the kitchen.
(a) Jane
(b) Jack
(c) Jill
(d) Mary
Answer:
(c) Jill

Question 3.
Jack and Jill regard their home as their little ______
(a) hut
(b) abode
(c) home
(d) nest
Answer:
(d) nest

Question 4.
Jane had given Jack and Jill a wedding gift of ______ pounds.
(a) 2
(b) 20
(c) 200
(d) 2000
Answer:
(c) 200

Question 5.
The possessions of Jack and Jill ______ Aunt Jane.
(a) gladdened
(b) worried
(c) teased
(d) annoyed
Answer:
(b) worried

Question 6.
Aunt Jane doubted if she had presented a cheque for ______ pounds by oversight as the wedding gift to Jack and Jill.
(a) 20,000
(b) 200
(c) 20
(d) 2000
Answer:
(d) 2000

Question 7.
Aunt Jane was under the impression that Jack and Jill were living on a/an ______ home.
(a) own
(b) rented
(c) leased
(d) old age
Answer:
(b) rented

Question 8.
Jack said that they don’t pay ______ for their home.
(a) EMI
(b) electricity bill
(c) water charges
(d) rent
Answer:
(d) rent

Question 9.
Aunt Jane was worried that if Jack and Jill did not pay the rent for the home ______
(a) they will be threatened
(b) owners will evict them
(c) they will be sued
(d) they will be ill-treated
Answer:
(b) owners will evict them

Question 10.
Jack and Jill don’t pay rent because the house is ______
(a) leased
(b) owned
(c) bought
(d) loaned
Answer:
(b) owned

Question 11.
Jill had become the owner by just making an initial payment of ______ pounds.
(a) 200
(b) 2000
(c) 10
(d) 50
Answer:
(c) 10

Question 12.
Jack and Jill thought it ______ to pay rent.
(a) frugal
(b) uneconomic
(c) unwise
(d) absurd
Answer:
(b) uneconomic

Question 13.
Jill supported Jack by explaining they could afford all the loans for Jack had a pay hike of ______ shillings.
(a) 50
(b) 100
(c) 5
(d) 10
Answer:
(c) 5

Question 14.
Jill admitted that the steering wheel and ______ of the tyres belonged to them.
(a) two
(b) three
(c) one
(d) some
Answer:
(c) one

Question 15.
Jack enjoyed all the pleasure of motoring for a mere down payment of ______ pounds.
(a) fifty
(b) hundred
(c) five
(d) twenty
Answer:
(c) five

Question 16.
Jane understood that Jack intended to pay the balance amount for the car through easy ______
(a) loans
(b) installments
(c) schemes
(d) ways
Answer:
(b) installments

Question 17.
Jack earned ______ pounds a month.
(a) ten
(b) six
(c) eight
(d) two
Answer:
(b) six

Samacheer Kalvi 11th English Solutions Supplementary Chapter 6 The Never - Never Nest

Question 18.
Jack had to pay ______ pounds of eight pence for EMI.
(a) four
(b) six
(c) seven
(d) eight
Answer:
(d) eight

Question 19.
As the EMI payments exceeded Jack’s salary, he steadily borrowed the balance amount every month from ______
(a) friends
(b) bank
(c) money lenders
(d) Thrift and Providence Trust Corporation
Answer:
(d) Thrift and Providence Trust Corporation

Question 20.
Aunt Jane had ______ as her motto in life.
(a) cash later
(b) cash down
(c) hire purchase
(d) buy now pay later
Answer:
(b) cash down

Question 21.
Aunt Jane gave Jack and Jill a cheque for ______ pounds so that they could have at least one item as their own.
(a) 20
(b) 40
(c) 200
(d) 10
Answer:
(d) 10

Question 22.
Jill sent the gift cheque immediately to ______ for the penultimate payment for medical services received during the delivery of the baby.
(a) Joseph
(b) husband
(c) Dr. Martin
(d) Annamalai
Answers
(c) Dr. Martin

II. Identify the speaker:

  1. Why, of course, Aunt Jane. You simply must have a radio set nowadays. – Jack
  2. What do you think of our little nest? – Jill
  3. Charming! Charming! Such a costly little room! – Aunt Jane
  4. Rent? Oh, no, we don’t pay rent – Jack
  5. We don’t pay rent because the house is ours. – Jack
  6. Why be Mr. Tenant when you can be Mr. Owner? – Jack
  7. And it’s so nice for me when Jack’s away at business. – Jill
  8. Oh, I should say the steering wheel and one of the tyres… and about two of the cylinders. – Jack
  9. Even so, you must be getting on very well to keeping a place like this. – Aunt Jane
  10. But, Jack if you don’t pay rent you’ll get turned out into the street. You’ve Jill and the baby to think of now, you know. – Aunt Jane
  11. Oh, nurse, I want you to run and post this for me. I’ll look after the body when you’re gone.- Jill
  12. But why waste money on the Doctor? Doctors don’t expect to be paid anyway. – Jack
  13. Oh, have you got a radiogram as well as a car and a piano? – Aunt Jane
  14. What! Travel in a car that has only one tyre and two thingummies! No thank you. I’ll take the bus. – Aunt Jane
  15. But that’s absurd! How can you pay seven pounds eight and eightpence out of six pounds? – Aunt Jane

III. Rearrange the sentences

Question 1.
(a) She wondered how it was possible with a limited income as low as six pounds a month.
(b) She did not understand how Jack was able to pay rent for such a lovely home.
(c) Aunt Jane visited Jack and Jill.
(d) She found all modem amenities like sofa, cosy bed, car, radiogram, piano and a lovely home.
(e) She was pleasantly surprised to see them leading a comfortable life.
Answer:
(a) Aunt Jane visited Jack and Jill.
(b) She was pleasantly surprised to see then leading a comfortable life.
(c) She found all modem amenities like sofa, cosy bed, car, radiogram, piano and a lovely home.
(d) She did not understand how Jack was able to pay sent for much a lovely home.
(e) She wondered how it was possible with a limited income as low as six pounds a month.

Question 2.
(a) Aunt Jane was worried.
(b) It slowly dawned on Aunt Jane that they were living beyond the means .
(c) Then Jack boasted of his smart plan of becoming Mr. owner of the house by just a down payment of ten pounds and the rest by easy installments.
(d) Jack said to Aunt Jane that they owed their comfortable living to her .
(e) She asked Jack and Jill if she had written 2000 instead of 200 pounds in the gift cheque.
Answer:
(a) Jack said to Aunt Jane that they owed their comfortable living to her
(b) Aunt Jane was worried
(c) She asked Jack and Jill if she had written 2000 instead of200 pounds in the gift cheque.
(d) Then Jack boasted of his smart plan of becoming Mr. Owner of the house by just a downpayment of ten pounds and the rest by easy installments.
(e) It slowly dawned on Aunt Jane that they were living beyond the means

Samacheer Kalvi 11th English Solutions Supplementary Chapter 6 The Never - Never Nest

Question 3.
(a) Jill sends the cheque to Mr. Martin so that she could call her baby “theirs’ with just one more instalment.
(b) She doesn’t want to sit on the furniture which does not belong to Jack.
(c) Aunt Jane is deeply disturbed to find out that Jack and Jill have now been addicted to “Buy now, pay later” culture.
(d) She realizes that they have not made complete payment for any item in the house.
(e) She gives them a gift cheque for ten pounds with serious advice to become a real owner of atleast one item in their house.
Answer:
(a) Aunt Jane is deeply disturbed to find out that Jack and Jill have not addicted to the “Buy now, pay later” culture.
(b) She realizes that they have not made complete payment for any item in the house.
(c) She doesn’t want to sit on the furniture which does not belong to Jack.
(d) She gives them a gift cheque for ten pounds with serious advice to become a real owner of atleast one item in their house.
(e) Jill sends the cheque to Mr. Martin so that she could call her baby “theirs’ with just one more installment.

IV. Read the following and answer the questions given below.

Question 1.
Jack: No, no, Aunt Jane. You misunderstood me. We don’t pay rent because the house is ours. Aunt Jane: YOURS?
Jill: Why, yes; you just pay ten pounds and it’s yours.
Jack: You see, Aunt Jane, we realized how uneconomic it is to go on paying rent year after year, when you can buy and enjoy a home of your own for ten pounds and a few quarterly payments, of course. Why be Mr.Tenant when you can be Mr. Owner?
Aunt Jane: I see. Yes, there’s something in that. Even so, you must be getting on very well to keep up a place like this.
Jill: Oh, he is, Aunt Jane. Why only last year he had a five-shilling rise—didn’t you, Jack? Jack (modestly): Of course that was nothing, really. I’m expecting ten this Christmas.
Aunt Jane (suddenly): Jack! I’ve just thought of something. That car—is it yours?
Jill: Of course it’s ours.
Aunt Jane : All yours?
Jack : Well, no. Not exactly all.

Question (a)
Why doesn’t Jack pay rent?
Answer:
Jack doesn’t pay rent because he has brought the house on EMI loan scheme.

Question (b)
How much had Jack paid for his house?
Answer:
Jack had made an initial payment of ten pounds and was paying the rest in instalments.

Question (c)
Was Aunt Jane pleased to know Jack’s promotion to Mr. Owner from Mr. Tenant? How?
Answer:
No, she was not pleased. Aunt Jane was a practical lady. She wondered how on earth Jack was maintaining such a lovely house with a lot of amenities.

Question (d)
How did Jill support her husband’s penny wise and pound foolish investments?
Answer:
Jill said that they were well off as Jack was given a pay hike of five shillings.

Question (e)
What was Jack’s response to Aunt Jane’s questions about the ownership of the car?
Answer:
He said that the car was his but not all the parts.

Question 2.
Aunt Jane: Now, I’m sorry if I sounded rude, but really I’m shocked to find the way (relenting a little) you’re living. I’ve never owed a penny in my life – cash down, that’s my motto and I want you to do the same. (She opens her handbag.) Now look, here’s a little cheque I was meaning to give you, anyway. (She hands it to Jill.) Suppose you take it and pay off just one of your bills –  so that you can say one thing at least really belongs to you.
Jill: Er – thank you. Aunt Jane. It’s very nice of you. (awkwardly)
Aunt Jane: There! Now I must be going, (patting her arm)
Jack: I’ll see you to the bus, anyway.
Jill: Good-bye, Aunt Jane – and thanks so much for the present.
Aunt Jane: Goodbye, my dear. (She and Jack go out. Jill looks at the cheque and (kissing her) exclaims ‘Ten pounds! ’ Then she hurries to the table, addresses an envelope, endorses the cheque, and slips it inside with a bill which she takes from the bag and seals the envelope. Then she rings the bell. In a moment the NURSE comes in with the baby in her arms.)
Jill: Oh, nurse. I want you to run and post this for me. I’ll look after baby while you’re gone.

Question (a)
Why was Aunt Jane shocked?
Answer:
Aunt Jane was shocked to find Jack and Jill living beyond their means.

Question (b)
What was Aunt Jane’s motto in life?
Answer:
Aunt Jane has never owed a penny in her life. ‘Cash down’ – that was her motto.

Question (c)
Why did Aunt Jane give Jack and Jill a cheque for ten pounds?
Answer:
Aunt Jane wanted Jack and Jill to pay atleast one of their bills so that they can become the owner of atleast one item in their house.

Question (d)
Who did Jane send the cheque?
Answer:
Jane sent the cheque to Dr. Martin.

Question (e)
Why did Jane decide to send the cheque to Mr. Martin?
Answer:
Jane believed that she could call the baby ‘theirs’ only if she could make one more instalment. This reveals her real distress.

The Never – Never Nest About the Author

Samacheer Kalvi 11th English Solutions Supplementary Chapter 6 The Never - Never Nest

Cedric Mount is a considerably distinguished playwright of his age. Me wrote some thoughtful plays, which include Twentieth Century Lullahy. “lo cut a long Short Story Short and Nature Abhors a Vacuum. His one-act plays are easy to perform, satirical, w itty and insightful. These one art plays expose the shams of contemporary society besides delicately admonishing the guilty.

The Never – Never Nest Summary

Aunt Jane is quite impressed with the small house Jack and Jill live in with all modem amenities. She is amazed at their acquisitions such as piano, refrigerator, radio and furniture. Aunt Jane had given only two hundred pounds as a wedding gift to Jack. But she finds that they own a house and all modem amenities. She is unable to conclude how on earth they accomplished all those comfort when Jack earned so little as six pounds. She is mdely shocked to know the house, fridge, furniture, car, piano and even the radio are all bought on EMIs. They have not made a complete payment for a single item. It sounded absurd that Jack is earning only six pounds a week. But he is paying seven pounds eight and eightpence towards EMI.

The rest is borrowed every month from Thrift and providence Trust corporation. It is so disgusting for Aunt Jane. She doesn’t want to sit on the furniture which belongs to someone else. She gives them a gift of 10 pounds with sharp advice that they should try to pay full amount for atleast one item and call it theirs. Jane refuses to travel in Jack’s EMI car for which he has paid for steering wheel and one tyre only. He accompanies Aunt Jane to the bus stop. Before he returns, Jill sends the cheque to Dr. Martin through her nlxrse. Jack is not happy. Jack says doctors never expect to be paid quickly. But she tells him that the baby will be their’s with just one more payment.

Samacheer Kalvi 11th English Solutions Supplementary Chapter 6 The Never - Never Nest

The Never – Never Nest Glossary

Textual:

  • absurd – ridiculously unreasonable and meaningless
  • cosy – comfortable
  • endorse – to make over to another
  • instalment – one of the parts into which a debt is divided when payment is made at intervals
  • lounge – a place in a home or public building for leisure activities, living room
  • motto – a short sentence or phrase that expresses a rule guiding the behaviour of
  • possessed – a particular person or group completely controlled by an evil spirit
  • propose – intend to do something
  • realise – to understand or become aware of
  • tartar – a person of irritable temper
  • thingummies – small articles the names of which are not remembered

Additional:

  • accomplished – achieved
  • acquisitions – possessions
  • amenities – facilities
  • disgusting – revolting

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 English Solutions Poem Chapter 6 The Hollow Crown

Students can Download English Poem 6 The Hollow Crown Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 11th English Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations. Learn the Samacheer Kalvi 11th English Grammar to enhance your grammar skills like reading comprehension, passage writing, parts of speech, tenses, passive and active voice, and many other concepts in no time.

Tamilnadu Samacheer Kalvi 11th English Solutions Poem Chapter 6 The Hollow Crown

Warm up

(a) Work with a partner take this short quiz to find out how well informed you are about history.

Samacheer Kalvi 11th English Solutions Poem Chapter 6 The Hollow Crown (from Richard II)

Question 1.
Name a few wars and battles you have read about.
Answer:
World War I, Indo-Pak War. Battle of Panipet War of Roses.

Samacheer Kalvi 11th English Solutions Poem Chapter 6 The Hollow Crown

Question 2.
What is the difference between a war and a battle?
Answer:
A war is a long drawn affair. The conflict may continue even for years. Battles are small segments of a big war.

Question 3.
Why do rulers wage wars and battles?
Answer:
Rulers are greedy. They want to expand their kingdoms. So, they wage wars and battles.

Question 4.
Is the outcome of a war always fair?
Answer:
No, the outcome of war is not always fair.

Question 5.
Do you think rulers understand the true meaning of life – in defeat or in victory?
Answer:
No, rulers involve a large number of people whose lives or deaths don’t matter for them. So, rulers usually don’t understand the true meaning of life.

Question 6.
Can you name a few kings and leaders who have fallen from glory to disgrace?
Answer:
Chandragupta Maurya / Rajputs and Nelson Mandela

(b) The Historical Background:

The poem is an extract from William Shakespeare’s play King Richard the Second. The play is based on true events that occurred towards the end of the 14th century. Richard II was crowned the King of England in the year 1367. He continued to be the British Monarch until 1399, when he was deposed by his cousin, Henry of Bolingbroke, who crowned himself King Henry the Fourth in the same year. Shakespeare’s play is a dramatic rendition of the last two years of King Richard IPs life.

In this brief span of time, he was ousted from his royal position and sent to prison, where he died in captivity. The following extract is set in the Coast of Wales. King Richard and some of his followers awaited the arrival of the Welsh army [after facing defeat at the hands of his cousin, Bolingbroke], of about 10000 warriors. But to their shock and surprise, they received the message that the army was not coming to their rescue. His followers tried to boost their King’s courage against the news, only in vain. When Richard came face to face with the reality of his terrible fate, he spoke the following verse, famously known as the “Hollow Crown” speech in theatrical circles. In it, King Richard is reminded of the power of Death that overshadows everything else, including the power of rulers, and renders them as powerless as any commoner at a moment’s notice.

Samacheer Kalvi 11th English The Hollow Crown Textual Questions

First, listen to a reading of the complete poem. Then, read silently and try to answer the questions briefly, based on your understanding. You may refer to the glossary given at the end of the monologue to help you.

Let’s talk of graves, of worms, and epitaphs,
Make dust our paper, and with rainy eyes
Write sorrow on the bosom of the earth.
Let’s choose executors and talk of wills.
And yet not so – for what can we bequeath
Save our deposed bodies to the ground?
Our lands, our lives, and all, are
Bolingbroke’s,

Samacheer Kalvi 11th English Solutions Poem Chapter 6 The Hollow Crown (from Richard II)

And nothing can we call our own but death;
And that small model of the barren earth
Which serves as paste and cover to our bones.
For God’s sake let us sit upon the ground
And tell sad stories of the death of kings:
How some have been depos’d, some slain in war,
Some haunted by the ghosts they have deposed,
Some poisoned by their wives, some sleeping kill’d,

Samacheer Kalvi 11th English Solutions Poem Chapter 6 The Hollow Crown (from Richard II)

All murdered – for within the hollow crown
That rounds the mortal temples of a king
Keeps Death his court, and there the antic sits,
Scoffing his state and grinning at his pomp,
Allowing him a breath, a little scene,
To monarchize, be fear’d, and kill with looks;
Infusing him with self and vain conceit,
As if this flesh which walls about our life
Were brass impregnable; and, humour’d thus,
Comes at the last, and with a little pin
Bores through his castle wall, and farewell king!
Cover your heads, and mock not flesh and blood
With solemn reverence; throw away respect,
Tradition, form, and ceremonious duty;
For you have but mistook me all this while.
I live with bread like you, feel want,
Taste grief, need friends – subjected thus,
How can you say to me, I am a king?

Question 1.
Pick out the phrase that suggests that King Riehard was sorrowful.
Answer:
The phrase “Talk of graves of worms and epitaphs” suggests that King Richard was sorrowful.

Samacheer Kalvi 11th English Solutions Poem Chapter 6 The Hollow Crown

Question 2.
Why does the King suggest that it is now time for his will to be executed?
Answer:
The King knows pretty well that he will be executed very soon by Bolingbroke. So, he wants his will to be executed.

Question 3.
What is the only thing we bequeath to our descendants?
Answer:
We bequeath only immovable property to our descendants.

Question 4.
What are the vanquished men left with?
Answer:
The vanquished men are left with sorrow and thoughts about death.

Question 5.
What does the ‘small model’ refer to here?
Answer:
The perishable human body stands as a ‘small model’ of the barren earth.

Question 6.
What does a monarch’s crown symbolize?
Answer:
Monarch’s crown symbolizes “empty power” because real power is vested with death

Question 7.
What mocks the ruler’s power and pomp?
Answer:
Death mocks the ruler’s power and pomp.

A. Fill in the blanks using the words given in the box to complete the summary of the poem:

barren-earth friends graves slain
rebellious rebellious worms grief
impregnable epitaphs death farewell
reverence king pin

King Richard the second had surrendered to his (a) _______ cousin, Bolingbroke. He experienced deep distress at the horror of his circumstances. In that desperate situation, he speaks of (b) _______ , (c) _______ , (d) and other things connected with death. He spoke of how people leave nothing behind and can call nothing their own, except for the small patch of (e) _______ where they will be buried. King Richard yielded to dejection and talked of all the different ways in which defeated kings suffer and how some had been deposed, (f) _______ in war, (g) _______ by their wives and so forth. He attributed this loss of lives to (h) _______ , who he personified as the jester who watches over the shoulder of every ruler, who mocks kings by allowing them to think their human flesh, was like (i) _______ brass. However, Death penetrates through the castle walls, silentlyand unnoticed like a sharp (j) _______ thus bidding (k) _______ to him and all his pride forever. Finally, Richard appealed to his soldiers not to mock his mere flesh and blood by showing (l) and respect to him. He added that he too needed bread to live, felt want, tasted (m) _______ and needed (n) _______ . He concluded thus, urging his men not to call him a (o) _______ as he was only human, just like the rest of them.
Answer:
(a) rebellious
(b) graves
(c) epitaphs
(d) worms
(e) barren earth
(f) slain
(g) poisoned
(h) death
(i) impregnable
(j) Pin
(k) farewell
(l) reverence
(m) grief
(n) friends
(o) king

B. The words used by Shakespeare find a place in the present-day conversations also. Here are a few examples of how these poetic, standardized English words could be used by common people in their regular speech.

(a) Fill in the blanks with appropriate words from the box and complete the statements suitably:

[bequeath, antics, monarchize, impregnable, hollow]

  1. Shravan never keeps his promises. His friends know that his words are ______
  2. The spectators died laughing at the ______ of the clown.
  3. The businesswoman wished to ______ all her riches to an orphanage, after her death.
  4. The fortress was ______ and could not be conquered by the enemies.
  5.  Alexander the Great, wished to conquer many lands and ______ the entire world.

Answer:

  1. hollow
  2. antics
  3. bequeath
  4. impregnable
  5. monarchise

(b) Complete the passage given below, with suitable words from the box:

farewell ceremonious deposed
reverence vain pomp
conceited sorrow scoffing

Lima, a (a) _______ and (b) _______ woman, kept (c) _______ at her colleagues and went on taxing them with hard labour. Though they were (d) _______ to her, she being their head, were offended and filled with (e) _______ It so happened, that Lima was (f) _______ from her high position due to a serious blunder she had committed. Lima, having lost all her (g) _______ and glory, realized how arrogant she had been. She gave up her pride and with (h) _______ sought an apology from everyone. She thus turned over a new Leaf and bid (i) _______ to them.
Answers
(a) vain
(b) conceited
(c) scoffing
(d) ceremonious
(e) sorrow
(f) deposed
(g) pomp
(h) reverence
(z) farewell

C. From your understanding of the poem, answer the following questions briefly in a sentence or two:

Question 1.
What do the three words,‘graves, worms, and epitaphs’,refer to?
Answer:
The three words graves, worms, and epitaphs’ refer to the deep sorrow of King Richard II who was captured by rebellious cousins Bolling broke.

Samacheer Kalvi 11th English Solutions Poem Chapter 6 The Hollow Crown

Question 2.
What does the executor mentioned in the poem do?
Answer:
An executor is one who implements the contents of a will.

Question 3.
Who is Bolingbroke? Is he a friend or foe?
Answer:
Boling broke is king Richard II’s rebellious cousin. He is a foe.

Question 4.
Are all deposed kings slain by the deposer?
Answer:
No, some, of the deposed kings are jailed and some are slain.

Question 5.
What does the crown of rulers stand for?
Answer
‘The crown of rulers stands for jester.

Question 6.
What hides within the crown and laughs at the king’s grandeur?
Answer:
Death hides within the crown and laughs at the king’s grandeur.

Question 7.
What does ‘flesh mean here?
Answer:
Flesh means body’s flesh. It stands for all perishable things.

Question 8.
What are the various functions and objects given up by a defeated king?
Answer:
A defeated king abdicates his crown. He parts with his scepter too. He hands over his right to rule the kingdom to the victorious king. He gives up the right to levy taxes on subjects. Fie also gives up his right and listens to the woes of ordinary subjects and solve them.

Question 9.
How does the king establish that he and his subjects are equal in the end?
Answer:
In the end, King Richard II pathetically explains that he is also an ordinary mortal with desires, need for friends and the compulsion to taste grief. Even a king has a cup of misery in his life.

Question 10.
Bring out King Richard’s feelings when he was defeated.
Answer:
King Richard started feeling distressed about his impending death. He uses the words graves, epitaphs, and worms. He realizes his possessions will be reduced to a patch of land. He recalls how kings get slain in battlefield or poisoned to death by their own spouses. The king feels he is also an ordinary mortal deceived by the jester’s‘ death’. He also needs to taste grief and needs the support of friends during distress.

D. Explain the following lines with reference to the context in about 5 to 8 lines:

Question (i)
“Our lands, our lives, and all are Bolingbroke’s,
And nothing can we call our own but death;”
Answer:
Reference:
These lines are taken from the Poem – “The Hollow Crown”, Poet – “William Shakespeare”.
Context:
Here the poet talks about the disowning of kind Richard II
Explanation:
Richard said that they have lost their lands, their lives, and all things by Boling Broke. They have nothing except their death.

Question (ii)
“All murdered – for within the hollow crown ‘
That rounds the mortal temples of a king
Keeps Death his court, …”
Answer:
Reference: These lines are from the poem “The Hollow Crown” by William Shakespeare. The poem is an excerpt from the play “Richard II”.
Context: The defeated king thinks about death which is looming large. He remembers how other kings had met with their death. He says these words while sharing his understanding of the power of death who rules men who wear the crowns.
Explanation: A king wears a crown as a symbol of his power over the country he rules. But the empty space within the crown houses death. In the empty space, death conducts his court and gives his verdict when it is time.
Comment: The life of the dead is placed in the memory of the living.

Samacheer Kalvi 11th English Solutions Poem Chapter 6 The Hollow Crown

Question (iii)
“Comes at the last, and with a little pin
Bores through his castle wall, and farewell king!”
Answer:
Context:
Here the poet talks about the temporary license to ‘Monarchise’
Explanation:
Richard said that the crown is empty in the middle and this shows the power of the ruler is not permanent. Anytime it may be lost.

Question (iv)
“How can you say to me, I am a king? ”
Reference: This lines is from the poem “The Hollow Crown” by William Shakespeare. The poem is an excerpt from the play “Richard II”.
Context: King Richard says these words to his loyal nobels when he talks about the power of death over monarchs.
Explanation: British subjects usually believe that a king is born with a divine right to rule. People respect his crown as a symbol of great power. After he is deposed from power, Henry II realizes the bitter truth that he is no way different from ordinary subjects. He also has wants, need for friends and the compulsion to taste grief. Nobody can escape death.
Comment: Death – the only thing inevitable in life.

Speaking Activity

E. Working with your partner, discuss the following adages and share your views with the class. You may need to give your ideas and justify your point of view. Remember to take turns while making your presentation/short speech.

Question (a)
War begets war.
Answer:
Mahatma Gandhi said, “If you are indictive and take an eye for an eye, the whole world will be blind”. Today most lethal weapons of mass destruction are being piled up in China, USA and North Korea, Russia and Iran. The leaders of these countries claim that balance of power is required in North and South. But weapons of mass destruction will not create conditions of peace. Peace has to be created by dialogues between countries. War always begets war.

Question (b)
Uneasy lies the head that wears a crown.
Answer:
Whoever is heading an organization, a team, of players, a country does have heavy responsibility. The leadership may give the person a social recognition but in day to day life, the responsibilities of a leader are really heavy. A captain of the army during Kargil war, found one of his soldiers wounded. The Kargil war was heading to a victory for India. The captain did not allow his junior officers to go and bring the wounded soldier. He went and received the bullets. Yet he pulled the wounded soldier to safety. He brought the wounded soldier to the bunk. While returning also he was shot many times. He dropped down dead. He had saved the wounded soldier and the subordinate officer at the cost of his life. Sometimes, there is a coldwar, people try to usurp power by secret dealings.

Aurangzeb killed many of his brothers to ascend to the throne. While in power, kings are really worried about the conspiracy being cooked by relatives to overthrow him. King’s wife poisons king to death. Kings heading battles get killed too. So, we should never be jealous of people in power. Each post or power carries its own stress and unresolved conflicts, occasionally resulting in depression too. Being the head of an army, or that of a country is not always a matter of pride or glory. The grandeur conceals pain, anxiety and ever fear of impending death.

F. Poetic Devices

(a) Read the poem once again carefully and identify the figure of speech that has been used in each of the following lines from the poem:

  1. “Let’s talk of graves, of worms, and epitaphs Make dust our paper, and with rainy eyes Write sorrow on the bosom of the earth”.
  2. (“And yet not so – for what can we bequeath Save our deposed bodies to the ground?”
  3. “Keeps Death his court, and there the antic sits,..
  4. “How can you say to me, I am a king?”
  5. “Scoffing his state and grinning at his pomp,…”
  6. “Bores through his castle wall, and farewell

Answer:

  1. Personification (Earth)
  2. Metaphor
  3. Personification
  4. Interrogation
  5. Personification
  6. Personification

(b) Pick out the words in alliteration from the following lines:

Question (i)
“Our lands, our lives, and all, are Bolingbroke’s,…”
Answer:
lands, fives

Question (ii)
“And tell sad stories of the death of kings:”
Answer:
sad, stories

Samacheer Kalvi 11th English Solutions Poem Chapter 6 The Hollow Crown

Question (iii)
“Comes at the last, and with a little pin…”
Answer:
last, little

G. Based on your reading of King Richard’s speech, answer the following questions in about 100 – 150 words each. You may add your own ideas if required to present and justify your point of view.

Question 1.
What are the causes for King Richard’s grief?
Answer:
King Richard II was a popular king. He had many nobles at the service. His rebellious cousin
Bolingbroke attacks him with 10,000 men on his side. He sends message to the Welsh King for . sending his army to defeat Bolingbroke. But to his shock, Welsh army is not sent. He realizes with alarm the terrible fate he would suffer in the hands of his foe and his most impending death in captivity. King Richard is reminded of the power of death that overshadows everything else. Death scoffs at the power of rulers. Losing the battle, non-receipt of Welsh army and the prospect being jailed and killed worries Richard II.

He realizes that in the hollow crown death had reigned him. Infact, death, a jester had misled him to believe that he was monarchising England. He can now own only a patch of barren land. He is not an impregnable castle of brass anymore. He is an ordinary mortal. He too needs friends and needs to taste grief and face death.

“Life and death are illusions. We are in a constant state of transformation.”

Question 2.
How are the eternal truths and wisdom brought to the reader here?
Answer:
Human’s glorious life gets reduced to graves, epitaphs and worms. Men is left with nothing but his mortal remains to gift to the earth. The earth only serves as a paste and cover to the dead bodies. Great kings too have had inglorious death. Duncan was killed in bed. Hamlet was poisoned to death. Macbeth was slain in the war. The death gives freedom to monarchs from monarchising the country.

The king realizes with a shudder that Death has occupied a prominent position right inside the crown. He scoffs at the pomp and show of the temporal kings. Even the most powerful monarch who feels as strong as a brass castle is brought down by just a pin prick of death. Death is a great leveller who makes kings believe that they are also ordinary mortals with wants, need for friends and the need to taste grief.

“Life is a brief intermission between Birth and Death. Enjoy it.”

Question 3.
Death has been cited to in many ways in this monologue. Identify the poetic devices used in those references.
Answer:
bequeath deposed bodies – Metaphor
small model of barren earth-Metaphor
hollow crown – Metaphor
antics – Personification
Dust our paper – Metaphor
scoffing his state grinning at his pomp – Personification

Samacheer Kalvi 11th English Solutions Poem Chapter 6 The Hollow Crown

Question 4.
Who does the future generations remember easily – the victor or the vanquished? Give reasons. Also, cite relevant references from King Richard’s speech.
Answer:
Unusually future generations remember victors. But there are rare instances of just rulers falling due to the conspiracy and greed of an aggressor. On such occasions, future generations remember the vanquished. A Shiva devotee king was very generous. His enemies entered his kingdom under the guise of Shiva devotees in saffron clothes and slew the king and captured his kingdom. Alexander, King Richard was a just ruler. He was loved by his subjects and loyal nobles. He was defeated by his rebellious cousin simply because he wanted to be a king. When Richard was thinking about the welfare of his subjects, Bolingbroke was secretly raising an army to dethrone him.

People who are mad after power resort to unjust means. So, British subjects respected and loved the vanquished but were helpless and defeated Porus who had fought so valiantly and wanted to be treated with respect befitting a king. Alexander himself respected him and returned his kingdom and sealed a life time friendship with him. From King Richard’s speech one understands that he was good at heart but in the strategy of war, he was not good. Like a crooked end of a straight walking stick, a ruler has to have some secret deals with neighbouring countries to be protected during crisis. Bolingbroke turned out to be a more assertive and Shrewd king. But people would remember a just and noble person more even if defeated.

“Nobility of spirit has more to do with Simplicity than Ostentation, Wisdom than Wealth, Commitment rather than Ambition.

The Hollow Crown About the Poet

Samacheer Kalvi 11th English Solutions Poem Chapter 6 The Hollow Crown (from Richard II)

William Shakespeare (1564 – 1616), an English poet and playwright is widely regarded as the greatest writer in English language and the world’s pre-eminent dramatist. He was born and brought up in Stratford-upon-Avon, Warwickshire. He wrote about 39 plays, 154 sonnets, two long narrative poems, and a few other verses. He was often called England’s National Poet and nicknamed the Bard of Avon. The first publishing of Shakespeare’s works is the ‘The First Folio’. Playwright Ben Johnson wrote a preface to this book including the quote ‘(Shakespeare) is not of an age, but for all time.’ His plays have been translated into every major living language and are constantly studied and performed throughout the world.

The Hollow Crown Summary

King Richard II surrenders to his rebellious cousin Bolingbroke. The King talks to the few loyal friends on the nature of temporal power and how death over takes everything and everybody. Under critical circumstances, King Richard II talk about graves, epitaphs and worms. Shakespeare portrays the fleeting nature of human glory. He explains how even monarchs leave nothing behind to call as their own except a small patch of land into which they will get buried. The dejected King talks on various ways Kings get killed. Some are slain in the battle field.

Some are poisoned to death by their own spouses. The Kings who believed their bodies to be forts or impregnable brass are shattered by just a pinprick. The whole castle wall, the human body, is gone. Death like a jester waits for the King. In fact, he only allows the King to act as if he is ruling and in control of everything. In fact, death is in supreme command. He chides his loyal friends who still believe that he is a monarch. He tells them that he is an ordinary mortal just like them with basic wants and the need to taste grief. He is humbled and realizes he can no more be called a King as he is powerless before the impending death.

The Hollow Crown Glossary

Textual:
antic – someone who attention through silly or funny acts (here a court jester)
bequeath – pass on something to the next generation by means Of a will
ceremonious – being very formal
deposed – removed from office or power
epitaph – short pieces of writing inscribed on tombstones in memory of the dead
executors – persons who put someone’s terms of will into effect
grinning – smiling wildly
impregnable – impossible to pass through
monarchize – rule , carry out the duties functions of a ruler
scoffing – expressing mockery
slain – kill

Additional:
critical – serious
humble – modest
jester – clown
monarch – king
portrays – describes
spouse – wife
temporal – temporary

Samacheer Kalvi 11th Chemistry Solutions Chapter 2 Quantum Mechanical Model of Atom

Students can Download Chemistry Chapter 2 Quantum Mechanical Model of Atom 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 2 Quantum Mechanical Model of Atom

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 8 Physical and Chemical Equilibrium

Students can Download Chemistry Chapter 8 Physical and Chemical Equilibrium 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 8 Physical and Chemical Equilibrium

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium Textual Evaluation Solved

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium Multiple Choice Questions

Question 1.
If Kb and Kf for a reversible reactions are 0.8 x 10-5 and 1.6 x 10-4 respectively, the value of the equilibrium constant is
(a) 20
(b) 0.2 x 10-1
(c) 0.05
(d) None of these
Answer:
(a) 20
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-110

Question 2.
At a given temperature and pressure, the equilibrium constant values for the equilibria
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-111
The relation between K1 and K2 is
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-112
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-113
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-114

Question 3.
The equilibrium constant for a reaction at room temperature is K1 and that at 700 K is K2. If K1 > K2 then
(a) The forward reaction is exothermic
(b) The forward reaction is endothermic
(c) The reaction does not attain equilibrium
(d) The reverse reaction is exothermic
Answer:
(a) The forward reaction is exothermic
Solution:
T1 = 25 + 273 = 298 K, T2 = 700 K
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-115
In this case, T2 > T1 and K1 > K2
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-116
ΔH° is – ve i.e., forward reaction is exothermic.

Question 4.
The formation of ammonia from N2(g) and H2(g) is a reversible reaction
N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g) + Heat
What is the effect of increase of temperature on this equilibrium reaction
(a) equilibrium is unaltered
(b) formation of ammonia is favoured
(c) equilibrium is shifted to the left
(d) reaction rate does not change
Answer:
(c) equilibrium is shifted to the left
Solution:
Increase in temperature, favours the endothermic reaction, given that formation of NH3 is exothermic i.e., the reverse reaction is endothermic.
∴ Increase in temperature, shift the equilibrium to left

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 5.
Solubility of carbon dioxide gas in cold water can be increased by ………….
(a) increase in pressure
(b) decrease in pressure
(c) increase in volume
(d) none of these
Answer:
(a) increase in pressure
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-117
increase in pressure, favours the forward reaction.

Question 6.
Which one of the following is incorrect statement?
(a) for a system at equilibrium, Q is always less than the equilibrium constant
(b) equilibrium can be attained from either side of the reaction
(c) presence of catalyst affects both the forward reaction and reverse reaction to the same extent
(d) Equilibrium constant varied with temperature
Answer:
(a) for a system at equilibrium, Q is always less than the equilibrium constant
Solution:
Correct statement is, for a system at equilibrium, Q = Keq

Question 7.
K1 and K2 are the equilibrium constants for the reactions respectively.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-118
What is the equilibrium constant for the reaction NO2(g) \(\rightleftharpoons\) \(\frac { 1 }{ 2 }\)N2(g) + O2(g)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-119
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-120
Solution:
Let equilibrium constant for the required reaction be K. Then,
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-121

Question 8.
In the equilibrium, 2A(g) \(\rightleftharpoons\) 2B(g) + C2(g) the equilibrium concentrations of A, B and C, at 400 K are 1 x 104 M, 2.0 x 10-3 M, 1.5 x 10-4 M respectively. The value of Kc. for the equilibrium at 400 K is
(a) 0.06
(b) 0.09
(c) 0.62
(d) 3 x 10-2
Answer:
(a) 0.06
Solution:
[A] = 1 x 10-4 M
[B] = 2 x 10-3 M
[C] = 1.5 x 10-4 M
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-122

Question 9.
An equilibrium constant of 3.2 x 10-6 for a reaction means, the equilibrium is ………
(a) largely towards forward direction
(b) largely towards reverse direction
(c) never established
(d) none of these
Answer:
(b) largely towards reverse direction
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-123

Question 10.
\(\frac { { K }_{ c } }{ { K }_{ p } }\) for the reaction, N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g) is ………
(a) \(\frac { 1 }{ RT }\)
(b) \(g\sqrt { RT } \)
(c) RT
(d) (RT)2
Answer:
(d) (RT)2
Solution:
N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g)
∆ng = 2 – 4 = – 2
Kp = Kc (RT)-2
\(\frac { { K }_{ c } }{ { K }_{ p } }\) = (RT)2

Question 11.
For the reaction, AB(g) \(\rightleftharpoons\) A(g) + B(g), at equilibrium, AB is 20% dissociated at a total pressure of R The equilibrium constant K is related to the total pressure by the expression ………..
(a) P = 24 Kp
(b) P = 8 Kp
(c) 24 p = Kp
(d) none of these
Answer:
(a) P =24 Kp
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-125

Question 12.
In which of the following equilibrium, K and K are not equal?
(a) 2NO(g) \(\rightleftharpoons\) N2(g) + O2(g)
(b) SO2(g) + NO2(g) \(\rightleftharpoons\) SO3(g) + NO(g)
(c) H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
(d) PCI5(g) \(\rightleftharpoons\) PCl3(g) + Cl2(g)
Answer:
(d) PCI5(g) \(\rightleftharpoons\) PCl3(g) + Cl2(g)
Solution:
For reactions given in options (a), (b) and (c) ∆ng = 0
For option (d) ∆ng = 2 – 1 = 1
∴ Kp = Kc (RT)

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 13.
If x is the fraction of PCI5 dissociated at equilibrium in the reaction,
PCl5 \(\rightleftharpoons\) PCl3 + Cl2
then starting with 0.5 moIe of PCI5 the total number of moles of reactants and products at equilibrium is
(a) 0.5 – x
(b) x + 0.5
(e) 2x + 0.5
(d) x ± 1
Answer:
(b) x + 0.5
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-126
Total no. of moles at equilibrium = 0.5 – x + x + x = 0.5 + x

Question 14.
The values of Kp1 and Kp2 for the reactions X \(\rightleftharpoons\) Y + Z and A \(\rightleftharpoons\) 2B are in the ratio 9 : 1 if degree of dissociation and initial concentration of X and A be equal then total pressure at equilibrium P1, and P2 are in the ratio
(a) 36 : 1
(b) 1: 1
(c) 3 : 1
(d) 1: 9
Answer:
(a) 36: 1
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-127
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-128

Question 15.
In the reaction, Fe(OH)3(s) \(\rightleftharpoons\) Fe3+(aq) + 3OH(aq), if the concentration of OH ions is decreased by ¼ times, then the equilibrium concentration of Fe3+ will …………
(a) not changed
(b) also decreased by ¼ times
(c) increase by 4 times
(d) increase by 64 times
Answer:
(d) increase by 64 times
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-129
[∵ Concentration of solids is constant)
When concentration of OH ions decreased by \(\frac { 1 }{ 4 }\) times, then
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-130
To maintain KC as constant, concentration of Fe3+ will increase by 64 times.

Question 16.
Consider the reaction where Kp = 0.5 at a particular temperature
PCI5(g) \(\rightleftharpoons\) PCI3(g) + Cl2(g) if the three gases are mixed in a container so that the partial pressure of each gas is initially 1 atm, then which one of the following is true?
(a) more PCI3 will be produced
(b) more Cl2 will be produced
(c) more PCI5 will be produced
(d) none of these
Answer:
(c) more PCI5 will be produced .
Solution:
Kp = 0.5
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-131
Q = \(\frac { 1 x 1 }{ 1 }\)
Q > Kp
∴Reverse reaction is favoured; i.e., more PCI5 will be produced.

Question 17.
Equimolar concentrations of H2 and I2 are heated to equilibrium in a 1 litre flask. What percentage of initial concentration of H2 has reacted at equilibrium if rate constant for both forward and reverse reactions are equal
(a) 33%
(b) 66%
(c)(33)2 0/0
(d) 16.5%
Answer:
(a) 33%
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-132
As degree of dissociation cannot be negative. Therefore, degree of dissociation
= \(\frac { a }{ 3 }\) x 100 = 33.33%

Question 18.
In a chemical equilibrium, the rate constant for the forward reaction is 2.5 x 1 and the equilibrium constant is 50. The rate constant for the reverse reaction is ……….
(a) 11.5
(b) 5
(c) 2 x 102
(d) 2 x 103
Answer:
(b) 5
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-133

Question 19.
Which of the following is not a general characteristic of equilibrium involving physical process
(a) Equilibrium is possible only in a closed system at a given temperature
(b) The opposing processes occur at the same rate and there is a dynamic but stable condition
(c) All the physical processes stop at equilibrium
(d) All measurable properties of the system remains constant
Answer:
(c) All the physical processes stop at equilibrium
Solution:
Correct statement – Physical processes occurs at the same rate at equilibrium.

Question 20.
For the fórmation of two moles of SO3(g) from SO2 and O2, the equilibrium constant is K1. The equilibrium constant for the dissociation of one mole of SO3 into SO2 and O2 is …………
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-134
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-135
Solution:
2SO2 +O2 \(\rightleftharpoons\) 2SO3
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-136
Dissociation of 1 mole of 2SO3
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-137

Question 21.
Match the equilibria with the corresponding conditions …………
(i) Liquid \(\rightleftharpoons\) Vapour – 1. Melting point
(ii) Solid \(\rightleftharpoons\) Liquid – 2. Saturated solution
(iii) Solid \(\rightleftharpoons\) Vapour – 3. Boiling point
(iv) Solute (s) \(\rightleftharpoons\) Solute (Solution) – 4. Sublimation point
5. Unsaturated solution
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-138
Answer:
(b) 3, 1, 4, 2

Question 22.
Consider the following reversible reaction at equilibrium, A + B \(\rightleftharpoons\) C, if the concentration of the reactants A and B are doubled, then the equilibrium constant will ………
(a) be doubled
(b) become one fourth
(c) be halved
(d) remain the same
Answer:
(d) remain the same
Solution:
A + B \(\rightleftharpoons\) C
KC = \(\frac { [C] }{ [A] [B] }\)
If [A] and [B] are doubled, [C] increases 4 times to maintain KC as constant.
∴ Equilibrium constant will remain the same.

Question 23.
[CO(H2O)6]2+ (aq) (pink) + 4C (aq) \(\rightleftharpoons\) [COCI4]2- (aq) (blue) + 6H2O (1) In the above reaction at equilibrium, the reaction mixture is blue in colour at room temperature. On cooling this mixture, it becomes pink in colour. On the basis of this information, which one
of the following is true?
(a) ∆H > 0 for the forward reaction
(b) ∆H = 0 for the reverse reaction
(c) ∆H < 0 for the forward reaction
(d) Sign of the ∆H cannot be predicted based on this information.
Answer: (a) ∆H > 0 for the forward reaction
Solution:
On cooling, reverse reaction predominates and the solution is pink in colour. Decrease in temperature, favours the reverse reaction i.e. reverse reaction is exothermic (∆H < 0)and for the forward reaction is endothermic (∆H > 0).

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 24.
The equilibrium constants of the following reactions are:
N2 + 3H2 \(\rightleftharpoons\) 2NH3 ; K1
N2 + O2 \(\rightleftharpoons\) 2NO ; K2
H2+ \(\frac { 1 }{ 2 }\)O2 \(\rightleftharpoons\) H2O ; K3
The equilibrium constant (K) for the reaction; Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-140 will be
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-141
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-167
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-143

Question 25.
A 20 litre container at 400 K contains CO2(g) at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of CO2 attains its maximum value will be.
Given that: SrCO3(s) \(\rightleftharpoons\) SrO(s) + CO2(g)
(a) 2 litre
(b) 5 litre
(c) 10 Litre
(d) 4 litre
Answer:
(b) 5 litre
Solution:
Given that Kp = 1.6 atm
V1 = 20 L
V2 = ?
T1 = 400 K
T2 = 400 K
Kp = Pco2
Pco2 = 1.6 atm
P1 = 0.4 atm
P2 = 1.6 atm
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-144

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium Short Answer Questions

Question 26.
If there is no change in concentration, why is the equilibrium state considered dynamic?
Answer:
At chemical equilibrium the rate of two opposing reactions are equal and the concentration of. reactants and products do not change with time. This condition is not static and is dynamic, because both the forward and reverse reactions are still occurring with the same rate and no macroscopic change is observed. So chemical equilibrium is in a state of dynamic equilibrium.

Question 27.
For a given reaction at a particular temperature, the equilibrium constant has constant value. Is the value of Q also constant? Explain.
Answer:
In the chemical reaction, as the reaction proceeds, there is a continuous change in the concentration of reactants and products and also the Q value until the reaction reaches the equilibrium. So even at particular temperature, Q is not constant. Even once the equilibrium is achieved then change in concentration of reactants or products, pressure, volume will change the value of Q.

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 28.
What is the relation between Kp and KC. Give one example for which Kp is equal to KC.
Answer:
Kp = Kc(RT)∆ng
H2(g) + I2(g) ≅ 2HI (g) ; ∆ng = 0 ;
Kp = Kc.

Question 29.
For a gaseous homogeneous reaction at equilibrium, the number of moles of products is greater than the number of moles of reactants. Is KC is larger or smaller than Kp?
Answer:
For a homogeneous reaction at equilibrium, number of moles of products (np) are greater than the number of moles of reactants (nR) then ∆ng = + ve
np > nR
∆ng = + ve
If ∆ng is ve, Kp value is greater than KC
Kp = KC. (RT)+ve
Kp > KC
Example: PCl5 \(\rightleftharpoons\) PCl3(g) + Cl2(g)
2 – 1 = 1
Kp = KC (RT)1
Kp > KC

Question 30.
When the numerical value of the reaction quotient (Q) is greater than the equilibrium constant (K), in which direction does the reaction proceed to reach equilibrium?
Answer:
Q > Kc
Reverse reaction is favoured.

Question 31.
For the reaction: A2(g) + B2(g) \(\rightleftharpoons\) 2AB(g); ∆H is – ve.
Answer:
The following molecular scenes represent different reaction mixture
(A – dark grey, B – light grey)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-145

  1. Calculate the equilibrium constant Kp and KC
  2. For the reaction mixture represented by scene (x), (y) the reaction proceed in which directions?
  3. What is the effect of increase in pressure for the mixture at equilibrium?

Answer:
1. Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-146
At equilibrium
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-147

2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-148

3.  Since ∆ng = 2 – 2 = 0, thus, pressure has no effect. So by increasing the pressure, equilibrium will not be affected.

Question 32.
State Le – Chatelier principle.
Answer:
“If a system at equilibrium is distributed, then the system shifts itself in a direction that nullifies the effect of that disturbance”.

Question 33.
Consider the following reactions

  1. H2(g) + l2(g) \(\rightleftharpoons\) 2HI(g)
  2. CaCO3(s) \(\rightleftharpoons\) CaO(s) + CO2(g)
  3. S(s) + 3F2(g) \(\rightleftharpoons\) SF6(g)

In each of the above reaction find out whether you have to increase (or) decrease the volume to increase the yield of the product.
Answer:
1.  H2(g) + l2(g) \(\rightleftharpoons\) 2HI(g)
In the above equilibrium reaction, the volume of gaseous molecules is equal on both sides. So increase or decrease the volume will not affect the equilibrium and there will be no change in the yield of the product.

2. Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-149
Volume is greater in the product side. By decreasing the pressure, the volume will increase thus, to get more of product CO2, the pressure should be decreased or volume should be increased.

3. Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-150
Volume is lesser on the product side. So by increasing the pressure, equilibrium shifts to the product side.

Question 34.
State law of mass action.
Answer:
The law states that “At any instant, the rate of a chemical reaction at a given temperature is directly proportional to the product of the active masses of the reactants at that instant”.
Rate ∝ [Reactant]x
Active mass = \(\left(\frac{\mathrm{n}}{\mathrm{V}}\right)\) mol dm-3 (or) mol L-1

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 35.
Explain how will you predict the direction of an equilibrium reaction.
Answer:

  1. A large value of KC indicates that the reaction reaches equilibrium with high product yield.
  2. A low value of KC indicates that the reaction reaches equilibrium with low product formed.
  3. In general, if the K is greater than I the reaction proceeds nearly to completion. If is less than 10-3, the reaction rarely proceeds.
  4. If K < 10-3, reverse reaction is favoured. If Kc > 103, forward reaction is favoured.

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium Long Answer Questions

Question 36.
Derive a general expression for the equilibrium constant Kp and KC for the reaction.
3H2(g) + N2(g) \(\rightleftharpoons\) 2NH3(g)
Answer:
In the formation of ammonia, ‘a’ moles of Nitrogen and ‘b’ moles of hydrogen gas are allowed to react in a container of volume of ‘V’. Let ‘x’ moles of nitrogen react with 3x moles of hydrogen to give 2x moles of ammonia.
N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-151
Applying law of mass action
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-152
Total number of moles at equilibrium n = a – x + b -3x + 2x = a + b – 2x
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-153
Total number of moles at equilibrium
n = a – x + b – 3x + 2x = a + b – 2x
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-154

Question 37.
Write a balanced chemical equation for an equilibrium reaction for which the equilibrium constant is given by expression?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-155

Question 38.
What is the effect of added inert gas on the reaction at equilibrium at constant volume?
Answer:
When an inert gas (i.e., a gas which does not react with any other species involved in equilibrium) is added to an equilibrium system at constant volume, the total number of moles of gases present in the container increases, that is, the total pressure of gases increases. The partial pressure of the reactants and the products is unchanged. Hence at constant volume, the addition of inert gas has no effect on the equilibrium.

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 39.
Derive the relation between Kp and KC.
Answer:
Consider a general reaction in which all reactants and products are ideal gases.
x A + y B \(\rightleftharpoons\)  lC+mD
The equilibrium constant KC is
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-156
The ideal gas equation is
PV = nRT or P = \(\frac { n }{ V }\) RT
Since,
Active mass = molar concentration = \(\frac { n }{ V }\)
P = Active mass x RT
Based on the above expression, the partial pressure of the reacants and products can be expressed as
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-157
On substitution in equation (2),
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-158
By comparing equation (1) and (4), we get
Kp = KC (RT) ∆ng
where ∆ng is the difference between the sum of a number of moles of products and the sum of number of moles of reactants in the gas phase.
(i) If ∆ng = 0, Kb = KC(RT)0
Kp = KC
Example: H2(g) + I2 \(\rightleftharpoons\) 2HI(g)

(ii) where
∆ng = +Ve
Kp = KC (RT)+ve
Kp = KC
Example: 2NH3(g) N2(g) + 3H2(g)

(iii) When
∆ng = -Ve
Kp = KC (RT)-ve
Kp < KC
Example: 2SO2(g) +O2(g) \(\rightleftharpoons\) 2SO3(g)

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 40.
One mole of PCl5 is heated in one litre closed container. If 0.6 mole of chlorine is found at equilibrium, calculate the value of the equilibrium constant.
Answer:
Given that
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-159

Question 41.
For the reaction: SrCO2(s) \(\rightleftharpoons\) SrO (s) + CO2(g), the value of equilibrium constant Kp = 2.2 x 10-4 at 1002K. Calculate KC for the reaction.
Answer:
For the reaction, SrCO2(s) \(\rightleftharpoons\) SrO (s) + CO2(g)
∆ng = 1 – 0 = 1
Kp = Kp (RT)
2.2 x 10-4 = KC (0.0821)(1002)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-160

Question 42.
To study the decomposition of hydrogen iodide, a student fills an evacuated 3 litre flask with 0.3 mol of HI gas and allows the reaction to proceed at 500°C. At equilibrium, he found the concentration of HI which is equal to 0.05 M. Calculate Kp and KC for this reaction.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-161
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-162
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-168

Question 43.
Oxidation of nitrogen monoxide was studied at 200°C with initial pressures of 1 atm NO and 1 atm of O2. At equilibrium partial pressure of oxygen is found to be 0.52 atm. Calculate Kp value.
Answer:
2NO(g) + O2(g) \(\rightleftharpoons\) 2NO2(g)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-164
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-165

Question 44.
1 mol of CH4, 1 mole of CS2 and 2 mol of H2S are 2 moI of H2 are mixed in a 500 mL flask. The equilibrium constant for the reaction KC = 4 x 10-2 mol2 lit-2. In which direction will the reaction proceed to reach equilibrium?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-166
∴ The reaction will proceed in the reverse direction to reach equilibrium.

Question 45.
At particular temperature KC = 4 x 10-2 for the reaction
H2S(g) \(\rightleftharpoons\) 2H2(g) + \(\frac { 1 }{ 2 }\) S2(g). Calculate KC for each of the following reaction
(i) H2S(g) \(\rightleftharpoons\) 2H2(g) + S2(g)
(ii) 3H2S(g) \(\rightleftharpoons\) 3H2(g) + \(\frac { 3 }{ 2 }\) S2(g)
Answer:
KC = 4 x 10-2 for the reaction
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-167
For the reaction 3H2S(g) \(\rightleftharpoons\) 3H2(g) + \(\frac { 3 }{ 2 }\) S2(g)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-168

Question 46.
28 g of nitrogen and 6 g of hydrogen were mixed in a 1 litre closed container. At equilibrium 17g NH3 was produced. Calculate the weight of nitrogen, hydrogen at equilibrium.
Answer:
Given
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-169
[NH3] = \((\frac { 17 }{ 17 })\) = 1 mol = 2x
x = 0.5 mol
At equilibrium, [N2] = 1 – x = 0.5 mol
[H2] = 3 – 3x = 3 – 3 (0.5) = 1.5 mol
Weight of N2 (no. of moles of N2) x molar mass of N2 = 0.5 x 28 = 14g
Weight of H2 = (no. of moles of H2) x molar mass of H2= 1.5 x 2 = 3g

Question 47.
The equilibrium for the dissociation of XY2 is given as,
2XY2 (g) \(\rightleftharpoons\) 2XY (g) + Y2(g).
If the degree of dissociation x is so small compared to one. Show that 2 Kp = Px3 where P is the total pressure and K is the dissociation equilibrium constant of XY2.
Answer:
2XY2 (g) \(\rightleftharpoons\) 2XY (g) + Y2(g)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-170
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-171
Question 48.
A sealed container was filled with I mol of A, (g), I mol B, (g) at 800 K and total pressure 1.00 bar. Calculate the amounts of the components in the mixture at equilibrium given that K = 1 for the reaction:
A2(g) + B2(g) \(\rightleftharpoons\) 2AB(g)
Answer:
A2(g) + B2(g) \(\rightleftharpoons\) 2AB(g)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-172
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-173
[B2]eq = 1 – x = 1 – 0.33 = 0.667
[AB]2 = 2x = 2 x 0.33 = 0.66
Question 49.
Deduce the Vant Hoff equation.
Answer:
This equation gives the quantitative temperature dependence of equilibrium constant K. The relation between standard free energy change ∆G° and equilibrium constant is
∆G° = – RT ln K  …………(1)
We know that,
∆G° = ∆H° – T∆S°        ………….(2)
Substituting (1) in equation (2)
– RT lnK = ∆H° – T∆S°

Rearranging, ln K = \(\frac { – ∆H° }{ RT }\) + \(\frac { ∆S° }{ R }\)   ………….(4)
Differentiating equation (3) with respect to temperature
d(lnK) ∆H° \(\frac { d(lnK) }{ dT }\) = \(\frac { { \triangle H }^{ 0 } }{ { RT }^{ 2 } }\)
Equation (4) is known as differential form of van’t Hoff equation. On integrating the equation 4, between T1 and T2 with their respective equilibrium constants and K2
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-174
The equation is known as an integrated form of van’t Hoff equation.

Question 50.
The equilibrium constant K for the reaction
N2(g) + 3H2(g) = 2NH3(g) is 8.19 x 102
at 298 K and 4.6 x 10-1 at 498 K. Calculate ∆H° for the reaction.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-175
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-176

Question 51.
The partial pressure of carbon dioxide in the reaction CaCO3(s) \(\rightleftharpoons\) CaO (s) + CO2(g) is 1.017 x 10-3 atm at 500° C. Calculate Kp, at 600° C for the reaction. ∆H for the reaction is 181 kJ mol-1 and does not change in the given range of temperature.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-177

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium In – Text problems Solved

Question 1.
Consider the following reaction
Fe3+(aq) + SCN(aq) [Fe(SCN)]2+(aq)
A solution is made with initial Fe3+, SCN concentration of 1 x 10-3 M and 8 x 10-4M respectively. At equilibrium [Fe(SCN)]2+ concentration is 2 x 10-4M. Calculate the value of equilibrium constant.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-1

Question 2.
The atmospheric oxidation of NO
2NO(g) + O2(g) \(\rightleftharpoons\) 2NO2(g)
was studied with initial pressure of 1 atm of NO and I atm of O2. At equilibrium, partial pressure of oxygen is 0.52 atm. Calculate Kp of the reaction.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-2
As,
1 – x = 0.52
x = 0. 48
= At equilibrium,
PNO = 1 – 2x = 1 – 2(0.48) = 0.04
PNO2 = 2x = 2(0.48) = 0.96
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-3

Question 3.
The following water gas shift reaction is an important industrial process for the production of hydrogen gas.
CO(g) + H2O(g) \(\rightleftharpoons\) CO2(g) + H2(g) At a given temperature Kp = 2.7.  If 0.13 moI of CO, 0.56 moI of water, 0.78 mol of CO2 and 0.28 mol of H2 are introduced into a 2L flask, find out in which direction must the reaction proceed to reach equilibrium.
Answer:
CO(g) + H2O(g) \(\rightleftharpoons\) CO2(g) + H2(g)
Given
Kp = 2.7
[CO] = 0.13 mol, [H2O] = 0.56 mol
[CO2] = 0.78 mol; [H2] = 0.28 mol
V = 2L
Kp = KC (RT)Δng
2.7 = KC(RT)°
KC = 2.7
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-1.2
Q = 3
Q > KC Hence, the reaction proceed in the reverse direction.

Question 4.
1 moI of PCl5, kept in a closed container of volume 1 dm3 and was allowed to attain equilibrium at 423 K. Calculate the equilibrium composition of reaction mixture. (The KC value for PCl5 dissociation at 423 K is 2)
Answer:
PCl5 \(\rightleftharpoons\) PCL3 + Cl2
Given that [PCl5]initial = 1 mol
V = 1 dm3
KC = 2
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-4
∴ Equilibrium concentration of
[PCl5]eq = \(\frac { 1-x }{ 1 }\) = 1 – 0.732 = 0.268 M
[PCl3]eq = \(\frac { x }{ 1 }\) = \(\frac { 0.732 }{ 1 }\) = 0.732
[Cl2]eq = \(\frac { x }{ 1 }\) = \(\frac { 0.732 }{ 1 }\) = 0.732

Question 5.
The equilibrium constant for the following reaction is 0.15 at 298 K and 1 atm pressure.
Answer:
N2O4(g) \(\rightleftharpoons\) 2NO2(g)
T1 = 298 K
Kp1 = 0.15
T2 = 1000C = 100 + 273 = 373 K
Kp2 = ?
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-5

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium In-Text problems Solved

Question 1.
One mole of H2 and one mole of I2 are allowed to attain equilibrium in 1 lit container. If the equilibrium mixture contains 0.4 mole of HI. Calculate the equilibrium constant.
Answer:
Given data: [H2] = I mole, [I2] = 1 mole
At equilibrium, [HI] = 0.4 mole, KC = ?
H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-6

Question 2.
The equilibrium concentrations of NH3, N2 and H2 are 1.8 x 102 M, 1.2 x 10-2 M and 3 x 10-2 M respectively. Calculate the equilibrium constant for the formation of NH3 from N2 and H2. [Hint: M = mol lit-1]
Answer:
Given data:
[NH3] 1.8 x 10-2 M, [N2] = 1.2 x 10-2M, [H2] 3 x 10-2 M, KC = ?
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-7

Question 3.
The equilibrium constant at 298 K for a reaction is 100.
A+ B \(\rightleftharpoons\) C + D
If the initial concentration of all the four species is 1 M, the equilibrium concentration of D (in mol lit-1) will be
Answer:
Given data: [A] = [B] = [C] = [D] = 1 M, KC = 100, [D]eq = ?
Let x be the number of moles reactants reacted
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-8

Question 4.
For an equilibrium reaction Kp = 0.0260 at 25° C and ?H = 32.4 kJ mor-1. Calculate Kp at 37° C.
Answer:
T1 = 25 + 273 = 298 K
T2 = 37 + 273 = 310 K
ΔH = 32.4 kJ mor-1 = 32400 J mol-1
R = 8.314 JK-1 mol-1
Kp1 = 0.0260
Kp2 = ?
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-9
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-10

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium Additional Questions 

I. Choose the correct answer.

Question 1.
For which of the following Kp is less than Kc?
a) N2O4 ⇌ 2NO2
b) N2 + 3H2 ⇌ 2NH3
c) H2 + I2 ⇌ 2HI
d) CO + H2O ⇌ CO2 + H2
Answer:
b) N2 + 3H2 ⇌ 2NH3

Question 2.
Which one of the following is an example of chemical equilibrium?
(a) 2NO(g) + O2(g) \(\rightleftharpoons\) 2NO2(g)
(b) I2(s) \(\rightleftharpoons\) I2(g)
(c) H2O(s) \(\rightleftharpoons\) H2O(1)
(d) NH2CI(s) \(\rightleftharpoons\) NH4CI(g)
Answer:
(a) 2NO(g) + O2(g) \(\rightleftharpoons\) 2NO2(g)
Solution:
All the other three are physical equilibrium. Only (a) is chemical equilibrium.

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 3.
For homogeneous gas reaction 4NH3 + 5O2 ⇌ 4NO + 6H2O. The equilibrium constant Kc has the unit
a) (concentration)1
b) (concentration)1
c) (concentration)9
d) (concentration)10
Answer:
a) (concentration)1

Question 4.
At chemical equilibrium,
(a) rate of forward reaction = rate of backward reaction
(b) rate of forward reaction > rate of backward reaction
(c) rate of forward reaction < rate of backward reaction
(d) rate of forward reaction = rate of backward reaction
Answer:
(a) rate of forward reaction = rate of backward reaction

Question 5.
Kp for the following reaction at 700 K is 1.3 × 10-3 atm-1. The Kc same temperature for the reaction. 2SO2 + O2 ⇌ 2SO3 will be
a) 1.1 × 10-2
b) 3.1 × 10-2
c) 5.2 × 10-2
d) 7.4 × 10-2
Answer:
d) 7.4 × 10-2

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 6.
Which of the following is an example of heterogeneous equilibrium?
(a) Synthesis of HI
(b) Dissociation of PCI5
(c) Acid hydrolysis of ester
(d) Decomposition of limestone
Answer:
(d) Decomposition of limestone
Solution:
CaCO3(s) → CaO(s) + CO2(g)
Here CO2 is in a gaseous state while CaCO3 and CaO are in a solid-state.

Question 7.|
For the system 3A + 2B ⇌ C, the expression for equilibrium constant is
a) \(\frac{[3 \mathrm{~A}][2 \mathrm{~B}]}{[\mathrm{C}]}\)

b) \(\frac{[\mathrm{C}]}{[3 \mathrm{~A}][2 \mathrm{~B}]}\)

c) \(\frac{[\mathrm{A}]^{3}[\mathrm{~B}]^{2}}{[\mathrm{C}]}\)

d) \(\frac{[\mathrm{C}]}{[\mathrm{A}]^{3}[\mathrm{~B}]^{2}}\)
Answer:
d) \(\frac{[\mathrm{C}]}{[\mathrm{A}]^{3}[\mathrm{~B}]^{2}}\)

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 8.
In the reaction, 2NH3(g) \(\rightleftharpoons\) N2(g) + 3H2(g)
(a) Kp = KC
(b) Kp < KC
(c) Kp > KC
(d) Kp = \(\frac { 1 }{ { K }_{ C } }\)
Answer:
(c) Kp > KC
Solution:
Kp = KC (RT)Δng and Δng = 4 – 2 = 2
∴ Kp = KC (RT)2 = Kp > KC

Question 9.
The equilibrium constant for the reaction N2(g) + O2(g) ⇌ 2NO(g) at temperature T is 4 × 10-4. The value of Kc for the reaction NO(g) ⇌ 1/2 N2(g) + 1/2 O2(g) at the same temperature is ___________.
a) 4 × 10-4
b) 50
c) 2.5 × 102
d) 0.02
Answer:
b) 50

Question 10.
In the equilibrium reaction CaCO3(s) \(\rightleftharpoons\) CaO(s) + CO2(g) whose concentration remains constant at a given temperature?
(a) CaO
(b) CO2
(e) CaCO3
(d) Both (a) and (c)
Answer:
(d) Both (a) and (c)
Solution:
The concentration of solids remains constant at a particular temperature.

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 11.
Consider the following equilibrium reaction and relate their equilibrium constants
(i) N2 + O2 \(\rightleftharpoons\) 2NO; K1
(ii) 2NO + O2 \(\rightleftharpoons\) 2NO2; K2
(iii) N2 + 2O2 \(\rightleftharpoons\) 2NO2; K3
(a) K3 = K2 = K1
(b) K1 x K3 = K2
(c) K1 x K2 = K3
(d) \(\frac { { K }_{ 1 } }{ { K }_{ 2 } }\) = K3
Answer:
(c) K1 x K2 = K3
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-11

Question 12.
The Kc for H2(g) + I2(g) ⇌ 2 HI(g), is 64. If the volume of the container is reduced to one – half of its original volume, the value of the equilibrium constant will be
a) +28
b) 64
c) 32
d) 16
Answer:
b) 64

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 13.
Find the Q value of the reaction H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g) at an instant where concentration of H2, I2 and HI are found to be 0.2 moI L-1, 0.2 mol L-1, and 0.6 moI L-1 respectively.
(a) 48
(b) 9
(c) 0.9
(d) 90
Answer:
(b) 9
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-12

Question 14.
For the reaction N2O4(g) \(\rightleftharpoons\) 2NO2(g) KC = 0.21 at 373 K. The concentrations of N2O4 and NO2 are found to be 0.125 mol dm-3 and 0.5 mol dm-3 respectively at a given temperature. Predict the direction of the reaction.
(a) At equilibrium
(b) reverse direction
(c) forward direction
(d) Both reverse and forward direction
Answer:
(b) reverse direction
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-13
KC = 0.21
Q = 2
Q > KC
Hence the reaction will proceed in the reverse direction

Question 15.
The equilibrium constant of the reaction H2(g) + I2(g) ⇌ 2 HI(g) is 64. If the volume of the container is reduced to one – fourth of its original volume, the value of the equilibrium constant will be
a) 16
b) 32
c) 64
d) 128
Answer:
c) 64

Question 16.
Statement I. In Haber’s process, NH3 is liquefied and removed.
Statement II. In the manufacture of NH3, liquefied and removal of NH3, keeps the reaction moving in forward direction.
(a) Statement I and II are correct and II is the correct explanation of I.
(b) Statement I and II are correct but II is not the correct explanation of I.
(c) Statement I is wrong but statement II is correct.
(d) Statement I is correct but statement II is wrong.
Answer:
(a) Statement I and II are correct and II is the correct explanation of I.
Solution:
Removal of NH3 will decrease its concentration which favours the production of NH3 according to Le – Chatelier’s principle.

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 17.
For the reaction 2 NOCl(g) ⇌ 2 NO(g) + Cl2(g), Kc at 427 is 3 × 10-6 L mol-1. The value of Kp is nearly
a) 7.50 × 10-5
b) 2.50 × 10-5
c) 2.50 × 10-4
d) 1.72 × 10-4
Answer:
d) 1.72 × 10-4

Question 18.
Among the following reactions which one has Kp = KC
(a) N2O4 \(\rightleftharpoons\) 2NO(g)
(b) 2SO2(g) + O2(g) \(\rightleftharpoons\) 2SO3(g)
(c) N2(g) + O2(g) \(\rightleftharpoons\) 2NO(g)
(d) N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g)
Answer:
(c) N2(g) + O2(g) \(\rightleftharpoons\) 2NO(g)
Solution:
Kp = KC. RTΔng
In equation N2(g) + O2(g) \(\rightleftharpoons\) 2NO(g)
Δng = 0
Kp = KC. RT°
Kp = KC

Question 19.
Statement I. Addition of inert gas at constant volume has no effect on the equilibrium.
Statement II. When an inert gas is added, the total number of moles of gases present in the container increases and total pressure also increases, the partial pressure of the products and reactants are unchanged.
(a) Statement I and II are correct but statement II is not the correct explanation of I.
(b) Statement I and II are correct and statement II is the correct explanation of 1.
(c) Statement I is correct but statement II is not correct.
(d) Statement I is wrong but statement II is correct.
Answer:
(b) Statement I and II are correct and statement II is the correct explanation of I.

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 20.
2NO2 ⇌ 2NO + O2, K = 1.6 × 10-12 NO + \(\frac{1}{2}\)O2 ⇌ NO2; K’ = ?
a) K’ = \(\frac{1}{\mathrm{~K}^{2}}\)
b) K’ = \(\frac{1}{\mathrm{~K}}\)
c) K’ = \(\frac{1}{\sqrt{\mathrm{K}}}\)
d) None of these
Answer:
c) K’ = \(\frac{1}{\sqrt{\mathrm{K}}}\)

Question 21.
The equilibrium expression, KC = [CO2] represents the reaction.
(a) C(s) + O2(g) \(\rightleftharpoons\) CO2(g)
(b) CaCO3(s) \(\rightleftharpoons\) CaO(s) + CO2(g)
(c) 2CO(g) + O2(g) \(\rightleftharpoons\) 2CO2(g)
(d) CaO(s) + CO2(g) \(\rightleftharpoons\) CaCO3(s)
Answer:
(b) CaCO3(s) \(\rightleftharpoons\) CaO(s) + CO2(g)

Question 22.
Hydrogen molecule (H2) can be dissociated into hydrogen atoms (H). Which one of the following changes will not increase the number of atoms present at equilibrium?
(a) adding H atoms
(b) increasing the temperature
(c) increasing the total pressure
(d) increasing the volume of the container
Answer:
(d) increasing the total pressure container
Solution:
It favours backward reaction i.e., formation of H2 molecule.

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 23.
What is the expression for Keq for the reaction, 2N2O(g) + O2(g) \(\rightleftharpoons\) 4NO(g)?
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-14
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-15

Question 24.
The equilibrium constant in a reversible reaction at a given temperature
a) depends on the initial concentration of the reactants
b) depends on the concentration of the products at equilibrium
c) does not depend on the initial concentrations
d) It is not characteristic of the reaction
Answer:
c) does not depend on the initial concentrations

Question 25.
H2 + S \(\rightleftharpoons\) H2S + energy. In this reversible reaction, select the factor which will shift the equilibrium to the right.
(a) adding heat
(b) adding H2S
(c) blocking hydrogen gas reaction
(d) removing hydrogen sulphide gas
Answer:
(a) removing hydrogen sulphide gas

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 26.
A chemical reaction is at equilibrium when
a) Reactants are completely transformed into products
b) The rates of forward and backward reactions are equal
c) Formation of products is minimized
d) Equal amounts of reactants and products are present
Answer:
b) The rates of forward and backward reactions are equal

Question 27.
A chemist dissolves an excess of BaSO4 in pure water at 25°C if its Ksp = 1 x 10-10. What is the concentration of barium in the water?
(a) 10-4 M
(b) 10-5 M
(c) 10-15 M
(a) 10-6 M
Answer:
(c) 10-15 M
Solution:
Ksp = [Ba2+] [SO42-]
1 x 10-10 = (x) (x)
10-5 = x

Question 28.
If in a mixture where Q = K, then what happens?
(a) the reaction shift towards products
(b) the reaction shift towards reactants
(c) nothing appears to happen, but forward and reverse reactions are continuing at the same rate
(d) nothing happens
Answer:
(c) nothing appears to happen, but forward and reverse reactions are continuing at the same rate

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 29.
If dissociation for reaction PCI5 \(\rightleftharpoons\) PCI3 + Cl2 is 20% at 1 atm pressure. Calculate the value of KC.
(a) 0.04
(b) 0.05
(c) 0.07
(d) 0.06
Answer:
(d) 0.05
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-18

Question 30.
What would be the value of Δng for the reaction NH4CI(s) \(\rightleftharpoons\) NH2(g) + HCI(g)?
(a) 1
(b) 0.5
(c) 2
(d) 1.5
Answer:
(c) 2
Solution:
Δng = np – nr = 2 – 0 = 2

Question 31.
The rate constants for forward and backward reactions of the hydrolysis of ester are 1.1 × 10-2 and 1.5 × 10-3 per minute respectively. The equilibrium constant for the reaction is ___________.
a) 4.33
b) 5.33
c) 6.33
d) 7.33
Answer:
d) 7.33

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 32.
At 500K, equilibrium constant KC for the following reaction is 5, \(\frac { 1 }{ 2 }\)H2(g) + \(\frac { 1 }{ 2 }\)I2(g) \(\rightleftharpoons\) HI(g) what would be the equilibrium constant KC. for the reaction 2HI(g) \(\rightleftharpoons\) H2(g) + I2(g)
(a) 0.44
(b) 0.04
(c) 25
(d) 2.5
Answer:
(b) 0.04
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-19

Question 33.
For the reaction 2NO2(g) \(\rightleftharpoons\) 2NO(g) + O2 (g), KC = 1.8 x 10-6 at 1850C. At the same temperature the value of KC for the reaction. NO(g) + \(\frac { 1 }{ 2 }\)O2 \(\rightleftharpoons\) NO2(g) is ……….
(a) 0.9 x 106
(b) 7.5 x 102
(c) 1.95 x 10-3
(d) 1.95 x 103
Answer:
(b) 7.5 x 102
Solution:
The reaction is reversed and halved.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-20

Question 34.
Which of the following reaction will be favoured at low pressure?
(a) N2 + O2 \(\rightleftharpoons\) 2NO
(b) H2 ± I2 \(\rightleftharpoons\) 2HI
(c) PCI5 \(\rightleftharpoons\) PCI3 + Cl2
(d) N2 + 3H2 \(\rightleftharpoons\) 2NH3
Answer:
(c) PCI5 \(\rightleftharpoons\) PCI3 + Cl2

Question 35.
For the reaction
CO(g) + \(\frac{1}{2}\)O2(g) ⇌ CO2(g), Kp/Kc is _______.
a) RT
b) (RT)-1
c) (RT)-1/2
d) (RT)1/2
Answer:
c) (RT)-1/2

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 36.
For the reaction PCI5(g) \(\rightleftharpoons\) PCI3(g) + Cl2(g) the forward reaction at constant temperature is favoured by …………………
(a) introducing an inert gas at constant volume
(b) introducing PCl3(g) at constant volume
(c) introducing PCl5(g) at constant volume
(d) introducing CI2(g) at constant volume
Answer:
(c) introducing PCI5(g) at constant volume

Question 37.
The equilibrium of the reaction N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g) will shift to product side when….
(a) Kp > 1
(b) Q < Kp
(c) Q = Kp
(d) Q = 2Kp
Answer:
(b) Q < Kp

Question 38.
NO2 is involved in the formation of smog and acid rain. A reaction that is important in the formation of NO2 is O3(g) + NO(g) \(\rightleftharpoons\) O2(g) + NO2(g) KC = 6.0 x 1034. If the air over a section of New Delhi contained 1.0 x 10-6 M of O3, 1.0 x 10-5 M of NO, 2.5 x 10-4 M of NO2 and 8.2 x 10-3 of O2, what can we conclude?
(a) there will be a tendency to form more NO and O2
(b) there will be a tendency to form more NO2 and O2
(c) there will be a tendency to form more NO2 and O3
(d) there will no tendency for change because the reaction is at equilibrium
Answer:
(b) there will be a tendency to form more NO2 and O2
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-21
As Q < KC the reaction will hive a tendency to move forward.

Question 39.
In a reaction A+B ⇌ C+D,the concentrations of A, B, C and D [in mole/lit] are 0.5, 0.8, 0.4 & 1.0 respectively. Th. equilibrium constant is
a) 0.1
b) 1.0
c) 10
d) ∞
Answer:
b) 1.0

Question 40.
In the reaction N2 + 3H2 \(\rightleftharpoons\) 2NH3 + x kCal, one mole of N2 reacts with. 3 moles of H2 at equilibrium. Then the value of a (degree of dissociation) is approximate……………… P is the pressure at equilibrium
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-22
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-23
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-24

II. Match the following.

Question 1.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-25
Answer:
(a) 2, 1, 4, 3

Question 2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-26
Answer:
(b) 2, 3, 4, 1

Question 3.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-27
Answer:
(a) 4, 1, 2, 3

Question 4.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-28
Answer:
(a) 3, 4, 1, 2

Question 5.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-29
Answer:
(b) 2, 3, 4, 1

III. Fill in the blanks.

Question 1.
Transport of oxygen by Hemoglobin in our body is ………… a reaction.
Answer:
reversible

Question 2.
The temperature at which the solid and liquid phases of a substance are at equilibrium is called …………
Answer:
Freezing point

Question 3.
The temperature at which the liquid and vapour phases are at equilibrium is called …………
Answer:
Condensation point

Question 4.
………… law is used to explain gas-solution equilibrium processes.
Answer:
Henry’s law

Question 5.
In the reaction 2H2(g) + O2(g) \(\rightleftharpoons\) 2H2O(g), the Kp value is equal to ………
Answer:
< KC
Solution:
Δng = 2 – 3 = -1
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-30

Qustion 6.
The expression of K for the reaction CO2(g) + H2O(l) \(\rightleftharpoons\) H+(aq) + HCO3(aq) is equal to …………..
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-31

Question 7.
The expression of K for the reversible reaction 2CO(g) \(\rightleftharpoons\) CO2(g) + C(s)
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-32

Question 8.
The Ang value for the reaction 2NO(g) + O2(g) \(\rightleftharpoons\) 2NO2(g) is …………….
Answer:
– 1
Solution:
Δng = np – nr = 2 – 3 = -1

Question 9.
The correct differential form of van’t Hoff equation is ………….
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-33

Question 10.
For the reaction H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g), the equilibrium constant K is
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-34

Question 11.
PCI5 is kept in a closed container at a temperature of 250K the equilibrium concentrations of PCI5, PCl3 and Cl2 are 0.045 moles L-1, 0.096 moles L-1, 0.096 moles L-1 respectively. The value of equilibrium constant for the reaction PCI5(g) \(\rightleftharpoons\) PCl3(g) + Cl2(g) will be ……………………
Answer:
0.205
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-35

Question 12.
Equilibrium constant changes with ………………..
Answer:
Both temperature and pressure

Question 13.
For the reaction 2HI(g) \(\rightleftharpoons\) H2(g) + I2(g) at 720 K, the equilibrium constant value is 50. The equilibrium constant for the reaction H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g) at the same temperature will be ………..
Answer:
0.02
Solution:
Forward reaction equilibrium constant K1 = 50
Reverse reaction equilibrium constant K2 = ?
K2 = \(\frac { 1 }{ { K }_{ 1 } }\) = \(\frac { 1 }{ 50 }\) = 0.02

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 14.
If equilibrium constant for the reaction N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g) at 298 K is 2.54, the value of equilibrium constant for the reaction \(\frac { 1 }{ 2 }\)N2 + \(\frac { 3 }{ 2 }\)H2 \(\rightleftharpoons\) NH3 will be ………….
Answer:
1.59
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-37

Question 15.
The chemical system at equilibrium is not affected by the addition of ……………..
Answer:
catalyst

Question 16.
A catalyst will increase the rate of a chemical reaction by lowering the ……………
Answer:
activation energy

Question 17.
In a closed system, A(s) \(\rightleftharpoons\) 3B(g) + 3C(g) If the partial pressure of C is doubled, then the partial pressure of B will be ………….. times the original value.
Answer:
\(\frac { 1 }{ 2\sqrt { 2 } }\)

Question 18.
Consider the following gaseous equilibria with equilibrium constants K1 and K2 respectively
SO2 + \(\frac { 1 }{ 2 }\)O2(g) \(\rightleftharpoons\) SO3(g) – K1
2SO3(g) \(\rightleftharpoons\) 2SO2(g) + O2(g) – K2
The equilibrium constants are related as ………….
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-39.1
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-39

Question 19.
K2 for the following reaction at 700 K is 1.3 x 10-3 atm-1. The K at same temperature for the reaction 2SO2(g) + O2(g) \(\rightleftharpoons\) 2SO3(g) will be ………….
Answer:
7.4 x 10-2
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-40

Question 20.
For the reaction PCI23(g) + CI2(g) \(\rightleftharpoons\) PCI5(g) at 250°C, the value of KC is 26 then the value of Kp on the same temperature will be ……………
Answer:
0.61
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-41

Question 21.
In the reaction N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g), the value of the equilibrium constant depends on ………………
Answer:
the temperature

Question 22.
K1 and K2 are velocity constant o forward and backward reactions. The equilibrium constant KC of the reaction is …………
Answer:
\(\frac { { K }_{ 1 } }{ { K }_{ 2 } }\)

Question 23.
The equilibrium constant of the reaction 3C2H2 \(\rightleftharpoons\) C6H6 is 4.0 at a temperature of T K. If the equilibrium concentration of C2H2 is 0.5 moles L-1, the concentration of C6H6 is …………..
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-42

Question 24.
In an equilibrium reaction for which ΔG0 = 0, the equilibrium constant K should be = ……………
Answer:
1

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 25.
The equilibrium constant for the reaction 2SO2(g) + O2(g) \(\rightleftharpoons\) 2SO3(g) is 5. If the equilibrium constant mixture contains equal moles of SO3 and SO2, the equilibrium partial pressure of O2,
gas is ……………..
Answer:
0.2 atm
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-43

Question 26.
In the reaction NH4CI(s) \(\rightleftharpoons\) NH3(g) + HCI(g) the value Offlg is Δng ………………….
Answer:
2
Solution:
Δng = np(g) – nr(g) = 2 – 0 = 2

IV. Choose the odd one out.

Question 1.
(a) seesaw
(b) tug – of – war
(c) sublimation of camphor
(d) Acid hydrolysis of an ester
Answer:
(d) Acid hydrolysis of an ester is a chemical equilibrium.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-44
Whereas a, b, c are examples of physical equilibrium.

Question 2.
(a) Synthesis of hydrogen iodide
(b) Decomposition of calcium carbonate
(c) Sublimation of iodine
(d) Dissociation of PCl5
Answer:
(c) Sublimation of iodine I2(s) \(\rightleftharpoons\) I2(g) is an example of physical equilibrium, whereas a, b, and dare examples of chemical equilibrium.

Question 3.
(a) Synthesis of HI
(b) Dissociation of PCl5
(e) Synthesis of NH3
(d) Decomposition of CaCO3
Answer:
(a) Decomposition of CaCO3
The decomposition of CaCO3 is an example of heterogeneous equilibrium.
CaCO3(s) \(\rightleftharpoons\) CaO(s) + CO2(g)
Where as a, b, c are examples of homogeneous equilibrium.

Question 4.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-46
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-47
Synthesis of SO3 is an example of homogeneous equilibrium whereas the others a, b and d are heterogeneous equilibrium.

Question 5.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-48
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-49
(b) is an example of heterogeneous equilibrium whereas the others a, c and d are homogeneous equilibrium.

V. Choose the correct pair.

Question 1.
(a) Q = KC : Reaction is in equilibrium state
(b) Q < KC : Reaction proceed in reverse direction
(c) Q > KC : Reaction proceed in ftrward direction
(d) Q = KC : Reaction proceed in both directions
Answer:
(a) Q = KC : Reaction is in equilibrium state

Question 2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-50
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-51

Question 3.
(a) Kp = KC : Synthesis of HI
(b) Kp > KC : Dissociation of PCl5
(c) Kp < KC : Synthesis of SO3
(d) Kp = KC : Synthesis of HI
Answer:
(a) Kp = KC : Synthesis of HI
H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g) : Δng = 2 – 2 = 0
ΔKp = KC

VI. Choose the incorrect pair.

Question 1.
(a) Acid hydrolysis of an ester – Homogeneous equilibrium
(b) Synthesis of Ammonia – Homogeneous equilibrium
(c) Decomposition of CaCO3 – Homogeneous equilibrium
(d) Synthesis of HI – Homogeneous equilibrium
Ans.
(c) Decomposition of CaCO3: Homogeneous equilibrium It is a heterogeneous equilibrium.

Question 2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-52
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-53

VII. Assertion and Reason.

Question 1.
Assertion (A): Chemical equilibrium is in a state of dynamic equilibrium.
Reason (R): At equilibrium, the forward and backward reactions are proceeding at the same rate and no macroscopic change is observed.
(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 2.
Assertion (A): In Haber’s process, NH3 is liquefied and removed.
Reason (R): Because the reaction keeps moving in the backward direction.
(a) Both (A) and (R) are correct but (R) is not the correct explanation 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.
Answer:
(b) Both (A) and (R) are correct and (R) is the correct explanation of(A).

Question 3.
Assertion (A): In the dissociation of PCI5 at constant pressure and temperature addition of helium at equilibrium increases the dissociation of PCl5.
Reason (R): Helium removes CI, from the field of action.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false
(d) Both A and R are false
Answer:
(d) Both A and R are false

VIII. Choose the incorrect statement.

Question 1.
(a) In equilibrium mixture of ice and water kept in perfectly insulted flask, mass of ice and water does not change with time.
(b) The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate
(c) On addition of catalyst the equilibrium constant value is not affected
(d) Equilibrium constant for a reaction with negative .H value decreases as the temperature increases.
Answer:
(b) The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate

Solution:
Oxalate ions of oxalic acid form a complex with ferric ions thus decrease its concentration thus, the concentration of red complex in the product decreases.

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium 2 Mark Questions and Answers.

I. Write brief answer to the following questions:

Question 1.
What are reversible reactions?
Answer:
Reactions in which the products can react to give back reactants under suitable conditions are called reversible reactions.

Question 2.
What are the different types of equilibrium? Explain with example?
Answer:
1. Physical equilibrium:
A system in which the amount of matter constituting different phases does not change with time is said to be in physical equilibrium.
H2O(s) \(\rightleftharpoons\) H2O(1). Solid-liquid equilibrium.

2. Chemical equilibrium:
Chemical reactions in which the forward and backward reactions are proceeding at the same rate and no macroscopic change is observed is said to be in chemical equilibrium. H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 3.
Explain the equilibrium involving dissolution of solid in liquid with a suitable example.
Answer:
When sugar is added to water at a particular temperature. it dissolves to form a sugar solution. When more sugar is added to that solution, a particular stage of sugar remains solid and results in the formation of a saturated solution. Here a dynamic equilibrium is established between the solute molecules in the solid phase and in the solution phase.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-54

Question 4.
What is equilibrium constant?
Answer:
The equilibrium constant Kc may be defined as the ratio of the product of the active masses of the products to that of the reactants.

Question 5.
What is meant by active mass? Give its unit.
Answer:
The active mass represents the molar concentration of the reactants (or) products.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-55

Question 6.
Show that Kp = KC with two example
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-56

Question 7.
Give two examples of equilibrium reactions where Kp > KC.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-57

Question 8.
When will be Kp < KC? Give two example.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-58

Question 9.
Write the KC for the reaction CO2(g) + H2O(l) \(\rightleftharpoons\) H+(aq) + HCO3(aq)
Answer:
H2O(l) is a pure liquid and its concentration remains constant.
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-59

Question 10.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-60 What is the value of K4 in A \(\rightleftharpoons\) D
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-61

Question 11.
Write the Kp and KC for the following reactions
(i) 2SO2(g) + O2(g) \(\rightleftharpoons\) 2SO3(g)
(ii)2CO(g) \(\rightleftharpoons\) CO2(g) + C(s)
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-62

Question 12.
Explain how the equilibrium constant K. predict the extent of a reaction.
Answer:
1. The value of equilibrium constant KC tells us the extent of the reaction i.e., it indicates how far the reaction has proceeded towards product formation at a given temperature.

2. A large value of KC indicates that the reaction reaches equilibrium with high product yield on the other hand, lower value of KC indicates that the reaction reaches equilibrium with low product yield.

3. If KC > 103 the reaction proceeds nearly to completion.

4. If KC < 10-3 the reaction rarely proceeds.

5. If the KC is in the range 10-3 to 103 significant amount of both reactants and products are present at equilibrium.

Question 13.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-63
Predict the extent of the above two reactions.
Answer:
In the reactions, decomposition of water at 500 K and oxidation 01 nitrogen at 1000 K, the value of KC is very less KC < 10-3. So reverse reaction is favoured.
∴ Products << reactants

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 14.
What is solid-liquid equilibrium? Give example.
Answer:

  • The equilibrium between solid and liquid at its melting point (or freezing point) at normal atmospheric pressure is called solid-liquid equilibrium.
  • If some ice cubes along with water are placed in a thermos flask (at 00 C and 1 atm), then there will be no change in the mass of ice and water.
  • At equilibrium,
    Rate of melting of ice = Rate of freezing of water
    H2O(s) ⇌ H2O(l)

Question 15.
What is the KC value for the formation of HCl at 700 K? Predict the extent of the reaction?
Answer:
H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g) at 700 K
KC = 57.0
10-3 <KC <103
So both forward and backward reaction make significant progress. Neither forward nor reverse reaction predominates.

Question 16.
What is the KC value of formation of HCI at 300 K? Explain it.
Answer:
H2(g) – Cl2(g) \(\rightleftharpoons\) 2HCI(g) at 300 K
K2 = 4 x 1031
KC > 103 So[products] >> [Reactant]
Reaction nearly goes to completion. So forward reaction is favoured.

Question 17.
2CO(g) + O2(g) \(\rightleftharpoons\) 2CO2(g) at 1000 K. What is the Kc this reaction? Predict the extent of this reaction.
Answer:
2CO(g) + O2(g) \(\rightleftharpoons\) 2CO2(g) at 1000 K
KC = 2.2 x 1022
KC > 103.
So [Products] >> [Reactants]
Reaction nearly goes to completion and forward reaction is favoured.

Question 18.
What is Law of mass action?
Answer:
“At any instant, the rate of a chemical reaction at a given temperature is directly proportional to the product of the active masses of the reactants”.
Rate ∝ [Reactants]x
[ ] = \(\frac{\mathrm{n}}{\mathrm{v}}\), {Unit is mol.dm-3 or mol.L-1}
where,
n is the number of moles and
V is the volume of the container (dm3 or L).

Question 19.
Explain the diagrammatic expression about the direction of the reaction.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-65
In (i) Q < KC, the reaction will proceed in forward direction.
In (ii) Q = KC, the reaction is in equilibrium state.
In (iii) Q> KC, the reaction will proceed in the reverse direction.

Question 20.
Explain the effect of a catalyst in an equilibrium reaction?
Answer:
Addition of catalyst does not affect the state of equilibrium. The catalyst increases the rate of both the forward and reverse reactions to the same extent. Hence it does not change the equilibrium composition of the reaction mixture.

Question 21.
For the following equilibrium. KC = 6.3 x 1014 at 1000K
NO(g) + O3(g) \(\rightleftharpoons\) NO2(g) + O2(g)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions what is KC for the reverse reaction?
Answer:
For the reverse reaction Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-66

Question 22.
Explain why pure liquids and solids can be ignored while writing the value of equilibrium constants.
Answer:
This is because molar concentration of a pure solid or liquid is independent of the amount present.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-67
Since density of pure liquid or solid is fixed and molar mass is also fixed, therefore molar concentration are constant.

Question 23.
A sample of Hl(g) is placed in a flask at a pressure of 0.2 atm. At equilibrium partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium?
2HI(g) \(\rightleftharpoons\) H2 (g) + I2(g)
Answer:
pHI = 0.04 atm. pH2 = 0.08 atm; pl2 = 0.08 atm
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-68

Question 24.
The equilibrium constant expression for a gas reaction is.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-69
Write the balanced chemical equation corresponding to this expression.
Answer:
Balanced chemical equation for the reaction is
4 NO(g) + 6 H2O(g) \(\rightleftharpoons\) 4NH3(g) + 5O2(g)

Question 25.
Predict which of the following will have an appreciable concentration of reactions and products:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-70
Answer:
The following conclusion can be drawn from the values of
(a) Since the value of K is very small, this means that the molar concentration of the products is very small as compared to that of the reactants.
(b) Since the value of K is quite large, this means that the molar concentration of the products is very large as compared to that of the reactants.
(c) Since the value of K is 1.8, this means that both the products and reactants have appreciable concentrations.

Question 26.
Write the equilibrium constant (KC) expression for the following reactions.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-71
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-72

Question 27.
For the equilibrium 2NOCI(g) \(\rightleftharpoons\) 2NO(g) + CI2(g) the value of the equilibrium constant KC is 3.75 x at 1069 K. Calculate the Kp for the reaction at this temperature?
Answer:
We know that Kp = Kc(RT)∆ng
For the above reaction, ∆ng = (2 + 1) – 2 = 1
Kp = 3.75 x 10-6 (0.0831 x 1069) = 3.3 x 10-4

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 28.
The value of Kc for the reaction 2A \(\rightleftharpoons\) B + C is 2 x 10-3. At a given time, the composition of reaction mixture is [A] [B] [C] = 3 x 10-4 M. In which direction the reaction will proceed?
Answer:
For the reaction, the reaction quotient Q is given by QC = [B] [C]/[A]2
as [A] = [B] = [C] = 3 x 10-4 M
Qc (3 x 10-4)(3 x 10-4)/(3 x 10-4)2 = l
as QC > KC so, the reaction will proceed in the reverse direction.

Question 29.
Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:
CH4(g) + H2O(g) \(\rightleftharpoons\) CO(g) + 3H2(g)
(a) Write an expression for Kp for the above reaction.
(b) How will the values of Kp and composition of equilibrium mixture be affected by
(i) increasing the pressure
(ii) increasing the temperature
(iii) using a catalyst
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-73
(b) (I) value of Kp will not change, equilibrium will shift in backward direction.
(ii) value of Kp will increase and reaction will proceed in forward direction.
(iii) no effect.

II. Answer the following questions:

Question 1.
Derive the Equilibrium constants Kp and K for the homogeneous reactions.
Answer:
Let us consider a reversible reaction,
xA + yB ⇌ lC + mD
where A and B are the reactants, C and D are the products, and x, y, l, and m are the coefficients of A, B, C, and D respectively.
Applying Law of mass action, the rate of the forward reaction,
rf α [A]x [B]y (or)
rf = kf [A]x [B]y
Similarly, the rate of the backward reaction,
rb a [C]l [D]m (or)
rb = k, [C]l [D]m
where kf and kb are proportionality constants
At equilibrium, Rate of forward reaction(rf) = Rate of backward reaction(rb)
kf α [A]x [B]y = kb a [C]l [D]m (or)
\(\frac{\mathrm{k}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{b}}}=\frac{[\mathrm{C}]^{l}[\mathrm{D}]^{m}}{[\mathrm{~A}]^{x}[\mathrm{~B}]^{y}}\) = Kc
where, Kc is the equilibrium constant in terms of concentration.

At a given temperature, the ratio of the product of active masses of products to that the reactants is a constant, known as equilibrium constant. Each concentration term is raised to the power which is equal to the respective stoichiometric coefficients in the balanced chemical equation. Later when we study chemical kinetics we will learn that this is only approximately true. If the reactants and products behave as gases, then the equilibrium constant in terms of partial pressures can be written as

\(\frac{\mathrm{P}_{\mathrm{C}}^{l} \times \mathrm{P}_{\mathrm{D}}^{m}}{\mathrm{P}_{\mathrm{A}}^{x} \times \mathrm{P}_{\mathrm{B}}^{y}}\)

where, Pc is the partial pressure of the gas, C and so on.

Question 2.
How is liquid – vapour equilibrium exist?
Answer:
1. Liquid water is in equilibrium with vapour at 373 K and I atm pressure in a closed vessel

2. Rate of evaporation = Rate of condensation

Question 3.
What is meant by boiling point and condensation point of the liquid?
Answer:
The temperature at which the liquid and vapour phases are at equilibrium is called the boiling point and condensation point of the liquid.

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 4.
Define melting point (or) freezing point of the substance.
Answer:
The temperature at which the solid and liquid phases of a substance are at equilibrium is called the melting point or freezing point of substance.

Question 5.
Illustrate the formation of solid – vapour equilibrium with suitable example.
Answer:
1. Consider a system in which the solid sublimes to vapour. e.g., I, (or) camphor.
2. When solid iodine is placed in a closed transparent vessel, after sometime, the vessel gets filled up with violet vapour due to sublimation of iodine.
3. Initially the intensity of the violet colour increases, after some time it decreases and finally it becomes constant as the following equilibrium is attained.
I2(s) \(\rightleftharpoons\) I2 (g)

Question 6.
Give three examples for solid vapour equilibrium.
Answer:
I2(s) \(\rightleftharpoons\) I2 (g)
Camphor (s) \(\rightleftharpoons\) Camphor (g)
NH4CI(s) \(\rightleftharpoons\) NH4CI(g)

Question 7.
Explain the following diagrams.
Diagram – 1
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-74
Answer:

  1. As the concentration of the products increases, more products collide and react in the backward direction.
  2. As the rate of the reverse reaction increases, the rate of the forward reaction decreases.
  3. Eventually the rate of both reactions becomes equal.

Diagram-II
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-75

  1. Concentration of reactants decreases with time initially and concentration of products increases with time.
  2. After sometime, equilibrium is reached i.e., concentration of reactants and products remains constant.

Question 8.
What are the types of chemical equilibrium? Explain with suitable example.
Answer:

  1. Chemical equilibrium is of two types:
    • Homogeneous equilibrium
    • Heterogeneous equilibrium.
  2. In a homogeneous equilibrium, all the reactants and products are in the same phase
    H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
    In the above equilibrium H2, I2 and HI are in the gaseous state.
  3. If the reactants and products of a reaction in equilibrium are in different phases, then it is calLed heterogeneous equilibrium.
    e.g., CaCO3(s) \(\rightleftharpoons\) CaO(s) + CO3(g)

Question 9.
Write the value of K and K equation for CaCO3(s) \(\rightleftharpoons\) CaO(s) + CO2(g)
Answer:
A pure solid always has the same concentration at a given temperature, as it does not expand to fill its container i.e., it has the same no. of moles of its volume. Therefore the concentration of pure solid is a constant. So the expression if K and K is K [CO2], K = PCO2

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 10.
Consider the following equilibrium reaction and relate their equilibrium constants

  1. N2 + O2 \(\rightleftharpoons\) 2NO, K1
  2. 2NO + O2 \(\rightleftharpoons\) 2NO2, K1
  3. N2 + 2O2 \(\rightleftharpoons\) 2NO2, K3

Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-77

Question 11.
Explain the effect of concentration in an equilibrium state?
Answer:
At equilibrium, the concentration of the reactants and products does not change. The addition of more reactants or products to the reacting system at equilibrium causes an increase in their respective concentration.

According to Le – Chatelier’s principle, the effect of increase in concentration of a substance is to shift the equilibrium in a direction that consumes the added substance. For example,
H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)

The addition of H2 or I2 to the equilibrium mixture, disturbs the equilibrium. In order to minimize the stress, the system shifts the reaction in a direction where H2 and I2 are consumed i.e., formation of additional HI would balance the effect of added reactant.

Hence the equilibrium shifts to the right (forward direction). i.e., the equilibrium is re – established. Similarly, removal of HI (Product) also favours forward reaction. If HI is added to the equilibrium mixture, the concentration of HI is increased and system proceeds in the reverse direction to nullify the effect of increase in concentration of HI.

Question 12.
Consider the reaction, N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g). Explain the effect of pressure on this equilibrium reaction.
Answer:
N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g)
In the above equilibrium, if the pressure is increased, the volume wiLl decreases. The system responds to this effect by reducing the number of gas molecules. i.e., it favours the formation of ammonia. If the pressure is reduced, the volume will increases. It favours the decomposition of ammonia.

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 13.
Why pressure has no effect on the synthesis of HI?
Answer:
(i) When the total number of moles of gaseous reactants and gaseous products are equal, the change in pressure has no effect on system at equilibrium.
H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
Here the number of moles of reactants and products are equal. So the pressure has no effect on such equilibrium with ∆ng = O.

Question 14.
Explain the effect of temperature on the following equilibrium reaction.
N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g) ∆H = – 92.2 kJ.
Answer:
In this equilibrium, N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g) ∆H = – 92.2 kJ.

Forward reaction is exothermic while the reverse reaction is endothermic. If the temperature of the system increased, the system responds by decomposing some of ammonia molecules to nitrogen and hydrogen by absorbing the supplied heat energy.

Similarly, the system responds to a drop in the temperature by forming more ammonia molecule from nitrogen and hydrogen which release heat energy.

Question 15.
How does oxygen exchanges between maternal and fetal blood in a pregnant woman?
Answer:
1. In a pregnant women, the oxygen supply for the fetus is provided by the maternal blood in the placenta where the blood vessels of both mother and fetus arc in close proximity. Both fetal and maternal hemoglobin binds tO oxygen reversibly as follows.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-78

2. In the above two equilibrium, the equilibrium constant value for the oxygenation of fetal hemoglobin is higher which is due to its higher affinity for oxygen compared to adult hemoglobin. Hence in placenta, the oxygen from the mother’s blood s effectively transferred to fetal hemoglobin.

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 16.
What is K for the following reaction in state of equilibrium?
2SO2(g) + O2 \(\rightleftharpoons\) 2SO3(g)
(Given:[SO2] = 0.6 M; [O2] = 0.82 M; and [SO3] = 1.90 M
Answer:
2SO2(g) + O2 \(\rightleftharpoons\) 2SO3(g)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-79

Question 17.
At a certain temperature and total pressure of 105Pa, iodine vapours contain 40% by volume of iodine atoms in the equilibrium 12(g) \(\rightleftharpoons\) 2I(g). Calculate Kp for the equilibrium.
Answer:
According to available data:
Total pressure of equilibrium mixture = 105 Pa
Partial pressure of iodine atoms (1) = \(\frac { 40 }{ 100 }\) x (105 Pa) = 0.4 x 105 Pa
Partial pressure of iodine molecules (I2) = \(\frac { 60 }{ 100 }\) x (105 Pa) = 0.6 X 105 Pa
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-80

Question 18.
A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant KC for the reaction
N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g) is 1.7 x 10-2
Is this reaction at equilibrium? if not. what is the direction of net rection?
Answer:
The reaction is: N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g)
Concentration quotient (Q ) =
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-81
The equilibrium constant (Kr) for the reaction = 1.7 x 10-2
As QC \(\neq\) KC; this means that the reaction is not in a state of equilibrium.

Question 19.
What is the effect of:

  1. addition of H2
  2. addition of CH3OH
  3. removal of CO
  4. removal of CH3OH

On the equilibrium 2H2(g) + CO(g) \(\rightleftharpoons\) CH3OH(g)
Answer:

  1. Equilibrium will be shifted in the forward direction.
  2. Equilibrium will be shifted in the backward direction.
  3. Equilibrium will be shifted in the backward direction.
  4. Equilibrium will be shifted in the forward direction.

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 20.
At 473 K, the equilibrium constant Kc for the decomposition of phosphorus pentachioride (PCl5) is 8.3 x 10-3 . If decomposition proceeds as:
Answer:
PCI5(g) \(\rightleftharpoons\) PCI3(g) + Cl2(g);
∆H = + 124.0 kJ mol-1

  1. Write an expression for K. for the reaction.
  2. What is the value of K for the reverse reaction at the same temperature.
  3. What would be the effect on KC if
    • More of PCI5 is added
    • Temperature is increased.

Answer:
1. The expression for Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-82
2.  For reverse reaction Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-83
3.
(I) By adding more of PCI5, value of Kc will remain constant because there is no change in temperature.
(ii) By increasing the temperature the forward reaction will be favoured since it is endotherniic in nature. Therefore, the value of equilibrium constant will increase.

Question 21.
Dihydrogen gas used in Haher’s process is produced by reacting methane from natural gas with high temperature stam. The first stage of two stage reaction involves the formation of CO and H2. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction.
Answer:
CO(g) + H2O(g) \(\rightleftharpoons\) CO2(g) + H2(g)
If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam so that PCO = PH2O = 4.0 bar., what will be the partial pressure of 2 at equilibrium? Kp = 0.1 at 400°C.
Answer:
Let the partial pressure of hydrogen (H2) at equilibrium point p bar
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-84

Question 22.
The value of KC for the reaction 3O2(g) \(\rightleftharpoons\) 2O3(g) is 2.0 x 10-50 So at 25°C. If equilibrium concentration of 0, in 25°C is 1.6 x 102 what is the concentration of O3?
Answer:
3O2(g) \(\rightleftharpoons\) 2O3(g)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-85

Question 23.
The reaction CO(g) + 3H2(g) \(\rightleftharpoons\) CH4(g) + H2O(g) is at equilibrium at 1300 K in a 1K flask. It also contain 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H2O and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant, KC for the reaction at the given temperature is 3.90.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-86

Question 24.
The following concentration were obtained for the formation of NH3 from N2 and H2 at equilibrium at 500 K.

  • [N2(g)] 1.5 x 10-2 M
  • [H2(g)] = 3.0 x 10-2 M
  • [NH3]= 1.2 x 10-2M.

Calculate equilibrium constant.
Answer:
N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g)
Calculate equilibrium constant
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-87

Question 25.

  1. In the reaction A+ B → C + D, what will happen to the equilibrium if concentration of A is increased?
  2. The equilibrium constant for a reaction is 2 x 10-23 at 25°C and 2 x 10-2 at 50°C. Is the reaction endothermic or exothermic?
  3. Mention at least three ways by which the concentration of SO3 can be increased in the following reaction in a state of equilibrium. 2SO2(g) + O2(g) \(\rightleftharpoons\) 2SO3(g)

Answer:

  1. The reaction will shift in the forward direction.
  2. Endothermic
  3. following reaction
    • increasing concentration of SO2
    • increasing pressure.
    • increasing concentration of oxygen.

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 26.
PCI5, PCI3 and CI2 are at equilibrium at 500 K and having concentration I.59M PCl3, l.59M
CI2 and 1.41M PCI5. Calculate K. for the reaction PCl5 \(\rightleftharpoons\) PCl3 + Cl2
Answer:
The equilibrium constant K. for the above reaction can be written as:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-88

Question 27.
Given the equilibrium
N2O4(g) \(\rightleftharpoons\) 2NO2(g) with Kp = 0.15 atm at 298 K
(a) What is Kp using pressure in torr?
(b) What is KC using units of moles per litre.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-89

III. Answer the following questions in detail:

1. Derive the values of equilibrium constants Kp and KC for a general reaction
x A + y B \(\rightleftharpoons\) lC + mD
Answer:
Let us consider a reversible reaction x A + y B \(\rightleftharpoons\) lC + mD
where A, B are the reactants C and D are the product and x, y. l and m are the stoichiometric coefficients of A. B, C and D respectively. Applying the law of mass action the rate of forward reaction.
rf ∝ [A]x [B]y or rf Kf [A]x [B]y
Similarly the rate of backward reaction
rb ∝ [C]l [D]m or rb = Kb [C]l [D]m
where Kf and Kb are proportionality constants.
At equilibrium, Rate of forward reaction (rf) = Rate of backward reaction (rb)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-90
where KC is the equilibrium constant in terms of concentration. At a given temperature, the ratio of the product of active masses of reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of the reactants is a constant known as equilibrium constant.

If the reactants and products of the above reaction are in gas phase, then the equilibrium constant can be written in terms of partial pressures
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-91
where PA, PB, PC and PD are the partial pressure of gases A, B, C and D respectively.

Question 2.
Derive the values of K and K for the synthesis of HI.
Answer:
H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
Let us consider the formation of HI in which ‘a’ moles of hydrogen, ‘b’ moles of iodine gas are allowed to react in an container of volume ‘V’. Let ‘x’ moles of each of H2, and I2, react together to form 2x moles of HI.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-92

Question 3.
Derive the values of K and K for dissociation of PCI5.
Answer:
Consider that ‘a’ moles of PCl5 is taken in container of volume ‘V’ Let x moles of PCI5 be dissociated into x moles of PCI3 and x moles of Cl2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-93
Applying law of mass action
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-94
Where ‘n’ is the total number of moles at equilibrium
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-95

Question 4.
At certain temperature and under a pressure of 4 atm, PCl5 is 10% dissociated. Calculate the pressure at which PCI5 will be 20% dissociated at temperature remaining constant.
Calculation of Kp
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-96
Total no. of moles in the equilibrium mixture = 1 – α + α + α = (1 + α) mol.
Let the total pressure of equilibrium mixture = ρ atm
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-97
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-98
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-99
calculation of P under new condition
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-100

IV. Numerical Problems

Question 1.
Find the value of K for each of the following equlibria from the value of K
(a) 2NOCI(g) \(\rightleftharpoons\) 2NO(g) + Cl2(g); Kp= 1.8 x 102 atm at 500 K
(b) CaCO3(s) \(\rightleftharpoons\) CaO(s) + CO2(g): Kp = 167 atm at 1073 K.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-102

Question 2.
What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICI was 0.78 M?
2ICI(g) \(\rightleftharpoons\) I2(g) + CI2(g); KC = 0.14
Answer:
Suppose at equilibrium, the molar concentration of both I2(g) and Cl2(g) is x mol L-1.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-103

Question 3.
Equilibrium constant K for the reaction. N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g) at 500 K is 0.061. At particular time, the analysis shows that the composition of the reaction mixture is 3.0 mol L-1 of N2; 2.0 mol L-1 of H2; 0.50 moI L-1 of NH3. Is the reaction at equilibrium?
Answer:
The given reaction is: N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g)
According to available data
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-104

Common Errors

  1. In writing Kp, KC values from the equations, students may confuse to write whether products or reactants in the numerator or in the denominator
  2. When Kp = KC.(RT)0 students may confuse.
  3. In writing An values, students will consider all reactants and products.
  4. Students are confused to understand the concept of Q value and KC value.
  5. In writing chemical equilibrium reaction, you may miss the physical states of reactants and products.
  6. ∆ngvalue calculation will go wrong if you consider all the reactants and products.
  7. Equilibrium constant value is calculated under equilibrium condition and reaction quotient value defined at sometimes at equilibrium condition wrongly by the students.
  8. Chemical equilibrium condition must be known. Students may wrongly write in an open vessel, the equilibrium take place.
  9. Equilibrium symbol, students may wrongly write as =

Rectifications

  1. Always we have to write Kp & KC as Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-105
  2. (Anything)° = 1. So Kp = KC.
  3. Only they have to consider gaseous reactants and gaseous products. Since solid and liquid are constant in their concentration.
  4. Q value is calculated under non equilibrium conditions. KC value is calculated under equilibrium condition.
  5. Physical states of reactants and products must be written as a subscript by the words s, l, g. (solid, liquid, gas)
  6. ∆ng = ngp – ngr You should consider only gaseous products and gaseous reactants.
  7. Reaction quotient Q value is calculated only under non-equilibrium conditions.
  8. Chemical equilibrium reactions are always take place at closed vessel only.
  9. Equilibrium reaction symbol is ⇔.

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

Samacheer Kalvi 11th English Solutions Poem Chapter 4 Macavity – The Mystery Cat

Students can Download English Lesson 4 Macavity – The Mystery Cat Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 11th English Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations. Learn the Samacheer Kalvi 11th English Grammar to enhance your grammar skills like reading comprehension, passage writing, parts of speech, tenses, passive and active voice, and many other concepts in no time.

Tamilnadu Samacheer Kalvi 11th English Solutions Poem Chapter 4 Macavity – The Mystery Cat

Warm Up

A. French proverb goes thus: ‘The dog may be wonderful prose, but only the cat is poetry.’ You may have observed that all animals possess a number of unique qualities. Fill in the columns with words and phrases associated with each of the following animals.

DOG CAT WOLF ELEPHANT

Answer:

DOG CAT WOLF ELEPHANT
Barks meows howls trumpets
open fight stealthy groups groups
aggressive pretends to be gentle rough gentle
faithful ungrateful wild civilized
fights for food quickly steals food predatory graceful

B. People admire some of these animal qualities. What are they? Have you noticed some of them in yourself or in others? Share your views with the class.

Cats are stealthy and walk without making noise.
My friend Vimal / Mala is cat-like. He / She surprises me often appearing suddenly. Jackal is clever and persuades others like Ratna / Raja to work for him. Whenever a difficult assignment is given, he / she praises the gifted student and gets his/ her work done. Elephant is graceful and known for strong memory. Murugan / Neela is very gentle and has terrific memory. One must be very cautious with such persons. Even a small hurt will be in their memory for long.

Samacheer Kalvi 11th English Macavity – The Mystery Cat Textual Questions

A. Based on your understanding of the poem, answer the following question in a sentence or two.

Macavity’s a Mystery Cat: he’s called the Hidden Paw
For he’s the master criminal who can defy the Law.
He’s the bafflement of Scotland Yard, the Flying Squad’s despair:
For when they reach the scene of crime – Macavity’s not there!

Macavity, Macavity, there’s no one like Macavity,
He’s broken every human law, he breaks the law of gravity.
His powers of levitation would make a fakir stare,
And when you reach the scene of crime – Macavity’s not there!

Samacheer Kalvi 11th English Solutions Poem Chapter 4 Macavity - The Mystery Cat

You may seek him in the basement, you may look up in the air
But I tell you once and once again, Macavity’s not there!
Macavity’s a ginger cat, he’s very tall and thin;
You would know him if you saw him, for his eyes are sunken in.

His brow is deeply lined with thought, his head is highly domed;
His coat is dusty from neglect, his whiskers are uncombed.
He sways his head from side to side, with movements like a snake;
And when you think he’s half asleep, he’s always wide awake.

Macavity, Macavity, there’s no one like Macavity,
For he’s a fiend in feline shape, a monster of depravity.
You may meet him in a by-street, you may see him in the square
But when a crime’s discovered, then Macavity’s not there!

Samacheer Kalvi 11th English Solutions Poem Chapter 4 Macavity - The Mystery Cat

He’s outwardly respectable. (They say he cheats at cards.)
And his footprints are not found in any file of Scotland Yard’s.
And when the larder’s looted, or the jewel-case is rifled,
Or when the milk is missing, or another Peke’s been stifled,

Or the greenhouse glass is broken, and the trellis past repair
Ay, there’s the wonder of the thing! Macavity’s not there!
And when the Foreign Office find a Treaty’s gone astray,
Or the Admiralty lose some plans and drawings by the way,

There may be a scrap of paper in the hall or on the stair
But it’s useless to investigate, Macavity’s not there!
And when the loss has been disclosed, the Secret Service say:
‘It must have been Macavity!’ but he’s a mile away.

You’ll be sure to find him resting, or a licking of his thumbs,
Or engaged in doing complicated long division sums.
Macavity, Macavity, there’s no one like Macavity,
There never was a Cat of such deceitfulness and suavity.

He always has an alibi, and one or two to spare:
At whatever time the deed took place, MACAVITY WASN’T THERE!
And they say that all the Cats whose wicked deeds are widely known,
(I might mention Mungojerrie, I might mention Griddlebone)

Are nothing more than agents for the Cat who all the time
Just controls their operations: the Napoleon of Crime

Question (i)
What is Macavity’s nickname?
Answer:
Macavity’s nickname is the Hidden Paw.

Question (ii)
Why is the Flying Squad frustrated?
Answer:
The Flying squad is frustrated because every time they rush to the spot of crime to arrest Macavity, he is not there.

Question (iii)
Which law does Macavity break?
Answer:
Macavity not only breaks the human law but also breaks the law of gravity.

Question (iv)
What makes the fakir stare in wonder?
Answer:
Macavity’s power of levitation makes the fakir stare in wonder.

Question (v)
Describe Macavity’s appearance.
Answer:
Macavity is a tall and thin ginger cat whose eyebrows are deep with lines. He has sunken eyes which gives him a devil look. His coat is untidy and his whiskers are uncombed.

Question (vi)
Where can you encounter Macavity?
Answer:
One may meet Macavity in a street or in the square. But he vanishes when a crime is committed.

Question (vii)
Why does the poet say Macavity is ‘outwardly’ respectable?
Answer:
The poet says that Macavity is outwardly’ respectable because he pretends to be a good one. But his actions disprove it.

Question (viii)
Why is Macavity called the ‘Napoleon of Crime’?
Answer:
He is agile and cautious. He is a monster of depravity

Samacheer Kalvi 11th English Solutions Poem Chapter 4 Macavity - The Mystery Cat

Question (ix)
Which two characters does the poet refer to as examples of wicked cats?
Answer:
Macavity will escape a mile away from the scene of the crime.

Question (x)
Mention any two qualities of Macavity.
Answer:
Similar to Napoleon he was a strategist and military leader. Macavity controlled the operations of all the wicked cats in London.

B. Read the poem once again and complete the summary using the words given in the box.

larder whiskers respectable
criminal devil thought
sunken division agents
detective alibi desperate
fakir qualities gravity
greenhouse

Macavity – The Mystery Cat’ is a humorous poem, where the poet T.S. Eliot describes the mysterious

(a) ______ of a shrewd vile cat. He commits a crime at every possible opportunity.He is an elusive master (b) ______ who leaves no evidence after he commits a crime.Even the Scotland Yard, the London (c) ______ agency is unable to arrest him. The Flying Squad is (d) ______ because every time they rush to the crime spot to seize Macavity, he is not there. He breaks the human law as well as the law of (e) ______ He baffles even a (f) ______ with his powers of levitation. Macavity appears tall and thin with (g) ______ eyes. He is always preoccupied with some serious (h) ______ His coat is dusty and his (i) ______ are unkempt. Macavity is a (f) ______ in the guise of a cat. He appears to be outwardly (k) ______ but his actions disprove it. Macavity loots the (l) ______ , ransacks the jewel-case, and breaks the (m) ______ glass but wonder of wonders he is not to be found anywhere there. He is always a mile away from the scene of crime, happily relaxing or doing difficult (n) ______ sums. He is clever at making up an (o) ______ every time he plots a crime. All the notorious cats are nothing but the (p) ______ of Macavity, the Napoleon of Crime.
Answer:
(a) qualities
(b) criminal
(c) detective
(d) baffled
(e) gravity
(f) fakir
(g) sunken
(h) thought
(i) whiskers
(j) fiend
(k) respectable
(p) agents
(0) larder
(m) greenhouse
(n) division
(o) alibi

C. Read the poem and answer the following in a short paragraph of 8 to 10 sentences each.

Question (i)
What are the mysterious ways in which Macavity acts?
Answer:
Macavity is an elusive master criminal who leaves no evidence after he commits a crime. He baffles Scotland Yard police and the flying squad as he disappears before their arrival to the scene of crime. He defies law of gravity and his powers of levitation make me fakir stare with wonder. Macavity loots the larder. He ransacks the jewel case. He is an elusive criminal who escapes from the spot of crime before the flying squad or Scotland Yard reach there. He breaks every human law and laws of gravity. He breaks greenhouse glasses. He steals stealthily into . the kitchen and empties milk. When an important treaty in the embassy is missing or. when the admiralty loses some plans or drawings, the investigation terms conclude that it is the work of Macavity. But Macavity, as a wonder of wonders, would be miles away relaxing somewhere.

Question (ii)
Give an account of Macavity’s destructive mischief.
Answer:
Macavity is a “Friend in feline shape”. He loots the food from the larder. Jewel’s case gets ransacked. Milk gets robbed. A Pekinese dog is stifled. Greenhouse glass is broken. The important agreement in the embassy is tom and important drawings in the admiralty are lost. The investigation team may find a scrap of paper in the stairs but Macavity always makes good his escape after doing all the above destructive activities. Besides, he is the Napoleon of crime controlling the operations of all cats in London.

Question (iii)
Describe the appearance and qualities of Macavity.
Answer:
Macavity is tall and thin. His eyes are sunken in. His brow is deeply lined with thought. His head is highly domed. His coat is dusty and his whiskers unkempt. He makes movements like a snake. He is a friend in feline shape. He is a monster of depravity. He is a cat of deceitfulness and suavity. When one thinks he is half-asleep, he is wide awake. He is an elusive, agile and phantom-like cat.

D. Read the given lines and answer the questions that follow.

(i) Macavity’s a Mystery Cat: he’s called the Hidden Paw…

Question (a)
Does the poet talk about a real cat?
Answer:
No, Macavity is not a real cat.

Question (b)
Why is he called the Hidden Paw?
Answer:
He is called a “ Hidden paw “ because even Scotland Yard is unable to arrest him after he commits any crime. He does not leave his footprints in the spot of crime.

(ii) He’s the bafflement of Scotland Yard, the Flying Squad’s despair: For when they reach the scene of crime Macavity’s not there!..

Question (a)
What is ‘Scotland Yard’?
Answer:
‘Scotland yard’ is the headquarters of London Metropolitan police service.

Question (b)
Why does the flying squad feel disappointed?
Answer:
The flying squad reaches the scene of crime very fast. But Macavity is not at all there. The flying squad is disappointed because they are unable to arrest the crafty criminal.

(iii) He sways his head from side to side, with movements like a snake;
And when you think he’s half asleep, he’s always wide awake…

Question (а)
Explain the comparison made here.
Answer:
The poet compares the movement of the cat to that of a snake. He employs a simile here.
The movement is quiet but swift.

Samacheer Kalvi 11th English Solutions Poem Chapter 4 Macavity - The Mystery Cat

Question (b)
What does he pretend to do?
Answer:
He pretends to be half asleep when he is fully awake. .

Question (c)
Identify the figure of speech in the first line.
Answer:
Simile.

(iv) For he’s a fiend in feline, shape, a monster of depravity.

Question (a)
How is the cat described in this line?
Answer:
Macavity is described as a “demon in the shape of a cat”.

Question (b)
Explain the phrase ‘monster of depravity’.
Answer:
Satan is called the master of depravity. T.S. Eliot calls Macavity, the master of depravity. He means that the cat is an embodiment of evil. He is wicked, all the time involved in doing something evil.

(v) And his footprints are not found in any file of Scotland Yard’s.

Question (а)
What seems to be a challenge for the Scotland Yard?
Answer:
Scotland Yard police, known for its efficiency to nab criminals in record time, is unable to link any crime to Macavity. He has an alibi whenever a crime is committed. Arresting Macavity with clinching evidence for his involvement in a crime is a challenge for Scotland Yard.

Question (b)
Why do they need his footprints?
Answer:
They need Macavity’s footprints to prove to the court of law that he was present at the scene of the crime.

(iv) It must have been Macavity!’ but he’s a mile away.

Question (a)
What is Macavity blamed for?
Answer:
Macavity is blamed for most of the crimes which leave the Scotland police and flying squad fuming and fretting because he just vanishes after every crime is committed. Besides, he leaves no proof or evidence behind.

Question (b)
Where is he?
Answer:
He is in a by-street or in the square when a crime is discovered. He always has one or two alibi.

(vii) There never was a Cat of such deceitfulness and suavity.

Question (a)
Which cat is being talked of here?
Answer:
Macavity is being talked of here.

Question (b)
How is he different from the rest?
Answer:
Other cats are lazy and just stay in the kitchen and take the food offered by their master. But Macavity is agile and defies law of the land and laws of gravity.. Despite doing all wicked things, he pretends to be innocent. So the poet claims one can never come across such a cat of “Deceitfulness and suavity”

E. Explain the following lines with reference to the context.

Question (i)
His powers of levitation would make a fakir stare
Answer:
Reference: These words are from the poem “Macavity – the mystery cat” written by T.S. Eliot. Context: T.S Eliot says, these words describing the skills of Macavity – The mystery cat. Explanation: Macavity does all kinds of mischiefs, petty thefts. He breaks things also. But before anyone could link the crime to Macavity he makes good his escape, floating in the air, jumping from building to building. His powers of levitation baffle even a fakir who has mystical powers.
Comment: The truth behind levitation is well brought out.

Question (ii)
And when you think he’s half asleep, he’s always Wide awake
Answer:
Reference: These words are from the poem “ Macavity – the mystery cat” written by T.S. Eliot.
Context: The poet says these words about the ability of the mysterious cat to hoodwink everyone.
Explanation: Macavity is a master of deceitfulness and suavity. When he appears to “be half-asleep with his half-closed eyes, he would be wide-awake. He is an enigma to everyone. Comment: Macavity is indeed a mystery.

Question (iii)
And his footprints are not found in any file of Scotland Yard’s
Answer:
Reference: These words are. from the poem “ Macavity the mystery cat” written by T.S. Eliot. Context: The poet says these words about the clever escape Macavity makes after every crime is committed.
Explanation: Scotland yard police is known all over the world for its capacity to investigate crimes and nab criminals in record time. But many crimes happen in London. Before Scotland Yard or the flying squad could reach the spot of crime, the criminal vanishes without leaving
any trace of the evidence. Scotland yard police wants to nab him with evidence. But his foot prints are nowhere to be found. So, Scotland Yard is unable to arrest Macavity.
Comment: The mysterious moves of Macavity stuns even the Scotland Yard.

Samacheer Kalvi 11th English Solutions Poem Chapter 4 Macavity - The Mystery Cat

Question (iv)
There may be a scrap of paper in the hall or on the stair.
But it’s useless to investigate…
Answer:
Reference: There words are from the poem “Macavity- the mysterious cat” written by T.S. Eliot .
Context: The poet says these words while discussing the leftover evidence of crime. Explanation: The Scotland yard police and the flying squads are guardians of peace in London. They keep watching. But whenever Macavity does a crime, he leaves before the police arrives. The clues like an empty larder, rifled jewel case, the disappearance of a treaty or drawings from the office or admiralty may lead to some shredded bits of paper lying on the floor or the staircase. But these bits of paper can’t help the police nab Macavity. They know it is the work of Macavity but they are helpless.
Comment: Not a trace is left behind by mysterious Macavity.

Question (v)
He always has an alibi, and one or two to spare
Answer:
Reference: These words are from the poem, “ Macavity – the mystery cat” written by T.S. Eliot.
Context: The poet says these words while describing the deceitful and clever nature of Macavity.
Explanation: Macavity breaks the laws of the land regularly. But gets away before the long arm. of the law reaches the spot of crime. He always has an alibi (one or more to spare) to escape from being caught. This proves his cleverness.
Comment: The wit of Macavity needs high commendation.

Additional Questions

Question (vi)
Macavity is a Mystery cat; he’s called the Hidden paw
Answer:
Reference: These words are from the poem, “Macavity – the mystery cat”.
Context: T.S. Eliot says these words while discussing the deceitfulness and the ability of the wicked cat to disappear room after a crime is committed.
Explanation: The poet describes the attributes of the mystery cat “ Macavity the cat is deceitful • and he baffles Scotland yard. The modus operandi of each unsolved crime points to Macavity only. But the lack of evidence like footprint prevents Scotland yard police from arresting him. Hence, he is called “The Hidden Law”.
Comment: Does Macavity truly have an unseen paw?

Question (vii)
Just controls their operation ; the Napoleon of Crime .
Answer:
Reference: These words are from the poem, “ Macavity – the mystery cat, written by T.S. Eliot.
Context: The poet says this about the daring acts of evil done by enigmatic villainous cat Macavity.
Explanation: Scotland yard police is unable to arrest Macavity as he leaves no evidence of crime he commits. He has many agents whose operations are controlled by him. Macavity is like military despot Napoleon, he guides all wicked cats in London unseen.
Comment: There is indeed no difference between Napoleon and Macavity.

F. Eliot has used many figures of speech to present the poem to the readers in an interesting way. He has attributed human qualities to a cat in this poem.

Question (i)
Identify the literary devices used in the following lines: .

Question (a)
He sways his head from side to side, with movements like a snake.
Answer:
simile

Samacheer Kalvi 11th English Solutions Poem Chapter 4 Macavity - The Mystery Cat

Question (b)
They say he cheats at cards.
Answer:
Personification

Additional:
(c) for he’s a friend in feline shape – Personification
(d) He’s outwardly respectable (they say he cheats at cards) – personification (The animal is attributed to human qualities)
(e) Just control their operations ; the Napoleon of Crime – Personification
(f) He always has an alibi and one or two to spare – Personification
(g) Or engaged in doing a complicated long division sums – Personification
(h) Are nothing more than agents for the Cat – Personification
(i) Just controls their operations the Napoleon or crime – metaphor

Question (ii)
Give four instances where the poet has used alliteration in the poem.
Answer:
milk, missing, larder’s, looted, sways, side to side, snake, break, broke – Alliteration

Question (iii)
What is the rhyme scheme used in the poem?
Answer:
Rhyme scheme of the poem is aa bb

(iv) Pick out all the pairs of rhyming words used in the poem.

Rhyming words in the poem
(a) Paw: Law
(b) despair: there
(c) Macavity: gravity
(d) stare: there
(e) air: there
(f) thin: in
(g) domed: uncombed
(h) snake: awake
(i) Macavity: depravity
(j) Square : their
(k) cards: yards
(l) rifled: stifled
(m) repair: there
(n) astray: way
(o) stair: there
(p) say: away
(q) thumbs : sums.
(r) Macavity: suavity
(s) spare: there
(t) known: bone
(u) time: crime

listening Activity

G. First, read the following sets of limericks with missing words. Now, listen to them being read out aloud by your teacher or played on the recorder. As you enjoy the absurd fun, complete the verse with what you hear. You may listen to them again, if required.

I
A wonderful bird is the (i) ______
His beak can hold more than his (ii) _______ can.
He can hold in his beak Enough food for a (iii) _______ !
But I’ll be darned if I know how the Peli-can?

II
There once was a (iv) _______ at the zoo
Who always had something to do
When it (iv) _______ him, you know,
To go to and fro.
He (v) _______ . it and went fro and to.

III
There once was a (vi) _______ little bunny
Who I thought was sweet and (vii) _______ .
He ate all the carrots,
And looked at the .(ix) _______
And that was my cute little (x) _______
Answers:
(i) pelican
(ii) belly
(iii) week
(iv) bear
(v) bored
(vi) reversed
(vii) cute
(viii) funny
(ix) Parrots
(x) bunny

Speaking Activity

H. Speaking Activity

Work with a partner. Read the following questions and share your views with the class. Have you heard of the phrase ‘cat’s paw’? The meaning is similar to that of ‘firing from the other’s shoulder’. ‘Cat’s paw’ refers to a person who is used unwittingly or unwillingly by another person to accomplish his own purpose.

Question (a)
This phrase originates from the fable ‘The Monkey and the Cat’. Explain how Macavity contradicts the phrase ‘cat’s paw’.
Answer:
Cat’s paw means being an incumbent in someone’s hand and do what the other person says. But Macavity is the master. He’s nicknamed the “hidden paw”. He is the Napoleon of crime controlling the operation of all the wicked cats in London. Macavity really contradicts “cat’s paw” absolutely.

Samacheer Kalvi 11th English Solutions Poem Chapter 4 Macavity - The Mystery Cat

Question (b)
‘When the mouse laughs at the cat, there is a hole nearby.’ Explain the meaning of this statement to your friends.
Answer:
Mouse knows the capacity of the cat to pounce on him and make a meal of him in just records. The mouse could dare to laugh at the cat when the scope of escape into the hole is bright.

Question (c)
Compose your own limericks on an elephant, a peacock and a butterfly. Read it put to your class.

1. Butterfly
A Spider awaits a butterfly
As he comes fluttering by
It’s caught in the silken trap
And straggles acts wings flap
Battling for survival under the blue skies

2. An elephant
There was a little elephant
To whom the river bank was forbidden .
But he went to the brink
Waiting for the crow to drink
And a bitter lesson when his flunk got bitten

3. A Peacock
“Joy is a peacock. It’s beauty so rare
A rainbow of colours that vibrantly flares
After the rain, brightly they come out
Into a fan-like form uniquely it creates
Never forgot, this vision, joyfully it illuminates”!

Macavity – The Mystery Cat About the Poet:

Samacheer Kalvi 11th English Solutions Poem Chapter 4 Macavity - The Mystery Cat

T.S Eliot (1888-1965) is an Essayist, play wright, literary and social critic. He moved out of USA to become a citizen of Great Britain at the age of 25. He renounced his American Passport when he was 39. He attracted widespread attention for his poem “The Love Song” of Alfred Prufrock (1915). It was seen on his Masterpiece of modern movement. It was followed by some of his best known poems including “The Wasteland” (1922), “The hollow men” (1915), “Ash Wednesday” (1930) and “Four quarters” (1943). During his stay in Harvard University, he developed a deep understanding of Indian philosophy and also studied Sanskrit. It is said that when he wrote the poem “The Wasteland”, doctors expressed concern over the health of his mind and strongly advised him to avoid writing.

But he wrote the poem in any paper that was found. It was Ezra Pound who brought order by editing the poem. He is well remembered for his plays such as “Murder in the Cathedral” (1935) and “The Cocktail Party” (1949). He was awarded the Nobel Prize for Literature in 1948 for his outstanding, pioneering contribution to modern poetry.

Macavity – The Mystery Cat Summary

“Macavity, The mystery cat”, is a humorous poem from a serious poet. The poet describes the atrocities done by Macavity. He commits crime after crime with impunity and without leaving any evidence. He looks for the opportune moment to commit the crime and gets away before the Scotland yard Police troops come in. Every time he does a crime, he is sensible enough to stay out of the spot of crime when investigators reach the spot of crime. He breaks every human law and even the law of gravity.

His powers of levitation startles even a fakir. He is always preoccupied with some serious thought. He has dirty coat and unkempt hair like modem gypsies. He is a hypocrite who appears to be respectable but does every mean act. He loots food, ransacks the jewel case and breaks the green house glass. But it is amazing that nobody is able to find him in the scene of crime. It looks as if all the vile cat in the area act as per his script. Like cold-blooded criminals, he always makes up an alibi every time he commits a criminal act. He is the chief of all wicked cats. He resembles “Napoleon of crime”, the chief of criminal act.

Macavity – The Mystery Cat Glossary

Textual:
Admiralty – a Government Department that administered the British Navy
alibi – a claim Of evidence that one was elsewhere when a Crime was committed
bafflement total confusion
deceitfulness – cunnmgness
defy – to resist or to challenge
depravity – evil quality; immorality
fakir – a holy person who live son alms and has the power to levitate
feline – cat
fiend – demon
Flying Squad – a police force ready to plunge into action
ginger – alert and cautious
larder – cupboard for storing food
levitation – the action of rising and floating in air
Peke – a Pekinese dog
rifle – ransack; to steal
Scotland Yard – the headquarters of London Metropolitan Police Service
stifle – suppress someone from acting; restrain
suavity – confidence and sophistication
trellis – wooden bar used as a support for creepers

Additional:
atrocities – cruelty; violation; wrongdoing
complicated – tough
criminal – law breaker
go astray – become useless; trangress
hypocrite – pretender, one who acts like adifferent person
impunity – doing wicked things without ever getting punished;unpunished
investigate – enquire
limerick – a humourous verse of five lines
startles – amazes
treaty – pact
vile – wicked