Category: Class 11

Immediately get Chapter Wise Tamilnadu State Board Solutions for 11th Economics to gain more marks and start Quick Revision of all Chapters. All Chapters Pdf is provided along with the Questions and Answers. You Can Dowload Samacheer Kalvi 11th Economics Book Solutions Questions and Answers for Chapter wise are given with a clear explanation. Tamilnadu State Board Solutions for 11th Economics Chapter 8 Indian Economy Before and After Independence Questions and Answers is for free of cost.

Tamilnadu Samacheer Kalvi 11th Economics Solutions Chapter 8 Indian Economy Before and After Independence

Tamilnadu State Board Solutions for 11th Economics Chapter 8 Indian Economy Before and After Independence Questions and Answers PDF has all given in Chapter Wise Section. Check Out daily basis with Tamilnadu State Board Solutions 11th Economics PDF will help to improve your score. Improve your level of accuracy to answer a question by reading with Samacheer Kalvi 11th Economics Book Solutions Questions and Answers PDF.

Samacheer Kalvi 11th Economics Indian Economy Before and After Independence Text Book Back Questions and Answers

Part – A
Multiple Choice Questions

Question 1.
The arrival of Vasco da Gama in Calicut, India
(a) 1498
(b) 1948
(c) 1689
(d) 1849
Answer:
(a) 1498

Question 2.
In 1614 Sir Thomas Roe was successful in getting permission from
(a) Akbar
(b) Shajakan
(c) Jahangir
(d) Noorjakhan
Answer:
(c) Jahangir

Question 3.
The power for governance of India was transferred from the East India Company (EIC) to the British crown in
(a) 1758
(b) 1858
(c) 1958
(d) 1658
Answer:
(b) 1858

Question 4.
Ryotwari system was initially introduced in _______
(a) Kerala
(b) Bengal
(c) Tamil Nadu
(d) Maharastra
Answer:
(c) Tamil Nadu

Question 5.
First World War started in the year _______
(a) 1914
(b) 1814
(c) 1941
(d) 1841
Answer:
(a) 1914

Question 6.
When did the Government of India declared its first Industrial Policy ?
(a) 1956
(b) 1991
(c) 1948
(d) 2000
Answer:
(c) 1948

Question 7.
The objective of the Industrial Policy 1956 was
(a) Develop heavy industries
(b) Develop agricultural sector only
(c) Develop private sector only
(d) Develop cottage industries only
Answer:
(a) Develop heavy industries

Question 8.
The industry which was de-reserved in 1993 ?
(a) Railways
(b) Mining of copper and zinc
(c) Atomic energy
(d) Atomic minerals
Answer:
(b) Mining of copper and zinc

Question 9.
The father of Green Revolution in India was _______
(a) M.S. Swaminathan
(b) Gandhi
(c) Visweswaraiah
(d) N.R. Viswanathan
Answer:
(a) M.S. Swaminathan

Question 10.
How many commercial banks were nationalised in 1969 ?
(a) 10
(b) 12
(c) 14
(d) 16
Answer:
(c) 14

Question 11.
The main objective of nationalisation of banks was _______
(a) Private social welfare
(b) Social welfare
(c) To earn
(d) Industries monopoly
Answer:
(b) Social welfare

Question 12.
The Planning Commission was setup in the year _______
(a) 1950
(b) 1955
(c) 1960
(d) 1952
Answer:
(a) 1950

Question 13.
In the first five year plan, the top priority was given to _______ Sector.
(a) Service
(b) Industrial
(c) Agriculture
(d) Bank
Answer:
(c) Agriculture

Question 14.
Tenth Five year plan period was _______
(a) 1992-1997
(b) 2002-2007
(c) 2007-2012
(d) 1997-2002
Answer:
(b) 2002-2007

Question 15.
According to HDR (2016), India ranked _______ out of 188 countries.
(a) 130
(b) 131
(c) 135
(d) 145
Answer:
(b) 131

Question 16.
Annual Plans formed in the year _______
(a) 1989-1991
(b) 1990-1992
(c) 2000-2001
(d) 1981-1983
Answer:
(b) 1990-1992

Question 17.
The Oldest large scale industry in India ________
(a) cotton
(b) jute
(c) steel
(d) cement
Answer:
(a) cotton

Question 18.
Human development index (HDI) was developed _________
(a) Jawaharlal Nehru
(b) M.K.Gandhi
(c) Amartiya Sen
(d) Tagore
Answer:
(c) Amartiya Sen

Question 19.
The main theme of the Twelth Five Year Plan __________
(a) faster and more inclusive growth
(b) growth with social Justice
(c) socialistic pattern of society
(d) faster, more inclusive and sustainable growth
Answer:
(d) faster, more inclusive and sustainable growth

Question 20.
The PQLI was developed by _______
(a) Planning Commission
(b) Nehru
(c) Morris
(d) Morris
Answer:
(c) Morris

Part – B
Answer the following questions in one or two sentences

Question 21.
What are the Phases of colonial exploitation of India?
Answer:
The period of merchant capital, the period of industrial capital, the period of finance capital.

Question 22.
Name out the different types of land tenure existed in India before Independence.
Answer:

1. Zamindari system
2. Mahalwari system
3. Ryotwari system

Question 23.
State the features that distinguish a land tenure system from other system.
Answer:
The land tenure system differs as

1. Who owns the land ?
2. Who cultivates the land ?
3. Who is responsible for paying the land revenue to the government ?

Question 24.
List out the weaknesses on Green Revolution.
Answer:

1. Indian agriculture was still a gamble of the monsoons.
2. This strategy needed heavy investment.
3. The income gap has increased
4. Increased unemployment among agricultural labourers.
5. Reduced the soil fertility and spoiled human wealth.

Question 25.
What are the objectives of Tenth five year plan ?
Answer:

1. Double the per capita income in the next 10 years.
2. Aimed to reduce the poverty ratio to 15% by 2012.
3. Growth target was 8.0% but it achieved only 7.2%

Question 26.
What is the difference between HDI and PQLI ?
Answer:

1. Human Development Index : It is constructed based on life expectancy index, Education index and GDP per capita.
2. Physical quality of life index : The PQLI is a measure to calculate the quality of life.

Question 27.
Mention the indicators which are used to calculate HDI.
Answer:

1. Life expectancy index.
2. Educational index.
3. GDP per capita.

Part – C
Answer the following questions in one Paragraph

Question 28.
Explain about the Period of Merchant Capital.
Answer:

1. The period of merchant capital was from 1757 to 1813.
2. The only aim of the East India company was to earn profit by establishing monopoly trade.
3. During this period, India had been considered as the best hunting ground for capital
4. The objective of monopoly trade was fulfilled by achieving political control.
5. The company administration succeeded in generating huge surpluses which were repatriated to England.

Question 29.
The Handicrafts declined in India in British Period. Why?
Answer:

1. Through discriminatory tariff policy, the British Government purposefully destroyed the handicrafts.
2. With the disappearance of Nawabs and Kings, there was no one to protect Indian handicrafts.
3. Indian products could not compete with machine made products.
4. Introduction of railways in India increased the domestic market for the British goods.

Question 30.
Elucidate the different types of land tenure system in colonial India.
Answer:
The three types of land tenure system are Zamindari, Mahalwari and Ryotwari system.

1. Zamindari system or the land – lords : This system was created in 1793,after the introduction of permanent settlement act. Under this system the landlords were declared as the owners of the land. They were responsible to pay the land revenue. The share of the rent to government is fixed at 10/11th and the balance as zamindar’s remuneration.

2. Mahalwari system or communal farming : The ownership of the land was maintained by the collective body of villagers which served as a unit of management. They distribute the land and collect revenue and pay it to the state.

3. Ryotwari or the owner – cultivator system : The ownership rights of use and control of land were held by the tiller himself.

Question 31.
State the reasons for nationalization of commercial banks.
Answer:

1. The main objective of the economic planning aimed at social welfare.
2. Before independence commercial banks were in the private sector.
3. These commercial banks failed in helping the government to achieve social objectives of planning.
4. Therefore, the government decided to nationalize 14 major commercial banks on 1969

Question 32.
Write any three objectives of Industrial Policy 1991.
Answer:

1. Ensuring rapid industrial development in a competitive environment.
2. Enchancing support to the small scale sector.
3. Providing more incentives for industrialisation of the backward areas.

Question 33.
Give a note on Twelfth Five Year Plan.
Answer:

1. Twelfth five year plan (2012 – 17).
2. Its main theme is “Faster more inclusive and sustainable growth”
3. Its growth rate target is 8%

Question 34.
What is PQLI ?
Answer:
Morris D Morris developed the physical quality of life index (PQLI). The PQLI is a measure to calculate the quality of life.
PQLI includes three indicators life expectancy, infant mortality rate and literacy rate. A scale of each indicator ranges from the number 1 to 100. Represents the worst performance and 100 is the best performance.
HDI includes income while PQLI do not. PQLI has only the physical aspects of like.

Part – D
Answer the following questions in about a page

Question 35.
Discuss about the Indian economy during British Period.
Answer:
Britain had exploited India over a period of two centuries of its colonial rule. On the basis of the form of colonial exploitation, economic historians have divided the whole period into three phases namely the period of merchant capital, period of industrial capital and the period of finance capital.

1. Period of merchant capital (1757 to 1813):
The only aim of the East India company was to earn profit by establishing monopoly trade. India was considered as the best hunting ground for capital. By attaining political power the objective of monopoly trade was fulfilled. The company administration succeeded in generating huge surpluses which were repatriated to England.

2. Period of industrial capital (1813 to 1858):
India had become a market for British textiles. Indians were exploited by fixing low price for exports and high price for imports. India’s traditional handicrafts were thrown out of gear.

3. Period of financial capital (Late 19th century – 1947) : Finance imperialism began to entrench. Britain decided to make massive investments in various fields by plundering Indian capital.

4. The land tenure system in India : Land tenure refers to the system of land ownership and managements.
The Zamindari system, the Mahalwari system and the Ryotwari system were the systems of land tenure introduced by British.

5. Problems of British rule : Their profit motives led to drain of resources from India. The handicraft industries were collapsed.
The British rule stunted the growth of Indian enterprise. Capital formation in India were retarded.

Question 36.
Explain the role of SSIs in economic development?
Answer:

1. SSIs provide employment:

• SSIs use labour intensive techniques thus reduce the problem of unemployment to a great extent.
• They provide employment to people in villages and unorganized sectors.
• The employment – capital ratio is high for the SSIs.

2. SSIs bring balanced regional development: SSIs remove regional disparities by industrializing rural and backward areas and bring balanced regional development.

3. Help in mobilization of local resources : SSIs help to mobilize and utilize local resources like small savings, entrepreneurial talent etc.,

4. Pave for optimization of capital: SSIs require less capital per unit of output. They function as a stabilizing force by providing output-capital ratio as well as high employment capital ratio.

5. Promote exports : SSIs earn valuable foreign exchange through exports from India.

6. Complement large scale industries : SSIs serve as ancillaries to large scale units.

7. Meet consumer demands : SSIs serves as an anti-inflationary force by providing goods of daily use.

8. Develop entrepreneurship : They promote self-employment and spirit of self-reliance in the society. They help to increase the per capita income. They help in distributing national income in more efficient and equitable manner.

Question 37.
Explain the objectives of nationalization of commercial banks.
Answer:

1. The main objective of nationalization was to attain social welfare. Sectors such as agriculture, small and village industries were in need of funds for their expansion and further economic development.
2. It helped to curb private monopolies in order to ensure a smooth supply of credit to socially desirable sections.
3. To encourage the banking habit among the rural population.
4. To reduce the regional imbalances where the banking facilities were not available.
5. After nationalization, new bank branches were opened in both rural and urban areas, and they created credit facilities mainly to the agriculture sector and its allied activities.

Question 38.
Describe the performance of 12th five year plan in India.
Answer:
Economic planning is the process in which the limited natural resources are used skillfully so as to achieve the desired goals.

1. First five year plan (1951 – 56) : Main focus was on the agricultural development it achieved the growth rate of 3.6%.

2. Second five year plan (1956 – 61) : Focus was on the industrial development of the country and achieved the growth rate of 4.1%.

3. Third five year plan (1961 – 66): To make the economy independent and to reach take off.

4. Plan holiday (1966 – 69): Equal priority was given to agriculture, its allied sectors and the industrial sectors during annual plans.

5. Fourth five year plan (1969 – 74): Growth with stability and progressive achievement of self reliance was the goal but the plan failed.

6. Fifth five year plan (1975 – 79): A successful plan prioritized agriculture and then industry and mines.

7. Rolling plan : This plan was started with an annual plan for 1978-79.

8. Sixth five year plan (1980 – 85): Based on investment yojana, its objective was poverty eradication and technological self reliance.

9. Seventh five year plan (1985 – 90): Establishment of the self sufficient economy and opportunities for productive employment. Private sector got the priority over public sector.

10. Annual plans : Two annual plans were formed in 1990 – 91 & 1991 -92.

11. Eighth five year plan (1992 – 97): Priority was given to development of the human resources. New economic policy was introduced.

12. Ninth five year plan (1997 – 02):

• Aimed to double the percapita income in the next 10 years.
• Aimed to reduce the poverty ratio to 15% by 2012.

13. Tenth five year plan (2002 – 07): Growth with justice and equity was the focus but the plan failed with growth rate of 5.6%. Aimed to double the percapita income in the next 10 years. Aimed to reduce the poverty ratio to 15% by 2012.

14. Eleventh five year plan (2007 – 12): Main theme was faster and more inclusive growth.

15. Twelfth five year plan (2012 – 17): Main theme was faster more inclusive and sustainable growth. The five year plans played a’very prominent role in the economic development of the country. These plans had guided the government as to how it should utilise scarce resources so that maximum benefits can be gained.

Samacheer Kalvi 11th Economics Indian Economy Before and After Independence Additional Questions and Answers

Part – A
Choose the best options

Question 1.
Lord Cornwallis introduced “Permanent Settlement Act in” _______
(a) 1793
(b) 1794
(c) 1795
(d) 1796
Answer:
(a) 1793

Question 2.
In the year _______ green revolution started in India.
(a) 1950
(b) 1951
(c) 1948
(d) 1960
Answer:
(d) 1960

Question 3.
KGN. Daber established Mumbai’s spinning and weaving co. in _______
(a) 1810
(b) 1854
(c) 1845
(d) 1948
Answer:
(b) 1854

Question 4.
The oil well of Digboi, Assam was dug in.
(a) 1889
(b) 1898
(c) 1988
(d) 1810
Answer:
(a) 1889

Question 5.
The period of twelfth five year plan _______
(a) 2010 – 15
(b) 2011 – 16
(c) 2012 – 17
(d) 2013 – 18
Answer:
(c) 2012 – 17

Question 6.
NITI Aayog replaced the planning commission in _______
(a) 2013
(b) 2015
(c) 2014
(d) 2016
Answer:
(b) 2015

Question 7.
Human development index is _______
(a) HDI
(b) UNDP
(c) PQLI
(d) HID
Answer:
(a) HDI

Question 8.
Physical Quality of Life Index (PQLI) was developed by _______
(a) HDI
(b) PQLI
(c) UNDP
(d) HID
Answer:
(c) UNDP

Question 9.
In _______ system the ownership rights of use and control of land were held by the tiller himself.
(a) Zamindari
(b) Mahalwari
(c) Ryotwari
(d) Green revolution
Answer:
(c) Ryotwari

Question 10.
The period of industrial capital is _______
(a) 1757 – 1813
(b) 1813 – 1858
(c) 1757 – 1858
(d) 19 th century
Answer:
(b) 1813 – 1858

Match the following and choose the answer using the codes given below

Question 1.

(a) 12 3 4
(b) 2 4 3 1
(c) 2 3 1 4
(d) 4 3 2 1
Answer:
(c) 2 3 1 4

Question 2.

(a) 12 3 4
(b) 2 1 4 3
(c) 4 3 2 1
(d) 2 3 1 4
Answer:
(b) 2 1 4 3

Choose the correct option

Question 3.
The period of merchant capital is _______
(a) 1842 – 1857
(b) 1918 – 1920
(c) 1757 – 181
(d) 1721 – 1838
Answer:
(c) 1757 – 181

Question 4.
How many types of land tenure system were there before independence ?
(a) 2
(b) 3
(c) 1
(d) 5
Answer:
(b) 3

Question 5.
The first Indian modernized cotton cloth mill was established _______
(a) Chennai
(b) Calcutta
(c) Mumbai
(d) Bengaluru
Answer:
(b) Calcutta

Fill in the blanks with suitable option given below

Question 6.
The plan holiday was _______
(a) 1951-53
(b) 1966-69
(c) 2001 -02
(d) 1976-79
Answer:
(b) 1966-69

Question 7.
The planning commission was replaced by _______
(a) NITI Aayog
(b) Planning group
(c) Finance commission
(d) None of the above
Answer:
(a) NITI Aayog

Question 8.
NITI Aayog was formed in _______
(a) 2001
(b) 2014
(c) 2015
(d) 2016
Answer:
(c) 2015

Choose the incorrect statement

Question 9.
(a) Ryotwari system was first established in Tamilnadu.
(b) Zamindari system was established by British East India Company
(c) Mahalwari system was also called as the owner – cultivator system
(d) Zamindari system was created in 1793
Answer:
(c) Mahalwari system was also called as the owner – cultivator system

Question 10.
(a) The investment of small service enterprises should be more than ? 10 lakhs.
(b) The investment of micro service enterprises should not exceed ?10 lakhs.
(c) The investment of medium service enterprises should be more than ?2 crores,
(d) The investment of medium service enterprises should not exceed ?10 crores.
Answer:
(d) The investment of medium service enterprises should not exceed ?10 crores.

Choose the incorrect pair

Question 11.

Answer:
(c) Vijaya nagar (iii) England

Question 12.

Answer:
(c) Serampur (iii) Iron and steel

Chooose the correct pair

Question 13.

Answer:
(a) Industrial policy (i) 1948

Question 14.

Answer:
(d) White revolution (iv) Milk

Analyse the reason for the following

Question 15.
Assertion (A) : Before the advent of the British, Indian economy was self sufficient.
Reason (R) : Before the advent of the British, India lived in village.
(a) Both (A) and (R) are true
(b) (A) is true (R) is false
(c) Both (A) and (R) are false.
(d) (A) is false (R) is true.
Answer:
(a) Both (A) and (R) are true

Choose the incorrect statement

Question 16.
(a) The industrial policy resolutions in India ushered a socialistic economy
(b) The high yielding varieties programme was called as pilot project of green revolution
(c) The aim of seventh five year plan is ‘Garibi Hatao’
(d) Finance commission replaced the planning commission.
Answer:
(b) The high yielding varieties programme was called as pilot project of green revolution

Question 17.
(a) Switzerland ranks first in HDI.
(b) PQLI includes income whereas HDI excludes income
(c) The process of transforming private assets to government ownership is called privatization
(d) The third plan was called as ‘Gadgil plan’.
Answer:
(d) The third plan was called as ‘Gadgil plan’.

Pick the odd one out

Question 18.
(a) The period of revenue capital
(b) The period of industrial capital
(c) The period of merchant capital
(d) The period of financial capital
Answer:
(c) The period of merchant capital

Question 19.
(a) Investment system
(b) Zamindari system
(c) Mahalwari system
(d) Ryotwari system
Answer:
(a) Investment system

Question 20.
(a) Black revolution
(b) Red revolution
(c) White revolution
(d) Violet revolution
Answer:
(d) Violet revolution

Part – B
Answer the following questions in one or two sentences

Question 1.
Name the classification of industries in India?
Answer:

1. Public sector
2. Public-cum-private sector
3. Controlled private sector
4. Private and co-operative sectors.

Question 2.
What is the resolution of industrial policy 1948 ?
Answer:
The first industrial policy was declared on 6th April 1948. Its main importance was introducing the system of mixed economy in India.

Question 3.
What is the resolution of industrial policy 1956 ?
Answer:
The policy was shaped by the mahalanobis model of growth with emphasis on heavy industries which would lead to higher growth path.

Question 4.
Write a note on micro manufacturing enterprises.
Answer:
In micro enterprises, the investment in plant and machinery does not exceed Rs. 25 lakhs.

Question 5.
Write a note on small manufacturing enterprises
Answer:
The investment in plant and machinery is more than twenty five lakh rupees but does not exceed Rs. 5 crores.

Question 6.
Write a note on medium manufacturing enterprises.
Answer:
The investment in plant and machinery is more than Rs. 5 crores but not exceeding Rs. 10 crores.

Question 7.
Name the classification of public sector banks.
Answer:

1. Nationalised Banks.
2. State Bank and its associates

Part – C
Answer the following questions in one Paragraph

Question 1.
Describe the problems of British rule in Indian economy ?
Answer:

1. The British rule stunted the growth of Indian enterprise. It retarded the capital formation in India.
2. The drain of wealth financed capital development in Britain.
3. Indian agricultural sector became stagnant and deteriorated.
4. Indian handicraft industries were collapsed.
5. The system of capitalist firms with profit motives led to drain of resources from India.

Question 2.
Mention the achievements of green revolution.
Answer:

1. The production of major cereals wheat and rice were boosted. India became food surplus exporting food grains to European countries.
2. It was confined to high yielding variety cereals.
3. Production of commercial crops were increased.
4. Per hectare productivity of all crops had increased due to better seeds.
5. The revolution had positive effect on development of industries manufacturing agricultural tools.
6. It brought prosperity to rural people.
7. Demand for labour increased.
8. Financial resources were provided by banks and co-operative societies.

Question 3.
Examine the requirements of second green revolution.
Answer:

1. Introduction of genetically modified (GM) seeds which double the per average production.
2. Contribution of private sector to market the usage of GM foods.
3. Government can play a key role in expediting irrigation schemes and managing water resources.
4. Linking of rivers to transfer surplus water to deficient areas.

Question 4.
Write a note on iron and steel industry.
Answer:

1. First steel industry at Kulti, near Jharia, West Bengal – Bengal iron works company in 1870.
2. First large scale steel plant TISCO at Jamshedpur in 1907.
3. Steel authority of India Ltd. (SAIL) was established in 1974 and was made responsible for the development of the steel industry.
4. India is the eighth largest steel producing country in the world.

Question 5.
Write a note on petrol and natural gas.
Answer:

1. First successful oil well was dug in India in 1889 at Digboi, Assam.
2. At present a number of regions with oil reserves have been identified and oil is being extracted in these regions.
3. For exploration purpose, Oil and Natural Gas Commission (ONGC) was established in 1956 at Dehradun, Uttarakhand.

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Immediately get Chapter Wise Tamilnadu State Board Solutions for 11th Economics to gain more marks and start Quick Revision of all Chapters. All Chapters Pdf is provided along with the Questions and Answers. You Can Dowload Samacheer Kalvi 11th Economics Book Solutions Questions and Answers for Chapter wise are given with a clear explanation. Tamilnadu State Board Solutions for 11th Economics Chapter 7 Indian Economy Questions and Answers is for free of cost.

Tamilnadu Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy

Tamilnadu State Board Solutions for 11th Economics Chapter 7 Indian Economy Questions and Answers PDF has all given in Chapter Wise Section. Check Out daily basis with Tamilnadu State Board Solutions 11th Economics PDF will help to improve your score. Improve your level of accuracy to answer a question by reading with Samacheer Kalvi 11th Economics Book Solutions Questions and Answers PDF.

Samacheer Kalvi 11th Economics Indian Economy Text Book Back Questions and Answers

Part – A
Multiple Choice Questions

Question 1.
The main gold mine region in Karnataka is _________
(a) Kolar
(b) Ramgiri
(c) Anantpur
(d) Cochin
Answer:
(a) Kolar

Question 2.
Economic growth of a country is measured by national income indicated by _________
(a) GNP
(b) GDP
(c) NNP
(d) Per capital income
Answer:
(b) GDP

Question 3.
Which one of the following is a developed nations ?
(a) Mexico
(b) Ghana
(c) France
(d) SriLanka
Answer:
(c) France

Question 4.
The position of Indian Economy among the other strongest economies in the world is _______
(a) Fourth
(b) Seventh
(c) Fifth
(d) Tenth
Answer:
(b) Seventh

Question 5.
Mixed economy means _________
(a) Private sectors and banks
(b) Co-existence of Public and Private sectors
(c) Public sectors and banks
(d) Public sectors only
Answer:
(b) Co-existence of Public and Private sectors

Question 6.
The weakness of Indian Economy is _________
(a) Economic disparities
(b) Mixed economy
(c) Urbanisation
(d) Adequate employment opportunities
Answer:
(a) Economic disparities

Question 7.
A scientific study of the characteristics of population is _________
(a) Topography
(b) Demography
(c) Geography
(d) Philosophy
Answer:
(b) Demography

Question 8.
The year 1961 is known as _________
(a) Year of small divide
(b) Year of Population Explosion
(c) Year of Urbanisation
(d) Year of Great Divide
Answer:
(b) Year of Population Explosion

Question 9.
In which year the population of India crossed one billion mark ?
(a) 2000
(b) 2001
(c) 2005
(d) 1991
Answer:
(b) 2001

Question 10.
The number of deaths per thousand population is called as ________
(a) Crude Death Rate
(b) Crude Birth Rate
(c) Crude Infant Rate
(d) Maternal Mortality Rate
Answer:
(a) Crude Death Rate

Question 11.
The number of births per thousand population is called as _________
(a) Crude death rate
(b) Mortality rate
(c) Morbidity rate
(d) Crude Birth Rate
Answer:
(d) Crude Birth Rate

Question 12.
Density of population = _________
(a) Land area / Total Population
(b) Land area / Employment
(c) Total Population / Land area of the region
(d) Total Population
Answer:
(c) Total Population / Land area of the region

Question 13.
Who introduced the National Development Council in India?
(a) Ambedkar
(b) Jawaharlal Nehru
(c) Radhakrishanan
(d) V.K.R.V. Rao
Answer:
(b) Jawaharlal Nehru

Question 14.
Who among the following propagated Gandhian Ecomomic thinkings.
(a) Jawaharlar Nehru
(b) VKRV Rao
(c) JC Kumarappa
(d) A.K.Sen
Answer:
(c) JC Kumarappa

Question 15.
The advocate of democratic socialism was
(a) Jawaharlal Nehru
(b) P.C. Mahalanobis
(c) Dr. Rajendra Prasad
(d) Indira Gandhi
Answer:
(a) Jawaharlal Nehru

Question 16.
Ambedkar the problem studied by in the context of Indian Economy is ________
(a) Small land holdings and their remedies
(b) Problem of Indian Currency
(c) Economics of socialism
(d) All of them
Answer:
(b) Problem of Indian Currency

Question 17.
Gandhian Economics is based on the Principle _________
(a) Socialistic idea
(b) Ethical foundation
(c) Gopala Krishna Gokhale
(d) Dadabhai Naoroji
Answer:
(b) Ethical foundation

Question 18.
V.K.R.V Rao was a student of _________
(a) J.M. Keynes
(b) Colin Clark
(c) Adam smith
(d) Alfred Marshal
Answer:
(a) J.M. Keynes

Question 19.
Amartya Kumara Sen received the Nobel prize in Economics in the year.
(a) 1998
(b) 2000
(c) 2008
(d) 2010
Answer:
(a) 1998

Question 20.
Thiruvalluvar economic ideas mainly dealt with _________
(a) Wealth
(b) Poverty is the curse in the society
(c) Agriculture
(d) All of them
Answer:
(d) All of them

Part – B
Answer the following questions in one or two sentences

Question 21.
Write the meaning of Economic Growth.
Answer:
A country’s economic growth is usually indicated by Gross Domestic Product (GDP). The GDP is the total monetary value of the goods and services produced by that country over a specific period of time, usually one year.

Question 22.
State any two features of developed economy.
Answer:

1. High National Income
2. High per capital Income

Question 23.
Write the short note on natural resources
Answer:
Natural resources are stock or reserve that can be drawn from nature. The major natural resources are land, forest, water, mineral and energy.

Question 24.
Point out any one feature of Indian Economy
Answer:
Indian economy is a mixed economy. In India both the private and public sectors coexist.

Question 25.
Give the meaning of non-renewable energy
Answer:
The sources of energy which cannot be renewed or re-used are called non-renewable energy.
(Eg.) Coal, Oil, Gas etc.

Question 26.
Give a short note on Sen’s ‘Choice of Technique’.
Answer:
In a labour surplus economy, generation of employment cannot be increased at the initial stage by the adaptation of capital intensive technique whatsen called as choice of technique.

Question 27.
List out the reasons for low per capital income as given by V.K.R.V. Rao.
Answer:

1. Uneconomic holdings.
2. Low levels of water availability for crops.
3. Excess population pressure on agriculture due to the absence of a large industrial sector.
4. Absence of capital.
5. Absence of autonomy in currency policy.

Part – C
Answer the following questions in one Paragraph

Question 28.
Define Economic Development.
Answer:
The level economic development is indicated not just by GDP, but by an increase in citizen’s quality of life or well-being. The quality of life is being assessed by several indices such as Human Development Index (HDI), Physical Quality of Life Index (PQLI) and Gross National Happiness Index (GNHI)

Question 29.
State Ambedkar’s Economic ideas on agricultural economics.
Answer:
Dr. B.R.Ambedkar was a versatile personality. In 1918, he published a paper “Small holding in India and their remedies” citing Adam Smith’s “Wealth of Nations”, he made a fine distinction between “Consolidation of holdings” and “Enlargement of holdings”.

Question 30.
Write on short note on village sarvodhaya.
Answer:
Village sarvodhaya is the concept of Gandhiji on the development of villages. To Gandhi India lives in villages. He was interested in developing the villages as self – sufficient units.

According to him, “Real India was to be found in villages and not in towns or cities”.
So he suggested the development of self-suffficient, self dependent villages.

Question 31.
Write the strategy of Jawahar lal Nehru in India’s planning.
Answer:

1. Jawahar lal Nehru was responsible for the introduction of planning in our country.
2. To Nehru, the plan was essentially an integrated approach for development.
3. He said the essence of planning is to find the best way to utilize all resources of manpower, of money and so on.
4. Planning for Nehru was essentially linked up with industrialization and eventual self reliance for the country’s economic growth on a self-accelerating growth.
5. Nehru carried through this basic strategy of planned development.

Question 32.
Write the V.K.R.V.Rao’s contribution on multiplier concept.
Answer:

1. Rao’s examination of the “interrelation between investment, income and multiplier in an under developed economy (1952)” was his major contribution to macroeconomic theory.
2. In it he asserts that keynesian multiplier principle remains inoperative in underdeveloped countries.
3. V.K.R.V. Rao was the best equipped of all keynes’ pupils.

Question 33.
Write a short note on Welfare Economics given by Amartya Sen.
Answer:

1. Amartya Sen was awarded the nobel prize for his contributions to welfare economics.
2. Sen’s major point has been that the distribution of income / consumption among the persons below the poverty line is to be taken into account.
3. Sen has focused on the poor, viewing them not as objects of pity requiring charitable hand-outs, but as disempowered folk needing empowerment in all aspects.

Question 34.
Explain Social infrastructure.
Answer:

1. Social infrastructure refers to those structures which are improving the quality of man power and contribute indirectly towards the growth of an economy.
2. These structures are outside the system of production and distribution.
3. The development of these social structures help in increasing the efficiency and productivity of man power.
   (Eg.) Schools, Colleges, Hospitals

Part – D
Answer the following questions in about a page

Question 35.
Explain strong features Indian economy
Answer:

1. India has a mixed economy : In India both private and public sectors coexist.
2. Agriculture plays the key role : Around 60% of the people in India depend upon agriculture for their livelihood.
3. An emerging market: India has a high potential for prospective growth which attracts FDI and FII.
4. Emerging economy : As a result of rapid economic growth Indian economy has a place among the G20 countries.
5. Fast growing economy : India has emerged as the world’s fastest growing economy in 2016-17 with 7.1% GDP next to China.
6. Fast growing service sector : The service sector, contributes a lion’s share of the GDP in India.
7. Large domestic consumption : Due to large domestic consumption the standard of living has considerably improved and life style has changed.
8. paid growth of urban areas : Improved connectivity in transport and communication, education and health have speeded up the pace of urbanization.
9. Stable macro economy : The current year’s economic survey represents the Indian economy to be a heaven of macroeconomic stability, resilence and optimism.
10. Demographic dividend : India is a pride owner of the maximum percentage of youth. This has invited foreign investments to the country and outsourcing opportunities.

Question 36.
Write the importance of mineral resources in India.
Answer:

1. Iron-ore : Hematite iron is mainly found in Chattisgarh, Jharkhand, Odisha, Goa and Karnataka. Magnetite is found in Western Coast of Karnataka. Some deposits of iron ore are also found in Kerala, Tamil Nadu and Andhra Pradesh.
2. Coal and Lignite : India ranks third in the coal production. The main centres are West-Bengal, Bihar, Madhya Pradesh, Maharashtra, Odisha and Andhra Pradesh. Bulk production comes from Bengal-Jharkhand coal fields. Lignite is found in Neyveli in Tamil Nadu.
3. Bauxite : Bauxite is a main source of aluminium. Major reserves are found in Odisha and Andhra Pradesh.
4. Mica : India stand first in sheet mica production. Andhra Pradesh, Jharkhand, Bihar and Rajasthan are major reserves.
5. Crude Oil : Oil is being explored in India at many places of Assam and Gujarat.
6. Gold : India possesses only a limited gold reserve.
7. Diamond : The total reserves of diamond is estimated at around 4582 carats.

Question 37.
Bring out Jawharlal Nehru’s contribution to the idea of economic development.
Answer:
Jawahar Lal Nehru was one of the Chief builders of modem India. He was a great patriot, thinker and statesman. His ideas of economic development are :

1. Democracy : He was a firm believer in democracy. He believed in free speech, civil liberty, adult franchise and the rule of law and parliamentary democracy.

2. Secularism : Secularism is another signal contribution of Nehru to India. There are so many religions in India but there is no domination by religious majority.

3. Planning : Nehru was responsible for the introduction of planning in our country. The plan was essentially an integrated approach for development. Planning for Nehru was essentially linked up with industrialization and eventual self-reliance for the country’s economic growth on a self-accelerating growth.

4. Advancement of Science : Nehru made a great contribution to the advancement of Science, research, technology and industrial development. In his period, many IITs and research institutions were established. He always insisted on scientific temper.

5. Democratic socialism : Nehru put the country on the road towards a socialistic pattern of society. Nehru’s socialism is democratic socialism.

Question 38.
Write a brief note on the Gandhian economic ideas.
Answer:
Gandhian economics is based on ethical foundations.
Salient features of Gandhian economic thought:

1. Village republics : Gandhi was interested in developing the villages as self-sufficient units.
2. On machinery : Gandhi described machinery as ‘Great Sin’.
3. Industrialism : Gandhi considered industrialism as a curse on mankind.
4. Decentralization : He advocated a decentralized economy.
5. Village sarvodaya : Gandhi suggested the development of self-sufficient, self-dependent villages.
6. Bread labour: Gandhi realized the dignity of human labour. Bread labour or body labour was the expression that Gandhi used to mean manual labour.
7. The doctrine of trusteeship : Trusteeship provides a means of transforming the present capitalist society into an egalitarian one.
8. On the food problem : Gandhi was against any sort of food controls. Food controls only create artificial scarcity.
9. On population : Gandhi was in favour of birth control through brahmacharya or self-control.
10. On prohibition : Gandhi advocated cent per cent prohibition.

Samacheer Kalvi 11th Economics Indian Economy Additional Questions and Answers

Part -A
Choose the best options

Question 1.
Third world country __________
(a) India
(b) China
(c) Bangladesh
(d) Srilanka
Answer:
(c) Bangladesh

Question 2.
__________ is the year of great divide.
(a) 1921
(b) 1918
(c) 1943
(d) 1969
Answer:
(a) 1921

Question 3.
__________ are BIMARU States.
(a) Tamil Nadu, Kerala, Karnataka, Madhya Pradesh
(b) Bihar, Madhya Pradesh, Rajasthan, Uttarpradesh
(c) West Bengal, Himalchalpradesh, Punjab, Delhi
(d) Andhra Pradesh, Odisa, Gujarat, Chattisgarh
Answer:
(b) Bihar, Madhya Pradesh, Rajasthan, Uttarpradesh

Question 4.
In India, Lignite is found in.
(a) New Delhi
(b) Kolkatta
(c) Ranji
(d) Neyveli
Answer:
(d) Neyveli

Question 5.
Indian educational system comprises of __________ stages.
(a) 6
(b) 12
(c) 3
(d) 4
Answer:
(a) 6

Question 6.
__________ is considered as the best pupil of J.M. Keynes
(a) A.K. Sen
(b) J.C.Kumarappa
(c) V.K.R.V. Rao
(d) Manmohan Singh
Answer:
(c) V.K.R.V. Rao

Question 7.
Indian Railways introduced first wifi facility in __________
(a) Chennai
(b) Calcutta
(c) Bengaluru
(d) Hyderabad
Answer:
(c) Bengaluru

Question 8.
National Harbour board was set up in __________
(a) 1947
(b) 1950
(c) 1948
(d) 1951
Answer:
(b) 1950

Question 9.
A study on poverty and famines and concept of entitlements and capability development was done by __________
(a) Adamsmith
(b) A.K.Sen
(c) Marshall
(d) V.K.R.V. Rao
Answer:
(b) A.K.Sen

Question 10.
Percapital income is __________
(a) National income / output
(b) National income / population
(c) National income / percapital output
(d) National income / male population
Answer:
(b) National income / population

Question 11.
‘India lives in villages’ said by __________
(a) V.K.R.V. Rao
(b) J.C. Kumarappa
(c) Mahatma Gandhi
(d) Jawaharlal Nehru
Answer:
(c) Mahatma Gandhi

Match the following and choose the answer using the codes given below

Question 1.

(a) 1 2 3 4
(b) 3 4 2 1
(c) 1 3 4 2
(d) 2 4 3 1
Answer:
(b) 3 4 2 1

Question 2.

(a) 4 1 2 3
(b) 3 4 2 1
(c) 2 4 3 1
(d) 1 2 4 3
Answer:
(a) 4 1 2 3

Question 3.

(a) 4 1 2 3
(b) 1 2 3 4
(c) 4 3 2 1
(d) 2 4 1 3
Answer:
(d) 2 4 1 3

Choose the odd one out

Question 4.
(a) Poverty and famine
(b) Choice of technique
(c) Economics of caste
(d) Concept of capability
Answer:
(c) Economics of caste

Question 5.
(a) Budget by assignment
(b) Budget by assigned revenue
(c) Budget by shared revenues
(d) Budget by fiscal revenues
Answer:
(d) Budget by fiscal revenues

Question 6.
(a) Net Domestic Product (NDP)
(b) Human Development Index (HDI)
(c) Physical Quality of Life Index (PQLI)
(d) Gross National Happiness Index (GNHI)
Answer:
(a) Net Domestic Product (NDP)

Choose the correct statement

Question 7.
(a) Indian economy is the seventh largest in the world
(b) Indian economy is a mixed economy
(c) India ranks third in terms of purchasing power parity
(d) The human capital of India is old
Answer:
(d) The human capital of India is old

Question 8.
(a) The annual addition of population in India equals the total population of Australia.
(b) There exists a huge economic disparity in the Indian economy
(c) Kerala has the highest birth rate
(d) Every 6th person in the world is an Indian
Answer:
(c) Kerala has the highest birth rate

Choose the incorrect statement

Question 9.

Answer:
(b) Demographic transition – (ii) 2001

Question 10.

Answer:
(c) 1971 – (iii) 1040

Choose the correct statement

Question 11.
(a) Nehru always insisted on scientific temper
(b) Gandhi advocated a centralized economy
(c) Valluvar has recommended an unbalanced budget
(d) V.K.R.V. Rao’s economic thought is coined as Gandhian Economics.
Answer:
(a) Nehru always insisted on scientific temper

Question 12.
(a) Until 1979, Education in India was in state list
(b) Education in India follows the 10+3 pattern
(c) Sex ratio refers to the number of females per 1000 males.
(d) India ranks fourth in coal production of the world.
Answer:
(c) Sex ratio refers to the number of females per 1000 males.

Analyse the reason for the following 

Question 13.
Assertion (A) : There exist a huge economic disparity in the Indian economy.
Reason (R) : The proportion of income and assets owned by top 10% of Indians goes on increasing.
(a) Both (A) and (R) are true, (R) is the correct explanation of (A)
(b) Both (A) and (R) are true, (R) is the not correct explanation of (A)
(c) (A) is true and (R) are false.
(d) (A) is false, but (R) is true.
Answer:
(a) Both (A) and (R) are true, (R) is the correct explanation of (A)

Question 14.
Assertion (A) : In Kerala the adult sex ratio is 1084 as on 2011.
Reason (R) : Kerala provides better status to women as compared to other states.
(a) Both (A) and (R) are true, (R) is not the correct explanation of (A)
(b) Both (A) and (R) are true, (R) is the correct explanation of (A)
(c) Both (A) and (R) are false.
(d) None of the above
Answer:
(b) Both (A) and (R) are true, (R) is the correct explanation of (A)

Fill in the blanks with suitable option given below

Question 15.
In India, Lignite is found in _____
(a) New Delhi
(b) kolkata
(c) Ranji
(d) Neyveli
Answer:
(d) Neyveli

Question 16.
Indian educational system comprises of _____ stages
(a) 12
(b) 3
(c) 6
(d) 4
Answer:
(c) 6

Question 17.
Indian railways introduced first wifi facility in _____
(a) Chennai
(b) Bengaluru
(c) Calcutta
(d) Hyderabad
Answer:
(b) Bengaluru

Choose the best option 

Question 18.
_____ is considered as the best pupil of J.M. Keynes
(a) A.K. Sen
(b) J.C.Kumarappa
(c) V.K.R.V.Rao
(d) Nehru
Answer:
(c) V.K.R.V.Rao

Question 19.
Third world country is _____
(a) India
(b) China
(c) Srilanka
(d) Bangladesh
Answer:
(d) Bangladesh

Question 20.
Per capital income is _____
(a) National income / output
(b) National income / population
(c) National income / per capital output
(d) National income / male population
Answer:
(b) National income / population

Part – B
Answer the following questions in one or two sentences 

Question 1.
What is Gross National Happiness Index?
Answer:
GNHI is an indicator of progress, which measures sustainable development, environmental conservation, promotion of culture and good governance.

Question 2.
State any two weakness of Indian Economy.
Answer:

1. Large population.
2. Inequality and poverty.

Question 3.
What is sex ratio ?
Answer:
Sex ratio refers to the number of females per 1,000 males.

Question 4.
Explain economic infrastructure.
Answer:
Economic infrastructure is the support system which helps in facilitating production

Question 5.
What is Gandhian Economics ?
Answer:
J.C. Kumarappa developed economic theories based on Gandhism. He developed a school of economic thought coined as Gandhian Economics.

Question 6.
Which are “BIMARU” states ?
Answer:

1. Bihar
2. Madhya Pradesh
3. Rajasthan
4. Uttarpradesh

Question 7.
What is year of small divide ?
Answer:
In 1951, population growth rate has come down from 1.33% to 1.25%. Hence it is known as year of small divide.

Part – C
Answer the following questions in one Paragraph

Question 1.
What are the aspects of demographic trends in India ?
Answer:

1. Size of population
2. Rate of growth.
3. Birth and death rates
4. Density of population
5. Sex – ratio
6. Life – expectancy at birth
7. Literacy ratio.

Question 2.
Explain Thiruvalluvar’s ideas on agriculture.
Answer:
According to Thiruvalluvar, agriculture is the most fundamental economic activity. They are the axle-pin of the world, for on their prosperity revolves prosperity of other sectors of other sectors of the economy. He says, ploughmen alone is the freemen of the soil. He believes that agriculture is superior to all other occupation.

Question 3.
Write short note on Birth rate and Death rate.
Answer:
Short Note On Birth Rate And Death Rate:

1. Birth Rate : It refers to the number of births per thousand of population.
2. Death Rate : It refers to the number of deaths per thousand of population.

Kerala has the lowest birth rate and Uttarpradesh has the highest birth rate. West Bengal has the lowest and Orissa has the highest death rates.

Part – D
Answer the following questions in about a page

Question 1.
Explain about health in India.
Answer:
Health in India is a state government responsibility. The central council of health and welfare formulates the various health care projects and health department reform policies. The administration of health industry is the responsibility of the ministry of health and welfare.

Health care in India has many forms. But, all medical systems are under one ministry AYUSH.
Health status is better in Kerala as compared to other states. India’s health status is poor compared to Sri Lanka.

Question 2.
Explain Thiruvalluvar’s economic ideas.
Answer:
The economic ideas of Thiruvalluvar are found in his immortal work, Thirukkural in its second part porutpal.

1. Factors of production : Thiruvalluvar has made many passing references about the factors of production, land, labour, capital, organisation.
2. Agriculture : Agriculture is the most fundamental economic activity. The prosperity of other sectors depends on the prosperity of agriculture. Valluvar believes that agriculture is superior to all other occupation.
3. Public finance : He elaborately explained public finance under the headings public revenue, financial administration and public expenditure.
4. Public expenditure : Valluvar has recommended a balanced budget. He advocates
   1. Defence
   2.  Public works
   3. Social Services
5. External assistance : Valluvar was against seeking external assistance. He advocated a self-sufficient economy.
6. Poverty and begging : According to him ‘poverty is the root cause of all other evils which would lead to ever-lasting sufferings.
7. Wealth: Valluvar has regarded wealth as only a means and not an end.
8. Welfare state : The important elements of a welfare state are
   1. Perfect health of the people without disease
   2. Abundant wealth
   3. Good crop
   4. Prosperity and happiness
   5. Full security for the people.

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Samacheer Kalvi 11th Economics Distribution Analysis Text Book Back Questions and Answers

Part – A
Multiple Choice Questions

Question 1.
In Economics, distribution of income is among the
(a) factors of production
(b) individual
(c) firms
(d) traders
Answer:
(a) factors of production

Question 2.
Theory of distribution is popularly known as ………………………..
(a) Theory of product – pricing
(b) Theory of factor – pricing
(c) Theory of wages
(d) Theory of Interest
Answer:
(b) Theory of factor-pricing

Question 3.
Rent is the reward for the use of
(a) capital
(b) labour
(c) land
(d) organization
Answer:
(c) land

Question 4.
The concept of ‘Quasi – Rent’ is associated with ………………………
(a) Ricardo
(b) Keynes
(c) Walker
(d) Marshall
Answer:
(d) Marshall

Question 5.
The Classical Theory of Rent was propounded by
(a) Ricardo
(b) Keynes
(c) Marshall
(d) Walker
Answer:
(a) Ricardo

Question 6.
‘Original and indestructible powers of the soil’ is the term used by ……………………….
(a) J.S.Mill
(b) Walker
(c) Clark
(d) Ricardo
Answer:
(d) Ricardo

Question 7.
The reward for labour is
(a) rent
(b) wage
(c) profit
(d) interest
Answer:
(b) wage

Question 8.
Money wages are also known as ……………………….
(a) Real wages
(b) Nominal wages
(c) Original wages
(d) Transfer wages
Answer:
(b) Nominal wages

Question 9.
Residual Claimant Theory is propounded by
(a) Keynes
(b) Walker
(c) Hawley
(d) Knight
Answer:
(b) Walker

Question 10.
The reward is given for the use of capital ………………………
(a) Rent
(b) Wage
(c) Interest
(d) Profit
Answer:
(c) Interest

Question 11.
Keynesian Theory of interest is popularly known as
(a) Abstinence Theory
(b) Liquidity Preference Theory
(c) Loanable Funds Theory
(d) Agio Theory
Answer:
(b) Liquidity Preference Theory

Question 12.
According to the Loanable Funds Theory, supply of loanable funds is equal to …………………………
(a) S + BC + DH + DI
(b) I + DS + DH + BM
(c) S+ DS + BM+ DI
(d) S + BM + DH + DS
Answer:
(a) S + BC + DH + DI

Question 13.
The concept of meeting unexpected expenditure according to Keynes is
(a) Transaction motive
(b) Precautionary motive
(c) Speculative motive
(d) Personal motive
Answer:
(b) Precautionary motive

Question 14.
The distribution of income or wealth of a country among the individuals are ………………………….
(a) Functional distribution
(b) Personal distribution
(c) Goods distribution
(d) Services distribution
Answer:
(b) Personal distribution

Question 15.
Profit is the reward for
(a) land
(b) organization
(c) capital
(d) labour
Answer:
(b) organization

Question 16.
Innovation Theory of profit was given by …………………………..
(a) Hawley
(b) Schumpeter
(c) Keynes
(d) Knight
Answer:
(b) Schumpeter

Question 17.
Quasi-rent arises in
(a) Man-made appliances
(b) Homemade items
(c) Imported items
(d) None of these
Answer:
(a) Man-made appliances

Question 18.
“Wages as a sum of money are paid under contract by an employer to a worker for service rendered”- Who said this?
(a) Benham
(b) Marshall
(c) Walker
(d) J.S.Mill
Answer:
(a) Benham

Question 19.
Abstinence Theory of Interest was propounded by
(a) Alfred Marshall
(b) N.W Senior
(c) Bohm-Bawerk
(d) Knut Wicksell
Answer:
(b) N.W Senior

Question 20.
Loanable Funds Theory of Interest is called as …………………………
(a) Classical Theory
(b) Modem Theory
(c) Traditional Theory
(d) Neo-Classical Theory
Answer:
(d) Neo-Classical Theory

Part – B
Answer the following questions in one or two sentences

Question 21.
What is meant by distribution?
Answer:
Distribution means division of income among the four factors of production as rent, wage, interest and profit.

Question 22.
Mention the types of distribution?
Answer:
Personal Distribution: Personal Distribution is the distribution of national income among the individuals.
Functional Distribution: Functional Distribution means the distribution of income among the four factors of production namely land, labour, capital and organization for their services in production process.

Question 23.
Define ‘Rent’.
Answer:
Rent is that part of payment made by a tenant to his landlords for the use of land only.

Question 24.
Distinguish between real and money wages.
Answer:
Real wages :
Wages paid in terms of goods and services. It refers to the purchasing power of money wages.

Money wages :
Nominal wages are referred to as wages paid in terms of money.

Question 25.
What do you mean by interest?
Answer:

1. Interest is the reward paid by the borrower to the lender for the use of capital.
2. Interest is the price paid for the use of capital in any market.
3. Generally speaking, interest is a payment made by a borrower to the lender for the money borrowed.

Question 26.
What is profit?
Answer:
Profit is the amount left with the entrepreneur after he has made payments for all the other factors of production.

Question 27.
State the meaning of liquidity preference?
Answer:

1. Liquidity preference means the preference of the people to hold wealth in the form of liquid cash rather than in other non – liquid assets like bonds, securities, bills of exchange, land, building, gold etc.
2. “ Liquidity Preference is the preference to have an amount of cash rather than of claims against others”.

Part – C
Answer the following questions in One Paragraph

Question 28.
What are the motives of demand for money?
Answer:
1. Transaction motive :
It relates to the desire of the people to hold cash for the current transactions.
Mt=f (y)

2. Precautionary motive :
It relates to the desire of the people to hold cash to meet unexpected or unforeseen expenditures.
Mp = f (y)

3. Speculative motive :
It relates to the desire of the people to hold cash in order to take advantage of market movements regarding future changes in the price of bonds and securities in the capital market.
Ms = f (i)

Question 29.
List out the kinds of wages:
Answer:
Wages are divided into four types.

1. Nominal Wages or Money Wages: Nominal wages are referred to the wages paid in terms of money.
2. Real Wages: Real wages are the wages paid in terms of goods and services. Hence, real wages are the purchasing power of money wages.
3. Piece Wages: Wages that are paid on the basis of quantum of work done.
4. Time Wages: Wages that are paid on the basis of the amount of time that the worker works.

Question 30.
Distinguish between rent and quasi-rent
Answer:
Rent:

1. Rent accrues to land
2. The supply of land is fixed forever
3. It enters into price

Quasi – rent:

1. Quasi-rent accrues to man-made appliances
2. The supply of man made appliances is fixed for a short period only
3. It does not enter into price.

Question 31.
Briefly explain the Subsistence Theory of Wages?
Answer:
The subsistence theory of wages:

1. Subsistence theory of wages is one of the oldest theories of wages.
2. According to this theory, wage must be equal to the subsistence level of the labourer and his family.
3. Subsistence means the minimum amount of food, clothing and shelter which workers and their family require for existence.
4. If workers are paid higher wages than the subsistence level, the workers would be better off and they will have large families.
5. Hence, the population would increase.
6. When the population increases, the supply of labourer would increase and therefore, wages will come down.
7. If wages are lower than the subsistence level, there would be a reduction in population and thereby the supply of labour falls and wages increase to the subsistence level.

Question 32.
State the Dynamic Theory of Profit.
Answer:

• J.B. Clark propounded this theory in 1900. To him, profit is the difference between price and cost of production of the commodity.
• Profit is the reward for dynamic changes in society. He points out that, profit cannot arise in a static society. In static society everything remains stationary.
• The following changes take place in a dynamic society.
  1. Population is increasing.
  2. Volume of capital is increasing.
  3. Methods of production are improving.
  4. Forms of industrial organization are changing.
  5. The wants of consumer are multiplying.

Question 33.
Describe briefly the Innovation Theory of Profit?
Answer:
Innovation Theory of Profit:

1. The innovation theory of profit was propounded by Joesph. A.Schumpeter.
2. Schumpeter says an entrepreneur is not only an undertaker of a business but also an innovator in the process of production.
3. Profit is the reward for “innovation”.
4. According to Schumpeter, an innovation may consist of the following:
   • Introduction of a new product.
   • Introduction of a new method of production.
   • Opening up of a new market.
   • Discovery of new raw materials
   • Reorganization of an industry / firm.

When any one of these innovations is introduced by an entrepreneur, it leads to reduction in the cost of production and thereby brings profit to an entrepreneur. To obtain profit continuously, the innovator needs to innovate continuously. The real innovators do so. Imitative entrepreneurs cannot innovate.

Question 34.
Write a note on the Risk-bearing Theory of Profit.
Answer:

• Risk – the bearing theory was propounded by F.B.Hawley in 1907.
• Profit is the reward for “risk-taking” in business.
• Every business involves some risks. So risk-taking is an essential function of the entrepreneur and is the basis of profit.
• Higher the risks, the greater are the profit.
• Profit induces entrepreneurs to undertake risks.

Part – D
Answer the following questions in about a page

Question 35.
Explain the Marginal Productivity Theory of Distribution.
Answer:
Marginal productivity theory of distribution was developed by Clark, Wickseed and Walras. This theory explains how the prices of various factors of production are determined. This theory is also known as “General theory of distribution” or “National dividend theory of distribution”.
Assumptions:

1. All the factors of production are homogenous and can be substituted for each other.
2. There is perfect competition in both factor and product market.
3. There is perfect mobility and full employment of factors of production.
4. This theory is applicable only in the long-run.
5. The entrepreneurs aim at profit maximization.
6. There is no government intervention and no technological change.

Explanation of the theory :
Each factor is rewarded according to its marginal productivity.

Marginal product:
The marginal product of a factor of production means the addition made to the total product by employment of an additional unit of that factor.

Marginal physical product (MPP) :
The MPP of a factor is the increment in the total product obtained by the employment of an additional unit of that factor.

Value of marginal product (VMP):
VMP = MPP x price

Statement of the theory:

1. The price of a factor of production depends upon its productivity.
2. The price of a factor is determined by and will be equal to marginal revenue product of that factor.
3. Under certain conditions, the price of a factor will be equal to both average and marginal products of that factor.

Marginal productivity under perfect competition:

X- axis represents factor units and Y-axis represents the factor price and revenue product.
MRP – Marginal Revenue Product Curve
ARP – Average Revenue Product Curve
AFC – Average Factor cost Curve
MFC – Marginal Factor cost Curve
AFC – is horizontal under perfect competition and MFC coincides with it.
When there is perfect competition in the factor market, the firm is in equilibrium only when MFC = MRP
At the point Q by employing ON units of factors and paying OP price (NQ) where MFC = MRP
At Q, MRP = ARP
Price of the factor (NQ) = Marginal revenue
Product (NQ) = Average revenue product (NQ)
There is no exploitation of factors under perfect competition.
Beyond the point Q the price paid to the factor is more than marginal revenue product and average revenue product, so employer do not employ the factors.

Question 36.
Illustrate the Ricardian Theory of Rent.
Answer:
The classical theory of rent is called “Ricardian theory of rent”.
Definition :
Rent is that portion of the produce of the earth which is paid to the landlord for the use of the original and indestructible powers of the soil.

Assumptions :

1. Land differs in fertility.
2. The law of diminishing returns operates in agriculture.
3. Rent depends upon fertility and location of land.
4. Theory assumes perfect competition and long period.
5. There is existence of marginal land or no-rent land.
6. Land has certain “Original and indestructible powers”.
7. Land, is used for cultivation only.
8. Most fertile lands are cultivated first.

Statement of the theory with illustration :
Assume that some people settle in newly discovered island. People will first cultivate the most fertile ‘A’ grade land. They produce 40 bags of paddy.

Suppose after some time if another group of people settle down in the same island. They cultivate ‘B’ grade land which produce 30 bags of paddy. Suppose yet another group of people settle down there they cultivate ‘C’ grade land. It produce 20 bags of paddy.

This surplus of ‘A’ grade land is now raised to 20 bags (40 – 20) and it is the ‘Economic Rent’ of ‘A’ grade land. The surplus of ‘B’ grade land is 10 bags (30 – 20). In ‘C’ grade land cost of production is equal to the price of its produce and it does not yield any rent (20 – 20). Hence ‘C’ grade land is called ‘no-rent land’ or marginal land.

The land which yields rent is called “intra-marginal land”.
Rent indicates the differential advantage of the superior land over the marginal land.
Ricardian Theory’ of Rent :

Diagrammatic illustration :
X – axis represents various grades of land and Y axis yield per acre (in bags). OA, AB and BC are the ‘A’, ‘B’, ‘C’ grade lands respectively.

The ‘C’ grade land is the no-rent land. A and B grade lands are “intra-marginal lands. The economic rent yielded by ‘A’ and ‘B’ grade lands is equal to the shaded area of their respective rectangles.

Criticisms :

1. The order of cultivation from most fertile to least fertile lands is historically wrong.
2. This theory assumes that rent does not enter into price. But in reality, rent enters into price.

Question 37.
Elucidate the Loanable Funds Theory of Interest.
Answer:

1. The loanable funds theory, also known as the “Neoclassical theory”. This theory was developed by Swedish economists like Wicksell, Bertilohlin, Viner, Gunnarmyrdal and others.
2. Interest is the price paid for the use of loanable funds.
3. The rate of interest is determined by the equilibrium between demand for and supply of loanable funds in the credit market.

Demand for loanable funds :

1. Demand for Investment (I)
2. Demand for Consumption (C)
3. Demand for Hoarding (H)

Supply of loanable funds :
1. Savings (S) :
Savings may be of two types, namely.

1. Savings planned by individuals are “ex-ante savings”. (Eg.) LIC premium
2. Unplanned savings are called “ex-post savings”

2. Bank credit:
Commercial banks create credit and supply loan able funds to the investors.

3. Dishoarding (DH) :
Dishoarding means bringing out the hoarded money into use and thus it constitutes a source of supply of loanable funds.

4. Disinvestment (DI):
Disinvestment is the opposite of investment. It means not providing sufficient funds for depreciation of equipment.

Equilibrium :
The rate of interest is determined by the equilibrium between the total demand for and the total supply of loanable funds.
Supply of loanable funds = S + BC + DH + DI
Demand for loanable funds = I + C + H

Explanation :
X-axis represents the demand for and supply of loanable funds, Y axis represents the rate of interest. The LS curve represents the total supply curve ofloanable funds. The LD curve represents the total demand for loanable funds. The LD and LS curves, intersect each other at the point “E” the equilibrium point. At this point OR rate of interest and OM is the amount ofloanable funds.

Criticisms :

1. Many factors have been included in this theory’. Still there are many more factors like
   • Asymmetric information
   • Moral Hazard.
2. It is very difficult to combine real factors with monetary factors.

Question 38.
Explain the Keynesian Theory of Interest.
Answer:
According to Keynes, interest is the reward for parting with liquidity’ for a specified period of time.
Meaning of liquidity preference:
Liquidity preference means the preference of the people to hold wealth in the form of liquid cash rather than in other non-liquid assets.
Motives of demand for money:
The three motives of liquidity preference are.

1. The transaction motive M_t = f (y)
2. The precautionary motive M_p = f (y)
3. The speculative motive M_s = f (i)

Determination of Rate of Interest:
According to Keynes, the rate of interest is determined by the demand for and the supply of money. The demand for money is liquidity preference. The supply of money is determined by the policies of the government and the central bank.

Equilibrium between demand and supply of money:
The equilibrium between liquidity preference and demand for money determine the rate of interest. In short-run, the supply of money is assumed to be constant.

LP is the liquidity preference curve. M₂ shows the supply curve of money to satisfy speculative motive. Both curves intersect at the point E which is the equilibrium. Here, rate of interest is I.

If liquidity preference increases from LP to L₁ P₁ the supply of money remains constant, the rate of interest increase from OI to OI₁

Suppose LP remains constant. If the supply of money is OM₂, the interest is OI₂ and if the supply of money is reduced from OM₂ to OM₂, the interest would increase from OM₂ to OM₄. If the supply of money is increased from OM₂ to OM₄ the interest would decrease from OI₂ to Ol₄

Criticisms :
1. This theory does not explain the existence of different interest rates prevailing in the market at the same time.
2. It explains interest rate only in the short – run.

Samacheer Kalvi 11th Economics Distribution Analysis Additional Questions and Answers

Part -A
Choose the best options

Question 1.
Marginal Productivity theory is the …………………….. theory of distribution.
(a) Average
(b) Marginal
(c) Liquidity preference theory
(d) General
Answer:
(d) General

Question 2.
“Interest is the price paid for the use of capital in any market” is said by_______
(a) Adam Smith
(b) Torrance
(c) Walker
(d) Marshall
Answer:
(d) Marshall

Question 3.
The standard of living of workers in a country depends upon the ……………………… wages.
(a) Real
(b) Effective
(c) Direct
(d) Elastic
Answer:
(a) Real

Question 4.
The author of agio-theory of interest.
(a) N.W. Senior
(b) Bohm Bawerk
(c) Walker
(d) Marshall
Answer:
(b) Bohm Bawerk

Question 5.
Who propounded the risk bearing theory of profit?
(a) Bohm – Bawerk
(b) F.B Hawley
(c) Marshall
(d) Walker
Answer:
(b) F.B Hawley

Question 6.
Dynamic theory of profit was propounded by_______
(a) Marshall
(b) J.B.Clark
(c) J.M. Keynes
(d) Walker
Answer:
(b) J.B.Clark

Question 7.
What is meant by MPS?
(a) Marginal production supply
(b) Marginal production sale
(c) Marginal production service
(d) Marginal propensity to save
Answer:
(d) Marginal propensity to save

Question 8.
Time preference theory was given by_______
(a) John Ray
(b) Bohm Bawerk
(c) F.B. Hawley
(d) Irving Fisher
Answer:
(d) Irving Fisher

Question 9.
The demand for labour is ………………………
(a) Effective demand
(b) Direct demand
(c) Derived demand
(d) Elastic demand
Answer:
(c) Derived demand

Question 10.
_______ is the author of modern theory of rent
(a) Joan Robinson
(b) Boulding
(c) Both
(d) None
Answer:
(c) Both

Match the following and choose the answer using the codes given below

Question 1.

(a) 3 4 2 1
(b) 4 3 1 2
(c) 1 2 3 4
(d) 2 3 4 1
Answer:
(c) 1 2 3 4

Question 2.

(a) 2 1 4 3
(b) 1 4 2 3
(c) 1 2 3 4
(d) 4 1 3 2
Answer:
(a) 2 1 4 3

Choose the incorrect pair

Question 3.

Answer:
(d) Total demand for money (iv) Mt + Mp

Question 4.

Answer:
(b) Wage is the residual portion (ii) J.S. Mill

Choose the correct statement

Question 5.
(a) The supply of land is fixed forever
(b) Quasi-rent accrues to land
(c) The supply of man-made appliances are fixed
(d) Quasi-rent enters price
Answer:
(a) The supply of land is fixed forever

Question 6.
(a) Rent is the reward for labour
(b) Profit is the reward for labour
(c) Wages are the reward for labour
(d) Interest is the reward for organisation
Answer:
(b) Profit is the reward for labour

Choose the odd one out

Question 7.
(a) Bank credit
(b) Hoarding
(c) Dishoarding
(d) Disinvestment
Answer:
(b) Hoarding

Question 8.
(a) Monopoly profit
(b) Wind fall profit
(c) Functional reward
(d) Reward for labourer
Answer:
(d) Reward for labourer

Question 9.
(a) Nominal wages
(b) Real wages
(c) Direct wages
(d) Piece wages
Answer:
(c) Direct wages

Question 10.
(a) Abstinence theory
(b) Loanable funds theory
(c) Liquidity preference theory
(d) Risk bearing theory
Answer:
(b) Loanable funds theory

Analyse the reason for the following

Question 11.
Assertion (A) : Demand for loanable funds depends on investment, consumption and hoarding.
Reason (R) : Supply of loanable funds depends on hoarding and investment.
(a) (A) and (R) are true, (R) is not the correct explanation of (A)
(b) (A) and (R) are true, (R) is the correct explanation of (A)
(c) (A) is true (R) are false.
(d) (A) is false (R) is true.
Answer:
(c) (A) is true (R) are false.

Question 12.
Assertion (A) : The rate of interest is determined by the demand for money and the supply of money.
Reason (R) : The demand for money is liquidity preference.
(a) Both (A) and (R) are true, (R) is not the correct explanation of (A)
(b) Both (A) and (R) are correct, (R) is the correct explanation of (A)
(c) Both (A) and (R) are false.
(d) (A) is false and (R) is true.
Answer:
(a) Both (A) and (R) are true, (R) is not the correct explanation of (A)

Question 13.
Assertion (A) : Only superior land get rent
Reason (R) : Rent arose on account of differences in the fertility of land.
(a) Both (A) and (R) are true
(b) Both (A) and (R) are false correct explanation of (A)
(c) (A) is true and (R) are false.
(d) (A) is false and (R) is true.
Answer:
(a) Both (A) and (R) are true

Choose the incorrect pair

Question 14.

Answer:
(d) Wage fund theory (iv) J.S. Mill

Question 15.

Answer:
(b) Wages (ii) Labour

Fill in the blanks with suitable option given below

Question 16.
F.A. Walker wrote a book ______ in 1875.
(a) Political economy
(b) Social economy
(c) Principles of economics
(d) Wealth of nations
Answer:
(a) Political economy

Question 17.
The author of Agio theory of interest______
(a) N.W. Seniors
(b) Bohm-Bawerk
(c) Walker
(d) Marshall
Answer:
(b) Bohm-Bawerk

Question 18.
______ is the produced means of production.
(a) Land
(b) Labour
(c) Capital
(d) Organisation
Answer:
(c) Capital

Choose the best option

Question 19.
The theory of factor prices is popularly known as the theory of
(a) Distribution
(b) Exchange
(c) Wages
(d) Profit
Answer:
(a) Distribution

Question 20.
______ is the author of the modern theory of rent
(a) Joan Robinson
(b) Boulding
(c) Both
(d) None
Answer:
(c) Both

Part – B
Answer the following questions in one or two sentences

Question 1.
What is the theory of distribution?
Answer:
The theory of factor prices is popularly known as the theory of distribution.

Question 2.
Define VMP?
Answer:

1. VMP – means Value of Marginal Product.
2. The Value of Marginal Product is obtained by multiplying the Marginal Physical Product of the factor by the price of product. Symbolically, VMP = MPP x Price.

Question 3.
Define : Marginal productivity theory of distribution.
Answer:
The theory states that price or the reward for any factor of production is equal to the marginal productivity of that factor.

Question 4.
What is marginal product ? Define marginal physical product ?
Answer:
The addition made to the total product by employment of an additional unit of that factor.

Question 5.
Define modern theory of rent.
Answer:
Rent is the difference between the actual earnings of a factor of production and its transfer earning.
Rent = Actual earning – Transfer earning

Question 6.
What is “Wage Fund“.
Answer:
According to Mill, “Every employer will keep a given amount of capital for payment to the workers”. It is known as “Wage Fund”.

Question 7.
Define Quasi Rent.
Answer:
“Quasi rent is the income derived from machines and other appliances made by man
QR = Total revenue – Total variable cost

Part – C
Answer the following questions in one Paragraph

Question 1.
Explain the types of distribution.
Answer:
1. Personal distribution :
It is the distribution of national income among individuals.

2. Functional distribution :
It means the distribution of income among the four factors of production for their services in the production process.

Question 2.
Explain the standard of living theory of wages.
Answer:
This theory of wages was developed by Torrance is an improved version of the subsistence theory of wage. According to this theory, the wage is equal to the standard of living of the workers. If the standard of living is high, wages will be high and vice versa. The standard of a living wage means the amount necessary to maintain the labourer in the standard of life to which he is accustomed.

Question 3.
Briefly explain the Abstinence theory or waiting theory?
Answer:

1. This theory was propounded by N.W.Senior.
2. According to Senior, capital is the result of saving.
3. Saving involves “ abstinence ” or “ Sacrifice
4. It is possible to save only if one abstains from present consumption.
5. Interest is the reward or compensation paid to the saver (capitalist) for his “ abstinence ” or “ sacrifice”.
6. Marshall accepted the Abstinence theory of interest.
7. According to him, interest is a reward for waiting.
8. Saving involves waiting.
9. Therefore, interest is the reward paid to the saver for his “ waiting ”.

Question 4.
State the theories of interest.
Answer:

1. Abstinence theory or waiting theory.
2. Agio theory or the psychological theory.
3. Loanable funds theory or the neoclassical theory.
4. Liquidity preference theory or the monetary theory.

Question 5.
What are the four sources of loanable funds.
Answer:
The supply of loanable funds depends upon the following four sources.

1. Savings (S)
2. Bank Credit (BC)
3. Dishoarding (DH)
4. Disinvestment (DI)

Question 6.
What are the kinds of profit.
Answer:
1. Monopoly profit:
Profit earned by the firm because of its monopoly control.

2. Windfall profit:
Some times, profit arises due to changes in price level. Profit is due to unforeseen factors.

3. Profit as functional reward :
Just like rent, wage and interest, profit is earned by the entrepreneur for his entrepreneurial function.

Question 7.
State the residual claimant theory of wage.
Answer:
This theory was propounded by the American economist F.A. Walkar in 1875, in his book political economy. According to this theory, wage is the residual portion after paying the remuneration of all the other three factors namely land capital and organization.

Criticisms :

1. This theory does not explain the role of trade unions can secure higher wage for workers.
2. Demand side of labour in the determination of wages needs to be considered.

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Samacheer Kalvi 11th Chemistry Basic Concepts of Organic Reactions Textual Evaluation Solved

I. Choose The Best Answer:

Question 1.
For the Following reactions –
(A) CH₃ CH₂ CH₂ Br + KOH → CH₃ – CH + KBr + H₂O
(B) (CH₃ )₃ CBr + KOH → (CH₃ )₃ COH + KBr
(C)
Which of the following statement is correct’?
(a) (A) is elimination, (B) and (C) are substitution
(b) (A) is substitution, (B) and (C) are elimination
(c) (A) and (B) are elimination and (C) is addition reaction
(d) (A) is elimination, B is substitution and (C) is addition reaction.
Answer:
(d) (A) is elimination, B is substitution and (C) is addition reaction.

Question 2.
What is the hybridisation state of benzyl carbonium ion?
(a) sp²
(b) spd²
(c) sp³
(d) sp²d
Answer:
(a) sp²

Question 3.
Decreasing order of nucleophilicity is –
(a) OH^–> RNH₂^– > ^–OCH₃> RNH₂
(b) NH₂^–> OH^–> ^–OCH₃> RNH₂
(c) NH,> CH₃O >OH^–> RNH₂
(d) CH₃O^–> NH₂^–> OH^–> RNH₂
Answer:
(b) NH₂^–> OH^–> ^–OCH₃> RNH₂₊

Question 4.
Which of the tòllowing species is not eLectrophilic in nature?
(a) Cl⁺
(b) BH₃
(c) H₃O⁺
(d) ⁺NO₂
Answer:
(c) H₃O⁺

Question 5.
Homolytic fission of covalent bond leads to the formation of –
(a) electrophile
(b) nucicophile
(c) Carbocation
(d) free radical
Answer:
(d) free radical

Question 6.
Hyper Conjugation is also known as –
(a) no bond resonance
(b) Baker – nathan effect
(c) both (a) and (b)
(d) none of these
Answer:
(c) both (a) and (b)

Question 7.
Which of the group has highest +I effect?
(a) CH₃^–
(b) CH₃CH₂^–
(c) (CH₃)₂ -CH-
(d) (CH₃)₃-C-
Answer:
(d) (CH₃)₃-C-

Question 8.
Which of the following species does not exert a resonance effect?
(a) C₆H₅OH
(b) C₆H₅Cl
(c) C₆H₅NH₂
(d) C₆H₅NH₃
Answer:
(d) C₆H₅NH₃

Question 9.
-I effect is shown by –
(a) -Cl
(b) -Br
(c) both (a) and (b)
(d) -CH₃
Answer:
(c) both (a) and (b)

Question 10.
Which of the following carbocation will be most stable?
(a) \({ PH }_{ 3 }\overset { + }{ C- } \)
(b) CH₃–\(\overset { + }{ C{ H }_{ 2 } } \)
(c) (CH₃)₂–\(\overset { + }{ C{ H } } \)
(d) CH₂ = CH-CH₃
Answer:
(a) \({ PH }_{ 3 }\overset { + }{ C- } \)

Question 11.
Assertion : Tertiary Carbocations are generally formed more easily than primary Carbocations.
Reason : Hyper conjugation as well as inductive effect due to additional alkyl group stabilize tertiary carbonium ions.
(a) both assertion and reason arc true and reason is the correct explanation of assertion.
(b) both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false
(d) Both assertion and reason are false
Answer:
(a) both assertion and reason are true and reason is the correct explanation of assertion.

Question 12.
Heterolytic fission of C-C bond results in the formation of –
(a) free radical
(b) Carbanion
(c) Carbocation
(d) Carbanion and Carbocation
Answer:
(e) Carbocation

Question 13.
Which of the following represent a set of nucleophiles?
(a) BF₃, H₂O, NH^2-
(b) AlCl₃, BF₃, NH₃
(c) CN^–, RCH^–₂, ROH
(d) H⁺, RNH⁺₂, CCl₂
Answer:
(c) CN^–, RCH^–₂, ROH

Question 14.
Which of the following species does not acts as a nucleophile?
(a) ROH
(b) ROR
(c) PCl₃
(d) BF₃
Answer:
(d) BF₃

Question 15.
The geometrical shape of carbocation is
(a) Linear
(b) tetrahedral
(c) Planar
(d) Pyramidal
Answer:
(c) Planar

II. Write brief answer to the following questions.

Question 16.
Write short notes on:
(a) Resonance
(b) Hyper conjugation
Answer:
(a) Resonance:
1. Certain organic compounds can be represented by more than one structure and they differ only in the position of bonding and lone pair of electrons. Such structures are called resonance structure and this phenomenon is called as resonance. This phenomenon is also called as mesomerism or mesomeric effect.

2. For example, the structure of aromatic compounds such as benzene and conjugated system like 1,3 butadiene cannot be represent by a single structure and their observed properties can be explained on the base of a resonance hybrid.

3. Resonance structure of benzene.

(I) and (II) are called as resonance hybridš of benzcnc.

4. For 1,3 butadiene:

(I), (II) and (III) are called as resonance hybrids of 1,3 butadiene.

(b) Hyper conjugation:
1. The denationalization of electrons of σ bond is called as hyper conjugation. it is a special stabilizing effect that results due to the interaction of electrons of a σ bond with the adjacent empty non-bonding P-orbitais resulting in an extended molecular orbital.

2. Hyper conjugation is a permanent effect.

3. For example, in propane. the σ- electrons of C-H bond of methyl group can be delocalised into the it- orbital of doubly bonded carbon as represented below.

Question 17.
What are electrophiles and nucleophiles? Give suitable examples for each.
Answer:
Electrophiles:
Electrophiles are reagents that are attracted towards negative charge or electron rich center. They are either positively charged ions or electron deficient neutral molecules.
Example:
CO₂, AlCl₃, BF₃, FeCl₃, NO⁺, NO⁺₂,, etc.

Nucleophiles:
Nucleophiles are reagents that has high affinity for electropositive centers. They possess an atom has a unshared pair of electrons. They are usually negatively charged ions or electron rich neutral molecules.
Example.
NH₃, R-NH₂, R-SH, H₂O, R-OH, CN^–. OH^– etc.

Question 18.
Show the heterolysis of covalent bond by using curved arrow notation and complete the following equations. Identify the nucleophile In each case.
1. CH₃-Br+KOH →
2. CH₃-OCH₃+HI →
Answer:

In this case, \(\overset { \ominus }{ Br } \) and \(\overset { \ominus }{ OH } \) are nucleophiles.

Another mechanism:
Step – I

Step-II

In this case, OCH₃ and -I are nucleophiles
Another mechanism:
Step – I

Step – II

Question 19.
Explain inductive effect with suitable example.
Inductive effect:
Answer:
1. it is defined as the change ¡n the polarization of a covalent bond due to the presence of adjacent bonded atoms or groups in the molecule. It is denoted as I-effect.

2. Atoms or groups which lose electron towards a carbon atom are said to have a +I effect.
Example:
CH₃-,(CH₃)₂ CH-,(CH₃)₂ C- etc.

3. Atoms or groups which draw electrons away from a carbon atom are said to have a -I effect.
Example:
-NO₂. -I, -Br, -OH, C₆H₅ etc.

4. For example, consider ethane and ethyl chloride. The C-C bond in ethane is non polar while the C-C bond in ethyl chloride is polar. We know that chlorine is more electronegative than carbon and hence it attracts the shared pair of electrons between C-Cl in ethyl chloride towards it self.

This develops a slight negative charge on chlorine and a slight positive charge on carbon to which chlorine is attached. To compensate it, the C₁ draws the shared pair of electron between itself and C₂. This polarization effect is called inductive effect.

Question 20.
Explain electromeric effect.
Electromeric effect:
Answer:
1. The electromeric effect refers to the polarity produced in a multiple bonded the compound when it is attacked by a reagent when a double or a triple bond is exposed to an attack by an electrophile the two π electrons which from the π bond are completely transferred to one atom or the other.

2. When a nucleophile approaches the carbonyl compound, the π-electrons between C and O is instantaneously shifted to the more electronegative oxygen. This make the carbon electron deficient and thus facilitating the formation of a new bond between the incoming nucleophile and the carbonyl carbon atom.

3. When an electrophile such as H approaches an alkene molecule their electrons are instantaneously shifted to the electrophiie and a new bond is formed between carbon and hydrogen. This makes the other carbon electron deficient and hence it acquires a positive charge.

4. This effect denotes as E-effect.

Question 21.
Give examples for the following types of organic reactions
1. – elimination
2. Electrophilic substitution.
Answer:
1. β-eIimination:
Elimination reactions involve the cleavage of a a bond and tórmation of air bond. A nucleophilic pair of electrons heads into a new it bond as a leaving group departs. This process is called 3-elimination because the bond Í to the nucleophilic pair of electrons breaks.
Example:
(a) n-propyl bromide on reaction with alcoholic KOH give propene. in this reaction hydrogen and Br are eliminated.

(b) Acid-catalysed dehydration of alcohols

2. Electrophilic substitution:
Substitution reactions when are brought about by electrophiles are called electrophilic substitution reaction.
Example:
(a) Nitration of benzene

(b) Bromination of benzene:

Samacheer Kalvi 11th Chemistry Basic Concepts of Organic ReactionsAdditional Questions Solved

I. Choose The Correct Answer.

Question 1.
Statement-I: All the organic molecules contain covalent bonds.
Statement-II: Organic molecules are formed by the mutual sharing of electrons between atoms.
(a) Statement-I and II are correct and statement-II is correct explanation of statement-I.
(b) Statement-I and II are correct but Statement-II is not correct explanation of statement-I.
(c) Statement-I is correct but statement-II is wrong.
(d) Statement-I is wrong but statement-II is correct.
Answer:
(a) Statement-I and II are correct and statement-II is correct explanation of statement-I.

Question 2.
Statement-I : Homolytic cleavage is symmetrical one.
Statement-II : A single covalent bond breaks and each of the bonded atoms retains one electron.
(a) Statement-I and II are correct and statement-II is correct explanation of statement-I.
(b) Statement-I and II are correct but statement-II is not correct explanation of statement-I.
(c) Statement-I is correct but statement-II is wrong.
(d) Statement-I is wrong but statement-II is correct.
Answer:
(a) Statement-I and II are correct and statement-Il is correct explanation of statement-I.

Question 3.
Which one of the following is an example for free radical initiators?
(a) Benzoyl peroxide
(b) Benzyl alcohol
(c) Benzyl acetate
(d) Benzaldehyde.
Answer:
(a) Benzoyl peroxide.

Question 4.
Which one of the following ¡s correct order of stability of alkyl free radicals?

Answer:

Question 5.
Statement-I: Ileterolytic cleavage is unsymmetrical one.
Statement-II: A covalent bond breaks and one of the bonded atom retains the bond pair of electrons.
(a) Statement-I and II are correct and statement-II is correct explanation of statement-I.
(b) Statement-I and II are correct but statement-II is not correct explanation of statement-I.
(c) Statement-I is correct but statement-II is wrong.
(d) Statement-I is wrong but statement-II is correct
Answer:
(a) Statement-I and II are correct and statement-II is correct explanation of statement-I.

Question 6.
Which of the following is correct order of the stability of carbocations?
(a) ⁺CH₃ > ⁺CH₂ CH₃> ⁺CH(CH₃)₂ > ⁺C(CH₃)₃
(b) ⁺CH₂CH₃ > ⁺CH₃ >⁺ CH(CH₃)₂ > ⁺C(CH₃)₃
(c) ⁺C(CH₃)₃ > ⁺CH(CH₃)₂ >⁺CH₂CH₃^–CH₃
(d) ⁺CH(CH₃)₂ > ⁺CH₃ > ⁺CH₂CH₃ > ⁺C(CH₃)₃
Answer:
(c) ⁺C(CH₃)₃ > ⁺CH(CH₃)₂ >⁺CH₂CH₃^–CH₃

Question 7.
Which one of the following is correct order of the stability of carbanions?
(a) ^–C(CH₃)₃> ^–CH(CH₃)₂> ^–CH₂-CH₃> ^–CH₃
(b) ^–CH₃> ^–CH₂-CH₃> ^–CH(CH₃)₂> ^–C(CH₃)₃
(c) ^–CH(CH₃), > ^–CH₃> ^–CH₂-CH₃ > ^–C(CH₃)₃
(d) ^–CH₂-CH₃ > ^–CH(CH₃)₂> ^–CH₃> ^–C(CH₃)₃
Answer:
(b) CH₃> CH₂-CH₃> CH(CH₃)₂> C(CH₃)₃

Question 8.
Which one of the following is not electrophile’?
(a) NH₃
(b) AlCl₃
(c) FeCl₃
(d) R-X
Answer:
(a) NH₃

Question 9.
Which one of the following is not nucleophile
(a) H₂O
(b) NH₃
(c) R-OH
(d) FeCl₃
Answer:
(d) FeCl₃

Question 10.
Which one of the following is positively charged electrophiles?
(a) CO₂
(b) AlCl₃
(c) BF₃
(d) RX
Answer:
(d) RX

Question 11.
Which one of the following is nucleophile?
(a) BF₃
(b) AlCl₃
(c) CO₂
(d) R-SH
Answer:
(d) R-SH

Question 12.
Which one of the following species has tendency to show-I effect?
(a) -CH₃
(b) -CH₂-CH₃
(c) -CH(CH₃)₂
(d) -C₆H₅
Answer:
(d) -C₆H₅

Question 13.
Which one of the following species has tendency to show +I effect?
(a) -NH₂
(b) -Cl
(c) -C₆H₅
(d) -CH₃
Answer:
(d) -CH₃

Question 14.
Which one of the following has strongest acidic character?
(a) HCOOH
(b) CH₃COOH
(c) CH₂ClCOOH
(d) CCl₃COOH
Answer:
(d) CCl₃COOH

Question 15.
Which one of the following has least acidic character?
(a) HCOOH
(b) CH₃COOH
(c) CH₂ClCOOH
(d) CCl₃COOH
Answer:
(b) CH₃COOH

Question 16.
Pick out the correct order of acid strength.
(a) CH₃-CH₂-COOH > CH₃COOH > CH₂CICOOH
(b) CH₃COOH > CH₃-CH₂-COOH >CH₂CICOOH
(c) CH₂ClCOOH > CH₃COOH > CH₃-CH₂-COOH
(d) CH₂ClCOOH > CH₃-CH₂-COOH > CH₃COOH
Answer:
(c) CH₂ClCOOH > CH₃COOH > CH₃—CH₂-COOH

Question 17.
Statement-I: Fluoro acetic acid is stronger acid than acetic acid
Statement-II: Fluorine has high electronegativity and it is facilitate to dissociate the OH bond easily.
(a) Statement-I and II are correct and statement-II is correct explanation of statement-I.
(b) Statement-I and II are correct but statement-II is not correct explanation of statement-I.
(c) Statement-I is correct but statement-II is wrong.
(d) Statement-I is wrong but statement-Il is correct
Answer:
(a) Statement-I and II are correct and statement-II is correct explanation of statement-I.

Question 18.
Which one of the following is an example for negative mesomeric effect?
(a) -SH
(b) -SR
(c) -NH₂
(d) -NO₂
Answer:
(d) -NO₂

Question 19.
Which one of the following electrophile used for nitration of benzene?
(a) \({ Br }^{ \oplus }\)
(b) \({ { NO }_{ 2 }^{ \ominus } }\)
(c) -NH₂
(d) \({ NO }^{ \ominus }\)
Answer:
(b) \({ { NO }_{ 2 }^{ \ominus } }\)

Question 20.
Identify the one which does not come under the organic addition reaction
(a) Hydration
(b) Dehydration
(c) Halogenation
(d) Hydro halogenation
Answer:
(b) Dehydration

Question 21.
Primary alcohols undergo which type of reaction to form alkenes?
(a) Elimination
(b) Oxidation
(c) Reduction
(d) Hydrolysis
Answer:
(a) Elimination

Question 22.
Identify the type of reaction
(a) Addition reaction
(b) Elimination reaction
(c) Reduction reaction
(d) Oxidation reaction
Answer:
(d) Oxidation reaction

II. Match the following.

Answer:

III. Fill in the blanks.
Question 1.
The slowest step in the mechanism determines ………..
Answer:
Rate of reaction

Question 2.
Homolytic cleavage occurs under the conditions of ………….
Answer:
High temperature.

Question 3.
During the cleavage of Azobisisobutyronitrile produces species ……….
Answer:
Free radical

Question 4.
The cleavage of C-Br bond in tert-butyl bromide leads to formation of ………..
Answer:
Carbocation

Question 5.
The cleavage of C-H bond in aldehydes Leads to formation of ………….
Answer:
Carbanion

Question 6.
Electron displacement occurring in saturated compounds along a carbon chain is termed as …………
Answer:
Inductive effect

Question 7.
The addition of H to alkene is an example of effect ……….
Answer:
+E

Question 8.
Example for positive mesomeric effect is ………..
Answer:
-OH

Question 9.
Acidity of phenol was explained by …………
Answer:
R-effect

Question 10.
Hydrolysis of alkyl halide is an example for ……….
Answer:
Nucleophilic substitution

Question 11.
4-hydroxy phenol reacts with acidified potassium dichromate to gives
Answer:
p-Benzoquinone

Question 12.
Enzyme present in apple is
Answer:
Poly phenol oxidase

Question 13.
Benzene reacts with H₂ in the presence of Pt to give
Answer:
Cyclohexane

Question 14.
Alcohol on refluxing with K₂Cr₂O₇ gives……..
Answer:
Carboxylic acid

Question 15.
Carbonyl compounds especially ketones undergo reduction to form ………..
Answer:
Secondary alcohols

Question 16.
Ethane undergo thrombolytic cleavage to form ………..
Answer:
Methyl free radical

IV. Choose the odd one out.
Question 1.
(a) NH₃
(b) H₂O
(c) CN^–
(d) RSH
Answer:
(c) CN^– It is a negatively charged nucleophile whereas others are neutral nucleophiles.

Question 2.
(a) CO₂
(b) RX
(c) MCl₃
(d) FeCl₃
Answer:
(b) RX. It is a positive charged electrophiles whereas others are neutral clectrophiles.

Question 3.
(a) -F
(b) – Cl
(c) – COOH
(d) CH₃O^–
Answer:
(d) CH₃O^–. It is a group carrying negative charge one electron donating (or) +I group whereas others are -I group.

Question 4.
(a) – C(CH₃)₃
(b) – COOK
(c) – CH(CH₃)₂
(d) – COO^–
Answer:
(b) – COOH. It has -I effect whereas other have +I effect.

Question 5.
(a) -OH
(b) -NH₂
(c) -SH
(d) -COOH
Answer:
(d) – COOH. It has negative mesomeric effect whereas others have positive mesomeric effect.

V. Choose the correct pair.
Question 1.
(a) Homolytic clevage : carbocation
(b) Homolytic clevage : carbanion
(c) Homolytic clevage : free radicals
(d) Heterolytic clevage : free radicals
Answer:
(c) Homolytic clevage : free radicals

Question 2.
(a) + C(CH₃)₃ > ⁺CH(CH₃)₂> ⁺CH₂-CH₃ > ⁺CH₃ : Relative stability of carbocations.
(b) – C(CH₃)₃ > ^–CH(CH₃)₂ > ^–CH₂-CH₃ > ^–CH₃ : Relative stability of carbanion.
(c) OH^–,RO^–, RCOO^– : Neutral nucleophile
(d) AlCl₃, BF₃, FeCl₃ : Positively charged nucleophile
Answer:
(a) + C(CH₃)₃ > ⁺CH(CH₃)₂> ⁺CH₂-CH₃ > ⁺CH₃ : Relative stability of carbocations.

Question 3.
(a) CH₃ – CH₂ – CH₂Br + Alcoholic KOH : Substitution reaction
(b) CH₃ – CH₂ – CH₂Br + Alcoholic KOH : Elimination reaction
(c) CH₃CHO + Acidic dichromate : Reduction reaction
(d) Benzene + Pt + H₂ : Oxidation reaction
Answer:
(b) CH₃ – CH₂ – CH₂Br + Alcoholic KOH : Elimination reaction

VI. Choose the incorrect pair.
Question 1.
(a) – OH,- SH, -NH₂ : Positive mesometic effect
(b) – NO₂ >CO, – COOH : Negative mesomeric effect
(c) – F, – Cl, – NO₂ : Electron withdrawing group
(d) – C(CH₃)₃, – CH(CH₃)₃ – CH₂ – CH₃ : Electron withdrawing group
Answer:
(c) – F, – Cl, – NO₂ : Electron withdrawing group

Question 2.
(a) NH₃ andAmines : Neutral nucleophile
(b) OH^– and RCOO^– : Negative nucleophile
(c) RX and H₃O₃ : Positive electrophile
(d) MCl₃, BF₃ : Negative electrophile
Answer:
(d) AlCl₃, BF₃ : Negative electrophile

Question 3.

Answer:

VII. Assertion and Reason.
Question 1.
Assertion (A) : Neutral molecule SnCl₄ can act as an electrophile.
Reason (R) : it has vaccant ‘d’ orbitais which can accommadate the electrons from others.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 2.
Assertion (A) : The C-C bond in ethane is non-polar while the C-C bond in ethyl chloride is polar.
Reason (R) : Chlorine is more electronegative than carbon and hence it attracts the shared pair of electron between C-C in ethyl chloride and it develops a negative charge on Cl and positive charge on Carbon.
(a) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 3.
Assertion (A): Phenol is more acidic than aliphatic alcohols.
Reason (R): The phenoxide ion is more stabilized than phenol by resonance effect and hence resonance favours ionization of phenol to form H and shows acidity.
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).

Question 4.
Assertion (A): The cut apple turns brown.
Reason (R): Cut apple exposes its cells to atmospheric oxygen and the oxidizes the phenolic compound present in it. Due to this enzymatic browning, the cut apple turns brown.
(a) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Samacheer Kalvi 11th Chemistry Basic Concepts of Organic Reactions 2 Marks Question and Answers

II. Answer briefly.

Question 1.
What are organic reactions?
Answer:
Substrate is an organic molecule reacts with reagent, which may be an organic, inorganic or any agent like heat, photon etc, that brings about the chemical change to form a product, this is known as organic reactions.

Question 2.
What is mechanism of the reaction?
Answer:
In the organic reactions, that series of simple steps which collectively represent the chemical change, from substrate to product, this is called as the mechanism of the reaction. The slowest step in the mechanism determines the overall rate of the reaction.

Question 3.
Mention the types of fission of a covalent bond?
Answer:
There are three types of fission of a covalent bond. They are,

• Homolytic cleavage
• Heterolytic cleavage

Question 4.
Explain the homolytic fission of a covalent bond?
Answer:

• Homolytic cleavage is the process in which a covalent bond breaks symmetrically in such way that each of the bonded atoms retains one electron.
• This type of cleavage occurs under high temperature or in the presence of UV light.
• in a compound containing non-polar covalent bond formed between atoms of similar electronegativity, in such molecules the cleavage of bonds results into free radicals.
• For example, ethane undergo homolytic fission to produce, two methyl free radicals.
  

Question 5.
What are free radical initiators?
Answer:
The types of reagents that promote homolytic cleavage in substrate are called as free radical initiators. They are short lived and are highly reactive.

Question 6.
Mention any two examples for free radical initiators?
Answer:
Two examples for free radical initiators

• Azobisisobutyronitnie (AIBN)
• Benzoyl peroxide

Question 7.
Explain the heterolytic fission of a covalent bond?
Answer:

• Heterolytic cleavage is the process in which a covalent bond breaks unsymmetrically such that one of the bonded atom retains the bond pair of electron.
• It results in the formation of a cation and an anion of the two bonded atoms the most electronegative.
• For example in tert-butyibromide, the C-Br bond is polar as bromine is more electronegative than carbon. Hence the C-Br undergoes heterolytic cleavage to form a terl-butyl carbocation and bromide anion.
  

Question 8.
What are carbocations?
Answer:
1. Let us consider the heterolytic fission of the bond C-X present in organic molecule. If the atom X has greater electronegativity than the carbon atom, the former takes away the bonding electron pair and becomes negatively charged while the carbon will lose its electron and thus acquire a positive charge.
2. Such a cationic species carrying a positive charge on carbon are known as carbocation or carhonitim ion.

Question 9.
What are carbanions?
Answer:
Let us consider the heterolytic fission of the bond C-X present in an organic molecule. if the carbon atom has greater electronegativity than the atom X, the former takes away the bonding electron pair and acquires a negative charge. The resulting carbon anion is known as carbanion.

Question 10.
Differentiate the carbocation and carbanion
Answer:
Carbocation:

• In a carbocation carbon bearing positive charge.
• Carbon bearing positive charge has sp2 hybridization.
• It has a planar structure.
• Example:
  CH₃⁺ (CH₃)₂\({ C }^{ \ominus }\)H, (CH₃)₃C⁺ etc.

Carbanion:

• In a carbanion carbon bearing negative charge.
• Carbon bearing negative charge has sp³ hybridization.
• It has a pyramidal structure.
• Example:
  CH₃–\({ C }^{ \ominus }\)H₂, CH₃-CHO, (CH₃)₂\({ C }^{ \ominus }\)H etc.

Question 11.
Identify which of the following are electrophiles and nucleophiles?

1. NH₃
2. AlCl₃
3. R-SH
4. R-X
5. R-O-R
6. BF₃

Answer:

Question 12.
How will you distinguish between electrophiles and nucleophiles?
Answer:
Electrophiles:

• They are electron deficient.
• They are cations.
• They are lewis acids.
• Accept an electron pair.
• Attack on electron rich sites.

Nucleophites:

• They are electron rich.
• They are anions.
• They are lewis bases.
• Donate an electron pair.
• Attack on electron deficient sites.

Question 13.
What are all the sources for human body that produces free radicals?
Answer:
Sources for human body produces free radicals,

• Human body is exposed to X-rays.
• Cigarette smoke.
• Industrial chemicals.
• Air pollutants.

Question 14.
In what way free radical affect the human body?
Answer:

• Free radicals can disrupt cell membranes.
• Increase the risk of many forms of cancer.
• Damage the interior lining of blood vessels.
• leads to a high risk of heart disease and stroke.

Question 15.
How to reduce the effect of free radicals?
Answer:

• Use vitamins and minerals to counter the effects of free radicals.
• Fruits contains antioxidants which decrease the effects of free radicals.

Question 16.
Identify which of the following shows +I and -I effect?

1. -NO₂
2. -SO₂H
3. -I
4. -OH
5. CH₂O-
6. CH₂–

Answer:

Question 17.
Why chloro acetic acid is stronger acid than acetic acid?
Chioro acetic acid:
Answer:

Chioro acetic acid has Cl-group and it has high electronegativity and shows -I effect. Therefore Cl-atom to faciLitate the dissociation of O-H bond very fastly, whereas in the case of acetic acid, has CH₃ group and it shows +I effect, theretòre dissociation of O-H bond will be more difficult. Thus chloro acetic acid is stronger acid than acetic acid.

Question 18.
Explain the positive and negative electromeric effects?
Answer:
1. When an electrophile such as W approaches an alkene molecule, the π electrons are instantaneously shifted to the electrophile and a new bond is formed between carbon and hydrogen. This makes the other carbon electron deficient and hence it acquires a positive charge.

2. When the it electron is transferred towards the attacking reagent, it is called positive electromeric (+E) effect.

3. Whèn the t electron is transferred away from the attacking reagent it is called negative electromenc (-E) effect.

For example : The attack of C\({ N }^{ \ominus }\) on a carbonyl carbon.

Question 19.
Write a short notes on positive mesomeric effect?
Answer:

• Positive mesomeric effect occurs, when the electrons move away from substituent attached
  to the conjugated system.
• It occurs, if the electron releasing substituents are attached to the conjugated system.
• The attached group has a tendency to release electrons through resonance, these electron
  releasing groups are usually denoted as +R or +M groups
• Examples: -OH, -SH, -OR, -SR, -NH₂ etc.

Question 20.
Write a short notes on negative mesomeric effect?
Answer:

• Negative mesomeric effect occurs, when the electrons move towards the substituent attached to the conjugated system.
• It occurs if the electron withdrawing substituents are attached to the conjugated system.
  The attached group has a tendency to withdraw electrons through resonance, these electron withdrawing groups are usually denoted as -R or -M groups.
• Examples:
  

Question 21.
What are addition reactions? Give an example.
Answer:
All organic compounds having double or triple bond adopt addition reactions in which two substances unite to form a single compound. During the addition reaction the hybridization of the substrate changes as only one bond breaks and two new bonds are formed.
Example:

Question 22.
What are elimination reaction? Give an example.
Answer:
In these reactions, two atoms or groups are removed from a molecule without being substituted by other atoms or groups. A new C-C double bond is formed between the carbon atoms to which the eliminated atoms or groups are previously attached. It is always accompanied with change in hybridization.
Example:

Question 23.
What are organic oxidation reactions? Give an example.
Answer:
Most of the oxidation reaction of organic compounds involves gain of oxygen or loss of hydrogen.
Example:

Question 24.
What are organic reduction reactions? Give an example.
Answer:
Most of the reduction reaction of organic compounds involves gain of hydrogen or loss of oxygen.
Example:

Question 25.
Why cut apple turns a brown colour?
Answer:

• Apples contains an enzyme called polyphenol oxidase (PPO) also known as tyrosinase.
• Cutting an apple exposes its cells to the atmospheric oxygen and oxidizes the phenolic compounds present in apples. This is called the “enzymatic browning” that turns a cut apple brown.
• In addition to apples, enzymatic browning is also evident in bananas, pears, avocados and even potatoes.

Question 26.
What are functional group inter conversions?
Answer:
Organic synthesis involves functional group inter conversions. A particular functional group can be converted into other functional group by reacting it with suitable reagents. For example, the carboxylic acid group (-COOH) present in organic acids can be transformed into -COCl by treating the acid with SOCl₂ reagent..

Question 27.
How will you convert alcohol into aldehyde?
Answer:
When primary alcohol reacts with acidified potassium dichromate to gives aldehyde.

Question 28.
What happen when nitrile undergoes acid hydrolysis?
Answer:
When alkyl nitrile undergoes acid hydrolysis to give amide, which on further hydrolysis to give carboxylic acid.

Question 29.
Complete the following reactions and identify the products?
Answer:

Question 30.
Predict the product for the following reaction.
Answer:

Answer:

Question 31.
How will you convert benzene into cyclohexane?
Answer:
Benzene undergo catalytic hydrogenation in presence of platinum to give cyclohexane.

Question 32.
Complete the reaction and name the product

Answer:

Question 33.
Identify the product and mention the type of organic reaction.
Answer:

This is one of the elimination reaction.

Question 34.
Complete the following reaction and identify the product.
Answer:

Question 35.
What are substitution reaction?
Answer:
In this reaction an atom or a group of atoms attached to a carbon atom is replaced by a new atom or a group of atoms.

Here, -Br is replaced by -OH group.

Question 36.
How will substitution reactions are classified?
Answer:
Substitution reactions can be classified as,

• Nucleophilic substitution reaction.
• Electrophilic substitution reaction.
• Free radical substitution reaction.

Question 37.
Draw the resonance structures for the following compounds?

Answer:

Question 38.
Identify the reagents shown in box in the following equations as nucleophiles or electrophiles.
Answer:

In (a) \(\overset { \ominus }{ OH } \) is nucleophile
In (b) \(\overset { \ominus }{ CN } \) is nucleophile
In (c) CH₃\(\overset { \ominus }{ CO } \) is electrophile

Question 39.
Classify the following reactions in one of the reaction type of organic reaction.
Answer:
(a) CH₃CH₂Br + \({ SH }^{ \ominus }\) → CH₃CH₂SH + \({ Br }^{ \ominus }\)
(b) (CH₃)₂C = CH₂ + HCI (CH₃)₂CCl – CH₃
(c) CH₃CH₂Br + \({ OH }^{ \ominus }\) → CH₂ = CH₂ + H₂O + \({ Br }^{ \ominus }\)
(a) Nucleophilic substitution reaction.
(b) Electrophilic addition reaction.
(c) Elimination reaction.

Question 40.
Which electron displacement effect explain the following correct orders of acidity of the carboxylic acids?
Answer:
CH₃CH₂COOH > (CH₃)₂ CHCOOH > (CH₃)₃C COOH
As the number of alkyl group increases, +I effect increases and it strengthen the O-H bond. i.e., O-H bond dissociation is very difficult.

Therefore +I effect explain the correct orders of acidity of the carboxylic acids.

Question 41.
Which of the two, NO₂CH₂CH₂\({O}^{ \ominus }\) or CH₃CH₂\({O}^{ \ominus }\) is expected to be more stable and why?
Answer:

(I) is more stable than (II) because NO₂ group has -I effect and hence it tends to disperse the -ve charge on the 0-atom. In contrast., CH₃CH, has +I effect. It therefore, tends to intensify the -ve charge and hence destabilizes it.

Samacheer Kalvi 11th Chemistry Basic Concepts of Organic Reactions 5 Marks Questions and Answers

Question 1.
Explain electron movement in organic reactions.
Answer:
All organic reactions can be understood by following the electron movements.

• Lone pair becomes a bonding pair.
• Bonding pair becomes a lone pair.
• A bond breaks and becomes another bond.

The electron movement depends on the nature of the substrate, reagent and the prevailing conditions.
Type – 1.
A lone Pair to a bonding pair

Type – 2.
A bonding pair to a lone Pair

Type – 3.
A bonding pair to an another bonding pair

Question 2.
How does inductive effect influence the reactivity and acidity of carboxylic acids?
Answer:
(a) Reactivity:

• When a highly electronegative atom such as halogen is attached to a carbon then it makes the C—X bond polar.
• In such cases the -I effect of halogen facilitates the attack of an incoming nucleophile at the polarized carbon and hence increases the reactivity.
  
• If a -I group is attacher neared to a carbonyl carbon, it decreases the availability of electron density on the carbonyl carbon and hence increases the rate of the nucleophilic addition reaction.

(b) Acidity of carboxylic acid:

• When a halogen atom is attached to the carbon which is neared to the carboxylic acid group, its -I effect withdraws the bonded electrons towards itself and makes the ionization of H+ easy.
• The acidity of various chioro acetic acid is in the following order. Cl₃C-COOH > Cl₂CHCOOH > ClCH₂COOH The strength of the acid increases with increase in the effect of the group attached to the carboxyl group.
• Similarly the following order of acidity in the carboxylic acids is due to the +1 effect of alkyl group. (CH₃)₃CCOOH < (CH₃)₂CHCOOH <CH₃COOH

Question 3.
Explain the acidic nature of phenol.
Answer:

• Resonance is useful in explaining certain properties such as acidic of phenol.
• The phenoxide ion is more stabilized than phenol by resonance effect.
• The resonance favours ionisation of phenol to form H+ and shows acidity.
  
  phenoxide ion resonance structures.
  
• The above structures shows that there is a charge separation in the resonance structure of phenol which needs energy whereas there is no such hybrid structures in the case of phenoxide ion. This increases stability accounts for the acidic character of phenol.

Question 4.
How does hyper conjugation effect explain the stability of alkenes?
Answer:

• The relative stability of various classes of carhonium ions may be explained by the number of no bond resonance structures that can be written for them.
• Such structures are arrived by shifting the bonding electrons from an adjacent C-H bond to the electron deficient carbon.
• in this way, the positive charge originally on carbon is dispersed to the hydrogen. This manner of electron release by assuming no bond character in the adjacent C-H bond is called hyper conjugation or Baker- Nathan effect.
• The greater the hyper conjugation, the greater will be the stability of the compound. The increasing order of stability can be shown as:
  
• Alkyl group increases in the C=C double bond carbon, hyper conjugation increases and stability of that organic compound also increases.

Question 5.
Explain the types of addition reactions?
Answer:
Addition reactions are classified into three types. They are,

1. Electrophi lic addition reaction
2. Nucleoph lije addition reaction
3. Free radical addition reaction

1. Electrophilic addition reaction:
An electrophilic addition reaction can be described as an addition reaction in which a reactant with multiple bonds as in a double or triple bond undergoes has it air bond broken and two new a bond are formed.

2. Nucleophilic addition reaction:
A nucleophilic addition reaction is an addition reaction where a chemical compound with an electron deficient or electrophilic double or triple bond, a it bond, reacts with a nucleophilic which is an electron rich reactant with the disappearance of the double bond and creation of two new single or a bonds.

3. Free radical addition reaction:
it is an addition reaction in organic chemistry involving free radicals. The addition may occur between a radical and a non radical or between two radicals.

Question 6.
Explain the types of substitution reaction?
Answer:
Substitution reactions are classified into three types. They are,

1. Nucleophilic substitution reaction.
2. Electrophilic substitution reaction.
3. Free radical substitution reaction.

1. Nucleophilic substitution reaction:
A nucleophilic substitution reaction in organic chemistry is a type of reaction where a nucleophilic gets attached to the positive charged atoms or molecules of the other substance. A good example of a nucleophilic substitution reaction is the hydrolysis of alkyl bromide, under the basic conditions where in the nucleophile is nothing but the base H. whereas the leaving group is the Br°. The reaction for the following is as give below.

2. Electrophilic substitution reaction:
The electrophilic substitution reaction involves the electrophiles. The electrophilic reactions occur mostly with the aromatic compounds. This types of substitution reaction arc basically defined as those chemical reactions where the electrophile replaces the functional group in a compound but not the hydrogen atom, sometimes hydrogen atom can be also replaced by electrophiles.
For example:

3. Free radical substitution reaction:
Free radical substitution reaction involving free radicals arc a reactive intermediate.
A – X + Y → A – Y + X
CH₄ + Cl → CH + HCl
CH₃ + Cl → CH₃Cl (methyl chloride)

Question 7.
For the following bond cleavages use curved-arrows to show the electron flow and classify each as homolytic or heterolytic fission. Identify reactive intermediate produced as free radical,, carbocation and carbanion?

Answer:

Question 8.
An organic compound (A) has a molecular formula C₂H₆O it is one of the primary alcohol. A reacts with acidified potassium dichromate to give B. B on further undergoes to oxidation reaction to give C. C on reacts with SOCl₃ to give D which ¡s chlorinated product. Identify A,B,C and D, explain with equation.
Answer:
1. C₂H₆O is CH₃-CH₂-OH which is a primary alcohol (A)
2. CH₃-CH₂-OH (A) reacts with H⁺/ K₂Cr₂O₇ to give acetaldehyde (B)

3. CH₃CHO (B) which on further oxidation to give acetic acid (C).

4. Acetic acid reacts with SOCl₂ to give acetyl chloride (D).

Question 9.
An organic compound (A) of a molecular formula C₂H₄ which is a simple alkene. A reacts with dil H₂S0₄ to give B. A again reacts with Cl₂ to give C. Identify AB and C and write the equations.
Answer:
1. C₂H₂ is CH₂=CH₂ is a simple alkene. A is ethylene.
2. Ethylene (A) reacts with dii H₂SO₄ to give ethanol (B)

3. Ethylene (A) reacts with Cl, to give 1.2 dichloro ethane (C)

Question 10.
Complete the reactions and identify the products.
Answer:

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Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry Textual Evaluation Solved

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry Multiple Choice Questions

Question 1.
Select the molecule which has only one ir bond.
(a) CH₃-CH=CH-CH₃
(b) CH3-CH=CH-CHO
(c) CH3-CH=CH-COOH
(d) All of these
Answer:
(a) CH₃-CH=CH-CH₃

Question 2.
In the hydrocarbon the state of hybridisation of carbon 1,2,3,4 and 7 are in the following sequence.
(a) sp, sp, sp³, sp², sp³
(b) sp², sp, sp³, sp², sp³
(c) sp, sp, sp², sp, sp³
(d) none of these
Answer:
(a) sp, sp, sp³, sp², sp³

Question 3.
The general formula for alkadiene is ……….
(a) C_nH_2n
(b) C_nH_2n-1
(c) C_nH_2n-2
(d) C_nH_n-2
Answer:
(c) C_nH_2n-2

Question 4.
Structure of the compound whose IUPAC name is 5, 6 – dimethylhept-2-ene is ……….

Answer:

Question 5.
The IUPAC name of the compound is …………

(a) 2,3 – Diemethyiheptane
(b) 3 – Methyl – 4 – ethyloctane
(c) 5 – ethyl – 6 – methyloctanc
(d) 4 – Ethyl -3 – methyloctane.
Answer:
(d) 4 – Ethyl – 3 – methyloctane.

Question 6.
Which one of the following names does not fit a real name?
(a) 3 – Methyl – 3 – hexanone
(b) 4- Methyl – 3 – hexanone
(c) 3 – Methyl – 3 – hexanol
(d) 2 – Methyl cyclo hexanone.
Answer:
(a) 3 – Methyl – 3 – hexanone

Question 7.
The TUPAC name of the compound CH3— CII = CH -C CH is ……………
(a) Pent – 4 – yn – 2 – ene
(b) Pent -3-en – 1- yne
(c) pent – 2 – en – 4 – yne
(d) Pent – 1 – yn – 3 – ene
Answer:
(b) Pent -3-en – 1- yne

Question 8.
IUPAC name of is ……….
(a) 3, 4, 4 – Trimethylheptane
(b) 2 – Ethyl – 3, 3 – dimethyl heptane
(c) 3, 4, 4 – Trimethyloctane
(d) 2 Butyl – 2 -methyl – 3 – ethyl-butane.
Answer:
(c) 3, 4, 4 – Trimethyloctane

Question 9.
The IUPAC name of is ………
(a) 2 – Hydroxypropionic acid
(b) 2, 4. 4 – Trimethylpent -3-ene
(c) Propan – 2 – ol 1 – oie acid
(d) 2, 2, 4 – Trimethylpent -2-ene
Answer:
(b) 2 – Hydroxy Propanoic acid

Question 10.
The IUPAC name of the compound is ………
(a) 3 – Ethyl – 2- hexene
(b) 3 – Propyl – 3. hexene
(c) 4 – Ethyl – 4 – hexene
(d) 3 – Propyl -2-hexenc
Answer:
(a) 3 – Ethyl – 2- hexene

Question 11.
The IUPAC name of the compound is ……..
(a) 2 – Hydroxypropionic acid
(b) 2 – Hydroxy Propanoic acid
(c) Propan 2 – ol – 1 – oic acid
(d) 1 – Carboxyethanol.
Answer:
(b) 2 – Hydroxy Propanoic acid

Question 12.
The IUPAC name of is ………
(a) 2 – Bromo – 3 – methylbutanoic acid
(b) 2 – methyl – 3 – bromobutanoic acid
(c) 3 – Bromo – 2 – methylbutanoic acid
(d) 3 – Bromo – 2. 3 – dimethyl propanoic acid.
Answer:
(c) 3 – Bromo – 2 – methylbutanoic acid

Question 13.
The structure of isobutyl group in an organic compound is ………

Answer:

Question 14.
The number of stereoisomers of 1, 2-dihydroxycyclopentane is ……..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 15.
Which of the following is optically active?
(a) 3 – Chioropentane
(b) 2 – Chioropropane
(c) Meso – tat-tat-ic acid
(d) Glucose
Answer:
(d) Glucose

Question 16.
The isomer of ethanol is ……….
(a) acetaldehyde
(b) dimethylether
(c) acetone
(d) methyl carbinol
Answer:
(b) dimethylether

Question 17.
How many cyclic and acyclic isomers are possible for the molecular formula C₃H₆O?
(a) 4
(b) 5
(c) 9
(d) 10
Answer:
(c) 9

Question 18.
Which one of the following shows functional isomerism?
(a) ethylene
(b) Propane
(c) ethanol
(d) CH₂Cl₂
Answer:
(c) ethanol

Question 19.
are ………
(a) resonating structure
(b) taulomers
(c) optical isomers
(d) conformers
Answer:
(b) tautomers

Question 20.
Nitrogen detection in an organic compound is earned out by Lassaigne’s test. The blue colour formed is due to the formation of ………….
(a) Fe₃[Fe(CN)₆]₂
(b) Fe₄[Fe(CN)₆]₃
(c) Fe₄[Fe(CN)₆]₂
(d) Fe₃[Fe(CN)₆]₃
Answer:
(b) Fe₄[Fe(CN)₆]₃

Question 21.
Lassaigne’s test for the detection of nitrogen fails in ………..
(a) H₂N -CO – NH. NH₂. HCl
(b) NH₂ – NH₂. HCl
(c) C₆H₅ – NH – NH₂. HCl
(d) C₆H₅CONH₂
Answer:
(c) C₆H₅ – NH – NH₂. HCl

Question 22.
Connect pair of compounds which give blue colouration/precipitate and white precipitate respectively, when their Lassaigne’s test is separately done.
(a) NH₂NH₂HCl and CICH₂ – CHO
(b) NH₂CS NH₂ and CH₃ – CH2Cl
(c) NH₂CH₂ COOH and NH₂CONH₂
(d) C₆H₅NH₂ and ClCH₂ – CHO
Answer:
(d) C₆H₅NH₂ and ClCH₂ – CHO

Question 23.
Sodium nitropruside reacts with suiphide ion to give a purple colour due to the formation of ………..
(a) [Fe(CN)₅N0]^3-
(b) [Fe(NO)₅CN]⁺
(c) [Fe(CN)₅NOS]^4-
(d) [Fe(CN)₅NOS]^3-
Answer:
(c) [Fe(CN)₅NOS]^4-

Question 24.
An organic compound weighing 0.15 g gave on carius estimation, 0.12 g of silver bromide. The percentage of bromine in the compound will be close to ……….
(a) 46%
(b) 34%
(c) 3.4%
(d) 4.6%
Answer:
(b) 34%

Question 25.
A sample of 0.5g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50 mL of 0.5M H₂SO₄. The remaining acid after neutralisation by ammonia consumed 80 mL of 0.5 M NaOH, The percentage of nitrogen in the organic compound is ……….
(a) 14%
(b) 28%
(c) 42%
(d) 56%
Answer:
(b)28%

Question 26.
In an organm compound, phosphorus is estimated as ……….
(a) Mg₂P₂O₇
(b) Mg₃(PO₄)₂
(c) H₃PO₄
(d) P₂O₅
Answer:
(a) Mg₂P₂O₇

Question 27.
Ortho and para-nitro phenol can be separated by ………….
(a) azeotropic distillation
(b) destructive distillation
(c) steam distillation
(d) cannot be separated
Answer:
(c) steam distillation

Question 28.
The purity of an organic compound is determined by …………
(a) Chromatography
(b) Crystallisation
(c) melting or boiling point
(d) both (a) and (c)
Answer:
(d) both (a) and (c)

Question 29.
A liquid which decomposes at its boiling point can be purified by …………
(a) distillation at atmospheric pressure
(b) distillation under reduced pressure
(c) fractional distillation
(d) steam distillation
Answer:
(b) distillation under reduced pressure

Question 30.
Assertion: is 3-carbethoxy -2- butenoicacid.
Reason: The principal functional group gets lowest number followed by double bond (or) triple bond.
(a) both the assertion and reason are true and the reason is the correct explanation of assertion.
(b) both assertion and reason are true and the reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d both the assertion and reason are false.
Answer:
(a) both the assertion and reason are true and the reason is the correct explanation of assertion.

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry Short Answer Questions

Question 31.
Give the general characteristics of organic compounds.
Answer:

• All organic compounds are covalent compounds of carbon and are insoluble in water and soluble in organic solvents.
• They are inflammable (except CCl₄).
• They possess low boiling and melting points due to their covalent nature.
• They are characterised by functional groups.
• They exhibit isomerism.

Question 32.
Describe the classification of organic compounds based on their structure.
classification of organic compounds based on the structure
Answer:

Question 33.
Write a note on homologous series.
Answer:

• A series of organic compounds each containing a characteristic functional group and the successive members differ from each other in molecular formula by a CH₂ group is called homologous series.
• e.g., Alkanes, Methane (CH₄) ethane (C₂H₆), Propane (C₃H₈) etc.
• Compounds of the homologous series are represented by a general formula. e.g., Alkanes: C₂H_2nAlkene: C_nH_2n
• They can be prepared by general methods.
• They show regular gradation in physical properties but have almost similar chemical properties.

Question 34.
What is meant by a functional group? Identify the functional group in the following compounds.
(a) acetaldehyde
(b) oxalic acid
(c) dimethyl ether
(d) methylamine
Answer:
1. A functional group ¡s an atom or a specific combination of bonded atoms that react in a characteristic way, irrespective of the organic molecule in which it is present.

Question 35.
Give the general formula for the following classes of organic compounds
(a) Aliphatic monohydric alcohol
(b) Aliphatic ketones
(c) Aliphatic amines.
Answer:
(a) Aliphatic monohydric alcohol – C_nH_2n+1 + OH
(b) Aliphatic ketones – C_nH_2nO
(c) Aliphatic amines – C₂H_2n+1NH₂

Question 36.
Write the molecular formula of the first six members of homologous series of nitro – alkanes.
Nitroalkanes:
Answer:

• CH₂NO₂ Nitromethane
• CH₂-CH₂NO₂ Nitroethane
• CH₃-CH₂-CH₂NO₂ 1- nitropropane
• CH₃-CH₂-CH₂-CH₂-NO₂ 1- nitrobutane
• CH₃-CH₂-CH₂-CH₂-CH₂-NO₂ 1 – nitropentane
• CH₂-CH₂-CH₂-CH₂-CH₂-CH₂-CH₂-NO₂ 1- nitrohexane

Question 37.
Write the molecular and possible structural formula of the first four members of homologous series of carhoxylic acids.
Answer:

Question 38.
Give the IUPAC names of the following compounds.
Answer:



(i) 2,3,5-tnmethylhexane
(ii) 2-bromo-3-methylbutane
(iii) methoxymethane
(iv) 2-hydroxybutanal
(v) buta-1,3-diene
(vi) 4-chioropent-2-yne
(vii) 1 -bromobut-2-ene
(viii) 5-oxohexanoic acid
(ix) 3-ethyl-4-ethenylheptane
(x) 2, 4, 4-trimethylpent-2-ene
(xi) 2- methyl-I -phenyipropan- I -amine
(xii) 2,2- dimethyl-4-oxopentanenitrile
(xiii) 2-ethoxypropane
(xiv) I -fluoro-4-methyl-2-nitrobenzene
(xv) 3-bromo-2-methylpentanal

Question 39.
Give the structure for the following compound.
Answer:
(i) 3 – ethyl – 2 methyl -1 – pentene
(ii) 1, 3, 5 – Tnmethyl cyclohex – 1 – ene
(iii) tertiary butyl iodide
(iv) 3 – Chlorobutanal
(y) 3 – Chlorobutanol .
(vi) 2 – Chloro – 2 – methyl propane
(vii) 2, 2-dimethyl- 1 – chioropropane
(viii) 3 – methylbut – 1 – ene
(ix) Butan – 2, 2 – diol
(x) Octane – 1 ,3 – diene
(xi) 1 ,5 – Dimethylcyclohexane
(xii) 2 – Chlorobut – 3 – ene
(xiii) 2 – methylbutan – 3 – ol
(xiv) acetaldehyde

Question 40.
Describe the reactions involved in the detection of nitrogen in an organic compound by Lassaigne method.
Answer:
Detection of Nitrogen:
The following reactions are involved in the detection of nitrogen with formation of prussian blue precipitate conforming the presence of nitrogen in an organic compound.

Question 41.
Give the principle involved in the estimation of halogen in an organic compound by Carius method.
Estimation of halogens:
Answer:
carius method:
1. A known mass of the organic compound is heated with fuming HNO₃ and AgNO₃.
2. C, H and S gets oxidised to CO₂, H₂O and SO₂ and halogen combines with AgNO₃ to form a precipitate of silver halide

3. The precipitate AgX is filtered, washed, dried and weighed.
4. From the mass of AgX and the mass of organic compound taken, the percentage of halogens are calculated.

Question 42.
Give a brief description of the principles of:
1. Fractional distillation
2. Column Chromatography
Answer:
1. Fractional distillation:
This method is used to purify and separate liquids present in the mixture having their boiling point close to each other. The process of separation of the components in liquid mixture at their respective boiling points in the form of vapours and the subsequent condensation of those vapours is called fractional distillation. This method is applied in distillation of petroleum. coal tar and crude oil.

2. Column chromatography:
(a) The principle behind chromatography is selective distribution of mixture of organic substances between two phases a stationary phase and a moving phase. The stationary phase can be a solid or liquid, while the moving phase is a liquid or a gas. when the stationary phase is a solid, the moving phase is a liquid or gas.

(b) If the stationary phase is solid, the basis is adsorption, and when it is a liquid, the basis is partition.

(c) Chromatography is defined as a technique for the separation of a mixture brought about by differential movement of the individual component through porous medium under the influence of moving solvent.

(d) In column chromatography, the above principle is carried out in a long glass column.

Question 43.
Explain paper chromatography.
Answer:
Paper chromatography:
1. It is an example of partition chromatography. A strip of paper acts as an adsorbent. This method involves continues differential partioning of components of a mixture between stationary and mobile phase. In paper chromatography, a special quality paper known as chromatographic paper is used. This paper act as a stationary phase.

2. A strip of chromatographic paper spotted at the base with the solution of the mixture is suspended in a suitable solvent which acts as the mobile phase. The solvent rises up and flows over the spot. The paper selectivity retains different components according to their different partition in the two phases where a chromatograrn is developed.

3. The spots of the separated coloured components are visible at different heights from the position of initial spots on the chromatogram. The spots of the separated colourless compounds may be observed either under ultraviolet light or by the use of an appropriate spray reagent.

Question 44.
Explain various types of constitutional isomerism (structural isomerism) in organic compounds.
Answer:
Constitutional isomers:
These isomers having the same molecular formula but differ in their bonding sequence. It is classified into 6 types:
1. Chain (or) nuclear (or) skeletal isomerism:
The phenomenon in which the isomers have similar molecular formula but differ in the nature of carbon skeleton (i.e., straight (or)
branched)
e.g., C₅H₁₂:

2. Position isomerism:
If different compounds belonging to same homologous series with the same molecular formula and carbon skeleton but differ in the position of substituent or functional group or an unsaturated linkage are said to exhibit position isomerism.
e.g., C₅H₁₀:

3. Functional isomerism:
Different compounds having same molecular formula but different functional groups are said to exhibit functional isomerism.
e.g., C₃H₆O:
(i) CH₃-CH₂-CH₂-CH=CH₂ propanal (Aldehyde group)

4. Metamerism:
This isomerism anses due to the unequal distribution of carbon atoms on either side of the functional group or different alkyl groups attached to either side of the same functional group and having same molecular formula.
e.g., C₄H₁₀O:

5. Tautomerism:
It is an isomerism in which a single compound exists in two readily inter convertible structures that differ markedly in the relative position of atleast one atomic nucleus generally hydrogen.
e.g., C₂H₄O:

6. Ring chain isomerism:
It is an isomerism in which compounds having same molecular formula but differ in terms of bonding of carbon atom to form open chain and cyclic structures.
e.g., C₂H₆:

Question 45.
Describe optical isomerism with suitable example.
Answer:
Optical isomerism:
1. Compounds having same physical and chemical property but differ only in the rotation of plane of polarised light are known as optical isomers and the phenomenon is known as optical isomerism.

2. Glucose have the ability to rotate the plane of plane polarised light and it is said to be an optically active compound and this property of any compound is called optical activity.

3. The optical isomer which rotates the plane of plane polarised light to the right or in clockwise direction is said to be dextrorotatory and is denoted by the sign (+).

4. The optical isomer which rotates the plane of plane polarised light to the left or in anti- clockwise direction is said to be laveo rotatory and is denoted by the sign (-).

5. Dextrorotatory compounds are represented as ‘d’ (or) by (+) sign and leave rotatory compounds are represented as l (or) by (-) sign.

6. The optical isomers which are non-superimposible mirror images of each other are called enantiomers.

Question 46.
Briefly explain geometrical isomerism in alkenes by considering 2- butene as an example.
Answer:
2-butene: Geometrical isomerism : CH₃  -CH = CH – CH₃

1. Geometrical isomers are the stereoisomers which have different arrangement of groups or atoms around a rigid framework of double bonds. This type of isomerism occurs due to restricted rotation of double bonds or about single bonds in cyclic compounds.

2. In 2-butene, the carbon-carbon double bond is sp2 hybridised. The carbon-carbon double bond consists of a a bond and a it bond. The presence of it bond lock the molecule in one position. Hence, rotation around C = C bond is not possible.

4. These two compounds are termed as geometrical isomers and are termed as cis and trans form.

5. The cis-isomer is the one in which two similar groups arc on the same side otthe double bond. The trans-isomer is that in which two similar groups are on the opposite side of the double bond. Hence, this type of isomerism is called cis-Irans isomerism.

Question 47.
0.30 g of a substance gives 0.88 g of carbon dioxide and 0.54 g of water calculate the percentage of carbon and hydrogen in it.
Answer:
Weight of organic compound = 0.30 g
Weight of carbon-dioxide = 0.88 g
Weight of water = 0.54 g
Percentage of hydrogen:
18 g of water contains 2 g of hydrogen
0.54 g of water contain = \(\frac {2}{18}\) × 0.54 g of hydrogen
% of hydrogen = \(\frac {2}{18}\) × \(\frac {0.54}{0.30}\) × 100 = \(\frac {2}{18}\) × \(\frac {54}{0.3}\)
% of H = 0.111 × 180 = 19.888 ≈ 20%

Percentage of carbon:
44 g of CO₂ contains 12 g of carbon
0.88 g of CO₂ contains = \(\frac {12}{44}\) × 0.88 g of carbon
% of carbon = \(\frac {12}{44}\) × \(\frac {0.88}{0.30}\) × 100 = \(\frac {12}{44}\) × \(\frac {88}{0.3}\) = \(\frac {24}{0.3}\)
% of carbon = 80 % .

Question 48.
The ammonia evolved form 0.20 g of an organic compound by kjeldahl method neutralised 15m1 of N/20 Sulphuric acid solution. Calculate the percentage of Nitrogen.
Answer:
Weight of organic compound = 0.20 g
Volume of sulphuric acid taken = 15 ml
Strength of sulphuric acid taken = \(\frac {N}{20}\) = 0.05 N
Percentage of nitrogen = \(\frac{14 \times \mathrm{NV}}{1000 \times 10} \times 100\)
= \(\frac{14 \times 0.05 \times 15}{1000 \times 0.20} \times 100\)
= \(\frac {1050}{200}\) = 5.25
% of nitrogen = 5.25%

Question 49.
0.32 g of an organic compound. after heating with fuming nitric acid and barium nitrate crystals is a scaled tube gave 0.466 g of barium sulphate. Determine the percentage of sulphur in the compound.
Answer:
Weight of organic compound = 0.32 g
Weight of BaSO₄ formed = 0.466 g
233 g of BaSO₄ contains = 32 g of sulphate
0.466 g of l3aSO₄ contain = \(\frac {32}{233}\) x \(\frac {0.466}{2.32}\) x 100
= \(\frac {32}{233}\) x \(\frac {46.6}{0.32}\) = 19.999 g of sulphur
% of sulphur = 20 %

Question 50.
024 g of an organic compound gave 0.287 g of silver chloride in the carius method. Calculate the percentage of chlorine in the compound.
Answer:
Weight of organic compound = 0.24 g
Weight of silver chloride = 0.287 g
143.5 g of AgCl contains = 35.5 g of Cl
0.287 g of AgCl contains = \(\frac {35.5}{143.5}\) x 0.287 g of Cl
% of chlorine = \(\frac {35.5}{143.5}\) x \(\frac {0.287}{0.24}\) x 100 = 29.58 %

Question 51.
In the estimation of nitrogen present in an organic compound by Dumas method 0.35 g yielded 20.7 mL of nitrogen at 15°C and 760 mm Hg pressure. Calculate the percentage of nitrogen ¡n the compound.
Answer:
Weight of organic compound = 0.35 g
Volume of moist nitrogen (V₁) = 20.7 ml = 20.7 x 10^-3 L
Temperature = T₁ = 15°C = 273 + 15°C = 288K
Pressure of moist nitrogen P₁ = 760 mmHg

V₀ = 19.62 x 10^-3L
Percentage of nitrogen = \(\frac {28}{22.4}\) x \(\frac{\mathrm{V}_{0}}{\mathrm{W}}\) x 100
= \(\frac {28}{22.4}\) x \(\frac{19.62 \times 10^{-3}}{0.35}\) x 100
= \(\frac {28}{22.4}\) x = \(\frac {19.62}{0.35}\) x 10^-1
= 56.05 x 10^-3 x 100 = 7.007%
Percentage of nitrogen = 7.007%

In Text Questions – Evaluate Yourself

Question 1.
Give two examples for each of the following type of organic compounds.
1. non-benzonoid aromatic
2. aromatic heterocyclic
3.  alicycic
4. aliphatic open chain.
Answer:
1. Non benzenoid aromatic compounds

2. Aromatic heterocyclic compounds

3. Alicyclic compounds

4. Aliphatic open chain compounds

• CH₃-CH₂-CH₂-CH₂-CH₃ n-pentane
• CH₃-CH₂-CH₂OH 1-propanol

Question 2.
Write structural formula for the following compounds
1. Cyclohexa-1, 4-diene
2. Ethynykyclohexane

Question 3.
Write structural formula for the following compounds
1. m – dinitrobenzene
2. p-dichlorobenzene
3. 1, 3, S- Trimethytbeuzene
Answer:

Question 4.
Write all the possible isomers of molecular formula C₄H₁₀O and identify the isomerisms found in them.
Answer:
C₄H₁₀O isomers:

Question 5.
0.2346 g of an organic compound containing C, H & O, o comhution giweb 0.2754 g of H₂O and 0.4488 g CO₂. Calculate the % composition of C, H & O in the organic compound.
Answer:
Weight of organic substance (w) = 0.2346 g
Weight of water (x) = 0.2754 g
Weight of CO₂ (y) = 0.4488 g
Percentage of carbon = \(\frac {12}{44}\)  x  \(\frac {y}{w}\) x 100
= \(\frac {12}{44}\) – \(\frac {0.4488}{0.2346}\) x 100 = 52.17%
Percentage of hydrogen = \(\frac {2}{18}\) x \(\frac {y}{w}\) x 100
= \(\frac {2}{18}\) x \(\frac {0.2754}{0.2346}\) x 100 = 13.04%
Percentage of oxygen = [100- (52.17 +13.04)] = 100 – 65.21 = 34.79%

Question 6.
0.16 g of an organic compound was heated in a carlus tube and H₂SO₄ acid formed was precipitated with BaCl₄. The mass of BaSO₄ was 0.35 g. Find the percentage of sulphur.
Answer:
Weight of organic substance (w) = 0.16 g
Weight of Barium sulphate (x) = 0.35 g
Percentage of Sulphur = \(\frac {32}{233}\) x \(\frac {x}{w}\) x 100
= \(\frac {32}{233}\) x \(\frac {0.35}{0.16}\) x 100 = 30.04%

Question 7.
0.185 g of an organic compound when treated with Conc. HNO₃ and silver nitrate gave 0.320 g of silver bromide. Calculate the % of bromine in the compound.
Answer:
Weight of organic substance (w) 0.185 g ;
Weight of silver bromide (x) = 0.320 g
Percentage of bromine = \(\frac {80}{188}\) x \(\frac {x}{w}\) x 100 = \(\frac {80}{188}\) x \(\frac {0.32}{0.185}\) x 100 = 73.6%

Question 8.
0.40 g of an iodo-substituted organic compound gave 0.235 g of Agi by carius method. Calculate the percentage of iodine in the compound. (Ag = 108, I = 127).
Answer:
Weight of organic substance (w) = 0.40 g
Weight of silver iodide (x) = 0.235 g
127 x 127 0.235
Percentage of iodine = \(\frac {127}{235}\) x \(\frac {x}{w}\) x 100
= \(\frac {x}{w}\) x \(\frac {0.235}{0.40}\) x 100 = 31.75%

Question 9.
0.33 g of an organic compound containing phosphorous gave 0.397 g of Mg₂P₂O₇ by the analysis. Calculate the percentage of P in the compound.
Answer:
Weight of organic compound = 0.33g ;
Weight of Mg,P,07 = 0.397g
222 g of Mg₂P₂O₇ contains 62 g of phosphorous.
0.397 g of Mg₂P₂O₇ will contain \(\frac {62}{222}\) x 0.397 g of P.
0.33 g of organic compound contains \(\frac {62}{222}\) x 0.397 g of P
100 g of organic compound will contain \(\frac {62}{222}\) x \(\frac {0.397}{0.33}\) x 100
= \(\frac {2,461.4}{73.26}\) = 33.59 %
Percentage of phosphorous = 33.59 %

Question 10.
0.3 g of an organic compound on Kjeldahl’s analysis gave enough ammonia to just neutralise 30 mL of 0.1N H₂SO₄. Calculate the percentage of nitrogen in the compound.
Answer:
Weight of organic compound (w) = 0.3 g
Strength of sulphuric acid used (N) = 0.1 N
Volume of sulphuric acid used (V) = 30 mL
30 ml of 0.1 N sulphuric acid 30 ml of 0.1 N ammonia
Percentage of nitrogen = \(\left(\frac{14 \times \mathrm{NV}}{1000 \times w}\right)\) x 100
= \(\left(\frac{14 \times 0.1 \times 30}{1000 \times 0.3}\right)\) x 100 = 14%

Example Problems

Question 1.
Classify the following compounds based on the structure
1. CH≡C-CH₂-C≡CH
2. CH₃-CH₂-CH₂-CH₂-CH₃


Answer:

1. Unsaturated open chain compound
2. Saturated open chain compound
3. Aromatic benzenoid compound
4. Alicyclic compound

Question 2.
0.26 g of an organic compound gave 0.039 g of water and 0.245 g of carbon dioxide on combustion. Calculate the percentage of C & H.
Answer:
Weight of organic compound = 0.26 g
Weight of water = 0.039 g
Weight of CO₂ = 0.245 g
Percentage of hydrogen:
18 g of water contains 2 g of hydrogen
0.039 g of water contains = \(\frac {2}{18}\) x \(\frac {0.039}{0.26}\) of H
% of hydrogen = \(\frac {0.039}{0.26}\) x \(\frac {2}{18}\) x 100 = 1.66%
Percentage of carbon:
44 g of CO₂ contains 12 g of C
0.245 g of CO₂ contains = \(\frac {12}{44}\) x \(\frac {0.245}{0.26}\) g of C
% of Carbon = \(\frac {12}{44}\) x \(\frac {0.245}{0.26}\) x 100 = 25.69%

Question 3.
In an estimation of sulphur by Carius method, 0.2 175 g of the substance gave 0.5825 g of BaSO₄, calculate the percentage composition of S ¡n the compound.
Answer:
Weight of organic compound = 0.2 175 g
Weight of BaSO₄ = 0.5825 g
233 g of BaSO₄ contains = 32 g of S
0.5825 g of BaSO₄ contains = \(\frac {32}{233}\) x \(\frac {0.5825}{0.2175}\) g of S
Percentage of S = \(\frac {32}{233}\) x \(\frac {0.5825}{0.2175}\) x 100 = 36.78%

Question 4.
0.284 g of an organic substance gave 0.287 g AgCl in Carius method for the estimation of halogen. Find the percentage of Cl in the compound.
Answer:
Weight of the organic substance = 0.284 g
Weight of AgCl = 0.287 g
143.5 g of AgCl contains 35.5 g of chlorine
0.287 g of AgCl Contains = \(\frac {35.5}{143.5}\) x \(\frac {0.287}{0.284}\)
% of chlorine = \(\frac {35.5}{143.5}\) x \(\frac {0.287}{0.284}\) x 100 = 24.98%

Question 5.
0.24 g of organic compound containing phosphorous gave 0.66 g of Mg₂P₂O₇ by the usual analysis. Calculate the percentage of phosphorous ¡n the compound
Answer:
Weight of an organic compound = 0.24 g
Weight of Mg₂P₂O₇ = 0.66 g
222 g of Mg₂P₂O₇ contains = 62 g of P
o. 66 g contains = \(\frac {62}{222}\) x 0.66 g of P
Percentage of P = \(\frac {62}{222}\) x \(\frac {0.66}{0.24}\) x 100 = 76.80%

Question 6.
0.1688 g when analysed by the Dumas method yield 31.7 mL of moist nitrogen measured at 14°C and 758mm mercury pressure. Determine the % of N in the substance (Aqueous tension at 14°C =12 mm of Hg).
Answer:
Weight of Organic compound = 0.168 g
Volume of moist nitrogen (V₁) = 31.7 ml. = 31.7 x 10^-3 L
Temperature (T₁) = 14°C = 14 + 273 = 287 K
Pressure of Moist nitrogen (P) = 758 mm Hg
Aqueous tension at 14°C = 12 mm of Hg
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{0} V_{0}}{T_{0}}\)
V₀ = \(\frac{746 \times 31.7 \times 10^{-3}}{287} \times \frac{273}{760}\)
V₀ = 29.58 x 10^-3 L
Percentage of nitrogen:

= 21.90 %

Question 7.
0.6 g of an organic compound was Kjeldhalised and NH₃ evolved was absorbed into 50 mL of semi-normal solution of H₂SO₄. The residual acid solution ws diluted with distilled water and the volume made up to 150 mL. 20 mL of this diluted solution required 35 mL of \(\frac {N}{2}\) NaOH solution for complete neutralisation. Calculate the % of N in the compound.
Answer:
Weight of Organic compound = 0.6 g
Volume of sulphuric acid taken = 50 mL
Strength of sulphuric acid taken = 0.5 N
20 ml of diluted solution of unreacted sulphuric acid was neutralised by 35 mL of 0.05 N Sodium hydroxide
Strength of the diluted sulphuric acid = \(\frac {35 x 0.05}{20}\) = 0.0875 N
Volume of the sulphuric acid remaining after reaction with ammonia = V₁ mL
Strength of H₂SO₄ = 0.5 N
Volume of the diluted H₂SO₄ = 150 mL
Strength of the diluted sulphuric acid = 0.0875 N
V₁ = \(\frac {150 x 0.087}{0.5}\) = 26.25 mL
Volume of H₂SO₂consumed by ammonia = 50 – 26.25 = 23.75 mL
23.75 mL of 0.5 N H₂SO₄ = 23.75 mL of 0.5N NH₃
The amount of Nitrogen present in the 0.6 g of organic compound
= \(\frac{14 \mathrm{g}}{1000 \mathrm{mL} \times 1 \mathrm{N}}\) x 23.75 x 0.5 N = 0.166 g
Percentage of Nitrogen \(\frac {0.166}{0.6}\) x 100 = 27.66%

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry Additional Questions Solved

I. Choose the correct answer.
Question 1.
Statement 1. The tendency of an atom to form a chain of bonds with the atoms of the same element is called catenation.
Statement 2. The high strength of C-C bond is responsible for its catenation property.
(a) Statement 1 & 2 are correct and statement 2 is the correct explanation of statement 1.
(b) Statement 1 & 2 arc correct but statement 215 not the correct explanation of statement 1.
(c) Statement 1 is correct but statement 2 is wrong.
(d) Statement 1 is wrong but statement 2 is correct.
Answer:
(a) Statement 1 & 2 are correct and statement 2 is the correct explanation of statement 1.

Question 2.
Which of the following is not an organic compound?
(a) DNA
(b) Lipid
(c) Glycogen
(d) Bronze
Answer:
(d) Bronze
Solution:
It is an alloy and a mixture of metals and all other are organic compounds.

Question 3.
Which of the following is an example of an organic reaction?
(a) Rusting of iron
(b) Combustion of magnesium
(c) Biochemical reactions
(d) All the above
Answer:
(c) Biochemical reactions

Question 4.
Which of the following is an example of heterocylic aromatic compound?
(a) THF
(b) Cyclopropane
(c) Pyridine
(d) Azulene
Answer:
(c) Pyridine

Question 5.
Which of the following is an example of non-benzenoid aromatic compound?
(a) Tolucnc
(b) Phenol
(c) Benzyl alcohol
(cl) azulene
Answer:
(d) azulene

Question 6.
Which of the following pair is an example of aromatic compounds?
(a) Benzene, Toluene
(b) Cyclopropane, Cyclobuane
(c) Pyridine, Pyrrole
(d) Propane, Butane
Answer:
(a) Benzene, Toluene

Question 7.
Which of the following is an example of carbocyclic alicyclic compound?
(a) Pyrrole
(b) Thiophene
(c) Cyclopropane
(d) Phenol
Answer:
(c) Cyclopropane

Question 8.
Which one of the following is the functional group of ketone?
(a)-CHO

(c) -O-
(d)-OH
Answer:

Question 9.
Which one of the following indicates isothiocyanate functional group?
(a) -NC
(b) -NCS
(c) -SCN
(d) -NCO
Answer:
(b) -NCS

Question 10.
Which of the following represent thiol?
(a) -SH
(b) -OH
(c) -SR
(d) -SCN
Answer:
(a) -SH

Question 11.
Which structure ¡s named as 3-chlorocyclobut-1-ene?

Answer:

Question 12.
Which one of the following is called 2-cyclobutyipropanal?

Answer:

Question 13.
Which one of the following is called cyclopentylbenzene’?

Answer:

Question 14.
Which one of the following is commonly called mesitylene?

Answer:

Question 15.
Which one of the following is called benzylchloride?
(a) C₆H₅CH₂Cl
(b) C₆H₅CHCl₂
(c) C₆H₅CCl₃
(d) C₆H₅Cl
Answer:
(a) C₆H₅CH₂Cl

Question 16.
Which of the following pair are called functional isomers?
(a) methyl propyl ether and diethyl ether
(b) 2-pentanone & 3-pentanone
(c) propanoic acid and methyl acetate
(d) I -butanol and 2-butanol
Answer:
(c) propanoic acid and methyl acetate

Question 17.
Which of the following does not show optical isomerism’?
(a) Glucose
(b) Tartane acid
(c) Lactic acid
(d) Methane
Answer:
(d) Methane

Question 18.
Which metal is used to prepare Lassaigne’s extract?
(a) Copper
(b) Sodium
(c) Aluminium
(d) Iron
Answer:
(b) Sodium

Question 19.
Which colour is formed in the Lassaigne’s test for nitrogen?
(a) Purple
(b) Black
(c) Prussian blue
(d) Violet
Answer:
(c) Prussian blue

Question 20.
Which one of the following is called feme ferrocyanide?
(a) Na₄[Fe(CN)₆]
(b) Na₄[Fe(CN)₆]₃
(c) Fe₄[Fe(CN)₆]
(d) Fe₄[Fe(CN)₆]₃
Answer:
(d) Fe₄[Fe(CN)₆]₃

Question 21.
What is the colour formed in Lassaigne’s test of an organic compound containing N and S?
(a) Prussian blue colour
(b) Blood red colour
(c) Purple colour
(d) Canary yellow colour
Answer:
(b) Blood red colour

Question 22.
Which one of the following is the formula of sodium nitroprusside?
(a) Na₄[Fe(CN)₅N0₅]
(b) Na₄[Fe(CN)₅SON]
(c) Na₄[Fe(CN)₆]
(d) Fe₄[Fe(CN)₆]₃
Answer;
(a) Na₄[Fe(CN)₅N0₅]

Question 23.
Identify the colour formed when Lassigne’s extract of sulphur containing organic compound is mixed with sodium nitroprusside solution?
(a) Prussian blue
(b) Black
(c) Green
(d) Purple
Answer:
(d) Purple

Question 24.
Which one of the following solutions are added to Lassaigne’s extract to identify halogens?
(a) Acetic acid + Lead acetate
(b) dil HNO₃ + AgNO₃
(c) Fe(OH)₂ + FeCl₃
(d) Na₂CO₃ + KNO₃
Answer:
(b) dil HNO₃ + AgNO₃

Question 25.
Which one of the following is not identified by Lassaigne’s test?
(a) nitrogen
(b) sulphur
(c) halogens
(d) phosphorous
Answer:
(d) phosphorous

Question 26.
Which one of the following test is used to detect phosphorous in an organic compound?
(a) Silver nitrate test
(b) Copper oxide test
(c) Ammonium molybdate test
(d) Lassaigne’s test
Answer:
(c) Ammonium molybdate test

Question 27.
Identify the colour formed in the test kr phosphorous using ammonium molybciate.
(a) Crimson red colour
(b) Deep violet colour
(c) Prussian blue colour
(d) Canary yellow colour
Answer:
(d) Canary yellow colour

Question 28.
Which of the following will absorb CO₂?
(a) Conc. H₂SO₄
(b) KOH
(c) HCl
(d) Copper
Answer:
(b) KOH

Question 29.
Which of the following is used as moisture absorbent?
(a) Potash
(b) Soda
(c) Conc. H₂SO₄
(d) Na₃CO₄
Answer:
(c) Conc. H₂SO₄

Question 30.
Which method is used to estimate sulphur?
(a) Lassaigne’s test
(b) Oxide test
(c) Carius method
(d) Kjedahl’s method
Answer:
(c) Cari us method

Question 31.
Which method is used to estimate nitrogen?
(a) Dumas method and Kjeldahl’s method
(b) Carius method & Oxide method
(c) Lassaignes test & Copper oxide test
(d) Ammonium molybdate test & Silver nitrate test
Answer:
(a) Dumas method and Kjeldahl’s method

Question 32.
Which of the following is not purified by sublimation method?
(a) Camphor
(b) Benzoic acid
(c) Naphthalene
(d) Nitrobenzene
Answer:
(d) nitrobenzene

Question 33.
Which of the following is used to decolourise the organic compounds?
(a) Chlorine
(b) Bleaching powder
(c) Animal charcoal
(d) Iodine
Answer:
(c) Animal charcoal

Question 34.
Which method is used to extract essential oils from plants and flowers?
(a) Crystallization
(b) Sublimation
(c) Steam distillation
(d) Differential extraction
Answer:
(c) Steam distillation

Question 35.
Which of the following is used as adsorbent?
(a) silica gel and alumina
(b) glass wool and cotton
(c) glass plate and paper
(d) glucose and fructose
Answer:
(a) silica gel and alumina

Question 36.
Which of the following compounds gives prussian blue colour in Lassaigne’s test?
(a) CH₄ and CH₃OH
(b) CH₃NH₂ and CH₃NO₂
(c) CH₃Cl and CHCl₃
(d) CH₃CHO and CH₃COCH₃
Answer:
(b) CH₃NH₂ and CH₃NO₂

Question 37.
Which of the following compounds gives curdy white precipitate in Lassaigne’s test?
(a) CH₃Br
(b) C₂H₅I
(c) CH₃Cl
(d) C₆H₅NO₂
Answer:
(c) CH₃Cl

Question 38.
Which one of the following is not used as air adsorbent in chromatography?
(a) Alumina
(b) Silica gel
(c) Magnesia
(d) Sucrose
Answer:
(d) Sucrose

Question 39.
The IUPAC name of ……..
(a) 2-methyl butanal
(b) butan-2-aldehyde
(c) 2-ethyipropanal
(d) 3-methyl isobutraldehyde
Answer:
(c) 2-ethyipropanal

Question 40.
Which of the following compounds will exhibit cis-trans isomerism?
Answer:
(a) 2-Buiene
(b) 2-Butyne
(c) 1-Butene
(d) 2-Butanol
Answer:
(a) 2-Butene

Question 41.
Which of the following sodium fusion extract of organic compound gives brilliant violet colour with sodium nitroprusside solution?
(a) Urea
(b) Thiourea
(c) Benzoic acid
(d) Aniline
Answer:
(b) Thiourea

Question 42.
Which of the following reagent is used to distinguish between halogens (Cl, Br, I) in an organic compound?
(a) NaOH
(b) FeCl₃
(c) H₂SO₄
(d) NH₄OH
Answer:
(d) NH₄OH

Question 43.
In which of the following, functional group isomerism is not possible?
(a) Alcohols
(b) Aldehydes
(c) Alkyl halides
(d) Cyanides
Answer:
(c) Alkyl halides

Question 44.
Which one of the following is used as a column in the separation of pigments of chlorophyll by chromatography technique?
(a) Petroleum ether
(b) CaCO₃
(c) Activated charcoal
(d) Ethanoic acid
Answer:
(b) CaCO₃

Question 45.
Which one of the following compound does not give Prussian blue colour in Lassaigne’s test?
(a) C₆H₅NH₂

(c) C₆H₅CONH₂
(d) C₆H₅COCl
Answer:
(d) C₆H₅COCl

Question 46.
Which one of the following shows geometrical isomerism?
(a) n-Butane
(b) 1-butene
(c) 2-butene
(d) butyne
Answer:
(c) 2-butene

Question 47.
Which one of the following shows functional group isomerism?
(a) Ethene
(b) Acetone
(c) Ethane
(a) Propane
Answer:
(b) Acetone

Question 48.
Which of the following pair gives curdy white precipitate and yellow precipitate respectivety in their Lassaignes test?
(a) C₂H₅I and C₂H₅Br
(b) C₂H₅NO₂ and C₂H₅NH₂
(c) C₆H₅Cl and CH₃
(d) CH₄ and CH₃OH
Answer:
(c) C₆H₅Cl and CH₃

II. Match the following.

Question 1.

Answer:

Question 2.

Answer:

Question 3.

Answer:

Question 4.

Answer:

Question 5.

Answer:

Question 6.

Answer:

Question 7.

Answer:

Question 8.

Answer:

III. Fill in the blanks.

Question 1.
The property catenation is possible in ……..
Answer:
Carbon.

Question 2.
Acetic acid was synthesised by ………
Answer:
Kolbe

Question 3.
Methane was synthesised in laboratory by ………
Answer:
Berthiot
CH₃

Question 4.
is an example of ………
Answer:
aromatic benzenoid compound

Question 5.
2-butene is an example of compound.
Answer:
unsaturated open chain

Question 6.
The IUPAC name of is …….
Answer:
2, 2, 5-trimethyl heptane

Question 7.
The IUPAC name of is ……..
Answer:
3, 4-diethyl, 4-methyiheptane

Question 8.
The name of is ……….
Answer:
1-cyclobutyl-2-cyclopropylethene

Question 9.
The name of
Answer:
2-(cyclobut-2-en-1-yl)-propanal

Question 10.
The name ofis …………
Answer:
2-cyclopentyipropanal

Question 11.
Esopentane and neopentane are the examples for ………
Answer:
chain isomerism

Question 12.
are called ……….
Answer:
functional isomers

Question 13.
Copper oxide test is used to detect ………..
Answer:
Carbon & Hydrogen

Question 14.
The formula of feme suiphocyanide is ………
Answer:
Fe(CNS)₃

Question 15.
In Lassaigne’s test for halogens, if colour of the precipitate is curdy white, the halogen present is ……….
Answer:
chlorine

Question 16.
The formula of ammonium phospho molybate is ……….
Answer:
(NH₄)₃.PO₄. MO₃

Question 17.
Silver nitrate test is used to detect the presence of ……….
Answer:
Halogens

Question 18.
During the estimation of carbon and hydrogen, presence of nitrogen can be avoided by using ………
Answer:
a spiral of copper

Question 19.
In Carius method, the sulphur in an organic compound is oxidised to ……….
Answer:
H₂SO₄

Question 20.
The method used to estimate nitrogen in foods and fertilisers is ………
Answer:
Kjeidahl’s method

Question 21.
The mixture of diethyl ether and ethanol can he purified by ………
Answer:
simple distillation

Question 22.
The method used to purify petroleum. coal-tar and crude oil is ………
Answer:
fractional distillation

Question 23.
The method used in the manufacture of aniline and turpentine is ……….
Answer:
steam distilation

Question 24.
The mixture of exhanol and water are separated by ………..
Answer:
azeotropic distillation

Question 25.
The different coloured constituents of chlorophyll are separated by ………
Answer:
chromatography

Question 26.
The large number of organic compounds is due to of carbon ………..
Answer:
catenation

Question 27.
The IUPAC name of the compound shown below is:

Answer:
1, 1-dichloropropane

Question 28.
name of this compound is ………..
Answer:
2-chloro-3-ethyl- 1, 4-pentadiene

Question 29.
Carboxylic acids are isomenc with ……….
Answer:
esters

Question 30.
Alcohols are isomeric with ……….
Answer:
ethers

Question 31.
The correct IUPAC name for the following structure is ……….

Answer:
5-hex-i -en-3-ol

Question 32.
The Prussian blue colour confirms the presence of nitrogen in an organic compound is due to the formation of ………
Answer:
Fe₄[Fe(CN)₆]I

Question 33.
The principle involved in paper chromatography is ……….
Answer:
partition

Question 34.
Steam distillation is used for the extraction of ……..
Answer:
essential oils

Question 35.
In chromatography, if the stationary phase is solid, the basis is ……….
Answer:
adsorption

Question 36.
In chromatography, if the stationary phase is liquid, the basis is ………
Answer:
partition

Question 37.
The isomer of acetaldehyde is ……….
Answer:
acetone

Question 38.
The general formula of alkyne is ……….
Answer:
C_nH_2n-2

Question 39.
The IUPAC name of (CH₃)₂CH-CH₂-CH(CH₃)₂-CH(CH₃)₂ is ……..
Answer:
2, 3, 5-trimethyl hexane

IV. Choose the odd one out.

Question 1.
(a) THF
(b) Pyridine
(c) Phenol
(d) Thiophen
Answer:
(c) Phenol. ft is a homocyclic compound whereas others are heterocyclic compounds.

Question 2.

Answer:

Question 3.
(a) Azulene
(b) Propane
(c) Butane
(d) Ethene
Answer:
(a) Azulene. It is a non benzenoid aromatic homocyclic compound whereas others are aliphatic compounds.

Question 4.
(a) Dyes
(b) Polymers
(c) Cosmetics
(d) Common salt
Answer:
(d) Common salt. It is an inorganic compound whereas others are organic compound.

Question 5.
(a) Renzene
(b) Water
(c) Ether
(d) Chloroform
Answer:
(b) Water. It is a polar solvent whereas others are non-polar solvents.

V Choose the correct pair.

Question 1.
(a) Benzene : Aliphatic compound
(b) Propane : Aromatic compound
(c) Pyridine : Heterocyclic compound
(d) Cyclohexane : Polycyclic compound
(c) Pyridine : Heterocyclic compound

Question 2.
(a)-OH : Ketone
(b)-CHO : Carboxylic acid

(d)-NO₂ : Amine
Answer:

Question 3.
(a) Organic compounds : inflammable
(b) Organic compounds : ionic compound
(c) Organic compounds : high boiling point and high melting point
(d) Organic compounds : soluble in water
Answer:
(a) Organic compounds : inflammable

Question 4.
(a) C_nH_2n+2 : C₂H₄
(b) C_nH_2n : C₃H₆
(c) C_nH_2n-2 : C₂H₆
(d) C_nH_2n+2 : C₃H₄
Answer:
(b) C_nH_2n : C₃H₆

Question 5.

Answer:

Question 6.
(a) n-pentane and iso pentane : position isomerism
(b) neopentane and n-pentane : chain isomerism
(c) propanal and propanone : position isomerism
(d) propañoic acid and methyl acetate : chain isomerism
Answer:
(b) neopentane and n-pentane : chain isomerism

VI. Choose The incorrect pair.

Question 1.
(a) Dumas method : Estimation of nitrogen
(b) Kjeldahls method : Estimation of nitrogen
(c) Carius method : Estimation of halogens
(d) Dumas method : Estimation of sulphur
Answer:
(d) Dumas method : Estimation of sulphur

Question 2.
(a)-CHO : Aldehyde
(b)-COOH : Carboxylic acid
(c)-NH₂ : Nitro group
(d)-O- : Ether
Answer:
(c)-NH₂ : Nitro group

Question 3.
(a) Benzene and nitro benzene : Distillation
(b) Coal tar and crude oil : Fractional distillation
(c) Aniline and turpentine : Steam distillation
(d) Naphthalcne and benzoic acid : Crystallization
Answer:
(d) Naphthalene and benzoic acid: Crystallization

Question 4.
(a) BaSO₄ : White colour precipitate
(b) Ag₂S : Black colour precipitate
(c) Fe(CNS)₃ : Prussian blue colour
(d) PbS : Black colour precipitate
Answer:
(c) Fe(CNS)₃ : Prussian blue colour

Question 5.
(a) propanal and propropane : Functional isomerism
(b) Nitrite fòrrn and nitro form : Tautomerism
(c) Pent- 1-ene and pcnt-2-ene : Chain isomerism
(d) Propanoic acid and methyl acetate: Functional isomerism
Answer:
(c) Pent-1-ene and pent-2-ene : Chain isomerism

VII. Assertion & Reason.

Question 1.
Assertion (A) : Carbon cannot form ionic bond.
Reason (R) : It is not possible for the carbon to form either C4t or C ions, as it requires large amount of energy.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct hut R is not the correct explanation of A.
(c) A is.correct but R is wrong.
(d) A is wrong but Ris correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 2.
Assertion (A) : Simple distillation can help in separating a mixture of propan-1-ol (boiling point 97°C) and propanone (boiling point 56°C).
Reason (R) : Liquids with a dîftèrence of more than 30°C in their boiling points can be separated by simple distillation.
(a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
(c) Assertion is correct but Reason is wrong.
(d) Assertion is wrong but Reason is correct.
Answer:
(a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Question 3.
Assertion (A) : Pent-1 -ene and pent-2-ene are position isomers.
Reason (R) : Position isomers ditTer in the position of functional group or substituent.
(a) Both assertion and reason are correct and reason is the correct explanation of assertion.
(b) Both assertion and reason are correct but reason is not the correct explanation of assertion.
(c) Assertion is correct but reason is wrong.
(d) Assertion is wrong but reason is correct.
Answer:
(a) Both assertion and reason are correct and reason is the correct explanation of assertion.

VIII. Choose the correct statement.

Question 1.
(a) All organic compounds are ionic compounds.
(b) All organic compounds have high boiling point and high melting point.
(c) Many of the organic compounds are inflammable.
(d) Organic compounds are mostly soluble in water.
Answer:
(c) Many of the organic compounds are inflammable.

Question 2.
(a) Propane is heterocyclic compound.
(b) Azulene is a non benzenoid and aromatic homocyclic compound.
(c) Pyridine is a homocyc lic compound.
(d) Cyclopropane is an aromatic compound.
Answer:
(b) Azulene is a non henzenoid and aromatic homocyclic compound.

Question 3.
(a) CH≡CH-CH₂-C≡CH is a saturated open chain compound.
(b) CH₃-CH₂-CH₂-CH₂-CH₃ is an aromatic benzenoid compound.
is an aromatic benzenoid compound.
Answer:
is an aromatic benzenoid compound.

Question 4.
(a) Organic compounds are covalent and generally insoluble in water.
(b) Organic compounds are ionic but generally soluble in water.
(c) Organic compounds non-inflammable
(d) Organic compounds do not show catenation.
Answer:
(a) Organic compounds are covalent and generally insoluble in water.

Question 5.
(a) Fe₄[Fe(CN)₆] is prussian blue precipitate.
(b) Ag₂S is a white precipitate.
(c) PbS is a blood red colour precipitate.
(d) BaSO4 is a black colour precipitate.
Answer:
(a) Fe₄[Fe(CN)₆] is prussian blue precipitate.

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry 2 Mark Question and answers

Write brief answer to the following questions:

Question 1.
What is catenation?
Answer:
The tendency of an atom to form a chain of bonds with the atoms of the same element is called catenation. The high strength of C-C bond is responsible for its catenation property.

Question 2.
Almost all compounds of carbon form covalent bonds. Give reason.
Answer:
Carbon (Z = 6) have electronic configuration of is² 2s² 2p². It is not possible for the carbon to form either C⁴⁺ or C^4- ions to attain the nearest noble gas configuration as it requires large amount of energy. This implies that carbon cannot form ionic bond. So almost in all compounds of carbon, it form four covalent bonds.

Question 3.
What is meant by functional group? Give two example.
Answer:
A functional group is an atom or a specific combination of bonded atoms that react in a characteristic way. irrespective of organic molecule in which it is present. The reaction of an organic compound takes place at the functional group.
e.g.. Alcohol -OH group
Ether -O- group

Question 4.
Classify the following compounds based on the structure.
(i) CH₂=CH-CH=CH₂
(ii) CH₃-CH₂-CH₂-CH₂-CH₂-CH₂-CH₃

Answer:
(i) Unsaturated open chain compound
(ii) Saturated open chain compound
(iii) Aromatic benzenoid compound
(iv) Alicylic compound

Question 5.
Give one example for each of the following type of organic compounds.

1. Non-benzeoid
2. Aromatic heterocclic
3. Alicyclic
4. Aliphatic open chain

Answer:
1. Non-benzenoid aromatic compound

2. Alicyclic compound

3. Aromatic heterocyclic Compound

4. Atiphatic open chain compound

Question 6.
Give two examples for each of the following type of organic compounds.

1. Aromatic bomocyclic compound
2. Aromatic heterocyclic compound

Answer:
1. Aromatic homocyclic compound

2. Aromatic heterocyclic compound

Question 7.
Write the functional group of the following compounds

1. Aldehyde
2. Ester
3. Ether
4. alcohol

Answer:

Question 8.
Write the functional group of –

1. cyanide
2. socyanide
3. cyanate
4. isocyanate

Answer:

1. cyanide -CN
2. isocyanide -NC
3. cyanate -CNO
4. isocyanate -NCO

Question 9.
Write the functional group of –

1. thiocyanate
2. isothiocyanate
3. thiols
4. thioether

Answer:

1. Thio cyanale -SCN
2. Isothiocyanate -NCS
3. thiols -SH
4. thicethers -S-

Question 10.
Write the IUPAC names of the following compounds.

Answer:

1. 3-methylpentane
2. 2, 2, 5-trimethylheptane

Question 11.
Write the IUPAC names of the following compounds.

Answer:

1. 3-ethyl-2-methyl pentane
2. 2-methylbutanal

Question 12.
What are the IUPAC names of the following compounds?


Answer:

1. 2-ethyl-but-4-ene-oic acid
2. 2, 2-dimethyl-hexanoic acid

Question 13.
Write the IUPAC names of the following compounds.

Answer:

1. 2-ethyl-3-propyl pentane-dioic acid
2. 3-methy-1 hexane

Question 14.
Predict the IUPAC names of the following compounds

Answer:

1. cyclo butane
2. cyclo pentane
3. cyclo hutene
4. cyclo octane

Question 15.

Answer:

1. Ethyl cyclobutanc
2. Methyl cyclohexane
3. Cyclo hexanol

Question 16.
Write the ¡UPAC names of the following compounds.


Answer:

1. 2-cyclobutyl propanol
2. 3-cyclohexyl pentan-2-one

Question 17.
Write the structural formula for the following compounds.
(i) cyclohexa- 1, 3-diene
(ii) methyl cyclopentane
Answer:

Question 18.
Draw the structures of –
(i) o-xylene
(ii) m-xylene
(iii) p-xylene
Answer:

Question 19.
Draw the structure of –
(i) Mesitylene
(ii) 1,2, 3-trimethyl benzene
Answer:

Question 20.
Write the structure of –
(i) p-dinitrobenzene
(ii) o-dichlorobenzene
Answer:

Question 21.
Draw the structure of –
(i) 2-cyclopentyl propanal
(ii) 2-(cyclo-but-cnyl) propanal
Answer:

Question 22.

Answer:

1. N, N-dimethylbenzene amine
2. N-ethyl-N-methylpropan-I-amine

Question 23.
Draw the structure of 4-hydroxy-3(2-hydroxy ethyl) hexanoic acid.
Answer:

Question 24.
Explain about dash line structure with a suitable example.
Answer:
The line bond structure is obtained by representing the two electron covalent bond by a dash or line (-) in a lewis structure. A single line or dash represents a single covalent bond.
e.g., n- propanal:

Question 25.
What ¡s meant by condensed structure? Explain with an example.
Answer:
The bond line structure can be further abbreviated by omitting all these dashes representing covalent bonds and by indicating the number of identical groups attached to an atom by a subscript. The resulting expression of the compound is called a condensed structural formula.
e.g., 1, 3-butadiene.

Question 26.
What are bond line structures? Give one example.
Answer:
Condensed structural formula is simplified in which only lines are used. In this type of representation of organic compounds, carbon and hydrogen atoms are not shown and the lines representing carbon-carbon bond are shown in zig zag fashion. The only atoms specially written are oxygen, chlorine, nitrogen etc. Example, Ten. butyl chloride
condensed structure

bond line structure

Question 27.
Define isomerism. Give example.
Answer:
Isomerism represents the existence of two or more compounds with the same molecular formula but different structure and properties. Compounds exhibiting this isomerism are called isomers.
e.g., C₂H₆O:

1. CH₃-CH₂OH Ethanol
2. CH₃-O-CH₃ Methoxyrnethane

Question 28.
Write the possible isomers for the formula C₅H₁₀ with their name and type of isomerism present in it.
Answer:
C₅H₁₀:

1. CH₃-CH₂-CH₂-CH = CH₂ (Pent-1-ene)
2. CH₃-CH₂-CH=CH-CH₃ (Pent-2-ene)
   The type of isomerism present above is position isomerism.

Question 29.
Write the possible isomers for the formula C₅H₁₀O with their name indicating position isomerism.
Answer:


Question 30.
Draw the functional isomers for the formula C₃H₆O₂ with their names.
Answer:
C₃H₆O₂:

1. CH₃-CH₂-COOH (Propanoic acid)
2. CH₃-COOCH₂ (Methyl acetate)

Question 31.
What is metamerism? Give an example.
Answer:
Metamerism is one kind of structural isomerism which arises due to the unequal distribution of carbon atoms on either side of the functional group or diffèrent alkyl groups attached to the either side of the same functional group and having the same molecular formula.
e.g., C₄H₁₀O:

Question 32.
What is tautomerism?
Answer:
It is a special type of functional isomerism in which single compound exists in two readily interconvertible structures that differ markedly in the relative position of atleast one atomic hydrogen. The two different structures are known as tautomers.

Question 33.
What ¡s meant by dyad system? Explain with example.
Answer:
In this system, hydrogen atom oscillates between two directly linked polyvalent atoms.

Question 34.
What is triad system? Give example.
Answer:
1. In this system hydrogen atom oscillates between three polyvalent atoms. It involves 1. 3-migration of hydrogen atom from one polyvalent atom to other with in the molecule.
2. The most important type of triad system is keto-enol tautomerism and the two groups of tautomers are keto form and enol form.

Question 35.
What is enolisation? What is labile form?
Answer:
Enolisation is a process in which keto form is converted into eno! form. Both tautomeric forms are equally stable. The less stable form is known as labile form.

Question 36.
Give the structures of Nitro-aci tautomerism.
Answer:

Question 37.
Explain ring chain isomerism with the formula C₄H₈.
Answer:
In ring chain isomerism, compounds have the same molecular formula but differ in terms of bonding of carbon atoms to form open chain and cyclic structures.
C₄H₈:

Question 38.
Define stereo-isomerism.
Answer:
The isomers which have same bond connectivity but different arrangement of groups or atoms in space are known as stereoisomers. This phenomenon is known as stereoisomerism.

Question 39.
Define geometrical isomerism with an example.
Answer:
Geometrical isomers are the stereoisomers which have different arrangement of groups or atoms around a rigid framework of double bonds. This type of isomerism occurs due to restricted rotation of double bonds or about single bonds in cyclic compounds.

Question 40.
Trans isomer is more stable than cis isomer. Justify this statement.
Answer:
Trans isomer is more stable than cis isomer. This is because in the cis isomer, the bulky groups are on the same side of the double bond. The steric repulsion of the groups makes the cis isomers less stable than the trans isomers in which bulky groups are on the opposite side.

Question 41.
Draw the cis, trans isomeric structures of 1, 3-butadiene.
Answer:

Question 42.
What are the condition for optical isomerism (or) enantiomerism.
Answer:
1. A carbon atom whose tetravalency in satisfied by four different substituents (atoms (or) groups) is called asymmetric carbon (or) chiral carbon. The optical isomer should have one or more chiral carbon to show optical activity.

2. The molecule possessing chiral carbon atom and is non-superimposable its own mirror image is said to be chiral ntolecule and the property is called chirality or dissymmetry.

Question 43.
How will you prepare Lassaigne’s extract?
Lassagine’s extract preparation:
Answer:

• A small piece of Na dried by pressing between the folds of filter paper is taken in a fusion tube and it is gently heated. When it melts to a shining globule, a pinch of organic compound is added.
• The tube is heated till reaction ceases arid become red hot. Then it is plunged in 50 ml of distilled water taken in a china dish and the bottom of the tube is broken by striking it against the dish.
• The contents of the dish is boiled for 10 minutes and then it is filtered. The filtrate is known as Lassaigne’s extract.

Question 44.
What Is the need for purification of organic compounds?
Answer:
In order to study the structure, physical properties, chemical properties and biological properties of organic compounds, they must be in the pure state. So organic compounds must be purified.

Question 45.
Define sublimation. Give two examples.
Answer:
The process of conversion of solid to vapour without melting or heating and on cooling the vapours getting back solids, such phenomenon is known as sublimation.
e.g., Naphthalene, Camphor.

Question 46.
Explain the process of chromatography in chlorophyll.
Answer:
The separation of different coloured constituents of chlorophyll is done by chromatography by M.S. Tswelt. He achieved it by passing a petroleum ether solution of chlorophyll present in leaves through a column of CaCO3 firmly packed into a narrow glass tube. Different components of the pigments got separated and lands to form zones of different colours.

Question 47.
Draw the first six members of the carboxylic acid homologous series.
Answer:

1. HCOOH
2. CH₃COOH
3. CH₃-CH₂-COOH
4. CH₃-CH₂-CH₂-COOH
5. CH₃-CH₂-CH₂-CH₂COOH
6. CH₃-CH₂-CH₂-CH₂-CH₂-COOH

Question 48.
Give the condensed formula and bond line formula of 2, 2,4- trimethylpentane.
Answer:
2, 2, 4 –  irimethylpentane
(CH₃)₃CCH₂CH(CH₃)₂-Condensed formula

Question 49.
Differentiate between the principle of estimation of nitrogen in an organic compound by

1. Dumas method
2. Kjeldahl’s method.

Answer:
1. Dumas method:
The organic compound is heated strongly with excess of CuO (Cupic Oxide) in an atmosphere of CO₂ where free nitrogen, CO₂ and H₂O are obtained.

2. Kjeldahl’s method:
A known mass of the organic compound is heated strongly with conc. H₂SO₄, a little amount of potassium sulphate and a little amount of mercury (as catalyst). As a result of reaction, the nitrogen present in the organic compound is converted to ammonium sulphate.

Question 50.
Explain the principle of paper chromatography.
Answer:
This is the simplest form of chromatography. Here a strip of paper acts as an adsorbent. It is based on the principle which is partly adsorption. The paper is made of cellulose fibres with molecules of water adsorbed on them. This acts as stationary phase. The mobile phase is the mixture of the components to be identified whose solution is prepared in a suitable solvent.

Question 51.
Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.
Answer:
Organic compound is fused with sodium metal so as to convert organic compounds into NaCN, Na₂S, NaX and Na₃PO₄. Since these are ionic compounds and become more reactive and thus can be easily tested by suitable reagents.

Question 52.
Name a suitable technique of separation of the components from a mixture of calcium, sulphate and camphor.
Answer:
Sublimation. Because camphor can sublime whereas CaSO₄ does not.

Question 53.
Explain, why an organic liquid vapourises at a temperature below its boiling point on steam distillation?
Answer:
It is because in steam distillation the sum of vapour pressure of organic compound and steam should be equal to atmospheric pressure.

Question 54.
Will CCl₄give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.
Answer:
No. CCl₄ is a completely non-polar covalent compound whereas AgNO₃ is ionic in nature. Therefore they are not expected to react and thus a white ppt. of silver chloride will not be formed.

Question 55.
Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?
Answer:
CO₂ is acidic in nature and therefore it reacts with the strong base KOH to form K₂CO₃:
2KOH + CO₂ – K₂CO₂ + H₂O

Question 56.
Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?
Answer:
Sulphur sodium extract is acidified with acetic acid because lead acetate is soluble and does not interfere with the test.
Pb(OCOCH₃)₂ + H₂SO₄ → PbSO₄-+ 2CH₃COOH

Question 57.
Why is an organic compound fused with sodium for testing nitrogen, halogens and sulphur?
Answer:
On Ilising with sodium metal the clements present in an organic compound are converted into sodium salts which are water soluble which can be filtered and detected by the respective tests.

Question 58.
Under what conditions can the process of team distillation is used?
Answer:
Steam distillation is used to purify the Liquids which are steam volatile and not miscible with water.

Question 59.

Answer:
1, 2-dichloropropane

Question 60.
Write bond-line formulas for: Isopropyl alcohol, 2, 3-dimethvlbutanal, Heptan-4-one.
Answer:

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry 3-Mark Questions

Question 1.
Write the functional group of the following comopounds:
(i) carboxylic acid
(ii) Acid anhydride
(iii) Acyichioride
(iv) Amide
(v) imines
(vi) Nitroso compound
Answer:

Question 2.
What are the general molecular formula and functional group of the following compounds?
Answer:
(i) Ilydrazines
(ii) Hydrazo compound
(iii) Imide
(iv) Phenols
(v) Amine
(vi) Nitroalkane
Answer:

Question 3.
Write the tUPAC names of the following compounds.

Answer:
(i) Pentan-2a1
(ii) Pentan-(2-ene-2-propyl)- i -oic acid
(iii) 4-methyl-i -cyanohexane

Question 4.

Answer:
(I) Hex-4-ene-2-oI
(ii) 3-ethyl-5-methyl heptane

Question 5.
Draw the structure of
(i) 1-ethyl-2-methyl cyclopentane
(ii) 1-ethyl-2, 3-dimethyl cyclohexane
(iii) 5-ethyl-2-methylcyclohcx- I -ene
Answer:

Question 6.
Draw the structures of:
(i) 2-cyclobutyl propane
(ii) 2-cyctopropyl butane
(iii) chiorocyclo but-2-eue
Answer:

Question 7.
Give the IUPAC name of the following compounds:

Answer:
(i) 2-(cyclo but-2-en-1-yl)-propanal
(ii) 4-(cyclopent-3-en- 1-yl )-3-methylbutanoic acid
(iii) 3-(3-nitro cyclopenryl)-prop-2-enoic acid

Question 8.
Write the IUPAC names of the following compounds

Answer:
(I) 2-(2-hydroxypropyl) cyclohexan- 1-01
(ii) CyclopentyÍ benzene
(iii) Cyclohexane carboxyl Ic acid

Question 9.
Draw the structure
(i) 1-(cyclo bytyl)-2 (cylopropyl) ethane
(ii) 2-carbamyl cyclobutane-1-carboxylic acid
Answer:

Question 10.
Draw the structures of:
(i) Bromohenzene
(ii) 1, 2-dichlorobenzene
(iii) 1-chloro-3-methvlbenzene
Answer:

Question 11.
Draw the structures of –
(i) Benzvl chloride
(ii) Benzal dichloride
(iii) Benzotrichloride
Answer:

Question 12.
Write the IUPAC names of the following compounds.

Answer:
(i) 3-methylpentane
(ii) 2, 2. 5-trimethylheptane
(iii) 2-methylbutanal

Question 13.
Write the IUPAC names of the following compounds.

Answer:
(i) 3-ethyl-2-methylpentane
(ii) 2-ethyl but-3-enoic acid
(iii) 2-forrnyl-2-methylheptanoic acid

Question 14.
Draw the structures of
(i) 3-methylpentanal
(ii) 5-hydroxy 2,2-dimethyl heptanoic acid
(iii) 2-ethyl-4-propy Ipentane-d ioic acid
Answer:

Question 15.

Answer:
(i) 3-methylhexane
(ii) 2-methylbutanal
(iii) 2-ethylbut-3-enoic acid

Question 16.
Give the IUPAC name Of –

Answer:
(i) 4-methyl/hexanenitrite
(ii) 2-methyl but-3-en-amide
(iii) I-Iex-4-en-2-ol

Question 17.
Draw the structures of –
(i) 3-ethyl-5-methylheptane
(ii) 3-ethyl-2-methylhexane
(iii) 2, 4-dimethylpent-2-ene
Answer:

Question 18.
Draw the structures of:
(i) 3-methylhcpta 1, 3, 5-triene
(ii) pent-1-yne
(iii) 2-methylpropan-2-oI
Answer:

Question 19.
Give the IUPAC name of the following compounds.

Answer:
(i) 4-mcthylpcntan-1-ol
(ii) 2, 2-dimethylpropan-1-ol
(iii) Propanoic acid

Question 20.
Draw the structural formula of:
(i) 4-methylpent-3-en-2rone
(ii) pent-I-yne-3-one
Answer:

Question 21.
Write the IUPAC names of the following compounds.
CH₃-CH₂-CH₂-NH-CH₃

Answer:
(i) N-methylpropan- 1 -amine
(ii) N-rnethylpropan-2-amine
(iii) N, N-dimethylpropan- 1-amine

Question 22.
Draw the structurai formula of the following compounds.
(i) N-cthyl-N-methylpropan-1-amine
(ii) N, N-dimethyl benzenamine
Answer:

Question 23.
Draw the complete structural formula, condensed structure and bond line structure of
(i) n-propanol
(ii) 1, 3-butadiene.
Answer:

Question 24.
Draw the dash line structure, condensed structure and bond line structure of 1, 3-dimethyl cyclopentane.
Answer:

Question 25.
What is wedge formula? Explain with suitable example.
Answer:
1. The simplest convention is solid and dashed wedge formula in which 3-D image of a molecule can be perceived from two dimensional picture.

2. In this representation, a tetrahedral molecule with four atoms or groups a, b, e and d bonded to it can be represented by wedge formula as follows.

3. A solid wedge
or a heavy line is used to indicate a bond projecting above the plane of the paper and dashed wedge
or a dashed line is used to depict the bond below the plane. The bonds lying in the plane of the paper are shown by normal lines.

Question 26.
Draw the fisher projection formula for tartaric acid.
Answer:

Question 27.
Explain the advantage of sawhorse projection formula over the fisher projection formula with an example.
Answer:
1. The fisher projection fonnula inadequately portrays the spatial relationship between ligands attached to the atoms. The sawhorse projection attempts to clarify the relative location of the groups.

2. In sawhorse projection formula, the bond between two carbon atoms ¡s drawn diagonally and slightly elongated. The lower left hand carbon is considered lying towards the front and the upper right hand carbon towards the back.

Question 28.
Explain about the Newmann projection formula with an example.
Answer:
1. In this method, the molecules are viewed from the front along the carbon-carbon bond axis.

2. The two carbon atom forming the G bond is represented by two circles. One behind the other so that only the front carbon is seen. The front carbon atom is shown by a point where as the carbon lying farther from the eye is represented by the origin of the circle.

3. Therefore the C-H bonds of the front carbon are depicted from the circle while the C H bonds of the back carbon are drawn from the circumferance of the circle with an angle of 120 to each other.

Question 29.
Write the possible isomers for the formula C₅H₁₂ with their names and structures.
Answer:

(i), (ii) and (iii) are chain isomers.

Question 30.
What are the possible isomers for the formula C₄H₉Cl? Give their structures and IUPAC names.
Answer:
C₄H9Cl:

Question 31.
Write the metamers for the formula C₅H₁₀O with their IUPAC names.
Answer:

Question 32.
Explain about the geometrical isomerism possible in oximes.
Answer:
1. Restricted rotation around C = N (oximes) gives rise to geometrical isomerism in oximes. Here syn and anti are used instead of cis and trans respectively.

2. In the syn isomer the H atom of a doubly bonded carbon and OH group of doubly bonded nitrogen lie on the same side of the double bond, while in the anti isomer, they lie on the opposite side of the double bond.

3. for e.g.,

Question 33.
What are enantiomers?
Answer:
1. An optically active substance may exist in two or more isomeric forms which have same physical and chemical properties but diflèr in terms of direction of rotation of plane polarised light, such optical isomers which rotate the plane polarised light with equal angle but in opposite directions are known as enantiomers and (he phenomenon is known as enantiomerism.

2. Isomers which are non-super impossible mirror images of each other are called enantiomers.

Question 34.
How would you detect sulphur?
Answer:
1. To a portion of the Lassaigne’s extract, freshly prepared sodium nitroprusside solution is added. If deep violet or purple colour is formed, the presence of sulphur is confirmed.

2. To another portion of Lassaignes extract, acetic acid and lead acetate solution are added. If black precipitate is formed, sulphur presence is confirmed.

Question 35.
Explain about the oxidation test for sulphur.
Answer:
1. Oxidation test:
The organic substances are fused with a mixture of KNO₃ and Na₂CO₃. The sulphur if present is oxidised to sulphate.
Na₂CO₃ + S + 3(O) Na₂SO₄ + CO₂

2. The fused mass is extracted with water, acidified with HCl and the BaCl₂ solution is added to it. A while precipitate indicates the presence of sulphur.

Question 36.
How would you detect the halogens in an organic compound?
Answer:

• To a portion of the Lassaigne’s filtrate. dii. HNO₃ is added, warmed gently and AgNO₃ solution is added.
• Appearance of curdy white precipitate soluble in ammonia solution indicates the presence of chlorine.
• Appearance of pale yellow precipitate sparingly soluble in ammonia solution indicates the presence of bromine.
• Appearance of yellow precipitate insoluble in ammonia solution indicates the presence of iodine.
  

Question 37.
Why nitric acid is added in the Lassaigne’s test for halogens?
Answer:
1. if N (or) S is present in the organic compound along with the halogen, we might obtain Na₂S and NaCN in the solution which interfere with the detection of the halogen in the AgNO₃ test.

2. Therefore we boil the Lassiagne’s extract with HNO₃ which decomposes NaCN and Na₂ S as follows:

Question 38.
Explain about the test for phosphorous in an organic compound.
Answer:

• A solid organic compound is strongly heated with a mixture of Na₂CO₃ and KNO₃. Phosphorous present in the compound is oxidised to sodium phosphate.
• The residue is extracted with water and boiled with cone. HNO₃. A solution of ammonium molyixiate is added to this solution.
• A canary yellow precipitate shows the presence of phosphorous.

Question 39.
Explain about principle and reactions involved in carius method of estimation of sulphur. Carius method:
Answer:
1. Principle:
A known mass of the organic substance is heated strongly with fuming HNO₃. C and H get oxidized to CO₂. and H₂O while sulphur is oxidised to sulphuric acid as follows.

2. The resulting solution is treated with excess of BaCl₂ solution, H₂SO₄ present in the solution is converted into BaSO₄. From the mass of BaSO4, the percentage of sulphur can be calculated.

Question 40.
Explain about the procedure and calculation behind the carius method of estimation of sulphur.
Answer:
Carius method:
(I) Procedure:
A known mass of the organic compound is taken in a clean carius tube and few mL of fuming HNO₃ is added and then the tube is sealed. It is then placed in an iron tube and bcatcd for 5 hours. The tube is allowed to cool and a hole is made to allow gases to escape. The carius tube is broken and the content collected in a beaker. Excess of BaCl, is added to the beaker. BaSO₄ furmed is converted to BaSO₄ (white ppt.) The precipitate is filtered. washed, dried and weight. From the mass of BaSO₄, percentage of S is calculated.

(ii) Calculation:
Mass of organic compound = Wg
Mass of BaSO₄ formed = r g
233 g of BaSO₄ contains 32 g of sulphur
∴ x g of BaSO₄ contain \(\frac {32}{233}\) x x g of sulphur

Question 41.
How are naphthalene and camphor purified?
Answer:
1. Naphthalcne, camphor and benzoic acid when heated, pass directly from solid to vapour without melting. On cooling the vapours will give back solid. This phenomenon is known as sublimation. This technique is used to purify naphthalene, camphor from non volatile impurities.

2. Substances to he purified is taken in a beaker. It is covered with a watch glass. The beaker is heated for a while and the resuJting vapours condense on the bottom of the watch glass. Then the watch glass is removed and the crystals are collected.

Question 42.
Explain how nitrobenzene and benzenc arc separated and purified. (or) How will you separate the mixture of diethyl ether and ethanol?
Answer:
Distillation:

• The process of distillation involves the impure liquid when boiled gives out vapour and the vapour so formed is collected and condensed to give back the pure liquid in the receiver.
• This method ¡s to purify liquids from non-volatile impurities and used for separating the constituents of a liquid mixture which differ in their boiling points.
• In this simple distillation process, liquids with large difference in boiling point (about 40K) and do not decompose under ordinary pressure can be purified.
• e.g., the mixture of C₆H₅NO₂ nitrohenzene (h.p. 484 K) and C₆H₆ benzene (b.p. 354 K) can be urified and separated. Similarly the mixture of diethyL ether (b.p. 308K) and ethyl alcohol (h.p. 351 K) can he purified and separated.

Question 43.
Explain about differential extraction.
Answer:

• The process of removing a substance from its aqueous solution by shaking with a suitable organic solvent is termed extraction.
• When an organic substance present as solution in water can be recovered from the solution by means of a separating funnel.
• The aqueous solution is taken in a separating funnel with little quantity of ether or chloroform (CHCl₃). The organic solvent immiscible with water will form a separate layer and the contents are shaken gently.
• The solute being more soluble in the organic solvent is transferred to it.
• The solvent layer is then separated by opening the tap of separating funnel and the substance is recovered.

Question 44.
Explain about the principle involved in chromatography. Give its types.
Answer:

1. The principle behind chromatography is selective distribution of the mixture of organic substances between two phases-a stationary phase and a moving phase. The stationary phase can be a solid or liquid while the moving phase is a liquid or a gas.
2. if the stationary phase is solid, the basis is adsorption and when it is a liquid, the basis is partition.
3. Chromatography is defined as technique for the separation of a mixture brought about by differential movement of the individual compound through porous medium under the influence of moving solvent.
4. The various methods of chromatography are:
   • Column chromatography (CC)
   • Thin layer chromatography (TLC)
   • Paper chromatography (PC)
   • Gas liquid chromatography (GLC)
   • Ion exchange chromatography

Question 45.
Describe about adsorption chromatography.
Answer:

• The principle involved is different compounds are adsorbed on an adsorbent to different degree.
• Silica gel and alumina are the commonly used adsorbent. The components of the mixture move by varying distances over the stationary phase.

Question 46.
What are hybridisation states of each carbon atom ¡n the following compounds?
Answer:

Question 47.
Give the IUPAC names of the following compounds.

Answer:
(a) Propylbenzene
(b) Methylpentanenitrile
(c) 2, 5-Dimethyllieptanc
(d) 3-Bromo-3-chloroheptane
(e) 3-Chioropropanal
(f) 2, 2-Dichloroethanol

Question 48.
Write the formulas for the first five members of each homologous series beginning with the following compound. CH₃COCH₃
Answer:
CH₃COCH₃
CH₃COCH₂CH₂CH₃
CH₃COCH₂CH₂CH₂CH₃
CH₃COCH₂CH₂CH₂CH₃
CH₃CO(CH₂)₄CH₃

Question 49.
Write the formulas for the first five members of each homologous series beginning with the following compound: H-CH=CH2
Answer:
H-CH=CH₂
CH₃CH=CH₂
CH₃CH₂CH=CH₂
CH₃CH₂CH₂CH=CH₂
CH₃CH₂CH₂CH₂CH=CH₂

Question 50.
Identify the functional groups ¡n the following compounds.
Answer:

Question 51.
What is the difference between distillation, distillation under reduced pressure and steam distillation?
Answer:
Distillation:
it is used in case of volatiLe liquid mixed with a non-volatile impurities.

Distillation under reduced pressure:
This method is used to purify such liquids which have very high boiling points and which decompose at or below their boiling points.

Steam distillation:
it is used to purify steam volatile liquids associated with water immiscible impurities.

Question 52.
(a) What is Lassaigne’s extract? Will NaCN give a positive Lassaigne’s test for nitrogen?
(b) Which colour will appear in the Lassaigne’s test if the compound contains both nitrogen and sulphur.
(c) Why is Lassaigne’s extract prepared in distilled water? Can we detect oxygen iii a compound by Lassaigne’s test?
Answer:
(a) When organic compounds is fused with sodium metal and then extracted by water it is called Lassaigne’s extract. Yes.
(b) Blood red colour.
(c) Lassaigne’s extract is prepared in distilled water since ta water contains Cl’ ions. No, oxygen cannot be detected by Lassaigne’s test.

Question 53.
0.3780 g of an organic compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine in the compound.
Answer:
Mass of the compound = 0.3780
Mass of silver chloride 0.5740 g

Question 54.
In an estimation of sulphur by Carius method, 0.468 of an organic sulphur compound gave 0.668 g of barium sulphate. Find the percentage of sulphur in the compound.
Answer:
Percentage of sulphur

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry 5 marks Questions and Answers

Question 1.
Explain about the steps involved in naming an organic compound as per IUPAC nomenclature.
Answer:
The following steps should be followed for naming an organic compound as per IUPAC nomenclature.

• Choose the longest carbon chain (Root word). Consider all other groups attached to this chain as substituents.
• Numbering of the longest carbon chain.
• Naming the substituents (prefixes) or (suffixes).
• Arrange the substituents in the alphabetical order.
• Write the name of the compound as below.

Prefix + Root word Primary suffix + r Secondary suffix

Question 2.
How will you detect the presence of carbon and hydrogen in an organic compound?
Answer:
Copper oxide test:
1. The organic substance is mixed with three times its weight of dry copper oxide by grinding. The mixture is placed in a hard glass test rube fitted with a bent delivery tube. The other end of which is dipping into lime water in an another test tube. The mixture is heated strongly.

2. 2CuO + C → CO₂ + 2Cu
CuO + 2H → H₂O + Cu

3. Thus if carbon is present, it is oxidized to CO₂ which turns lime water milky. If hydrogen is also present, it will be oxidized to water and condenses in small droplets on the cooler wall of the test tube and inside the bulb. Water is collected on white anhydrous CuSO₄ which turns blue.

4. This confirms the presence of C and H in the compound.

Question 3.
Explain about lassaigne’s test for detection of nitrogen ¡n an organic compound.
Answer:
I step:
Preparation of sodium fusion extract:
A small piece of Na dried by pressing between the folds of filter paper is taken in a fusion tube and it is heated. When it melts to a shining globule, a pinch of organic compound is added to it. The tube is then heated till the reaction ceases and becomes red hot.

Then the test tube is plunged in about 50 ml of distilled water taken in a china dish and break the bottom of the tube by striking against the dish. The contents of the dish is boiled for about 10 minutes and then filtered. This filtrate is known as lassaigne’s extract (or) sodium fusion extract.

II step :
Test for Nitrogen:
If Nitrogen is present, it gets converted to sodium cyanide which reacts with freshly prepared ferrous sulphate and ferric ion followed by cone. HCl and gives a Prussian blue colour (or) green coloured precipitate, it confirms the presence of nitrogen. HCl is added to dissolve the greenish precipitate of ferrous hydroxide produced by the action of NaOH on FeSO4 which would otherwise mark the Prussian blue precipitate.

Reactions involved:
Na + C + N → NaCN
FeSO₄ + 2NaOH → Fe(OH)₂ + Na₂SO₄


if both N & S are present. a blood red colour is obtained due to the following reactions

Question 4.
Explain about the estimation of carbon and hydrogen.
1. Principle:
A known weigIt of organic substance is brunt in excess of oxygen and the carbon and hydrogen present in it are oxidised to CO₂ and H₂O respectively.

The weight of carbon dioxide and water thus formed are determined and the amount of carbon and hydrogen in the organic substance are calculated.

2. Description of the apparatus:

(a) The oxygen supply (b) combustion tube (c) Absorption tube

Oxygen supply:
To remove the moisture from oxygen. it is allowed to bubble through sulphuric acid and then passed through a U-tube containing sodalime to remove CO₂. The oxygen gas free from moisture and CO₂ enters the combustion tube.

Combustion tube:
A hard glass tube open at both ends used for the combustion. It contains (i) an oxidized copper gauze to prevent the backward diffusion of the products of combustion (ii) a porcelain boat containing a known weight of the organic substance (iii) coarse copper oxide on either side and (iv) an oxidized copper gauze placed towards the end of the combustion tube. The combustion tube is heated by a gas burner.

Absorption apparatus:
The combustion products containing moisture and CO₂ are then passed through the absorption apparatus which consists of(i) a weighed U-tube packed with pumice soaked in conc. H₂SO₄ to absorb water (ii) a set of bulbs containing a Strong solution of KOH to absorb CO₂ and finally (iii) a guard tube filled with anhydrous CaCl₂ to prevent the entry of moisture from atmosphere.

3. Procedure:
The combustion tube is heated strongly to dry its content. It is then cooled and connected to absorption apparatus. The other end of the combustion tube is open for a while and the boat containing weighed organic substance is introduced. The tube is again heated strongly till all the substance in the boat is burnt away. This takes about 2 hours. Finally a strong current of oxygen is passed. Then the U-tube and potash bulbs are then
detached and increase in weight of each of them is determined.

4. Calculation:
Weight of organic substance = W g
Increase in weight of H₂O = x g
Increase in weight of CO₂ = y g
18g of H₂O contains 2 g of hydrogen
∴ x g of H₂O contain \(\frac {1}{2}\) × x g of hydrogen.
∴ Percentage of hydrogen = (\(\frac {2}{18}\) x \(\frac {x}{w}\) x 100
44g of CO₂ contains 12 g of carbon
∴ y g of CO₂ contain \(\frac {12}{44}\) × y g of carbon
∴ Percentage of carbon = (\(\frac {12}{44}\) × \(\frac {12}{44}\) × 100)%

Question 5.
Explain about the estimation of halogens by carius method.
Answer:
Carius method:
A known mass of the substance is taken along with fuming HNO₃ and AgNO₃ taken in a clean carius tube. The open end of the carius tube is sealed and placed in a iron tube for 5 hours in a range at 530 to 540 K. Then the tube is allowed to cool and a small hole is inade in the tube to allow the gases to excape. The tube is broken and the precipitate is filtered, washed, dried and weighed. From the mass of AgX produced percentage of halogen in the organic compound is calculated.

Calculation:
Weight of the organic compound = Wg
Weight of AgCl = a g
143.5 g AgCl contains 35.5g of Cl
a g of AgCl contain \(\frac {35}{143.5}\) x 9 g of Cl
Wg of organic compound contains \(\frac {35}{143.5}\) × 9 g of Cl
% of chlorine = (\(\frac {35}{143.5}\) × \(\frac {a}{w}\) × 100)%
Weight of silver bromide = b g
188 g of AgBr contains 80 g of Br
b g of AgBr contain = \(\frac {80}{188}\) × b g of Br
% of Bromine = (\(\frac {80}{188}\) × \(\frac {b}{w}\) × 100)
Weight of silver iodide = c g
235 g of AgI contains 127 g of I
c g of AgI contain = \(\frac {80}{188}\) × c g of I
% of Iodine = (\(\frac {80}{188}\) × \(\frac {c}{w}\) × 100)

Question 6.
How will you estimate phosphorous in an organic compound?
Answer:
Carius method:
Procedure:
A known mass of organic compound (wg) containing phosphorus is heated with fuming HNO₃ in a sealed tube where C is converted into CO₂ and H to H₂O. Phosphorous present in the organic compound is oxidised to phosphoric acid which is precipitated as ammonium phospho molybdate by heating with conc. HNO₃ and by adding ammonium molybdate.

The precipitate of ammonium phrospho molybdate is filtered, washed, dried and weighed.

Calculation:
Mass of organic compound = Wg
Mass of ammonium phospho molybdate = x g
Molar mass of ammonium phospho molybdate 1877 g
1877 g of ammonium phospho molybdate contains 31 g of phosphorous
x g of ammonium phospho molybdate contain = \(\frac {31}{1877}\) × x g of phosphorous
% of phosphorous = (\(\frac {31}{1877}\) × \(\frac {x}{w}\) × 100)% of phosphorous
In an alternate method, phosphoric acid is precipitated as magnesium-ammoniumphosphate by adding magnesia mixture. The ppt. is washed dried and ignited to get magnesium pyrophosphate which is washed, dried and weighed.

Weight of magnesium pyrophosphate = y g
Molar mass of magnesium pyrophosphate = 222 g
222 g of magnesium pyrophosphate contains 62 g of P
y g of magnesium pyrophosphate contain = \(\frac {62}{222}\) × y g of P
% of phosphorous = (\(\frac {31}{1877}\) × \(\frac {y}{w}\) × 100)%

Question 7.
Explain Dumas method of estimation of nitrogen.
Answer:
Dumas method:
Principle:
This method is based on the fact that nitrogeneous compound when heated with cupric oxide in an atmosphere of CO₂ yields free nitrogen.
Traces of nitrogen are reduced to elemental nitrogen by passing over heated copper spiral.
Description of the apparatus:

CO₂ Generator:
CO₂ needed in this process’is prepared by heating magnetite or sodium bicarbonate contained in a hard glass tube (or) by the action of dil. HCl on marble in a kipps apparatus. The gas is passed through the combustion tube after dried by bubbling through cone. H₂SO₄.

Combustion tube:
The combustion tube is heated in a furnace is charged with (a) A roll of oxidised copper gauze to prevent the back diffusion of products of combustion and to heat the organic substance mixed with CuO by radiation (b) a weighed amount of organic substance mixed with excess of CuO (c) a layer of CitO packed in about 2/3 length of the tube and kept in position by loose asbestos plug on either side and (d) a reduced copper piral which reduces any oxides of nitrogen formed during combustion of nitrogen.

Schiff’s nitromctc:
The nitrogen gas obtained by the decomposition of the substance in the combustion tube is mixed with considerable excess of CO₂. It is estimated by passing nitro meter when CO₂ is absorbed by KOH and the nitrogen gas gets collected in the upper part of the graduated tube.

Calculation:
Weight of the substance taken Wg
Volume of nitrogen = V₁L
Room temperature = T₁K
Atmospheric pressure = p mm Hg
Aqueous tension al room temperature P’ nun of Hg
Pressure of dry nitrogen = P—P’ = P’₁ mm Fig
P₀, V₀ and T₀ be the pressure, volume and temperature respectively of dry nitrogen at S.T.P.
Then, \(\frac{P_{0} V_{0}}{T_{0}}=\frac{P_{1} V_{1}}{T_{1}}\)
V₀ = \(\frac{P_{1} V_{1}}{T_{1}} \times \frac{T_{0}}{P_{0}}\)
V₀ = \(\frac{P_{1} V_{1}}{T_{1}} \times \frac{273 \mathrm{K}}{760 \mathrm{mm} \mathrm{Hg}}\)
22.4 L of N₂ at STP weigh 28 g of N₂
V₀L of N₂ at STP weigh \(\frac {28}{22.4}\) x V₀
W g of organic compound contain \(\frac {28}{22.4}\) x V₀ g of N₂
100 g of organic contain \(\frac {28}{22.4}\) x \(\frac{V_{o}}{w}\) x 100 = % of Nitrogen

Question 8.
Explain Kjeldahl’s method.
Answer:
Principle:
This method is based on the fact that an organic compound containing nitrogen is heated with cone. H₂SO₄. the nitrogen is conerted Lo ammonium sulphate. The resultant liquid is heated with excess of alkali and then liberated ammonia gas is absorbed in excess of standard acid. The amount of ammonia (nitrogen) is determined by finding the amount of acid neutralised by back titration with same standard alkali.

Procedure:
A weighed quantity of the substance 0.3 to 0.5g is placed in a special long necked Kjeldahl flask made of pyrex glass. About 25 ml of cone. H₂SO₄ together with a little K₂SO₄ and CuSO₄ [catalyst] are added to il. the flask is loosely stoppered by a glass bulb and heated gently in an inclined position.

The heating is continued till the brown colour of the liquid disappears leaving the content clear as before. At this point all the nitrogen is converted to ammonium sulphate. The kjeldahl flask is cooled aiid its contents are diluted with distilled water and carefully transferred into a I litre round bottom flask. An excess NaOH is poured down the side of the flask and it is filled with a kjeldhals trap and a water condenser.

The lower end of the condenser dips in a measured volume of excess of \(\frac {N}{20}\) H₂SO₄ solution. The liquid in the round bottom flask is heated and liberated ammonia is distilled to sulphuric acid. When no more ammonia passes over (test the distillate with red litmus) the receiver is removed. The excess of acid is then determined by titration with alkali, using phenolphthalein as the indicator.

Calculation:
Weight of the substance = Wg
Volume of H₂SO₄ required for the complete neutralisation of evolved NH₃ = V ml
Strength H₂SO₄ used to neutralise NH₃ = N.
Let the volume and strength of NH₃ formed are V₁ and N₁ respectively.
V₁N₁ = VN
The amount of nitrogen present in W g of organic compound = \(\frac{14 \times \mathrm{NV}}{1 \times 1000 \times w}}\)
Percentage of nitrogen = \(\frac{14 \times \mathrm{NV}}{1000 \times w} \times 100\) = \(\frac {1.4NV}{w}\) %

Question 9.
Explain the various steps involved in crystallization method.
Answer:
Most solid organic compounds are purified by crystallization method. This process is carried out by the following steps.
1. Selection of solvent:
rganic substances being covalent do not dissolve in water, hence selection of suitable solvent becomes necessary. Hence the powdered organic substance is taken in a test tube and the solvent is added little by little with constant stirring and heating, till the amount added is sufficient to dissolve the organic compound.

If the solid dissolves upon heating and throws out maximum crystals on cooling, then the solvent is suitable. This process is repeated with benzene, ether, acetone and alcohol various solvent. till the most suitable one is sorted out.

2. Preparation of solution:
The organic compound is dissolved in minimum quantity of suitable solvent small amount of animal charcoal can he added to decolonize any colored substance. The solution may be prepared by heating over a wire gauze or water bath.

3. Filtration of hot solution:
The hot solution so obtained is filtered through a fluted filter paper placed in a funnel.

4. Crystallization:
The hot filtrate is then allowed to cool. Most of the impurities are removed on the filter paper. the pure solid substance separate as crystal. If the rate of crystallization slow, it is induced either by scratching the walls of the beaker with a glass rod or by adding a few crystals of pure compounds to the solution.

5. Isolation and drying of crystals:
The crystals are separated from the mother liquor by filtration is done under reduced pressure using a Buchner funnel. Finally the crystals are washed with small amount of pure cold solvent and dried.

Question 10.
Explain about steam distillation (or) How is essential oils are recovered from plants and flowers.
Steam distillation:
This method is applicable for solids and liquids. If the compound to be steam distilled and it should not decompose at any steam temperature should have a fairly high vapour pressure at 273 K, it should be insoluble in water and the impurities present should be non-volatile.

The impure liquid along with little water is taken in a round bottomed flask which is connected to a boiler on one side and water condenser on the other side, the flask is kept in a slanting position so that no droplets of the mixture will enter into the condenser on the brisk boiling and bubbling of steam.

The mixture in the flask is heated and then a current of steam passed into it. The vapours of the compound mix up with the steam and escape into the condenser. The condensate obtained is a mixture of water and organic compound which can be separated. This method is used to recover essential oils from plants and flowers also in the manufacture of aniline and turpentine oil.

Question 11.
Explain about azeotropic distillation.
Answer:
The mixture of liquids that cannot be separated by fractional distillation can be purified by azeotropic distillation. The mixture are called azeotropes. These azeotropes are constant boiling mixture which distill as a single component at a fixed temperature for example ethanol and water in the ratio of 95.87: 4.13. In this method, the presence of a third component C₆H₆, CCl₄, ether, glycol glycerol which act as dehydrating agent depress the partial pressure of one component of azeotropic mixture and raises the boiling point of that component and thus the other component will distil over. Substance like C₆H₆, CCl₄ have low b.pt. and reduce the partial vapour pressure of alcohol more than that of water while substance like glycerol and glycol have high boiling point and reduce the partial vapour pressure of water more than that of alcohol.

Question 12.
Explain about thin layer chromatography.
Answer:
A sheet of glass is coated with a thin layer of adsorbent (cellulose, silica gel (or) Alumina). This sheet of glass is called chromplate or thin layer chromatography plate. After drying the plate, a drop of the mixture is placed just above one edge and the plate is then placed in a closed jar containing eluant (solvent) .

The eluant is drawn up the adsorbent layer by capillary action. The components of the mixture move up along with the eluent to different distances depending upon their degree of adsorption of each component of the mixture. It is expressed in terms of its retention factor (R_f) value.

The spots of coloured compounds are visible on TLC plate due to their original colour. The colourless compounds are viewed under Uy light or in another method using Iodine crystals or by using appropriate reagent.

Question 13.
An organic compound contains 69% carbon and 4.9% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this compound is subjected to complete combustion.
Answer:
Step I.
Calculation of mass of CO₂ produced
Mass of compound = 0.20 g
Percentage of carbon = 69 g

Mass of CO₂ formed = \(\frac{69 \times 44 \times(0.20 \mathrm{g})}{12 \times 100}\) = 0.506g

Step II.
Calculation of mass of H₂O produced
Mass of compound = 0.20 g
Percentage of hydrogen = 4.8%

Mass of H₂O formed = \(\frac{4.8 \times 18 \times(0.20 \mathrm{g})}{2 \times 100}\) = 0.0864 g

Question 14.
0.50 g of an organic compound was Kjeldahlished. The ammonia eolved was passed in 50cm³ of IN H₂SO₂. The residual acid required 60 cm³ of N/2 NaOH solution. Calculate the percentage of nitrogen in the compound.
Answer:
Step 1.
Calculation of volume of unused acid
Volume of NaOli solution required = 60 cm³
Normality of NaOH solution = \(\frac {1}{2}\) N
Normality of H₂SO₄ solution = \(\frac {1}{N}\)
Volume of unused acid can be calculated by applying normality equation

1 x V = \(\frac {1}{2}\) x 60 = 30cm³

Step II.
Calculation of volume of acid used
Volume of acid added = 50 cm³
Volume of unused acid = 30 cm³
Volume of acid used = (50 – 30) = 20 cm³

Step III.
Calculation of percentage of nitrogen
mass of compound = 0.05 g
Volume of acid used = 20 cm³
Normality of acid used = 1 N

percentage of nitrogen = \(\frac{1.4 \times 20 \times 1}{0.50}=56 \%\)

Question 15.
In a Dumas nitrogen estimation method, 0.30 g of an organic compound gave 50 cm3 of N₂ collected at 300 K and 715 mm Hg pressure. Calculate the percentage composition of nitrogen ¡n the compound. (‘apour pressure of water at 300 K is 15 mm Hg)
Answer:
P₁ = 715 – 15 = 700mm Hg, P₁ = 760 mm Hg
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)
T₁ = 300 K, T₂ = 273K, V₁ = 50 cm³, V₂ = ?
V₂ = \(\frac{700 \times 50 \times 273}{300 \times 760}\) = 41.9 cm³
% of N = \(\frac {28}{22400}\) x 41.9 x \(\frac {100}{W}\) = 17.46%

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Samacheer Kalvi 11th Chemistry Chapter 10 Chemical Bonding Textual Evaluation Solved

Samacheer Kalvi 11th Chemistry Chemical Bonding Multiple Choice Questions

Question 1.
In which of the following compounds does the central atom obey the octet rule?
(a) XeF₄
(b) AICI₃
(c) SF₆
(d) SO₂
Answer:
(d) SCl₂
Solution:

Hence in (d) SCl₂ octet rule is followed

Question 2.
In the molecule O_A = C = O_B the formal charge on O_A, C and O_B are respectively.
(a) – 1, 0, +1
(b) +1, 0, – 1
(c) – 2, 0, +2
(d) 0, 0, 0
Answer:
(d) 0, 0, 0

Formal charge of
Formal charge of C = 4 – \(\left( 0+\frac { 8 }{ 2 } \right)\) = 4 – 4 = 0

Question 3.
Which of the following is electron deficient?
(a) PH₃
(b) (CH₃)₂
(c) BH₃
(d) NH₃
Answer:
(c) BH₃
Solution:
– electron rich,
CH₃ – CH₃ – Covalent neutral molecule,
BH₃ – electron deficient

Question 4.
Which of the following molecule contain no π bond?
(a) SO₂
(b) NO₂
(c) CO₂
(d) H₂O
Answer:
(d) H₂O
Solution:

Water (H₂O) contains only σ bonds and no π bonds.

Question 5.
The ratio of number of sigma (σ) and pi (π) bonds in 2- butynal is …………..
(a) 8/3
(b) 5/3
(c) 8/2
(d) 9/2
Answer:
(d) 9/2
Solution:

no. of σ bonds = 8 [4C – H; 3C – C; 1C— O]
no.of π bonds = 3 [2C – C; 1C – O]
∴ratio = \(\frac { 8 }{ 3 }\)

Question 6.
Which one of the following is the likely bond angles of sulphur tetrafluoride molecule?
(a) 120°, 80°
(b) 109°. 28
(c) 90°
(d) 89°, 117°
Answer:
(d) 89°, 117°

Solution:
Normal bond angle in regular trigonal bipyramidal are 90° and 120°. Due to l.p – b.p repulsion, bond angle is reduced to 89°, 117° option (d).

Question 7.
Assertion: Oxygen molecule is paramagnetic.
Reason: It has two unpaired electron in its bonding molecular orbital.
(a) both assertion and reason are true and reason is the correct explanation of assertion
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) Both assertion and reason are false
Answer:
(c) assertion is true but reason is false
Correct statement: Oxygen molecule is paramagnetic
Correct Reason: It has two unpaired electrons in its antibonding molecular orbital.

Question 8.
According to Valence bond theory, a bond between two atoms is formed when ……………….
(a) fully filled atomic orbitals overlap
(b) half filled atomic orbitals overlap
(c) non-bonding atomic orbitals overlap
(d) empty atomic orbitals overlap
Answer:
(b) half filled atomic orbitals overlap

Question 9.
In CIF₃, NF₃ and BF₃ molecules the chlorine, nitrogen and boron atoms are …………………….
(a) sp³ hybridised
(b) sp³, sp³ and sp² respectively
(c) sp² hybridised
(d) sp³d, sp³ and sp² hybridised respectively
Answer:
(d) sp³d, sp³ and sp² hybridised respectively
Solution:
CIF₃ – sp³d hybridisation
NF₃ – sp³ hybridisation
BF₃ – sp² hybridisation

Question 10.
When one s and three p orbitais hybridise,
(a) four equivalent orbitais at 90⁰ to each other will be formed
(b) four equivalent orbitais at 109⁰ 28’ to each other will be formed.
(c) four equivalent orbitals, that are lying the same plane will be formed
(d) none of these
Answer:
(b) four equivalent orbitals at 109° 28′ to each other will be formed.

Question 11.
Which of these represents the correct order of their increasing bond order?
(a) C₂ < C₂^2- < O₂
(b) C₂^2- < C₂⁺ < O² < O₂^2-
(c) O₂^2- < O₂⁺ < O₂ < C₂^2-
(d) O₂^2- < C₂⁺ < O₂ < C₂^2-
Answer:
(d) O₂^2- < C₂⁺ < O₂ < C₂^2-
Solution:
bond order = (n_b – n_a)
bond order of O₂^2- = \(\frac { 1 }{ 2 }\) (8 – 6) = 1
bond order of C₂⁺ = \(\frac { 1 }{ 2 }\) (5 – 2) = 1.5
bond order of O₂ = \(\frac { 1 }{ 2 }\) (8 – 4) = 2
bond order of C₂^2- = \(\frac { 1 }{ 2 }\) (8 – 2) = 3

Question 12.
Hybridisation of central atom in PCl₅ involves the mixing of orbitais.
(a) s, p_x, p_y, d_x², d_x²-y²
(b) s, p_x, p_y, p_xy, d_x²-y²
(c) s, p_x, p_y, p_z, d_x²-y²
(d) p_x, p_y, p_xy, d_x²-y²
Answer:
(c) s, p_x, p_y, p_z, d_x²-y²
Solution:
PCl₅ – sp³d hybridisation s, p_x, p_y, p_z, d_x²-y²

Question 13.
The correct order of O – O bond length in hydrogen peroxide, ozone and oxygen is ……………..
(a) H₂O₂ > O₃ > O₂
(b) O₂ > O₃ > H₂O
(c) O₂ > H₂O₂ > O₃
(d) O₃ > O₂ > H₂O₂
Answer:
(b) O₂ > O₃ > H₂O₂
Solution:
The bond order for O₂, O₃ and H₂O₂ decreases in the order 2 > 1.5 > 1

Question 14.
Which one of the following is diamagnetic?
(a) O₂
(b) O₂^2-
(c) O₂²⁺
(d) None of these
Answer:
(b) O₂^2-
Solution:
O₂^2- is diamagnetic. Additional two electrons are paired in anti-bonding molecular orbits π*_2py and π*_2pz

Question 15.
Bond order of a species is 2.5 and the number of electrons are in its bonding molecular orbital is found to be 8. The no. of electrons in its anti-bonding molecular orbital is ………………….
(a) three
(b) four
(c) zero
(d) cannot be calculated form the given information.
Answer:
(a) three
Solution:
Bond order = \(\frac { 1 }{ 2 }\) (n_b – n_a)
2.5 = \(\frac { 1 }{ 2 }\) (8 – n_a)
⇒ 5 = 8
⇒ n_a = 8 – 5 = 3

Question 16.
Shape and hybridisation of IF₅ are ………….
(a) Trigonal bipyramidal, sp³d²
(b) Trigonal bipyramidal, sp³d
(c) Squai2e pyramidal, sp³d²
(d) Octahedral, sp³d²
Answer:
(c) Square pyramidal, sp³d²
Solution:
IF₅ – 5 bond pair + 1 lone pair
∴ hybridisation sp³d²

Question 17.
Pick out the incorrect statement from the following.
(a) sp³ hybrid orbitais are equivalent and are at an angle of 1 09°28’ with each other.
(b) dsp² hybrid orbitais are equivalent and bond angle between any two of them is 900.
(c) All five sp³d hybrid orbitais arc not equivalent. Out of these five sp³d hybrid orbitais, three are at an angle of 120°, remaining two are perpendicular to the plane containing the other three
(d) none of these
Answer:
(c) All five sp³d hybrid orbitals are not equivalent. Out of these five sp³d hybrid orbitals, three are at an angle of 120° remaining two are perpendicular to the plane containing the other three.

Question 18.
The molecules having same hybridisation, shape and number of lone pairs of electrons are ………………..
(a) SeF₄, XeO₂F₂
(b) SF₄, XeF₂
(c) XeOF₄, TeF₄
(d) SeCI₄, XeF₄
Answer:
(a) SeF₄, XeO₂F₂
Solution:
SeF₄, XeO₂F₂ – sp³d hybridisation
T – shaped, one lone pair on central atom.

Question 19.
In which of the following molecules / ions BF₃, NO₂^–, H₂O the centrai atom is sp² hybridised?
(a) NO₂^– and H₂O
(b) NO₂^– and H₂O
(c) BF₃ and NO₂
(d) BF₃ and NH₂^–
Answer:
(c) BF₃ and NO₂
Solution:
H₂O – Central atom sp³ hybridised
NO₂^– – Central atom sp² hybridised
BF₃ – Central atom sp² hybridised
NH₂^–  – Central atom sp³ hybridised

Question 20.
Some of the following properties of two species, NO₃^– and H₃O⁺ are described below. Which one of them is correct?
(a) dissimilar in hybridisation for the central atom with different structure.
(b) isostnictural with same hybridisation for the Central atom.
(c) different hybridisation for the central atom with same structure
(d) none of these
Answer:
(a) dissimilar in hybridisation for the central atom with different structure.
Solution:
NO₃^– – sp² hybridisation, planar
H₃O⁺ – sp³ hybridisation, pyramidal

Question 21.
The types of hybridisation on the five carbon atom from right to left in the, 2,3 pentadiene.
(a) sp³, sp², sp, sp², sp³
(b) sp³, sp, sp, sp, sp³
(c) sp², sp, sp², sp , sp³
(d) sp³, sp³, sp², sp³, sp³
Answer:
(a) sp³, sp², sp, sp², sp³
Solution:

Question 22.
XeF₂ is isostructural with ………..
(a) SbCl₂
(b) BaCl₂
(c) TeF₂
(d) ICl₂^–
Answer:
(d) ICl₂^–
Solution:
XeF₂ is isostructural with ICI₂^–

Question 23.
The percentage of s-character of the hybrid orbitais in methane, ethane, ethene and ethyne are respectively ………………..
(a) 25, 25, 33.3, 50
(b) 50, 50, 33.3, 25
(c) 50, 25, 33.3, 50
(d) 50, 25, 25. 50
Answer:
(a) 25, 25, 33.3, 50
Solution:

Question 24.
Of the following molecules, which have shape similar to carbon dioxide?
(a) SnCI₂
(b) NO₂
(c) C₂H₂
(d) All of these
Answer:
(c) C₂H₂
Solution:
CO₂ – Linear
C₂H₂ – Linear

Question 25.
According to VSEPR theory, the repulsion between different parts of electrons obey the order …………………
(a) l.p – l.p > b.p – b.p > l.p – b.p
(b) b.p – b.p > b.p – 1.p > l.p – b.p
(c) l.p – l.p > b.p – l.p > b.p – b.p
(d) b.p – b.p > l.p – l.p > b.p – l.p
Answer:
(c) l.p – l.p > b.p – l.p > b.p – b.p

Question 26.
Shape of CIF₃ is ……………………..
(a) Planar triangular
(b) Pyramidal
(c) ‘T’ Shaped
(d) none of these
Answer:
(c) ‘T’ Shaped
Solution:
dF₃ – sp³d hybridisation

Question 27.
Non- Zero dipole moment is shown by …………………
(a) CO₂
(b) p – dichlorobenzene
(c) carbon tetrachloride
(d) water
Answer:
(d) water
Solution:

Question 28.
Which of the following conditions is not correct for resonating structures?
(a) the contributing structure must have the same number of unpaired electrons.
(b) the contributing structures should have similar energies.
(c) the resonance hybrid should have higher energy than any of the contributing structure.
(d) none of these
Answer:
(c) the resonance hybrid should have higher energy than any of the contributing structure.
Solution:
Correct statement is – the resonance hybrid should have lower energy than any of the contributing structure.

Question 29.
Among the following, the compound that contains, ionic, covalent and coordinate linkage is …………………
(a) NH₄Cl
(b) NH₃
(c) NaCl
(d) none of these
Answer:
(a) NH₄Cl

Question 30.
CaO and NaCl have the same crystal structure and approximately the same radii. If U is the lattice energy of NaCl, the approximate lattice energy of CaO is ………….
(a) U
(b) 2U
(c) U/2
(d) 4U
Answer:
(d) 4U

Samacheer Kalvi 11th Chemistry Chemical Bonding Short Answer Questions.

Question 31.
Define the following

1. Bond order
2. Hybridisation
3. a- bond

Answer:
1. Bond order:
Bond order
The number of bonds formed between the two bonded atoms in a molecule is called bond order.

2. Hybridisation:
It is a process of mixing of atomic orbitais of the same atom with. comparable energy to form equal number of new equivalent orbitais with same energy. The resultant orbitais are called hybridised orbitais and they possess maximum symmetry and definite orientation in space so as to minimise the force of repulsion between their electrons.

3. σ – bond:
When two atomic orbitais overlap linearly along the axis, the resultant bond is called sigma (σ) bond.

Question 32.
What is a pi bond?
Answer:
Pi – bond:
When two atomic orbitals overlap sideways, the resultant covalent bond is called a pi (π) bond.

Question 33.
In CH₄, NH₃ and H₂O, the central atom undergoes sp³ hybridisation-yet their bond angles are different, why?
Answer:
1. In CH₄, NH₃ and H₂O the central atom undergoes sp³ hybridisation. But their bond angles are different due to the presence of lone pair of electrons.

2. It can be explained by VSEPR theory. According to this theory, even though the hybridisation is same, the repulsive force between the bond pairs and lone pairs arc not same.

3. Bond pair-Bond pair < Bond pair – Lone pair < Lone pair – Lone pair So due to the varying repulsive force the bond pairs and lone pairs are distorted from regular geometry and organise themselves in such a way that repulsion will be minimum and stability will be maximum.

4. In case of CH₄, there are 4 bond pairs and no lone pair of electrons. So it remains in its regular geometry. i.e., tetrahedral with bond angle 109° 28’.

5. H₂O has 2 bond pairs and 2 lone pairs. There is large repulsion between lp – lp. Again repulsion between lp – bp is more than that of 2 bond pairs. So 2 bonds are more restricted to form inverted V shape (or) bent shape molecule with a bond angle of 104° 35’.

6. NH₃ has 3 bond pairs and 1 lone pair. There is repulsion between lp – bp. So 3 bonds are more restricted to form pyramidal shape with bond angle equal to 107° 18’.

Question 34.
Explain sp² hybridisation in BF₃
Answer:
1. sp² hybridisation in boron trifluoride – Boron atom – B. Electronic configuration [H₂]2s²2p².

2. In boron, the s orbital and two p orbitals in the valence shell hybridises to generate three equivalent sp² orbitais. These 3 orbitaIs lie in the same xy plane and the angle between any two orbitals is equal to 120°.

3. The 3 sp² hybridised orbitais of boron now overlap with the 2p_z orbitais of fluorine (3 atoms). This overlap takes place along the axis.

Question 35.
Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O₂ is paramagnetic.
Answer:

1. Electronic configuration of O atom is is 1s² 2s² 2P⁴
2. Electronic configuration of O, molecule is
   
3. Bond order =
4. Molecule has two unpaired electrons, hence it is paramagnetic.

Question 36.
Draw MO diagram of CO and calculate its bond order.
Answer:

1. Electronic configuration of C atom: 1s² 2s² 2p²
   Electronic configuration of O atom: 1s² 2s² 2p⁴
2. Electronic configuration of CO molecule is: σ1s² σ*1s² σ2s² σ*2² π2p_y² π2p_z² σ2p_x²
3. Bond order
4. Molecule has no unpaired electron, hence it is diamagnetic.

Question 37.
What do you understand by Linear combination of atomic orbitais in MO theory.
Answer:
Linear combination of atomic orbitais (LCAO):
1. The wave functions for the molecular orbitais can be obtained by solving Schrodinger wave equation for the molecule. Since solving Schrodinger wave equation is too complex, a most common method linear combination of atomic orbitais (LCAO) is used to obtain wave function for molecular orbitals.

2. Atomic orbitais are represented by wave functions ψ. Consider two atomic orbitals represented by the wave functions ψ_A and ψ_B with comparable energy that combines to form two molecular orbitals.

3. One is bonding molecular orbitai (ψ bonding) and the other is anti-bonding molecular orbital (ψ anti-bonding).

4. The wave function for molecular orbitais, ψ_A and ψ_B can be obtained by the LCAO as shown below:
ψ_bonding = ψ_A + ψ_B
ψ_anti-bonding = ψ_A – ψ_B

5. The formation of bonding molecular orbital can be considered as the result of constructive interference of the atomic orbitais and the formation of anti-bonding molecular orbital can be the result of the destructive interference of the atomic orbitais.

6. The formation of two molecular orbitals from two is orbitals is show below.

Constructive interaction:
The two is orbitals are in phase and have the same signs.

Destructive interaction:
The two is orbitals are out of phase and have opposite signs

Question 38.
Discuss the formation of N₂ molecule using MO Theory.
Answer:

1. Electronic configuration of N atom 1s² 2s² 2p³.
2. Electronic configuration of N, molecule is: σ1s² σ*1s² σ2s² σ*2² π2p_y² π2p_z² σ2p_x²
3. Bond order
4. Molecule has no unpaired electrons hence, it is diamagnetic.

Question 39.
What is dipole moment?
Answer:
1. The polarity of a covalent bond can be measured in terms of dipole moment which is defined as: µ = q x 2d, where µ is the dipole moment, q is the charge, 2d is the distance between the two charges.

2. The dipole moment is a vector quantity and the direction of the dipole moment points from the negative charge to positive charge.

3. The unit of dipole moment is Coulomb metre (C m). It is usually expressed in Debye unit (D).

4. 1 Debye = 3.336 x 10^-30 Cm

Question 40.
Linear form of carbon dioxide molecule has two polar bonds. yet the molecule has zero dipole moment, why?
Answer:

1. The linear form of carbon dioxide has zero dipole moment, even though it has two polar bonds.
2. In CO₂, there are two polar bonds [C = O], which have dipole moments that are equal in magnitude but have opposite direction.
3. Hence the net dipole moment of the CO₂ is µ = µ₁ + µ₂ = µ₁ +( – µ₁) = 0
4. 
5. In this case

Question 41.
Draw the Lewis structures for the following species.

1. NO₃^–
2. SO₄^2-
3. HNO₃
4. O₃

Answer:
1. NO₃^–

2. SO₄^2-

3. HNO₃

4. O₃

Question 42.
Explain the bond formation in BeCl₂ and MgCl₂. BeCl₂ bond formation:
Answer:
1. Electronic confiuration of Be(Z = 4) is 1s² 2s² and electronic configuration of Cl (Z = 17) is 1s² 2s² 2p⁶ 3s² 3p⁵.

2. Beryllium has 2 electrons in its valence shell and chlorine atoms (2) have 7 electrons in their valence shell.

3. By losing two electrons, Beryllium attains the inert gas configuration of Helium and becomes a dipositive cation, Be²⁺ and each chlorine atom accepts one electron to become (Cl^–) uninegative anion and attains the stable electronic configuration of Argon.

4. Then Be²⁺ combine with 2Cl^– ions to form an ionic crystal in which they are held together by electrostatic attractive forces.

5. During the formation of 1 mole of BeCl₂, the amount of energy released is – 468 kJ/mol. This favours the formation of BeCl, and its stabilisation.

MgCI₂ bond formation:
1. Electronic configuration of Mg (z = 12) is 1s² 2s² 2p⁶ 3s².
Electronic configuration of Cl (z = 17) is 1s² 2p⁶ 3p⁶ 3p⁵

2. Magnesium has 2 electrons in its valence shell and chlorine has 7 electrons in its valence shell.

3. By losing two electrons, magnesium attains the inert gas configuration of Neon and becomes a dipositive cation (Mg²⁺) and two chlorine atoms accept these electrons to become two uninegative anions [2Cl^–] by attaining the stable inert gas configuration of Argon.

4. These ions, Mg²⁺ and 2Cl^– combine to form an ionic crystal in which they are held together by electrostatic attractive forces.

5. The energy released during the formation of 1 mole of MgCl₂ is – 783 kJ/mole. This favours the formation of MgCI₂ and its stabilisation.

Question 43.
Which bond is stronger or π? Why?
Answer:
1. Sigma bonds (σ) are stronger than Pi bonds (π). Because, sigma bonds are formed from bonding orbitals directly between the nuclei of the bonding atoms, resulting in greater overlap and a strong sigma bond (axial overlapping).

2. π bonds results from overlap of atomic orbitals that are in contact through two areas of overlap (lateral overlapping). Pi bonds are more diffused bonds than sigma bonds.

Question 44.
Define bond energy.
Answer:
Bond energy:
Bond energy (or) Bond enthalpy is defined as the minimum amount of energy required to break one mole of a bond in molecules in their gaseous state. The unit of bond energy is kJ mol^-1