## Online Education MCQ Questions for Class 10 Economics Chapter 4 Globalisation and the Indian Economy with Answers

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## Online Education for Globalisation and the Indian Economy Class 10 MCQs Questions with Answers

Class 10 Economics Chapter 4 MCQ With Answers Question 1.
Which one of the following organisations lag stress on liberalisation of foreign trade and foreign investment?
(a) International Monetary Fund
(b) International Labour Organisation
(c) World Health Organisation

Explanation: The World Trade Organisation was created with an aim to regulate and monitor if appropriate rules were being implemented by the countries.

Globalisation And The Indian Economy Class 10 MCQ Question 2.
Choose the correctly matched pair:
(a) Sundaram Fasteners – Indian MNC
(b) Ranbaxy-Chinese MNC
(c) Ford Motors-British MNC
(d) Xiaomi-Korean MNC
(a) Sundaram Fasteners-Indian MNC
Explanation: Ranbaxy-Indian MNC
Ford Motors-American MNC
Xiaomi-Chinese MNC

Globalisation Class 10 MCQ Question 3.
Entry of MNCs in a domestic market may prove harmful for
(a) Large scale producers
(b) Domestic producers
(c) Native Bankers
(d) Exporters of goods
(b) Domestic producers

Explanation: The entry of MNC into the domestic market will prove fruitful to native bankers, large scale producers as they will be involved in the manufacturing and exporters as export will improve. The domestic producers alone will be harmed as it might close down their production.

Globalisation MCQ Class 10 Question 4.
Which one of the following has been major source of foreign investment in the IT industry?
(a) Bharat Heavy Electricals Limited
(b) Oil India Limited
(c) Steel Authority of India Limited

Explanation: The other choices are all manufacturing units.

Which among Mex, RCH2X, R2CHX , and R3CX is most reactive towards sn2 reaction​.

Globalisation And The Indian Economy MCQ Question 5.
Which one of the following types of countries has been more benefited from Globalisation?
(a) Rich countries
(b) Poor countries
(c) Developing countries
(d) Developed countries
(c) Developing countries

Explanation: Employment has been generated to the youth in many developing countries, which means better quality of living.

MCQ On Globalisation Class 10 Question 6.
Which one of the following Indian industries has been hit hard by globalisation?
(a) IT
(b) Toy making
(c) Jute
(d) Cement
(b) Toy making

MCQ Of Globalisation Class 10 Question 7.
Removing barriers on or restrictions set by the government is known as:
(a) Globalisation
(b) Privatisation
(c) Nationalisation
(d) Liberalisation
(d) Liberalisation

Globalization Class 10 MCQ Question 8.
Ford Motors set up its first plant in India at
(a) Kolkata
(b) Mumbai
(c) Chennai
(d) Delhi

Class 10 Globalisation MCQ Question 9.
…………. has helped most in the spread of production of services.
(a) Manufacturing Industry
(b) Automobile Sector
(c) Call centres
(d) IT Sector
(d) IT Sector

Globalisation Refers To MCQ Question 10.
“MNCs keep in mind certain factors before setting up production”. What of the following statements is wrong when that assertion is concerned?
(a) MNCs keep the availability of cheap skilled and unskilled labour in mind.
(b) Proximity to markets is another important factor when MNCs are established.
(c) Presence of a large number of local competitors also changes the opinions of people while establishing an MNC.
(d) Favourable government policies are important for establishment of an MNC.
(c) Presence of a large number of local competitors also changes the opinions of people while establishing an MNC. Explanation: MNCs are not interested in the local competitors as most of their products are for the export market.

Globalization Refers To MCQ Question 11.
…………… have led to a huge reduction in port handling costs and increased the speed with which exports can reach markets.
(a) Containers
(b) Airports
(c) Vessels with Stands
(d) Ships
(a) Containers

Globalization MCQs Class 10 Question 12.
(a) the slowdown in the agricultural sector
(b) the slowdown in demands of domestic goods in the country
(c) the slowdown in the industrial sector
(d) slowdown in services
(b) the slowdown in demands of domestic goods in the country

MCQ Of Globalisation And Indian Economy Question 13.
Which of the following country is NOT in the list of countries whom Ford Motors exporting its cars?
(a) South Africa
(b) Mexico
(c) Brazil
(d) All of these
(d) All of these

Explanation: Ford Motors, an American company, is one of the world’s largest automobile manufacturers with production spread over 26 countries of the world.

Related Theory
Ford Motors came to India in 1995, invested Rs.1700 crore to set up a plant near Chennai in collaboration with Mahindra and Mahindra, manufacturer of jeeps and trucks in India.

MCQ Globalisation Class 10 Question 14.
Which of the following company is the largest producer of edible oil in India?
(a) Parakh Foods
(b) Cargill Foods
(c) Mahindra and Mahindra
(d) None of the above
(b) Cargill Foods

Explanation: Cargill Foods is a very large American MNC, producing 5 million pouches of edible oil in India daily.

Related Theory
Cargill has bought over Parakh Foods which had built a large marketing network in various parts of India, also it had four oil refineries whose control has now shifted to Cargill.

Class 10 Economics Chapter 4 MCQ Question 15.
Which of the following statement is NOT correct about a multinational company?
(a) It set up new factories for production.
(c) It forms partnerships with local companies.
(d) It does not have new technology and large huge money with it.
(d) It does not have new technology and huge money with it.

Globalization And The Indian Economy Class 10 MCQ Question 16.
Which one of the following companies is NOT a multinational company?
(a) Tata Motors
(b) Reliance India Ltd.
(c) Ranbaxy
(d) Infosys
(b) Reliance India Ltd.

Explanation: Globalisation has enabled some large Indian companies to emerge as multinationals themselves. For example, Tata Motor (automobiles), Infosys (IT), Ranbaxy (medicines), Asian Paints (Paints), Sundaram Fasteners (nuts and bolts) are some Indian companies which are spreading their operations worldwide.

Question 17.
With what objective was World Trade Organisation set up?
(b) To allow free trade for all
(c) To establish rules regarding international trade
(d) All of these

Question 18.
Which of the following statement clearly defines what an MNC is?
(a) A large company that owns or controls production in more than one nation.
(b) A large company that owns or controls raw material for production.
(c) A large company that deals in automobiles.
(d) All of these.
(a) A large company that owns or controls production in more than one nation.

Question 19.
The most common route for investments by MNCs in countries around the world is to:
(a) set up new factories
(c) form partnerships with local companies
(d) none of these

Question 20.
Which of the following attributes are NOT determined by an MNC?
(a) Price, Quality, delivery and raw material
(b) Price, labour conditions, quality, delivery
(c) Delivery and price only
(d) Raw material and transport cost
(b) Price, labour conditions, quality, delivery

Question 21.
Large MNCs in developed countries place orders for production with small producers which include:
(a) Garments
(b) Footwear
(c) Sports
(d) All of these
(d) All of these

Identify

Question 22.
Identify the following organization on basis of the hints given.
(1) It was established with the aim to liberalise international trade.
(2) 164 countries are its members.

Correct and Rewrite/ True-False
State whether the following statements are True or False. If false, correct the statement.

Question 23.
The removal of barriers or restrictions by the government is known as globalisation.
The removal of barriers or restrictions by the government is called liberalisation. Explanation: Liberalisation helps to remove all types of barriers to allow easy and affordable exchange of ideas, culture and even goods between countries.

Related Theory
Globalisation is the process of rapid integration or interconnection between countries by exchange of trade, ideas, culture, information and even markets.

Question 24.
International Monetary Fund (IMF) is an organisation whose aim is to liberalise international trade. [CBSE 2020,15]
WTO is an organisation whose aim is to liberalise international trade.

Explanation: World Trade Organisation (WTO) forms uniform rules and regulations to make trade simpler across the world.

Related Theory
International Monetary Fund was created to support economic stabilisation of Second World War economies. Nations were hurt and affected and hence needed funding and support to develop again.

Uniform Distribution Calculator solves two basic tasks.

Fill in the blanks with suitable information:

Question 25.
………… results in connecting the markets or integration of markets in different countries.

Explanation: ForeignTrade helps manufacturers to find markets in various countries. It results in tax on imports which governments can use to increase or decrease foreign trade and to decide what kind of goods and how much of each, should come into the country.

Question 26.
………………… is the process of rapid integration or interconnection between countries.
Globalisation

Question 27.
………. refers to some restrictions including the tax on Imports which governments can use to increase or decrease foreign trade and to decide what kind of goods and how much of each, shouLd come into the country.

Explanation: Indian government put barriers on foreign trad and foreign investment after independence to protect the producers within the country from foreign competition.

Question 28.
………… is the freedom of markets and abolition of state imposed restrictions on the movement of goods.
Liberalisation

Question 29.
……….. is an international organisation whose aim is to liberalise international trade:
(a) WTO
(b) UNDP
(c) World Bank
(d) MNC

Match the Columns Choose the correct pairs:

Question 30.
Match the following Indian companies from column A with the products these companies deal in from column B:

 Column A (Indian Companies) Column B (Products these companies deal in) (a) Tata Motors (i) IT (b) Infosys (ii) Automobiles (c) Ranbaxy (iii) Paints (d) Asian Paints (iv) Medicines

 Column A (Indian Companies) Column B (Products these companies deal in) (a) Tata Motors (ii) Automobiles (b) Infosys (i) IT (c) Ranbaxy (iv) Medicines (d) Asian Paints (iii) Paints

Explanation: Globalisation has enabled these large Indian companies to emerge as MNCs. Another company that has become an MNC is Sundaram Fasteners which deals in nuts and bolts.

Question 31.
Match the following key terms given in column A with their meanings in column B:

 Column A (Key Terms) Column B (Meanings) (a) Globalisation (i) Investment by multinationals (b) Liberalisation (ii) Controlling foreign trade (c) Tax barrier (iii) Integration between economies (d) Foreign investment (iv) Removing trade barriers

### Assertion Reasoning questions Class 10 Economics Chapter 4

In each of the following questions, a statemant of Assertion (A) is given followed by a corresponding statement of Reason (R). Select the correct answer to codes (a), (b) (c) or (d) as given below:
(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 correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.

Question 32.
Assertion (A): People also can play an important role in the struggle for fair globalisation.
Reason (R): Governments have done exceptionally well in implementing the rules of WTO.
(c) (A) is correct but (R) is wrong.

Explanation: Massive campaigns and representation by people’s organisations have influenced important decisions relating to trade and investments at the WTO.

Question 33.
Assertion (A): Not everyone has benefited from Globalisation equally.
Reason (R): People with education, skill and wealth have made the best use of the new opportunities while others have not been able to make the best use of it.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Question 34.
Assertion (A): Several of the top Indian companies have been able to benefit from the increased competition.
Reason (R): They have been able to get more customers using the existing technology.
(c) (A) is correct but (R) is wrong.

Explanation: Greater competition has forced the companies to invest in newer technologies and more infrastructure, along with better quality products which has helped these companies to gain a better consumer base with help from the competition.

Question 35.
Assertion (A): Foreign companies are demanding more rigidity in labour laws.
Reason (R): Flexible Labour laws give better chances for companies to profit.

(Competency Based Questions (CBQs))

Question 1.
Mr. Samson buys a company and then proceeds to buy new machines and furniture for the new company. He buys all his new employees clothes or new uniforms. He buys them new computers and redecorates the entire office.
Which one of the following statements is correct considering the source?
(a) Mr Samson loaned the money to this new office.
(b) Mr Samson spent money on social customs.
(c) Mr Samson invested this money because he bought the furniture, uniforms and machines.
(d) Mr Samson traded this money.
(c) Mr Samson invested this money because he bought the furniture, uniforms and machines.

Question 2.
A news magazine published for London readers is to be designed and printed in Delhi.
The text of the magazine is sent through the Internet to the Delhi office. The designers in the Delhi office get orders on how to design the magazine from the office in London using telecommunication facilities.
Which of the following phenomena helps these companies to execute these activities?
(c) Globalisation
(d) Investment

Question 3.
Choose the correct statement about factors of globalisation in India:
(I) Improvement in transportation techno¬logy.
(II) Liberalisation of foreign trade and foreign investment.
(III) Favourable rules of World Trade Organisation towards India in comparison to developed countries.
Choose the correct options from the codes given below:
(a) Only (I) and (II)
(b) Only (I) and (III)
(c) Only (II) and (III)
(d) Only (III)
(a) Only (I) and (II)

Explanation: WTO is supposed to allow free trade for all. However, in practice, it is seen that developed countries have retained trade barriers unfairly. On the other hand, WTO rules have forced developing countries to remove trade barriers. It is not partial in favour of India because it is an international organisation initiated by developed countries.

Question 4.
The above evidence indicates that not everyone has benefited from gLobaiisation. People with education, skill and wealth have made the best use of the new opportunities. On the other hand, there are many people who have not shared the benefits. Since globalisation is now a reality, the question is how to make globalisation more ‘fair’? Fair globalisation would create opportunities for all, and also ensure that the benefits of globalisation are shared better.
Identify which of the following statements is not true about Globalization.
(a) Globalisation is the process of rapid integration between countries.
(b) It is movement of more and more investments and technology between countries.
(c) Globalisation is introduction of restrictions or barriers by the government.
(d) It is movement of more and more goods and services between countries.
(c) Globalisation is introduction of restrictions or barriers by the government.

Explanation: Globalisation is the process of rapid integration or interconnection between the countries. Through Globalisation more and more goods, services, greater foreign investments, technology and movement of people are taking place between countries. Globalisation doesn’t call for restrictions or barriers. It rather brings the countries in closer contact with each other.

Question 5.
Ford Motors, an American company, is one of the world’s largest automobile manufacturers with production spread over 26 countries of the world. Ford Motors came to India in 1995 and spent Bs. 1700 crore to set up a large plant near Chennai. This was done in collaboration with Mahindra and Mahindra, a major Indian manufacturer of jeeps and trucks. By the year 2017, Ford Motors was selling 88,000 cars in the Indian markets, while another 1,81,000 cars were exported from India to South Africa, Mexico, Brazil and the United States of America. The company wants to develop Ford India as a component supplying base for its other plants across the globe.
Answer the following MCQs by choosing the most appropriate option:
(A) Fill in the blank with an appropriate option:
Ford Motors is a Company.
(a) Uni-national
(b) Bi-national
(c) Regional
(d) Multi-national
(d) Multi-national

Explanation: Its production units are present in various countries and hence it is a MNC.

(B) Which of the following local companies did Ford Motors collaborate with?
(a) Hero motors
(b) Honda and Suzuki
(c) Mahindra and Mahindra
(d) Volkswagen

(C) Which of the following benefits are availed by the local company on collaborating with Ford Motors?
(a) The demand for cars from local companies in India rises.
(b) Ford Motors supplies new technology to the local company.
(c) Ford Motors takes over the local company.
(d) Ford Motors changes the management of local companies.
(b) Ford Motors suppiies new technology to the local companies.

(D) Which of the following statements best describe the benefit to Indian producers in this statement in simple words?
“By the year 2017, Ford Motors was selling 88,000 cars in the Indian markets, while another 1,81,000 cars were exported from india.”
(a) The demand of products after the joint collaboration has decreased.
(b) The demand of products after the joint collaboration has increased.
(c) The demand of products after the joint collaboration has remained the same.
(d) The demand of products after the joint collaboration has completely shifted.
(b) The demand of products after the joint collaboration has increased.

Question 6.
Even more remarkable have been the developments in information and communication technology. In recent times, technology in the areas of telecommunications, computers, and the Internet has been changing rapidly. Telecommunication facilities (telegraph, telephone including mobile phones, fax) are used to contact one another around the world, to access information instantly, and to communicate from remote areas. This has been facilitated by satellite communication devices.

As you would be aware, computers have now entered almost every field of activity. You might have also ventured into the amazing world of the internet, where you can obtain and share information on almost anything you want to know. The Internet also allows us to send instant electronic mail (e-mail) and talk (voice-mail) across the world at negligible costs.
Answer the following MCQs by choosing the most appropriate option:
(A) Which of the following factors have experienced a change due to the Internet?
(a) All the information in the world can be accessed at the click of a finger.
(b) One can fly from one country to another.
(c) One can establish a telephonic contact between a man in India and Alaska.
(d) One can cook by watching a show on TV.

(B) Which of the following ways has the IT sector helped in?
(a) In production of goods
(b) In production of services
(c) In export of goods
(d) In export of services
(b) In production of services

Explanation: Information and communication technology (or IT in short) has played a major role in spreading out production of services like Banking, teaching, hospitality across countries.

(C) Fill in the blank with an appropriate option:
…………………. is not a Telecommunication facility?
(a) Telegraph
(b) Mobiles
(c) Emails
(d) A cars
(d) A Car

Explanation: It is a transportation facility.

(D) Complete the following:
Boating- Development in transportation facilities; E- Banking
(a) Development in Transport
(c) Development in Production
(d) Development in IT

Question 7.
Having assured themselves of these conditions, MNCs set up factories and offices for production. The money that is spent to buy assets such as land, building, machines and other equipment is called investment. Investment made by MNCs is called foreign investment. Any investment is made with the hope that these assets will earn profits.

At times, MNCs set up production jointly with some of the local companies of these countries. The benefit to the local company of such joint production is two-fold. First, MNCs can provide money for additional investments, like buying new machines for faster production. Second, MNCs might bring with them the latest technology for production.
Answer the following MCQs by choosing the most appropriate option:
(A) Statements are given below. Find the correct ones which suit the statement given in the questions.
Investment means spending on:
(I) factories and industrial building
(II) machines
(III) equipments
(IV) gold ornaments for employees
(V) Uniforms
(VI) Vehicles for the company
(a) (I) & (IV)
(b) (I) & (III)
(c) (II), (III) & (IV)
(d) (I), (II), (III), (V) & (VI)
(d) (I), (II), (III), (V) & (VI)

(B) FDI (Foreign Direct Investment) attracted by Globalisation in India comes from the:
(a) World Bank
(b) Multinational companies
(c) foreign governments
(d) Indian residents
(b) Multinational companies

(C) How does Joint Production with Zara help a local company called Style Quotient?
(a) It brings better technology and production methods to Style Quotient.
(b) Zara gets a new manager.
(c) Style Quotient gets a better transport vehicle.
(d) Employees at Style Quotient get discounts at Zara Store.
(a) It brings better technology and production methods to Style Quotient.

(D) Which one of the following is a major benefit (to the MNC) of joint production between a local company and a Multinational Company?
(a) MNC are able to earn more customers.
(b) MNC can control the increase in the price of a good.
(c) MNC can expand its office physically.
(d) MNC can promote the local company.
(a) MNC are able to earn more customers

Question 8.
35-year old Sushila has spent many years as a worker in the garment export industry of Delhi. She was employed as a ‘permanent worker’ entitled to health insurance, provident fund, overtime at a double rate, when Sushila’s factory closed in the late 1990s. After searching for a job for six months, she finally got a job 30 km away from where she lives. Even after working in this factory for Several years, she is a temporary worker and earns less than half of what she was earning earlier. Sushila leaves her house every morning, seven days a week at 7:30 a.m. and returns at 10 p.m. A day off from work means no wage. She has none of the benefits she used to get earlier. Factories closer to her home have widely fluctuating orders and therefore pay even less. Answer the following MCQs by choosing the most appropriate option:
(A) Has Globalisation been favourable to Sushila and other workers like her?
(a) Yes, there are new jobs
(b) No, there is greater unemployment because people now employ workers on a seasonal basis.
(c) There is no change in employment opportunities.
(d) Workers are not affected by Globalisation.
(b) No, there is greater unemployment because people now employ workers on a seasonal basis.

(B) Fill in the blank with an appropriate option:
Globalisation so far has been more in favour of
(a) Developed countries
(b) Underdeveloped countries
(c) Poor countries
(d) It has had an equal effect.
(a) Developed countries

Explanation: They have been able to expand their markets and their consumer bases.

(C) Which one of the following has benefited least because of globalisation in India?
(a) Industrial Sector
(b) Agriculture Sector
(c) Service Sector
(d) IT Sector
(b) Agriculture Sector

(D) Which of the following statements are not true about Sushila’s employment terms?
(a) She gets more benefits like Medical leave due to Globalisation.
(b) She has to work longer than usual.
(c) Her wages have doubled.
(a) She gets more benefits like Medical leave due to Globalisation.
Explanation: all her existing terms have been cancelled and now she has no extra benefits.

Question 9.
Read the sources given below and answer the questions that follow: [CBSE 2020,14] Source A – Production across countries
Until the middle of the twentieth century, production was largely organised within countries. What crossed the boundaries of these countries were raw material, food stuff and finished products. Colonies such as India exported raw materials and food stuff and imported finished goods. Trade was the main channel connecting distant countries. This was before large companies called multinational corporations (MNCs) emerged on the scene. Source B – Foreign trade and integration of markets

Foreign trade creates an opportunity for the producers to reach beyond the domestic
markets, i.e., markets of their own countries. Producers can sell their produce not only in markets located within the country but can also compete in markets located in other countries of the world. Similarly, for the buyers, import of goods produced in another country is one way of expanding the choice of goods beyond what is domestically produced.

Source C – Impact of globalisation in India Globalisation and greater competition among producers – both local and foreign producers – has been of advantage to consumers, particularly the well-off sections in the urban areas. There is greater choice before these consumers who now enjoy improved quality and lower prices for several products. As a result, these people today, enjoy much higher standards of living than was possible earlier. Source A – Production across countries
(A) How are MNCs a major force in connecting the countries of the world?
Source B – Foreign trade and integration of markets
MNCs act as a major force in connecting various countries of the world.
They interact with Local producers in various counties across the globe to expand their production and markets which results in connection of widely dispersed locations on countries.

(B) How does foreign trade become a main channel in connecting countries?
Source C – Impact of globalisation in India
Foreign trade helps to integrate various markets with one another through the means of domestic and international producers who sells the same product in domestic and international markets. The markets of all countries garner similar products and are integrated together.

(C) How is globalisation beneficial for consumers?
Globalisation is beneficial for consumers in the following ways:

1. High quality goods are available at lower prices.
2. A wide range of choice is available for the consumers.

Question 10.
Read the sources given below and answer the questions related to them :
Source A – Globalisation and the Indian economy
An consumers in today’s world, some of us have a wide choice of goods and services before us. The latest models of digital cameras, mobile phones and televisions made by the leading manufacturers of the world are within our reach. Every season, new models of automobiles can be seen on Indian roads. Source B – Foreign trade and integration of markets
Foreign trade creates an opportunity for the producers to reach beyond the domestic markets, i.e., markets of their own countries. Producers can sell their produce not only in markets located within the country but can also compete in markets located in other countries of the world. Similarly, for the buyers, the import of goods produced in another country is one way of expanding the choice of goods beyond what is domestically produced. Source C – The struggle for fair globalisation In the past few years, massive campaigns and representation by people’s organisations has influenced important decisions relating to trade and investments at the World Trade Organisation. This has demonstrated that people can also play an important role in the struggle for fair globalisation.
Source A – Globalisation and the Indian economy
(A) How is the impact of globalisation visible on consumers?
Source B – Foreign trade and integration of markets

(B) How does foreign trade integrate markets? Explain.
Source C – The struggle for fair globalisation

(C) How do people play an important role in the struggle for fair globalisation? Explain.

Question 1.
State any one example for a trade barrier?
Taxes and Import duties

Question 2.
Why did the Indian government remove barriers to a large extent on foreign trade and foreign investment?
They removed the trade barriers to help the native industries and boost their trade by letting the foreign competition in the country’s market.

Question 3.
What attracts the foreign investment?

Question 4.
Due to which reason the latest models of different items are available within our reach?
Foreign Trade integrates different markets and helps the countries to make products from across the world available to their citizens.

Question 5.
Why are MNC setting them Customer Care Centres in India?
India has a lot of educated youth who can speak English well and thereby are the perfect candidates to be customer care professionals.

Question 6.
Differentiate between multinational corporations and domestic companies.
Multinational corporations operate in two or more countries while domestic companies restrict their operations to a single country.

Related Theory:
Multinational companies move to other countries and expand their business for various reasons. They also give a tough competition to domestic companies.

Question 7.
What is the basic function of foreign trade?

Question 8.
The MNC’s of a country sets up a production jointly with the local company of other country. State any one benefit of this joint production to the local company.
Benefits of Joint Production:

1. MNC can provide money for additional investment.
2. MNC might bring latest technology for production, (any 1 to be mentioned)

Question 9.
Distinguish between investment and foreign investment.
When land, assets or buildings are purchased and kept hold of, they are considered as investments, whereas when multinational corporations (MNC’s) invest money or finances in a project of another company in another country, it is considered as foreign investment.

Question 10.
Due to which reason are the latest models of different items available within our reach?
The latest models of different items are available within our reach due to ‘globalisation.

Question 11.
Define globalisation.
Globalisation is defined as the integration between different countries through foreign trade and foreign investments by various muLti-national corporations (MNCs).

Related Theory:
Globalisation has helped a lot in interlinking economies and markets across the world. Now, people are able to sell their produce in the foreign markets also.

Question 12.
Where do MNCs prefer to set up their offices and factories?
MNCs set up offices and factories for production in regions where they can get cheap labour and other resources.

Explanation: Every MNC wants to attain maximum profit off their business. One of the easiest way to do is to reduce the cost of production by either updating the infrastructure, to include better tecnnology, or by employing cheap labour, finding cheaper sources of energy to be used.

Question 13.
How has globalisation enabled the national companies to emerge as Multinational Companies (MNCs)?
Globalisation enabled the national companies to emerge as MNCs:
New opportunities have been provided to the companies/ Helped in collaboration with foreign companies/ Technological and production inputs.

Question 14.
Which is the most common route for investment by MNCs in countries around the world?
The easiest way for MNCs is to invest around the world is to buy up local companies and then to expand production by controlling the means.

Question 15.
Define the term Liberalisation.
Liberalisation refers to the removal or partial or complete loosening of the restrictions set by the government on foreign investment and trade. It also eases up the process of import and export.

Question 16.
Name some Indian companies which are now emerging as Multi National Corporations.
Tata motors, Infosys, Ranbaxy, Asian paints and Sundaram fasteners are some Indian companies which are spreading their operations worldwide.

Explanation: MNCs or Multi National Corporations are companies which are based in various countries and their market and production are both expanded across various countries and continents.

Question 17.
Define the term ‘investment’.
The money that is spent to buy assets such as land, building, machines and other equipment is called investment.

Explanation: Any investment is made with the hope that these assets will earn profits.

Related Theory:
Investment made by MNCs is called foreign investment.

Question 18.
In recent years, the central and state governments in India are taking special steps to attract foreign companies to invest in India. Companies who set up production units in the SEZs do not have to pay taxes for an initial period of five years.

Question 19.
What are SEZs?
SEZs are Special Economic Zones or Industrial Zones set up to attract foreign investment by government of India.

Related Theory:
SEZs have world class infrastructural facilities and other facilities like tax rebates, flexibility in labour laws to attract investors.

## Online Education MCQ Questions for Class 8 Hindi Chapter 3 बस की यात्रा with Answers

Check the below Online Education NCERT MCQ Questions for Class 8 Hindi Vasant Chapter 3 बस की यात्रा with Answers Pdf free download. MCQ Questions for Class 8 Hindi with Answers were prepared based on the latest exam pattern. We have provided बस की यात्रा Class 8 Hindi MCQs Questions with Answers to help students understand the concept very well. https://ncertmcq.com/mcq-questions-for-class-8-hindi-with-answers/

Students can also read NCERT Solutions for Class 8 Hindi Chapter 3 Questions and Answers at LearnInsta. Here all questions are solved with a detailed explanation, It will help to score more marks in your examinations.

## Online Education for बस की यात्रा Class 8 MCQs Questions with Answers

Bus Ki Yatra MCQ Class 8 Question 1.
कुल कितने लोग शाम की बस से यात्रा करने वाले थे?
(a) तीन
(b) चार
(c) पाँच
(d) छह

Class 8 Hindi Chapter 3 MCQ Question 2.
इस पाठ के लेखक कौन हैं?
(a) भगवती चरण वर्मा
(b) राम दरश मित्र
(c) कामतानाथ
(d) हरिशंकर परसाई

Bas Ki Yatra MCQ Class 8 Question 3.
पन्ना से सतना के लिए बस कितनी देर बाद मिलती है?
(a) आधा घंटा
(b) एक घंटे बाद
(c) दो घंटे बाद
(d) प्रातः काल

MCQ Questions For Class 8 Hindi Chapter 3 Question 4.
यह बस कहाँ की ट्रेन मिला देती है?
(a) सतना की
(b) पन्ना की
(c) जबलपुर की
(d) भोपाल की

Class 8 Hindi Ch 3 MCQ Question 5.
उस बस में कंपनी के कौन सवार थे?
(a) चौकीदार
(b) हिस्सेदार
(c) दावेदार
(d) इनमें से कोई नहीं

बस की यात्रा MCQ Class 8 Question 6.
इस पाठ में गांधी जी के किस आंदोलन का उल्लेख है?
(a) असहयोग आंदोलन
(b) सविनय अवज्ञा आंदोलन
(c) उपर्युक्त दोनों
(d) इनमें से कोई नहीं

Ncert Class 8 Hindi Chapter 3 MCQ Question 7.
लेखक हरे-भरे पेड़ों को क्या समझता था?
(a) जीवनदाता
(b) मित्र
(c) शत्रु
(d) शुभचिंतक

Ch 3 Hindi Class 8 MCQ Question 8.
‘समझदार आदमी’ रेखांकित शब्द क्या है?
(a) संज्ञा
(b) सर्वनाम
(c) क्रिया
(d) विशेषण

Class 8 Chapter 3 Hindi MCQ Question 9.
‘उत्सर्ग’ शब्द कैसा है?
(a) तत्सम
(b) तद्भव
(c) देशज
(d) विदेशी

Hindi Class 8 Chapter 3 MCQ Question 10.
‘फर्स्ट क्लास’ शब्द निम्नलिखित में से किस प्रकार का शब्द है-
(a) आगत
(b) तत्सम
(c) देशज
(d) तद्भव

(1)

हम पाँच मित्रों ने तय किया कि शाम चार बजे की बस से चलें। पन्ना से इसी कंपनी की बस सतना के लिए घंटे भर बाद मिलती है जो जबलपुर की ट्रेन मिला देती है। सुबह घर पहुँच जाएँगे। हम में से दो को सुबह काम पर हाज़िर होना था इसीजिए वापसी का यही रास्ता अपनाना ज़रूरी था। लोगों ने सलाह दी कि समझदार आदमी इस शाम वाली बस से सफ़र नहीं करते। क्या रास्ते में डाकू. मिलते हैं ? नहीं, बस डाकिन है।
बस को देखा तो श्रद्धा उमड़ पड़ी। खूब वयोवृद्ध थी। सदियों के अनुभव के निशान लिए हुए थी। लोग इसलिए इससे सफ़र नहीं करना चाहते कि वृद्धावस्था में इसे कष्ट होगा।

Bus Ki Yatra Class 8 MCQ Question 1.
उपर्युक्त गद्यांश के पाठ का नाम और लेखक का नाम लिखिए।
(a) गद्यांश के पाठ का नाम- बस की यात्रा
लेखक का नाम- हरिशंकर परसाई।
(b) लेखक- कामतानाथ, पाठ-लाख की चूड़ियाँ
(c) पाठ- भगवान के डाकिए, लेखक- रामधारी सिंह दिनकर
(d) पाठ- कामचोर, इस्मत चुगलाई।

(a) गद्यांश के पाठ का नाम- बस की यात्रा
लेखक का नाम- हरिशंकर परसाई।

Class 8 Ch 3 Hindi MCQ Question 2.
लेखक के मन में बस को देखकर कैसा भाव उमड़ा?
(a) प्रेम
(b) श्रद्धा
(c) दया
(d) इनमें से कोई नहीं

Bas Ki Yatra Class 8 MCQ Question 3.
लेखक और उसके मित्रों को कहाँ जाना था?
(a) सतना
(b) जबलपुर
(c) पन्ना
(d) रायगढ़

Chapter 3 Hindi Class 8 MCQ Question 4.
यात्री इस बस में सफ़र क्यों नहीं करना चाहते थे?
(a) क्योंकि बस के चलने के आसार ही दिखाई नहीं देते थे।
(b) क्योंकि बस में सीटें बहुत कम थी।
(c) क्योंकि बस अपनी जर्जर अवस्था के कारण नहीं धोखा दे सकती थी।
(d) इनमें से कोई नहीं

Answer: (c) क्योंकि बस अपनी जर्जर अवस्था के कारण नहीं धोखा दे सकती थी।

Class 8 Hindi Chapter 3 Extra Questions Question 5.
यह बस कहाँ की ट्रेन से मिला देती है?
(a) सतना की
(b) जबलपुर की
(c) पन्ना की
(d) भोपाल की

Class 8 Hindi Bus Ki Yatra MCQ Question 6.
समझदार आदमी में समझदार शब्द है
(a) संज्ञा की
(b) सर्वनाम की उत्तर
(c) क्रिया की
(d) विशेषण की

(2)

इंजन सचमुच स्टार्ट हो गया। ऐसा, जैसे सारी बस ही इंजन है और हम इंजन के भीतर बैठे हैं। काँच बहुत कम बचे थे। जो बचे थे, उनसे हमें बचना था। हम फ़ौरन खिड़की से दूर सरक गए। इंजन चल रहा था। हमें लग रहा था कि हमारी सीट के नीचे इंजन है। बस सचमुच चल पड़ी और हमें लगा कि यह गांधी जी के असहयोग और सविनय अवज्ञा आंदोलनों के वक्त अवश्य जवान रही होगी। उसे ट्रेनिंग मिल चुकी थी। हर हिस्सा दूसरे से असहयोग कर रहा था। पूरी बस सविनय अवज्ञा आंदोलन के दौर से गुजर रही थी। सीट का बॉडी से असहयोग चल रहा था। कभी लगता सीट बॉडी को छोड़कर आगे निकल गई है। कभी लगता कि सीट को छोड़कर बाडी आगे भागी जा रही है। आठ-दस मील चलने पर सारे भेदभाव मिट गए। यह समझ में नहीं आता था कि सीट पर हम बैठे हैं या सीट हम पर बैठी है।

Question 1.
इंजन सचमुच स्टार्ट हो गया वाक्य में लेखक का कहने का अभिप्राय क्या है?
(a) बस की सुंदर स्थिति के कारण
(b) ड्राइवर के दयनीय स्थिति को देखकर
(c) बस की दशा और पहलीबार में ही स्टार्ट होने के कारण
(d) बस की हालत को देखकर

Answer: (d) बस की हालत को देखकर

Question 2.
लेखक को ऐसा क्यों लग रहा था कि हम इंजन के भीतर बैठे हैं?
(a) सरदी के कारण
(b) परेशानी के कारण
(c) उनके उम्र के कारण
(d) शोर और कंपन के कारण

Answer: (d) शोर और कंपन के कारण

Question 3.
उपरोक्त गद्यांश में गांधी जी के किस आंदोलन का वर्णन है?
(a) असहयोग आंदोलन
(b) सविनय अवज्ञा आंदोलन
(c) उपर्युक्त दोनों
(d) कोई नहीं

Question 4.
गद्यांश में बस की दशा के बारे में क्या पता चलता था?
(a) बस अच्छी हालत में थी
(b) बस की हालत दयनीय थी
(c) बस खराब थी
(d) पता नहीं

Answer: (b) बस की हालत दयनीय थी

Question 5.
आठ-दस मील के बाद बस की चाल में क्या परिवर्तन आया?
(a) बस का टायर खराब हो गया
(b) बस खराब हो गई
(c) बस सहज हो गई, बिलकुल आराम से चलने लगी
(d) बस काफ़ी तेज़ी से चलने लगी।

Answer: (c) बस सहज हो गई, बिलकुल आराम से चलने लगी

(3)

बस की रफ्तार अब पंद्रह-बीस मील हो गई थी। मुझे उसके किसी हिस्से पर भरोसा नहीं था। ब्रेक फेल हो सकता है, स्टीयरिंग टूट सकता है। प्रकृति के दृश्य बहुत लुभावने थे। दोनों तरफ़ हरे-भरे पेड़ थे जिन पर पक्षी बैठे थे। मैं हर पेड़ को अपना दुश्मन समझ रहा था। जो भी पेड़ आता, डर लगता कि इससे बस टकराएगी। वह निकल जाता तो दूसरे पेड़ का इंतज़ार करता। झील दिखती तो सोचता कि इसमें बस गोता लगा जाएगी।

Question 1.
अब बस किस रफ्तार से चल रही थी?
(a) पंद्रह से बीस मील प्रति घंटा
(b) आठ से दस मील रफ़्तार
(c) दस से बारह मील रफ़्तार
(d) चार से पाँच मील रफ्तार

Answer: (a) पंद्रह से बीस मील प्रति घंटा

Question 2.
लेखक को बस पर भरोसा क्यों नहीं रहा?
(a) क्योंकि बस का ड्राइवर नशे में था
(b) क्योंकि रास्ता काफ़ी खराब था
(c) क्योंकि लेखक ने सोच लिया कि बस का कभी भी ब्रेक फेल हो सकता है, या स्टीयरिंग टूट सकता है।

Answer: (c) क्योंकि लेखक ने सोच लिया कि बस का कभी भी ब्रेक फेल हो सकता है, या स्टीयरिंग टूट सकता है।

Question 3.
लेखक पेड़ों को अपना शत्रु क्यों समझ रहे थे?
(a) पेड़ों के हरे-भरे होने के कारण
(b) पेड़ों के कारण रास्ता न दिखने के कारण
(c) पेड़ों से बस टकराने के भय के कारण
(d) अत्यधिक छायादार होने के कारण

Answer: (c) पेड़ों से बस टकराने के भय के कारण

Question 4.
लेखक को बस डूबने का डर कहाँ सताने लगा?
(a) पुलिया पर
(b) नदी में
(c) झील में
(d) समुद्र में

Question 5.
गद्यांश में लेखक ने सड़क के दोनों किनारे का दृश्य कैसे प्रस्तुत किया है?
(a) दोनों ओर हरे-भरे पेड़ थे जिस पर पक्षी बैठे थे।
(b) दोनों तरफ़ नदियाँ बह रही थीं।
(c) चारों तरफ़ काले-काले बादल आसमान में छाए थे।
(d) दोनों तरफ़ झीलें ही झीलें थीं।

Answer: (a) दोनों ओर हरे-भरे पेड़ थे जिस पर पक्षी बैठे थे।

(4)

एक पुलिया के ऊपर पहुँचे ही थे कि एक टायर फिस्स करके बैठ गया। वह बहुत ज़ोर से हिलकर थम गई। अगर स्पीड में होती तो उछलकर नाले में गिर जाती। मैंने उस कंपनी के हिस्सेदार की तरफ़ पहली बार श्रद्धाभाव से देखा। वह टायरों की हालत जानते हैं फिर भी जान हथेली पर लेकर इसी बस से सफ़र कर रहे हैं। उत्सर्ग की ऐसी भावना दुर्लभ है। सोचा, इस आदमी के साहस और बलिदान भावना का सही उपयोग नहीं हो रहा है। इसे तो किसी क्रांतिकारी आंदोलन का नेता होना चाहिए। अगर बस नाले में गिर पड़ती और हम सब मर जाते तो देवता बाँहें पसारे उसका इंतज़ार करते।

Question 1.
बस कहाँ खराब हो गई?
(a) झील के पास
(b) एक गाँव में
(c) पुलिया पर
(d) पुल के नीचे

Question 2.
लेखक ने बस कंपनी के हिस्सेदार को किस भाव से देखा?
(a) घृणा से
(b) श्रद्धा से
(c) प्यार से
(d) उपेक्षा से

Question 3.
किसके साहस और बलिदान की भावना का दुरुपयोग हो रहा था?
(a) यात्रियों की
(b) बस ड्राइवर की
(c) कंपनी के हिस्सेदारों की
(d) कंडक्टर की

Answer: (c) कंपनी के हिस्सेदारों की

Question 4.
लेखक के अनुसार क्रांति नेता में कौन से गुण होने चाहिए।
(a) ईमानदार और त्यागी
(b) सच्चाई और साहस
(c) त्याग और परोपकार
(d) साहस और बलिदान

We hope the given NCERT MCQ Questions for Class 8 Hindi Vasant Chapter 3 बस की यात्रा with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 8 Hindi बस की यात्रा MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

## Online Education for CBSE Maths Lab Manual Class 10 Activities Solutions

Online Education Maths Lab Manual Class 10: The student working in Maths Lab Manual Class 10 CBSE Pdf must rely to a large extent on methods of enquiry, trial and error in combination with his own curiosity, intuition, and ingenuity; nowhere else in any subject is rigorous proof is so often preceded by the patient, plodding experiments. If going occasionally becomes slow and difficult, one can take comfort in that a CBSE Maths Lab Manual Class 10 Free Download PDF available to increase his pace and make the subject enjoyable.

## Online Education for CBSE Class 10 Maths Lab Manual Activities Solutions

Our treatment is structured with a revised list of activities and projects so as to facilitates students and teachers to go through maths activities Chapter wise.

Viva-voce Questions (Very Short Answer Type Questions) and Multiple Choke Questions (MCQ) are incorporated at the end of each activity to check the basics of the activity.

Maths lab manual class 10 activities will serve the needs of students and teachers alike by

• Encouraging the students to carry out activities in a systematic manner, and
• Helping the teachers to evaluate the student’s creative skills.

The activities contained in the book have been chosen from the child’s own environment and will make both the learning and teaching of mathematics not only more fruitful but also much more interesting and lively.

### Lab Manual Class 10 Maths Activities Solutions

1. HCF of Two Numbers
2. Zeroes of Polynomials
3. System of Linear Equations
4. Basic Proportionality Theorem
5. Pythagoras Theorem
6. (A) A System of Similar Triangles
(B) Areas of Two Similar Triangles
7. Median
8. Mode
9. Arithmetic Progression (I)
10. Arithmetic Progression (II)
11. Arithmetic Progression (III)
12. Verification of Lengths of Tangents
13. Area of a Circle
14. Areas of Sectors
15. Comparison of Surface Areas
16. Comparison of Volumes of Right Circular Cylinders
17. Volume of a Right Circular Cylinder
18. A Right Circular Cone
19. Surface Area of a Cone
20. Volume of a Cone
21. Surface Area of a Sphere
22. Volume of a Sphere
23. Centroid of a Triangle
24. (A) Idea of Probability
(B) Determination of Experimental Probability
25. Measurement of Height Using Clinometer

### Syllabus for Lab Manual Class 10 Maths

 Subjects 80 Marks (Board Examination) 20 Marks (Internal Assessment) The student has to secure 33% marks out of 80 marks in each subject The student has to secure 33% marks out of overall 20 marks earmarked in each subject Periodic Test (10 Marks) Notebook Submission (5 Marks) Subject Enrichment Activity (5 Marks) (i) (ii) (ill) Mathematics Class-X Examination will be conducted for 80 marks in each subject covering 100% syllabus of the subject of Class-X only. Marks and Grades both will be awarded for individual subjects. 9-point grading will be the same as followed by the Board. Periodic written Test, restricted to three in each subject in an Academic Year. Average of the best two tests to be taken for final marks submission. This will cover • Regularity • Assignment Completion • Neatness and upkeep of notebook Maths Lab Practical

(i) Periodic Test (10 marks)
The school should conduct three periodic written tests in the entire academic year and the average of the best two will be taken. The schools have the autonomy to make their own schedule. However, for the purpose of gradient learning, three tests may be held as one being the mid-term test and the other the two being pre mid and post mid-term with a portion of syllabus cumulatively covered. The gradually increasing portion of contents would prepare students to acquire confidence for appearing in the Board examination with a 100% syllabus. The school will take the average of the best two tests for final marks submission.

(ii) Notebook Submission (5 marks)
Notebook submission as a part of internal assessment is aimed at enhancing the seriousness of students towards preparing notes for the topics being taught in the classroom as well as assignments. This also addresses the critical aspect of regularity, punctuality, neatness, and notebook upkeep.

(iii) Subject Enrichment Activities (5 marks)
These are subject-specific application activities aimed at enrichment of understanding and skill development. These activities are to be recorded internally by respective subject teachers.

## Online Education for Unseen Passage with Questions and Answers

In Online Education The students are required to have a thorough study and understanding of the given comprehension/Unseen passage which may consist of one or more than one paragraphs. Also, the Unseen passages with questions and answers following them constitute one of the highest weightage in the exam. The main purpose of this activity is to test the reading ability of the students and their intellectual skills.

You can refer to NCERT Solutions to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

## Online Education Unseen Passage for Class 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12

### Short Unseen Passage with Questions and Answers in English

Tips for solving comprehension passages:

• Focus on the relevant details and underline them with a pen or a pencil.
• Read the questions carefully and go back to the passage to find the answers.
• The answers are generally in a logical sequence.
• To find answers to the vocabulary-based questions like synonyms etc., replace the word with the meaning. If you find that it is the same in meaning, the answer is correct.
• To find the correct option in Multiple Choice Questions, go through all the options. Re-read the passage and then tick the correct option.

## Extra Questions for Class 9 Science with Answers

Download NCERT Extra Questions for Class 9 Science: Here we are providing NCERT Extra Questions for Class 9 Science with Solutions Answers Chapter Wise Pdf free download. Students can get Class 9 Science NCERT Solutions, CBSE Class 9 Science Important Extra Questions and Answers designed by subject expert teachers.

NCERT Class 9 Science Extra Questions is the most useful study material for CBSE board students. These solved important questions are asked previously in the CBSE class 9 science board examination. Practicing with these NCERT Class 9th Science Extra Questions will improve your knowledge and problem-solving capabilities on science topics and make you feel confident in the board exams as well as competitive exams.

## NCERT Extra Questions for Class 9 Science with Answers Pdf

We hope the given CBSE NCERT Chapter Wise Extra Questions for Class 9 Science with answers and solutions will help you. If you have any query regarding CBSE Class 9 Science Important Extra Questions and Answers, drop a comment below and we will get back to you at the earliest.

## NCERT Books for Class 12, 11, 10, 9, 8, 7, 6 , 5, 4, 3, 2, and 1

#### एन सी ई आर टी कक्षा १२ की किताबें हिंदी में

NCERT Class 12 Physics Part 1 Book एन सी ई आर टी कक्षा १२ भौतिकी भाग 1
NCERT Class 12 Physics Part 2 Book एन सी ई आर टी कक्षा १२ भौतिकी भाग २
NCERT Class 12 Chemistry Part 1 Book एन सी ई आर टी कक्षा १२ रसायन विज्ञान भाग 1
NCERT Class 12 Chemistry Part 2 Book एन सी ई आर टी कक्षा १२ रसायन विज्ञान भाग २
NCERT Class 12 Maths Part 1 Book एन सी ई आर टी कक्षा १२ गणित भाग 1
NCERT Class 12 Maths Part 2 Book एन सी ई आर टी कक्षा १२ गणित भाग २
NCERT Class 12 Biology Book एन सी ई आर टी कक्षा १२ जीवविज्ञान
NCERT Class 12 Accountancy Books NCERT Class 12 Accountancy Books in Hindi
NCERT Class 12 Accountancy 1 Book एन सी ई आर टी कक्षा १२ लेखाशास्त्र भाग 1
NCERT Class 12 Accountancy 2 Book एन सी ई आर टी कक्षा १२ लेखाशास्त्र भाग २
NCERT Class 12 Business Studies Books NCERT Class 12 Business Studies Books in Hindi
NCERT Class 12 Business Studies 1 एन सी ई आर टी कक्षा १२ व्यवसाय अध्ययन भाग 1
NCERT Class 12 Business Studies 2 एन सी ई आर टी कक्षा १२ व्यवसाय अध्ययन भाग २
NCERT Class 12 Economics Books NCERT Class 12 Economics Books in Hindi
NCERT Book Introductory Microeconomics एन सी ई आर टी कक्षा १२ परिचयात्मक व्यष्टि अर्थशास्त्र
NCERT Book Introductory Macroeconomics एन सी ई आर टी कक्षा १२ परिचयात्मक सूक्ष्म अर्थशास्त्र
NCERT Class 12 Geography Books NCERT Class 12 Geography Books in Hindi
NCERT Book Fundamental of Human Geography एन सी ई आर टी कक्षा १२ मानव भूगोल के मूल सिद्धांत
NCERT Book India- People and Economy एन सी ई आर टी कक्षा १२ भारत लोग और अर्थव्यवस्था (भूगोल )
NCERT Book Practical Working Geography Part II एन सी ई आर टी कक्षा १२ भूगोल में परिजयोजनात्मक प्रयोगात्मक कार्य
NCERT Class 12 History Books NCERT Class 12 History Books in Hindi
NCERT Book Themes in Indian History 1 एन सी ई आर टी कक्षा १२ भारतीय इतिहास के कुछ विषय भाग 1
NCERT Book Themes in Indian History 2 एन सी ई आर टी कक्षा १२ भारतीय इतिहास के कुछ विषय भाग २
NCERT Book Themes in Indian History 3 एन सी ई आर टी कक्षा १२ भारतीय इतिहास के कुछ विषय भाग ३
NCERT Class 12 Political Science Books NCERT Class 12 Political Science Books in Hindi
NCERT Book Contemporary World Politics एन सी ई आर टी कक्षा १२ समकालीन विश्व राजनीति
NCERT Book Political Science 2 स्वतंत्र भारत में राजनीति भाग २
NCERT Class 12 Psychology Books NCERT Class 12 Psychology Books in Hindi
NCERT Book Psychology एन सी ई आर टी कक्षा १२ मनोविज्ञान
NCERT Class 12 Sociology Books NCERT Class 12 Sociology Books in Hindi
NCERT Book Indian Society एन सी ई आर टी कक्षा १२ भारतीय समाज
NCERT Book Social Change and Development India एन सी ई आर टी कक्षा १२ भारत में सामाजिक परिवर्तन और विकास
NCERT Class 12 Hindi Books
एन सी ई आर टी कक्षा १२ अंतरा
एन सी ई आर टी कक्षा १२ अंतराल भाग 2
एन सी ई आर टी कक्षा १२ आरोह
एन सी ई आर टी कक्षा १२ वितान
NCERT Class 12 English Books
NCERT Book Flamingo
NCERT Book Kaleidoscope
NCERT Book Vistas

#### एन सी ई आर टी कक्षा ११ की किताबें हिंदी में

NCERT Class 11 Physics Part 1 Book एन सी ई आर टी कक्षा ११ भौतिकी भाग 1
NCERT Class 11 Physics Part 2 Book एन सी ई आर टी कक्षा ११ भौतिकी भाग २
NCERT Class 11 Chemistry Part 1 Book एन सी ई आर टी कक्षा ११ रसायन विज्ञान भाग 1
NCERT Class 11 Chemistry Part 2 Book एन सी ई आर टी कक्षा ११ रसायन विज्ञान भाग २
NCERT Class 11 Maths Book एन सी ई आर टी कक्षा ११ गणित
NCERT Class 11 Biology Book एन सी ई आर टी कक्षा ११ जीवविज्ञान
NCERT Class 11 Accountancy Books NCERT Class 11 Accountancy Books in Hindi
NCERT Financial Accounting- 1 एन सी ई आर टी कक्षा ११ लेखाशास्त्र भाग 1
NCERT Accountancy- 2 एन सी ई आर टी कक्षा ११ लेखाशास्त्र भाग २
NCERT Class 11 Business Studies Books NCERT Class 11 Business Studies Books in Hindi
NCERT Business Studies एन सी ई आर टी कक्षा ११ व्यवसाय अध्ययन
NCERT Class 11 Economics Books NCERT Class 11 Economics Books in Hindi
NCERT Indian Economic Development एन सी ई आर टी कक्षा ११ भारतीय अर्थव्यवस्था का विकास
NCERT Class 11 Geography Books NCERT Class 11 Geography Books in Hindi
NCERT Class 11 Indian Physical Environment एन सी ई आर टी कक्षा ११ भारतीय भौतिक पर्यावरण
NCERT Class 11 Practical Work in Geography एन सी ई आर टी कक्षा ११ भूगोल में प्रयोगात्मक कार्य भाग 1
NCERT Class 11 Fundamentals of Physical Geography एन सी ई आर टी कक्षा ११ भौतिक भूगोल के मूल सिद्धांत
NCERT Class 11 History Books NCERT Class 11 History Books in Hindi
NCERT Themes in World History एन सी ई आर टी कक्षा ११ विश्व इतिहास के कुछ विषय
NCERT Class 11 Political Science Books NCERT Class 11 Political Science Books in Hindi
NCERT Indian Constitution at Work एन सी ई आर टी कक्षा ११ भारत का संविधान सिद्धांत और व्यवहार
NCERT Political Theory एन सी ई आर टी कक्षा ११ राजनीति सिद्धांत
NCERT Class 11 Psychology Books NCERT Class 11 Psychology Books in Hindi
NCERT Introduction to Psychology एन सी ई आर टी कक्षा ११ मनोविज्ञान
NCERT Class 11 Sociology Books NCERT Class 11 Sociology Books in Hindi
NCERT Introducing Sociology एन सी ई आर टी कक्षा ११ समाजशास्त्र भाग 1
NCERT Introducing Sociology एन सी ई आर टी कक्षा ११ समाज का बोध
NCERT Class 11 Economics Books NCERT Class 11 Economics Books in Hindi
NCERT Indian Economic Development एन सी ई आर टी कक्षा ११ भारतीय अर्थव्यवस्था का विकास
NCERT Class 11 Hindi Books
एन सी ई आर टी कक्षा ११ अंतरा
एन सी ई आर टी कक्षा ११ अंतराल
एन सी ई आर टी कक्षा ११ आरोह
एन सी ई आर टी कक्षा ११ वितान
NCERT Class 11 English Books
NCERT Hornbill
NCERT Woven Words

Students can also check NCERT Solutions here.

#### एन सी ई आर टी कक्षा १० की किताबें हिंदी में

NCERT Class 10 Science Book एन सी ई आर टी कक्षा १0 विज्ञान
NCERT Class 10 Maths Book एन सी ई आर टी कक्षा १0 गणित
NCERT Class 10 Social Science Books NCERT Class 10 Social Science Books
NCERT Contemporary India एन सी ई आर टी कक्षा १0 भारत और समकालीन विश्व भाग २
NCERT India and the Contemporary World-II एन सी ई आर टी कक्षा १0 आर्थिक विकास की समझ
NCERT Understanding Economic Development एन सी ई आर टी कक्षा १0 समकालीन भारत
NCERT Democratic Politics-II एन सी ई आर टी कक्षा १0 लोकतान्त्रिक राजनीति
NCERT Class 10 Hindi Books
एन सी ई आर टी कक्षा १0 कृतिका
एन सी ई आर टी कक्षा १0 क्षितिज
एन सी ई आर टी कक्षा १0 संचयन भाग २
एन सी ई आर टी कक्षा १0 स्पर्श
NCERT Class 10 English Books
NCERT First Flight
NCERT Footprints Without Feet Supplementary Reader

#### एन सी ई आर टी कक्षा ९ की किताबें हिंदी में

NCERT Class 9 Science Book एन सी ई आर टी कक्षा ९ विज्ञान
NCERT Class 9 Maths Book एन सी ई आर टी कक्षा ९ गणित
NCERT Class 9 Social Science Books NCERT Class 9 Social Science Books in Hindi
NCERT Contemporary India एन सी ई आर टी कक्षा ९ भारत और समकालीन विश्व भाग- I
NCERT India and the Contemporary World-I एन सी ई आर टी कक्षा ९ अर्थशास्त्र
NCERT Economics एन सी ई आर टी कक्षा ९ समकालीन भारत
NCERT Democratic Politics एन सी ई आर टी कक्षा ९ लोकतान्त्रिक राजनीति
NCERT Class 9 Hindi Books
एन सी ई आर टी कक्षा ९ कृतिका
एन सी ई आर टी कक्षा ९ क्षितिज
एन सी ई आर टी कक्षा ९ संचयन
एन सी ई आर टी कक्षा ९ स्पर्श
NCERT Class 9 English Books
NCERT BeeHive English Textbook
NCERT Words and Expressions- I

#### एन सी ई आर टी कक्षा ८ की किताबें हिंदी में

NCERT Class 8 Science Book एन सी ई आर टी कक्षा ८ विज्ञान
NCERT Class 8 Maths Book एन सी ई आर टी कक्षा ८ गणित
NCERT Class 8 Social Science Books NCERT Class 8 Social Science Books in Hindi
NCERT Our Past-III: Part 1 एन सी ई आर टी कक्षा ८ हमारे अतीत III(Itihas)
NCERT Our Past- III: Part 2 एन सी ई आर टी कक्षा ८ हमारे अतीत भाग २
NCERT Social and Political Life एन सी ई आर टी कक्षा ८ सामाजिक एवं राजनीतिक जीवन
NCERT Resource and Development (Geography) एन सी ई आर टी कक्षा ८ संसाधन एवं विकास (भूगोल )
NCERT Class 8 Hindi Books
एन सी ई आर टी कक्षा ८ भारत की खोज
एन सी ई आर टी कक्षा ८ दूर्वा
एन सी ई आर टी कक्षा ८ वसंत
NCERT Class 8 English Books
NCERT HoneyDew
NCERT It So Happened

#### एन सी ई आर टी कक्षा ७ की किताबें हिंदी में

NCERT Class 7 Science Book एन सी ई आर टी कक्षा ७ विज्ञान
NCERT Class 7 Maths Book एन सी ई आर टी कक्षा ७ गणित
NCERT Class 7 Social Science Books NCERT Class 7 Social Science Books in Hindi
NCERT Our Past-II एन सी ई आर टी कक्षा ७ हमारे अतीत II
NCERT Social and Political Life-II एन सी ई आर टी कक्षा ७ हमारा पर्यावरण
NCERT Our Environment एन सी ई आर टी कक्षा ७ सामाजिक एवं राजनीतिक जीवन
NCERT Class 7 Hindi Books
एन सी ई आर टी कक्षा ७ दूर्वा
एन सी ई आर टी कक्षा ७ महाभारत
एन सी ई आर टी कक्षा ७ वसंत
NCERT Class 7 English Books
NCERT Honeycomb

#### एन सी ई आर टी कक्षा ६ की किताबें हिंदी में

NCERT Class 6 Science Book एन सी ई आर टी कक्षा ६ विज्ञान
NCERT Class 6 Maths Book एन सी ई आर टी कक्षा ६ गणित
NCERT Class 6 Social Science Books NCERT Class 6 Social Science Books in Hindi
NCERT Our Pasts-I एन सी ई आर टी कक्षा ६ हमारे अतीत I
NCERT The Earth Our Habitat एन सी ई आर टी कक्षा ६ पृथ्वी हमारा आवास (भूगोल )
NCERT Social and Political Life-I एन सी ई आर टी कक्षा ६ सामाजिक एवं राजनीतिक जीवन
NCERT Class 6 Hindi Books
एन सी ई आर टी कक्षा ६ बाल राम कथा
न सी ई आर टी कक्षा ६ दूर्वा
एन सी ई आर टी कक्षा ६ वसंत
NCERT Class 6 English Books
NCERT A Pact With The Sun
NCERT HoneySuckle

#### एन सी ई आर टी कक्षा ५ की किताबें हिंदी में

NCERT Class 5 Maths Magic Book एन सी ई आर टी कक्षा ५ गणित
NCERT Class 5 Environmental Studies Books NCERT Class 5 Environmental Studies Books in Hindi
NCERT Looking Around एन सी ई आर टी कक्षा ५ आस पास
NCERT Class 5 Hindi Book
एन सी ई आर टी कक्षा ५ रिमझिम
NCERT Class 5 English Book
NCERT Marigold

#### एन सी ई आर टी कक्षा ४ की किताबें हिंदी में

NCERT Class 4 Maths Book – Magic एन सी ई आर टी कक्षा ४ गणित का जादू
NCERT Class 4 Environmental Studies Books NCERT Class 4 Environmental Studies Books in Hindi
NCERT Looking Around EVS एन सी ई आर टी कक्षा ४ आस पास
NCERT Class 4 Hindi Book
एन सी ई आर टी कक्षा ४ रिमझिम
NCERT Class 4 English Book
NCERT Marigold

#### एन सी ई आर टी कक्षा ३ की किताबें हिंदी में

NCERT Class 3 Maths Book एन सी ई आर टी कक्षा ३ गणित
NCERT Class 3 Environmental Studies Books NCERT Class 3 Environmental Studies Books in Hindi
NCERT Looking Around एन सी ई आर टी कक्षा ३ आस पास
NCERT Class 3 Hindi Book
एन सी ई आर टी कक्षा ३ रिमझिम
NCERT Class 3 English Book
NCERT Marigold

NCERT Class 2 Maths Book – Magic
एन सी ई आर टी कक्षा २ गणित
NCERT Class 2 Hindi Book
एन सी ई आर टी कक्षा २ रिमझिम
NCERT Class 2 English Book
NCERT Marigold
NCERT Raindrops

Maths Class 1 Maths Book – Magic
एन सी ई आर टी कक्षा १ गणित का जादू
NCERT Class 1 Hindi Book
एन सी ई आर टी कक्षा १ रिमझिम
NCERT Class 1 English Books
Marigold
Raindrops

## NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry

NCERT Solutions for Class 12 Chemistry Chapter 3 provides detailed solutions for the questions asked in the textbook. The solutions are provided by subject experts and the students can refer to these to prepare well for the exams. NCERT Solutions are a guide to the students appearing in different boards like MP board, UP board, CBSE, Gujarat board, etc.

Chemistry Class 12 Chapter 3 Electrochemistry is very important from the examination point of view. All the analytical and conceptual details are provided in the NCERT Solutions Class 12 Chapter 3 that will help the students to score well.

 Board CBSE Textbook NCERT Class Class 12 Subject Chemistry Chapter Chapter 3 Chapter Name Electrochemistry Number of Questions Solved 33 Category NCERT Solutions

## NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry

The production of electricity from chemical reactions and the use of electrical energy to bring non-spontaneous chemical transformations is known as electrochemistry. Both the theory and the practical portions are very important in this chapter.

Class 12 Chemistry chapter 3 Electrochemistry explains different types of cells and the differences between the two. Important concepts such as Gibb’s energy and equilibrium constant are also discussed here.

NCERT IN-TEXT QUESTIONS

Question 1.
How would you determine the standard electrode potential of the system ; Mg2+/Mg ?
n order to determine E° value of Mg2+/Mg electrode, an electrochemical cell is set up in which a Mg electrode dipped in 1 M MgSO4 solution acts as one half cell (oxidation half cell) while the standard hydrogen electrode acts as the other half cell (reduction half cell). The deflection of voltmeter placed in the cell circuit is towards the Mg electrode indicating the flow of current. The cell may be represented as :
Mg/Mg2+ (1 M) || H+(l M)/H2(1 atm), Pt
The reading as given by voltmeter gives $${ E }_{ cell }^{ \circ }$$

The expected value of standard electrode potential (E°) = -2-36 V.

Question 2.
Can you store copper sulphate solution in a zinc pot ?
No, it is not possible. The E° values of the copper and zinc electrodes are as follows :
Zn2+(aq) + 2e → Zn(s) ; E° = – 0·76 V
Cu2+(aq) + 2e → Cu(s) ; E° = + 0·34 V
This shows that zinc is a stronger reducing agent than copper. It will lose electrons to Cu2+ ions and redox reaction will immediately set in.
Zn(s) + Cu2+ (aq) → Zn2+(aq) + Cu(s)
Thus, copper sulphate solution cannot be stored in zinc pot.

Question 3.
Consultthe table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.
Oxidation of Fe2+ converts it to Fe3+, i.e.,Fe2+ –>Fe3+ +e ; E°ox= – 0.77 V Only those substances can oxidise Fe2+ to Fe3+ which are stronger oxidizing agents and have positive reduction potentials greater than 0.77 V, so that EMF of the cell reaction is positive. This is so for elements lying below Fe3+/Fe2+ in the series ex: Br2, Cl2 and F2.

Question 4.
Calculate the potential of hydrogen electrode in contact with a solution with pH equal to 10.
For hydrogen electrode, H+ + e → 1/2H2
Applying Nernst equation,

Question 5.
Calculate e.m.f. of the cell in which the following reaction takes place
Ni(s) + 2Ag+(0·002M) → Ni2+(0·160M) + 2Ag(s) Given that $${ E }_{ cell }^{ \circ }$$ = 1.05 V. (C.B.S.E. Outside Delhi 2015)

Question 6.
The cell in which the following reaction occurs
2Fe3+(aq) + 2I(aq) → 2Fe2+(aq) + I2(s) has $${ E }_{ cell }^{ \circ }$$ = 0-236 V at 298 K.
Calculate standard Gibbs energy and equilibrium constant for the reaction.
The two half reactions are :
2Fe3+ + 2e → 2Fe2+ and 2I → I2 + 2e
For the above reaction, n = 2

Question 7.
Why does the conductivity of a solution decrease with dilution?
Conductivity of a solution is the conductance of ions present in a unit volume of the solutions. On dilution, no. of ions per unit volume decreases. Hence, the conductivity decreases.

Question 8.
Suggest a way to determine the $${ A }_{ m }^{ \circ }$$ for water.
Water (H2O) is a weak electrolyte. Its molar conductance at infinite dilution i.e., $${ A }_{ m }^{ \circ }$$ can be determined in terms of $${ A }_{ m }^{ \circ }$$ for strong electrolytes. This is in accordance with Kohlrausch’s Law.

Question 9.
The molar conductance of 0·025 mol L-1of methanoic acid is 46·15 cm2 mol-1. Calculate its degree of dissociation and dissociation constant. Given λ°(H+) = 349·6 S cm2 mol-1 and λ°(HCOO) = 54·6 S cm2 mol-1.

Question 10.
If a current of 0·5 ampere flows through a metallic wire for 2 hours, then how many electrons flow through the wire ?
Quantity of charge (Q) passed = Current in amperes x Time in seconds = (0·5 A) X (2 x 60 x 60 s)
= 3600 As = 3600 C
No. of electrons flowing through the wire by passing a charge of one faraday (96500 C) = 6·022 x 1023
No. of electrons flowing through the wire by passing a charge of 3600 C

Question 11.
Suggest a list of metals that are extracted electrolytically.
Na, Ca, Mg and Al

Question 12.
Consider the reaction :
Cr2$${ O }_{ 7 }^{ 2- }$$ + 14H+ + 6e → 2Cr3+ + 7H2O.
What is the quantity of electricity in coulombs needed to reduce 1 mole of Cr2$${ O }_{ 7 }^{ 2- }$$ ions ?
The quantity of electricity in coulombs is 6 F or 6 x 96500 C = 5·76 x 105 C.

Question 13.
Write the chemistry of recharging the lead storage battery highlighting all the materials that are involved during recharging.
Chemical reactions while recharging :
Chemical reactions while recharging :
2PbSO4 + 2H2O → PbO2 + Pb + 2H2SO4
Electricity is passed through the electrolyte PbSO4 which is converted into PbO2 and Pb.
Recharging is possible in this case because the PbSO4 formed during discharging is a solid and sticks to the electrodes. Therefore, it can either take up or give electrons during recharging.

Question 14.
Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Methane and Methanol.

Question 15.
Explain how rusting of iron can be envisaged as the setting up of an electrochemical cell.
Iron (Fe) is involved in the redox-reaction that is carried in the electrochemical cell which is set up. As a result, it slowly dissolves and the metal surface gets rusted or corroded.
The redox-reaction may be described as follows :
At anode: Fe (s) undergoes oxidation to release electrons
F2(s) → Fe2+(aq) + 2e–              ….(oxidation)
At cathode: The electrons which are released participate in the reduction reaction and combine with H+ ions released from carbonic acid (H2CO3) formed by the combination of CO2 and H2O present.

NCERT EXERCISE

Question 1.
Arrange the following metals in the order in which they displace each other from the solution of their salts: Al, Cu, Fe, Mg and Zn.
Mg, Al, Zn, Fe, Cu, Ag.

Question 2.
Given the standard electrode potentials
K+/K = – 2·93 V, Ag+/Ag = 0·80 V
Hg2+/Hg = 0·79 V ; Mg2+/Mg = – 2·37V, Cr3+/Cr = – 0·74 V
Arrange these metals in increasing order of their reducing power.
Less the electrode potential more will be the reducing power.
Ag < Hg < Cr < Mg < K.

Question 3.
Depict the galvanic cell in which the reaction
Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s) takes place. Further show :
(i) which electrode is negatively charged ?
(ii) the carriers of the current in the cell.
(iii) individual reaction at each electrode. (C.B.S.E. Delhi 2008)
The galvanic cell in which the given reaction takes place is depicted as:
Zn(s) | Zn2+ (aq) || Ag+ (aq) | Ag(s)
(i) Zn electrode (anode) is negatively charged
(ii) Tons are carriers of current in the cell and in the external circuit, current from silver to Zinc.
(iii) The reaction taking place at the anode is given by,
Zn(s) -H → Zn2+(aq) + 2e
The reaction taking place at the cathode is given
Ag++ e → Ag(s)

Question 4.
Calculate the standard cell potentials of galvanic cell in which the following reactions take place

Question 5.
Write the Nernst Equation and calculate e.m.f. of the following cells at 298 K :
(i) Mg(s) | Mg2+ (0·001 M) || Cu2+ (0·0001 M) | Cu(s) (C.B.S.E. Delhi 2008, 2013)
(ii) Fe(s) | Fe2+ (0·001 M) || H+ (1 M) | H2(g) (1 bar) | Pt(s)
(iii) Sn(s) | Sn2+ (0·050 M) || H+ (0·02 M) | H2(g) (1 bar) | Pt(s) (C.B.S.E. Outside Delhi 2013, 2015)
(iv) Pt(s) | Br2(l) | Br (0·010 M) || H+ (0·030 M) | H2(g) (1 bar) | Pt(s)

Question 6.
In the button cell widely used in watches and in other devices, the following reaction takes place:
Zn (s) + Ag2O (s) + H2O (l) → Zn2+ (aq) + 2Ag (s) + 2OH (aq)
Determine E° and ∆G° for the reaction. (C.B.S.B. Delhi 2005, Outside Delhi 2006 Supp., 2008, C.B.S.E. Sample Paper 2010)

Question 7.
Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
The conductivity of a solution is defined as the conductance of a solution 1 cm in length and the area of cross-section cm2.1 is represented by K.

Conductivity always decreases with a decrease in concentration both for weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases with a decrease in concentration.

Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length.

Molar conductivity increases with a decrease in concentration. This is because the total volume of the solution containing one mole of the electrolyte increases on dilution.

Question 8.
The conductivity of 0·20 M solution of KCl at 298 K is 0·0248 S cm-1. Calculate its molar conductivity.
(C.B.S.E. Delhi 2008, 2013)

Question 9.
The resistance of a conductivity cell containing 0·001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if the conductivity of 0·001M KCl solution at 298 K is 0·146 x 10-3 S cm-2? (C.B.S.E. Outside Delhi 2007, 2008, 2013)

Question 10.
The conductivity of sodium chloride solution at 298 K has been determined at different concentrations and results are given below :

Calculate molar conductivity for all the concentrations and draw a plot between $${ A }_{ m }^{ c }$$ and $$\sqrt { c }$$. Find the value $${ A }_{ m }^{ \circ }$$ from the graph.
$$\frac { 1S{ cm }^{ -1 } }{ 100S{ m }^{ -1 } } =1$$ (unit conversion factor)

$${ A }_{ m }^{ \circ }$$ can be obtained on extrapolation to zero concentration along Y-axis. It is 124·0Scm2mol-1.

Question 11.
The conductivity of 0·00241 M acetic acid is 7·896 x 10-5 S cm-1. Calculate the molar conductivity. If A° for acetic acid is 390·5 S cm2 mol-1, what is its dissociation constant? (C.B.S.E. Delhi 2008, C.B.S.E. Outside Delhi 2016)

Question 12.
How much charge is required for the reduction of :
(i) 1 mol of Al3+ to Al
(ii) 1 mol of Cu2+ to Cu
(iii) 1 mol of Mn$${ O }_{ 4 }^{ – }$$ to Mn2+.

Question 13.
How much electricity in terms of Faraday is required to produce.
(i) 20.0 g of Ca from molten CaCl2?
(ii) 40.0 g of Al from molten Al203?

Question 14.
How much electricity is required in coulomb for the oxidation of
(i) 1 mol of H2O to O2
(ii) 1 mol of FeO to Fe2O3.

For the oxidation of two moles of FeO, charge required = 2 F
For the oxidation of one mole of FeO, charge required = 1 F = 96500 C.

Question 15.
A solution of Ni(NO3)2 is electrolyzed between platinum electrodes using a current of 5·0 ampere for 20 minutes. What weight of Ni will be produced at the cathode? (Atomic mass of Ni = 58·7). (Jharkhand Board 2009)

Question 16.
Three electrolytic cells A, B, and C containing electrolytes zinc sulphate, silver nitrate, and copper sulphate respectively, were connected in series. A steady current of 1·50 ampere was passed through them until 1·45 g of silver was deposited at the cathode of cell B. How long did the current flow? What weight of copper and of zinc were deposited? (Atomic mass of Cu = 63·5 ; Zn = 65·3; Ag = 108) (C.B.S.E. Outside Delhi 2008, Jharkhand Board 2010)

Question 17.
Predict if the reaction between the following is feasible:

The reaction is feasible if the EMF of the cell reaction is positive.

Question 18.
Predict the products of electrolysis of each of the following :
(i) An aqueous solution of AgNO3 using silver electrodes.
(ii) An aqueous solution of AgNO3 using platinum electrodes.
(iii) A dilute solution of H2SO4 using platinum electrodes. (C.B.S.E. Outside Delhi 2007)
(iv) An aqueous solution of CuCl2 using platinum electrodes. (C. B. S. E. Sample Paper 2010)
(i) An aqueous solution of AgNO3 using silver electrodes :
Both AgNO3 and water will ionise in aqueous solution

At cathode: Ag+ ions with less discharge potential are reduced in preference to H+ ions which will remain in solution. As a result, silver will be deposited at cathode.
Ag+ (aq) + e → Ag (deposited)
At anode: An equivalent amount of silver will be oxidised to Ag+ ions by releasing electrons.

(ii) An aqueous solution of AgNO3 using platinum electrodes:

In this case, the platinum electrodes are the non-attackable electrodes. On passing current the following changes will occur at the electrodes.
At cathode: Ag+ ions will be reduced to Ag which will get deposited at the cathode.
At anode: Both $${ NO }_{ 3 }^{ – }$$ and OH ions will migrate. But OH ions with less discharge potential will be oxidised in preference to $${ NO }_{ 3 }^{ – }$$ ions which will remain in solution.

Thus, as a result of electrolysis, silver is deposited on the cathode while O2 is evolved at the anode. The solution will be acidic due to the presence of HNO3.
(iii) A dilute solution of H2SO4 using platinum electrodes:
On passing current, both acid and water will ionise as follows:

At cathode:
H+ (aq) ions will migrate to the cathode and will be reduced to H2.

Thus, H2 (g) will be evolved at the cathode.
At anode: OH ions will be released in preference to $${ SO }_{ 4 }^{ 2- }$$ ions because their discharge potential is less. They will be oxidized as follows:

Thus, O2 (g) will be evolved at the anode. The solution will be acidic and will contain H2SO4.
(iv) An aqueous solution of CuCl2 using platinum electrodes :
The electrolysis proceeds in the same manner as discussed in the case of AgNO3 solution. Both CuCl2 and H2O will ionise as follows :

At cathode:
Cu2+ ions will be reduced in preference to H+ ions and copper will be deposited at the cathode

Cu2+ (aq) + 2e → Cu (deposited)

At anode: C ions will be discharged in preference to OH ions which will remain in solution.

Cl → Cl + e; Cl + Cl → Cl2 (g)

Thus, Cl2 will be evolved at the anode.

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## NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds

NCERT Class 12 Chemistry Solutions for Chapter 9 Coordination Compounds provides in-depth knowledge of the concepts given in the chapter. It helps the students prepare well for boards as well as competitive exams. It also helps the students strengthen their basics for advanced concepts.

The students appearing for UP board, MP board, CBSE, Gujarat board, Maharashtra board, etc. can refer to these NCERT Solutions and score well in the examination.

 Board CBSE Textbook NCERT Class Class 12 Subject Chemistry Chapter Chapter 9 Chapter Name Coordination Compounds Number of Questions Solved 43 Category NCERT Solutions

## NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds

Complex compounds are known as coordination compounds. It is an important chapter from the examination perspective. The students need to be thorough with the concepts such as properties, examples and applications of coordination compounds.

A thorough knowledge of the basic concepts helps the students while studying the advanced concepts. This not only helps them during boards and home exams but also during competitive exams.

NCERT IN-TEXT QUESTIONS

Question 1.
Write the formulas of the following coordination compounds:
(a) tetraamminediaquacobalt (III) chloride
(b) potassium tetracyanonickelate (II)
(c) tris (ethane-1, 2-diamine) chromium (III) chloride
(d) amminebromidochloridonitrito-N-platinate (II)
(e) dichlorobis (ethane-1, 2-diamine) platinum (TV) nitrate
(f) iron (III) hexacyanoferrate (II).
(a) [Co(NH3)4(H2O)2]Cl3
(b) K2[Ni(CN)4]
(c) [Cr(en)3]Cl3
(d) [Pt(NH3)BrCl(NO2)]
(e) [PtCl2(en)2](NO3)2
(f) Fe4[Fe(CN)6]3.

Question 2.
Write the IUPAC names of the following coordination compounds:

1. [Co(NH3)6]Cl3
2. [Co(NH3)5Cl]Cl2
3. K3[Fe(CN)6l
4. K3lFe(C2O4)3]
5. K2[PdCl4]
6. [Pt(NH3)2Cl(NH2CH3)]Cl

1. Hexaamminecobalt (III) chloride
2. Pentaamminechloridecobalt (III) chloride
3. Potassium hexacyanoferrate (III)
4. Potassiumtrioxalatoferrate(III)
6. Diamminechloride (methylamine) platinum (II) chloride

Question 3.
Give the types of isomerism exhibited by the following complexes and draw the structures of these isomers:
(a) K[Cr(H2O)2(C2O4)2]
(b) [Co(en)3]Cl3
(c) [CO(NH3)5(NO2)] (NO3)2
(d) [Pt(NH3) (H2O)Cl2] (C.B.S.E. Outside Delhi 2013)
(a) K[Cr(H2O)2(C2O4)2] or K[Cr(H2O)2(OX)2]
(i) It exists as geometrical isomers :

(ii) The cis isomer can also exist as pair of optical isomers.

(b) The complex can exist as optical isomers. For structure, consult section 9.7.
(c) The complex can exist as pair of ionisation isomers as well as linkage isomers.
Ionisation isomers: [Co(NH3)5(NO2)] (NO3)2 and [Co(NH3)5(NO3)](NO2)(NO3)
(d) The complex can exist as pair of geometrical isomers.

Question 4.
Give evidence to show that [Co(NH3)5Cl]SO4 and [CO(NH3)5SO4]Cl exist as ionisation isomers.
Dissolve both the complexes separately in water. First add a few drops of BaCl2 solution to both these complexes. Only one of these will give white precipitate with the solution indicating that SO4 is not a part of complex entity. If exists as an anion.

Now again add a few drops of AgN03 solution to both these complexes. Only one of these will give white precipitate with the solution indicating that in this case Cl is not a part of complex entity. It exists as anion.

This shows that the complexes exist as pair of ionisation isomers.

Question 5.
Explain on the basis of valence bond theory that [Ni(CN)4]2- ion with square planar structure is diamagnetic and the[Ni(CN)4]2- ion with tetrahedral geometry is paramagnetic.

Question 6.
[MCl4]2- is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why ? (C.B.S.E. Outside Delhi2012)
In the complex [NiCl4]2-, Ni is in + 2 oxidation state and has the configuration 3d840. The CL ion being a weak ligand cannot pair the two unpaired electrons present in 3d orbitals. This means that 3d orbitals are not involved in hybridisation. The complex is sp3 hybridised (tetrahedral) and is paramagnetic in nature. In the other complex [Ni(CO)4], the oxidation state of Ni is zero and electronic configuration is 3d84s2. In the presence of the ligand CO, the 4s electrons shift to the two half filled 3d orbitals and make all the electrons paired. The valence 4s and 3p orbitals are involved in hybridisation. The complex is tetrahedral but diamagnetic. For more details, consult text part.

Question 7.
[Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3- is weakly paramagnetic. Explain.
Outer electronic configuration of iron (Z = 26) in-ground state is 3d64s2. Iron in this complex is in +3 oxidation state. Iron achieves +3 oxidation state by the loss of two 4s electrons and one 3d electron. The resulting Fe3+ ion has outer electronic configuration of 3d5. Since H2O is not a strong field ligand, it is unable to cause electron pairing.

Outer electronic configuration of iron (Z=26) in ground state is 3d64s2. Iron in this complex is in a +3 oxidation state. Iron achieves +3 oxidation state by the loss of two 4s electrons and one 3d electron. The resulting Fe3+ ion has an outer electronic configuration as 3d5. CN ion is a strong field ligand

Question 8.
Explain [Co(NH3)6]3+ is an inner orbital complex while [Ni(NH3)6]2+ is an outer orbital complex.
In the complex [Co(NH3)6]3+, the oxidation state of cobalt is +3 and has 3d6 configuration. In the presence of NH3 molecules (ligands), two 3d electrons pair up and two 3d orbitals remain empty. Since six ligands are to be accommodated the hybridisation of the metal ion is d2sp3.

As inner d-electrons are involved, the complex is inner orbital complex and is diamagnetic in nature.
In the complex [Ni(NH3)6]3 + , the oxidation state of Ni is +2 and has 3d8 configuration. Since six NH3 molecules (ligands) are to be accomodated, the hybridisation of metal ion is sp3d2. This means that 4d orbitals are involved in the hybridisation.

The complex is paramagnetic as well as outer orbital complex since outer (4d) electrons are involved in the hybridisation.

Question 9.
Predict the number of unpaired electrons in the square planar [Pt(CN)4]2- ion.
The element Pt(Z = 78) is present in group 10 with electronic configuration 5d96s1. The divalent cation Pt2+ has 5d8 configuration.

For square planar complex, Pt (II) is in dsp2 hybridisation state. To achieve this, the two unpaired electrons present in 5d orbitals get paired. The complex has, therefore, no unpaired electrons.

Question 10.
The hexaquomanganese(II) ion contains five unpaired electrons, while the hexacyano ion contains only one unpaired electron. Explain using Crystal Field Theory.

Question 11.
Calculate the overall complex dissociation equilibrium constant for [Cu(NH3)4]2+ ion, given that p4 for the complex is 2·1 x 1013.
The dissociation constant is the reciprocal of overall stability constant (β4)

NCERT EXERCISE

Question 1.
Explain bonding in co-ordination compounds in terms of Werner’s postulates.
Werner’s coordination theory: Alfred Werner gave his co-ordination theory in 1893. The important postulates of this theory are:
(i) All metals in atomic or ionic form exhibit two types of valencies in coordination compounds :
(a) Primary or principal or ionic valency (—–),
(b) Secondary or auxiliary or nonionic valency (—).
The primary valency is ionizable and it is shown by dotted lines. The secondary valency is non-ionizable and is shown by a continuous line.
(ii) Primary valency represents oxidation states of a metal atom or ion and secondary valency represents the co-ordination number of metal ion which is fixed for a particular atom.
(iii) The primary valencies are satisfied by negative ions whereas the secondary valencies may be satisfied either by negative ions (e.g., Cl, Br, CN etc.) or neutral molecules (ag., H2O).
(iv) Secondary valencies are directed towards a fixed position in space.
(v) Every element tends to satisfy both its primary and secondary valencies. For this purpose a negative ion may often act a dual behaviour i.e., it may satisfy primary as well as secondary valency (—–,-).
Example : (i) Luteo cobaltic chloride CoCl3.6NH3 or [Co(NH3)6]Cl3.
(ii) Purpureo cobaltic chloride COC13.5NH3 or [Co(NH3)5Cl]Cl2

Question 2.
FeSO4 solution mixed with (NH4)2SO4solution in 1:1 molar ratio gives the test of, Fe2+ion but CuSO4solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu2+ ion. Explain why.
FeSO4 solution mixed with (NH4),SO4 solution in 1 : 1 molar ratio forms a double salt, FeS04 (NH4)2SO4-6H2O (Mohr’s salt) which ionizes in the solution to give Fe2+ions. Hence it gives the tests of Fe2+ ions. CuSO4 solution mixed with aqueous ammonia in 1:4 molar ratio forms a complex salt, with the formula [CU(NH3)4]SO4. The complex ion [Cu(NH3)4]2+ does not ionize to give Cu2+ ions. Hence, it does not give the tests of Cu2+ ion.

Question 3.
Explain with two examples each of the following:
coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heterolepic.
(a) Co-ordination Entity: A coordination complex or entity is the species enclosed in square bracket. It is also called co-ordinate sphere. It contains in it a certain metal atom or ion to which a fixed number of neutral molecules or ions capable of donating electron pairs are linked with co-ordinate bonds. e.g. [COCl3(NH3)3]

(b) Ligand: Ligands are the electron donor molecules or ions which may be either neutral, or anionic (sometimes cationic as well) and are linked to the central metal atom or ion by co-ordinate bonds also called dative bonds. In fact, the electrons needed for the bond are provided by the ligands. e.g. H2NCH2CH2NH2 or N(CH2CH2NH2)3

(c) Co-ordination Number: The coordination number [C.N.] of a metal ion in a complex can be defined as the number of ligand or donor atoms to which the metal is directly bonded. For example, in the complex ions, [PtCl6]2- and [Ni(NH3)4]2+, the coordination number of Pt and Ni are 6 and 4 respectively. Similarly, in the complex ions, [Fe(C2O4)3]3- and [CO(en)3]3+,the coordination number of both, Fe and Co, is 6 because C2O42-and en (ethane-1,2-diamine) are bidentate ligands.

(d) Coordination polyhedron : The spatial arrangement of the ligand atoms which are directly attached to the central atom/ion defines a coordination polyhedron about the central atom. The most common coordination polyhedra are octahedral, square planar and tetrahedral. For example, [CO(NH3)6]3+ is octahedral, [Ni(CO)4] is tetrahedral and [PtCl4]2- is square planar.

(e) Homoleptic and heteroleptic complexe : Complexes in which a metal is bound to only one kind of donor groups, e.g., [CO(NH3)6]3+, are known as homoleptic. Complexes in which a metal is bound to more than one kind of donor group, e.g., [CO(NH3)4Cl2]+, are known as heteroleptic.

Question 4.
What is meant by unidentate, didentate and ambidentate ligands ? Give two examples for each.
Unidentate ligands are those which bind to the metal ion through a single donor atom, e.g., Cl , H2O.

Bidentate ligands are those which bind to the metal ion through two donor atoms. e.g., ethane-1,2-diamine (H2NCH2CH2NH2), oxalate (C2O42-) ion.

Ambidentate ligands are those which can bind to metal ion through two different donor atoms, e.g., NO2 and SCN ion.

Question 5.
Specify the oxidation numbers of metals in the following co-ordination entities :
(a) [Co(H2O)(CN)(en2)]2+
(b) [PtCl4]2-
(c) [Cr(NH3)3Cl3]
(d) [CoBr2(en)2]+
(e) K3[Fe(CN)6].
(a) O.N. of Co : x + 0 + (-1) + 2(0)= + 2 or x = + 2+ 1 = + 3
(b) O.N. of Pt : x + 4 (-1) =-2 or x =-2 + 4 = + 2
(c) O.N. of Cr : x + 3(0) + 3(- 1) = 0 or x = + 3
(d) O.N. of Co : x + 2(- 1) + 2(0) = + 1 or x = + 1 + 2 = + 3
(e) O.N. of Fe : x + 6 (- 1) = – 3 or x = – 3 + 6 = + 3

Question 6.
Using IUPAC norms write the formulas for the following:
(i)Tetrahydroxozincate(Il)
(iii)Diamminedichlorido platinum (II)
(iv)Potassium tetracyanonickelate (II)
(v)Pentaamminenitrito-O-cobalt(III)
(vi)Ilexaamminccobalt (III) sulphate
(vii)Potassium tri(oxalato) chromate (III)
(yiii)Hexaammineplatinum (IV)
(ix)Tetrabromidocuprate(II)
(x) Pentaamminenitrito-N-cobalt (III)
(i)[Zn(OH)4]2- (ii)K2[PdCl4]
(iii)[Pt(NH3)2Cl2]
(iv)K2[Ni(CN)4]
(v)[Co(NH3)5(ONO)]2+
(vi)[Co(NH3)6]2(SO4)3
(vii)K3[Cr(C2O4)3]
(viii)[Pt(NH3)6]4+
(ix)[CuBr4]2-
(x)[Co(NH3)5(N02)]2+

Question 7.
Using IUPAC norms write the systematic names of the following :
(a) [CO(NH3)6]Cl3
(b) [CO(NH3)4Cl(NO2)]Cl
(c) [Ni(NH3)6]Cl2
(d) [Pt(NH3)2Cl(NH2CH3)]Cl
(e) [Mn(H2O)6]2+
(f) [NiCl4]2-
(g) [Co(en)3]3+
(h) [Ti(H2O)6]3+
(i) [Ni(CO)4]. (Jharkhand Board 2015)
(a) hexamminecobalt(III) chloride
(b) tetramminechloriodonitrito-N-cobalt(III) chloride
(c) hexaamminenickel(II) chloride
(d) diamminechlorido (methaneamine) platinum(II) chloride
(e) hexaaquamanganese(II) ion
(f) tetrachloriodonickelate(II) ion
(g) tris(ethane-l, 2-diammine) cobalt(III) ion
(h) hexaaquatitanium(III) ion
(i) tetracarbonylnickel (0).

Question 8.
List various types of isomerism possible for coordination compounds, giving an example of each.
(i) Geometrical isomerism: The isomer in which similar ligands occupy adjacent positions is referred to as cis isomer and the isomer in which similar ligands occupy opposite positions is referred to as trans isomer. Therefore, this type of isomerism is also known as cis-trans isomerism.

(ii) Optical isomerism: The isomer which rotates the plane of polarised light to the right is called dextro rotatory and designated as d- and the one which rotates the plane of polarised light to the left is called laevo rotatory and designated as l. These optical isomers have identical physical and chemical properties except their behaviour towards the plane polarised light.

(iii) Linkage isomerism: Linkage isomerism arises in a coordination compound containing ambidentate ligand. A simple example is provided by complexes containing the thiocyanate ligand, NCS, which may bind through the nitrogen to give M-N CS or through sulphur to give M-SCN. This behaviour was seen in the complex [CO(NH3)5(NO2)]Cl2, which is obtained as the red form, in which the nitrite ligand is bound through oxygen ( ONO), and as the yellow form, in which the nitrite ligand is bound through nitrogen (-NO2).

(iv) Coordination isomerism : This type of isomerism arises from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex. An example is provided by [Co(NH3)6] [Cr(CN)6], in which the NH3 ligands are bound to CO3+ and the CN ligands to Cr3+. In its coordination isomer [Cr(NH3)6][CO(CN)6], the NH3 ligands are bound to Cr3+ and the CN ligands to CO3+.

(v) Ionisation isomerism : This form of isomerism arises when the counter ion in a complex salt is itself a potential ligand and can displace a ligand which can then become the counter ion. An example is provided by the ionisation isomers [CO(NH3)5SO4]Br and [CO(NH3)5Br]SO4.

(vi) Solvate isomerism : This form of isomerism is known as ‘hydrate isomerism in case where water is involved as a solvent. This is similar to ionisation isomerism. Solvate isomers differ by whether or not a solvent molecule is directly bonded to the metal ion or merely present as free solvent molecules in the crystal lattice. An example is provided by the aqua complex [Cr(H2O)6]Cl3 (violet) and its solvate isomer [Cr(H2O)5Cl]Cl2 H2O (grey green).

Question 9.
How many geometrical isomers are possible in the following coordination entities ?
(a) [Cr(C2O4)3]3-
(b) [Co(NH3)3Cl3]
(a) [Cr(C2O4)2]3- : No geometrical isomerism is possible.
(b) [CO(NH3)3Cl3] : Two geometrical isomers : fac and mer

Question 10.
Draw the structures of optical isomers of :
(a) [Cr(C2O4)3]3-
(b) [PtCl2(en)2]2+
(c) [Cr(NH3)2Cl2(en)]+ (C.B.S.E. Foreign 2015)

Question 11.
Draw all the isomers (geometrical and optical) of :
(a) [CoCl2(en)2]+
(b) [CoNH3Cl(en)2]2+
(c) [Co(NH3)2Cl2(en)]+
(a)

(b)

(c)

Question 12.
Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)]. How many of these will exhibit optical isomerism ?
There are three geometrical isomers.

Optical isomerism is generally not shown by the square planar complexes with CN = 4.

Question 13.
Aqueous copper sulphate solution (blue in colour) gives (a) green precipitate with aqueous potassium fluoride and (b) a bright green solution with aqueous potassium chloride solutions. Explain these experimental results.
Aqueous solution of copper sulphate which is blue in colour exists as [Cu(H2O)4]SO4 and gives [Cu(H2O)4]2+ in solution. It is a labile complex entity in which the ligands H2O get easily replaced by F ions of KF and by Cl ions of KCl.

Question 14.
What is the coordination entity formed when excess aqueous KCN is added to an aqueous solution of copper sulphate? Why is that no precipitate of copper sulphide is obtained when H2S(g) is passed through the solution?
When an excess aqueous KCN is added to an aqueous solution of CuSO4, Potassiumtetra-cyanocuprate (II) is formed. When H2S(g) is passed through the above solution, no precipitate of copper sulphide is obtained because CN ions are strong ligands so the complex [Cu(CN)4]2- is very stable. As Cu2+ ions are not available so CuS precipitate is not formed.

Question 15.
Discuss the nature of bonding in the following co-ordination complexes on the basis of valence bond theory :
(a) [Fe(CN)6]4-
(b) [FeF6]3-
(c) [Co(C2O4)3]3-
(d) [CoF6]3-
(a) Hexacyanoferrate(II) ion [Fe(CN)6]4-: Iron in this complex is in +2 oxidation state. Iron achieves + 2 oxidation state by the loss of two 4s electrons. The resulting Fe2+ ion has outer electronic configuration of 3d6.

(b) Hexafluoriodoferrate(II) ion [FeF6]3-: Iron in this complex is in +2 oxidation state. Iron achieves + 2 oxidation state by the loss of two 4s electrons. The resulting Fe2+ ion has outer electronic configuration of 3d6.

(c)

(d) Hexafluorocobaltate(III) [CoF6]3-: Cobalt ion in the complex is in + 3 oxidation state. Cobalt achieves + 3 oxidation state by the loss of two 4s electrons and one 3d electron. The resulting Co3 + ion has outer electronic configuration of 3d6.

Question 16.
Draw figure to show the splitting of d-orbitals in octahedral crystal field.
Let us assume that the six ligands are positioned symmetrically along the Cartesian axes, with a metal atom at the origin.

As the ligands approach, first there is an increase in energy of d-orbitals relative to that of the free ion just as would be the case in a spherical field.

The orbitals lying along the axes (dz2 and dx2 – y2) get repelled more strongly than dxy, dyz and dzx orbitals which have lobes directed between the axes.

The dz2 and dx2-y2 orbitals get raised in energy and dxy dyz, dxz orbitals are lowered in energy relative to the average energy in the spherical crystal field. Thus, the degenerate set of d-orbitals get split into two sets: the lower energy orbitals set, t2g, and the higher energy orbitals set, eg. The energy is separated by ∆0

Question 17.
What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Spectrochemical series: The arrangement of ligands in order of their increasing field strengths i.e. increasing crystal field splitting energy (CFSE) values is called spectrochemical series, which is as follows:
I < Br < SCN < Cl < S2- < F < OH < C2O2-4 < H2O < NCS < edta-4 < NH3 < en < CN < CO.
Difference between weak field ligand and a strong field ligand: The ligand with a small value of CFSE (∆0) are called weak field ligands whereas those with a large value of CFSE are called strong field ligands.

Question 18.
What is crystal field splitting energy ? How does the magnitude of ∆0 decide the actual configuration of d- orbitals in a coordination entity ?
The degenerate d-orbitals (in a spherical field environment) split two-level i.e. eg and t2g in the presence of ligands. The splitting of the degenerate orbitals in the presence of ligands is called crystal field splitting and the energy difference between the two levels (e and t2g) is called the crystal field splitting energy. It is denoted by ∆o. After the orbitals have split, the filling of the electrons takes place. After 1 electron (each) has filled in the three t2g orbitals, the filling of the electrons takes place in 2 ways.

It can enter the orbital (giving) rise to t3g eg, like electronic configuration on the pairing of the electrons can take place in the t2g orbitals (giving rise to t42g eg0 like electronic configuration). If the ∆o value of a ligand is less than the pairing energy, then the electrons enter the eg orbital. On the other hand, if the ∆o value of a ligand is more than the pairing energy, then the electrons enter the t2g orbitals.

Question 19.
[Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2- is diamagnetic. Explain why?
(i) Hexaamminechromium(III) ion [Cr(NH3)6]3+: Outer electronic configuration of chromium (Z=24) in ground state is 3d24s1 and in this complex, it is in the +3 oxidation state. Chromium achieves +3 oxidation state by the loss of one 4s electron and two 3d-electrons. The resulting Cr3+ ion has outer electronic configuration of 3d3.

(ii) Tetracyanonickelate (II) ion [Ni(CN)4]2-: Outer electronic configuration of nickel (Z = 28) in ground state is 3d84s2. Nickel in this complex is in + 2 oxidation state. It achieves + 2 oxidation state by the loss of the two 4s-electrons. The resulting Ni2+ ion has outer electronic configuration of 3d8.

Question 20.
A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2- is colourless. Explain. (C.B.S.E. Delhi 2017)
Formation of [Ni(H2O)6]2+

Since H2P molecules represent weak field ligands, they do not cause any electron pairing. As a result the complex has two unpaired electrons and is coloured. The d-d transitions absorb radiations corresponding to a red light and the complementary colour emitted is green.
Formation of [Ni(CN)4]2-. For the details of the structure of the complex,

Since the complex has no unpaired electrons there is no scope for any d-d transition. The complex is therefore, colourless.

Question 21.
[Fe(CN)6]4- and [Fe(H2O)6]2+ are of different cdlours in dilute solutions. Why?
In both the complexes, Fe is in +2 state with the configuration 3d6 i.e., it has four unpaired electrons. As the ligand H2O and CN possess different crystal field splitting energy (∆0), they absorb different components of the visible light (VIBGYOR) for the transition. Hence, the transmitted colours are different.

Question 22.
Discuss the nature of bonding in metal carbonyls.
The metal-carbon bond in metal carbonyls possess both s and p character. The M-C σ bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal. The M-C π bond is formed by the donation of a pair of electrons from a filled d-orbital of metal into the vacant antibonding π* orbital of carbon monoxide. The metal to ligand bonding creates a synergic effect which strengthens the bond between CO and the metal.

Question 23.
Give the oxidation state, d-orbital occupation and coordination number of the central metal ion in the following complexes :
(a) K3[CO(C2O4)3]
(b) (NH4)2[CoF4]
(C) Cis – [CrCl22(en)2]Cl
(d) [Mn(H2O)6]SO4
(a) OS = + 3, CN = 6, d-orbital occupation is 3d6 $${ t }_{ 2g }^{ 6 }{ e }_{ g }^{ 0 }$$,
(b) OS = + 2, CN = 4, 3d7 ($${ t }_{ 2g }^{ 5 }{ e }_{ g }^{ 2 }$$),
(c) OS = + 3, CN = 6, 3d3 ($${ t }_{ 2g }^{ 3 }$$),
(d) OS = + 2, CN = 6, 3d6 ($${ t }_{ 2g }^{ 3 }{ e }_{ g }^{ 2 }$$).

Question 24.
Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex :
(a) K[Cr(H2O)2(C2O4)2]3H2O
(b) [CrCl3(py)3]
(c) K4[Mn(CN)6]
(d) [Co(NH3)5Cl]Cl2
(e) Cs[FeCl4]
(a) IUPAC name : potassium diaquadioxalatochromate (III) hydrate.
O.S. of Cr = + 3 ; 3d3 ($${ t }_{ 2g }^{ 3 }{ e }_{ g }^{ 0 }$$) CN = 6 ; shape = octahedral, three unpaired electrons.
Magnetic moment (μ) = $$\sqrt { n\left( n+2 \right) } =\sqrt { 3\times 5 } =\sqrt { 15 } =3\cdot 87BM$$
(b) IUPAC name: trichloridotripyridinechromium (III) O.S. of Cr = + 3; 3 d3 ($${ t }_{ 2g }^{ 3 }{ e }_{ g }^{ 0 }$$) CN = 6
shape = octahedral ; three unpaired electrons.
Magnetic moment (μ) = $$\sqrt { n\left( n+2 \right) } =\sqrt { 3\times 5 } =3\cdot 87BM$$
(c) IUPAC name : potassiumhexacyanomanganate (II) O.S. of Mn = + 2 ; 3d5 ($${ t }_{ 2g }^{ 5 }{ e }_{ g }^{ 0 }$$), CN = 6, shape = octahedral; one unpaired electron.
Magnetic moment (μ) = $$\sqrt { n\left( n+2 \right) } =\sqrt { 1\times 3 } =\sqrt { 3 } =1\cdot 73BM$$
(d) IUPAC name : pentaamminechloridocobalt (III) chloride
O.S. of Co = + 3 ; 3d6 ($${ t }_{ 2g }^{ 6 }{ e }_{ g }^{ 0 }$$), CN = 6
shape = octahedral; zero unpaired electron. Magnetic moment (μ) = 0
(e) IUPAC name : cesium tetrachloridoferrate (III)
O.S. of Fe = + 3 ; 3d5 ($${ e }^{ 2 }{ t }_{ 2 }^{ 3 }$$), CN = 4.
shape = tetrahedral ; five unpaired electrons.
Magnetic moment (μ) = $$\sqrt { n\left( n+2 \right) } =\sqrt { 5\times 7 } =\sqrt { 35 } =5\cdot 92BM$$

Question 25.
What is meant by stability of a coordination compound in solution ? State the factors which govern stability of complexes.
The stability of a complex in solution refers to the degree of association between the two species involved in the state of equilibrium. The magnitude of the (stability or formation) equilibrium constant for the association, quantitatively expresses the stability. Thus, if we have a reaction of the type :

then, the larger the stability constant, the higher is the proportion of ML4 that exists in the solution. Free metal ions rarely exist in the solution so that M will usually be surrounded by solvent molecules which will compete with the ligand molecules, L, and be successively replaced by them. For simplicity, we generally ignore these solvent molecules and write four stability constants as follows :

Factors affecting stability of complexes :

1. The smaller the size of the cation, the greater will be the stability of the complex e.g., Fe3+ forms a more stable complex than Fe2+.
2. The greater the charge on the central metal ion, the more stable will be the complex e.g., Pt4+ forms a more stable complex than Pt2+.
3. Stronger the ligand, more stable will be the complex formed e.g., CN forms more stable complex then NH3.

Question 26.
What is meant by chelate effect ? Give an example.
When a ligand attaches to the metal ion in a manner that form’s a ring, then the metal-ligand association is found to be more stable. In other words, we can say that complexes containing chelate ring more stable than complexes without rings. This is known as the chelate effect.
Examples: EDTA, DMG, etc.

Question 27.
Discuss briefly giving an example in each case the role of coordination compounds in:
(i) biological systems
(ii) medicinal chemistry
(iii) analytical chemistry and
(iv) extraction/ metallurgy of metals.
(i) Role of coordination compounds in biological systems:
We know that photosynthesis is possible by the presence of chlorophyll pigment. This pigment is a coordination compound of magnesium. In the human biological system, several coordination compounds play important roles. For example, the oxygen – carrier of blood, i.e hemoglobin is a coordination compound of iron.

(ii) Role of coordination compounds in Medicinal chemistry: Certain coordination compounds of platinum (for example cis-platin) are used for inhibiting the growth of tumors.

(iii) Role of coordination compounds in analytical chemistry: During salt analysis, a number of basic radicals are detected with the help of the colour changes they exhibit with different reagents. These colour changes are a result of the coordination compounds or complexes that the basic radicals form with different ligands.

(iv) Role of coordination compounds in interaction or metallurgy of metals: The process of extraction of some of the metals from their ores involves the formation of complexes. For example in an aqueous solution, gold combines with cyanide ions to form [Au (CN)2]. From this solution, gold is later extracted by the addition of Zn metal.

Question 28.
How many ions are produced from the complex Co(NH3)6Cl2 in solution?
(a) 6
(b) 4
(c) 3
(d) 2.
The complex will dissociate in aqueous solution to give three ions

Therefore, (c) is the correct answer.

Question 29.
Amongst the following ions which one has the highest magnetic movement value?
(a) [Cr(H2O)6]3+
(b) [Fe(H2O)6]2+
(c) [Zn(H2O)6]2+
The oxidation states of the metals in the complexes along with the electronic configuration are given:
(a) Cr3+ : 3d3 configuration ; unpaired electrons = 3
(b) Fe2+ : 3d6 configuration ; unpaired electrons = 4
(c) Zn2+ : 3d10 configuration ; unpaired electrons = 0
The complex (b) with maximum number of unpaired electrons has the highest magnetic moment. Therefore, (b) is the correct answer.

Question 30.
The oxidation number of cobalt in K[Co(CO)4] is
(a) + 1
(b) + 3
(c) – 1
(d) – 3.
O.N. of Co : x + 4(0) = -1 or x = -1. Therefore, (c) is the correct answer.

Question 31.
Amongst the following, the most stable complex is :
(a) [Fe(H2O)6]3+
(b) [Fe(NH3)6]3+
(c) [Fe(C2O4)3]3-
(d) [FeCl6]3-.
In all the complexes, Fe is in + 3 oxidation state. However, the complex (c) is a chelate because three $${ C }_{ 2 }{ O }_{ 4 }^{ 2- }$$ ions act as the chelating ligands. Thus, the most stable complex is (c).

Question 32.
What will be the correct order for the wavelengths of absorption in the visible region for the following : [Ni(NO2)6]4-,[Ni(NH3)6]2+,[Ni(H2O)6]2+.
In all the complexes, the metal ion is the same (Ni2+). The increasing field strengths of the ligands present as per electrochemical series are in the order :
H2O < NH3 < $${ NO }_{ 2 }^{ – }$$
The energies absorbed for excitation will be in the order :
[Ni(H2O)6]2+ < [Ni(NH3)6]2+ < [Ni(NO2)6]4-
As E = hc/λ i.e., E ∝ 1/λ; the wavelengths absorbed will be in the opposite order.

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## NCERT Solutions for Class 12 Chemistry Chapter 1 The Solid State

NCERT Solutions Class 12 Chemistry Chapter 1 contains the solved questions and answers provided in the textbook. The answers are provided by subject experts and hence the students can refer to these solutions for better preparations. The explanations are provided in an easy language which the students find easy to understand. The diagrammatic representations make the explanations even clearer.

The NCERT Solutions for Class 12 Chemistry Chapter 1 not only helps the students prepare for their board exams, but also prepares them for competitive exams. The solutions strengthen the conceptual knowledge of the students that clarifies even the minute doubts.

 Board CBSE Textbook NCERT Class Class 12 Subject Chemistry Chapter Chapter 1 Chapter Name The Solid State Number of Questions Solved 50 Category NCERT Solutions

## NCERT Solutions for Class 12 Chemistry Chapter 1 The Solid State

“The Solid State” is an important chapter in Chemistry from the examination perspective. We are surrounded by different states of matter. Most of the matter around us is in solid state. The NCERT Solutions for Class 12 Chemistry Chapter 1 explain the structure, classification and properties of solids. The explanation of the structure of solids states the correlation between the structure and properties of solids.

NCERT IN-TEXT QUESTIONS

Question 1.
Why are solids rigid?
Solids are rigid because the constituent particles are very closely packed. They don’t have any translatory movement
and can only oscillate about their mean positions.

Question 2.
Why do solids have a definite volume?
The constituent particles of a solid have fixed positions and are not free to move about, i.e., they possess rigidity. That is why they have definite volume.

Question 3.
Classify the following as amorphous and crystalline solids; polyurethane, naphthalene, benzoic acid, Teflon,
potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper.
Amorphous solids: Polyurethane, naphthalene, Teflon, cellophane, polyvinyl chloride, fiberglass. Crystalline solids: Benzoic acid, potassium nitrate, copper.

Question 4.
Why is glass considered a supercooled liquid?
Glass is an amorphous solid. Like liquids, it has a tendency to flow, though very slowly. This can be seen from the glass panes of windows or doors of very old buildings which are thicker at the bottom than at the top. Therefore, glass is considered as a supercooled liquid.

Question 5.
The Refractive index of a solid is observed to have the same value along with all the directions. Comment on the nature of the solid. Would it show cleavage property?
As the solid has the same refractive index along with all the directions, it is isotropic in nature and is, therefore, an amorphous solid. It is not expected to show a clean cleavage when cut with a special type of knife. It will break into pieces with irregular surfaces.

Question 6.
Classify .the following solids in different categories based on the nature of intermolecular forces operating in them: Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon carbide
Potassium sulphate = Ionic Tin=Metallic.
Benzene = Molecular (non-polar)
Urea=Molecular (polar).
Ammonia=Molecular (H-bonded)
Water = Molecular (H-bonded)
Zinc sulphide = Ionic Graphite=Covalent Rubidipm Metallic Argon = Molecular (non-polar)
Silicon Carbide=Covalent

Question 7.
A solid substance ‘A’ is a very hard and electrical insulator both in the solid-state as well as in the molten state. It has also the very high melting point. Is the solid metal like silver or network solid like silicon carbide (SiC)?
Since the solid behaves as an insulator even in the molten state, it cannot be metal like silver. Therefore, it is a covalent or network solid like SiC.

Question 8.
Ionic solids conduct electricity in molten state but not in solid state. Explain
In a solid-state, the ions cannot move, they are held by strong electrostatic forces of attraction. So, ionic solids do not conduct electricity in the solid state. However, in the molten state, they dissociate to give tree ions and hence conduct electricity.

Question 9.
What types of solids are electrical conductors, malleable and ductile? (C.B.S.E. Outside Delhi 2013)
Metallic solids exhibit these characteristics. Their atoms are linked to one another by metallic bonds.

Question 10.
Give the significance of a ‘lattice point’.
Each lattice point represents one constituent particle of the solid. This constituent particle may be an atom, a molecule or an ion.

Question 11.
Name the parameters which characterize a unit cell.
A unit cell is characterized by two types of parameters. These are edges (a, b, c) which may or may not be mutually perpendicular, and angles between the edges (α, β, and γ).
(i) Edges or edge lengths. The edges a, b and c represent the dimensions of the unit cell in space along the three axes. The edges may or may not be mutually perpendicular.
(ii) Angles between the edges. There are three angles between the edges. These are denoted as α (between b and c), β (between a and c), and γ (between a and b). Thus, a unit cell may be characterized by six parameters as shown in Fig. 1.12. The various types of crystal systems differ with respect to edge lengths as well as angles between the edges.

Question 12.
Distinguish between
(i) Hexagonal and monoclinic unit cells
(ii) Face-centred and end-centered unit cells.

(i)

 Hexagonal unit cell Monoclinic unit cell a=b≠c a≠b≠c α = β = 90° α = γ = 90° γ = 120° β ≠ 90°

(ii)

 Face-centred unit cell End-centred unit cell A Face-centred unit cell the constituent particles are present at the corners and one at the centre of each face. An End-centred unit cell contains particles at the corners and one at the centre of any two opposite faces. Total no of particles in a face centered unit cell= 4 Total no. of particles in an end centered unit cell = 2

Question 13.
Explain how many portions of an atom located at the
(i) corner and
(ii) body centre of a cubic unit cell is a part of the neighbouring unit cell.
(i) An atom located at the corner is shared by eight unit cells. Therefore, its contribution to a particular unit cell is 1/8.
(ii) An atom located at the body of the unit cell is not shared by any unit cell. It belongs to one particular unit cell only.

Question 14.
What is the two-dimensional coordination number of a molecule in a square close-packed layer?
In 2D, square close-packed layer, an atom touches 4 nearest neighbouring atoms. Hence, its CN=4

Question 15.
A compound forms a hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?
No. of atoms in 0.5 mole of the compound = 0.5 x N0 = 0.5 x 6.022 x 1023 = 3.011 x 1023
No. of octahedral voids = No. of atoms = 3.011 x 1023
No. of tetrahedral voids = 2 x 3.011 x 1023 = 6.022 x 1023
Total no. of voids = (3.011 + 6.022) x 10223 = 9.033 x 1023

Question 16.
A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy l/3rd of tetrahedral voids. What is the formula of the compound?
Atoms of N from ccp, therefore, if the lattice points are n, then
No.of atoms of N = n
No. of oct voids = n
No. of td voids = 2n= 2 x 1n/3 = 2n/3
∴ Formula of the compound is: M : N
2/3 n : n
2n: 3n
2: 3
i.e., M2N3

Question 17.
Which of the following lattices has the highest packing efficiency :
(i) simple cubic
(ii) body-centered cubic and
(iii) hexagonal close-packed lattice?
The packing efficiency of the different types of arrangement is :
(i) Simple cubic = 52.4%
(ii) Body-centred cubic = 68%
(iii) Hexagonal close-packed = 74%
his means that hexagonal close-packed arrangement has the maximum packing efficiency (74%).

Question 18.
An element with a molar mass 2.7 x 10-2 kg  mol-1 forms a cubic unit cell with an edge length of 405 pm. If its density is 2.7 x 103 kg m-3, what is the nature of the cubic unit cell?

Question 19.
What type of defect can arise when a solid is heated? Which physical property is affected by it and in what
way?
When a solid is heated, some atoms or ions may leave the crystal lattice. As a result, vacancies are created and this leads to vacancy defects in the crystalline solid. Since the number of atoms/ions per unit volume decreases, the vacancy defects lead to a decrease in the density.

Question 20.
What type of stoichiometric defect is shown by:
(i) ZnS
(ii) AgBr
(i) ZnS shows Frenkel defect
(ii) AgBr shows Frenkel as well as Schottky defect.

Question 21.
Explain how vacancies are introduced in the ionic solid when a cation of higher valence is added as an impurity to it.
Let us consider an ionic solid sodium chloride (Na+Cl) to which a small amount of strontium chloride (SrCl2) has been added to act as an impurity. Since the crystal as a whole is to remain electrically neutral, two Na+ ions have to leave their sites to create two vacancies. Out of these, one will be occupied by Sr2+ ion while the other will be vacant. Thus, vacancies will be created in the ionic solid. When a cation of higher valency is added as an impurity in the ionic solid, some of the sites of the original cations are occupied by the cations of higher valency. Each cation of higher valency replaces two or more original cations and occupies the site of one original cation and the other site(s) remains vacant.

Question 22.
Ionic solids which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.
Let us illustrate by sodium chloride (Na+Cl) crystals. Upon heating in the atmosphere of sodium (Na) vapours, sodium atoms get deposited on the surface of the crystals. The Cl ions from the crystal lattice leave their sites and diffuse into the surface. They tend to combine with sodium atoms present in the vapours which in turn get ionised to form Na+ ions by releasing electrons. The latter is trapped by the anionic vacancies created by Clions in order to maintain the crystals electrically neutral. Now, the electrons absorb radiations corresponding to a certain colour from white light and start vibrating. They emit radiations corresponding to yellow colour. That is how the crystals of sodium chloride develop yellow colour. These electrons are called F-centres because these are responsible for colours (In German, F = Farbe meaning colour).

Question 23.
A group 14 element is to be converted into an n-type semiconductor by doping it with a suitable impurity. To which group should the impurity element belong?
n-type semiconductors are conducting due to the presence of excess negatively charged electrons. In order to convert group 14 elements (e.g. Si, Ge) into n-type semi-conductors, doping is done with some elements of group 15 (e.g. P, As)

Question 24.
What type of substances would make better permanent magnets; ferromagnetic or ferrimagnetic? Justify your answer. (C.B.S.E. Outside Delhi 2013)
Ferromagnetic substances make better permanent magnets than ferrimagnetic substances. The metal ions of a ferromagnetic substance are grouped into small regions known as domains and these are randomly oriented. Under the influence of the applied magnetic field, all domains are oriented in the direction of the magnetic field and as a result, a strong magnetic field is produced. The ferromagnetic substance behaves as a magnet. This characteristic of the domains persists even when the external magnetic field is removed. This imparts permanent magnetic character to these substances. However, this property is lacking in ferrimagnetic substances. Therefore ferromagnetic substances are better magnets.

NCERT EXERCISE

Question 1.
Define the term ‘amorphous’. Give a few examples of amorphous solids.
Amorphous solids are those solids in which the constituent particles may have short-range order but do not have a long-range order. They have irregular shapes and are isotropic in nature. They do not undergo a clean cleavage. They do not have sharp melting points or definite heat of fusion. E.g.: Glass, rubber, and plastics.

Question 2.
What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?
Glass is made up of Si04 tetrahedral units. These constituent particles have short-range order only. Quartz is also made up of Si04 tetrahedral units. On heating it softens and melts over a wide range of temperature. It is a crystalline solid having long-range ordered structure. It has a sharp melting point. Quartz can be converted into glass by first melting and then rapidly cooling it.

Question 3.
Classify each of the following solids as ionic, metallic, molecular, network (covalent), or amorphous:
(a) Tetra phosphorus decoxide (P4O10)
(b) Graphite
(c) Brass
(d) Ammonium phosphate (NH4)3PO4
(e) SiC
(f) Rb
(g)l2
(h) LiBr
(i) P4
(j) Si
(k) Plastic.
(a) Molecular solid
(b) Covalent (Net-work) solid
(c) Metallic solid
(d) Ionic solid
(e) Covalent solid (Network)
(f) Metallic solid
(g) Molecular solid
(h) lonic solid
(i) Molecular solid
(j) Covalent solid
(k) Amorphous solid.

Question 4.
(a) What is meant by the term coordination number?
(b) What is the coordination number of atoms
(i) in a cubic close-packed structure
(ii) in a body-centered cubic structure?
(i) The number of nearest neighbours of a particle in its close packing is called its coordination number.
(ii) (a) 12, (b) 8.

Question 5.
How can you determine the atomic mass of an unknown metal if you know its density and the dimensions of its unit cell? Explain.
Let the edge length of a unit cell = a
Density = d
Molar mass = M
The volume of the unit cell = a3
Mass of the unit cell = No. of atoms in unit cell x Mass of each atom = Z × m

Question 6.
(a) Stability of a crystal is reflected in the magnitude of the melting point. Comment.
(b) Collect the melting point of

• Ice
• ethyl alcohol
• diethyl ether
• methane from a data book. What can you say about intermolecular forces between the molecules?

(a) Higher the melting point, greater are tire “forces holding the constituent particles together and hence greater is the stability.

(b) The intermolecular forces in water and ethyl alcohol are mainly hydrogen bonding. The higher melting point of water than alcohol shows that hydrogen bonding in ethyl alcohol molecules is not as strong as in water molecules. Diethyl ether is a polar molecule. The intermolecular forces present in them are the dipole-dipole attraction. Methane is a non-polar molecule. The only forces present in them are the weak Vander Waal’s forces (London / dispersion forces).

Question 7.
How will you distinguish between the following pairs of terms:
(i) Cubic close packing and hexagonal close packing?
(ii) Crystal lattice and unit cell?
(iii) Tetrahedral void and octahedral void?
(i) Cubic close packing: When the third layer is placed over the second layer in such a way that the spheres cover the octahedral voids, a layer different from first (A) and second (B) is produced. If we continue packing in this manner, then packing is obtained where the spheres in every fourth layer will vertically aligned. This pattern of packing spheres is called the ABCABC pattern or cubic close packing.

Hexagonal close-packing: When a third layer is placed over the second layer in such a manner that the spheres cover the tetrahedral void, a three-dimensional close packing is obtained where the spheres in every third or alternate layer are vertically aligned. If we continue packing in this manner, then the packing obtained would be called ABAB pattern or hexagonal close packing.

(ii) Crystal lattice: It is a regular arrangement of the constituent particles (i?.e., ions, atoms or molecules) of a crystal in three-dimensional space.
Unit cell: The smallest three-dimensional portion of a complete space lattice which when repeated over and over again in different directions produces the complete crystal lattice is called the unit cell.

(iii) Tetrahedral void: A simple) the triangular void is a crystal is surrounded by four spheres and is called a tetrahedral void.
Octahedral void: A double triangular void is surrounded by six spheres and is called an octahedral void.

Question 8.
How many lattice points are there in one unit cell of each of the following lattices
(a) face-centered-cubic
(b) face centred tetragonal
(c) body-centered cubic?
(i) In the face-centered cubic arrangement, a number of lattice points are = 8 (at comers) + 6 (at face centres)
Lattice points per unit cell = 8 × $$\frac { 1 }{ 8 }$$ + 6 × $$\frac { 1 }{ 2 }$$ = 4.

(ii) In face centred tetragonal, number of lattice points are = 8 (at comers) + 6 (at face centres)
Lattice points per unit cell = 8 × $$\frac { 1 }{ 8 }$$ + 6 × $$\frac { 1 }{ 2 }$$ = 4

(iii) In body centred cubic arrangement, number of lattice points are = 8 (at comers) + 1 (at body centres)
Lattice points per unit cell = 8 × $$\frac { 1 }{ 8 }$$ + 1 = 2

Question 9.
Explain:
(i) The basis of similarities and differences between metallic and ionic crystals.
(ii) Ionic solids are hard and brittle.
(i) Basis of similarities. The basis of similarities between the metallic and ionic crystals are the presence of strong electrostatic forces of attraction. These are present among the ions in the ionic crystals and among the kernels and
valence electrons in the metallic crystals. That is why both metals and ionic compounds are good conductors of electricity and have high melting points.
Basis of differences. The basis of differences in the absence of mobility of ions in the ionic crystals while the same is present in the valence electrons and kernels in the case of metallic crystals. As a consequence, the ionic compounds conduct electricity only in the molten state while the metals can do so even in the solid-state.
(ii) The ionic solids are hard and brittle because of strong electrostatic forces of attraction which are present in the oppositely charged ions.
The ionic solids are hard because of the presence of strong inter-ionic forces of attraction in the oppositely charged ions. These ions are arranged in three-dimensional space. The ionic solids are brittle because the ionic bond
is non-directional.

Question 10.
Calculate the efficiency of packing in the case of metal crystal for:
(i) Simple cubic
(ii) Body centered cubic
(iii) Face centered cubic (with the assumption that the atoms are touching each other). (C.B.S.E. Outside Delhi 2011) Answer:
(i) Simple cubic: We know that in a simple cubic unit cell, there is one atom (or one sphere) per unit cell. If r is the radius of the sphere, the volume occupied by one sphere present in the unit cell = 4/3πr3.

(ii) Body-centred cubic: We know that a body-centered cubic unit cell has 2 spheres (atoms) per unit cell. If r is the radius of the sphere Volume of one sphere = 4/3πrsup>3

(iii) Face centred cubic: We know that a face centered cubic unit cell (fcc) contains four spheres (or atoms) per unit cell.

Question 11.
Silver crystallizes in a face centered cubic lattice with all the atoms at the lattice points. The length of the edge of
the unit cell as determined by X-ray diffraction studies is found to be 4.077 x 10-8 cm. The density of silver is 10.5 g cm-3. Calculate the atomic mass of silver. (C.B.S.E. Sample Paper 2012)(Uttarakhand Board 2015)

Question 12.
A cubic solid is made up of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body centre. What is the formula of the compound? What are the coordination numbers of P and Q?
As atom Q are present at the 8 comers of the cube, therefore, number of atoms of Q in the unit cell = 8 × $$\frac { 1 }{ 8 }$$ = 1.
As atoms P are present at the body centre, therefore a number of atoms P in the unit cell = 1.
∴ The formula of the compound = PQ
Co-ordination number of each P and Q = 8.

Question 13.
Niobium crystallizes in a body-centered cubic structure. If the density is 8.55 g cm-1, calculate the atomic radius of niobium given that the atomic mass of niobium is 93 g mol-1. (C.B.S.E. Delhi 2008)
Step I. Calculation of edge length of unit cell.
No. of particles in b.c.c. type unit cell (Z) = 2
Atomic mass of the element (M) = 93 g mol-1

Step II. Calculation of radius of unit cell

Question 14.
If the radius of octahedral void is r and the radius of the atom in close packing is R, derive the relation between r and R. (C.B.S.E. Sample Paper 2017)
Let length of each side of the square is a and the radii of the void and the sphere are r and R respectively. Consider the right angled triangle ABC.

∴ Radius of octahedral void is 0.414R(or 41.4% as compared to that of the sphere).

Question 15.
Copper crystallises into a foc lattice with edge length 3•61 x 10-8 cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm. (C.B.S.E. Delhi 2009 Comptt.)

The calculated value of the density is nearly the same as the measured value.

Question 16.
Analysis shows that nickel oxide has formula Nin.os 01.00. What fraction of nickel exists as Ni2+ and as Ni3+ ions ?
The ratio of Ni and O atoms in pure nickel oxide (NiO) = 1:1
Let x be the no. of Ni (II) atoms replaced by Ni (III) atoms in the oxide.
∴ No. of Ni (II) atoms present = (0.98 – x)
Since the oxide is neutral in nature,
Charge on Ni atoms = Charge on oxygen atoms
2(0.98 – x) + 3x = 2
1.96 – 2x + 3x = 2
x = 2 – 1.96 = 0.04

% of Ni (II) atoms in nickel oxide = 100 – 4:01 = 95.99%

Question 17.
What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.
Those solids which have intermediate conductivities ranging from 10-6 to 104 ohm-1 m-1 are classified as semiconductors. As the temperature rises, there is a rise in conductivity value because electrons from the valence band jump to the conduction band.

(i) n-type semiconductor: When a silicon or germanium crystal is doped with group 15 elements like P or As, the dopant atom forms four covalent bonds like Si or Ge atom but the fifth electron, not used in bonding, becomes delocalized and continues its share towards electrical conduction. Thus silicon or germanium doped with P or As is called an H-type semiconductor, a-indicative of negative since it is the electron that conducts electricity.

(ii) p-type semiconductor: When silicon or germanium is doped with group 13 elements like B or Al, the dopant is present only with three valence electrons. An electron vacancy or a hole is created at the place of the missing fourth electron. Here, this hole moves throughout the crystal-like a positive charge giving rise to electrical conductivity. Thus Si or Ge doped with B or Al is called p-tvpe semiconductor, p stands for the positive hole since it is the positive hole that is responsible for conduction.

Question 18.
Non-stoichiometric cuprous oxide (Cu2O) can be prepared in the laboratory. In this oxide, the copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor?
The ratio less than 2: 1 in Cu2O shows that some cuprous (Cu+) ions have been replaced by cupric (Cu2+) ions. To maintain electrical neutrality, every two Cu+ ions will be replaced by one Cu2+ ion thereby creating a hole. As conduction will be due to the presence of this positive hole, hence it is a p-type semiconductor.

Question 19.
Ferric oxide crystallizes in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of ferric oxide.
There is one octahedral hole for each atom in the hexagonal close-packed arrangement.
If the number of oxide ions (O2-) per unit cell is 1, then the number of Fe3+ ions = 2/3 x octahedral holes = 2/3 x 1 = 2/3.
Thus, the formula of the compound = Fe2/3O1, or Fe2O3.

Question 20.
Classify each of the following as being either a p-type or an n-type semiconductor.
(i) Ge doped with In
(ii) B doped with Si.
(i) Ge belongs to group 14 and In belongs to group 13, therefore an electron-deficient hole is created and hence it is an n-type semiconductor.
(ii) B belongs to group 13 and Si belongs to group 14, therefore there will be a free electron and hence it is an n-type semiconductor.

Question 21.
Gold (atomic radius = 0.144 nm) crystallizes in a face-centered unit cell. What is the length of a side of the cell?
For fee lattice, edge length,
a = 2 √2 x 0.144 nm = 0.407 nm

Question 22.
In terms of Band Theory, what is the difference
(i) between a conductor and an insulator
(ii) between a conductor and semi-conductor?
The variation in the electrical conductivity of the solids can be explained with the help of the band theory.

(i) In insulators, the energy gaps are very large and the no electron jump is feasible from the valence band to the conduction band. The energy gaps also called forbidden zones. The insulators, therefore, do not conduct electricity.
(ii) In semi-conductors, there is a small energy gap between the valence band and conduction band. However, some electrons may jump to the conduction band and these semiconductors can exhibit a little electrical conductivity.

Question 23.
Explain the following terms with suitable examples.
(i) Schottky defect,
(ii) Frenkel defect,
(iii) Interstitials,
(iv) F-centres.
(i) Schottky defect: This rises because certain ions are missing from the crystal lattice and vacancies or holes are created at their respective positions. Since a crystal is electrically neutral, the number of such missing cations (A+) and anions (B) must be the same. e.g., KCl, NaCl, KBr, etc.

(ii) Frenkel defect:
It results when certain ions leave their normal sites and occupy positions elsewhere in the crystal lattice. Holes are created at their respective positions. Since cations are smaller in size as compared to anions normally these are involved in Frenkel defect. e.g., AgBr, ZnS, etc.

(iii) Interstitials:
This defect is noticed when constituent particles (atoms or molecules) occupy the interstitial sites in the crystal lattice. As a result, the number of particles per unit volume increases and so the density of the solid.

(iv) F-centres:
These are the anionic sites occupied by unpaired electrons. F-centres impart colour to crystals. The colour results by the excitation of electrons when they absorb energy from the visible light falling on the crystal.

Question 24.
Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm.
(a) What is the length of the side of the unit cell?
(b) How many unit cells are there in 1.00 cm? of aluminum? (C.B.S.E. Outside Delhi 2013)
Step I. Calculation of length of side of the unit cell
For f.c.c. unit cell, a = $$2\sqrt { 2 } r$$ = $$2\sqrt { 2 }$$ (125pm) = 2 x 1.4142 x (125 pm) = 354 pm.
Step II. Calculation of no. of unit cells in 1:00 cm3 of aluminium.
Volume of one unit cell = (354 pm)3 = (354 x 10-10 cm)3 = 44174155 x 10-30 cc.

Question 25.
If NaCl is doped with 10-3 mol % of SrCl2, what is the concentration of cation vacancies?

Question 26.
Explain the following with suitable examples.
(a) Ferromagnetism
(b) Piezoelectric effect
(c) Paramagnetism
(d) Ferrimagnetism
(e) Antifluoride structure
(f) 12 – 16 and 13 – 15 compounds.
(a) Ferromagnetism: A few solids like iron. cobalt, nickel, gadolinium and CeO2 are attracted very strongly by magnetic fields. These are known as ferromagnetic solids. Apart from that, they can be even permanently magnetised or become permanent magnet. e.g., Fe, Ni, Co and CrO2

(b) Piezoelectric effect: A dielectric crystal which has a resultant dipole moment can produce electricity or show the electrical property when external pressure is applied. Such a crystal is known as piezoelectric crystal and this property is called piezoelectricity or pressure electricity. e.g., PbZrO2, Nh4H2PO4 etc.

(c) Paramagnetism: These are the solids attracted by a magnet. Actually, the atoms of the elements present have certain unpaired electrons. Their spins or magnetic moments may lead to magnetic characters. Many transition metals such as Co, Ni, Fe, Cu, etc. and their ions are paramagnetic. e.g., O2, Cu2+, Fe3+, etc

(d) Ferrimagnetism: They have certain resultant magnetic moment or magnetic character which is of permanent nature. However, ferrimanetic solids are less magnetic than ferromagnetic solids. For example, magnetic oxide of iron (Fe3O4) and ferrites with general formula MFe2O4. e.g., Fe3O4

(e) Antifluoride structure: In this structure, the positions of the cations and anions as compared to fluorite structure get reversed i.e. the smaller cations occupy the position of fluoride ions while the anions with bigger size occupy the positions of calcium ions. e.g., Li2O, K2O, Rb2O and Rb2S.

(f) 12 – 16 and 13 – 15 compounds: A large variety of solid-state materials have been prepared by the combination of elements belonging to groups 13 and 15 or group 12 and 16. A few examples of compounds 13-15 combinations are InSb, Alp and GaAs. Similarly, compounds resulting from 12 – 16 combinations are AdS, CdSe, HgTe.

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## NCERT Solutions for Class 12 Chemistry Chapter 7 The p-Block Elements

Class 12 Chemistry NCERT Solutions for Chapter 7 The p-Block Elements is the best guide for the students appearing for boards and competitive exams. The solutions contain answers to the questions provided in the textbook. These help in strengthening all the concepts related to chapter 7 that help in better preparations.

NCERT Solutions not only help the students appearing in different boards but also the one’s appearing in the competitive exams. The students can practice the solutions provided by the subject experts.

 Board CBSE Textbook NCERT Class Class 12 Subject Chemistry Chapter Chapter 7 Chapter Name The p-Block Elements Number of Questions Solved 74 Category NCERT Solutions

## NCERT Solutions for Class 12 Chemistry Chapter 7 The p-Block Elements

Class 12 Chemistry Chapter 7 explains the p block elements and their properties, The elements in group 15, 16 and 17 are discussed along with their properties. Various concepts such as electronegativity, chemical and physical properties, ionization therapy, etc. are discussed in detail.

This chapter is important from examination perspective and the students are advised to go through the NCERT Solutions for better practice. The solutions are provided along with the diagrams for better understanding.

NCERT IN-TEXT QUESTIONS

Question 1.
Why are pentahalides more covalent than trihalides in the members of the nitrogen family?
The electronic configuration of the elements of nitrogen family (group 15) is ns2p3. Because of the inert pair effect, the valence s-electrons cannot be released easily for the bond formation. This means that the elements can form trivalent cation (E3+) by releasing valence p-electrons while it is difficult to form pentavalent cation (E5+). Under the circumstances, if all the five valence electrons are to be involved in the bond formation, the compounds showing pentavalency or +5 oxidation state must be of covalent nature. This is particularly the case in the higher members (Sb, Bi) of the family where the inert pair effect is quite prominent.

Question 2.
Why is BiH3 the strongest reducing agent amongst all the hydrides of Group 15 elements?
This is because as we move down the group, the size increases, as a result, length of E-H bond increases and its strength decreases, so that the bond can be broken easily to release H2 gas. Hence, BiH3 is the strongest reducing agent.

Question 3.
Why is N2 less reactive at room temperature?
In the nitrogen molecule (N2), two atoms of nitrogen are linked by triple bond (N = N). Due to small atomic size of the element (atomic radius = 70
pm), the bond dissociation enthalpy is very high (946 kJ mol1). This means that it is quite difficult to cleave or break the triple bond at room temperature. As a result, N2 is less reactive at room temperature.

Question 4.
Mention the conditions required for the maximum yield of ammonia.
In Haber’s process, ammonia is formed by the following reaction.

According to Le-chatelier’s principle, the favourable conditions for the maximum yield of ammonia are :
(i) Low temperature : But optimum temperature of 700 K is necessary to keep the forward reaction in progress.
(ii) High pressure : Pressure to the extent of about 200 atm is required.
(iii) Catalyst & promoter : In order to achieve the early attainment of equilibrium, iron oxide acts as catalyst. Along with that; K2O, Al2O3 or Mo metal may act as the promoter to increase the efficiency of the catalyst.

Question 5.
How does ammonia react with blue solution having Cu2+ ions ?
When ammonia gas is passed through blue solution containing Cu2+ ions to give a soluble complex with deep blue colour.

Question 6.
What is the covalency of nitrogen in N2O5?
The covalent structure of nitrogen pentoxide (N2O5) is given. Since the nitrogen atom has shared four electron pairs, its covalency is four in the molecule of N2O5.

Question 7.
Bond angle in PH4+ is higher than that in PH3. Why?
P in PH3 is sp3-hybridized with 3 bond pairs and one lone pair around P. Due to stronger lp-bp repulsions than bp-bp repulsions, tetrahedral angle decreases from 109°28′ to 93.6°. As a result, PH3 is pyramidal. In PH4+, there are 4 bp’s and no lone pair. As a result, there are only identical bp-bp repulsions so that PH4+ assumes tetrahedral geometry and the bond angle is 109°28′.Hence, bond angle of PH4+ > bond angle of PH3

Question 8.
What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO2?
A mixture of sodium hypophosphite and phosphine gas is formed upon heating the reaction mixture in an inert atmosphere.

Question 9.
What happens when PCl5 is heated?
Upon heating, PCl5 dissociates to give molecules of PCl3 and Cl2. Actually, the two P—Cl (a) bonds with more bond length break away from the molecule leaving three P— Cl(e) bonds attached to the central P atom since these are more firmly linked
PCl5 $$\underrightarrow { heat }$$ PCl3 + Cl2

Question 10.
Write the balanced equation for the hydrolytic reaction of PCl5 in heavy water.
With heavy water (D2O) ; phosphorus pentachloride (PCl5 reacts as follows :

Question 11.
What is the basicity of H3PO4 ?
The acid is tribasic since it has three P—OH bonds which can release H+ ions.

Question 12.
What happens when phosphorus acid (H3PO3) is heated ? (C.B.S.E. 2008)
In phosphorus acid (H3PO3), central atom P is in +3 oxidation state. Upon heating, it gives a mixture PH 3 (P in -3 oxidation state) and H3PO4 (P in +5 oxidation state). This means that phosphorus acid undergoes disproportionation reaction.

Question 13.
List the important sources of sulphur.
Combined sulphur exists as sulphates, such as gypsum, epsom, baryte and sulphides such as galena, zinc blende, copper pyrites, etc. Traces of sulphur occur as hydrogen sulphide in volcanoes. Few organic materials like eggs, proteins, garlic, onion, mustard, hair and wool contain sulphur. 0.03 – 0.1% sulphur is present in the earth’s crust.

Question 14.
Write the order of the thermal stability of the hydrides of group 16 elements.
The order of thermal stability of hydride is :
H2O > H2S > H2Se > H2Te
This is related to the bond dissociation enthalpies of the E—H bonds where E stands for the element.
E—H bond :                                           O—H S—H Se—H Te—H
Bond dissociation enthalpy : (kJ mol-1) 463    347      276  238
Based on bond dissociation enthalpy, H2O is maximum stable thermally while H2Te is the least stable.

Question 15.
Why is H2O a liquid and H2S a gas?

In H2O, the electronegativity difference between 0(3·5) and H(2·1) is more than difference between S(2·5) and H(21) in H2S. As a result, O—H bond is more polar than S—H bond. This leads to inter molecular hydrogen bonding in H20 molecules while it is almost absent in the molecules of H2S. The H2O molecules get associated and consequently exist as liquid (water). The association in H2S molecules is negligible and it exists as a gas at room temperature.

Question 16.
Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe
Pt being a noble metal does not react with oxygen directly. In contrast, Zn, Ti and Fe are active metals and hence they react with oxygen directly to form their oxides.

Question 17.
Complete the following reactions :
(i) C2H4 + O2
(ii) Al + O2
(i) C2H4 + 3O2 $$\underrightarrow { heat }$$ 2CO2 + 2H2O
(ii) 4Al + 3O2 $$\underrightarrow { heat }$$ 2Al2O3

Question 18.
Why does O3 act as a powerful oxidising agent ? (C.B.S.E. 2013)
Upon heating, ozone (O3) readily decomposes to give molecular oxygen (O2) which is more stable along with nascent oxygen (O). The released nascent oxygen readily takes part in oxidation reactions. Therefore, ozone acts as a powerful oxidising agent.
O3 $$\underrightarrow { heat }$$ O2 + O (Nascent)

Question 19.
How is ozone estimated quantitatively ?
When ozone reacts with an excess of KI solution buffered with a borate buffer (pH = 9.2), iodine is liberated which can be titrated against standard solution of sodium thiosulphate. This is used as a method of estimation of ozone quantitatively.

Question 20.
What happens when sulphur dioxide gas is passed through an aqueous solution of Fe(III) salt ?
The gas is a reducing agent and reduces a Fe(III) salt to Fe(II) salt.

Question 21.
Comment on the nature of two S—O bonds formed in SO2 molecule. Are the two bonds in the molecule equal ?
The two S—O bonds in the molecule are equal with bond length equal to 143 pm. This means that SO2 molecule exhibits two resonating structures as shown below.

Question 22.
How is presence of SO2 detected ?
Ans.
Presence of SO2 is detected by bringing a paper dipped in acidified potassium dichromate near the gas. If the paper turns green, it shows the presence of SO2 gas.

Question 23.
Mention three areas in which H2SO4 plays an important role.
(i) It is used in the manufacture of fertilizers such as (NH4)2 SO4 , calcium superphosphate.
(ii)It is used as an electrolyte in storage batteries.
(iii)It is used in petroleum refining, detergent industry and in the manufacture of paints, pigments and dyes.

Question 24.
Write the conditions to maximise the yield of sulphuric acid by Contact process.
Catalytic oxidation of sulphur dioxide into sulphur trioxide. Sulphur dixocide is oxidised to sulphur trioxide with air in the presence of V2O5 or platinised asbestos acting as the catalyst

Question 25.
Why is $${ K }_{ { a }_{ 2 } }$$ < $${ K }_{ { a }_{ 1 } }$$ for H2SO4 in water ?
H2SO4 is a very strong acid in water largely because of its first ionisation to H3O+ and HSO4– The ionisation of HSO4 to H3O+ and SO42- is very very small. That is why, $${ K }_{ { a }_{ 2 } }$$ < $${ K }_{ { a }_{ 1 } }$$.

Question 26.
Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F2 and Cl2.
Fluorine is a better oxidising agent than chlorine because E°F2/F- is higher than E°Cl2/Cl- It is mainly due to low bond dissociation energy, high hydration energy and lower electron gain enthalpy, non-availability of d-orbitals in valence shell, that results in higher reduction potential of F2 than chlorine.

Question 27.
Give two examples to show the anomalous behaviour of fluorine.
The anomalous behaviour of fluorine, the first member of the halogen family as compared to the rest of the members is due to its very small size, very high electronegativity and absence of vacant d-orbitals in the valence shell. It is supported by the following points.
(i) Fluorine shows only negative oxidation state of -1 in its compounds. The other members exhibit both positive and negative oxidation states.
(ii) Fluorine forms hexafluoride with sulphur (SF6). No other member of the family forms hexahalide with sulphur.

Question 28.
Sea is the greatest source of some halogens. Comment.
The name halogen is a Greek Word meaning lsea salt forming’. Sea is a major source of a members of halogens particularly chlorine, bromine and iodine and they exist as the soluble salts of sodium, potassium, calcium, magnesium etc. The deposits of dried up sea water contain sodium chloride and carnallite (KClMgCl2.6H2O). Sea weeds contain nearly 0-5 percent of iodine. Similarly Chile saltpeter contains about 0-2% of sodium iodate (NaIO3).

Question 29.
Give reason for the bleaching action of Cl2.
Bleaching by chlorine occurs in the presence of moisture. In fact, it releases nascent oxygen on reacting with HO which carries bleaching. Since the reaction cannot be reversed, the bleaching by chlorine is permanent.
Chlorine bleaches by oxidation Cl2 + H2O → HCl + HOCl → HCl + [O]
The nascent oxygen reacts with dye to make it colourless.

Question 30.
Name two poisonous gases which can be prepared from chlorine gas.
Two poisonous gases are phosgene and mustard gas.

Question 31.
Why is ICl more reactive than I2 ? (C.B.S.E. Outside Delhi 2012)
The reactivity of ICl is due to its polar nature (I—Cl). Iodine (I2) being non-polar is comparatively less reactive chemically.

Question 32.
Why is helium used in diving apparatus?
A mixture of helium and oxygen does not cause pain due to very low solubility of helium in blood as compared to nitrogen.

Question 33.
Balance the equation : XeF6 + H2O → XeO2F2 +HF (H.P. Board 2013)

Question 34.
Why has it been difficult to study the chemistry of radon ?
Radon (Rn) is a radioactive element with very short half period of 3-82 days. Therefore, it becomes quite difficult to study the details about the chemistry of the element.

NCERT EXERCISE

Question 1.
Discuss the general characteristics of group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity.
(i) Electronic configuration: All the elements in group 15 have 5 valence electrons. Their general electronic configuration is ns2np3

(ii) Oxidation states:
All these elements have 5 valence electrons and require three more electrons to complete octets. However, gaining electrons is very difficult as the nucleus will have to attract three more electrons. This can take place only with nitrogen as it is the smallest in size and the distance between the nucleus and the valance shell is relatively small. The remaining elements of this group show a formal oxidation state of -3 in their covalent compounds.

In addition to the -3 state, N and P also show -1 and.-2 oxidation states. All the elements present in this group show +3 and +5 oxidation states. However, the stability of the +5 oxidation state decreases down the group, whereas the stability of+3 oxidation state increases. This happens because of the inert pair effect.

(iii) Ionization energy and electronegativity First ionization energy decreases on moving down a group. This is because of increasing atomic sizes. As we move down a group, electronegativity decreases, owing to an increase in size.

(iv) Atomic size:
On moving down a group, the atomic size increases. This increase in the atomic size is attributed to an increase in the number of shells.

Question 2.
Why does the reactivity of nitrogen differ from phosphorus?
N2 exist as a diatomic molecule containing triple bond between two N-atoms. Due to the presence —of the triple bond between p  the two N-atoms, the bond dissociation energy is large (941 .4 kJ mol-1 ). As a result of this N2 is inert and unreactive whereas, phosphorus exists as a tetratomic molecule, containing P – P single bond. Due to the presence of single bond, the bond dissociation energy is weaker (213 kJmol-1 ) than N a N triple bond (941 .4 kJ mol-1 ) and moreover due to presence of angular strain in P4 tetrahedra. As a result of this, phosphorus is much more reactive than nitrogen.

Question 3.
Discuss the trends in chemical reactivity of group 15 elements.
(a) Hydrides: The hybrids are covalent with pyramidal structures  and the central atom is sp3 hybrilised. The presence of the lone pair of electrons on the central atom distorts the geometry of the molecules and the bond angle less than that of a regular tetrahedron.

(b) Halides:
Elements of group 15 form two types of halides viz. trihalides and pentahalides. The halides are predominantly basic (Lewis bases) in nature and have lone pair of electrons (central atom is sp3 hybridized). The pentahalides are thermally less stable than the trihalides.

(c) Oxides: All the elements of this group form two types of oxides ie., M2O3 and M2O5 and are called trioxides and pentoxides

Question 4.
Why does NH3 form hydrogen bonding while PH3 does not?
The N—H bond in ammonia is quite polar on account of the electronegativity difference of N (3·0) and H (2·1). On the contrary, P—H bond in phosphine is almost non-polar because both P and H atoms have almost same electronegativity (21). Due to polarity, intermolecular hydrogen bonding is present in the molecules of ammonia but not in those of phosphine.

Question 5.
How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved.
Laboratory preparation: Dinitrogen is prepared in the laboratory by heating a solution containing an equivalent amount of sodium nitrite and ammonium chloride.

Question 6.
How is ammonia manufactured industrially?
Ammonia is prepared on a commercial scale by Haber’s Process from dinitrogen and dihydrogen by the following chemical reaction
N2 + 3H2 ⇌ 2NH3; ∆fH° = – 46.1 kJ mon-1
Dihydrogen needed for the commercial preparation of ammonia is obtained by the electrolysis of water while dinitrogen is obtained from the liquefied air as a fractional distillation. The two gases are purified and also dried.
These are compressed to about 200 atmosphere pressure and are then led into the catalyst chamber packed with the catalyst and the promoter.

Question 7.
Illustrate how copper metal gives different products on reaction with HNO3.
Anhydrous nitric acid is colourless, fuming, and pungent-smelling liquid. However, it acquires a yellowish-brown colour in the presence of sunlight. HNO3 decomposes to give NO2 gas which dissolves and imparts its yellowish-brown colour.

On strong heating, the acid decomposes to give NO2 and O2

Question 8.
Give the resonating structures of NO2 and N2O5.

Question 9.
The HNH angle value is higher than those of HPH, HAsH, and HSbH angles; why?

The above trend in the H-M-H bond angle can be explained on the basis of the electronegativity tif central atom. Since nitrogen is highly electronegative, there is high electron density around nitrogen. This causes greater repulsion between the electron pairs around nitrogen, resulting in maximum bond angle. We know that electronegativity decreases on moving down a group consequently. The repulsive interactions between the electron pairs decrease, thereby decreasing the HMH bond angle.

Question 10.
Why does R3P = O exist but R3N = O does not (R = alkyl group)?
Nitrogen does not contain d-orbitals. As a result, it cannot expand its covalency beyond four and cannot form pπ – dπ multiple bonds. In contrast, P contains the d-orbitals, and can expand its covalency beyond 4 and can form pπ-dπ multiple bonds. Hence R3P = O exist but R3N = O does not.

Question 11.
Explain why is NH3 basic while PH3 is feebly basic in nature. (C.B.S.E. Outside Delhi 2008, 2009, Jharkhand Board 2009)
NH3 is distinctly basic while BiH3 is feebly basic.
Nitrogen has a small size due to which the lone pair of electrons are concentrated in a small region. This means that the charge density per unit volume is high. On moving down a group, the size of the central atom increases, and the charge gets distributed over a large area decreasing the electron density. Hence the electron-donating capacity of group 15 elements hydrides decreases on moving down the group.

Question 12.
Nitrogen exists as a diatomic molecule (N2) while phosphorus a tetra-atomic molecule (P4). Why?
Nitrogen is diatomic gaseous molecule at ordinary temperature due to its ability to form pπ – pπ multiple bonds. The molecule has one σ and two π – bonds. Phosphorus exists as discrete tetratomic tetrahedral molecules as these are not capable of forming multiple bonds due to repulsion between non-bonded electrons of the inner core.

Question 13.
Write the main difference between the properties of white and red phosphorus. (C.B.S.E. Delhi 2012)

Question 14.
Why does nitrogen show catenation properties less than phosphorus?
The catenation properties depend upon the strength of the element-element bond. The N-N bond strength is much weaker (due to the repulsion of lone pairs on nitrogen because of its small size) than the P-P bond strength, therefore, nitrogen shows catenation less than phosphorus.

Question 15.
Give one disproportionation reaction of phosphorus acid (H3PO3).
Upon heating to about 573 K, phosphorus acid undergoes disproportionation as follows :

Question 16.
Can PCl5 act as oxidising as well as a reducing agent? Justify.
In general, the molecules of a substance can behave as a reducing agent if the central atom is in a position to increase its oxidation number. Similarly, they can act as an oxidising agent if the central atom is in a position to decrease its oxidation number. Now, the maximum oxidation state or oxidation number of phosphorus (P) is +5. It cannot increase the same but at the same time can decrease its oxidation number. In PCl5, oxidation number of P is already +5. It therefore, cannot act as a reducing agent. However, it can behave as an oxidising agent in certain reactions in which its oxidation number decreases. For example,

Question 17.
Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation states and hydride formation.
The members of the oxygen family are placed in group 16 of p-block. Their inclusion in the same group is justified on the basis of the following characteristics.
1. Electronic configuration. Members of the family have ns2p4 configuration. Their group (10 + 6) is 16.
2. Oxidation states. With exception of oxygen which exhibits -2 oxidation state in its compounds (OF2 and H2O2 are exceptions), rest of the members of the family show variable oxidation states (-2, +2, +4, +6) in their compounds.
3. Hydride formation. All the members of the family form covalent hydrides (MH2) in which the central atom is sp3 hybridised. These have angular structures and their characteristics show regular gradation.
FeS + H2SO4 (dil.) → FeSO4 + H2S
Na2Se + H2SO4 (dil.) → Na2SO4 + H2Se

Question 18.
Why is dioxygen a gas while sulphur is a solid? (C.B.S.E. Delhi 2013)
The oxygen atom has a tendency to form pπ-pπ multiple bonding due to its small size and high electronegativity. As a result, oxygen exists as a diatomic molecule. These mol¬ecules are held together by weak van der Waal’s forces of attraction which can be easily overcome by collisions of the molecules at room temperature. Therefore, O2 is the gas at room temperature.

Sulphur, on the other hand, because of its bigger size and lower electronegativity, does not form pπ-pπ multiple bonds. Instead, it prefers to form an S-S single bond and form polyatomic complex molecules having eight atoms per molecule (S8) and have puckered ring structure. Therefore, S atoms are strongly held together and it exists as a solid.

Question 19.
Knowing the electron gain enthalpy values for O → O and O → O2- as – 141 kJ mol-1 and 702 kJ mol-1 respectively, how can you account for the formation of a large number of oxides having O2- species and not O ?
The stability of an ionic compound depends on its lattice energy. The more the lattice energy of a compound, the more stable it will be. Lattice energy is directly proportional to the charge carried by an ion. When a metal combines with oxygen, the lattice energy of the oxide involving O2- ion is much more than the oxide involving O ion. Hence, we can say that the formation of O2- is energetically more favourable than the formation of O

Question 20.
Which aerosols deplete ozone?
Aerosols or chlorofluorocarbons (CFC’s) such as freon (CCl2F2) deplete ozone layer by supplying chlorine-free radicals (Cl) which convert ozone into oxygen.

Question 21.
Describe the manufacture of H2SO4 by Contact process.
Sulphuric acid is manufactured by the Contact process which involves three steps :

1. burning of sulphur or sulphide ores in air to generate SO2.
2. conversion of SO2 to SO3 by the reaction with oxygen in the presr nee of a catalyst (V2O5), and
3. absorption of SO3 in H2SO4 to give oleum (H2S2O7).

A flow diagram for the manufacture of sulphuric acid is shown in the figure. The SO2 produced is purified by removing dust and other impurities such as arsenic compounds. The key step in the manufacture of H2SO4 is the catalytic oxidation of SO2 with O2 to give SO3 in the presence of V2O5 (catalyst).

The reaction is exothermic, reversible and the forward reaction leads to a decrease in volume. Therefore, low temperature and high pressure are favourable conditions for maximum yield. But the temperature should not be very low otherwise rate of reaction will become slow.

In practice, the plant is operated at a pressure of 2 bar and a temperature of 720 K. The SO3 gas from the catalytic converter is absorbed in concentrated H2SO4 to produce oleum. Dilution of oleum with water gives H2SO4 of the desired concentration. In the industry, two steps are carried out simultaneously to make the process a continuous one and also to reduce the cost.

Question 22.
How is SO2 an air pollutant?

• It combines with water vapour present in the atmosphere to form sulphuric acid. This causes acid rain. Acid rain damages soil, plants and buildings, especially those made of marble.
• Even in very low concentrations, SO2 causes irritation in the respiratory tract. It causes throat and eye irritation and can also affect the larynx to cause breathlessness.
• It is extremely harmful to plants. Plants exposed to sulphur dioxide for a long time lose colour from their leaves. This condition is known as chlorosis. This happens because the formation of chlorophyll is affected by the presence of SO2.

Question 23.
Why are halogens strong oxidising agents?
The general electronic configuration of halogens is np5, where n = 2-6. Thus, halogens need only one more e- to complete their octet and to attain the stable noble gas configuration. Also, halogens are highly electronegative with low dissociation energies and high negative electron gain enthalpies. Therefore, they have a high tendency to gain an election. Hence, they act as strong oxidizing agents.

Question 24.
Explain why does fluorine form only one oxoacid (HOF).
The members of the halogen family with the exception of fluorine show variable oxidation states due to the availability of rf-orbitals for the bond formation. They form a number of oxoacids such as HOX, HOXO, HOXO2, and HOXO3. However, fluorine which is highly electronegative and has no rf-orbitals, forms only one oxoacid (HOF) in which its oxidation state is +1.

Question 25.
Explain why in spite of nearly the same electronegativity, nitrogen is involved in hydrogen bonding while chlorine is not.
Both chlorine and oxygen have almost the same electronegativity values, but chlorine rarely forms hydrogen bonding. This is because in comparison to chlorine, oxygen has a smaller size and as a result, a higher electron density per unit volume.

Question 26.
Write two uses of ClO2.
ClO2 is a strong oxidising agent. Therefore,
(i) It acts as bleaching agent for paper pulp in paper industry and in textile industry.
(ii) It acts as germicide for disinfecting water.

Question 27.
Why are halogens coloured?
All the halogens are coloured in nature. The colour deepens with the increase in the atomic number of the element from fluorine to iodine.

The cause of the colour is due to the absorption of energy from the visible light by the molecules for the excitation of outer non -bonded electrons to higher energy levels. The excitation energy depends upon the size of the atom. Fluorine has the smallest size and the force of attraction between the nucleus and electrons is very large.

Question 28.
Write the reactions of F2 and Cl2 with water.

Question 29.
How can you prepare Cl2 from HCl and HCl from Cl2? Write chemical equations only.
(i) HCl can be oxidised to chlorine with the help of a number of oxidising agents like MnO2, KMnO4, K2Cr2O7 etc.
Mn0O2 + 4HCl → MnCl2 + Cl2 + 2H2O
(ii) Cl2 can be reduced to HCl by reacting with H2 in the presence of sunlight. The gas on passing through water dissolves to form hydrochloric acid

Question 30.
What inspired N. Bartlett for carrying out the reaction between Xe and PtF6?
The X-ray study of the compound has shown it be a crystalline solid consisting of O2+ and (PtF6] ions. In this reaction, PtF6 has oxidised O2 to O2+ ion. Bartlett through than PtF6 should Xe to xe+ since first ionisation enthalpy of xenon (1176 kJ mol-1) is quite close to that of O2 (1180 kJ mol-1).

Question 31.
What is the oxidation state of phosphorus in the following?
(a) H3PO3
(b) PCl3
(c) Ca3P2
(d) Na3PO4
(e) POF3

Question 32.
Write balanced equations for the following :
(i) NaCl is heated with sulphuric acid in the presence of MnO2
(iii) Chlorine gas is passed into a solution of Nal in water.

Question 33.
How are xenon fluorides XeF2, XeF4 and XeF6 obtained ?
Xenon forms three binary fluorides, XeF2, XeF4 and XeF6 by the direct reaction of elements under appropriate experimental conditions.

Question 34.
With what neutral molecule is ClO isoelectronic? Is that molecule a Lewis base?
OF2 and ClF are isoelectronic to ClO, out of which ClF is a Lewis base.

Question 35.
How are XeO3 and XeOF4 prepared?
Preparation of XeO3. By complete hydrolysis of XeF6.
XeF6 + 3H2O → XeO3+ 6HF.
Preparation ofXeOF4. By partial hydrolysis of XeF6.
XeF6 + H2O → XeOF4+ 2HF.

Question 36
Arrange the following in the order of property indicated for each set:
(i) F2, Cl2, Br2, I2 – increasing bond dissociation enthalpy.
Bond dissociation energy usually decreases on moving down a group as the atomic size increases. However, the bond dissociation energy of F2 is lower than that of Cl2 and Br2. This is due to the small atomic size of fluorine, Thus, the increasing order for bond dissociation energy among halogens is as follows:
I2 < F2 < Br2 < Cl2

(ii) HF, HCl, HBr, HI – increasing acid strength.
HF < HCl < HBr < HI
The bond dissociation energy of HX molecules where X – F, Cl, Br, I, decreases with an increase in the atomic size. Since H – I bond is the weakest, HI is the strongest acid.

(iii) NH3, PH3, ASH3, SbH3, BiH3 – increasing base strength.
BiH3 < SbH3 < AsH3 < PH3 < NH3 On moving from nitrogen to bismuth, the size of the atom increases while the electron density on the atom decreases. Thus, the basic strength decreases.

Question 37.
Which one of the following does not exist?
(i) XeOF4
(ii) NeF2
(iii) XeF2
(iv) XeF6.
NeF2 does not exist because the element Ne(Z = 10) with 1s22s22p6 does not have vacant 2d orbitals. As such, there is no scope of any electron promotion even by a highly electronegative element.

Question 38.
Give the formula and describe the structure of a noble gas species which is isostructural with :
(i) $$IC{ l }_{ 4 }^{ – }$$
(ii) $$IB{ r }_{ 2 }^{ – }$$
(iii) $$Br{ O }_{ 3 }^{ – }$$

(i) Structure of $$IC{ l }_{ 4 }^{ – }$$. The central I atom has in all 8 electrons (7 valence electrons +1 due to negative charge). Out of these, it shares 4 electrons with four atoms of Cl and the remaining four electrons constitute two lone pairs. In all, there are six pairs. The structure of the ion must be octahedral or distorted square planar in order to minimise the forces of repulsion among the two electrons pairs. $$IC{ l }_{ 4 }^{ – }$$ has (7 + 4 x 7 + 1) = 36 valence electrons and it iso-electronic and iso-structural with XeF4 (8 + 4 x 7) which has also 36 valence electrons.

(ii) Structure of $$IB{ r }_{ 2 }^{ – }$$.
In $$IB{ r }_{ 2 }^{ – }$$ ion, the central I atom has 8 valence electrons (7 + 1). Out of these, it shares 2 electrons with two atoms of Br and the remaining 6 electrons constitute three lone pairs. In all, there are five pairs. The structure of the ion must be trigonal bipyramidal or linear in order to minimise the force of repulsion among the three lone electrons pairs. IBrf has (7 + 2 x 7 + 1) = 22 valence electrons and is isoelectronic as well as iso-structural with noble gas species XeF2 which has also 22 (8 + 2 x 7) electrons.

(iii) Structure of $$Br{ O }_{ 3 }^{ – }$$ ion. In $$Br{ O }_{ 3 }^{ – }$$ ion, the central Br atom has 8 valence electrons (7 + 1). Out of these, it shares 4 with two atoms of O forming Br=0 bonds. Out of the remaining four electrons, 2 are donated to the third O atom and account for its negative charge. The remaining 2 electrons constitute one lone pair. In order to minimise the force of repulsion, the structure of $$Br{ O }_{ 3 }^{ – }$$ ion must be pyramidal. $$Br{ O }_{ 3 }^{ – }$$ ion has (7 + 3 x 6 + 1) = 26 valence electrons and is isoelectronic as well as iso-structural with noble gas species Xe03 which has also 26 (8 + 3 x 6) electrons.

Question 39.
Why do noble gases have comparatively large atomic sizes?
Noble gases do not form molecules. In the case of noble gases. the atomic radii correspond to Van der Waa1s radii. On the other hand, the atomic radii of other elements correspond to their covalent radii. By definition, Yan der Waal’s radii are larger than covalent radii. It is for this reason that noble gases are very large in size as compared to other atoms belonging to the same period.

Question 40.
List the uses of neon and argon gases.
Uses of Neon:

• It consists of a long tube fitted with electrodes at both ends.
• On filling the tube with neon gas and passing electric discharge of about 1000 volt potential, a bright red light is produced.
• Different colours can be obtained by mixing neon with other gases.
• For producing blue or green light neon is mixed with mercury vapours.

Uses of Argon:

• Like helium, it is also used to create an inert atmosphere in welding aluminium and stainless steel.
• It is filled in electric bulbs along with 25% nitrogen.
• It is also used in radio valves.
• Argon alone or its mixture with neon is used in tubes for producing lights of different colours.

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## NCERT Solutions for Class 12 Chemistry Chapter 4 Chemical Kinetics

NCERT Solutions for Class 12 Chemistry Chapter 4 provides excellent solutions for the questions asked in the textbook. The step wise solutions and diagrammatic representations make the concepts easy to understand. The subject experts have given the best explanations to the queries. The students appearing for the board exams or competitive exams can refer to these for better preparations.

CBSE, MP board, UP board, Gujarat board, etc. have the NCERT Solutions for reference in the curriculum. Chemistry is an important subject and requires conceptual analysis. The detailed explanations in the NCERT Solutions for Class 12 Chemistry Chapter 4 will help the students to score well in the examination.

 Board CBSE Textbook NCERT Class Class 12 Subject Chemistry Chapter Chapter 4 Chapter Name Chemical Kinetics Number of Questions Solved 39 Category NCERT Solutions

## NCERT Solutions for Class 12 Chemistry Chapter 4 Chemical Kinetics

Class 12 Chemistry chapter 4 Chemical Kinetics is an important chapter and is often asked in the examination. Chemical kinetics helps to understand the chemical reactions. This chapter explains all about the rate of reaction and the factors determining the rate of reaction.

NCERT IN-TEXT QUESTIONS

Question 1.
For a reaction R → P, the concentration of a reactant changes from 0·03 M to 0·02 M in 25 minutes. Calculate the average rate of the reaction using the units of seconds.
For a reaction, R → P

Question 2.
In a reaction, 2A → Products, the concentration of A decreases from 0·5 mol L-1 to 0·4 mol L-1 in 10 minute. Calculate the rate during this interval.
For the reaction: 2A → Products

Question 4.
Fora reaction,A+B —> Product; the rate law is given by, r =k [ A]1/2 [B]2. What is the order of the reaction?
Order of reaction. = 1/2+ 2 = 21/2 or 2.5

Question 4.
The conversion of the molecules X to Y follows second order kinetics. If concentration of X is increased to three times, how will it affect the rate of formation of Y ?
For the reaction X → Y
Reaction rate (r) = k[X]2
If the concentration be increased to three times, then
Reaction rate (r’) = k [3X]2
$$\frac { { r }^{ ‘ } }{ r } =\frac { k\left[ 3X \right] ^{ 2 } }{ k\left[ X \right] ^{ 2 } } =9$$

Question 5.
A first order reaction has rate constant of 1·15 x 10-3 s-1. How long will 5 g of this reactant take to reduce to 3g?
For the first order reaction,

Question 6.
Time required to decompose SO2Cl2 to half of its initial concentration is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
For the first order reaction ;
Rate constant (k) = $$\frac { 0.693 }{ { t }_{ 1/2 } } =\frac { 0.693 }{ \left( 60min \right) }$$
= $$\frac { 0.693 }{ \left( 60\times 60s \right) } =1.925\times { 10 }^{ -4 }{ s }^{ -1 }$$

Question 7.
What will be the effect of temperature on rate constant?
With the rise in temperature by 10°, the rate constant of a reaction is nearly doubled. The dependence of rate constant on temperature is given the Arrhenius equation, k = A e-Ea/RT where A is the Arrhenius constant and Ea is activation energy of the reaction.

Question 8.
In general, it is observed that the rate of a chemical reaction doubles with every 10° rise in temperature. If the generalisation holds for a reaction in the temperature range 295 K to 305 K, what would be the value of activation energy for the reaction? (C.B.S.E. Delhi2005 Supp., Pb. Board2007)
According to Arrhenius equation,

Question 9.
The activation energy for the reaction, 2HI(g) → H2(g) + I2(g) is 208·5 kJ mol-1. Calculate fraction of molecules of reactants having energy equal to or greater than activation energy.
The fraction of the molecules (x) having energy equal to or more than activation energy may be calculated as follows :

NCERT Exercise

Question 1.
The rate expression for the following reactions determine the order of reaction and the dimensions of the rate constant.

(a) 2
(b) 2
(c) 3/2
(d) 1

Question 2.
For the reaction, 2A + B → A2 B, the rate = k [AJ[B]2 with k = 2.0 x 10-6 mol-2 L2 s-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L-1, [B] = 0.2 mol L-1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L-1.
Initial rate of reaction = k [A] [B]2
= (20 x 10-6 mol-2 s-1) (0.1 mol L-1) (0.2 mol L-1)2 = 8 x 10-9molL-1 s-1.
When [A] is reduced from 010 mol L-1 to 0.06 molL-1, i.e., 0.04 mol L-1 of A has reacted, the concentration of B reacted, is = 1/2 x 0.04 mol L-1 = 0.02 mol L-1
Concentration of B, remained after reaction with A = 0.2 – 0-02=0.18 mol L-1
Now, rate=(20 x 10-6 mol-2 L2 s-1) (0.06 mol L-1) (0.18 molL-1)2
= 3-89 x 10-9mol L-1 s-1

Question 3.
The rate of decomposition of NH3 on the platinum surface is zero order. What is the rate of production of N2 and H2 if k = 2·5 x 10-4 Ms-1? (C.B.S.E. Delhi 2008)

Question 4.
The decomposition of dimethyl ether leads to the formation of CH4, H2, and CO, and the reaction rate is given by the expression:
rate = k [CH3OCH3]3/2
The rate of reaction is followed by increase in pressure in a close vessel and the rate can also be expressed in terms of partial pressure of dimethyl ether :
rate = k [pCH3OCH3]3/2
If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constant ?

Question 5.
Mention the factors that affect the rate of a chemical reaction.

• Concentration of reactants
• Temperature
• Nature of reactants and products
• Presence of catalysts.

Question 6.
A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is
(i) doubled
(ii) reduced to 1/2? (C.B.S.E. Outside Delhi 2008, 2009)
Let the reaction be; A → Products
Reaction rate (r) = k [A]2 (for second order reaction)
(i) When concentration is doubled, the rate of reaction may be expressed as :
Reaction rate (r’) = k [2A]2

reaction rate becomes four times.
(ii) When concentration is reduced to half, the rate of reaction may be expressed as :

reaction rate will be reduced to 1/4.

Question 7.
What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively?
Increasing the temperature on decreasing the activation energy will result in an increase in the rate of reaction and an exponential increase in the rate constant. On increasing the temperature the fraction of molecules which collide with energy greater than Ea increases and hence the rate constant (exponentially)
K = A -ea/RT, quantitative representation of temperature effect on rate constant.

Question 8.
In pseudo-first-order hydrolysis of ester in water, the following results were obtained.

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.

Question 9.
A reaction is first order in A and second order in B
(i) Write differential rate equation.
(ii) How is rate affected when the concentration of B is tripled?
(iii) How is rate affected when the concentration of both A and B are doubled? (C.B.S.E. Outside Delhi 2010, 2013)

Question 10.
In a reaction between A and B, the initial rate of reaction was measured for different initial concentration of A and B as given ahead :

What is the order of reaction with respect to A and B?

Question 11.
The following data were obtained at 300 K for the reaction 2A + B → C + D:

Calculate the rate of formation of D when [A] = 0·5 mol L-1 and [B] = 0·2 mol L-1.

Question 12.
The reaction between A and B is first order with respect to A and zero-order with respect to B. Fill in the blanks in the following table:

The rate equation for the reaction is: r = k [A]1 [B]0
(i) Comparing experiments I and II,

Thus, the concentration of A in experiment II is 0·2 M
(ii) Comparing experiments II and in.
When the concentration of A is made double, the reaction rate will also become twice.
∴ Rate of reaction in experiment III is 8·0 x 10-2
(iii) Comparing experiments I and IV.
Since the reaction rates are the same in both the experiments, the molar concentration of A in experiment IV must be the same as in experiment I i. e., it must be 0·1 M.

Question 13.
Calculate the half-life of the first-order reaction from their rate constants given as
(a) 200 s-1
(b) 2 min-1
(c) 4 year-1.

Question 14.
The half-life for the radioactive decay of 14C is 5730 Y. An archaeological artifact contained wood had only 80% of the 14 C found in a living tree. Estimate the age of the sample. (C.B.S.E. Delhi 2008)

Question 15.
The experimental data for decomposition of N2O2 [2N2O5 → 4NO2 + O2] in gas phase at 318 K are given below :

(a) Plot [N2O5] against t
(b) Find the half-life period for the reaction
(c) Draw a graph between log [N2O5] and t
(d) What is rate law?
(e) Calculate the rate constant
(f) Calculate the half-life period from k and compare it with (b).
The available data is:

(b) Initial cone. of N2O5 = 1·63 x 10-2M. Half of initial cone. = 1/2 x (1·63 x 10-2 M) = 0·815 x 10-2 M Time corresponding to half of inital concentration (t/2) from the plot (a) = 1400 (s) approximately
(c) The graph of log [N2O5] Vs. time has been plotted.
(d) Since the graph between log [N2O5] and time is a straight line the reaction is of the first order
The rate equation : rate (r) = k[N2O5]

Question 16.
The rate constant for the first-order reaction is 60 s-1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value? (C.B.S.E. Delhi 2013)
For the first-order reaction

Question 17.
During a nuclear explosion, one of the products is 90Sr with a half period of 28·1 Y. If 1 pg of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically?

Question 18.
For a first-order reaction, show that the time required for 99% completion is twice the time required for the completion of 90% of the reaction.

Question 19.
A first-order reaction takes 40 minutes for 30% decomposition. Calculate its half-life period. (C. B. S. E. Outside Delhi 2013)

Question 20.
For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data is obtained.

Calculate the rate constant.
The decomposition reaction is of gaseous nature and the expression of the rate equation for the reaction is :

Question 21.
The following data were obtained during the first-order thermal decomposition of SO2Cl2 at a constant volume.
SO2Cl2 (g) → SO2 (g) + Cl2(g)

Calculate the rate of the reaction when the total pressure is 0·65 atm. (C.B.S.E. Sample Paper 2011)

Question 22.
The rate constant for the decomposition of N2O5 at various temperatures is given below :

Draw a graph between ln k and 1/T and calculate the values of A and Ea. Predict the rate constant at 30° and 50°C.
To draw the plot of log K versus 1/T, we can re-write the given data as follows :

Question 23.
The rate constant for the decomposition of a hydrocarbon is 2.418 × 10-5 s-1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of the pre-exponential factor.

Question 24.
Consider a certain reaction A → Products with k = 2·0 x 10-2 s-1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1·0 mol L-1.
For the first-order reaction :

Question 25.
Sucrose decomposes in acid solution into glucose and fructose according to the first-order rate law with t1/2 = 3·0 hrs. What fraction of the sample of sucrose remains after 8 hours? (C.B.S.E. Sample Paper 2011)

Question 26.
The decomposition of a hydrocarbon follows the equation :
k = (4·5 x 1011 s-1)e-28000k/T.
Calculate the energy of activation (Ea).

Question 27.
The rate constant for the first-order decomposition of H2O2 is given by the following equation:
log k = 14·34 – 1·25 x 104K/T.
Calculate the Ea for the reaction. At what temperature will the half-life period be 256 minutes?

Question 28.
The decomposition of A into the product has a value of k as 4·5 x 103 s-1 at 10°C and energy of activation 60 kJ mol-1. At what temperature would k be 1·5 x 104 s-1? (C.B.S.E. Sample Paper 2011)