Maharashtra Board 11th OCM Important Questions Chapter 7 Business Environment

Balbharti Maharashtra State Board 11th OCM Important Questions Chapter 7 Business Environment Important Questions and Answers.

Maharashtra State Board 11th Commerce OCM Important Questions Chapter 7 Business Environment

Select the correct option and rewrite the sentence

Question 1.
……………….. indicates the importance of the business environment.
(a) Identifying opportunities and threats
(b) Identifying firm’s strength and weakness
(c) both (a) and (b)
Answer:
(c) both (a) and (b)

Question 2.
……………….. is an example of the social environment.
(a) Gross Domestic Product
(b) Consumer Protection, Act
(c) Customs and beliefs
Answer:
(c) Customs and beliefs

Maharashtra Board 11th OCM Important Questions Chapter 7 Business Environment

Question 3.
Liberalisation means ………………..
(a) reduced government controls and restrictions
(b) integration of economies
(c) policy of planned disinvestment
Answer:
(a) reduced government controls and restrictions

Question 4.
……………….. explains the impact of Government policy changes on business and industry.
(a) More demanding customers
(b) License system
(c) More industries for public sector
Answer:
(a) More demanding customers

Question 5.
……………….. environment consist of the methods and techniques adopted for production.
(a) Legal
(b) Political
(c) Technological
Answer:
(c) Technological

Question 6.
The policies decided by shareholders, managers, labour unions and the employees are known as ………………..
(a) Internal environment factors
(b) External environment factors
(c) International factors
Answer:
(a) Internal environment factors

Question 7.
……………….. is very powerful because powers such as policy making, law making, budget approving, etc. are in its hands.
(a) Government
(b) Judiciary
(c) Legislature
Answer:
(c) Legislature

Question 8.
……………….. economy is known as free market economy.
(a) Socialist
(b) Capitalist
(c) Mixed
Answer:
(b) Capitalist

Question 9.
……………….. does not involve an equal division of existing wealth among the people.
(a) Capitalist Economy
(b) Socialist Economy
(c) Mixed Economy
Answer:
(b) Socialist Economy

Question 10.
Globalisation includes ………………..
(a) Minimising high import tariffs
(b) Introduction of selective competition in the reserved area
(c) Reduction in physical restrictions on import and import duties.
Answer:
(a) Minimising high import tariffs

Question 11.
Changes in the fiscal policy of the Government is an aspect of ……………….. environment.
(a) Political
(b) Social
(c) Economic
Answer:
(c) Economic

Question 12.
Changes in the judicial system is an aspect of ……………….. environment.
(a) Economic
(b) Social
(c) Legal
Answer:
(c) Legal

Question 13.
Privatisation implies reduction in the role of ……………….. sector.
(a) Public
(b) Private
(c) Joint
Answer:
(a) Public

Question 14.
Level of literacy is an aspect of ……………….. environment.
(a) Economic
(b) Social
(c) Political
Answer:
(b) Social

Maharashtra Board 11th OCM Important Questions Chapter 7 Business Environment

Question 15.
Changes in the ruling government is an aspect of ……………….. environment.
(a) Economic
(b) Social
(c) Political
Answer:
(c) Political

Match the pairs

Question 1.

Part A Part B
(a) Globalisation (1) Reduced role of Public sector
(b) Privatisation (2) Digital India
(c) Liberalisation (3) Competition Act, 2002
(d) Technological Environment (4) Mining of coal
(e) Natural Environment (5) Integration of world economies
(6) Increased role of public sector
(7) Exercising controls
(8) Occupation of people
(9) Differentiation of economies
(10) Removing unnecessary controls

Answer:

Part A Part B
(a) Globalisation (5) Integration of world economies
(b) Privatisation (1) Reduced role of Public sector
(c) Liberalisation (10) Removing unnecessary controls
(d) Technological Environment (2) Digital India
(e) Natural Environment (4) Mining of coal

Question 2.

Part A Part B
(a) Internal Environment (1) Stability and peace in the country
(b) Social Environment (2) Value System
(c) Political Environment (3) Monetary Policy
(d) Economic Environment (4) Indian Contract Act, 1872 (2017)
(e) Legal Environment (5) Customs and Traditions
(6) Digital India

Answer:

Part A Part B
(a) Internal Environment (2) Value System
(b) Social Environment (5) Customs and Traditions
(c) Political Environment (1) Stability and peace in the country
(d) Economic Environment (3) Monetary Policy
(e) Legal Environment (4) Indian Contract Act, 1872 (2017)

Question 3.

Part A Part B
(a) Privatisation (1) Management
(b) Internal Environment (2) National economy
(c) Liberalisation (3) Values and demographic trends
(d) Political Environment (4) Disinvestment
(e) Globalisation (5) Delicensing
(f) Legal Environment (6) World Economy
(g) Social Environment (7) National Income Distribution
(h) Economic Environment (8) Judicial System

Answer:

Part A Part B
(a) Privatisation (4) Disinvestment
(b) Internal Environment (1) Management
(c) Liberalisation (5) Delicensing
(d) Political Environment (2) National economy
(e) Globalisation (6) World Economy
(f) Legal Environment (8) Judicial System
(g) Social Environment (3) Values and demographic trends
(h) Economic Environment (7) National Income Distribution

Write a word or a term or a phrase which can substitute each of the following statements

Question 1.
Controllable factor of the business environment
Answer:
Internal environment

Question 2.
The environment of business that includes economic condition, policies and system, basic philosophy, national income, money supply, trade cycle economic development.
Answer:
Economic environment

Maharashtra Board 11th OCM Important Questions Chapter 7 Business Environment

Question 3.
The environment of business that includes stability and peace in the country, specific attitudes of selected representative.
Answer:
Political environment

Question 4.
The environment of business which consist of factors such as climate, minerals, soil, landforms, rivers and oceans etc.
Answer:
Natural environment

Question 5.
The environment of business which consist of the methods and techniques adopted for production.
Answer:
Technological environment

Question 6.
The environment of business which regulates and controls within the legal framework of a country.
Answer:
Legal environment

Question 7.
The process of eliminating unnecessary controls and restrictions for smooth functioning of business.
Answer:
liberalization

Question 8.
The process of integration of national economy with world through cross country information, ideas, technologies, goods, services and people.
Answer:
Globalization

Question 9.
The process of reducing the involvement of state or public sectors by involving of private sector in economic activities.
Answer:
Privatization.

State whether the following statements are True or False

Question 1.
Internal factors of business environment are uncontrollable factors.
Answer:
False

Question 2.
Business and Environment are independent.
Answer:
False

Question 3.
Social Environment is not important for business.
Answer:
False

Maharashtra Board 11th OCM Important Questions Chapter 7 Business Environment

Question 11.
Business Environment consists of internal as well as external factors.
Answer:
True

Question 4.
All business from are affected by the government and its policies.
Answer:
True

Question 5.
Liberalization of Indian economy resulted into easy entry of foreign business in India.
Answer:
True

Question 6.
Privatisation is an effective tool for restriction and reforming the public sector.
Answer:
True

Find the odd word out

Question 1.
Legislature, Government Shareholders.
Answer:
Shareholders

Question 2.
Social, Employees, Legal.
Answer:
Employees

Question 3.
Judiciary, Social values, Traditions.
Answer:
Judiciary

Question 4.
Value system, Human Resources, Marketing Resources.
Answer:
Marketing Resources

Question 5.
Liberalization, Socialization, Globalization.
Answer:
Socialization,

Question 6.
Technological, Weather, Climate.
Answer:
Technological

Question 7.
Social aspects, Economical aspect, Traditions.
Answer:
Economical aspect

Question 8.
Capitalist Economy, Mixed Economy, Indian Economy.
Answer:
Indian Economy

Maharashtra Board 11th OCM Important Questions Chapter 7 Business Environment

Question 9.
Indian Contract Act 1872, Indian Partnership Act 1932, Industrial Dispute Act 1947.
Answer:
Indian Partnership Act, 1932

Question 10.
Natural Environment, Economic Environment, Political Environment.
Answer:
Natural Environment.

Complete the sentence

Question 1.
The business environment constitute threats to the firm and offers ………………… for potential market.
Answer:
opportunities

Question 2.
Every society develops its own ………………… which means how the members of that society between and interact with each other.
Answer:
culture

Question 3.
Indian political system comprises of three vital institution like …………………, ………………… and Judiciary.
Answer:
Legislature, Government

Question 4.
………………… changes are fast and to keep pace with it, business should be ever alert.
Answer:
Technological

Question 5.
All nation would be interdependent in the ………………… economy.
Answer:
Global.

Select the correct option

Question 1.
Which analysis will enable managers to deal with business challenges?
(Environmental / Data / Political)
Answer:
Environmental

Question 2.
Which factor of Business Environment are regarded as uncontrollable?
(Internal / External)
Answer:
External

Question 3.
Trade and commerce between two nations are dependent on which factors?
(Social / Geographic / Technological)
Answer:
Geographic

Maharashtra Board 11th OCM Important Questions Chapter 7 Business Environment

Question 4.
Which is the most powerful institution in the Indian Political system?
(Legislature / Government / Judiciary)
Answer:
Legislature

Question 5.
In which economic system factors of productions are organized .managed and oriented by government?
(Capitalist / Socialist / Mixed)
Answer:
Socialist

Question 6.
Which economic system is followed in India?
(Capitalist / Socialist / Mixed)
Answer:
Mixed

Question 7.
Which of the following has created on economically interdependent international environment?
(Liberalisation / Privatisation / Globalisation)
Answer:
Globalisation

Correct the underline word and rewrite the following sentence

Question 1.
Internal factors of Business Environment are uncontrollable.
Answer:
External factors of Business Environment are uncontrollable.

Question 2.
External factor of business are controllable.
Answer:
Internal factor of business are controllable.

Question 3.
Vision, mission and objectives of business are related to External factors.
Answer:
Vision, mission and objectives of business are related to Internal factors.

Question 4.
Social trends and social values are part of economic environment.
Answer:
Social trends and social values are part of social environment.

Question 5.
Factors like climate, minerals, soil, river, etc. are elements of political environment.
Answer:
Factors like climate, minerals, soil, river, etc. are elements of natural environment.

Question 6.
Socialist Economy is also known as free market economy.
Answer:
Capitalist Economy is also known as free market economy.

Question 7.
GST is part of political environment.
Answer:
GST is part of legal environment.

Question 8.
Privatisation has resulted into easy of foreign business in India.
Answer:
Liberalisation has resulted into easy of foreign business in India.

Question 9.
Globalisation is an effective tool for restricting and reforming the public sector.
Answer:
Privatisation is an effective tool for restricting and reforming the public sector.

Question 10.
Liberalisation means integration of natural economy with world economic.
Answer:
Globalisation means integration of natural economy with world economic.

Study the following case/situation and express your opinion.

1. Just after declaration of Lok Sabha 2014 result, the Bombay Stock Exchange Sensex rose by 1000 points in a day.

Question 1.
Identify the environmental factor which led to this rise giving reason.
Answer:
Rise in stock market index (sensex) is economic environment. But it has done due to political environment (Lok Sabha Election Results)

Question 2.
Explain any two other dimensions of the environment.
Answer:
Two other dimensions of the environment are (a) Legal environment and (b) Technological environment

2. Government of India is seriously thinking to allow oil marketing under public sector, by fixing their own price for petroleum products.

Question 1.
Name the economic reform related to this decision.
Answer:
Privatization is the economic reform related to this decision.

Question 2.
Name other two reforms recommended in new economic policy.
Answer:
The other two reforms recommended in new Economic policy are Liberalization and Globalization.

3. Today it is common to see computerized information Kiosks and Internet / Word Wide Web multimedia pages highlighting the virtues of products.

Question 1.
Name the dimension of business environment highlighted in this statement.
Answer:
Technological environment

Maharashtra Board 11th OCM Important Questions Chapter 7 Business Environment

4. An advertisement of baby product giving information to the potential buyers that “Mother Milk is the best”.

Question 1.
Name the component of business environment.
Answer:
Legal Environment.

5. Zero Defect Zero Effect(ZED) is a slogan coined by the PM of India, with reference to the Make in India initiative launched by the government. The Zed model aims to make India the manufacturer of the world and present the products in the global market.

Before the. initiative was launched, the limits of Foreign Direct Investment(FDI) in various sector had been released. The application for licenses was made available Online and various other norms and production were also considered by the government.

Question 1.
From the above, identify the major components of New Economic Policy of 1991.
Answer:
Liberalization and Globalization.

Question 2.
In the light of the above discussion explain the component of NEP 1991.
Answer:

  1. Liberalization : It refers to the process of eliminating unnecessary controls and restrictions for smooth functioning of business.
  2. Privatization : It refers to reduce the involvement of state or public sectors by involving of private sector in economic activities.
  3. Globalization : It refers to integration of national economy and societies through cross country flows of information, ideas, technologies, goods, services, capital, finance and people.

6. It is interesting to know that the menu items of multinational food chains are customized to suit the general palates of the local people in the region. When it started business in India in the year 1996, the company went through a complete location strategy. Me Donald’s changed its product menu to accommodate the vegetable burger given the large population. It priced the products by close to 15%. Also, the Mc-Aloo Tikki burger ‘is not available anywhere but in the Indian outlets of Me Donalds.

Question 1.
Identify the relevant feature of business environment discussed in the case above.
Answer:
Relativity is the feature of business environment discusses in the case above.

Question 2.
State the other features of business environment.
Answer:

  1. The features of business environment are:
  2. Determining opportunities and threats
  3. Identifying firm’s strength and weakness
  4. Continuous learning
  5. Image Building
  6. Meeting Competition
  7. Giving Direction for growth

7. Ultra paint company which is manufacturing paints has been enjoying a prominent market position as it manufactured best quality paints, made timely payment of taxes to government. It assembled various inputs like finance, machines, raw material, etc. from its environment. But since last year it has been dumping its untreated possessions waste on the river bank which has created many health problems for the people. As a result, the court passed an order to seal the manufacturing unit of the company.

Question 1.
State the importance of business environment highlighted in the above case.
Answer:
Business environment helps in tapping useful resources.

Maharashtra Board 11th OCM Important Questions Chapter 7 Business Environment

Question 2.
Identify any two dimension of business environment mentioned in the above case and comment.
Answer:

  1. Social environment : “It has been dumping untreated possession of waste on river bank problems” for people. This shows that it is at the cost of human life as their life is affected.
  2. Legal Environment : “The court passed on order to seal the company”. It shows that through court interference the company was sealed.

Question 3.
State any two values which have been overlooked by the company.
Answer:
Company overlooked Environment protection, concern for human life, respect for law and order and social responsibility.

Maharashtra Board 11th OCM Important Questions Chapter 8 Introduction to Management

Balbharti Maharashtra State Board 11th OCM Important Questions Chapter 8 Introduction to Management Important Questions and Answers.

Maharashtra State Board 11th Commerce OCM Important Questions Chapter 8 Introduction to Management

Select the correct option and rewrite the sentences

Question 1.
Management is a …………………. process.
(a) stable
(b) continuous
(c) one time
Answer:
(b) continuous

Question 2.
Formulating the plan is the main function of …………………. level management.
(a) top
(b) middle
(c) lower
Answer:
(a) top

Maharashtra Board 11th OCM Important Questions Chapter 8 Introduction to Management

Question 3.
Important, decisions in the business are made by ………………….
(a) management
(b) administration
(c) organisation.
Answer:
(b) administration

Question 4.
Management is a profession because it has ………………….
(a) specialised knowledge
(b) training facilities
(c) code of conduct.
Answer:
(a) specialised knowledge

Question 5.
According to views expressed by different management authorities, management and administration are ………………….
(a) synonymous
(b) different
(c) synonymous and different.
Answer:
(c) synonymous and different.

Question 6.
Management is an …………………. to co-ordinate groun efforts attaining the cherished goals of the business.
(a) activity
(b) process
(c) objective.
Answer:
(a) activity

Question 7.
…………………. is the basic and motivating factor of management.
(a) Human element
(b) Capital
(c) Direction.
Answer:
(a) Human element

Question 8.
Top management does …………………. and …………………. function.
(a) supervisory
(b) operative
(c) determination and administration.
Answer:
(c) determination and administration.

Question 9.
Management is a …………………. activity.
(a) group
(b) individual
(c) social.
Answer:
(a) group

Question 10.
Management is a …………………. science.
(a) social
(b) physical
(c) natural.
Answer:
(a) social

Question 11.
There are …………………. levels of management in a organisation.
(a) two
(b) four
(c) three.
Answer:
(c) three.

Question 12.
Management means to ………………….
(a) order
(b) co-ordinate
(c) help.
Answer:
(b) co-ordinate

Question 13.
Management motivates the emplovees to ………………….
(a) work hard
(b) take initiative
(c) earn more.
Answer:
(a) work hard

Question 14.
Management ensures the optimum use of ………………..
(a) blackmoney
(b) resources
(c) international market.
Answer:
(b) resources

Match the pairs

Question 1.

Part A Part B
(a) Management (1) Administration function
(b) Policy Making (2) Human knowledge
(c) Thinking Function (3) Top level management
(d) Doing Function (4) Lower level management
(e) Top level management (5) Middle level management
(6) Management function
(7) Knowledge of Administration
(8) Board of Directors

Answer:

Part A Part B
(a) Management (2) Human knowledge
(b) Policy Making (3) Top level management
(c) Thinking Function (1) Administration function
(d) Doing Function (6) Management function
(e) Top level management (8) Board of Directors

Maharashtra Board 11th OCM Important Questions Chapter 8 Introduction to Management

Question 2.

Part A Part B
(a) Management (1) Regular practice
(b) Management as an Art (2) Can be seen
(c) Top Level Management (3) Administration
(d) Management as a Profession (4) Universal in nature
(e) Lower Level Management (5) Code of conduct
(6) Evaluate performance
(7) Foreman
(8) Selective
(9) Directing subordinates to complete task
(10) Cause effect relationship

Answer:

Part A Part B
(a) Management (4) Universal in nature
(b) Management as an Art (1) Regular practice
(c) Top Level Management (6) Evaluate performance
(d) Management as a Profession (5) Code of conduct
(e) Lower Level Management (7) Foreman

Question 3.

Part A Part B
(a) Systematic Body of knowledge (1) Top level management
(b) Lower level management (2) Management as an art.
(c) Management as universal process (3) Board of Directors.
(d) Management as profession (4) Practice oriented
(e) Direction and control. (5) Service motive
(6) Technical skills
(7) Applicable to all organization.
(8) Management as a science.
(9) Applicable to business organization

Answer:

Part A Part B
(a) Systematic Body of knowledge (8) Management as a science.
(b) Lower level management (6) Technical skills
(c) Management as universal process (7) Applicable to all organization.
(d) Management as profession (5) Service motive
(e) Direction and control. (1) Top level management

Give one word / phrase / term

Question 1.
The activity which aims at achieving a definite goal.
Answer:
Management

Question 2.
Level of management requiring analytical or conceptual skill.
Answer:
Top level management

Question 3.
Another name for lower level management.
Answer:
Supervisory or Operational level

Question 4.
Level of management requiring human skills.
Answer:
Middle level management

Question 5.
Level of management which performs administrative functions.
Answer:
Top level management

Question 6.
Level of management that takes policy decision.
Answer:
Top level management

Question 7.
The highest level of management in the hierarchy.
Answer:
Top level management

Question 8.
The level which executes the policy.
Answer:
Middle level management

Question 9.
The level of management which requires few people.
Answer:
Top level management

Question 10.
The level of management which requires large number of persons.
Answer:
Lower level management

Question 11.
Levels of management which consists of supervisors and foremen.
Answer:
Lower level management

Maharashtra Board 11th OCM Important Questions Chapter 8 Introduction to Management

Question 12.
Level of management which consists of departmental managers.
Answer:
Middle level management

Question 13.
Level of management which consists of chairman and board of directors.
Answer:
Top level management

Question 14.
State the level of management in which Class-I executives are involved.
Answer:
Top level management

Question 15.
The art of directing and inspiring people.
Answer:
Management

Question 16.
The application of skill and knowledge to bring the desired result.
Answer:
Management as an art

Question 17.
Level of management that evaluates the performance of various departments.
Answer:
Top level management.

State True or False

Question 1.
There are no separate levels of management in an organization.
Answer:
False

Question 2.
Management aims at achieving goals for its organization.
Answer:
True

Question 3.
Management is an universal process.
Answer:
True

Question 4.
Management aims at achieving goals for its organization.
Answer:
True

Question 5.
Management is a profession.
Answer:
True

Question 6.
Management is a continuous process.
Answer:
True

Question 7.
Management aims at achieving business objectives.
Answer:
True

Question 8.
Objective of management is to obtain minimum output and maximum input.
Answer:
False

Question 9.
Management has to maintain good relations with all.
Answer:
True

Question 10.
Management is a perfect art.
Answer:
False

Maharashtra Board 11th OCM Important Questions Chapter 8 Introduction to Management

Question 11.
Management is neither a pure science nor a fine art.
Answer:
True

Question 12.
Profession do not have any ethics or code of conduct.
Answer:
False

Question 13.
Board of Directors are the part of lower management.
Answer:
False

Question 14.
The top level management is administrative in nature.
Answer:
True

Question 15.
The managers receive orders and instructions from lower level of management.
Answer:
False

Question 16.
Middle level management interprets and communicates the policies to top level management.
Answer:
False

Question 17.
Management is an art, a science and a profession.
Answer:
True

Question 18.
We cannot make any changes in the field of management.
Answer:
False

Question 19.
The professional manager does not need special knowledge and skill.
Answer:
False

Question 20.
Management is required to conduct business as well as non-business activity.
Answer:
True

Question 21.
Management helps in using the resources properly.
Answer:
True

Question 22.
Management is required even for a very small business.
Answer:
True

Question 23.
Management is an art only.
Answer:
False

Question 24.
Management cannot be considered as a business.
Answer:
True

Question 25.
Top level management performs the function of planning.
Answer:
True

Question 26.
Lower level management is engaged in the policy making.
Answer:
False

Question 27.
Levels of management are done according to the size of the organization.
Answer:
True

Find the odd one

Question 1.
Characteristics of Management Goal-oriented, Group Activity, Tangible, Continuous process.
Answer:
Tangible.

Complete the sentences

Question 1.
Administration is the involvement of ………………. management.
Answer:
Top

Question 2.
………………. deals with setting, seeking and reaching objectives.
Answer:
Management

Question 3.
Management has ………………. objectives.
Answer:
pre-determined

Maharashtra Board 11th OCM Important Questions Chapter 8 Introduction to Management

Question 4.
Supervisor level of management is directly linked to the ………………. functions of the firm.
Answer:
Routine

Question 5.
Lower level management is also known as ………………. management.
Answer:
Supervisory

Question 6.
………………. is an art of getting things done through others.
Answer:
Management

Question 7.
Generally there are ………………. levels of management in the organisation.
Answer:
Three

Question 8.
Management is ………………. oriented.
Answer:
Goal

Question 9.
Management is a ………………. science.
Answer:
Social

Question 10.
At the ………………. level of management, personnel are few.
Answer:
Top

Question 11.
Management means to ……………….
Answer:
Co-ordinate

Question 12.
Management has to satisfy unlimited want through limited ……………….
Answer:
Means

Question 13.
Management is an ………………. of getting work done through others.
Answer:
Art

Question 14.
Management is co-ordination of ………………. and physical resources.
Answer:
Human

Question 15.
A professional charges ………………. for his services.
Answer:
Fees

Question 16.
Management is done to achieve ……………….
Answer:
Goal/objective

Question 17.
Management is getting things done through ……………….
Answer:
Others

Question 18.
The word management is derived from Italian word ……………….
Answer:
Maneggiare

Question 19.
Implementation of policies is the function of ………………. level management.
Answer:
Lower

Question 20.
Every profession has a ………………. of conduct.
Answer:
Code

Question 21.
Management is derived from ………………. words.
Answer:
Two Latin.

Select the correct option

Question 1.
(Levels of Management, Lower Level Management, Management, Code of conduct, Board of Directors).

Part A Part B
(1) Continuous Process —————
(2) ————— Top Level Management
(3) Top, Middle, Lower —————
(4) Profession —————
(5) ————- Lower Level Management

Answer:

Part A Part B
(1) Continuous Process Management
(2) Board of Directors Top Level Management
(3) Top, Middle, Lower Levels of Management
(4) Profession Code of conduct
(5) Supervisory Level Lower Level Management

Answer in one sentence

Question 1.
What is Italian word for ‘manage’.
Answer:
Italian word for ‘manage’ is ‘maneggiare’ meaning to handle.

Maharashtra Board 11th OCM Important Questions Chapter 8 Introduction to Management

Question 2.
Name the executive related to top level management.
Answer:
Executives related to top level management are – Board of Directors, President, Chief Executive Officer, Directors, etc.

Question 3.
Name the executives related to middle level management.
Answer:
Executives related to middle level management are Finance manger, Production manager, Sales manager, etc.

Correct the underlined word and rewrite the following sentences

Question 1.
Management is tangible in nature.
Answer:
Management is intangible in nature.

Question 2.
Management is one time process.
Answer:
Management is continuous process.

Question 3.
Administration is an art of getting things done through others.
Answer:
Management is an art of getting things done through others.

Question 4.
Management is static in character.
Answer:
Management is dynamic in character.

Question 5.
Top Level Management decides short term objectives of the organization.
Answer:
Top Level Management decides long term objectives of the organization.

Question 6.
Finance Manager is an example of lower level management.
Answer:
Superintendent is an example of lower level management.

Question 7.
Drawing, Cycling is a science.
Answer:
Drawing, Cycling is an art.

Study the following case/situation and express your opinion

1. Shankar Patil is an advocate and he is running his profession in a very good manner. His younger brother Prashant Patil has done very beautiful and artistic decoration of his brother’s office with seating arrangement and symbol of Justice as his hobby. Vishwas Patil, older brother of Shankar Patil is a science graduate and doing farming in best way. He exports the farm production.

On the basis of above information:
Question 1.
Identify the source of income of Shankar Patil.
Answer:
Shankar Patil is a lawyer by profession. He is practicing as a Lawyer / an advocate. He render his services to his clients and against that charges fee from them. So, ‘fee’ is the income source of Shankar Patil.

Question 2.
What is the qualification of Prashant Patil as a decorator?
Answer:
Prashant Patil may be or may not be a professional interior decorator. His hobby is decorating the office / houses. He may be an artist and an art does not require any formal education.

Maharashtra Board 11th OCM Important Questions Chapter 8 Introduction to Management

Question 3.
Mention two features of profession.
Answer:
Following are the features of profession:
1. Formal Education : Profession needs formal education. Today managers are qualified with management diploma or degrees. They take formal training from management schools which help them to work with perfection.

2. Restricted Entry : The entry in any profession is restricted. It needs a formal education. For example- Lawyer, Doctor, Interior designs, Architect, Chartered Accountant, etc. Such compulsion is not there for becoming a manager.

Justify the following statements

Question 1.
Management is an universal process.
Answer:

  1. Management is applicable to all organization whether religious, social or cultural, educational or governmental or non-governmental.
  2. All kind of organization have accepted management, as it is required everywhere.
  3. Every activity which has to be performed by a group of persons requires effective planning, organizing, co-ordinating, and controlling.
  4. Thus, management is an universal process.

Question 2.
There are levels of management in an organization.
Answer:

  1. There are three levels of management in an organization.
  2. The top level which consists of chairman, board of directors, etc takes care of formulating the board policies of the organization.
  3. The middle level does the work of implementing and executing the orders to the lower level.
  4. The lower level management does the actual and execution and operating activities.
  5. All the three levels are important and they are dependent on each other.
  6. Thus, all three levels are required for effective operation of the organization.

Question 3.
Management is a profession.
Answer:

  1. It is a specialized body of knowledge and managers are required to have professional skill and training
    to be efficient.
  2. It has certain ethical standards.
  3. Its motive is service oriented.
  4. It requires training which is formal.
  5. Thus, Management is considered as a profession.

Question 4.
Management is neither a pure science nor a fine art.
Answer:

  1. Management is said to be both an art and a science, as science teaches one to know and art teaches what to do.
  2. Management is a science as it is a body of knowledge and generalisation that represents the truth and facts, but it is not a pure science like Physics, Chemistry and Biology.
  3. Management is an art of applying the knowledge, but not pure fine arts like painting, dancing, etc.
  4. Thus, Management is neither a pure science nor a fine art.

Question 5.
Profession has ethics or code of conduct.
Answer:

  1. Profession is well-defined as an occupation which requires specialized skills and ethical standards.
  2. Professional managers have to follow code of conduct and ethical standards laid down by the All India Management in India.
  3. If they do not follow the code of conduct they can be derecognised and disqualified from the profession.
  4. Thus, Profession has ethics and code of conduct.

Maharashtra Board 11th OCM Important Questions Chapter 8 Introduction to Management

Question 6.
Management is an art, a science and a profession.
Answer:

  1. Management is an art because every manager has to apply his skills, intelligence, creativity for the efficient execution of management activities.
  2. It is considered a science because it adopts the cause and effect relationship.
  3. It has a scientific approach but deals in human behaviour, therefore called social science.
  4. Management is also considered a profession because it is a specialized body of knowledge and managers are required to have professional skills and training to be efficient.
  5. Thus, Management is an art, a science, and a profession.

Attempt the following

Question 1.
State the functions of lower-level management.
Answer:
The functions of lower-level management are as follows:

  1. Planning day to day working.
  2. Acting as a link between the middle level of management and workers.
  3. Carrying out the instructions given by the middle-level management.
  4. Delegating duties to workers, inspecting and supervising their work.
  5. Attending to the problems of workers and making arrangements for their training and development.
  6. To maintain discipline and good human relations among the workers.
  7. Image building of the enterprise before workers and creating a sense of belongingness.
  8. Ensuring the safety of workers, machines, tools, etc.
  9. Making arrangements of materials and tools and preventing the misuse and wastage of such material.
  10. Providing feedback information about the worker’s problems to the higher authorities.

Maharashtra Board 11th OCM Important Questions Chapter 5 Forms of Business Organisation – II

Balbharti Maharashtra State Board 11th OCM Important Questions Chapter 5 Forms of Business Organisation – II Important Questions and Answers.

Maharashtra State Board 11th Commerce OCM Important Questions Chapter 5 Forms of Business Organisation – II

Select the correct option and rewrite the sentence

Question 1.
Multinational corporations have a network of subsidiaries in ………………
(a) several countries
(b) home countries
(c) state only
Answer:
(a) several countries

Question 2.
A multinational corporation operates in ………………..
(a) only one country
(b) developed countries
(c) all over the world
Answer:
(c) all over the world.

Maharashtra Board 11th OCM Important Questions Chapter 5 Forms of Business Organisation – II

Question 3.
A multinational corporation is also known as ……………….
(a) global corporation
(b) regional corporation
(c) national corporation
Answer:
(a) global corporation

Question 4.
Government company is registered under …………………
(a) Special legislation
(b) Companies Act, 2013
(c) Banking Act 1949
Answer:
(b) Companies Act, 2013

Question 5.
Private sector enterprise is owned, managed and controlled by the …………………
(a) Individuals
(b) Government
(c) Board of directors
Answer:
(a) Individuals

Question 6.
Public sector enterprise is owned, managed and controlled by …………………
(a) individuals
(b) Government
(c) Board of Directors
Answer:
(b) Government

Question 7.
Private Enterprises invest their funds in …………………
(a) areas of social benefit
(b) areas where return is maximum
(c) any area
Answer:
(b) areas where return is maximum

Question 8.
Atleast ………………… of the capital of Government company is held by the Government.
(a) 51%
(b) 71%
(c) 50%
Answer:
(a) 51%

Question 9.
Public enterprises are financed by …………………
(a) Government
(b) private businessmen
(c) political leaders
Answer:
(a) Government

Question 10.
The nature and power of a statutory corporation is defined in …………………
(a) Indian Companies Act, 2013
(b) Special Statute
(c) Prospectus
Answer:
(b) Special Statute

Question 11.
……………….. is formed by a Special Act of Parliament.
(a) Private Enterprise
(b) Government Company
(c) Statutory Corporation
Answer:
(c) Statutory Corporation

Question 12.
Public enterprises are generally not runned on …………………
(a) service motive
(b) commercial lines
(c) profit motive
Answer:
(c) profit motive

Question 13.
………………… is an example of Public Corporation.
(a) Post and Telegraph
(b) LIC
(c) BHEL
Answer:
(b) LIC

Question 14.
………………… is an example of departmental undertaking.
(a) LIC
(b) Reliance
(c) Post and Telegraph
Answer:
(c) Post and Telegraph

Maharashtra Board 11th OCM Important Questions Chapter 5 Forms of Business Organisation – II

Question 15.
………………… is the oldest form of Public Sector Organisation.
(a) MNC
(b) Government Company
(c) Departmental Undertaking
Answer:
(c) Departmental Undertaking

Question 16.
Departmental organisations are accountable to ………………… for all their actions.
(a) Government
(b) Ministry
(c) Public
Answer:
(b) Ministry

Question 17.
The technological gap between the developed and developing countries is bridged by the …………………
(a) MNC
(b) Government Company
(c) Private enterprises
Answer:
(a) MNC

Match the pairs

Question 1.

Part A Part B
(a) Departmental Organisation (1) Private sector
(b) Statutory Company (2) Operates globally
(c) Multinational Corporation (3) Profit oriented
(d) Public sector Organisation (4) Objective based agreement
(e) Government Company (5) Service oriented
(6) Established by passing a special act
(7) 51% paid up capital by Government
(8) Sole trading
(9) Owned by government
(10) Temporary partnership

Answer:

Part A Part B
(a) Departmental Organisation (9) Owned by government
(b) Statutory Company (6) Established by passing a special act
(c) Multinational Corporation (2) Operates globally
(d) Public sector Organisation (5) Service oriented
(e) Government Company (7) 51% paid up capital by Government

Question 2.

Part A Part B
(a) Railways (1) Statutory Corporation
(b) Government Company (2) Decentralisation of control
(c) Multinational Company (3) Not answerable to the Legislature
(d) Unit Trust of India (4) National Thermal Power Corporation (NTPC)
(e) Statutory Corporation (5) Departmental organisation
(6) Centralised control of their branches
(7) Answerable to the Legislature
(8) Bajaj Electricals Pvt. Ltd.
(9) Owned by private individuals

Answer:

Part A Part B
(a) Railways (5) Departmental organisation
(b) Government Company (4) National Thermal Power Corporation (NTPC)
(c) Multinational Company (6) Centralised control of their branches
(d) Unit Trust of India (1) Statutory Corporation
(e) Statutory Corporation (7) Answerable to the Legislature

Give one word / phrase / term

Question 1.
A company which is incorporated in one country and has business units in several countries.
Answer:
Multinational Corporation

Question 2.
A company where 51% of paid up capital is held by the Central or State government.
Answer:
Government company

Question 3.
Autonomous corporate body created by Special Act of Parliament.
Answer:
Statutory Corporation.

State True or False

Question 1.
Public enterprises are owned, managed and controlled by public.
Answer:
False

Question 2.
Multinational companies have branches throughout the world.
Answer:
True

Maharashtra Board 11th OCM Important Questions Chapter 5 Forms of Business Organisation – II

Question 3.
Public enterprises have public accountability.
Answer:
True

Question 14.
Departmental organisations have right to borrow funds from outside.
Answer:
False

Question 5.
The main aim of private sector is to render services to the public.
Answer:
False

Question 6.
Government Company is registered under special legislation.
Answer:
False

Question 7.
Statutory Corporations need not follow the Companies Act, 2013.
Answer:
True

Question 8.
Government Companies enjoy legal status.
Answer:
True

Question 9.
Government interference is more in public sector enterprises.
Answer:
True

Question 10.
Multinational Companies help the host companies.
Answer:
True

Find the odd word out

Question 1.
National Thermal Power Corporations, Bharat Heavy Electricals Limited, Hindustan Machine Tools, LIC.
Answer:
LIC

Question 2.
Bata India, RBI, Infosys, Tata Motors.
Answer:
RBI

Question 3.
Sole Trading Concern, Joint Hindu Family Firm, Statutory Corporation, Joint Stock Company.
Answer:
Statutory Corporation

Question 4.
Unit Trust of India, Shipping Corporation of India, Indian Space Research Organisation, Yes Bank.
Answer:
Yes Bank

Question 5.
Samsung, IBM, Amazon, Patanjali.
Answer:
Patanjali

Question 6.
National Tiger conservation Authority, Securities and Exchange Board of India, State Bank of India, Indian Post and Telegraphs.
Answer:
SBI

Question 7.
Flipkart, Google, HP, Canara Bank.
Answer:
Canara Bank

Question 8.
Oriental Bank, Vijaya Bank, Andhra Bank, HDFC Bank.
Answer:
HDFC Bank

Maharashtra Board 11th OCM Important Questions Chapter 5 Forms of Business Organisation – II

Question 9.
MTNL, BSNL, Vodofone, Air India.
Answer:
Air India

Question 10.
Air India, Indian Airlines, Spicejet.
Answer:
Spicejet

Question 11.
Tata, Reliance, Airtel, BSNL.
Answer:
BSNL

Question 12.
Coca Cola India, Hindustan Unilever, IBM, SAIL.
Answer:
SAIL

Question 13.
Railways, Post and Telegraph, Defence, RBI.
Answer:
RBI.

Complete the sentences

Question 1.
The main aim of private sector enterprises is to earn …………………
Answer:
profit

Question 2.
The main aim of public sector enterprises is to render …………………
Answer:
service

Question 3.
A government company may be a private limited company or a …………………
Answer:
public limited company

Question 4.
A company formed by passing a special act in the parliament or legislature is ………………… company.
Answer:
statutory

Question 5.
MNCs have the need of ………………… capital.
Answer:
huge

Question 6.
Profit maximization is the ultimate goal of ………………… sector.
Answer:
private

Question 7.
Departmental organisations are accountable to ………………… for all their actions.
Answer:
ministry (Parliament)

Question 8.
Departmental undertakings are a source of revenue to ………………… treasury.
Answer:
Government

Maharashtra Board 11th OCM Important Questions Chapter 5 Forms of Business Organisation – II

Question 9.
Reserve Bank of India is an example of …………………
Answer:
Statutory Corporation

Question 10.
A company which is incorporated in one country and has business units in several countries is called …………………
Answer:
Multinational-Corporation

Question 11.
Post and Telegraph is an example of …………………
Answer:
Departmental Undertaking.

Select the correct option

Question 1.
(Multinational Corporation, Private Sector Organisations, Departmental Organisations, Statutory Corporation, Government Company)

Part A Part B
(1) 51% paid up capital held by Central Government …………………
(2) ……………….. Autonomous Corporate Body
(3) Owned by Private individuals …………………
(4) ……………….. Operates Globally
(5) ………………. Controlled by Concerned Ministry

Answer:

Part A Part B
(1) 51% paid up capital held by Central Government Government Company
(2) Statutory Corporation Autonomous Corporate Body
(3) Owned by Private individuals Private Sector Organisation
(4) Multinational Corporation Operates Globally
(5) Departmental Organisations Controlled by Concerned Ministry

Question 2.
(Statutory Corporation, Government Company, Multinational Corporation, Departmental Organisation)

Part A Part B
(1) BATA India ………………….
(2) …………………. Post Office, Railways, etc.
(3) Reserve Bank of India …………………
(4) …………………. BHEL

Answer:

Part A Part B
(1) BATA India Multinational Corporation
(2) Departmental Organisation Post Office, Railways, etc.
(3) Reserve Bank of India Statutory Corporation
(4) Government Company BHEL

Correct the underlined word and rewrite the following sentences

Question 1.
Multinational Corporation is relatively free from Red tapism.
Answer:
Statutory Corporation is relatively free from Red tapism.

Question 2.
Infosvs is an example of Government Company.
Answer:
HMT is an example of Government Company.

Maharashtra Board 11th OCM Important Questions Chapter 5 Forms of Business Organisation – II

Question 3.
Private Sector aims at providing services to customers.
Answer:
Private Sector aims at profit maximisation.

Question 4.
Private sector is owned by Government.
Answer:
Private sector is owned by individuals.

Question 5.
Defence Industry is an example of Statutory Corporation.
Answer:
Reserve Bank of India is an example of Statutory Corporation.

Question 6.
Departmental Organisation is financed by shareholders.
Answer:
Departmental Organisation is financed by annual budget.

Question 7.
Departmental Organisation has separate Legal entity.
Answer:
Government Company has separate Legal entity.

Question 8.
MNC operates at national level.
Answer:
MNC operates at global level.

Justify the following statements

Question 1.
Departmental organisations do not have right to borrow funds from outside.
Ans:

  1. Department organisations is financed by annual appropriation from the treasury and its revenue also goes into Treasury.
  2. This enterprise cannot borrow money from the public as enterprise is financed by the government and hence the government controls its affairs. Hence, answerable for its performance to the Parliament.
  3. Government consent is required to borrow from public, if need arises.
  4. Thus, Departmental organisations have no right to borrow funds from outside.

Question 2.
Statutory Corporations need not follow the Companies Act, 2013.
Ans:

  1. Statutory corporation is formed or created by passing a Special Act in Parliament or State legislature.
  2. The Act or Special Statute which is passed in the Parliament defines it powers, objectives, aims, duties, functions, rules and regulations governing its employees and its relationship with the Government department.
  3. Statutory Corporation prepare its own budgets and recruits employees on the terms and conditions determined by its Board of Directors.
  4. As Statutory Corporation is regulated as per the provisions made in Special Act of Parliament or State Legislature, it need not follow the provisions and directives given as per the Indian Companies Act, 2013. They need not have a Memorandum of Association or Articles as they have to strictly follow the Special statute.
  5. Thus, Statutory corporation need not follow the Companies Act.

Question 3.
Government Companies enjoy legal status.
Answer:

  1. Government Company is one in which not less than 51% of the share capital is held by the Central Government or State Government or part by Centre and part by State.
  2. Government Companies are registered under Indian Companies Act, 2013 and its working is as per the Act and hence has distinct legal entity apart from the government,
  3. Being a company registered under 2013 Act, it has all features like common seal, limited liability, stability, etc. It can enter into contracts and acquire property in its own name.
  4. Thus, Government Company enjoys legal status or management is in the hands of Board of Directors and ownership is with Government.

Maharashtra Board 11th OCM Important Questions Chapter 5 Forms of Business Organisation – II

Question 4.
Government interference is more in public sector enterprises.
Answer:

  1. Public enterprises are accountable to the public and Government as the money is financed by the annual budget.
  2. Public sector is welfare oriented and not profit oriented as public sector carries out work which is essential in the larger interest of public like making roads, bridges, flyovers, etc.
  3. Various Parliamentary committees are set up to scrutinise the working of public enterprises and the reports are tabled before the Parliament.
  4. The PSE’s do not enjoy financial autonomy, operational or administrative autonomy.
  5. Thus, more government interference in public sector enterprises.

Question 5.
Multinational Companies help the host companies.
Answer:

  1. MNC’s offer growth opportunities for domestic firms. They help domestic producers to enter the global markets through their international network of production and marketing.
  2. MNC’s provide ready markets for raw materials or semi-finished goods produced by the local producers in the host countries.
  3. The entry of MNC’s in the domestic market tends to break the monopoly of the selected few domestic companies. Due to this consumers in the host countries are benefited.
  4. MNC’s help host countries by providing new sophisticated technology.
  5. Thus, due to their global presence and superior marketing skills, MNC’s are in a position to help the host companies.

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 7 Company Meetings – I

Balbharti Maharashtra State Board Class 11 Secretarial Practice Important Questions Chapter 7 Company Meetings – I Important Questions and Answers.

Maharashtra State Board 11th Secretarial Practice Important Questions Chapter 7 Company Meetings – I

1A. Select the correct answer from the options given below and rewrite the statements.

Question 1.
The business to be transacted at the meeting is called _____________
(a) Agenda
(b) Minutes
(c) Notice
Answer:
(a) Agenda

Question 2.
A proxy form must be deposited in the registered office of the company at least _____________ hours before the meeting.
(a) 48
(b) 24
(c) 12
Answer:
(a) 48

Question 3.
A copy of the special resolution must be filed with the registrar within _____________ days of the passing resolution.
(a) 30
(b) 60
(c) 90
Answer:
(a) 30

Question 4.
Quorum for a private company is _____________ members.
(a) 15
(b) 5
(c) 2
Answer:
(c) 2

Question 5.
Point of order can be raised by _____________
(a) Chairman
(b) Secretary
(c) Member
Answer:
(c) Member

Question 6.
One share one vote’ is a voting method following under _____________
(a) voting by show of hands
(b) voting by ballot
(c) voting by-poll
Answer:
(c) voting by-poll

Question 7.
A motion which is changed due to an amendment is called _____________ motion.
(a) Formal
(b) Substantive
(c) Closure
Answer:
(b) Substantive

1B. Match the pairs.

Question 1.

Group ‘A’ Group ‘B’
(a) Agenda (1) Alteration in the original motion
(b) Amendment (2) Extra vote for chairman
(c) Point of Order (3) Rejected a motion
(d) Casting Vote (4) Objection raised by a member in the meeting
(e) Resolution (5) List of items to be transacted at the meeting
(6) Alteration in a Resolution
(7) Suggestion given by the member in the meeting
(8) Accepted Motion
(9) Only important items to be discussed at the meeting
(10) Extra vote for Directors

Answer:

Group ‘A’ Group ‘B’
(a) Agenda (5) List of items to be transacted at the meeting
(b) Amendment (1) Alteration in the original motion
(c) Point of Order (4) Objection raised by a member in the meeting
(d) Casting Vote (2) Extra vote for chairman
(e) Resolution (8) Accepted Motion

1C. Write a word or a term or a phrase that can substitute each of the following statements.

Question 1.
A list of items to be transacted at the meeting.
Answer:
Agenda

Question 2.
The minimum number of members to be present for a valid meeting.
Answer:
Quorum

Question 3.
The person who presides over the meeting.
Answer:
Chairman

Question 4.
A proposal is put before the meeting for discussion and decision.
Answer:
Motion

Question 5.
A motion finally passed at the meeting.
Answer:
Resolution

Question 6.
An alteration was proposed to the original motion.
Answer:
Amendment

Question 7.
A motion which brings to the notice of the Chairman the irregularities at the meeting.
Answer:
Point of Order

Question 8.
A person appointed by a member to attend and vote at the meeting.
Answer:
Proxy

Question 9.
An extra vote for the Chairman in case of tie-in votes.
Answer:
Casting vote

1D. State whether the following statements are True or False.

Question 1.
A proxy can vote under the poll method.
Answer:
True

Question 2.
Notice of the meeting must be given 14 days in advance.
Answer:
False

Question 3.
An agenda is sent along with the notice.
Answer:
True

Question 4.
Secretary has to sign the minutes book.
Answer:
False

Question 5.
Motion is a final decision of the meeting.
Answer:
False

Question 6.
Alteration can be done only by adding some new words.
Answer:
False

Question 7.
Point of Order is an objection raised by a member.
Answer:
True

Question 8.
Chairman can exercise casting votes in case of tie-in votes.
Answer:
True

Question 9.
A proxy cannot speak at the meeting.
Answer:
True

Question 10.
Secrecy can be maintained under voting by the show of hands method.
Answer:
False

Question 11.
A proxy need not be a member of the company.
Answer:
True

1E. Find the odd one.

Question 1.
Ordinary, Resolution, Special Resolution, Formal Motion.
Answer:
Formal Motion

Question 2.
Voting by voice, Voting by Ballot, Casting vote.
Answer:
Casting vote

Question 3.
Formal Motion, Substantive Motion, Resolution.
Answer:
Resolution

Question 4.
Voting by Poll, Voting by show of hands, Virtual voting.
Answer:
Virtual voting

Question 5.
Bare Statement Agenda, Draft Minutes Agenda, Minutes.
Answer:
Minutes

1F. Complete the sentences.

Question 1.
A person who presides over the meeting is known as _____________
Answer:
Chairman

Question 2.
Agenda is sent along with _____________
Answer:
Notice

Question 3.
Quorum for a private company is _____________
Answer:
2 members

Question 4.
A person responsible for proper conduct and to maintain order in the meeting is _____________
Answer:
Chairman

Question 5.
A motion which is changed due to amendment is called _____________
Answer:
Substantive motion

Question 6.
A resolution passed by simple majority is called _____________
Answer:
Ordinary Resolution

Question 7.
A special resolution, within 30 days of its passing must be filed with _____________
Answer:
Registrar of Companies

Question 8.
A representative of a member in a meeting is called _____________
Answer:
Proxy

1G. Select the correct option from the bracket.

Question 1.

Group ‘A’ Group ‘B’
(1) Voting by voice ……………………..
(2) ………………… 21 clear days
(3) Up to 1000 shareholders  ……………………
(4) …………………….. 15 days
(5) More than 5000 shareholders …………………….
(6) …………………… Writing
(7) Voting by Poll …………………….
(8) …………………… Objection
(9) Resolution by circulation ……………………
(10) ………………… Chairman

(Casting vote, Point of order, Directors, Motion, Secret method, Quorum is 5 members, Notice, Yes or No, Quorum is 30 members, Minutes)
Answer:

Group ‘A’ Group ‘B’
(1) Voting by voice Yes or No
(2) Notice 21 clear days
(3) Up to 1000 shareholders Quorum is 5 members
(4) Minutes 15 days
(5) More than 5000 shareholders Quorum is 30 members
(6) Motion Writing
(7) Voting by-poll Secret method
(8) Point of order Objection
(9) Resolution by circulation Directors
(10) Casting vote Chairman

1H. Answer in one sentence.

Question 1.
What is a quorum?
Answer:
Quorum is the minimum number of members required to be present for a valid meeting.

Question 2.
What is a proxy?
Answer:
A proxy is a person, who can attend and vote at the meeting on behalf of an absent member.

Question 3.
What is motion?
Answer:
A motion is a proposal put before the meeting for discussion and decision.

Question 4.
What is Resolution?
Answer:
A motion accepted in a meeting is called Resolution.

1I. Correct the underlined word and rewrite the following sentences.

Question 1.
The minimum number of members required to be present at a meeting’s called proxy.
Answer:
The minimum number of members required to be present at a meeting’s called Quorum.

Question 2.
Director draft s the minutes of a meeting.
Answer:
Secretary draft s the minutes of a meeting.

Question 3.
Voting by show of hands is a capitalistic method.
Answer:
Voting by show of hands is a democratic method.

Question 4.
The ordinary resolution requires a 75% majority of votes.
Answer:
A special resolution requires a 75% majority of votes.

Question 5.
Agenda is a record of business transacted at the meeting.
Answer:
Minutes is a record of business transacted at the meeting.

Question 6.
A notice period of general meeting needs 7 clear days.
Answer:
A notice period of general meeting needs 21 clear days.

1J. Arrange in proper order.

Question 1.
(a) Chairman
(b) Agenda
(c) Amendment
Answer:
(a) Agenda
(b) Chairman
(c) Amendment

Question 2.
(a) Point of order
(b) Quorum
(c) Motion
Answer:
(a) Quorum
(b) Motion
(c) Point of order

2. Distinguish between the following.

Question 1.
Motion and Amendment
Answer:

Basis Motion Amendment
1. Meaning Motion is a written proposal placed before the meeting for decision and discussion. An amendment is an alteration or modification proposed to the original motion.
2. Purpose The main purpose of the motion is to discuss and to make proper decisions. The main purpose of an amendment is to revise or modify the main motion.
3. Right to speak Every member can speak only once either in favour or against the motion. Only those members who have not spoken on the original motion can speak on an amendment.
4. Effect When a motion is accepted by the majority in the meeting, it becomes a resolution. When an amendment is approved by the majority, it becomes a part of the resolution.
5. Mover Mover of the motion can speak twice on his own motions. Mover of the amendment can speak only once on his own amendment.

Question 2.
Ordinary Resolution and Special Resolution
Answer:

Basis Ordinary Resolution Special Resolution
1. Meaning The resolution which is passed by a simple majority of votes is called Ordinary resolution. The resolution which is passed by 3/4th majority of votes is called a Special resolution.
2. Majority It requires more than 50% of the majority. It requires at least a 75% majority of votes.
3. Filing It is not filed with the Registrar of companies. It is filed with the Registrar of companies within 30 days of passing the resolution.
4. Business transacted It is necessary for routine and ordinary business. It is necessary for special business.
5. Examples Appointment of auditors, declaration of dividend, the election of directors, etc. Change in the name of the company, alteration in object clause, reduction in spare capital, etc.

3. Answer in brief.

Question 1.
State the importance of the Agenda.
Answer:
Agenda is a list of items to be discussed or things to be done or the business to be transacted at the meeting. It is sent along with the notice. A secretary prepares the agenda in consultation with the Chairman.
Importance of Agenda:

  • Members come to know about the matter to be discussed at the meeting.
  • Members can come well prepared for the meeting, as the agenda is sent along with notice (21 clear days) before the meeting.
  • The agenda helps the chairman to conduct the meeting efficiently.
  • Items can be discussed as per the order mentioned in the agenda.
  • All items are discussed at the meeting. Routine matters are discussed first, followed by special matters.

Question 2.
State the provisions regarding Quorum.
Answer:
Quorum is the minimum number of members required to be present for transacting a valid business. Without a quorum, the proceeding of the meeting becomes invalid. The quorum should be present throughout the meeting i.e. from beginning till the end of the meeting. Secretary must check the quorum before the commencement of the meeting.

Provision relating to Quorum (Companies Act, 2013)
(i) Quorum for General Meeting:

In the case of a Public Company: Quorum depends on a number of shareholders.

No. of shareholders Quorum
Up to 1000 5 members
1000 – 5000 15 members
More than 5000 30 members

In the case of a Private Company: Two members personally present shall be the quorum for the meeting.

(ii) Quorum for Board Meeting:
According to the provision in the Companies Act, a quorum for the board meeting is l/3rd of its total directors or two directors whichever is higher. Any fraction should be rounded to one.

4. Justify the following statements.

Question 1.
Resolution cannot be amended
Answer:
(i) When a motion is accepted in a meeting it becomes a resolution.
(ii) When a motion is under discussion it can be amended.
(iii) Resolution is a final decision, it cannot be changed.
(iv) Motion is passed in the meeting after a required discussion on a particular matter.
(v) A resolution cannot be amended as it is a final decision taken in the meeting.
Thus, the resolution cannot be amended.

Question 2.
Quorum is required throughout the meeting.
Answer:

  • a minimum number of members required for a valid meeting is a quorum.
  • Members present in the beginning should be present during the entire period of the meeting.
  • If the quorum is not present, the meeting stands adjourned for the next week same day, same time, and same place.
  • In the absence of a quorum, if any meeting is conducted it is considered an invalid meeting.
  • If any resolution is passed in the absence of a quorum, then such resolution is considered an invalid and illegal resolution.
  • Thus, a quorum is required throughout the meeting.

Question 3.
The chairman has the right of casting vote.
Answer:

  • A casting vote is a special type of vote given only by the Chairman of the meeting.
  • It is an additional vote given to the Chairman in case of equality of votes in favour and against a motion.
  • The chairman can use his vote only in the case of a tie.
  • This means when the votes cast for and against the resolution are equal.
  • It is a decisive vote as the final decision depends on the manner in which it is used by the Chairman.
  • The purpose of the casting vote is to have some definite decision on the matter for discussion before the meeting.
  • As per the provision in the Articles of Association, this vote can be exercised only by the chairman of the meeting.
  • Hence, the chairman has the right of casting vote.
  • Thus, the chairman has the right of casting vote.

Question 4.
Minutes of a meeting, once approved cannot be changed.
Answer:

  • Minutes are the factual and official records of the proceedings of the meeting in the form of decisions and resolutions.
  • Minutes are the legal evidence of the proceedings conducted in a meeting.
  • They cannot be changed once they are finalized and confirmed by the Chairman.
  • If any rectification is to be done and is necessary, it is carried out by passing a resolution in the next meeting and recording the same in the minute’s book.
  • Thus, the minutes of a meeting, once approved cannot be changed.

Question 5.
The agenda is useful to the chairman of the meeting.
Answer:

  • Agenda means a list of items to be done at the meeting.
  • It is a programme of the meeting.
  • It is prepared by the secretary in consultation with the chairman.
  • It is a guideline to the chairman of the meeting.
  • It enables the chairman to conduct the business of the meeting in an orderly manner.
  • Routine matters which can be discussed and decided in fewer times are arranged first in the serial order, after that special business or time-consuming matters.
  • This makes it easy for the chairman to go through it and discuss the matters.
  • Thus, the agenda is useful to the chairman of the meeting.

5. Answer the following questions.

Question 1.
Explain the kind of motion.
Answer:
A motion is a proposal put before the meeting for discussion and decision. A person who proposes a motion is called a proposer or a mover of a motion.
Kinds of Motion:

  • Formal Motion
  • Substantive Motion

(i) Formal Motion:
Formal motions are moved for the purpose of preventing or delaying or speeding up discussion on a motion.
Types of Formal Motion:
(a) Closure Motion:
This motion is moved when sufficient time is spent on the discussion of a particular motion. Any member can propose that ‘The question be now put’ to vote. This is a closure motion. The main object of this motion is to avoid waste of time and to arrive at a quick decision. If this motion is put to vote and if the majority approves, no further discussion is permitted.

(b) Previous Question Motion:
The main purpose of this motion is to prevent discussion on the main motion. When a member feels that it is unwise to consider the main motion, they may move the previous question. The wording of this motion is ‘The question be not now put’. If it is carried, the discussion on the main motion is dropped. If the previous question is lost, the original motion is put to vote.

(c) Next Business Motion:
The purpose of the motion is similar to the previous question motion. The wording of the motion is ‘The meeting to proceed to next business. Such a motion is moved when a member feels that the main motion under discussion is of little importance and other items of importance remain to be transacted. If it is carried, the original motion is dropped at once and the meeting proceeds to the next business.

(d) Adjournment of Debate Motion:
The main aim of this motion is to postpone the debate. The wordings of this motion are ‘The debate on the subject is adjourned’. The mover of the motion feels that some extra information is needed for further discussion and the discussion should be delayed for some time. If this motion is carried, the debate will be adjourned. If it is lost, the debate continues.

(e) Adjournment of Meeting:
The main aim of this motion is to postpone the meeting for a particular period or indefinite period. The wording of this motion is ‘The meeting be now adjourned’. If it is carried the meeting is postponed to a future date. If it is lost, the meeting continues.

(ii) Substantive Motion:
A motion that is changed due to an amendment is called a substantive motion. When an amendment is passed, it is incorporated in the original motion and the substantive motion is put to vote. If it is passed, it becomes a resolution.

Question 2.
Explain the essentials of notice of a meeting.
Answer:
Notice is an advance intimation given by the company informing the day, date, time, and place of meeting and business to be transacted at the meeting. It is given in writing to all those who are entitled to receive it.
(i) Essentials of the notice of a meeting:

  • Nature and type of meeting.
  • The exact day, date, time, and place of the meeting.
  • The agenda of the meeting.
  • A statement of a member entitled to attend the meeting and appoint a proxy.
  • Any intention to pass a special resolution must be mentioned in the notice.
  • Statutory note and explanation, if any special business is to be transacted.

(ii) Proper authority to send notice:
The Board of Directors is the proper authority to send the notice. Under exceptional circumstances, members of the National Company Law Tribunal or Central Government may send a notice for the meeting.

(iii) Proper authority to receive notice:

  • All shareholders in case of the shareholders meeting.
  • Auditors, in case of Annual General Meeting.
  • The legal representative of the deceased or insolvent person.
  • First name in Register of members, in case of joint holders.
  • All directors in case of Board Meeting and General Meeting.

(iv) Period of Notice:
In the case of a general meeting, notice must be sent 21 clear days before the actual meeting (day of sending and day of meeting are excluded). In the case of a Board meeting, 7 days’ notice is required.

(v) Modes of serving notice:

  • It can be sent either personally or by ordinary post at the registered address of the member in India.
  • If no registered address in India, then it can be sent to the address, given by the members.
  • It can be sent by registered post if requested by the member. Charges for the same are to be paid in advance.
  • It is to be advertised in all leading newspapers, (English and regional language) where the registered office is situated.
  • It can also be sent through electronic mode i.e. email.

(vi) A statement to be sent along with notice:
In case any special business is to be transacted in the general meeting, an explanatory statement to that effect is to be annexed to the notice.

(vii) Omission to give notice:
If notice is not served to one or more members deliberately, the meeting stands invalid.

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 6 Directors and Key Managerial Personnel of a Company

Balbharti Maharashtra State Board Class 11 Secretarial Practice Important Questions Chapter 6 Directors and Key Managerial Personnel of a Company Important Questions and Answers.

Maharashtra State Board 11th Secretarial Practice Important Questions Chapter 6 Directors and Key Managerial Personnel of a Company

1A. Select the correct answer from the options given below and rewrite the statements.

Question 1.
DIN means __________
(a) Director Identification Number
(b) Secretary Identification Number
(c) Doctor Identification Number
Answer:
(a) Director Identification Number

Question 2.
Private company requires __________ Directors.
(a) 3
(b) 2
(c) 1
Answer:
(b) 2

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 6 Directors and Key Managerial Personnel of a Company

Question 3.
COO means __________
(a) Chief Operating Officer
(b) Chief Financial Officer
(c) Chief Organization Officer
Answer:
(a) Chief Operating Officer

Question 4.
CS means __________
(a) Company Secretary
(b) Company Standard
(c) Chief Store Manager
Answer:
(a) Company Secretary

1B. Match the pairs.

Question 1.

Group ‘A’ Group ‘B’
(a) DIN (1) Whole-time Director
(b) Executive Director (2) Chief Executive Officer
(c) Key Managerial Personnel (3) Alternate Director
(d) Company Secretary (4) Director Identification Number
(e) Chief Financial Officer (5) Member of ICWA
(6) Officer responsible for Company’s finance
(7) Member of ICSI
(8) Officer responsible for Company’s Management

Answer:

Group ‘A’ Group ‘B’
(a) DIN (4) Director Identification Number
(b) Executive Director (1) Whole-time Director
(c) Key Managerial Personnel (2) Chief Executive Officer
(d) Company Secretary (7) Member of ICSI
(e) Chief Financial Officer (6) Officer responsible for Company’s finance

1C. Write a word or a term or a phrase that can substitute each of the following statements.

Question 1.
DIN means.
Answer:
Director Identification Number

Question 2.
This is an audit that checks the compliance of the company.
Answer:
Secretarial Audit

Question 3.
It aims to achieve and integrate corporate practices to all companies.
Answer:
Secretarial Standard

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 6 Directors and Key Managerial Personnel of a Company

Question 4.
As a principal officer of the company.
Answer:
Secretary

Question 5.
A Director who is in Whole Time Employment of the company.
Answer:
Whole Time Director

Question 6.
A first secretary appointed by the promoter.
Answer:
Pro-tem Secretary

Question 7.
Written proceeding of the meeting.
Answer:
Minutes

Question 8.
A person holding shares of nominal value ₹ 20,000.
Answer:
Small shareholders

Question 9.
Shares are required to be bought by the directors.
Answer:
Qualification shares

Question 10.
The person nominated on the Board by a dominant shareholder.
Answer:
Nominee Director

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 6 Directors and Key Managerial Personnel of a Company

Question 11.
The director is in full-time employment with the company.
Answer:
Executive Director or Whole Time Director

Question 12.
Director designated as ‘Outside Director’.
Answer:
Nonexecutive Director

Question 13.
Fees paid to Directors for attending a Board meeting
Answer:
Sitting fees

1D. State whether the following statements are True or False.

Question 1.
A Directors include whole-time employment of company is called WTD.
Answer:
True

Question 2.
DIN is required for Directorship.
Answer:
True

Question 3.
Secretarial audit checks the compliance of the company.
Answer:
True

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 6 Directors and Key Managerial Personnel of a Company

Question 4.
Secretary is not the principal officer of the company.
Answer:
False

Question 5.
Not providing guidance to the Board of Directors of the company is a general duty of the secretary.
Answer:
False

1E. Find the odd one.

Question 1.
Directors as Agents, as managing partners, As a company secretary.
Answer:
As a company secretary

Question 2.
First Directors, Casual vacancy, Audit.
Answer:
Audit

Question 3.
Chief Executive Officer, Chief Financial Officer, Promoter.
Answer:
Promoter

1F. Complete the sentences.

Question 1.
Statutory duties of secretary is to make all meetings __________
Answer:
minutes

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 6 Directors and Key Managerial Personnel of a Company

Question 2.
There is no prohibition on having more than one in a company __________
Answer:
Managing Director

Question 3.
CFO is __________
Answer:
Chief Financial Officer

Question 4.
WTD means __________
Answer:
Whole Time Director

Question 5.
KMP is __________
Answer:
Key Managerial Personnel

1G. Select the correct option from the bracket.

Question 1.

Group ‘A’ Group ‘B’
(1) Managing Director …………………………..
(2) ……………………. ICSI passed
(3) Manager …………………………..
(4) Secretarial Audit Checks the Legislations

(Substantial powers, Fulltime employee, Secretary, Checks the legislations)
Answer:

Group ‘A’ Group ‘B’
(1) Managing Director Substantial powers
(2) Secretary ICSI passed
(3) Manager Full-time employee
(4) Secretarial Audit Checks the Legislations

Question 2.

Group ‘A’ Group ‘B’
(1) First Director ……………………….
(2) ……………………. Key Managerial Personnel
(3) Pro-tem Secretary ……………………….
(4) Secretarial Standards ……………………….

(First Secretary, Appointed by the Promoter, Formulated by ICSI, Whole Time Director)
Answer:

Group ‘A’ Group ‘B’
(1) First Director Appointed by the Promoter
(2) Whole Time Director Key Managerial Personnel
(3) Pro-tem Secretary First Secretary
(4) Secretarial Standards Formulated by ICSI

1H. Answer in one sentence.

Question 1.
What do you mean by Nominee Director?
Answer:
The person nominated on the board by a major/dominant shareholder is known as the nominee director.

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 6 Directors and Key Managerial Personnel of a Company

Question 2.
What do you mean by Independent Director?
Answer:
Independent Director is a director other than managing director, whole-time director, or a nominee director.

Question 3.
Which Companies have to appoint a Woman Director?
Answer:
Every listed company and the public company has a paid-up share capital of Rs. 100 crores or more and whose turnover is Rs. 300 crores or more, have to appoint a Woman Director.

Question 4.
What do you mean by Alternate Director?
Answer:
It means a director who is nominated by the board to act in the place of a director in his absence.

Question 5.
Who is in full-time employment with the company?
Answer:
The executive director (WTD) is in full-time employment with the company.

Question 6.
What do you mean by Qualification shares?
Answer:
Qualification shares are those shares that are to be bought by the intending director, so as to become a director of a company.

1I. Correct the underlined word and rewrite the following sentences.

Question 1.
RIN is the unique identification number of the director.
Answer:
DIN is the unique identification number of the director.

Question 2.
Whole-time Director is not involved in the day-to-day management of the company.
Answer:
Non-Executive Director is not involved in the day-to-day management of the company.

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 6 Directors and Key Managerial Personnel of a Company

Question 3.
A company having paid-up share capital of ₹ 10 crores or more has to appoint a Woman Director.
Answer:
A company having paid-up share capital of ₹ 100 crores or more has to appoint a Woman Director.

Question 4.
Big Shareholder’s Directors is not a mandatory appointment.
Answer:
Small Shareholder’s Directors is not a mandatory appointment.

Question 5.
Directors are regarded as an employee of the company.
Answer:
Directors are regarded as an elected representatives of shareholders of the company.

Question 6.
A small shareholder is a person holding shares of a maximum of ₹ 50,000.
Answer:
A small shareholder is a person holding shares of a maximum of ₹ 20,000.

Question 7.
The manager needs to be the director of the company.
Answer:
The managing director needs to be the director of the company.

2. Explain the following Terms/Concepts.

Question 1.
Pro-tem Secretary
Answer:
The first secretary appointed by the promoters of the company is termed as Pro-tem Secretary. He may or may not be appointed as a regular secretary by the Board.

Question 2.
Sitting Fees
Answer:
Fees paid to the director for attending Board or Committee meeting is called as Sitting fees. It may extend up to Rs. one lakh also.

3. Answer in brief.

Question 1.
State the importance of DIN.
Answer:
Importance of DIN:

  • It helps the investors to take accurate and appropriate decisions, as through DIN they get to know the composition of top management of the company.
  • It helps to handle the problems arising due to the company creating fraud after raising capital from the investors.
  • It helps to detect and handle offenses committed by a particular Director.

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 6 Directors and Key Managerial Personnel of a Company

Question 2.
What are the general duties of a Company Secretary?
Answer:
General duties depend on the size and nature of the business.
Some of the general duties are as follows:

  • To provide guidance to the Board and Director with respect to their duties, responsibilities, and powers.
  • To represent before different regulators and authorities and discharge different duties under the Act.
  • To assist the Board in the conduct of the affairs of the company.
  • To assist and advise the Board in ensuring good corporate governance.
  • To perform all the duties that may be assigned by the Board from time to time.

Question 3.
State the rights of a Company Secretary.
Answer:
Rights of a Company Secretary: Rights are given to the Secretary by the Companies Act, Board of Directors, and the Shareholders.
The rights of the Company Secretary are given below:

  • Right to control and supervise the working of his department.
  • Right to be indemnified by the company if any loss is suffered by Secretary while performing or discharging his duties.
  • Right to sign a document requiring authentication.
  • Right to get remuneration as an employee of the company.

Maharashtra Board Class 11 Secretarial Practice Important Questions Chapter 6 Directors and Key Managerial Personnel of a Company

Question 4.
State the importance of Secretarial Audit?
Answer:
Importance of Secretarial Audit:

  • It provides a mechanism that monitors the compliance requirements.
  • It detects errors and mistakes in compliance with companies’ rules and regulations mechanism.
  • It prevents the company from the risk and losses due to non-compliance.
  • It builds the confidence of regulators, management, and stakeholders.
  • Investors feel relaxed that the company is following a disciplined approach towards management.

Maharashtra Board 11th OCM Important Questions Chapter 4 Forms of Business Organisation – I

Balbharti Maharashtra State Board 11th OCM Important Questions Chapter 4 Forms of Business Organisation – I Important Questions and Answers.

Maharashtra State Board 11th Commerce OCM Important Questions Chapter 4 Forms of Business Organisation – I

Select the Correct option and rewrite the sentence

Question 1.
A sole trading concern is a ………………. form of organization.
(a) private sector
(b) public sector
(c) none of these
Answer:
(a) private sector

Question 2.
A sole trading concern has ………………. liability.
(a) limited
(b) unlimited
(c) restricted
Answer:
(b) unlimited

Maharashtra Board 11th OCM Important Questions Chapter 4 Forms of Business Organisation – I

Question 3.
Indian Partnership Act was passed in the year ………………..
(a) 1923
(b) 1956
(c) 1932
Answer:
(c) 1932

Question 4.
Minimum number of persons to start a public limited company is ……………..
(a) 2
(b) 7
(c) 10
Answer:
(b) 7

Question 5.
The minimum number of directors in a private company is ……………….
(a) 3
(b) 2
(c) 5
Answer:
(b) 2

Question 6.
Business organisation which is controlled by Hindu Succession Act is known as …………….
(a) Joint Stock Company
(b) Partnership Firm
(c) Joint Hindu Family Firm
Answer:
(c) Joint Hindu Family Firm

Question 7.
Limited managerial skill is the ……………… of Joint Hindu family business.
(a) feature
(b) limitation
(c) advantage
Answer:
(b) limitation

Question 8.
The Karta in Joint Hindu Family business has ……………… liability.
(a) unlimited
(b) limited
(c) joint
Answer:
(a) unlimited

Question 9.
The maximum number of partners for a firm carrying on banking business is ………………..
(a) Ten
(b) Twenty
(c) Seven
Answer:
(a) Ten

Question 10.
In partnership firm the liability of partners is …………….
(a) limited
(b) unlimited
(c) none of above
Answer:
(b) unlimited

Question 11.
In a partnership firm every partner is the principal as well as the ……………….
(a) agent
(b) karta
(c) partner
Answer:
(a) agent

Question 12.
In a co-operative society the principle followed is ………………
(a) one share one vote
(b) one man one vote
(c) no vote
Answer:
(b) one man one vote

Question 13.
From the point of continuity ………………… business organisation is the most suitable.
(a) company
(b) partnership
(c) sole proprietorship
Answer:
(a) company

Question 14.
A sole trading concern is ……………….. form of business organisation.
(a) latest
(b) oldest
(c) recent
Answer:
(b) oldest

Question 15.
Members of a partnership firm are individually known as ………………..
(a) partner
(b) agent
(c) co-parcener
Answer:
(a) partner

Question 16.
Joint Hindu Family business exists in …………….. only.
(a) India
(b) England
(c) all the countries
Answer:
(a) India

Question 17.
The Karta is ………………. liable for business debts.
(a) not
(b) jointly
(c) personally
Answer:
(c) personally

Question 18.
Management of a Co-operative Society is ………………….
(a) personal
(b) joint
(c) democratic
Answer:
(c) democratic

Question 19.
Rules for working of a Co-operative Society are called …………………
(a) Bye-laws
(b) Articles of Association
(c) Memorandum
Answer:
(a) Bye-laws

Question 20.
The nature …………………. of organisation is service oriented.
(a) Sole Trader
(b) Joint Stock Company
(c) Co-operative Society
Answer:
(c) Co-operative Society

Match the pairs

Question 1.

Group A Group B
(a) Registration of a Company (1) International Market
(b) Sole Trading Concern (2) Minimum no. of members is 10
(c) Co-operative society (3) Voluntary
(d) Joint Hindu Family Firm (4) Partners
(e) Active partners (5) Local market
(6) Co-parceners
(7) Compulsory
(8) Minimum no. of member is 5
(9) Who lends his name
(10) Takes active part in business

Answer:

Group A Group B
(a) Registration of a Company (7) Compulsory
(b) Sole Trading Concern (5) Local market
(c) Co-operative society (2) Minimum no. of members is 10
(d) Joint Hindu Family Firm (6) Co-parceners
(e) Active partners (10) Takes active part in business

Question 2.

Group A Group B
(a) Indian Companies Act (1) 1912
(b) Indian Co-operative Society’s Act (2) Joint Hindu Family Firm
(c) The Hindu Succession Act (3) Sole Trading Concern
(d) Indian Partnership Act (4) 1907
(e) No Separate Act (5) 1923
(6) 1947
(7) 2013
(8) 1956
(9) 1932
(10) 1958

Answer:

Group A Group B
(a) Indian Companies Act (7) 2013
(b) Indian Co-operative Society’s Act (1) 1912
(c) The Hindu Succession Act (8) 1956
(d) Indian Partnership Act (9) 1932
(e) No Separate Act (3) Sole Trading Concern

Give one word/phrase/term.

Question 1.
Persons who enter into partnership business.
Answer:
Partners

Question 2.
The rules and regulations laid down by the managing committee of a co-operative society.
Answer:
Bye-laws

Maharashtra Board 11th OCM Important Questions Chapter 4 Forms of Business Organisation – I

Question 3.
The commercial organization which ensures maximum secrecy.
Answer:
Sole trading concern

Question 4.
Kind of partnership formed under the Indian Partnership Act.
Answer:
General Partnership

Question 5.
Another name for a sleeping partner.
Answer:
Dormant partner

Question 6.
Name the kind of company in which there is non-transferability of shares.
Answer:
Private Limited Company

Question 7.
The business organization which has minimum ten members.
Answer:
Co-operative organization

Question 8.
The type of commercial organization established for providing services to its members.
Answer:
Co-operative organization

Question 9.
The rules and regulations laid down by the managing committee of a co-operative organization.
Answer:
Bye-laws

Question 10.
An artificial person created by law.
Answer:
Joint Stock Company

Question 11.
The seniormost family member of Joint Hindu Family Firm.
Answer:
Karta

Question 12.
The members of the Joint Hindu Family firm.
Answer:
Co-parceners

Question 13.
A partner in partnership firm who takes active participation in day to day work.
Answer:
Active partner

Question 14.
A business organization with a single owner.
Answer:
Sole Trading Concern

Question 15.
A partnership agreement in writing.
Answer:
Partnership Deed

Question 16.
The members of the Joint Hindu Family Firm.
Answer:
Co-parceners

Question 17.
The principle of the co-operative societies.
Answer:
Each of all, all for each

Question 18.
An organization which enjoys a separate legal existence.
Answer:
Joint Stock Company

Question 19.
A symbol of a company’s incorporate existence substitutes for its signature.
Answer:
Common seal

Question 20.
A company which need not obtain Trading Certificate.
Answer:
Private Limited Company

Question 21.
A company which can start business after obtaining Certificate of Incorporation.
Answer:
Private Limited Company.

State True or False

Question 1.
The members of the Joint Hindu Family Firm are called partners.
Answer:
False

Question 2.
The principle of one member one vote is followed in a Joint Stock Company.
Answer:
False

Question 3.
A public limited company can start business after getting the Certificate of Incorporation.
Answer:
False

Question 4.
There is no limit for membership in Joint Hindu Family Business.
Answer:
True

Question 5.
The liability of a Karta in a Joint Hindu Family Firm is limited.
Answer:
False

Question 6.
The maximum number of members in Joint Hindu Family is 20.
Answer:
False

Question 7.
Company form of organization has developed before industrial revolution.
Answer:
False

Question 8.
Shareholders can manage the business.
Answer:
False

Question 9.
The ownership and management are not separate in Joint Stock Company.
Answer:
False

Maharashtra Board 11th OCM Important Questions Chapter 4 Forms of Business Organisation – I

Question 10.
The main purpose of a co-operative organization is to earn profit.
Answer:
False

Find the odd one

Question 1.
Sole Trading Concern, Partnership Firm, Joint Stock Company, Co-operative Society.
Answer:
Co-operative Society

Question 2.
Sleeping partner, Quasi partner, Co-parcener, Sub-partner.
Answer:
Co-parcener

Question 3.
Partnership at will, Partnership for a particular period, Limited Liability Partnership, Partnership for a particular venture.
Answer:
Limited Liability Partnership

Question 4.
Nominal partner, Active partner, Dormant partner, Minor as a partner.
Answer:
Minor as a partner

Question 5.
Grocery shop, Beauty parlour, Gift shop, Reliance Ltd.
Answer:
Reliance Ltd.

Question 6.
Hindustan Petroleum, Amul India, Mahindra & Mahindra, Renault India.
Answer:
Amul India

Question 7.
Karta, Partnership Deed, Co-parceners, Hindu Law.
Answer:
Partnership Deed

Question 8.
One Share One Vote, Service motive, Common seal, Perpetual succession.
Answer:
Service motive

Question 9.
One man show, Sole trader, Co-parceners, Single ownership.
Answer:
Co-parceners

Question 10.
Joint management, Principal agent relationship, Separate legal entity, Joint ownership.
Answer:
Separate legal entity

Question 11.
Partnership at will, Partnership for particular period, Partnership for particular venture, Limited Liability Partnership.
Answer:
Limited Liability Partnership

Question 12.
Mitakshara, Shareholders, Dayabhaga, Karta.
Answer:
Shareholder

Question 13.
One Member One Vote, Open membership, Prospectus, Service motive.
Answer:
Prospectus

Complete the sentences

Question 1.
The address of registered office of a company is mentioned in the …………………
Answer:
Memorandum of Association

Question 2.
A Sole Trading concern is best suitable for ……………………
Answer:
self employment

Question 3.
A Joint Stock company enjoys continuous and ………………… life.
Answer:
stable.

Complete the following table

Question 1.
(Consumer co-operatives, Partnership for a fixed period, Sole trading concern, Karta, Secret partner.)

Group A Group B
(i) Apana Bazar …………………
(ii) ………………… takes utmost care of business in a Joint Hindu Family Firm
(iii) Partnership formed for 6 months …………………
(iv) ………………… is not known to third parties in a partnership.
(v) No sharing of profits and risks is a feature of …………………

Answer:

Group A Group B
(i) Apana Bazar Consumer Co-operatives
(ii) Karta takes utmost care of business in a Joint Hindu Family Firm
(iii) Partnership formed for 6 months Partnership for a fixed period
(iv) Secret partner is not known to third parties in a partnership.
(v) No sharing of profits and risks is a feature of Sole Trading Concern

Maharashtra Board 11th OCM Important Questions Chapter 4 Forms of Business Organisation – I

Question 2.
(Sole trading concern, Joint Hindu Family Business, Minimum 2 directors, Partnership Firm, Co-operative Societies.)

Group A Group B
(i) Private Company …………………
(ii) ………………… Beauty parlour
(iii) Principal-Agent relationship …………………
(iv) Tax Concession …………………
(v)  ………………… Partition of business

Answer:

Group A Group B
(i) Private Company Minimum 2 directors
(ii)  Sole Trading Concern Beauty parlour
(iii) Principal-Agent relationship Partnership Firm
(iv) Tax Concession Co-operative Societies
(v)  Joint Hindu Family Business Partition of business

Question 3.
(Partnership Firm, Sole Trading Concern, Partnership firm, Co-operative Societies, Private company.)

Group A Group B
(i) ………………… Minimum 7 members
(ii) ………………… Minimum paid up capital Rs 1,00,000
(iii) Owned and controlled by one person …………………
(iv) Registration Compulsory in Maharashtra …………………
(v)  ………………… Democratic Management

Answer:

Group A Group B
(i) Public Company Minimum 7 members
(ii) Private Company Minimum paid up capital Rs 1,00,000
(iii) Owned and controlled by one person Sole Trading Concern
(iv) Registration Compulsory in Maharashtra Partnership firm
(v)  Co-operative societies Democratic Management

Answer in one sentences

Question 1.
What is the meaning of unlimited liability?
Answer:
Unlimited liability means if a businessman becomes insolvent and if his business assets are insufficient to pay off his business liabilities, he will have to use his private property in order to pay off his business debts.

Question 2.
Who has unlimited liability in the Joint Hindu Family Firm?
Answer:
The Karta has unlimited liability in the Joint Hindu Family Firm.

Question 3.
Who are the Board of Directors of a Company?
Answer:
Elected representatives of the shareholders are called the Board of Directors.

Question 4.
What are the members of the Joint Hindu Family called?
Answer:
The members of the Joint Hindu Family are called as Co-parceners.

Question 5.
What are Co-operating farming societies?
Answer:
Co-operative farming societies are the voluntary associations of small farmers who come together to obtain the benefits of large scale farming.

Question 6.
What are marketing co-operatives?
Answer:
Marketing co-operatives are the voluntary associations of independent producers who are willing to sell their products output at a remunerative prices.

Question 7.
What are credit co-operatives?
Answer:
Credit co-operatives are the co-operative societies which are formed by the poor people to provide financial help and to develop the habit of savings among the members.

Maharashtra Board 11th OCM Important Questions Chapter 4 Forms of Business Organisation – I

Question 8.
What is partnership deed?
Answer:
Partnership deed is an agreement in writing which includes terms and conditions of partnership firm.

Question 9.
Who is dormant partner?
Answer:
A dormant partner is a partner in a partnership firm who contribute capital, shares profit and contributes to the losses of the firm, but does not take active participation in the working of the firm.

Correct the underlined word and rewrite the following sentences.

Question 1.
The partnership business has less capital as compared to Sole Trading concern.
Answer:
The partnership business has more capital as compared to Sole Trading concern.

Question 2.
Joint Hindu Family Business exists outside India.
Answer:
Joint Hindu Family Business exists only in India.

Question 3.
The liability of the co-parceners is unlimited in a Joint Hindu Family firm.
Answer:
The liability of the co-parceners is limited in a Joint Hindu Family firm.

Question 4.
Co-operative society is a compulsory association of individuals to provide services to its members.
Answer:
Co-operative society is a voluntary association of individuals to provide services to its members.

Question 5.
Formation of a Joint Stock company is easy.
Answer:
Formation of a Joint Stock company is complicated.

Question 6.
There is more scope for expansion in a Co-operative society.
Answer:
There is less scope for expansion in a Co-operative society.

Question 7.
In a Public company, there are minimum 2 fTwol Director’s.
Answer:
In a Public company, there are minimum 3 (Three) Director’s.

Question 8.
The registration of a Sole Trading concern is compulsory.
Answer:
The registration of a sole trading concern is not compulsory.

Explain the following terms/concepts

Question 1.
Active Partner.
Answer:

  1. A partner who takes active participation in day-to-day working of the firm, is called as active partner.
  2. He is also called as ordinary/general/ actual/ working partner.
  3. He acts as a manager, organizer, advisor and controller of the firm.

Question 2.
Dormant Partner
Answer:

  1. A partner who does not take active participation in day-to-day working of the firm, is called as dormant partner or sleeping partner.
  2. He contributes to the capital of the firm and also shares profit or loss of the business firm.
  3. He is liable for the debts of the firm.
  4. He cannot find the business firm or third party by his acts.

Question 3.
Unlimited Liability
Answer:

  1. It means that there is no distinction made between personal property and business property of a person. It further means that if business assets are not sufficient to pay off liabilities of a business than personal property of a person holding business will be attached to business property.
  2. Sole trader in case of sole trading concern, partners in case of partnership firm, Karta in case of Joint Hindu Family Business, has unlimited liabilities.

Question 4.
Principal – Agent Relationship
Answer:

  1. This term is related to partnership firm of organizations.
  2. In partnership firm all partners are the joint owners of the business and all takes part in the management of the business.
  3. Hence, every partner acts in dual role in partnership firm i.e.
    (a) When within the firm they act as ‘Principal’ and
    (b) When they deal with outsiders they act as an ‘Agent’.

Question 5.
Producer’s Co-operatives
Answer:

  1. They are voluntary associations of small producers and artisans, who come together to face the competition and increase the productions.
  2. Producers Co-operatives are of 2 types – (a) Industrial Service Co-operatives and (b) Manufacturing Co-operatives.
  3. Such co-operatives sells the output of members in the market and then profits are distributed among the members.
  4. They provide raw materials, implements tools, technical guidance to members on easy terms.

Question 6.
Housing Co-operatives
Answer:

  1. Housing Co-operatives are owned by the residents.
  2. Such societies purchases the land, develops it, constructs houses, etc. and allot them to members.
  3. They arrange loans from Government and Non-Government sources.

Maharashtra Board 11th OCM Important Questions Chapter 4 Forms of Business Organisation – I

Question 7.
Common Seal
Answer:

  1. A company being an artificial person cannot sign on its own.
  2. So, the law requires, every company to have its seal, bearing its name on it.
  3. Common seal is a symbol of company’s incorporate existence. It is like a signature of the company.
  4. It has to be affixed on all the important documents of the company.
  5. It has to be witnessed by a Company Secretary and minimum 2 directors.

Study the following case/situation and express your opinion

1. Mr. Jumbo a Chartered Accountant by profession and Mrs. Timbo, an Architect by profession running a firm namely ‘Buildsheet’ in Nagpur.

Question 1.
Identify business in above example.
Answer:
Partnership firm

Question 2.
Comment on it?
Answer:
It is a voluntary association of two persons with a common objective in mind.

Question 3.
Which two professions are mentioned here?
Answer:
A Chartered Accountant and an Architect.

2. Ram and Rahim have formed a Joint Stock company to produce and sell toys with a paid up capital of Rs 2,00,000.

Question 1.
Name the type of company.
Answer:
Private Limited Company.

Question 2.
List two features of the company.
Answer:
(a) Minimum 2 members are required to form a private company.
(b) Minimum paid up capital required is Rs 1,00,000.

3. Rahul is the only owner of his restaurant. All the assets belongs to him.

Question 1.
Name the type of business organisation.
Answer:
Sole Trading Concern.

Question 2.
State two features.
Answer:
(a) Single ownership
(b) No separate entity

4. Ajit Brothers is a firm formed for the specific purpose of constructing a bridge.

Question 1.
Name this type of business.
Answer:
Partnership for a particular venture or particular partnership.

Question 2.
When does this business dissolve?
Answer:
This business will dissolve on the completion of the venture i.e. after completing the bridge.

5. Employees Union of Hindustan Mills start a retail shop to provide cloth at reasonable price to its members.

Question 1.
Identify the form of business organisation.
Answer:
The type of organisation is Co-operative society.

Question 2.
State two features of the organisation.
Answer:
(a) Voluntary membership
(b) Service motive

6. Teachers of different colleges in Mumbai form an association to buy land and construct flats for them.

Question 1.
Name the form of business organisation.
Answer:
Co-operative Society

Question 2.
Is the association of members voluntary or compulsory.
Answer:
The association is voluntary, as it is one of the features of the Co-operative Society.

7. Mr. Shyam is a shareholder of Coal India Limited. He is in need of money. He transfers his shares to another person.

Question 1.
Name the kind of company.
Answer:
Public Limited Company

Question 2.
Can he transfer his shares.
Answer:
Yes, he can transfer his shares to anyone as it is one of the main features of a Public Limited Company.

8. If the business is a sole trading concern with a capital of Rs 25,000, but the debts are amounting to Rs 50,000.

Question 1.
Is the entrepreneur liable to pay his creditors.
Answer:
Yes, he has to pay to his creditors.

Question 2.
Give reasons.
Answer:
A sole trading concern has unlimited liability therefore, he has to pay his creditors by selling off his private property.

9. Amul – the taste of India started 65 years ago in 1946 in Anand.

Question 1.
Name the type of business organisation.
Answer:
Co-operative Society

Question 2.
What was the mission for starting AMUL.
Answer:
AMUL was started with a mission to stop the exploitation of farmers by middlemen.

Distinguish between the following

Question 1.
Joint Hindu Family Firm and Sole Trading Concern.
Answer:

Joint Hindu Family Sole Trading Concern
Meaning In Joint Hindu Family Firm, the Joint Hindu Family conducts business according to Hindu Laws. Sole Trading Concern is owned and controlled by one person.
Formation Joint Hindu Family Firm comes into existence by operation of Hindu Law. Sole Trading Concern can be formed easily. It is started as soon as the owner decides.
Number of Members Membership of the firm depends upon the birth and death in the family. There is no limit on membership. Sole Trading Concern is owned by a single person.
Sharing of Profits/Losses The profits and losses shared between Karta and Co-parceners. The sole trader alone enjoys all the profits of business. He alone bears the losses of business.
Liability Karta has unlimited liability. The liability of Co-parceners is limited. The sole trader has unlimited liability.
Act Joint Hindu Family Firm follows the Hindu Succession Act, 1956. There is no special legislation governing the sole trading concern.
Stability Comparatively more stability because if Karta dies, the next seniormost family member can take over the business. There is no stability as life of the organization depends upon the survival of the sole trader.
Capital More finance is available with the firm. Therefore, there is more capital. Sole trader has limited funds. Therefore, the concern has limited capital.
Ownership Assets and property of business is jointly owned by family members. Assets and properties of business are owned by the sole trader.
Efforts and Rewards There is no direct relation between efforts and rewards. There is a direct relation between efforts and rewards.

Question 2.
Partnership Firm and Joint Stock Company.
Answer:

Partnership Firm Joint Stock Company
Meaning Partnership firm is controlled by two or more persons called as partners. Joint Stock Company is an incorporated voluntary association created by law, having an independent legal status, owned by shareholders and managed by Directors.
Act Partnership firm is governed by Indian Partnership Act, 1932. Joint Stock Companies are governed by Indian Companies Act, 2013.
Membership Minimum two partners and Maximum 50 partners. Private company-
Minimum – 2
Maximum – 200Public company-
Minimum – 7
Maximum – No limit
Liability The liability of partners is unlimited, joint and several. The liability of shareholders is limited. Up to the unpaid amount of shares held by them.
Ownership & Management Partners are owners and managers of the firm. Joint Stock Company is owned by shareholders of the company and managed by Board of Directors.
Formation The formation is simple and less costly. Only an oral or written agreement is needed between partners. Formation of a Joint Stock Company is. difficult, costly and time-consuming.
Stability A partnership firm does not enjoy a stable life. Life of the firm is affected by death, insolvency or lunacy of even one partner. Joint Stock Company has a long and stable life. It has perpetual succession. Life of the company is not affected by death, insolvency or lunacy of any member or Director.
Registration Registration of a partnership firm is optional. It is compulsory in the State of Maharashtra. Registration of a Joint Stock Company is compulsory.
Legal Status A partnership firm has no legal status that is distinct from its members. A Joint Stock Company has a separate legal status.
Transfer of Shares A partner can transfer his share only with the consent of other partners. Shares of a public company are freely transferable.
Authority of Members A partner has implied authority. He is an agent of the firm. A shareholder has no implied authority.
Winding up A partnership firm can be easily dissolved. There are not much legal formalities for winding up the firm. A Joint Stock Company can be dissolved only by the court. A proper procedure has to be followed for dissolving the company.

Question 3.
Sole Trading Concern and Joint Stock Company.
Answer:

Sole Trading Concern Joint Stock Company
Meaning Sole proprietorship is owned, managed and controlled by one person. Joint Stock Company is an incorporated association created by law, having an independent legal status, owned by shareholders and managed by board of directors.
Formation Sole trading concern can be formed easily. It is started as soon as the owner decides. Formation of a Joint Stock Company is difficult, costly and time-consuming. There are many legal formalities needed to form a joint stock company.
Liability The sole trader has unlimited liability. Shareholders have limited liability. It depends upon unpaid value of shares.
Numbers of Member Sole trading concern is owned by a single person. The minimum number of persons needed to start a private company is two and maximum is fifty. Minimum number of public company is seven and maximum is unlimited.
Registration Sole trading concern need not be registered. Joint Stock Company has to be compulsorily registered.
Act There is no Act for a Sole Trading Concern. Companies have to follow Indian Companies Act, 2013.
Legal Status A Sole Trading Concern does not have a legal status. Joint Stock Company has a legal status.
Secrecy It is possible to have maximum business secrecy in a Sole Trading Concern. Business Secrecy cannot be maintained’. Books of accounts have to be kept open for inspection to shareholders.
Capital As there is only one owner, less capital is available for business. Capital is collected from shareholders in the form of shares. Huge capital can be collected for business.
Stability A Sole Trading Concern does not have a stable life. Life of the sole trader and his concern are one and the same. The company enjoys a long stable life.

Answer in brief

Question 1.
State any four merits of Sole Trading Concern.
Answer:
1. Easy Formation : A sole trading concern can be easily formed. Not much legal formalities are needed to start a sole trading concern. A person who is competent to enter into contract can start a sole trading concern.

2. Prompt Decision Making: A sole trader can take all decisions by himself. There is no external interference. As he can take quick decisions, prompt actions is also possible. This Organisation is suitable for business where quick decisions have to be taken. Ex. Share Trading.

3. Flexibility : The sole trader can make changes in business easily. He can expand or curtail business activity, whenever needed. He is the only person to take decisions in business. There is a lot of flexibility in sole trading concern.

4. Minimum Government Regulations : Sole trading concern need not follow any Special Act. There are not much legal formalities needed for forming and closing a sole trading concern. Only the general law of the country has to be followed. It states that a person doing sole trading business:

  • must be a major (above 18 years)
  • must be of sound mind
  • must not be disqualified by any law to do business.

Maharashtra Board 11th OCM Important Questions Chapter 4 Forms of Business Organisation – I

Question 2.
Explain any four demerits of Sole Trading Concern.
Answer:
1. Limited Capital : The sole trader is the only owner of business. He alone contributes capital for business. Capital is in the form of savings and borrowings from friends and relatives. In his individual capacity he cannot raise loans from banks to a large extent. The growth and expansion of business is affected because of limited capital. This is disadvantage to business.

2. Limited Managerial Ability : There is only one person to manage business. Inspite of his knowledge, the sole trader is not able to manage business efficiently. The sole trader may not be an expert in all fields like finance, advertising, management etc. Therefore, management ability is limited.

3. Unlimited Liability : The liability of the sole trader is unlimited. There is no difference between personal property and business property of sole trader. But unlimited liability prevents further expansion. He does not take risk in business. He is overcautious in taking decision which affects progress of business.

4. No Stability : A sole trading concern does not have a continuous existence. It does not have a stable life. Life of the organisation depends upon the life of the sole trader. In case the sole trader dies, becomes insolvent or insane the business has to be closed down.

Question 3.
What are the features of Joint Hindu Family Business ? (Any 4)
Answer:
1. Formation : Joint Hindu Family Business is created by operation of Hindu Law. There is no agreement between members. Each member becomes the co-parcener in the family business by birth.

2. Membership : Joint Hindu Family business has two types of partners.
(a) Karta : The seniormost male member of the family.
(b) Co-parceners (Copartners) : Other family members.
There is no limit on maximum number of members. In the state of Maharashtra, even girls born in the family become co-parceners of the firm. The number of members keep on changing depending on birth and death in the family.

3. Joint Ownership : The property of the firm is jointly owned by three generations of the family. Karta is the custodian of the joint property. All family members own the assets of business.

4. Sharing of Profits / Losses : Ratio of sharing of profits or losses of Karta and co-parceners is not given in the Hindu Act. The ratio of profit sharing keeps on changing depending on birth and death in the family.

Question 4.
Give any four demerits of Joint Hindu Family Business.
Answer:
(i) Limited Resources : Business is done by the joint family. So capital brought in by the family is limited. The capacity to raise funds is also limited as liability of co-parceners is limited. This affects growth and expansion of business.

(ii) Limited Managerial Ability : The management of business is totally taken care of by Karta. So the entire business depends upon the ability of the Karta. He does not even consult co-parceners for decision making. The Karta may not be an expert in all areas of business. Due to inadequate funds it is not possible to hire experts. Therefore, Joint Hindu Family Firm faces the disadvantage of limited managerial ability.

(iii) Unlimited Liability : The liability of Karta is unlimited. His decision making process becomes very slow. He is too cautious and unwilling to take risk in business. As there is no difference between personal property and business property of Karta he is afraid to take bold decisions. This affects the progress of business.

(iv) Limited Expansion : Even if Karta takes a lot of interest in business, the firm has the disadvantage of limited growth. Two reasons i.e., limited capital and limited managerial ability results in limited growth of business.

Question 5.
State any four merits of partnership firm.
Answer:
(i) Easy Formation : It is easy to form a partnership firm. Only an agreement is needed to start the firm. The agreement between partners may be an oral or written one. A written agreement is always preferred to avoid disputes in future. Not many legal formalities are needed for forming or closing business. Even the procedure for registration is simple.

(ii) Large Capital: Many partners come together to start a firm. Partners have joint responsibility to collect capital for the firm. As number of persons are more than sole trading concern, large capital can be collected. If business has to be expanded more capital has to be collected.

(iii) Business Secrecy : More secrecy can be maintained in business. Books of accounts need not be published. There is no legal requirement to publish annual report. Information is not leaked out to competitors, which is an advantage to the firm. If partners want they can keep secrets to themselves.

(iv) Rational Decisions : Decisions are taken collectively by partners. Two heads are always better than one.. Partners discuss can take decisions, which are always balanced. Partners have unlimited liability therefore they are careful in taking decisions.

Justify the following statements

Question 1.
There is no limit for membership in Joint Hindu Family Business.
Answer:

  1. Joint Hindu Family Firm is jointly owned by all family members.
  2. When a Hindu Joint Family does business and passes it on from one generation to another it is called a Joint Hindu Family business.
  3. There are two types of members in Joint Hindu Family Firm – Karta and Co-parceners.
  4. The membership of Joint Hindu Family Firm is unlimited.
  5. Membership is not limited because it depends upon the death and birth in the family. All the persons born in the family become a member of the family by birth.
  6. A person becomes a member even by adoption into the family.
  7. Thus, there is no maximum limit for membership in a Joint Hindu Family Business.

Question 2.
The membership of a co-operative society is voluntary.
Answer:

  1. Co-operative society is a voluntary association of persons formed for the purpose of promoting the interest of its members.
  2. There is open membership i.e., membership is open to all. No difference is made on language, religion, caste etc. A person becomes a member by purchasing shares of the organization.
  3. There is complete freedom about becoming or not becoming the member of the organization.
  4. A person can join on his own free will. There is no compulsion for becoming a member.
  5. He can leave the organization by giving due notice as and when he desires.
  6. The opportunity of becoming a member is given to all which means co-operative organization is a voluntary organization.
  7. Thus, the membership of a co-operative society is voluntary.

Question 3.
In partnership, the liability of every partner of the firm is unlimited.
Answer:

  1. Partnership is an association of persons competent to enter into contracts. The aim is to carry lawful business.
  2. All partnerships in India follow the Indian Partnership Act of 1932.
  3. Accordingly the liability of every partner is unlimited.
  4. Partners have joint and several liability. It means each partner is individually liable and jointly liable for the debts of the firm.
  5. There is no difference between personal property and business property of the partners. In case business assets are not sufficient to pay debts, the personal property of each and every partner can be used.
  6. If one partner is declared insolvent, his liability has to be borne by solvent partners.
  7. Thus, in partnership, the liability of every partner of the firm is unlimited.

Maharashtra Board 11th OCM Important Questions Chapter 4 Forms of Business Organisation – I

Question 4.
The owner of the sole proprietorship is the sole decision maker of his business.
Answer:

  1. A sole trader does business alone.
  2. He invests his own capital, skill and intelligence in the management of business.
  3. A sole trader is the only manager of business. So he is the sole decision maker of his business.
  4. He is not answerable to anyone and need not consult anyone for doing business.
  5. There is no external interference.
  6. Therefore, he can take quick decisions and prompt actions and take advantage of profit – oriented situations. Thus, sole trader is the only decision maker and therefore has complete control on business.

Question 5.
Sole Trading Concern is easiest to form.
Answer:

  1. Sole trading concern is owned by only one person.
  2. A very Simple process is needed to start a sole trading concern.
  3. There is no complicated legal formalities to start a sole trading concern, as in the case of joint stock company.
  4. A licence from appropriate authority is needed.
  5. Only the general law of the country has to be followed for starting business.
  6. A person must be of sound mind, a major and must not be disqualified by any law to do business.
  7. Thus, sole trading concern is the easiest form of organization to be started.

Maharashtra Board 11th OCM Important Questions Chapter 6 Institutes Supporting Business

Balbharti Maharashtra State Board 11th OCM Important Questions Chapter 6 Institutes Supporting Business Important Questions and Answers.

Maharashtra State Board 11th Commerce OCM Important Questions Chapter 6 Institutes Supporting Business

Select the correct option and rewrite the sentence

Question 1.
SIDBI has its head office at …………………
(a) Bihar
(b) Mumbai
(c) Lucknow
Answer:
(c) Lucknow

Question 2.
Dr. Muhammad Yunus from ………………… is known as ‘Father of Micro finance.
(a) Afghanistan
(b) Bangladesh
(c) India
Answer:
(b) Bangladesh

Maharashtra Board 11th OCM Important Questions Chapter 6 Institutes Supporting Business

Question 3.
The World Bank Group has set ………………… goals to be achieved by 2030.
(a) 4
(b) 3
(c) 2
Answer:
(c) 2

Question 4.
………………… is the lifeblood of a business.
(a) Finance
(b) Raw materials
(c) Labour
Answer:
(a) Finance

Question 5.
KVIC has its head office at …………………
(a) Mumbai
(b) Kolkata
(c) Delhi
Answer:
(a) Mumbai

Question 6.
………………… portal was launched by SIDBI.
(a) Maza-Mitra
(b) Aamcha Mitra
(c) Udyami Mitra
Answer:
(c) Udyami Mitra

Question 7.
………………… work as knowledge partner for State and Central Government.
(a) Udyami Mirta
(b) Start-up Mitra
(c) Wake-up Mitra
Answer:
(b) Start-up Mitra

Question 8.
………………… provides credit for rural development.
(a) SIDBI
(b) NABARD
(c) RRBs.
Answer:
(b) NABARD

Match the pairs

Question 1.

Part A Part B
(i) NABARD (a) Heart of business
(ii) Finance (b) Advisor and mentor to MSMEs
(iii) SIDBI (c) Life blood of business
(iv) MUDRA (d) March 17, 2016
(v) Startup Mitra (e) Recommendations to RBI
(f) Funding the unfunded enterprises
(g) March 17, 2014
(h) Recommendation to Commercial Banks

Answer:

Part A Part B
(i) NABARD (e) Recommendations to RBI
(ii) Finance (c) Life blood of business
(iii) SIDBI (b) Advisor and mentor to MSMEs
(iv) MUDRA (f) Funding the unfunded enterprises
(v) Startup Mitra (d) March 17, 2016

Question 2.

Part A Part B
(i) SIDBI (a) Maza Mitra
(ii) Green, Blue and White Revolution (b) RRB’s
(iii) Khadi (c) Symbol of dependence
(iv) Women’s Self Help Group (d) NABARD
(v) World Bank (e) Washington D.C.
(f) Udyami Mitra
(g) Symbol of self reliance
(h) Great Britain
(i) Mahila Bachat Gat

Answer:

Part A Part B
(i) SIDBI (f) Udyami Mitra
(ii) Green, Blue and White Revolution (d) NABARD
(iii) Khadi (g) Symbol of self reliance
(iv) Women’s Self Help Group (i) Mahila Bachat Gat
(v) World Bank (e) Washington D.C.

Give one word/phrase/term

Question 1.
Symbol and spirit of self reliance in pre-independence India.
Answer:
Khadi

Question 2.
Head office of KVIC.
Answer:
Mumbai

Maharashtra Board 11th OCM Important Questions Chapter 6 Institutes Supporting Business

Question 3.
A statutory body formed to provide livelihood to artisans in villages of India.
Answer:
Khadi and Village Industries Commission (KVIC)

Question 4.
A document signed by KVIC with National Institute of Design (NID).
Answer:
Memorandum of Understanding (MoU)

Question 5.
Economist of Bangladesh who jointly won Nobel Peace Prize.
Answer:
Dr. Muhammad Yunus

Question 6.
A refined model of Bangladesh model to eradicate poverty and empower women.
Answer:
Indian Self Help Group Model

Question 7.
A portal to improve accessibility of credit and handholding services to MSMEs.
Answer:
Udyami Mitra.

State True or False

Question 1.
SIDBI has set a goal to end extreme poverty by 2030.
Answer:
False

Question 2.
The World Bank has set a goal to promote shared prospering by fostering the income growth.
Answer:
True

Question 3.
Success of business depends on optimum utilization of capital.
Answer:
True

Question 4.
SIDBI provide debt financing to government.
Answer:
False

Question 5.
Dr. Ashfak Yunus founded the Grameen Bank in India.
Answer:
False

Question 6.
Body of Governors is the supreme policy making body of the World Bank.
Answer:
True

Find the odd word out

Question 1.
Features of KVIC-
Rural Development; Employment Generation; Entrepreneurship Development; Mutual Trust.
Answer:
Mutual Trust

Question 2.
Institutes supporting business-
SIDBI; IDBI; NABARD; KVIC.
Answer:
IDBI

Maharashtra Board 11th OCM Important Questions Chapter 6 Institutes Supporting Business

Question 3.
Features of SIDBI-
Sustainable Development; Services to MSMEs; Recommendations to RBI; Digital Initiatives.
Answer:
Recommendations to RBI

Question 4.
Terms related to SIDBI-
SIDBI startup mitra; Aamcha Mitra; Udyami Mitra; Micro Finance.
Answer:
Aamcha Mitra

Question 5.
Terms related to NABARD-
Apex Rural Bank; Recommendations to RBI; Green, Blue and White Revolution; Empowerment of women.
Answer:
Empowerment of women.

Question 6.
Woman’s Self Help Groups-
Urban Bank; Mahila Bachat Gat; Dr. Muhammad Yunus; Grameen Bank.
Answer:
Urban Bank.

Select the correct option

(Provides credit to rural development, Women Empowerment, Multilateral Investment Guarantee Agency (MIGA), KVIC, SIDBI)

Part A Part B
(i) World Bank ————-
(ii) ————— Achievement of National Goals
(iii) NABARD ——————-
(iv) Self Help Group (SHG) —————–
(v) ————— Promote Khadi and Village Industries

Answer:

Part A Part B
(i) World Bank Multilateral Investment Guarantee Agency (MIF)
(ii) SIDBI Achievement of National Goals
(iii) NABARD Provided credit to rural development
(iv) Self Help Group (SHG) Women Empowerment
(v) KVIC Promote Khadi and Village Industries

Answer in one sentence

Question 1.
Why do firm need finance ?
Answer:
Firms need finance to start business, to run business effectively and to expand business.

Question 2.
Name the institutes supporting business.
Answer:
The institutes supporting business are – SIDBI, NABARD, KVIC, Women’s Self Help Group and World Bank.

Question 3.
Where is the head office of SIDBI?
Answer:
The head office of SIDBI is at Lucknow.

Question 4.
Who resolves the financial and non-financial hurdles of MSMEs ?
Answer:
SIDBI resolves the financial and non-financial hurdles of MSMEs.

Maharashtra Board 11th OCM Important Questions Chapter 6 Institutes Supporting Business

Question 5.
Who are the focused domains for SIDBI?
Answer:
Micro, Small and Medium Enterprises (MSME’s ) are the focused domains for SIDBI.

Question 6.
Who helds the present ownership of SIDBI ?
Answer:
The present ownership of SIDBI is held by 34 public sector institutes, through its 15 regional offices and 84 branches.

Question 7.
Who acts as an advisor and mentor for MSMEs ?
Answer:
SDIBI act as an advisor and mentor for MSMEs.

Question 8.
Name the associates and subsidiaries of SIDBI.
Answer:
The associates and subsidiaries of SIDBI are – SVCL, CGTMSE, SMERA, ISTSL, ISARC and MUDRA.

Question 9.
What is the aim of ‘MUDRA’?
Answer:
The main aim of ‘MUDRA’ an associate of SIDBI is “funding the unfunded” micro enterprises in the country.

Question 10.
What are the different forms of finance offered by SIDBI?
Answer:
The different forms of finance offered by SIDBI are – direct finance, indirect finance and micro finance.

Question 11.
Name the two digital initiatives launched by SIDBI.
Answer:
The two digital initiatives launched by SIDBI are – SIDBI start-up Mitra and Udyami Mitra.

Question 12.
Who monitors the flow of rural credit in India?
Answer:
NABARD monitors the flow of rural credit in India.

Question 13.
What is the objective of KVIC?
Answer:
The objective of KVIC is to utilise the available raw materials and human skills to generate non-farm employment opportunities in the rural areas of India.

Question 14.
Who is the supreme policy making body of World Bank?
Answer:
Board of Governors is the supreme policy making body of World Bank.

Question 15.
What are the goals of World Bank to be achieved by, 2030?
Answer:
The two goals of World Bank to be achieved by 2030 is to end extreme poverty and promote shared prosperity.

Correct the Underlined word and rewrite the following sentences

Question 1.
SIDBI provides refinance to Banking Companies only.
Answer:
SIDBI provides refinance to Banking and Non-Banking Financial Comnanies.

Maharashtra Board 11th OCM Important Questions Chapter 6 Institutes Supporting Business

Question 2.
KVIC works as an advisor and mentor to MSMEs.
Answer:
SIDBI works as an advisor and mentor to MSMEs.

Question 3.
NABARD is working toward sustainable development of MSMEs in India.
Answer:
SIDBI is working toward sustainable development of MSMEs in India.

Question 4.
Udvami Mitra works as knowledge partner for State and Central Government.
Answer:
SIDBI Startup Mitra works as knowledge partner for State and Central Government.

Question 5.
SIDBI Startup Mitra was launched to improve accessibility of credit services to MSMEs.
Answer:
Udvami Mitra was launched to improve accessibility of credit services to MSMEs.

Question 6.
The apex institution for agricultural finance is SIDBI.
Answer:
The apex institution for agricultural finance is NABARD.

Question 7.
NABARD monitors the flow of Urban Credit in India.
Answer:
NABARD monitors the flow of Rural Credit in India.

Question 8.
World Bank provides recommendation to RBI on issue of licences to Co-operative Bank.
Answer:
NABARD provides recommendation to RBI on issue of licences to Co-operative Bank.

Question 9.
KVIC has its head office at Kolkata.
Answer:
KVIC has its head office at Mumbai.

Question 10.
Khadi Gramudvog works for planning, promotion and production of Khadi.
Answer:
KVIC works for planning, promotion and production of Khadi.

Question 11.
The social objective of KVIC is producing saleable articles.
Answer:
The economic objective of KVIC is producing saleable articles.

Question 12.
The wider objectives of KVIC is creating dependence amongst the poor.
Answer:
The wider objectives of KVIC is creating self-reliance amongst the poor.

Question 13.
KVIC signed Memorandum of Association with National Institute of design.
Answer:
KVIC signed Memorandum of Understanding with National Institute of design.

Question 14.
KVIC promote conventional energy for sustainable development.
Answer:
KVIC promote non-conventional energy for sustainable development.

Question 15.
“Helping Yourself” and ‘Unity is Strength’ is the principle of SHG.
Answer:
“Helping each other” and ‘Unity is Strength’ is the principle of SHG.

Question 16.
SHG – Bank Linkage programme was initiated by SIDBI since 1992.
Answer:
SHG – Bank Linkage programme was initiated by NABARD since 1992.

Question 17.
State Bank of India accepted SHG strategy as an alternative credit model.
Answer:
Reserve Bank of India accepted SHG strategy as an alternative credit model.

Question 18.
SHG is an involuntary association.
Answer:
SHG is a voluntary association.

Explain the following terms/concepts

Question 1.
SIDBI
Answer:

  1. Small Industrial Development Bank of India (SIDBI) was established on 2nd April, 1990.
  2. At present, the ownership of SIDBI is held by 34 public sector institutes.
  3. Main focus of SIDBI is to resolve the financial and non-financial hurdles of MSMEs.
  4. SIDBI helps for technological upgradation and modernisation of existing MSMEs.

Question 2.
SIDBI Startup Mitra
Answer:

  1. It is a digital initiative set up by SIDBI.
  2. It tries to bring all stakeholders, investors, new entrepreneurs, banks, mentors, etc on one platform.
  3. It helps in financing and development of new entrepreneurs.
  4. It also acts as a knowledge partner for State and Central Government.

Maharashtra Board 11th OCM Important Questions Chapter 6 Institutes Supporting Business

Question 3.
NABARD
Answer:

  1. It came into existence on 12th July, 1982 under the special Act of Parliament It is counted as an apex institution for agricultural finance.
  2. The main objective of NABARD is to provide and regulate credit and other facilities for the promotion of agriculture, small scale industries, cottage and village industries, etc.
  3. It provides refinancing facilities to State Co-operative Bank (SCBs), Land Development Bank (LDBs), Regional Rural Banks (RRBs) and other approved financial institutions.

Answer in brief

Question 1.
Explain the Digital initiative steps of SDBI.
Answer:
The digital initiative steps of the SDBI are as follows:
SIDBI Startup Mitra and Udyami Mitra:
(a) SIDBI Startup Mitra:
It brings together all stakeholders, start-up entrepreneurs, incubators, investors, industry bodies, mentors and advisors and banks at one platform. It helps in financing and development of new entrepreneurs. It also works as knowledge partner for State and Central Government.

(b) Udyami Mitra:
This is the portal launched by SIDBI to improve accessibility of credit and handholding services to MSME’s. Under this portal entrepreneurs can apply for loan without physically visiting any bank branches. The entrepreneurs can select and apply for preferred banks, they can select suitable branch, track their application status an avail multiple loan benefits.

Question 2.
Explain the role of NABARD in national development.
Answer:
Role of NABARD in national development:

  1. NABARD is an apex bank providing credit facilities to institute working in agricultural finance.
  2. It provides storage facilities for agricultural commodities by promoting development of warehousing facilities.
  3. It also promote the export of agricultural commodities.
  4. It plays an important role in sustainable development of the country through Green, Blue and White revolution.
  5. It refinances commercial and co-operative banks during natural calamities like drought, crop failure, etc. so that farmers can recover from the difficult period.

Question 3.
How does KVIC help in marketing and promotion?
Answer:
Role of KVIC in marketing and promotion:

  1. KVIC holds exhibitions, seminars and lectures in college and universities to spread the knowledge of KVIC products.
  2. It also launches a marketing development plan to generate interest, awareness and attraction amongst the masses.
  3. It takes all effort to improve the quality, packaging and marketing of a product.
  4. Government also supports KVIC, so that its product becomes competitive.

Maharashtra Board 11th OCM Important Questions Chapter 6 Institutes Supporting Business

Question 4.
Explain the formation of Womens SHGs.
Answer:
Formation of Women’s SHG’s.

  1. It is an informal group of homogeneous individuals generally formed by a NGO or by team of government.
  2. It is recognized by the Government, hence it does not require any registration under any Act.
  3. They have their well defined rules and bye-laws to run the group.
  4. Regular meetings of this groups are held and records are properly maintained.

Maharashtra Board 11th OCM Important Questions Chapter 3 Small Scale Industry and Business

Balbharti Maharashtra State Board 11th OCM Important Questions Chapter 3 Small Scale Industry and Business Important Questions and Answers.

Maharashtra State Board 11th Commerce OCM Important Questions Chapter 3 Small Scale Industry and Business

Select the Correct option and rewrite the sentence

Question 1.
Small Scale Industries play an important role in ……………. countries.
(a) developed
(b) developing
(c) advanced
Answer:
(b) developing

Question 2.
Traditionally a small scale industries are those using power with less than ……………… employees.
(a) 20
(b) 30
(c) 50
Answer:
(c) 50

Maharashtra Board 11th OCM Important Questions Chapter 3 Small Scale Industry and Business

Question 3.
……………….. is second largest industry which creates huge employment.
(a) SSI
(b) Reliance Ltd.
(c) Indian Railways
Answer:
(a) SSI

Question 4.
……………….. of people from rural to urban is reduced by small scale industries.
(a) Immigrations
(b) Migrations
(c) Re-migrations
Answer:
(b) Migrations

Question 5.
Small Scale Industries are ……………….. in operation.
(a) rigid
(b) flexible
(c) static
Answer:
(b) flexible

Question 6.
Small Scale Industries uses ……………….. technology.
(a) outdated
(b) advanced
(c) future
Answer:
(a) outdated

Question 7.
………………. resources are used by SSI.
(a) Foreign
(b) City
(c) Domestic
Answer:
(c) Domestic

Match the pairs

Question 1.

Group A Group B
(a) Small Manufacturing Sector (1) Investment more than Rs 2 crores but does not exceed Rs 5 crores
(b) Modern Small Scale Industries (2) SSI
(c) Large Employment (3) Capital problem
(d) High absenteeism (4) Investment does not exceed Rs 25 lakhs
(e) Medium Services Sector (5) Electric Appliances
(6) Labour problem
(7) Investment does not exceed Rs 10 lakhs
(8) Sericulture
(9) L&T

Answer:

Group A Group B
(a) Small Manufacturing Sector (4) Investment does not exceed Rs 25 lakhs
(b) Modern Small Scale Industries (5) Electric Appliances
(c) Large Employment (2) SSI
(d) High absenteeism (6) Labour problem
(e) Medium Services Sector (1) Investment more than Rs 2 crores but does not exceed Rs 5 crores

Give one word/phrase/term

Question 1.
Tiruppur a city located in state of Tamil Nadu is famous for which export.
Answer:
Textile export

Question 2.
Industries which requires low gestation period.
Answer:
Small Scale Industries

Maharashtra Board 11th OCM Important Questions Chapter 3 Small Scale Industry and Business

Question 3.
Part of rural industry based on agricultural products.
Answer:
Agro based industries

Question 4.
Moving of people from one place to another in search of job.
Answer:
Migration

Question 5.
Regular interruption of electricity.
Answer:
Load shedding

Question 6.
Life blood of the enterprise.
Answer:
Finance

Question 7.
Assessment of a project.
Answer:
Project Appraisal

Question 8.
Procedure of recruitment, training and placement.
Answer:
Staffing.

State True or False

Question 1.
Small Scale Industries increases import of raw materials from foreign countries.
Answer:
False

Question 2.
Tiruppur contribute to export textiles and earn valuable foreign exchange for India.
Answer:
True

Question 3.
Japan became major economic power because of many small entrepreneurs after second world war.
Answer:
True

Question 4.
Nearly 60% of the industrial exports are contributed by SSI.
Answer:
False

Maharashtra Board 11th OCM Important Questions Chapter 3 Small Scale Industry and Business

Question 5.
Small Scale Industries lead to inequalities of income, wealth and power.
Answer:
False

Question 6.
Small Scale Industries are more flexible.
Answer:
True

Find the odd one

Question 1.
Manufacturing Sector, Services Sector, Construction Sector
Answer:
Construction Sector

Question 2.
Does not exceed ?10 lakhs, Does not exceed Rs 25 lakhs, Does not exceed Rs 2 crores.
Answer:
Does not exceed Rs 25 lakhs

Question 3.
Job opportunities, Regional Balance, Increases Migration
Answer:
Increases Migration

Question 4.
Labour Problem, Marketing problem, Problem of transport, Domestic problem
Answer:
Domestic problem

Question 5.
Selection of product, Selection of place, Selection of House
Answer:
Selection of House.

Complete the sentences

Question 1.
Problem of is reduced or minimized by SSI.
Answer:
migration

Question 2.
Gestation period is low for industries.
Answer:
Small Scale

Question 3.
means the assessment of a project.
Answer:
Project appraisal

Question 4.
SSI provides raw material, semi finished goods to
Answer:
Large Scale

Maharashtra Board 11th OCM Important Questions Chapter 3 Small Scale Industry and Business

Question 5.
place from Tamil Nadu export textiles.
Answer:
Tiruppur

Question 6.
method of production is used by SSI.
Answer:
Old and outdated.

Select the correct option

(Modern Small Scale Industries, Challenges before Small Scale Industries, Traditional Small Scale Industries, Importance of Small Scale Industries)

Group A Group B
(1) Hand loom, Handicraft, Coir, Sericulture ——————-
(2) —————– Bicycles parts, sewing machines, Blades, Razor, Electric Appliances, Spare parts
(3) Job opportunities, Regional balance, Reduce migration, Maximum use of Natural Resources ——————-
(4) ————– Load-shedding, voltage fluctuation.

Answer:

Group A Group B
(1) Hand loom, Handicraft, Coir, Sericulture Traditional Small Scale Industries
(2) Modern Small Scale Industries Bicycles parts, sewing machines, Blades, Razor, Electric Appliances, Spare parts
(3) Job opportunities, Regional balance, Reduce migration, Maximum use of Natural Resources Importance of Small Scale Industries
(4) Challenges before Small Scale Industries Load-shedding, voltage fluctuation.

Answer in one sentences

Question 1.
What are the challenges before Small Scale industries?
Answer:
Old and outdated technology, finance, marketing, load shedding, voltage fluctuation, sickness, etc. are the challenges before Small Scale Industries.

Question 2.
What are the sectors of business classification as per SSI?
Answer:
There are two sectors of business as per SSI i.e.

  1. Manufacturing sector
  2. Services sector.

Question 3.
What is the classification of Small Scale Industries?
Answer:
Small Scale Industries are classified into two i.e. Traditional Small Scale Industries and Modern Small Scale Industries.

Maharashtra Board 11th OCM Important Questions Chapter 3 Small Scale Industry and Business

Correct the underlined word and rewrite the following sentences.

Question 1.
SSI is first largest industry which creates huge employment.
Answer:
SSI is second largest industry which creates huge employment.

Question 2.
Migration of people from rural to urban can be increase due to SSI.
Answer:
Migration of people from rural to urban can be reduced due to SSI.

Question 3.
Many Small Scale Industries provide finished goods to large scale industries.
Answer:
Many Small Scale Industries provide semi finished goods to large scale industries.

Question 4.
India is facing labour scarcity problem.
Answer:
India is facing labour problem.

Question 5.
Nearly 60% of industrial exports are contributed by SSI.
Answer:
Nearly 40% of industrial exports are contributed by SSI.

Study the following case/situation and express your opinion.

1. Mr. Rahim is keenly interested in starting Small Scale business. But he is worried about the challenges before Small Scale business. Guide him about the solutions to face these challenges.

Question 1.
Guidelines to start a small scale business:
Answer:
1. Method of Production : It is said that small scale industries use old methods of production Mr. Rahim can improve the method of production by using new technology. He can purchase the new technology at affordable rate and give training to the workers so that they can improve the skills and in turn increase the production.

2. Finance : Rahim can get information about various schemes given by government to raise finance. He can also compare the rate of interest of various financial institutes and take loan at a lower rate of interest.

3. Raw Material : Rahim can start business in area near by to the availability of raw material so that he can save the cost of transportation of raw material. If the cost of raw material decreases he will not have to compromise on the quantity and quality of raw material.

4. Labour Problem : Rahim can give training to the workers so that they can increase the productivity. Skilled workers can work more efficiently. Rahim should also provide with job security so that workers do not leave the job and experienced workers can increase productivity.

5. Marketing Problem : Rahim should have thorough study of need of the customer in the market. He should have all market updates so that he can make changes according to the requirement of customers. He can do market survey with use of latest technology i.e. Internet and WhatsApp. He can make use of these technology for advertising of his product also.

Maharashtra Board 11th OCM Important Questions Chapter 3 Small Scale Industry and Business

6. Transportation : As stated in point no. (c) Rahim can start business in area nearby to the availability of raw material so that he can save the cost of transportation of raw material. He can also keep the cost of the product less and then add the cost of transportation afterwards so that cost of his product will look less.

7. Sickness : Rahim will have to take care of the credit system which he is going to follow. It is advisable that he sells the goods on cash basis and not on credit basis. If he gives it on credit he should be good at follow up. While purchasing raw material he can take it on credit basis so that he gets adequate working capital to use. If he gets sufficient working capital he can produce goods on time and the industry will not become sick.

8. Globalisation : Rahim is to be aware about the various competitive products and other similar products which are there in the market. He will have to be updated with the latest things and latest prices and make changes in his small unit regularly as and when required.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 1.
What are alcohols? How are they classified?
Answer:
Alcohols are the hydroxy derivatives of hydrocarbons in which one or more hydrogen atoms are replaced by hydroxyl group.

Examples : CH3 – OH methyl alcohol, CH3 – CH2 – OH ethyl alcohol. Depending on the basis of hydroxyl groups present in a molecule, alcohols are classified into monohydric, dihydric, trihydric and polyhydric alcohols.

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 1

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 2.
What are monohydric alcohols? How are they classified?
Answer:
Alcohols having only one hydroxyl group in their molecules are called monohydric alcohols. Monohydric alcohols are classified according to the type of hybridization of the carbon atom to which the hydroxyl group is attached.

(1) Alcohols containing Csp3 – OH bond : In these alcohols -OH group is attached to a sp3 – hybridised carbon atom of alkyl group. These alcohols are represented as R-OH. They are further classified as primary, secondary and tertiary alcohols in which – OH group is attached to primary, secondary and tertiary carbon atoms respectively.

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 2

(a) Allylic alcohols : In these alcohols -OH group is attached to a sp3 -hybridised carbon atom next to the carbon-carbon double bond i.e., to allylic carbon. Allylic alcohols may be primary, secondary and tertiary alcohols.

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 3

(b) Benzylic alcohols : In these alcohols -OH group is attached to a sp3 -hybridised carbon atom next to an aromatic ring. Benzylic alcohols may be primary, secondary and tertiary alcohols.

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 4

(2) Alcohols containing Csp3 – OH bond: In these alcohols -OH group is attached to a sp2 – hybridised carbon atom,
i.e., vinylic carbon. These alcohols are also called vinylic alcohols.
e.g., CH2 = CH – OH vinyl alcohol.

Question 3.
What are phenols (carbolic acids)?
Answer:
Hydroxy derivatives of aromatic hydrocarbons in which the hydroxyl group is directly attached to the aromatic ring are called phenols.
Examples:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 5

Question 4.
What is phenol? OR Define carbolic acid.
Answer:
The hydroxy derivative of benzene in which the OH group is directly attached to benzene ring is called phenol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 6

Question 5.
How are phenols classified? Give suitable examples.
Answer:
Phenols are classified on the basis of number of hydroxyl (- OH) groups present in a molecule of phenol.
(1) Monohydric phenols : Phenols contain one hydroxyl group in their molecule.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 8

(2) Dihydric Phenols : Phenols contain two hydroxyl groups in their molecule.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 9

(3) Trihydric phenols : Phenols contain three hydroxyl groups in their molecule.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 10

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 6.
What are ethers? How are ethers classified?
Answer:
They are alkoxy derivatives of alkanes in which a hydrogen atom of alkane (R – H) is replaced by alkoxy group ( – O – R) and divalent oxygen atom is attached to two alkyl groups or two aryl groups or one alkyl and one aryl group. Ethers are organic oxides R – O – Ar, Ar – O – Ar.
E.g. R – O – R or C2H5 – O – C2H5.

Ethers are classified into two groups as follows :
(1) Simple or symmetrical ether : The ethers in which both alkyl (or aryl) groups attached to the oxygen atom are same are called simple ethers.
E.g. (R – O – R), CH3 – O – CH3, dimethyl ether; C6H5 – O -C6H5 diphenyl ether.

(2) Mixed or unsymmetrical ethers : The ethers in which the two alkyl (or aryl) groups attached to the oxygen atom are different are called mixed ethers.
E.g. (R – O – R), CH3 – O – C2H5 ethyl methyl ether; C2H5 – O – C6H5 ethyl phenyl ether.

Question 7.
What is the general formula of ethers?
Answer:
The general formula of ethers is CnH2n + 2 O. For example, dimethyl ether CH3 – O – CH3 has molecular formula C2H6O.

Common nomenclature system :
(1) In this system, monohydric alcohols (R — OH) are named as alkyl alcohols.
(2) According to the attachment of hydroxyl group to a carbon atom they are named with prefixes as n-(normal or primary) alcohol, sec-(secondary) alcohol, tert-(tertiary) alcohol.
(3) Alcohols with two hydroxyl groups are named as glycols.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 12

Carbinol system : In this system alcohols are considered as derivatives of methyl alcohol which is called carbinol. The alkyl group attached to the carbon carrying – OH group are named in alphabetical order. Then the suffix carbinol is added.

For example :Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 13

IUPAC system of nomenclature :

  • In this system, alcohols are named as alkanols.
  • A longest continuous chain of carbon atoms containing – OH group is chosen as a parent hydrocarbon and alcohol is considered as a hydroxy derivative of this alkane.
  • The carbon atoms are numbered from a terminal carbon atom nearest to a carbon atom attached to – OH group so that the position of – OH group is indicated by the lowest locant.
  • ‘e’ of an alkane is preplaced by ‘oT, giving alkanol. The number of OH groups is indicated by prefix, di, tri, etc. before ‘oT. The positions of -OH groups are indicated by appropriate locants.
  • The different substituents are arranged in the alphabetical order, and their positions are indicated by proper numbers.
  • Their names are hyphened on either sides except the last substituent.
  • For cyclic alcohols are named by using prefix cyclo to the parent alkane considering – OH group attached to carbon atom C – 1.

Question 8.
Give common and IUPAC names for the following :
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 14
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 15

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 9.
Classify the following alcohols as primary, secondary and tertiary and write their IUPAC names.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 16

Question 10.
Give the structures of the following :
(1) Ethanol
(2) Propan-l-ol
(3) 2-Methylbutan-l-ol
(4) Pentan-3-ol
(5) Butan-2-ol
(6) 3-Methylbutan-2-ol
(7) Hexan-l-ol
(8) 3-Methylpentan-3-ol
(9) 2-Methylpropan-2-ol
(10) Butan-l-ol (H-Butyl alcohol)
(11) 2-Methylpropan-l-ol
(12) 2,3-Dimethylbutan-l-ol
(13) 2, 3-Dimethylbutan-2-oI
(14) 2-MethyIhexan-l-ol
(15) 2,2,3-Trimethylpentan-3-ol
(16) 2,3,3-Trimethylbutan-2-ol
(17) 3-EthyI-4-methylpentan-l-ol.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 17
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 18

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 11.
Give the IUPAC names of the following alcohols. Classify them as primary (1°), secondary (2°) and tertiary (3°) alcohols. Identify allylic and benzylic alcohols amongst them.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 19
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 20

Question 12.
Write all the possible structural isomers of alcohol having molecular formula C6H140. Give their IUPAC names. Classify them as primary, secondary and tertiary alcohols. Identify optically active alcohols amongst them.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 21
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 22

The optical isomers of C6H14O are (2), (3), (6), (7), (12).

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 13.
Give the IUPAC names of the following compounds :
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 23

Question 14.
Write IUPAC names of following alcohols :
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 24

Question 15.
Write the structures of following alcohols : (1 mark each)
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 25

The IUPAC system name of phenol is benzenol. The common name phenol is also accepted by IUPAC. The common names have prefixes ortho, meta and para in substituted phenols. IUPAC system uses the locant 2-, 3-, 4-, etc. to indicate the positions of substituents.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 16.
Write the IUPAC names of the following compounds :
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 26
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 27

Question 17.
Write IUPAC name of the following compounds :
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 28

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 18.
Give the structures and IUPAC names of isomeric phenols represented by the molecular formula C8H10O.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 29

Common and IUPAC system of nomenclature of ethers

In the common system of nomenclature, the ethers are named by writing names of the alkyl groups attached to the oxygen atom in alphabetical order and word ether is added. If two alkyl groups are same, prefix di- is used. According too the IUPAC system of nomenclature, ethers are named as alkoxyalkanes. The larger alkyl group is considered to be parent alkane. The name of the smaller alkane is prefixed by the name of alkoxy group and its locant.

Question 19.
Give common name and JUPAC name for the I11owing ethers :
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 30

Question 20.
Give the IUPAC name of the following ethers.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 31

Question 23.
Give the structures and IUPAC names of all metameric ethers represented by formula C5H12O.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 38

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 24.
How many isomeric compounds can be represented by formula C4H10O?
Answer:
A compound with the molecular formula C4H10O can show two functional isomers as a monohydric alcohol and ether.

Isomers of C4H10O as alcohol :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 39
Isomers of C4H10O as ether :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 40

Hence, the total number of isomers of C4H10O are seven

Question 25.
Write the structural formula and IUPAC names of all possible isomers of the compound with molecular formula C3H8O.
Answer:
Possible isomers of C3H8O with structural formulae and IUPAC names :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 41

Question 26.
Write structures of alcoholic and ether isomers of a compound having molecular formula C7H8O.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 42

Question 27.
How is alkyl halide converted into alcohol by using
(1) Aqueous NaOH (or KOH),
(2) Moist silver oxide?
Answer:
(1) When an alkyl halide (R – X), is boiled with aqueous NaOH (or KOH) an alcohol is obtained,
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 43
(2) Alkyl halide when heated with moist Ag2O, undergoes hydrolysis and forms an alcohol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 44

Question 28.
How are following compounds prepared by hydrolysis of alkyl halides
(1) Ethanol
(2) Isopropyl alcohol
(3) tert-butyl alcohol
(4) methyl alcohol
(5) butan-2-ol?
Answer:
(1) Ethanol When ethyl bromide (bromoethane) is refluxed with aqueous potassium hydroxide, ethyl alcohol is formed. The reaction is called a hydrolysis reaction.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 45

(2) Isopropyl alcohol : When isopropyl bromide (2-bromopropane) is boiled with aqueous potassium hydroxide, isopropyl alcohol is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 46

(3) Tert-butyl alcohol : When tert-butyl chloride is refluxed with aqueous potassium hydroxide, tert-butyl alcohol is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 47

(4) Methyl alcohol : When methyl bromide (bromomethane) is heated with aq KOH, it is hydrolysed to methyl alcohol (methanol).
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 48

(5) Butan-2-ol : When 2-Chlorobutane is boiled with aqueous KOH, Butan-2-ol is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 49

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 29.
How are following compounds prepared from alkyl halides using moist silver oxide?
(1) Ethanol
(2) Propan-2-ol.
Answer: ‘
(1) Bromoethane (C2H5Br) when boiled with moist Ag2O undergoes hydrolysis and forms C2H5OH.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 50

(2) When 2-chloropropane is boiled with moist Ag20, propan-2-ol is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 51

Question 30.
What is hydration of alkenes or olefins? How is it carried out? Explain with an example.
OR
How are alcohols prepared from alkenes?
Answer:
The addition of a water molecule across the double bond in an alkene is called hydration of alkenes or olefins.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 52

Hydration does not take place directly. It is carried out by passing an alkene through cold and concentrated H2SO4 which forms deliquescent solid, alkyl hydrogen sulphate, which when boiled with water forms an alcohol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 53

Question 31.
How are the following compounds obtained by hydration of alkenes :
(1) Ethyl alcohol
(2) Isopropyl alcohol
(3) Tert-butyl alcohol?
Answer:
(1) When ethene is passed through cold 98 % H2SO4, ethyl hydrogen sulphate is formed, which on heating with water gives ethanol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 54

(2) Propene with cold 80% H2SO4 gives isopropyl hydrogen sulphate which further on boiling with water gives isopropyl alcohol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 55

(3) 2-Methylpropene (isobutylene) directly reacts with 50 % H2SO4 giving tert-butyl hydrogen sulphate, which when heated with water gives tert-butyl alcohol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 56

Question 32.
Identify C in the following reaction.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 57
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 58

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 33.
Explain hydroboration-oxidation of alkene.
Answer:
When diborane is treated with alkene in the presence of tetrahydrofuran (THF) solvent, an addition product trialkyl borane is formed. Trialkyl borane is then oxidised with alkaline peroxide forms primary alcohol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 62

The addition of diborane to the double bond takes place in such a way that the boron gots attached to the less substituted carbon. The overall reaction gives Anti-Markovnikov’s product from unsymmetrical alkenes.

Question 34.
How is propan-l-ol prepared using diborane?
Answer:
When diborane is treated with propene, in the presence of THF an addition product tripropyl borane is formed. Tripropyl borane is then oxidised to propan-l-ol using hydrogen peroxide in the presence of dil NaOH.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 63
The addition of diborane to the double bond takes place in such a way that the boron gets attached to the less substituted carbon. The alcohol formed by the addition of water to the alkene in a way opposite to the Markovnikov’s rule.

Question 35.
How is ethanol prepared using diborane?
Answer:
When diborane is treated with ethene in the presence of THF an addition product triethylborane is formed.Triethylborane is then oxidised with hydrogen peroxide to form ethyl alcohol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 64

Question 36.
Predict the major product when 2-methylbut-2-ene is converted into an alcohol in each of the following methods :
(1) acid catalysed hydration
(2) hydroboration by BH3 – THF complex.
Answer:
(1) Acid catalysed hydration :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 66
(2) Hydroboration by BH3 – THF complex :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 67

Question 37.
How are alcohols prepared from aldehydes and ketones?
OR
How are the following compounds obtained using Ni as catalyst and at high temperature
(1) Ethanol
(2) Propan-2-ol?
Answer:
Aldehyde and ketones are carbonyl compounds containing a carbonyl group  C = O. The reduction of the carbonyl group gives an alcohol.

(1) Primary alcohols are prepared by the reduction of aldehydes.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 68

Example : When acetaldehyde is reduced with hydrogen in the presence of nickel as catalyst and at high temperature, ethyl alcohol (ethanol) is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 69

(2) Secondary alcohols are prepared by the reduction of ketones.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 70

Example : When acetone is reduced with hydrogen in the presence of nickel as catalyst and at high temperature, isopropyl alcohol (propan-2-ol) is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 71

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 38.
How will you convert carboxylic acids and esters to primary alcohols? Explain with suitable examples.
Answer:
Carboxylic acids and esters are not easily reduced by catalytic hydrogenation or by NaBH4. However, relatively more reactive and an expensive LiAlH4 is used to convert carboxylic acids and esters to primary alcohols. When acetic acid is reduced in the presence of LiAlH4 and followed by their acid hydrolysis, ethyl alcohol is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 72

When ethyl acetate is reduced in the presence of LiA1H4 and followed by their acid hydrolysis. n-propyl alcohol and ethyl alcohol is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 73

[Since LiA1H4 is an expensive reagent. commercially acids are reduced to alcohol by converting them to esters, followed by their reduction. (catalytic hydrogenation)
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 74

Question 39.
How is Crotoflyl alcohol obtained from crotonaldehyde?
Answer:
When crotonaldehyde is reduced in the presence of lithium aluminium hydride, the produc obtained is hydrolysed to give crotonyl alcohol. Here. LiA1H4 does not reduce carbon-carbon double bond.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 75

Question 40.
WrIte the structure of aldehyde that yields
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 79
Answer:
The structure of aldehyde:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 80

Question 41.
How are the following compounds prepared using Grignard reagent
(1) Ethanol
(2) Propan-l-ol
(3) Propan-2-ol
(4) 2-Methyl propan-2-ol?|
Answer:
(1) Ethanol : Formaldehyde on reaction with Grignard reagent, CH3 – Mg – I in dry ether forms a complex which on further hydrolysis with dilute HCl forms ethanol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 81

(2) Propanol-l-ol : Formaldehyde on reaction with Grignard reagent, C2H5 – Mg – I in dry ether forms a complex which on further hydrolysis with dilute HC1 forms Propan-l-ol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 82

(3) Propan-2-ol : An acetaldehyde on reaction with Grignard reagent in dry ether forms a complex which on further hydrolysis with dilute acid HC1, forms propan-2-ol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 83
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 84

(4) 2-Methyl propan-2-ol : Acetone on reaction with Grignard reagent in dry ether forms a complex which on further hydrolysis with dilute acid HCl, forms a tertiary butyl alcohol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 85

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 42.
Give a mechanism of following reaction.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 86
Answer:
Grignand reagent reacts with aldehyde or ketone to form an adduct which on hydrolysis with dil. acid gives the corresponding alcohol.

In the first step, the nucleophilic addition of Grigard reagent to the carbonyl group resulting in the formation of an adduct, which on hydrolysis yields an alcohol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 87

Question 43.
Write the structure of carbonyl compounds that can be converted by reduction methods into following alcohols :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 89
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 90
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 91

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 92
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 93

Question 44.
Using Grignard reagent, suggest synthesis of following alcohols from aldehydes or ketones. Wherever possible, suggest more than one combination.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 94
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 95
(b) Synthesis of propan- 1-01:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 96
(c) Synthesis of butan-2-ol:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 97
(d) Synthesis of 2-methylhexan-2-oI:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 98

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 45.
How will you obtain butan-2-ol from
(1) Propanal
(2) butan-2-one
(3) but-2-ene?
Answer:
(1) Propanal : When propanal is treated with methyl magnesium iodide in the presence of dry ether, a complex is formed, which on acid hydrolysis butan 2-ol is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 99
(2) Butan-2-one: When butan-2-one is hydrogenated at 413 K in the presence of catalyst finely divided nickel butan-2-ol is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 100
(3) But-2-ene : When but-2-ene is passed through cold concentrated sulphuric acid, isobutyl hydrogen sulphate is formed. Isobutyl hydrogen sulphate on heating with water gives butan-2-ol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 101

Question 46.
Write the structure of aldehyde, carboxylic acid and ester that yield the following alcohol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 102
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 103

Question 47.
How will you prepare?
(1) 2-Ntethylbutan-1-oI from an alkene.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 104

(2) CycIoheyImethanoI from a Grignard reagent.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 105

(3) 1-Phenyl ethanol from acelaldehyde.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 106

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 48.
How are following conversions brought about?
(1) Benzyl chloride to benzyl alcohol. (NCERT)
OR
How is benzyl alcohol prepared from benzyl chloride?
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 107

(2) Benzyl alcohol to Benzoic acid.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 108

(3) 1-Ethyl cyclohexanol from cyclohexanone.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 109

Question 49.
How is phenol (carbolic acid) prepared from chlorobenzene (Dow’s process)? OR
Write chemical reaction for the preparation of phenol from chlorobenzene.
Answer:
Preparation of phenol from chlorobenzene (Dow’s process) : Chlorobenzene is fused with NaOH at about 623 K under a pressure of about 150 atmospheres (1.5 x 107 Nm-2), when sodium phenoxide is formed. Sodium phenoxide is acidified with dil.HCl to obtain phenol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 110

Question 51.
How is phenol (carbolic acid) prepared from benzene sulphonic acid?
Answer:
Preparation of phenol from benzene sulphonic acid : Benzene sulphonic acid is neutralized with the requisite quantity of soda ash (Na2CO3) or NaOH and the solution is evaporated to obtain sodium benzene sulphonate salt. Dry sodium benzene sulphonate is fused with an excess of caustic soda (NaOH) at about 573 K when sodium phenoxide is formed. The fused mass of sodium phenoxide on treatment with dilute HC1 gives phenol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 112

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 52.
How is phenol (carbolic acid) prepared from aniline (diazotization)?
OR
How is carbolic acid prepared from amino benzene?
Answer:
Preparation of phenol from aniline (diazotization) : When aniline is treated with sodium nitrite and hydrochloric acid (NaNO2 + HC1) at low temperature (0°C – 5°C), benzene diazonium chloride is formed. This reaction is called diazotization. An aqueous solution of benzene diazonium chloride on warming with water or dil. H2S04 gives phenol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 113

Question 54.
Describe the physical properties of alcohols and phenols.
Answer:
The properties of alcohols and phenols are mainly due to the hydroxyl group.

(1) Nature of intermolecular forces : Due to presence of – OH groups, alcohols and phenols are polar molecules. The polar – OH groups are held together by the strong intermolecular forces i.e. hydrogen bonding.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 114

(2) Physical State : Lower alcohols are colourless, toxic liquids having characteristic alcoholic odour. Pure phenol is colourless, toxic, low melting solid having characteristic carbolic or phenolic odour.

(3) Boiling Points : The boiling points of alcohols and phenols increase with increase in their molecular mass.
Methyl alcohol – 65 °C
Phenol -182° C
n-Butyl alcohol-118 °C
o-nitrophenol-217 °C

(4) Solubility : Solubility of alcohols and phenols in water due to their ability to form intermolecular hydrogen bonding.

Question 55.
Arrange the following compounds in order of their increasing boiling points.
Butan-2-ol, ethanol, pentan-l-ol, butan -l-ol, propan-l-ol, methanol.
Answer:
Methanol, ethanol, propan:l-ol, butan-2-ol, butan-1 -ol pentan-l-ol.

Question 56.
Explain the following :
(1) Ethanol has higher boiling point than ethane.
Answer:
(1) The hydroxyl group in alcohols is highly polar. The H-atom has partial positive charge and the oxygen atom has partial -ve charge. The hydroxyl group in ethanol is extensively hydrogen bonded.

(2) Large number of ethanol molecules associated together by intermolecular hydrogen bonding. The energy required to separate the molecules by breaking hydrogen bond into vapour state is higher. This results in increasing the boiling point of ethanol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 116

(3) On the other hand, in ethane (alkane) there is no hydrogen bonding between the molecules. The molecules of ethane are held together by weak van der Waals forces of attraction. Hence, these molecules can be easily separated. Thus, ethanol has higher boiling point than that of ethane.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

(2) Methanol is more soluble in water than propan-l-ol.
Answer:
(1) Methanol being lower members of alcohols is more soluble in water, but as the size of an alkyl group or molecular weight of alcohol increases the solubility decreases.

(2) The solubility of methanol in water is due to polar characters of alcohols (R – O -H ) and water (H – O – H).

(3) The solubility of methanol is due to the formation of intermolecular hydrogen bonding between polar molecules of methyl alcohol and water. Hence, methyl alcohol is a associated liquid.

(4) In methyl alcohol, size of methyl group being very small, -OH group constitutes major part of the molecule giving more solubility. As size of alkyl group increases, the non-polar character increases the solubility decreases. Hence, methanol is more soluble in water than propan-l-ol.

Question 57.
Which of the following pair is more acidic and why?
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 270
Answer:
Due to high electronegativity of sp2-hybridized carbon, electron density on oxygen in phenol (I) decreases. This increases the polarity of O – H bond and results in more ionization of phenol than that of cyclohexyl alcohol (II). Therefore, phenol is more acidic than cyclohexyl alcohol.

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 120
Answer:
In p-nitrophenol, nitro group (N02) is an electron-withdrawing group present at ortho position which enhances the acidic strength (- I effect). The O – H bond is under strain and release of proton (H+) becomes easily. Hence, o-nitrophenol is more acidic than phenol.

Question 58.
Draw intramolecular hydrogen bonding structures in the following compounds :
(a) o-nitrophenol
(b) o-hydroxy benzoic acid.
Answer:
(a) o-nitrophenol :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 121
(b) o-hydroxy benzoic acid :

Question 59.
Explain laboratory test of alcohols and phenols.
Answer:
Laboratory test : Aqueous solution of alcohols and phenols can be tested with litmus paper. Aqueous solution of alcohols is neutral to litmus (neither blue nor red litmus change colour). Aqueous solutions of phenols turn blue litmus red. Thus, phenols have acidic character.

Question 62.
Write the action of aq NaOH on phenol and the product obtained is acidified.
Answer:
Phenols dissolve in aqueous NaOH by forming water soluble sodium phenoxide and are reprecipitated/ regenerated as phenols on acidification with HC1.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 124

Question 63.
Explain the action of sodium on ethanol.
Answer:
When cthanol is treated with sodium metal, sodium ethoxide is formed and hydrogen gas is liberated.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 125
Liberation of H2 gas is used to detect the presence of alcoholic -OH group of a molecule.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 64.
How are the following compounds obtained from alcohols using HCI:
(1) C2H5CI
(2) Isopropyl chloride
(3) tert-butyl chloride?
Answer:
(1) Ethyl alcohol, C2H5OH in the presence of Lucas reagent (ZnCl2 + HCI conc.) forms ethyl chloride.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 126

(2) Isopropyl alcohol reacts with Lucas reagent forms isopropyl chloride.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 127

(3) Tert-butyl alcohol reacts with Lucas reagent forms tert-butyl chloride
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 128

Question 65.
What is Lucas reagent? What are its uses?
Answer:
Lucas reagent is composed of a mixture of concentrated HCl and Lewis acid. anhydrous ZnCI2.

It is used to prepare alkyl chlorides and distinguish between 10, 2° and 3° alcohols.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 129

Question 66.
How can alcohols be distinguished with the help of Lucas reagent?
Answer:
Lucas reagent is a mixture of concentrated HCI and anhydrous ZnCl2. It is used to distinguish between primary, secondary and tertiary alcohols.
Alcohol with Lucas reagent forms an alkyl chloride, R -Cl which is insoluble and gives cloudiness and forms separate layer.

The time required for cloudiness to appear is based on the type of the alcohol and its reactivity :

(1) Primary alcohol : It reacts with the Lucas reagent very slowly and on heating forms alkyl chloride. The cloudiness and separation of a layer takes place after a long time on heating.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 130

(2) Secondary alcohol : It reacts with the Lucas reagent much faster to form alkyl chloride. The cloudiness and separation of layer takes place slowly.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 131

(3) Tertiary alcohol : It reacts immediately with the Lucas reagent at room temperature to form alkyl chloride. The cloudiness and separation of layer takes place instantaneously.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 132

Question 67.
What is esterificatlon? how Is an ester obtained from alcohol or phenol?
Answer:
(1) When an alcohol or phenol is heated with a carboxylic acid in the presence of conc.sulphuric acid an ester is obtained. The reaction is called esterification. This is reversible reaction and the formation of an ester is favoured
using excess of alcohol in the presence of conc. H2SO4.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 138

(2) Alcohols and phenols react with acid anhydrides in presence of acid catalyst to form eser.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 139

(3) The reaction of alcohol and phenols with acid chloride is carried out in the presence of pyridine (base), which neutralizes HCl.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 140

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 68.
Explain the action of the following on ethanol :
(1) Acetic acid
(2) Acetic anhydride
(3) Acetyl chloride.
Answer:
(1) Acetic acid : When ethanol is treated with acetic acid in the presence of cone, sulphuric acid, ethyl acetate (ester) is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 141

(2) Acetic anhydride : When ethanol is treated with acetic anhydride in the presence of cone, sulphuric acid, ethyl acetate (ester) is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 142

(3) Acetyl chloride : When ethanol is treated with acetyl chloride in the presence of pyridine, ethyl acetate (ester) is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 143

(Pyridine neutralises HC1 formed during reaction)

Question 69.
Explain the action of the following phenol :
(1) Acetic acid
(2) Acetic anhydride
(3) Acetyl chloride
Answer:
(1) Acetic acid : When phenol is treated with acetic acid in the presence of cone, sulphuric acid, (ester) is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 144
(2) Acetic anhydride : When phenol is treated with acetic anhydride in the presence of cone, sulphuric acid, (ester) is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 145
(3) Acetyl chloride : When phenol is treated with acetyl chloride in the presence of cone, sulphuric acid, (ester) is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 146
(Pyridine neutralizes HC1 formed during reaction.)

Question 70.
What is the action of acetic anhydride on salicyclic acid?
Answer:
When acetic anhydride is treated with salicyclic acid in presence of glacial acetic acid, acetyl salicyclic acid (aspirin) is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 147
Aspirin is a common analgesic, antipyretic drug. Reactions involving breaking of C – O bond of alcohol :

Question 71.
How are the following compounds prepared by using HBr from corresponding alcohols :
(1) Ethyl bromide
(2) Isopropyl bromide
(3) Tert-butyl bromide (2-Methyl-propan-2-ol)?
Answer:
(1) When ethyl alcohol is heated with HBr, ethyl bromide is formed. (HBr is prepared in situ by adding NaBr to HCl or H2SO4)
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 148

(2) Isopropyl alcohol on heating with HBr forms isopropyl bromide.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 149

(3) Tert-butyl alcohol on heating with HBr forms tert-butyl bromide.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 150

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 72.
Explain the action of hydroiodic acid on the following :
(1) Propan-2-ol
(2) 3-Methyl butan-2-ol.
Answer:
(1) Propan-2-ol : When propan-2-ol is heated with hydroiodic acid, 2-iodopropane is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 151

(2) 3-Methyl butan-2-ol : When 3-Methyl butan-2-ol is heated with hydroiodic acid, 2-Iodo-2-methyl butane is obtained. Here, secondary alcohol is converted into a tertiary alkyl halide.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 152

Question 73.
Describe the action of PCl3 on
(1) ethanol
(2) Propan-l-ol
(3) Propan-2-ol.
Answer:
(1) Ethanol : When ethanol is treated with PCl3, ethyl chloride is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 153

(2) Propan-l-ol : When propan-l-ol is treated with PCl3, n-propyl chloride is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 154

(3) Propan-2-ol : When Propan-2-ol is treated with PCl3, 2-Chloropropane is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 155

Question 74.
Describe the action of PCl5 on
(1) ethanol
(2) propan-2-ol.
Answer:
(1) Ethanol : When ethanol is treated with PCl5, ethyl chloride is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 156

(2) Propan-2-ol : When propan-2-ol is treated with PCl5, 2-chloropropane is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 157

Question 75.
Describe the action of SOCl2 on
(1) ethanol
(2) propan-l-ol.
Answer:
(1) Ethanol : When ethanol is treated with SOCl2 in the presence of pyridine, ethyl chloride is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 158

(2) Propan-l-ol : When propan-l-ol is treated with SOCl2 in the presence of pyridine, w-propyl chloride is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 159

Question 76.
Explain dehydration of alcohols.
OR
What is dehydration of alcohols? Give the chemical reactions showing dehydration of primary (1°), secondary (2°) and tertiary (3°) alcohols.
Answer:
Removal of water from an alcohol is called dehydration of alcohol. Alcohols having a /i-hydrogen is heated with dehydrating agents like concentrated H2SO4 (or H3PO4 or P2O5 or Al2O3). The ease of dehydration of alcohols is in the following order : tert-alcohol (3°) > secondary (2°) > primary (1°)
(1) Primary (1°) alcohol is dehydrated by heating it with 95% H2SO4 at 453 K.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 160
(2) Secondary alcohol (2°) is dehydrated by heating with 60% H2SO4 at 373 K.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 161
(3) A ternary alcohol can be easily dehydrated by heating with 20% H2SO4 at 363 K.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 162

An alcohol can be dehydrated by passing vapours alcohols over heated alumina (Al2O3).
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 163

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 77.
Explain oxidation of primary and secondary alcohols.
Answer:
(1) Primary alcohol on oxidation with CrO3 forms aldehyde. However, a better reagent to bring about this oxidation is PCC (pyridinium chlorochromate).
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 169

(2) Secondary alcohol on oxidation with chromic anhydride (CrO3) forms ketone.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 170

Question 78.
What is the action of acidified K2Cr2O7 on the following :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 171
How are following alcohols distinguished :
(1) ethyl alcohol
(2) isopropyl alcohol
(3) tert-butyl alcohol?
OR
How will you distinguish primary, secondary and tertiary alcohols by oxidation process?
Answer:
Primary, secondary and tertiary alcohols are distinguished on the basis of their oxidation products, when their oxidation is carried out using K2Cr2O7 and dil. H2SO4. Acidified potassium dichromate, K2Cr2O7 is an oxidising agent.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 172

(1) Ethyl alcohol is a primary alcohol and on oxidation, it first forms acetaldehyde, which on further oxidation forms acetic acid. In this both, aldehyde and acid have same number of carbon atoms.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 173

(2) Isopropyl alcohol is a secondary alcohol and on oxidation it gives a ketone, acetone with the same number of carbon atoms. Acetone resists further oxidation as it involves breaking of C-C bond.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 174

(3) The oxidation of tert-butyl alcohol is difficult, since it does not have a-hydrogen atom. It is oxidised by using acidic and stronger oxidising agents like KMn04, CrO3 at high temperature, which dehydrate tertiary alcohol to alkene and then oxidise it to a ketone, acetone with less number of carbon atoms.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 175

Question 79.
What happens when vapours of primary, secondary and tertiary alcohols are passed over heated copper at 573 K?
Answer:
When vapours of primary, secondary and tertiary alcohols are passed over heated copper at 573 K, dehydrogena¬tion of primary and secondary alcohol takes place while tertiary alcohols undergo dehydrogenation to given an alkene.

Primary alcohol :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 176

Secondary alcohol :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 177

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Reactions of phenols :

Question 80.
(1) Explain the action of bromine in carbon disulphide (CS2) on phenol (carbolic acid). OR Give equation of the reaction of bromine in CS2 with phenol.
Answer:
When phenol is stirred at a low temperature with bromine dissolved in a polar solvent such as carbon disulphide or CCl4 at (273 K), a mixture of o-bromophenol and p-bromophenol is formed, p-bromophenol is the major product.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 179

Question 81.
Explain the action of bromine water on phenol (carbolic acid).
OR
Name the reagent used in the bromiriation of phenol to 2, 4, 6 tribromophenol.
Answer:
When phenol is treated with bromine water, a yellowish white precipitate of 2, 4, 6 – tribromophenol is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 180

Question 82.
Explain the action of dilute nitric acid on phenol (carbolic acid).
OR
Give equation of the reaction of dilute HNO3 with phenol.
Answer:
When phenol is treated with dilute nitric acid, a mixture of o-nitrophenol and p-nitrophenol is formed. In this reaction, p-nitrophenol is formed as the major product.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 181

Question 83.
Explain the action of cone, nitric acid (nitrating mixture) on phenol (carbolic acid).
OR
How is phenol converted into picric acid?
Answer:
When phenol is warmed with a mixture of cone, nitric acid and cone, sulphuric acid (a nitrating mixture or the mixed acid), 2, 4, 6 – trinitrophenol, commonly called picric acid, is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 182

Question 84.
Explain the action of concentrated sulphuric acid on phenol at different temperatures.
Answer:
(a) At room temperature : When phenol is treated with cone. H2SO4 at room temperature (about 300 K), o-phenol sulphonic acid is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 183
(b) At 373 K : When phenol is treated with cone. H2SO4 at about 373 K, p-phenol sulphonic acid is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 184

Question 85.
Explain (1) Kolbe’s reaction (2) Reimer-Tiemann reaction.
Answer:
(1) Kolbe’s reaction : When phenol reacts with sodium hydroxide, sodium phenoxide is obtained. Phenoxide ion being more reactive than phenol towards electrophilic substitution. Phenoxide undergoes electrophilic substitution with carbon dioxide at 398 K under pressure of 6 atm (a weak electrophile) forms salicylic acid as major product.

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 185

(2) Reimer-Tiemann reaction : Phenol is heated with chloroform along with aqueous NaOH, this is followed by acidification with dil. HC1 when salicyladehyde (2-hydroxy benzaldehyde) is formed as the major product, which can be separated from p-isomer by steam distillation. The stability of o-isomer is due to intramolecular hydrogen bonding.

Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 186

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 86.
Explain the action of zinc dust on phenol.
OR
How is phenol converted into benzene?
Answer:
When phenol is heated with zinc dust, benzene is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 187

Question 87.
Explain the action of chromic acid on phenol.
OR
How is phenol converted into benzoquinone?
Answer:
When phenol is oxidised by chromic acid, a diketone, p-benzoquinone is formed. It is a conjugated diketone.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 188

Question 88.
Explain catalytic hydrogenation of phenol.
Answer:
When a mixture of vapours of phenol and hydrogen is passed over nickel catalyst at 433 K., cyclohexanol is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 189

Question 89.
How is diethyl ether (ethoxyethane) obtained from alcohol?
Answer:
When excess of ethyl alcohol is distilled with concentrated sulphuric acid (H2SO4) at 413 K, diethyl ether is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 190

Question 91.
Write a note on Williamson’s synthesis.
OR
How are ethers prepared from Alkyl halides?
OR
How are simple ethers and mixed ethers prepared by Williamson’s synthesis?
Answer:
Williamson’s synthesis : When an alkyl halide (R – X) is heated with sodium alkoxide (R – O – Na), an ether is obtained, this reaction is known as Williamson’s synthesis. This method is used to prepare simple (or symmetrical) ethers and mixed (unsymmetrical) ethers.

Sodium alkoxide is obtained by a reaction of sodium with an alcohol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 194

(A) Simple (Symmetrical) ether : When an alkyl halide and sodium alkoxide having similar alkyl groups are heated, symmetrical ether is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 195
Sodium ethoxide on heating with ethyl bromide gives diethyl ether.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 196

(B) Mixed (Unsymmetrical) ether : When an alkyl halide and sodium alkoxide or sodium phenoxide having different alkyl groups are heated, unsymmetrical ether (dialkyl ethers or alkyl aryl ether) is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 197
Sodium ethoxide on heating with methyl bromide gives ethyl methyl ether.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 198
Sodium phenoxide on heating with ethyl bromide gives ethyl phenyl ether.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 199

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 92.
How is anisole obtained from phenol?
OR
How is methoxy benzene prepared from carbolic acid?
Answer:
Phenol reacts with sodium hydroxide, sodium phenoxide is formed. When sodium phenoxide is heated with methyl iodide, anisole is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 200

Question 93.
Explain the limitations for the preparation of unsymmetrical ethers.
OR
What care is to be taken in the preparation of unsymmetrical ethers by Williamson’s synthesis? Explain.
OR
Illustrate with examples the limitations of Williamson’s synthesis for the preparation of certain types of ethers.
Answer:
(1) In the preparation of unsymmetrical ethers by Williamson’s synthesis, the proper choice of the reactants namely alkyl halide and sodium alkoxide is necessary.

(2) The best yield of unsymmetrical ether is obtained when primary alkyl halide and tertiary alkoxide are heated, since primary alkyl halides are more susceptible to SN2 reaction.

(3) Secondary or tertiary alkyl halide undergo a and fi (halogen and hydrogen) elimination reaction giving an alkene instead of an ether since a carbon atom is sterically hindered by bulky alkyl groups.

(4) For example : t-butyl methyl ether can be synthesised by reaction of methyl bromide with sodium t-butoxide.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 201

Question 94.
State physical properties of ethers.
Answer:
(1) Dimethyl ether and ethyl methyl ether are gases. Other ethers are colourless liquids with pleasant odour.
(2) Lower ethers are highly volatile and highly inflammable substances.
(3) Boiling points of ethers show gradual increase with the increase in molecular mass.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 205
(4) The solubility/miscibility of ethers in water is similar to that of alcohols of comparable molecular mass.

Question 95.
Explain, ethers posses a small net dipole moment.
Answer:
In ethers, Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 206 bond angle is 110° and not 180°, bond dipole moments of the two C – O bonds do not cancel each other, therefore, ethers possess a smal net dipole moment, (for example, dipole moment of diethyl ether is 1.18 D)
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 207

Question 96.
Explain, the solubility/miscibility of ethers in water is similar to that of alcohols of comparable molecular mass.
Answer:
The solubility/miscibility of ethers in water is similar to that of alcohols of comparable molecular mass. This is because ethers can form hydrogen bonds with water through ethereal oxygen.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 208

For example, diethyl ether and n-butyl alcohol have respective miscibilities of 7.5 and 9 g per 100 g of water.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 97.
Explain laboratory test for ethers.
Answer:
Ethers are neutral compounds in aqueous medium. Ethers do not react with bases, cold dilute acids, reducing agents, oxidizing agents and active metals. However, ethers dissolve in cold concentrated H2SO4 due to formation of oxonium salts.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 209
This property distinguishes ethers from hydrocarbons.

Question 98.
What is the action of atmospheric oxygen on diethyl ether?
Answer:
When atmospheric oxygen combines with diethyl ether, peroxide of diethyl ether is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 210

Question 99.
What is the action of dilute sulphuric acid on
(1) Dimethyl ether
(2) Diethyl ether
(3) Ethyl methyl ether?
(4) Anisole?
OR
Explain hydrolysis of ethers.
Answer:
Simple ethers on heating with dilute sulphuric acid under pressure undergoes hydrolysis to give alcohol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 211
(1) Dimethyl hydrolysis on hydrolysis give methanol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 212
(2) Diethyl ether on hydrolysis give ethanol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 213
(3) A mixed ether on heating with dii. H2SO4 under pressure undergoes hydrolysis to give mixture of two different alcohols.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 214
Ethyl methyl ether on hydrolysis give a mixture of ethanol and methanol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 215
(4) Anisole on hydrolysis give a mixture of phenol and methanol.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 216

Question 100.
State the combustion products of diethyl ether.
Answer:
The combustion products of diethyl ether are CO2 and H2O.

Question 101.
What is the action of phosphorus pentachloride on
(1) Diethyl ether
(2) Ethyl methyl ether
(3) Methyl phenyl ether (anisole)?
Answer:
(1) Diethyl ether : When diethyl ether is heated with ethyl methyl ether, ethyl chloride is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 217
(2) Ethyl methyl ether : When ethyl methyl ether is heated with phosphorus pentachloride, a mixture of ethyl chloride and methyl chloride is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 218
(3) Methyl phenyl ether (Anisole) : When methyl phenyl ether is heated with phosphorus pentachloride, a mixture of methyl chloride, chlorobenzene is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 219

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 102.
Describe the action of hot concentrated HI on
(i) Dialkyl ether
(ii) Alkyl aryl ether.
Answer:
Ether reacts with excess of hot concentrated hydrogen halide to give two alkyl halide molecules.
R – O – R + HX → RX + R – OH
R – OH + HX → RX + H20
Alkyl aryl ether reacts with hot concentrated hydrogen halide to give phenol and alkyl halide.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 220
Ethers with two different alkyl groups react with hot.conc. HI to give alkyl halides.
R – O – R’+ 2H – X → R – X + R’ – X + H2O
The order of reactivity of HX is HI > HBr > HC1

Question 103.
What is the action of hot HI on isopropyl methyl ether?
Answer:
When isopropyl methyl ether is treated with excess of hot hydroiodic acid, a mixture of isopropyl iodide and methyl iodide is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 221

Question 104.
Describe the action of hydroiodic acid on the following :
(1) Diethyl ether.
Answer:
When diethyl ether (ethoxy ethane) is treated with hydroiodic acid, a mixture of ethanol and ethyl iodide is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 222

If excess of hydroiodic acid is available, then ethyl alcohol further reacts with hydroiodic acid at higher temperature to form ethyl iodide and water.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 223

(2) Ethyl methyl ether.
Answer:
When ethyl methyl (methoxy ethane) is treated with hydroiodic acid, a mixture of ethyl alcohol and methyl iodide is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 224

If excess of hydroiodic acid is available, then ethyl alcohol further reacts with hydroiodic acid at higher temperature to form ethyl iodide and water.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 225

(3) Methyl n-propyl ether.
Answer:
When methyl n-propyl ether (1-Methoxy propane) is treated with hydroiodic acid, a mixture of n-propyl alcohol and methyl iodide is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 226

If excess of hydroiodic acid is available then n-propyl alcohol further reacts with hydroiodic acid at higher temperature to form n-propyl iodide and water.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 227

(4) Methyl phenyl ether (anisole).
Answer:
When methyl phenyl ether (anisole) is treated with hydroiodic acid, phenol and methyl iodide is formed. Here, phenol does not react further with HI because – OH group is attached to sp2-hybridised carbon atom and it cannot be replaced by iodide (nucleophile).
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 228

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 105.
Draw the resonance structures of aromatic ethers.
Answer:
The alkoxy group in aromatic ether is a ring activating and ortho-, paradirecting group toward electrophilic aromatic substitution.

Resonance structures :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 234
+ R Effect of – OR group results in increased electron density at the para- and two ortho-posotions (see resonance structures II, III and IV).

Question 106.
Describe the action of bromine in acetic acid on anisole.
OR
Write the equation of the reaction of bromination of anisole in ethanoic acid medium.
Answer:
When anisole is treated with bromine in acetic acid, /i-bromoanisole (major product) is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 235

Question 107.
Describe the action of methyl chloride on anisole (Friedel-Crafts reaction).
OR
Write the equation of Friedel-Crafts reaction-alkylation of anisole.
Answer:
When anisole is treated with alkyl halide in the presence of anhydrous aluminium chloride (a Lewis acid) as catalyst, 4-Methoxy toluene is formed as major product. The alkyl groups are introduced at -ortho and -para positions in anisole, the reaction is known as Friedel-Crafts alkylation reaction.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 236

Question 108.
Describe the action of Acetyl chloride on anisole (Friedel-Crafts acylation).
OR
Write the equation of the reaction Friedel-Crafts acylation of anisole.
OR
Write a note on Friedel-Crafts acylation.
Answer:
When anisole is treated with acetyl chloride in the presence of anhydrous aluminium chloride (a Lewis acid), 4-Methoxy acetophenone (major product) is obtained. The acetyl groups are introduced at -ortho and -para positions in anisole, the reaction is known as Friedel Craft’s acylation reaction.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 237

Question 109.
Describe the action of cone. HNO3 on anisole.
OR
Write the equation of nitration of anisole.
Answer:
When anisole is reacted with nitrating mixture (cone. HNO3 + cone. Fl2SO4), a mixture of p-nitroanisole and o-nitroanisole is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 238

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 110.
An organic compound with the formula C4H10O3 shows properties of ether and alcohol. When treated with an excess of HBr yields only one compound 1,2 dibromomethane. Write structural formula of ether and that of alcohol.
Answer:
When C4H10O3 is treated with excess of HBr, a single compound 1,2-dibromoethane is formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 239

Question 111.
An organic compound ‘A’ having molecular formula C4H10O does not react with sodium metal. On hydrolysis with dilute H2SO4 it gives only one organic compound ‘B’ The compound B on heating with red phosphorus and iodine gives compound ‘C’. The compound C can also be obtained from compound ‘A’ on heating with excess HI. Identify the compounds A, B and C.
Answer:
(1) The organic compound ‘A’ with molecular formula C4H10O may be an alcohol or ether.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 240
(2) Since ‘A’ does not react with sodium (Na), it is not an alcohol. Hence ‘A’ may be an ether.
(3) Since an ether ‘A’ on hydrolysis with dilute H2SO4 gives only one compound ‘B’, the compound ‘A’ must be a symmetrical (simple) ether. Hence compound ‘A’ may be, C2H5 – O – C2H5Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 241

Question 112.
How will you affect the following two-step conversions? Diethyl ether into n-butane :
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 242

Question 114.
Answer in one sentence/word.

(1) Name the alcohol that is used to make propan-2-one.
Answer:
The alcohol used to make propan-2-one is iso-propyl alcohol.

(2) Which is the first oxidation product of secondary alcohol?
Answer:
The first oxidation product of secondary alcohol is ketone.

(3) Name the alcohol that is used to make acetic acid.
Answer:
The alcohol that is used to make acetic acid is ethyl alcohol.

(4) Write the structure of cyclohexane-1, 4-diol.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 246

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

(9) Which of the following isomers is more volatile : o-nitrophenol or p-nitrophenol?
Answer:
The isomer o-nitrophenol with lower boiling point is more volatile.

(10) Identify the product of the following reaction.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 247
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 248

(11) Identify the product obtained by industrial synthesis of carboxylation of phenoxide ion followed by acidification.
Answer:
The product is phenol
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 249

(12) What is the product A obtained in the following reaction?
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 250
Answer:
The product is phenolMaharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 251

(13) Which positions are occupied by – NO2 group during nitration of carbolic acid?
Answer:
– o – (ortho) and – p – (para) positions are occupied by – NO2 group during nitration of carbolic acid.

(14) Write the name of reactants used for the preparation of ethyl-tert-butyl ether.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 252
are used for the preparation of ethyl-tert-butyl ether.

(15) The product formed in the reaction of reverse of dehydration of alcohol is
Answer:
The product formed in the reaction of reverse of dehydration of alcohol is alkene.

(16) Which rule is obeyed by hydroboration oxidation process?
Answer:
The rule obeyed by hydroboration oxidation process is opposite to the Markovnikov’s rule.

(17) Name the reagents for the complete hydroboration-oxidation reaction in step 1 and step 2.
Answer:
Step 1 : Diborane
Step 2 : Hydrogen peroxide and dil. NaOH.

(18) Write the name of the test by which methanol can be distinguished from ethanol.
Answer:
Iodoform test by which methanol can be distinguished from ethanol.

(19) Write the name of reactant used for preparation of phenol, which gives byproduct used as solvent.
Answer:
Reactant used in the preparation of phenol :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 253

(20) Ether is a good solvent for Grignard reagent. Which property makes it a good solvent?
Answer:
Ether has a low polarity, this property makes it a good solvent.

(21) The C – O – C bond angle in dimethyl ether is
Answer:
The C – O – C bond angle in dimethyl ether is 110°.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Question 115.
State the uses of methyl alcohol.
Answer:
(1) Methyl alcohol is used as an industrial solvent for dissolving oils, fats, gums, etc.
(2) It is used for dry cleaning and preparation of perfumes and varnishes.
(3) It is used as antifreeze agent for automobile radiators at low temperature.
(4) It is used in the preparation of methyl chloride, dimethyl sulphate and formaldehyde.
(5) It is used to denature ethyl alcohol.

Question 116.
State the uses of ethyl alcohol. OR Write two uses of ethyl alcohol.
Answer:
(1) Ethyl alcohol is used as solvent for dyes, oils, perfumes, cosmetics and drugs.
(2) A mixture of 10-20% ethyl alcohol with petrol is used as motor fuel.
(3) A mixture of ethyl alcohol and calcium acetate in gel form is used as solid fuel.
(4) It is widely used in beverages.
(5) Since ethyl alcohol has low freezing point, it is used in thermometer.
(6) It is an effective tropical antiseptic therefore it is used in many mouth washes.
(7) It kills micro-organisms on wound surface and in the mouth but its low toxicity does not kill the cells of the skin or mouth tissues.
(8) It is used in the preparation of chloroform, iodoform, acetic acid and ethers.
(9) It is used as fuel.

Question 117.
Give the important uses of phenol.
OR
Write two uses of phenol.
Answer:
(1) Phenol is used in the preparation of phenol-formaldehyde polymer which is used in a plastic bakelite.
(2) It is used in the preparation of phenol-phthalein-an indicator and in certain dyes.
(3) It is used in the preparation of drugs such as salol, aspirin, etc.
(4) It is used in the preparation of dettol, which is an antiseptic.
(5) It is used in the preparation of 2,4-dichlorophenoxy acetic acid which is used as selective weed killer.
(6) It is used to prepare picric acid which is used as explosive.

Multiple Choice Questions

Question 118.
Select and write the most appropriate answer from the given alternatives for each sub-question:

1. Which one of the following is a tertiary alcohol?
(a) Pentan-l-ol
(b) Pentan-2-ol
(c) 2-Methylpentan-2-ol
(d) 3-Methylpentan-2-ol
Answer:
(c) 2-Methylpentan-2-ol

2. Which of the following is a primary alcohol?
(a) 3-ethyl-3-hexanol
(b) 2-butanol
(c) 3-methyl-l-butanol
(d) 1-hexanol
Answer:
(d) 1-hexanol

3. The molecular formula C4H10O represents
(a) aldehydes
(b) alcohols
(c) ethers
(d) both (b) and (c)
Answer:
(d) both (b) and (c)

4. The general formula of primary, secondary and tertiary alcohols is
(a) CnH2nOH
(b) CnH2n-1OH
(c) CnH2n+1OH
(d) CnHn+1OH
Answer:
(c) CnH2n+1OH

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

5. Which of the following alcohols cannot be prepared by hydration of the corresponding alkene?
(a) Ethanol
(b) Propan- l-ol
(c) Propan-2-ol
(d) 2-Methylpropan-2-ol
Answer:
(b) Propan- l-ol

6. Which of the following compounds when treated with CH3MgI in dry ether followed by the hydroly¬sis, will give Propan-2-ol?
(a) HCHO
(b) CH3CHO
(c) CH3CH2OH
(d) CH3COCH3
Answer:
(b) CH3CHO

7. To prepare 3-Ethylpentan-3-ol, the reagents needed are
(a) CH3CH2MgBr + CH3COCH2CH3
(b) CH3MgBr + CH3CH2CH2COCH2CH3
(c) CH3CH2MgBr + CH3CH2COCH2CH3
(d) CH3CH2CH2MgBr + CH3COCH2CH3
Answer:
(c) CH3CH2MgBr + CH3CH2COCH2CH3

8. How is 1-propanol obtained?
(a) Using propanal
(b) Using propanone
(c) Using propene
(d) All of these
Answer:
(a) Using propanal

9.
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 260
Answer:
(a)

10. Ketone on reduction gives
(a) 1° alcohol
(b) 2° alcohol
(c) 3° alcohol
(d) all of these
Answer:
(b) 2° alcohol

11. Primary alcohols are prepared by catalytic hydro-genation of aldehydes in presence of ……………………………… as a catalyst
(a) aluminium bromide
(b) fluoroboric acid
(c) dry ether
(d) palladium
Answer:
(d) palladium

12. Lower member of alcohols are
(a) insoluble in water
(b) soluble in water
(c) insoluble in acetaldehyde
(d) insoluble in petrol
Answer:
(b) soluble in water

13. Which of the following compounds contain hydro¬gen bonds?
(a) Ethane
(b) Ethanol
(c) Methoxymethane
(d) Ethylene
Answer:
(b) Ethanol

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

14. Which of the following is a trihydric alcohol?
(a) n-propyl alcohol
(b) Glycerol
(c) Glycol
(d) Glycine
Answer:
(b) Glycerol

15. Aldehydes are first oxidation products of
(a) primary alcohols
(b) secondary alcohols
(c) tertiary alcohols
(d) carboxylic acids
Answer:
(a) primary alcohols

16. The oxidation product of alcohol depends on
(a) – OH group of an alcohol
(b) number of carbon atoms in alcohol
(c) number of hydrogen atoms attached to hydroxyl bearing carbon
(d) all of these
Answer:
(c) number of hydrogen atoms attached to hydroxyl bearing carbon

17. The order of ease of oxidation is
(a) primary > secondary > tertiary
(b) primary < secondary < tertiary
(c) primary > tertiary > secondary
(d) secondary > tertiary > primary
Answer:
(b) primary < secondary < tertiary

18. Ethyl alcohol \(\frac{\text { conc. } \mathrm{H}_{2} \mathrm{SO}_{4}}{443 \mathrm{~K}} \mathrm{~A} \stackrel{\mathrm{HBr}}{\longrightarrow}\) Ethyl bromide. Identify A.
(a) Ethyl hydrogen sulphate
(b) Ethylene
(c) Isopropyl hydrogen sulphate
(d) Acetic acid
Answer:
(b) Ethylene

19. One mole of PCl5 reacts with one mole of ethyl alcohol to give
(a) 1 mole Cl2
(b) 1/2 mole Cl2
(c) 1 mole HCl
(d) 1/2 mole HCl
Answer:
(c) 1 mole HCl

20. When ethyl alcohol is reacted with sodium metal and the compound so formed is treated with ethyl bromide, the product formed is
(a) an acid
(b) an alkane
(c) an ether
(d) an ester
Answer:
(c) an ether

21. Dehydration occurs at the lowest temperature and concentration for
(a) methyl alcohol
(b) n-propyl alcohol
(c) iso-propyl alcohol
(d) tert-butyl alcohol
Answer:
(d) tert-butyl alcohol

22. One mole of sodium when reacts with one mole of methyl alcohol, gives
(a) one mole of oxygen
(b) one mole of hydrogen
(c) half mole of hydrogen
(d) half mole of oxygen
Answer:
(c) half mole of hydrogen

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

23. A compound ‘X’(C3H8O) on oxidation gives com-pound ‘ Y’(C3H6O2), hence compound ‘X’ must be
(a) a ketone
(b) an aldehyde
(c) a primary alcohol
(d) a secondary alcohol
Answer:
(c) a primary alcohol

24. The toxicity of alcohols
(a) increases with an increase in their molecular weight
(b) decreases with an increase in their molecular weight
(c) increases with a decrease in their molecular weight
(d) does not depend on their molecular weight
Answer:
(a) increases with an increase in their molecular weight

25. The structural formula of 2-phenyl ethanol is
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 261
Answer:
(c)

26. Compound that fails to give effervescence with NaHCO3 is
(a) C6H5COO
(b) CH3COOH
(c) C6H5OH
(d) Picric acid
Answer:
(c)

27. Phenol is a bifunctional compound because
(a) it gives reactions of hydroxyl group (-OH) as well as aromatic ring
(b) it is a strong acid
(c) it is insoluble in NaOH
(d) it is readily soluble in water
Answer:
(a) it gives reactions of hydroxyl group (-OH) as well as aromatic ring

28. Phenol gives characteristic colour with
(a) iodine solution
(b) bromine water
(c) ammonium hydroxide
(d) aqueous ferric chloride solution
Answer:
(d) aqueous ferric chloride solution

29. Ethanol and phenol are distinguished from each other by the action of
(a) neutral ferrous chloride
(b) neutral ferric chloride
(c) ferric hydroxide
(d) ferrous hydroxide
Answer:
(b) neutral ferric chloride

30. Which of the following compounds is used to prepare bakelite resin?
(a) Acetaldehyde
(b) Ethanol
(c) Phenol
(d) Methyl amine
Answer:
(d) Methyl amine

31. Increasing order of acid strength among p-methoxy-phenol, p-methyl phenol and p-nitro- phenol is as
(a) p-nitrophenol > p-methoxyphenol > p-methylphenol
(b) p-methylphenol > p-methoxyphenol > p-nitrophenol
(c) p-nitrophenol > p-methylphenol > p-methoxyphenol
(d) p-methoxyphenol > p-methylphenol > p-nitrophenol
Answer:
(d) p-methoxyphenol > p-methylphenol > p-nitrophenol

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

32. The IUPAC name of the compound is Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 262
(a) 3,3-dimethyl-l-hydroxy cyclohexane
(b) l,l-dimethyl-3-cyclohexanol
(c) 3,3-dimethyl-1-cyclohexanol
(d) l,l-dimethyl-3-hydroxy cyclohexane
Answer:
(c) 3,3-dimethyl-1-cyclohexanol

33. Which of the following is the most reactive towards electrophilic attack?
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 263
Answer:
(a)

34. In the following reaction, Ethanol \(\stackrel{\mathrm{PBr}_{3}}{\longrightarrow} \mathrm{A}\)\(\stackrel{\text { Alc-KOH }}{\longrightarrow} \text { B } \frac{\text { (i) } \mathrm{H}_{2} \mathrm{SO}_{4} \text {, room temp. }}{\text { (ii) } \mathrm{H}_{2} \mathrm{O}, \text { heat }} \mathrm{C} \text {. }\). The product C is
(a) CH3 – CH2 – O – CH2 – CH3
(b) CH3 – CH2 – OSO3H
(c) CH3CH2OH
(d) CH2 = CH2
Answer:
(c) CH3CH2OH

35. What is the general formula of ethers?
(a) CnH2nO
(b) CnH2n+2O
(c) CnH2n-1O
(d) CnH2n-2O
Answer:
(b) CnH2n+2O

36. The Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 264 linkage is present in
(a) proteins
(b) ketones
(c) ethers
(d) aldehydes
Answer:
(c) ethers

37. Methoxy ethane is the functional isomer of
(a) CH3CHOHCH3
(b) CH3CH2CH2OH
(c) CH3 – O – CH3
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

38. An oxygen atom in ether is
(a) sp2-hybridized
(b) sp3-hybridized
(c) sp-hybridized
(d) sp2-d-hybridized
Answer:
(b) sp3-hybridized

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

39. C5H12O represents
(a) only an alcohol
(b) only an aldehyde
(c) only an ether
(d) an alcohol and an ether
Answer:
(d) an alcohol and an ether

40. Ether molecules are
(a) tetrahedral
(b) angular
(c) pyramidal
(d) diagonal
Answer:
(b) angular

41. Which one of the following compounds is not isomeric with Ethoxypropane?
(a) 1-Methoxypropane
(b) 2-Methoxypropane
(c) 2-Methylpropane-2-ol
(d) 2-Methylbutan-2-ol
Answer:
(d) 2-Methylbutan-2-ol

42. The IUPAC name of C2H5 – O – CH2 – CH(CH3)2 is
(a) 1-Ethoxy-1-butane
(b) 2-Ethoxy-2-butane
(c) l-Ethoxy-2-methylpropane
(d) 3-Ethoxy-2-methylpropane
Answer:
(c) l-Ethoxy-2-methylpropane

43. The IUPAC name of CH3OC6H5 is
(a) methoxy phenyl ether
(b) phenoxy methane
(c) methoxy benzene
(d) methyl phenyl ether
Answer:
(c) methoxy benzene

44. In CH3(CH2)3 – O -CH3, the parent hydrocarbon of large alkyl group is
(a) n-butane
(b) butane
(c) pentane
(d) n-pentane
Answer:
(a) n-butane

45. Which one of the following alkyl halide gives best yield in Williamson’s synthesis?
(a) CH3 – Br
(b) CH3 – CH(Br) – CH3
(c) CH2 – CH(Br) – CH3 – CH3
(d) (CH3)3 C – Br
Answer:
(a) CH3 – Br

46. Which one of the following ethers cannot be pre¬pared by using diazomethane?
(a) Dimethyl ether
(b) Diethyl ether
(c) Ethyl methyl ether
(d) t-Butyl methyl ether
Answer:
(b) Diethyl ether

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

47. The continuous etherification is carried out at
(a) 473 K
(b) 413 K
(c) 498 K
(d) 403 K
Answer:
(b) 413 K

48. Williamson’s synthesis is used for the preparation of
(a) only unsymmetrical ethers
(b) only symmetrical ethers
(c) both symmetrical and unsymmetrical ethers
(d) only methyl ethers
Answer:
(c) both symmetrical and unsymmetrical ethers

49. Which of the following ethers on hydrolysis gives two different products that are successive members of a homologous series?
(a) Methoxymethane
(b) Ethoxythane
(c) Methoxyethane
(d) 2-Methoxypropane
Answer:
(c) Methoxyethane

50. Which one of the following compounds dissolves in hot dilute sulphuric acid but does not react with sodium metal?
(a) Ethyl bromide
(b) Acetic acid
(c) Ethyl alcohol
(d) Diethyl ether
Answer:
(d) Diethyl ether

51. Identify ‘A’ in the following reaction :
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 271
Answer:
(c)

52. The C-O-C bond angle in an ether is
(a) 180
(b) 90°
(c) 104.5°
(d) 109.5°
Answer:
(d) 109.5°

53. Ether free from moisture and alcohol is known as ,
(a) dry ether
(b) absolute ether
(c) pure ether
(d) spirit ether
Answer:
(b) absolute ether

54. The geometry of ether is similar to
(a) ammonia
(b) methane
(c) water
(d) ethyne
Answer:
(c) water

55. Sodium metal does not react with
Answer:
C2H5 – O – C2H3

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

56. Ethers have boiling points
(a) lower than those of alkanes of comparable molecular masses
(b) higher than those of isomeric alcohols
(c) lower than those of isomeric alcohols
(d) higher than those of alkanes of comparable molecular masses
Answer:

57. Ether on hydrolysis gives
(a) aldehyde
(b) alcohol
(c) acid
(d) ester
Answer:
(c) acid

58. Which of the following reactions represent Williamson’s reaction?
(a) R – O – R’ + HI →
\(\text { (b) } \mathrm{R}^{\prime}-\mathrm{OH}+\stackrel{\mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow}\)

(d) R – O – Na + R’X →
Answer:
(c)

59. Which of the following reactions represent the continuous etherification process?
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 265
Answer:
(b)

60. Ethers on complete combustion produce
(a) an alcohol and water
(b) an alkene and water
(c) an alkane and an alkene
(d) carbon dioxide and water
Answer:
(d) carbon dioxide and water

61. Diethyl ether is used as a solvent in many organic reactions because it
(a) is liquid at room temperature
(b) has lower boiling point
(c) contains divalent oxygen atom
(d) is inert in nature
Answer:
(d) is inert in nature

62. Which of the following reagents can be prepared using ether as a solvent?
(a) Tollen’s reagent
(b) Grignard’s reagent
(c) Schiff’s reagent
(d) Millon’s reagent
Answer:
(b) Grignard’s reagent

63. Diethyl ether is used as
(a) a hypnotic
(b) an antiseptic
(c) an anaesthetic
(d) an antipyretic
Answer:
(c) an anaesthetic

64. To obtain fuel, diethyl ether is mixed with
(a) ester
(b) ethanal
(c) ethanol
(d) 2-propanone
Answer:
(c) ethanol

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

65. Which of the following alcohols is prepared by acid catalyzed hydration of alkenes?
(a) Butan-l-ol
(b) Propan-l-ol
(c) Ethanol
(d) Methanol
Answer:
(c) Ethanol

66. Which of the following alcohols can be prepared by direct hydration of corresponding alkene in presence of 50 % sulphuric acid?
(a) Butan-l-ol
(b) Butan-2-ol
(c) 2-Methylpropan-1 -ol
(d) 2-Methylpropan-2-ol
Answer:
(d) 2-Methylpropan-2-ol

67. Which of the following alcohols cannot be prepared by reduction of carbonyl compounds?
(a) Pentan-l-ol
(b) Pentan-2-ol
(c) 2-Methylpentan-2-ol
(d) 3-Methylpentan-2-ol
Answer:
(c) 2-Methylpentan-2-ol

68. Which of the following conversions explains the acidic nature of alcohols?
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 268
Answer:
(b)

69. Which of the following compounds gives 3-ethyl- pentan-3-ol by the action of ethyl magnesium iodide followed by acid hydrolysis?
(a) Propanone
(b) Butanone
(c) Pentan-2-one
(d) Pentan-3-one
Answer:
(d) Pentan-3-one

70. Benzyl phenyl ether reacts with hydrogen bromide to give
(a) benzyl bromide and phenol
(b) benzyl alcohol and bromobenzene
(c) benzyl bromide and bromobenzene
(d) benzyl alcohol and phenol
Answer:
(a) benzyl bromide and phenol

71. Ethers are considered as
(a) monoalkyl derivatives of water
(b) divalent oxygen atom is attached to two alkyl groups
(c) alkyl derivatives of fatty acids
(d) condensation products of acid and alcohol
Answer:
(b) divalent oxygen atom is attached to two alkyl groups

72. Which of the following compounds is not isomeric with ethoxyethane?
(a) 1-methoxypropane
(b) 2-methoxypropane
(c) 2-methylpropan-2-ol
(d) 2-methylbutan-2-ol
Answer:
(d) 2-methylbutan-2-ol

73. Which of the following compounds dissolves in hot dilute sulphuric acid but does not reacts with sodium metal?
(a) Ethyl bromide
(b) Acetic acid
(c) Ethyl alcohol
(d) Diethyl ether
Answer:
(d) Diethyl ether

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

74. Which of the following alcohol will have the fastest rate of dehydration?
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 266
Answer:
(c)

75. The phenol having lowest acidity is
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 267
Answer:
(b)

76. Which of the following reagents is best for the following conversion?
Maharashtra Board Class 12 Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 269
Answer:
(a)

77. 3-Methyl butane-2-ol on heating with HI gives
(a) 2 -iodo-3-methyl butane
(b) 2-iodo-2-methyl butane
(c) l-iodo-3-methyl butane
(d) l-iodo-2-methyl butane
Answer:
(b) 2-iodo-2-methyl butane

78. In phenol carbon atom attached to -OH group undergoes –
(a) sp3-hybridisation
(b) sp-hybridisation
(c) sp2-hybridisation
(d) No hybridisation
Answer:
(c) sp2-hybridisation

79. Which among the following reducing agents is ‘not’ used to reduce acetaldehyde to ethyl alcohol?
(a) Na-Hg and water
(b) Zn-Hg and cone. HCl
(c) H2-Raney Ni
(d) Li-A1H4/H+
Answer:
(b) Zn-Hg and cone. HCl

80. Identify the weakest acidic compound amongst the following :
(a) p-nitrophenol
(b) p-chlorophenol
(c) p-cresol
(d) p-aminophenol
Answer:
(d) p-aminophenol

Maharashtra Board Class 12 Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

81. Natalite is a mixture of
(a) diethyl ether and methanol
(b) diethyl ether and ethanol
(c) dimethyl ether and methanol
(d) dimethyl ether and ethanol
Answer:
(b) diethyl ether and ethanol

82. The alcohol used in thermometers is
(a) methanol
(b) ethanol
(c) propanol
(d) butanol
Answer:
(b) ethanol

83. Which of the following is the first oxidation product of secondary alcohol?
(a) Alkene
(b) Aldehyde
(c) Ketone
(d) Carboxylic acid
Answer:
(c) Ketone

84. When phenol is heated with cone. HNO3 in presence of cone. H2SO4 it yields
(a) o-nitrophenol
(b) p-nitrophenol
(c) 2, 4, 6-trinitrophenol
(d) m-nitrophenol
Answer:
(c) 2, 4, 6-trinitrophenol

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 1 Solid State Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 1 Solid State

Question 1.
What are the physical states of matter? How can they be changed into one another?
Answer:

  • There are three physical states of matter namely solid, liquid and gas.
  • They differ in intermolecular or interatomic or interionic forces which are strongest in the solid-state.
  • By raising the temperature of solids to their melting point, solids are converted into liquids while heating liquids to their boiling points, they can be converted into vapour or gaseous state.
  • On the contrary, by cooling the gases to very low temperature and subjecting to high pressure they can be transformed into liquid which on further cooling can be transformed into solid-state.

The equilibrium existing between three states of matter may be represented as,
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 1

Question 2.
What are the constituents of solids?
Answer:
The smallest constituent particles of various solids are atoms, ions or molecules. All such small constituents are referred to as ‘particles’.

Question 3.
What are the characteristic properties of solids?
Answer:

  • The solid state of matter is characterised by strong interparticle forces of attraction.
  • There is regularity and periodicity in the arrangement of constituent particles of solid.
  • Generally solids are hard, incompressible and rigid except some solids like Na, K, P which are soft.
  • The constituent particles of solids like molecules, atoms or ions have fixed stationary positions in solid and can only oscillate about their equilibrium or mean positions. Hence, they have fixed shape and cannot be poured like liquids.
  • Crystalline solids have sharp melting points and they melt at a definite temperature. Amorphous solids do not have sharp melting points.
  • They are anisotropic or isotropic.

Question 4.
Give classification of solids.
Answer:
Depending on orderly arrangement of the constituent particles, the solids are classified into two types :

  • Crystalline solids. For example, diamond, NaCl, K2SO4, etc.
  • Amorphous solids or non-crystalline solids. For example, tar, glass, plastics, rubber, butter, etc.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

Question 5.
Define :
(1) Crystalline solid.
(2) Amorphous solid.
Answer:
(1) Crystalline solid : A homogeneous solid in which the constituent particles like atoms, ions or molecules are arranged in a definite repeating pattern throughout the solid is called crystalline solid. For example, NaCl, KNO3, etc.

(2) Amorphous solid : A substance which appears like solid but does not have perfectly ordered crystalline structure and no regular arrangement of constituent particles in structure is called amorphous solid. For example, glass, rubber, plastics, etc.

Question 6.
Define the term anisotropy.
OR
Define and explain the term anisotropy.
Answer:
Anisotropy : The ability of crystalline solids to change their physical properties when measured in different directions is called anisotropy.

Explanation : This property is due to different arrangement of constituents in different directions. Different types of particles fall on the way of measurements in different directions. Hence the composition of crystalline solid changes with directions changing their physical properties.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 2
Fig. 1.1 : Anisotropy in crystals : Different arrangements of constituent particles about different directions, AB, CD and EF.

Question 7.
Define and explain isotropy.
Answer:
Isotropy : The ability of amorphous solids to exhibit identical physical properties even though measured in different directions is called isotropy.

Explanation : This property arises because there is no long range order of regular pattern of arrangement in them and hence the arrangement is irregular along all the directions. Therefore the magnitude of any physical property would be identical along all directions.

Question 8.
Why does crystalline solid show different refractive indices in different directions ?
Answer:

  1. Crystalline solid has long range order of regular pattern of arrangement which repeats periodically over entire crystal.
  2. Within the given pattern, the arrangements of different atoms or ions or molecules is different in different directions. Hence the properties like refractive indices in the different directions are different.

This shows that the crystalline solids are anisotropic in nature.

Question 9.
Explain the properties of amorphous solids.
Answer:

  1. The constituent particles in amorphous solids are arranged randomly.
  2. They have short range ordered structure.
  3. Amorphous solids are called supercooled liquids having very high viscosity.
  4. They do not have sharp melting points and they melt gradually over a temperature interval.
  5. Amorphous substances appear like solids but they do not have perfectly ordered crystalline structure, hence they are not real solids. Therefore they are pseudo solids.
  6. They are isotropic and exhibit the same magnitude of any property in every direction.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

Question 10.
What is isomorphism ?
Answer:
Isomorphism : A phenomenon in which two or more crystalline substances show same crystalline structure is called isomorphism and the crystals are said to be isomorphous. For example, NaNO3 and CaCO3. They have atomic ratios 1 : 1 : 3.

Question 11.
What is isomorphous?
Answer:
Isomorphous : When two or more crystalline substances have the same crystalline structure, they are said to be isomorphous. For example, NaF and MgO, NaNO3 and CaCO3.

Question 12.
What is polymorphism ?
Answer:
Polymorphism : A phenomenon in which when a single substance crystallises in two or more forms under different conditions of solidification is called polymorphism and the substance is called polymorphous. For example, calcite and oragonite are two forms of CaCO3.

Question 13.
What is polymorphous ?
Answer:
Polymorphous: A single substance which crystallises in two or more forms under different conditions of solidification is called polymorphous. Polymorphic forms of an element are called allotropic forms or allotropes. For example, carbon exists as diamond and graphite, or sulphur exists in rhombic and monoclinic allotropic forms.

Question 14.
Identify isomorphous and polymorphous substances in the following :
K2SO4, graphite, β-quartz, Na2SeO4, CaCO3, diamond, cristobalite, CsNO3.
Answer:

Isomorphous Polymorphous
K2SO4, Na2SeO4

CaCO3, CsNO3

Graphite, diamond

Β-quartz, cristobalite

Question 15.
Why does a crystalline solid has a sharp melting point ?
Answer:

  1. Crystalline solid is a homogeneous solid and it has long range order of regular pattern of arrangement which repeats periodically over entire crystal.
  2. The interatomic or intermolecular forces are identical, hence the thermal energy required to break the regular structure by overcoming the intermolecular forces is uniform throughout.
  3. Hence the heat and temperature needed to melt the solid are same, and therefore solids have sharp melting points.

Question 16.
Amorphous solids do not have sharp melting points. Explain.
Answer:

  1. Amorphous solids do not have perfectly ordered crystalline structure.
  2. They have short range order of regular pattern hence periodically repeating regular pattern is over a short distance.
  3. The thermal energy required to break the structure and separate constituent particles is not uniform.
  4. Hence the temperature needed to melt the solid is not same, therefore amorphous solids do not have sharp melting points but melt over a range of temperature.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

Question 17.
Give examples of (1) crystalline solids and (2) amorphous solids.
Answer:
(1) Crystalline solids : Metallic solids (Cu, Fe, etc.) crystalline salts (NaCl, K2SO4, etc.)
(2) Amorphous solids : Glass, plastics, rubber, etc.

Question 18.
Distinguish between crystalline solids and amorphous solids.
Answer:
Crystalline solids:

  1. They have definite characteristic geometrical shape.
  2. They have long range order of regular pattern of arrangement of constituent particles.
  3. They are true solids.
  4. They have sharp melting points.
  5. They are anisotropic in nature.
  6. They have definite heat of fusion.

Amorphous solids:

  1. They have irregular shape.
  2. They have short range order of regular pattern of arrangement of constituent particles.
  3. They are pseudo solids or supercooled liquids.
  4. They do not have sharp melting points.
  5. They are isotropic in nature.
  6. They do not have definite heat of fusion.

Question 19.
How are crystalline solids classified ?
Answer:
Crystalline solids are classified as follows :

  1. Ionic crystals
  2. Covalent network crystals
  3. Molecular crystals
  4. Metallic crystals

Question 20.
What are crystalline solids?
Answer:
The solids in which the constituent particles are charged ions namely cations and anions held together by electrostatic force of attraction are called crystalline solids.

Question 21.
What are the characteristics of ionic crystals ?
Answer:
The characteristics of ionic crystals are as follows :

  1. The constituents of ionic crystals are charged ions namely cations and anions. They differ in ionic size.
  2. The ions in these crystals are held by strong electrostatic force of attraction.
  3. Ionic crystals have high melting points and they are hard and brittle.
  4. In solid state they are nonconductors of electricity but they are good conductors when melted or dissolved in water.
  5. In aqueous solution they dissociate forming ions.
  6. Example : NaCl, KCl. CaF2, K2SO4, etc.

Question 22.
Explain why ionic solids are hard and brittle.
Answer:

  1. In ionic crystalline solids, constituent particles are positively charged cations and negatively charged anions placed at alternate lattice points.
  2. The ions are held by strong coulombic electrostatic forces of attraction compensating opposite forces. Hence they are hard.
  3. Since there are no free electrons, they are not malleable and on applying a shearing force, ionic crystals break into small units. Hence they are brittle.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

Question 23.
What are covalent network crystals ?
Answer:
The crystals in which the constituent particles are atoms linked by covalent bonds forming a continuous network are called covalent network crystals. For example, diamond, quartz.

Question 24.
What are the characteristics of covalent network crystals ?
Answer:
The characteristics of covalent network crystals are as follows :

  • The constituent particles in these solids are atoms.
  • The atoms in these crystals are held by covalent bonds forming a rigid three dimensional network which gives a giant molecule. Hence, the entire crystal is a single molecule.
  • These crystals are very hard (or hardest) and most incompressible.
  • They have high melting points and boiling points.
  • Since the electrons are localised they are poor conductors of heat and electricity.
  • Example : Quartz (SiO2), diamond.

Question 25.
Give the examples of network solids.
Answer:
The examples of covalent network solids are as follows :
Quartz (SiO2), diamond, boron nitride carborundum.

Question 26.
What are allotropes ?
Answer:
Allotropes : When a substance exists in two or more forms then they are called allotropes. They are polymorphous. For example, carbon has allotropes diamond and graphite.

Question 27.
What are molecular crystals ?
Answer:
The crystals in which the constituent particles are molecules (or unbonded single atoms) of the same substance held together by intermolecular forces of attraction. For example solidified Cl2, CO2, etc.

Question 28.
What are the characteristics of molecular crystals ?
Answer:
The characteristics of molecular crystals are as follows :

  1. The constituent particles of these solids are molecules (or unbonded single atoms) of the same substance.
  2. The atoms within the molecules are bonded by covalent bonds.
  3. The molecules are held together by intermolecular forces of attraction.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

Question 29.
What are intermolecular forces of attraction involved in molecular crystals ?
Answer:
The intermolecular forces involved in molecular crystals are as follows :
(1) Weak dipole-dipole interactions :
The solids constituting polar molecules like HCl, H2O, SO2, etc. which possess permanent dipole moment involve weak dipole-dipole interactions.

(2) Very weak dispersion or London forces :
The solids consisting of nonpolar molecules like CH4, H2, etc. involve weak dispersion forces. They are also involved in monoatomic solids like Ar, Ne.

(3) Intermolecular Hydrogen bonds :

  • In this crystalline solids, the constituent particles are the molecules which contain hydrogen atom linked to highly electronegative atom like F, O or N.
  • In these, molecules are held by hydrogen bonds in which H atom of one molecule is bonded to electronegative atom (like F, N or O) of another molecule.
  • Since hydrogen bonding is weak, these solids have very low melting points and generally at room temperature they exist in the liquid or gaseous state.
  • They are non-conductors of electricity.

Question 30.
What are metallic crystals ?
Answer:
These are crystalline solids formed by atoms of the same metallic element held together by metallic bonds.

Question 31.
What are the characteristics of metallic crystals ?
Answer:

  1. Metallic crystals are solids formed by atoms of the same metallic element held together by metallic bonds.
  2. Metallic crystals have high melting point and boiling point.
  3. Metals are malleable and can be hammered into thin sheets.
  4. Metals are ductile and can be drawn into thin wires.
  5. Metals are good conductors of heat and electricity.
  6. Examples are Cu, Ag, Au, Ni, etc.

Question 32.
Classify the following solids into different types :
(i) Plastic (ii) P4 molecule (iii) S8 molecule (iv) Iodine molecule (v) Tetra phosphorus decoxide (vi) Ammonium phosphate (vii) Brass (viii) Rubidium (ix) Graphite (x) Diamond (xi) NaCl (xii) Silicon.
Answer:

Solid Type
(i) Plastic Covalent network crystal
(ii) P4 molecule Covalent network crystal
(iii) S8 molecule Covalent network crystal
(iv) Iodine molecule Covalent network crystal
(v) Tetra phosphorus decoxide Covalent network crystal
(vi) Ammonium phosphate Ionic crystal
(vii) Brass Metalic crystal
(viii) Rubidium Metalic crystal
(ix) Graphite Covalent crystal
(x) Diamond Covalent crystal
(xi) NaCl Ionic crystal
(xii) Silicon Covalent crystal

Question 33.
Mention the types of the following solids :
(i) CaF2 (ii) SiC (iii) Ice (iv) SO2 (v) CaCO3 (vi) ZnS (vii) HCl (viii) CO2
Answer:

Solid Type
(i) CaF2 Ionic crystal
(ii) SiC Covalent crystal
(iii) Ice Hydrogen bonded molecular crystal
(iv) SO2 Molecular crystal
(v) CaCO3 Ionic crystal
(vi) ZnS Ionic crystal
(vii) HCl Polar molecular crystal
(viii) CO2 Non-polar molecular crystal

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

Question 34.
What is a giant solid ?
Answer:
Covalent solid formed by covalent bonds between neighbouring constituent atoms of non-metallic solid is called a giant solid. For example, graphite.

Question 35.
What is lattice ?
Answer:
Lattice is a geometrical arrangement of points in a three dimensional periodic array.

Question 36.
What is crystal lattice (space lattice) ?
Answer:
Crystal lattice (space lattice) : A regular arrangement of the constituent particles (atoms, ions or molecules) of a crystalline solid having similar environment in three dimensional space is called crystal lattice or space lattice.

Question 37.
What is a lattice point?
Answer:
Lattice point : A position occupied by a crystal constituent particle like an atom, ion or a molecule in the crystal lattice is called lattice point or lattice site.
OR
Any point at the intersection of the lines in the unit cell occupied by a constituent particle like an atom, an ion or a molecule in the crystalline solid is called a lattice point.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

Question 38.
What are the parameters of a unit cell ?
Answer:
A unit cell is characterised by following parameters :
(1) Edges or edge lengths : The intersection of two faces of crystal lattice is called as edge. The three edges denoted by a, b and c represent the dimensions (lengths) of the unit cell along three axes. These edges may or may not be mutually perpendicular.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 3

(2) Angles between the edges (or planes) : There are three angles between the edges of the unit cell represented as α, β and γ.

  • The angle α is between edges b and c.
  • The angle β is between edges a and c.
  • The angle γ is between edges a and b.

The crystal is defined with the help of these parameters of its unit cell.

Question 39.
Represent space lattice and unit cell diagrammatically.
Answer:
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 4

Question 40.
What do you understand by the basis of crystal lattice ?
Answer:

  • A crystal structure is formed by attaching a constituent particle to lattice points.
  • The constituent particles attached to the lattice points form the basis of the crystal lattice.
  • The crystal structure is obtained by attaching a basis to each of the lattice points.

This is represented by the following schematic equation :
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 5

Question 41.
What are the types of unit cells?
Answer:
Basically unit cells are of two types as follows :

  1. Primitive unit cells : The unit cells in which the constituent particles like atoms, ions or molecules are present only at the corners of the unit cell are called primitive unit cells or simple unit cells.
  2. Body-centred unit cell : A unit cell in which the constituent particles are present at the corners as well as at its body-centre is called body-centred unit cell.
  3. Face-centred unit cell : A unit cell in which the constituent particles are present at the corners as well as at the centre of each face is called face-centred unit cell or cubic close packed (CCP) unit cell.
  4. Base-centred unit cell : A unit cell in which the constituent particles are present at the corners as well as at the centres of two opposite faces is called end-centred unit cell.

Question 42.
Explain briefly crystal systems.
Answer:
(1) The constituent particles like atoms, ions or molecules of the crystal can be arranged in seven different ways changing edges (a, b, c) and angles (α, β, γ) and accordingly they form seven systems or types of the crystal.

(2) These seven crystal system are named as :
(a) Cubic system, (b) Tetragonal system (c) Orthorhombic system (d) Rhombohedral system (e) Monoclinic system (f) Triclinic system (g) Hexagonal system.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

Question 43.
What are Bravais lattices ?
Answer:

  1. There are seven crystal systems according to the edges (a. b, c) and angles (α, β, γ).
  2. The constituents of the crystal may be present at corners, face centres, body centres, edge centres and voids.
  3. By mathematical analysis, it has been proved that only fourteen different kinds of space lattices are possible.
  4. Hence there are fourteen different ways of arrangement of the lattice basis.
  5. These fourteen lattices of seven crystal systems are called Bravais lattices.

Question 44.
Explain Bravais lattices of a cubic system.
OR
Explain unit cells of a cubic system.
Answer:
Cubic lattice : For this, edges are a = b = c and angles are α = β = γ = 90°. In this cubic system, there are three Bravais lattices.
(1) Simple (or primitive) cubic unit cell (SCC) : In this unit cell, atoms are present only at 8 corners of the cube.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 6

(2) Body-centred cubic unit cell (BCC) : In this, atoms are present at 8 corners along with one additional atom at the body-centre of the cube.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 7

(3) Face-centred cubic unit cell (FCC) : In this unit cell, atoms are present at 8 comers and at 6 face centres.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 8

Question 45.
Give the number of lattice points in one unit cell of the following crystal structures :
(1) Simple cubic
(2) Face-centred cubic
(3) Body-centred cubic.
Answer:
Lattice points in one unit cell represent the positions of atoms, ions or molecules in the unit cell.
(1) Simple cubic unit cell : In this primitive unit cell, the lattice points are at 8 corners of the unit cell. Hence there are 8 lattice points.
(2) Face-centred cubic unit cell : In this unit cell, the lattice points are at 8 comers and 6 face centres.
(In cubic close packing unit cell, the lattice points are also at edge centres and body centre.)
(3) Body-centred cubic unit cell : In this, the lattice points are at 8 comers and one at body centre.

Question 46.
Find the number of atoms per unit cell in the following crystal structures.
(1) Simple cubic unit cell
(2) Body-centred cubic unit cell
(3) Face-centred cubic cell.
Answer:
(1) Number of atoms in primitive simple cubic (scc) unit cell :
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 9
In simple or primitive cubic unit cell, there are 8 atoms at 8 corners. Each corner contributes 1/8th atom to the unit cell.
∴ Number of atoms present in the unit cell = \(\frac {1}{8}\) × 8 = 1
Hence the volume of the unit cell is equal to the volume of one atom.

(2) Number of atoms in body-centred cubic (bcc) unit cell:
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 58
In this unit cell, there are 8 atoms at 8 corners and one additional atom at the body centre. Each corner contributes 1/8th atom, to the unit cell, hence due to 8 corners.
Number of atoms = 8 × \(\frac {1}{8}\)
= 1 atom.
An atom at the body centre wholly belongs to the unit cell.
∴ Total number of atoms present in bcc unit cell = 1 + 1 = 2.
Hence the volume of unit cell is equal to the volume of two atoms.

(3) Number of atoms in face-centred cubic (fcc) unit cell :
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 10
In this unit cell, there are 8 atoms at 8 comers and 6 atoms at 6 face centres. Each corner contributes 1/8th atom to the unit cell, hence due to 8 corners,
Number of atoms = \(\frac {1}{8}\) × 8 = 1.
Each face centre contributes half of the atom to the unit cell, hence due to 6 face centres,
Number of atoms = \(\frac {1}{2}\) × 6 = 3.
∴ Total number of atoms present in fee unit cell = 1 + 3 = 4.
Hence the volume of the unit cell is equal to the volume of four atoms.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

Question 47.
Obtain a relation for the density of the unit cell and radius of atom or sphere for the following :
(1) Simple cubic (scc) crystal
(2) Body centred cubic (bcc) crystal
(3) Face centred cubic (fcc) crystal.
Answer:
(1) Consider a unit cell of a simple cubic crystal. It has 8 atoms at 8 corners of the unit cell.
∴ Total number of atoms in unit cell = \(\frac {1}{8}\) × 8 = 1
If a is the length of edge of cubic unit cell and r is the radius of the atom, then r = a/2 or a = 2r.
Volume of the unit cell = a3 = (2r)3 = 8r3
If M is atomic mass of the element, then mass of one atom is M/NA where NA is Avogadro number. If there are V atoms in one unit cell then,
Mass of unit cell = n × Mass of one atom = n × \(\frac{M}{N_{\mathrm{A}}}\)
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 11
Since there is one atom present in one unit cell,
ρ = \(\frac{M}{N_{\mathrm{A}} \times a^{3}}\)

(2) Consider a unit cell of body centred cubic (bcc) crystal. It has 8 atoms at 8 comers and one additional atom at the centre of body of unit cell.
Number of atoms due to 8 corners = \(\frac {1}{8}\) × 8 = 1
Body centred atom, wholly belong to the unit cell. Hence total number of atoms in the unit cell is two. If M is atomic mass of an element then M/NA is mass of one atom where NA is Avogadro number.
Mass of unit cell = Mass of 2 atoms in unit cell = 2 M/NA
If a is the edge length or lattice parameter then,
Volume of cubic unit cell = a3
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 12

(3) Consider a unit cell of face centred cubic (fcc) crystal.
It has 8 atoms at 8 comers and 6 atoms at 6 face centres.
∴ Total number of atoms in unit cell = \(\frac {1}{8}\) × 8 + \(\frac {1}{2}\) × 6 = 1 + 3= 4
If M is the atomic mass of an element, then mass of one atom is M/NA, where NA is Avogadro number. Mass of unit cell = Mass of 4 atoms
= \(4 \times \frac{M}{N_{\mathrm{A}}}\)
If a is the edge length of this cubic unit cell then,
volume of unit cell = a3
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 13

Question 48.
What is coordination number? What is its significance?
Answer:
(1) Coordination number : The number of the closest neighbouring constituent particles like atoms, ions or molecules which are in contact with a particular particle or an atom in the crystal lattice is called coordination number of that particle.
(In the crystal lattice, all atoms may have same or different coordination numbers.)

(2) The magnitude of the coordination number is a measure of compactness of spheres in close-packed structures.

(3) The higher the coordination number, the closer are the spheres to each other.

Question 49.
Explain linear packing in one direction.
OR
Explain close packing in one dimension.
Answer:
Linear packing in one direction or close packing in one dimension :
The constituent particles of the crystal may be of varying shapes but for better understanding we consider particles as hard spheres of equal size.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 14
There is only one way of arranging or packing spheres placed in a horizontal row touching one another. Since one sphere is in contact with two neighbouring spheres except the end atoms, the coordination number in this arrangement is two. This packing may be in any one direction x, y or z.

Question 50.
Explain the following :
Planar packing arrangement of spheres.
OR
Close packing in two dimensions.
OR
AAAA type and ABAB type of two dimensional arrangement.
OR
(i) Square close packing
(ii) Hexagonal close packing.
Answer:
Two dimensional close packing crystal structure can be generated by placing one dimensional linear crystal structure over another to form multiple layers. This staking of linear rows may be taking place in two different ways giving two different two dimensional structures as follows :

(i) AAAA type two dimensional close packing or square close packing :
In this arrangement, various one dimensional rows are placed on one over other so that each sphere in one row is over the another sphere of another row forming planar structure. In this, spheres have horizontal as well as vertical alignment. All the rows of spheres are identical in planar structure. All crests as well as all the depressions or troughs formed by the arrangement are also aligned.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 15

If the first row is labelled as A type, then second and all subsequent rows are also identical, hence are of A type. Therefore this planar two dimensional close packing is called AAAA type packing.

In this arrangement, each sphere is in contact with (or touching) four other spheres around it, hence the coordination number of each sphere is four and the packing is called two dimensional or planar square close packing. In this, packing efficiency is 52.4%.

(ii) ABAB type two dimensional packing or hexag-onal close packing :
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 16

In this arrangement, crests of the spheres of one row are placed into the depressions or troughs formed between adjacent spheres of next row. This arrangement is repeated consecutively throughout.

In this arrangement, crests of the spheres of one row are in contact with depressions or troughs of next row.

If one row of spheres is labelled as A then the next row will be B, third row will again be A, fourth row B and so on. Hence this planar or two dimensional close packing is called ABAB… type packing.

In this arrangement, each sphere is in contact with six other spheres around it hence the coordination number of each sphere is six and the packing is called two dimensional or planar hexagonal close packing. In this, the packing efficiency is 60.4% which is more than linear close packing.

Question 51.
Explain close packing in three dimensions.
Answer:
Close packing in three dimensions :
Three dimensional crystal structures are obtained by stacking of two dimensional layers. Simple cubic lattice is obtained by stacking of two dimensional square layers.

The stacking of two dimensional hexagonal close packed layers gives two structures namely hexagonal close packed (hcp) structure and face centred (fcc) structure.

(i) Stacking of square close packed layers :
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 17

In this arrangement, the two dimensional AAAA type square closed packed layers are placed one over the other in such a way that the crests of all spheres are in contact with successive layers in all directions. All spheres of different layers are perfectly aligned horizontally and vertically forming unit cells having primitive or simple cubic structure. Since all the layers are identical and if each layer is labelled as layer A, then whole three dimensional crystal lattice will be of AAAA… type.

Each sphere is in contact with six surrounded spheres, hence the co-ordination number of each sphere is six.

(ii) Stacking of two hexagonal close packed layers :
A close packed three dimensional structure can be generated by arranging hexagonal close packed layers in a particular manner.

In this the spheres of second layer are placed in the depression of the first layer.

In this if first layer is labelled as A then second layer is labelled as B since they are aligned differently.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 18

In this, all triangular voids of the first layers are not covered by the spheres of the second layer. The triangular voids which are covered by second layer spheres generate tetrahedral void which is surrounded by four spheres. The triangular voids in one layer have above them triangular voids of successive layers.

The overlapping triangular voids from two layers together form an octahedral void which is surrounded by six spheres.

(iii) Placing third hexagonal close packed layer :
(a) Hexagonal close packing (hcp) : If the crests of spheres of third layer are placed on the triangular shaped tetrahedral voids C of the second layer, then three dimensional closest packing structure is obtained in which the spheres of third layer lie directly above the spheres of first layer, i.e., first and third layers are identical. Following same placing of layers, fourth layer will be identical to second layer.

If the first layer is labelled A and second layer B, then the arrangement of packing will be of ABAB type. This is also called hexagonal close packing (hcp) as shown in the figure. In this, packing efficiency is 74%. The coordination number of each sphere is 12. The metals Be. Mg, Zn, Cd crystallise in HCP crystalline structure.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 19

(b) Cubic close packing (ccp) : If the crests of spheres of third layer are placed in the positions of tetrahedral void ‘a’ having apex upward of first layers, then the third layer will not be identical to the first, and may be labelled as C, which is different from A and B layers. Fourth layer may be arranged above third layer such that the spheres are aligned, so that the first layer and fourth layer are identical, second and fifth layers are identical and so on.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 20
If first, second and third layers are labelled as A, B and C respectively then the arrangement of packing will be ABCABC type. This is also called cubic close packing (ccp) as shown in the figure. This is similar to face centred cubic (fcc) packing.

In this, arrangement packing efficiency is 74% and the coordination number of each sphere is 12.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

Question 52.
Explain tetrahedral void.
Answer:
(1) Tetrahedral void : The vacant space or void among four constituent particles having tetrahedral arrangement in the crystal lattice is called tetrahedral void.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 21
The arrangement of four spheres around the void is tetrahedral. A tetrahedral void is formed when a triangular void made by three coplanar spheres is in contact with fourth sphere above or below it.

(2) Characteristics of tetrahedral void :

  • The volume of the void is much smaller than that of atom or sphere.
  • Larger the size of sphere, more is the size of void.
  • If R is the radius of the constituent atom, then the radius of the tetrahedral void is 0.225 R.
  • Coordination number of tetrahedral void is four.
  • There are two tetrahedral voids per sphere, in the crystal lattice. If the number of closed packed spheres is N then the number of tetrahedral voids is 2N.

Question 53.
Explain octahedral void.
Answer:
(1) Octahedral void : The vacant space or void at the centre of six spheres (or atoms) which are placed octahedrally is called octahedral void.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 22

(2) Characteristics of octahedral void :

  • The volume of the void is small.
  • There is one octahedral void at the body centre and twelve octahedral void positions at twelve edge centres.
  • If R is the radius of constituent atom, then the radius of the octahedral void is 0.414 R.
  • The coordination number of octahedral void is six.
  • There is one octahedral void per sphere in the crystal lattice. If the number of closed packed spheres is N then the number of octahedral voids is N.

Question 54.
What are number of voids per atom in hep and ccp ?
Answer:
The tetrahedral and octahedral voids occur in hep and ccp/fcc structures. There are two tetrahedral voids associated with each atom. The number of octahedral voids is half that of tetrahedral voids. Thus, there is one octahedral void per atom.

Question 55.
What is packing efficiency?
Answer:
(1) Packing efficiency : It is the fraction of a percentage of the total space (of the unit cell) occupied by the particles (spheres).
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 23
( 2) The magnitude of packing efficiency gives a measure of how tightly particles are packed together.

Question 56.
Calculate packing efficiency in body-centred cubic lattice.
Answer:
Step 1 : Radius of sphere :
In the unit cell of body-centred cubic lattice, there are 8 atoms at 8 corners and one atom at the centre of the cube.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 24

The atoms are in contact along the body diagonal BF. Let a be the edge length and r the radius of an atom.
Consider a triangle BCE.
BE2 = BC2 + CE2 = a2 + a2 = 2a2
Consider triangle BEF,
BF2 = BE2 + EF2 = 2a2 + a2 = 3a2
BF = \(\sqrt{3}\)a.
From figure, BF = 4r
∴ 4r = \(\sqrt{3}\)a
∴ r = \(\frac{\sqrt{3}}{4} a\)

Step 2 : Volume of sphere :
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 25

Step 3 : Total volume of particles :
The unit cell of bcc structure contains 2 particles.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 26

Step 4 : Packing efficiency
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 27
∴ Packing efficiency = 68%
∴ Percentage of void space = 100 – 68
= 32%

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

Question 57.
Calculate packing efficiency in face-centred cubic lattice.
Answer:
Step 1 : Radius of sphere :
In the unit cell of face-centred cubic lattice, there 8 atoms at 8 corners and 6 atoms at 6 face centres. Consider the face ABCD.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 28

The atoms are in contact along the face diagonal BD.
Let a be the edge length and r, the radius of an atom.
Consider a triangle BCD.
BD2 = BC2 + CD2
= a2 + a2 = 2a
∴ BD = \(\sqrt{2} a\)
From figure, BD = 4r
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 29

Step 3 : Total volume of particles :
The unit cell of fee crystal lattice contains 4 particles.
∴ Volume occupied by 4 particles = \(4 \times \frac{\pi a^{3}}{12 \sqrt{2}}\)
= \(\frac{\pi a^{3}}{3 \sqrt{2}}\)

Step 4 : Packing efficiency :
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 30
∴ Packing efficiency = 74%
∴ Percentage of void space = 100 – 74
= 26%

Edge length and particle parameters in cubic system
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 31a

Coordination number and packing efficiency in systems

Lattice Coordination number of atoms Packing efficiency
1. scc 6 : four in the same layer, one directly above and one directly below 52.4%
2. bcc 8 : four in the layer below and four in the layer above 68%
3. fcc/ccp/hcp 12 : six in its own layer, three above and three below 74%

Question 58.
Calculate the number of particles and unit cells in ‘x’ g of metallic crystal.
Answer:
Consider ‘x’ gram of a metallic crystal of molar mass M and density ρ. If the unit cell of the crystal has edge length ‘a’ then, volume of unit cell = a3.
Mass of one metal atom = \(\frac{M}{N_{\mathrm{A}}}\). If ‘n’ number of atoms are present in one unit cell then
Mass of unit cell = \(\frac{n \times M}{N_{\mathrm{A}}}\)
If ‘a’ is the edge length of the unit cell then, Volume of unit cell = a3
Density of unit cell = ρ = \(\frac{n \times M}{N_{\mathrm{A}}} \times \frac{1}{a^{3}}\)
∴ M = \(\frac{\rho \times N_{\mathrm{A}} \times a^{3}}{n}\)
∵ Molar mass M gram contains NA particles
∴ x gram contains \(\frac{x \times N_{\mathrm{A}}}{M}\) particles.
Substituting the value M,

(i) Number of particles in x g crystal
= \(\frac{x \times N_{\mathrm{A}}}{\rho \times N_{\mathrm{A}} \times a^{3} / n}\)
= \(\frac{x \times n}{\rho \times a^{3}}\) particles

(ii) Number of unit cells in x g crystal :
∵ n particles are present in 1 unit cell
∴ \(\frac{x \times n}{\rho \times a^{3}}\) are present in, \(\frac{x \times n}{\rho \times a^{3}} \times \frac{1}{n}\)
= \(\frac{x}{\rho \times a^{3}}\) unit cells

(iii) Number of unit cells in V volume of crystal = \(\frac{V}{a^{3}}\)

Alternative method :
Consider ‘x’ g metal of atomic mass M g mol-1.
Number of moles of metal = \(\frac{x}{M}\)
(a) Number of atoms (particles) of metal = \(\frac{x}{M} \times N_{\mathrm{A}}\)
(b) If unit cell contains ‘n’ atoms,
Number of unit cells = \(\frac{x}{M} \times \frac{N_{\mathrm{A}}}{n}\)
(c) If ‘a’ is the edge length then,
Volume of unit cells = a3
∴ Number of unit cells in V volume of crystal = \(\frac{V}{a^{3}}\).

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

Solved Examples 1.5 – 4.7

Question 59.
Solve the following :
[1 m = 10 dm = 100 cm = 109 nm = 1012 pm, 1Å = 10-8 cm = 100 pm]

(1) A cubic unit cell of a crystal consists of atoms of A and B elements. Atoms of A occupy corners of the unit cell while one B atom is present at the body centre. Determine the formula of the crystalline compound.
Solution :
Given : Atoms of A are at the comers while atom B is at the body centre of the cubic unit cell.
Since \(\frac {1}{8}\)th atom is contributed at each comer and there are 8 comers in unit cell, number of atoms of A due to comers is \(\frac {1}{8}\) × 8 = 1.

In addition there is one atom of B at body centre. Hence this unit cell contains one atom of each, A and B, therefore the formula of the compound is AB.
Ans. The formula of the compound = AB.

(2) Atoms C and D form fee crystalline structure. Atom C is present at the corners of the cube and D is at the face centres of the cube. What is the formula of the compound ?
Solution :
Given : Crystal has fee structure.
Atoms C are at 8 comers while atoms D are at 6 face centres of the cubic unit cell.
At the comer, \(\frac {1}{8}\)th of each C atom is present while at each face centre, half of each D atom is present.
Number of C atoms = \(\frac {1}{8}\) × 8 = 1.
Number of D atoms = \(\frac {1}{8}\) × 6 = 3.
Thus unit cell contains one C atom and three D atoms.
Hence the formula of the compound is CD3.
Ans. Formula of the compound = CD3.

(3) A cubic unit cell contains atoms A at the corners, atoms B at face centres and atom C at the body centre. What is the formula of the crystalline compound?
Solution :
Given : Atoms A are at 8 comers, atoms B at the 6 face centres and one atom C at body centre.
Total number of atoms of A = \(\frac {1}{8}\) × 8 = 1.
Total number of atoms of B = \(\frac {1}{2}\) × 6 = 3.
One atom of C at the body centre.
Therefore the unit cell contains one atom of A, three atoms of B and one atom of C.
Hence the formula of the compound is AB3C.
Ans. The formula of the crystalline compound is AB3C.

(4) An element A and B constitute bcc type crystalline structure. Element A occupies body centre position and B is at the corners of cube. What is the formula of the compound? What are the coordination numbers of A and B ?
Solution :
Given : Crystalline structure is bcc type.
Atoms A are at 8 comers and atom B is at body centre.
∴ Number of atoms of A in a unit cell
= \(\frac {1}{8}\) × 8 = 1.
Number of atom B in a unit cell = 1.
Since unit cell contains one atom each of A and B, the formula of the compound is AB.
The coordination number of an atom A at comer is 8.
The coordination number of an atom B at body centre is 8.
Answer. Formula of the compound = AB.
Coordination number of A = 8
Coordination number of B = 8.

(5) Mention the number of atoms in the following unit cells :
(a) scc (b) bcc (c) fcc (d) hcp.
Answer:

Unit cell Number of atoms
(a) scc 1
(b) bcc 2
(c) fcc 4
(d) hcp 3

(6) 0.1 mole of Buckminster fullerene of molar mass 720 gmol-1 contains how many Kg of carbon ?
Solution :
Molar mass of fullerene, C60 = 720 gmol-1
∵ Mass of 1 mole of fullerene = 720 g = 0.72 kg
∴ Mass of 0.1 mole of fullerene = 0.72 × 0.1
= 0.072 kg
Ans. 0.072 kg.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

(7) In a cubic crystalline structure of zinc blende (ZnS), sulphide ions are at corners and face centres while zinc ions occupy half of tetrahedral voids.
Find in the unit cell :
(i) Number of Zn2+ ions
(ii) Number of S2- ions
(iii) Number of ZnS molecules
(iv) Molecular formula of zinc blende.
Solution :
Given : S2- ions are at 8 comers and 6 face centres.
Zn2+ ions occupy half of tetrahedral voids.
Number of S2- ions in unit cell
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 32
= 1 + 3 = 4

Cubic unit cell has 8 tetrahedral voids. Since half of them are occupied by Zn2+ ions, there are 4Zn2+ ions in the unit cell.
Hence number of zinc sulphide (ZnS) molecules is 4.
Molecular formula of zinc blende is ZnS.
Ans. (i) Number of Zn2+ ions = 4
(ii) Number of S2- ions = 4
(iii) Number of ZnS molecules = 4
(iv) Molecular formula of zinc blende = ZnS.

(8) In a crystalline compound, atoms A occupy ccp lattices while atoms B occupy 2/3 rd tetrahedral voids. What is the formula of the compound ?
Solution :
In ccp unit, lattice points are 8 comers and 6 face centres where atoms A are present.
∴ Number of A atoms
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 33
In cubic unit cell, there are 8 tetrahedral voids.
Hence,
Number of B atoms = \(\frac {2}{3}\) × 8 = \(\frac {16}{3}\).
Hence the formula should be A4B16/3 or A12B16 or A3B4.
Ans. The formula of the compound is A3B4.

(9) A compound forms hep structure. What is the number of (a) octahedral voids, (b) tetrahedral voids, (c) total voids formed in 0.2 mol of the compound?
Solution :
Number of atoms in 0.2 mol of the compound
= 0.2 × NA = 0.2 × 6.022 × 1023
= 1.2044 × 1023 atoms
(a) Number of octahedral voids
= Number of atoms
= 1.2044 × 1023

(b) Number of tetrahedral voids
= 2 × Number of atoms
= 2 × 1.2044 × 1023
= 2.4088 × 1023

(c) Total number of voids
= 1.2044 × 1023 + 2.4088 × 1023
= 3.6132 × 1023
Ans. (a) Number of octahedral voids
= 1.2044 × 1023
(b) Number of tetrahedral voids = 2.4088 × 1023
(c) Total number of voids = 3.6132 × 1023

(10) Copper crystallises into a fcc structure and the unit cell has length of edge 3.61 × 10-8 cm. Calculate the density of copper. Atomic mass of copper is 63.5 g mol-1.
Solution :
Given : Crystalline structure of Cu is fcc.
Edge length = a = 3.61 × 10-8 cm
Atomic mass of Cu = 63.5 g mol-1
Avogadro number = 6.022 × 1023 mol-1
Density = d = ?

In fcc structure, there are 8 Cu atoms at 8 comers and 6 Cu atoms at 6 face centres.
∴ Total number of Cu atoms
= \(\frac {1}{8}\) × 8 + \(\frac {1}{2}\) × 6 = 1 + 3
= 4
Mass of one Cu atom
= \(\frac{63.5}{6.022 \times 10^{23}}\) = 1.054 × 10-22 g
∴ Mass of 4 Cu atoms = 4 × 1.054 × 10-22
= 4.216 × 10-22 g
Mass of unit cell = Mass of 4 Cu atoms
= 4.216 × 10-22g
Volume of unit cell = a3 = (3.61 × 10-8)3
= 4.7 × 10-24 cm3
Density of unit cell = \(\frac{\text { Mass of unit cell }}{\text { Volume of unit cell }}\)
∴ ρ = \(\frac{4.216 \times 10^{-22}}{4.7 \times 10^{-23}}\) = 8.97 g cm-3
Ans. Density of Cu = 8.97 g cm-3.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

(11) Niobium is found to crystallise with bcc structure and found to have density of 8.55 g cm-3 (OR 8.55 kg m-3). Determine the atomic radius of niobium if its atomic mass is 93.
Solution :
Given : Density of Niobium (Nb) crystal = 8.55 g cm-3
Crystalline stmeture is bcc.
Atomic mass of Nb = 93 g mol-1
Avogadro number = NA = 6.022 × 1023 mol-1
Atomic radius of Niobium = ?
In bcc unit cell, there are 8 atoms at 8 comers and 1 atom at the body centre.
∴ Number of Nb atoms = \(\frac {1}{8}\) × 8 + 1 = 1 + 1 = 2.
Mass of one Nb atom = \(\frac{93}{6.022 \times 10^{23}}\) = 1.544 × 10-22
∴ Mass of 2 Nb atoms = 2 × 1.544 × 10-22 = 3.088 × 10-22 g
Mass of unit cell
= Mass of 2Nb atoms = 3.088 × 10-22 g
If a is edge length of bcc unit cell, volume of unit cell = a3
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 34
= 0.361 × 10-22 cm3 = 36.1 × 10-24 cm3
∴ a = (36.1 × 10-24)-1/3 = 3.3 × 10-8 cm
If r is the radius of 1 Nb atom, then in bcc structure
r = \(\frac{\sqrt{3}}{4} a=\frac{\sqrt{3}}{4}\) × 3.3 × 10-8
= 1.43 × 10-8 cm
= 1.43 × 10-10 m
= 1.43 × 10-10 × 109 nm
= 0.143 nm
Ans. Atomic radius of niobium atom = 0.143 nm

(12) Gold occurs as face centred cube and has a density of 19.30 kg dm-3. Calculate atomic radius of gold. (Molar mass of Au = 197)
Solution :
Given : Density of Au = 19.3 kg dm-3
Molar mass = 197 g mol-1
Avogadro constant = NA = 6.022 × 1023 mol-1
Atomic radius of Au = ?
In fcc unit cell, there are 8 atoms of Au at 8 comers and 6 atoms at 6 face centres.
Number of Au atoms in the unit cell = \(\frac {1}{8}\) × 8 + \(\frac {1}{2}\) × 6
= 4 atoms
Mass of 1 Au atom = \(\frac{197}{6.022 \times 10^{23}}\) = 3.271 × 10-22 g
∴ Mass of 4 Au atoms = 4 × 3.271 × 10-22 g = 1.308 g × 10-21 g
∴ Mass of unit cell = 1.308 × 10-21 g
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 35
= 6.777 × 10-24 dm3
= 6.777 × 10-23 cm3
∴ a = (6.777 × 10-23)1/3 = (67.77 × 10-24)-1/3
= 4.077 × 10-8 cm
If r is the radius of Au atom, then for fcc unit cell,
r = \(\frac{a}{2 \sqrt{2}}\)
= \(\frac{4.077 \times 10^{-8}}{2 \sqrt{2}}\) = 1.442 x 10-8 cm = 144.2 pm
Ans. Radius of Au atom = 144.2 pm.

(13) A compound is formed by two elements X and Y. The atoms of Y form ccp structure. The atoms of A occupy \(\frac {1}{3}\) of tetrahedral voids. Find the formula of the compound.
Solution :
In ccp structure, Y atoms are present at 8 comers and 6 face-centres of the ccp structure.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 36
There are 8 tetrahedral voids of which \(\frac {1}{3}\)rd are occupied by atoms X. Hence unit cell has \(\frac {8}{3}\) atoms of X and 4 atoms of Y. The formula will be,
X8/3 Y4 or X8Y12 or X2Y3
Ans. Formula of the compound = X2Y3.

(14) A metal crystallises into two cubic faces namely face centered (FCC) and body centered (BCC), whose unit cell edge lengths are 3.5 Å and 3.0 Å respectively. Find the ratio of the densities of FCC and BCC.
Solution :
Given : Edge length of unit cell of fcc metal = 3.5 Å
= 3.5 × 10-8 cm
Edge length of unit cell of bcc metal = 3 Å = 3 × 10-8 cm
Density d = \(\frac{n \times \mathbf{M}}{a^{3} \times \mathbf{N}_{\mathrm{A}}}\)
where, n = Number of Fe atoms in the unit cell
M = Atomic mass of metal
a = Edge length of unit cell
NA = Avogadro number
∴ For fcc unit cell = n = 4
For bcc unit cell = n = 2
∴ \(\frac{\text { Density of fcc unit cell }}{\text { Density of bcc unit cell }}=\frac{d_{\mathrm{fcc}}}{d_{\mathrm{bcc}}}\)
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 37

(15) The density of silver having atomic mass 107.8 gram mol-1 is 10.8 gram cm-3. If the edge length of cubic unit cell is 4.05 × 10-8 cm, find the number of silver atoms in the unit cell. (NA = 6.022 × 1023, 1Å = 10-8 cm)
Solution :
Given : d = 10.8 g cm-3
M= 107.8 g mol-1
a = 4.05 × 10-8 cm
Number of Ag atoms in unit cell = n = ?
Mass of one Ag atom = \(\frac{107.8}{6.022 \times 10^{23}}\)
= 1.79 × 10-22 g
If there are n atoms, then
Mass of unit cell = n × 1.79 × 10-22 g
Volume of unit cell = a3 = (4.05 × 10-8)3
= 6.643 × 10-23 cm3
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 38
Ans. Number of silver atoms (Ag) atoms in unit cell = 4.

(16) Aluminium having atomic mass 27 g mol-1 crystallises in face centred packed cubic crystal. Find the number of Al atoms in 10 g aluminium. How many unit cells will be present in it?
Solution :
Given : Atomic mass of Al = 27 g mol-1
Mass of Al = 10 g
Avogadro number = NA = 6.022 × 1023 mol-1
Number of Al atoms = ?
Number of unit cells = ?
1 gram atom of Al = 27 g Al contains 6.022 × 1023
Al atoms
∴ Number of Al atoms in 10 g
= \(\frac{10 \times 6.022 \times 10^{23}}{27}\)
= 2.23 × 1023
In fcc structure, each unit cell contains 4Al atoms.
∴ Number of unit cells = \(\frac{2.23 \times 10^{23}}{4}\)
= 5.575 × 1022
Ans. Number of Al atoms = 2.23 × 1023
Number of unit cells = 5.575 × 1022.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

(17) The density of iron crystal is 8.54 gram cm-3. If the edge length of unit cell is 2.8 Å and atomic mass is 56 gram mol-1, find the number of atoms in the unit cell. What is the type of crystal ?
Solution :
Given :
Density of Fe crystal = d = 8.54 g cm-3
a = 2.8 Å = 2.8 × 10-8 cm
Atomic mass = M = 56 g mol-1
Number of atoms in unit cell, n = ?
Mass of one atom = \(\frac{56}{6.022 \times 10^{23}}\) = 9.3 × 10-23 g
Volume of unit cell = a3 = (2.8 × 10-8)3
= 2.195 × 10-23 cm3
If there are n atoms in the unit cell, then
Mass of unit cell = n × 9.3 × 10-23 g
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 39
Since unit cell contains 2 atoms, the crystal has bcc structure.
Ans. Number of atoms in unit cell = 2
Type of crystal = bcc

Question 60.
What is meant by crystal defects or imperfections?
Answer:

  1. Defect in crystalline structure : Any deviation from orderly and stoichiometrically perfect arrangement of atoms, ions or molecules in the crystal lattice is called a defect in the crystalline structure.
  2. Defects are created during the crystallisation process. If the process occurs at faster rate, the defects are more.
  3. The properties of solids are affected due to imperfactions.

Question 61.
Mention the types of defects in the solids or crystal structures.
Answer:
The defects in crystalline solids are of two types viz., (1) Point defect and (2) Line defect.
(1) Point defects are further classified as :
(a) Vacancy defect or Schottky defect
(b) Interstitial defect or Frenkel defect
(c) Impurity defect :

This is further classified as

  • Substitution impurity defect
  • Interstitial impurity defect.

(2) Line defects are further classified as

  • Edge dislocation
  • Screw dislocation.

Question 62.
What are point defects?
Answer:
Point defects : These defects arise due to irregularities produced in the arrangement of basis of lattice points in crystalline solids.

Question 63.
What are major classes of point defects ?
Answer:
There are three major classes of point defects : stoichiometric point defects, impurity defects and nonstoichiometric point defects.

There are four types of stoichiometric point defects as vacancy defect, self interstitial defect, Schottky defect and Frenkel defect.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

Question 64.
Explain vacancy defect.
Answer:
Vacancy defect :
(1) During crystallisation, some of regular sites in solid remain unoccupied and the missing particle creates a vacancy defect.
(2) The defect can be developed by heating the substance.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 40
(3) The mass of solid decreases due to absence of particles in regular sites.
(4) Since the volume remains the same the density of the substance decreases.

Question 65.
Explain self interstitial defect in elemental solid.
Answer:
Self Interstitial defect in elemental solid : The empty spaces or voids in between the particles at lattice points represent interstitial defective sites or self interstitial defects.

This defect arises in the following two ways :
(1) An extra particle occupies the empty interstitial space. This extra particle is similar to those already present in the crystal.

(2) (i) A particle gets shifted from its original regular site to an empty interstitial space in the crystal.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 41
(ii) This displacement of a particle produces a vacancy defect at its regular site.
(iii) This defect is referred to as a combination of vacancy defect and self interstitial defect.
(iv) Since there is neither loss or gain in mass of the substance, the density of it remains unchanged.

Question 66.
How does Schottky defect arise?
Answer:
(1) Schottky defect arises in ionic solids due to missing of equal number of cations and anions from their regular positions in the crystal lattice creating vacancies.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 42
(2) There arises formation of two holes per loss of ion pair.

(3) Conditions for formation of Schottky defect :
Characteristics of ionic solids showing Schottky defect :

  • High degree of ionic character
  • High coordination number of anion
  • Small difference between ionic size or radii of cation and anion. The ionic ratio \(\frac{r_{\text {cation }}}{r_{\text {anion }}}\) is not below unity.

Question 67.
How does Frenkel defect arise?
Answer:
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 43

  • Frenkel defect : This defect arises when an ion of an ionic compound is missing from its regular site and occupies interstitial vacant position between lattice points.
  • Cations have smaller size than anions, hence generally cations occupy the interstitial sites.
  • This creates a vacancy defect at its original position and interstitial defect at new position.
  • Frenkel defect is regarded as the combination of interstitial defect and vacancy defect.

Conditions for the formation of Frenkel defect :

  • This defect arises in ionic compounds with a large difference between the sizes of cation and anion.
  • The ionic compounds must have ions with low coordination number.

Question 68.
What are the consequences of Frenkel defect ?
Answer:
Consequences of Frenkel defect :

  • Since there is no loss of ions from the crystal lattice, the density of the solid remains unchanged.
  • The crystal remains electrically neutral.
  • This defect is observed in ZnS, AgCl, AgBr, AgI, CaF2, etc.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

Question 69.
Explain nonstoichiometric defect.
Answer:
Nonstoichiometric defect : This defect arises when the ratio of number of one kind of atoms to that of other kind of atoms (or ratio of number of cations to anions) becomes different from the actual stoichiometric formula. This involves the change in stoichiometry of the compound.

There are two types of nonstoichiometric defects as follows :
(1) Metal deficiency defect : This defect arises in compounds of metal which show variable oxidation states. In some metal crystals, positive metal ions are missing from their regular lattice sites. The extra negative charge is balanced by cations of the same metal with higher oxidation state than that of missing cation at site.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 44
Consider a crystal of NiO. When one Ni2+ is missed from its lattice point, it creates a vacant site.

The deficiency of two positive charges is compensated by two Ni3+ ions at other lattice points of Ni2+ ions and the composition of NiO crystal becomes Ni0.97O1.0.

(2) Metal excess defect : There are two types of metal excess defect as follows :
(a) Presence of a neutral atom or an extra positive ion at interstitial position :
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 45
There are two types or ways of metal excess defects in ZnO. In the first case, Zn atom is present in the interstitial space as shown in figure.

(b) Metal ions and electrons at interstitial sites :
The second case arises when ZnO is heated, Zn2+ and electrons are obtained,
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 46
The excess Zn2+ ions are trapped in interstitial sites in the crystal lattice. Electrons occupy interstitial sites by diffusing into the interstitial sites.
In above both cases, the nonstoichiometric formula of ZnO is Zn1 + x O1.0

Question 70.
Explain defects due to anion vacancies.
OR
Explain colour of crystals or F centres.
Answer:

  • The defect due to anion vacancies imparts colour to the colourless crystal.
  • When a colourless crystal of NaCl is heated in the atmosphere of sodium vapour, the sodium atoms are deposited on the crystal surface.
  • Due to diffusion of Cl ions to the crystal surface vacancies are created at their regular sites.
  • These diffused Cl ions combine with Na atoms on the surface forming NaCl along with releasing electrons from sodium atoms. Na + Cl → NaCl + e

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 47
The released electrons diffuse into the crystal and occupy vacant sites of anions Cl in the crystal.
The anion vacant sites occupied by electrons are called F-centres or colour centres. Due to colour centres NaCl shows yellow colour.
Now NaCl crystal has excess of Na atoms having nonstoichiometric formula Na1+x Cl1.0.

Question 71.
How are solids classified according to electrical conductivity?
Answer:
According to electrical conductivity, solids are classified as follows :
(1) Conductors :

  • The solids having electrical conductivity in the range of 104 to 107 Ohm-1m-1 are called conductors.
  • The examples of conductors are metals and electrolytes.
  • Electrical conductivity in metals is due to free movement of electrons while electrolytes conduct electricity due to migration ions.

(2) Insulators :

  • Solids having very low electrical conductivity in the range of 10-20 to 10-10 Ohm-1 m-1 are called insulators.
  • The examples of insulators are nonmetals and molecular solids.

(3) Semiconductors :

  • Solids having conductivity in the range of 10-6 to 104 Ohm-1m-1 are semiconductors.
  • The conductivity range is intermediate between conductors and insulators.
  • The examples of semiconductors are silicon and germanium.

Question 72.
Explain band theory.
OR
Explain the origin of electrical properties in solids.
Answer:
(1) Metals are good conductors of heat and electricity. This is explained on the basis of band theory which involves the presence of free electrons.

(2) According to band theory, the atomic orbitals of metal atoms overlap to form molecular orbitals which are spread all over the crystal structure.

(3) The energy difference between adjacent molecular orbitals decreases as the number of molecular orbitals increases and when it becomes very less, the orbitals merge into one another forming continuous bands which extent over the entire crystal.

(4) There are two types of bands of molecular orbitals as follows :

  • Valence band : The atomic orbitals with filled electrons from the inner shells form valence bands, where there are no free mobile electrons since they are involved in bonding.
  • Conduction band : Atomic orbitals which are partially filled or empty on overlapping form closely placed molecular orbitals giving conduction bands where electrons are delocalised and can conduct, heat and electricity.

(5) Band gap :

  • The energy difference between valence band and conduction band is called band gap.
  • Band gap decides whether electrons from valence band can be promoted to vacant conduction band or not.
  • The conductors like metals have very small or no band gap and electron can be promoted by thermal energy. The nonconductors have large band gap. The insulators have very large band gap.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 48

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

Question 73.
Metals are good conductors of electricity. Explain.
Answer:
(1) Metals are good conductors of electricity since the outermost electrons of atoms in metallic crystal occupy conduction bands.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 49
(2) The number of electrons in conduction bands is very large.
(3) The conduction bands in metals can be labelled as ‘s’ band, overlapping s and p bands etc. according to overlapping of orbitals.
(4) This results in delocalisation of outermost electrons forming metal ions. Hence, this is analogous to metal cations immersed in the sea of electrons.

Question 74.
Why does metallic conductivity decrease by increasing temperature?
Answer:

  1. In metals a large number of outermost electrons of atoms occupy conduction bands.
  2. Band formation in metals results in delocalisation of outermost electrons forming metal ions or cations.
  3. The metallic cations occupying crystal lattice sites vibrate about mean positions.
  4. As temperature increases the vibrational motion increases which interrupts flow of electrons decreasing electrical conductivity.

Question 75.
Explain insulators.
Answer:
(1) In insulators the valence band is completely filled by electronics while conduction band is empty.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 50
(2) The energy gap between valence band and conduction band in insulator is very large.
(3) Thermal energy is not sufficient to promote electrons from valence band to conduction band.
(4) Therefore, the conduction band in insulator remains vacant and does not allow the conduction of electricity.

Question 76.
What are semiconductors ? Mention the types of semiconductors.
Answer:
(1) Semiconductors : The substances like silicon, germanium which have poor electrical conductance at low temperature but the conductance increases with the increase in temperature are called semiconductors.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 51
(2) Their conductivity lies between metallic conductors and insulators.
(3) The energy difference between valence band and conduction band is relatively small, hence the electrons from valence band can be excited to conduction band by heating.

(4) Types of semiconductors : There are two types of semiconductors :
(a) Intrinsic semiconductor
(b) Extrinsic semiconductor

(a) Intrinsic semiconductor :

  • A pure semiconductor material like pure Si, Ge which have a very low but finite electrical conductivity is called intrinsic semiconductor.
  • The electrical conductivity of a semiconductor increases with the increase in temperature.

(b) Extrinsic semiconductor :

  • Semiconductor doped with different element is called extrinsic semiconductor.
  • By doping with elements like Ga or P, the electrical conductivity is increased.

Question 77.
Explain extrinsic semiconductor and doping.
Answer:
(1) A semiconductor obtained by doping intrinsic semiconductor with elements of third group and fifth group is called extrinsic semiconductor.
(2) This extrinsic semiconductor has higher electrical conductivity than pure intrinsic semiconductor.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 52

(3) There ate two types of extrinsic semiconductors:
(A) n-type semiconductors:
(i) n-type semiconductor contains increased number of electrons in the conduction band.
(ii) When Si semiconductor is doped with 15th group element phosphorus, P, the new atoms occupy some vacant sites in the lattice in place of Si atoms.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 53
(iii) P has five valence electrons, out of which four are involved in covalent bonding with neigh-bouring Si atoms while one electrons remains free and delocalised.
(iv) These free electrons increase the electrical conductivity of the semiconductor.
(v) The semiconductors with extra non-bonding free electrons are called n-type semiconductors.

(B) p-type semiconductor :
(i) p-type semiconductor is obtained by doping a pure semiconductor by an element of 13th group like B.
(ii) 13th group element has less number of valence electrons. When pure Si is doped with B atoms, these atoms occupy Si lattice points.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 54
(iii) Boron (5B) has only 3 valence electrons which form covalent bonds with the neighbouring Si atoms, while one bond has shortage of one electron.
(iv) This creates a vacancy or a hole, hence the electron from neighbouring Si atom jumps into this hole creating a vacancy in itself. This process continues, i.e., positive holes move in one direction while electrons moves in opposite direction.
(v) Due to electron deficient positions, this semiconductor is called p-type semiconductor.
(vi) When p-type semiconductor is connected to the external source of electricity, electrons from neighbouring silicon atoms jump into the holes so that electrons move towards positive electrode and holes migrate towards negative electrode.
(vii) Hence electrical conduction in p-type semiconductor is due to electrons and holes.

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

Question 78.
What are the uses of semiconductors ?
Answer:
The uses of semiconductors are as follows :

  1. They are used in transistors, digital computers and cameras.
  2. They are used in solar cells and television sets.
  3. By combining n-type and p-type semiconductors, n-p junctions are formed which are effectively used in rectifiers or to convert light energy into electrical energy.

Question 79.
Classify the following semiconductors into n or p-type.
(i) B doped with Si
(ii) As doped with Si
(iii) P doped with Si
(vi) Ge doped with In.
Answer:

Semiconductor Type
(i) B doped with Si p-type
(ii) As doped with Si n-type
(iii) P doped with Si n-type
(iv) Ge doped with In p-type

Question 80.
Explain the origin of magnetic properties in solids.
Answer:
(1) The magnetic properties of a substance arise due to the presence of electrons in their atoms or molecules.

(2) The electrons while revolving around the nucleus in various orbits, also spin around their own axes. Both these motions of electrons result in generating magnetic field and magnetic moments. Hence electron be haves as a tiny magnet.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 55

(3) An atomic orbital can accommodate maximum two electrons with opposite spins, clockwise and anticlockwise. The degenerate orbitals like p, d and f orbitals can accommodate electrons with same spins until they are half filled.

(4) When a substance contains one or more unpaired electrons spinning in same direction, then their magnetic moments and magnetic properties add and the substance is said to be paramagnetic.
Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 56
If a substance contains all electrons paired then their spins are balanced and magnetic moments and magnetic properties are cancelled and the substance is said to be diamagnetic.

Question 81.
Explain diamagnetism.
Answer:
(1) The magnetic properties of a substance arise due to presence of the electrons.

(2) An electron while revolving around the nucleus, also spins around its own axis and generates a magnetic moment and a magnetic property.

(3) If an atom or a molecule of the substance contains all electrons paired, spinning clockwise and anticlockwise, their magnetic moments and magnetic properties get cancelled. Hence they oppose and repel the applied magnetic field. This phenomenon is called diamagnetism and the substance is said to be diamagnetic.Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State 57
For example : Zn, Cd, H2O, NaCl, etc.

Multiple Choice Questions

Question 82.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. A substance which on cutting will give irregular cleavage is
(a) glass
(b) KBr
(c) ZnS
(d) NaCl
Answer:
(a) glass

2. A solid which has definite heat of fusion is
(a) plastic
(b) CaCl2
(c) glass
(d) soda lime glass
Answer:
(b) CaCl2

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

3. In solids the constituent particles may be
(a) atoms
(b) ions
(c) molecules
(d) any one of the above three
Answer:
(d) any one of the above three

4. A single substance that exists in two or more forms is called
(a) polymorphous
(b) amorphous
(c) isomorphous
(d) monomorphous
Answer:
(a) polymorphous

5. Graphite is a
(a) metallic crystal
(b) covalent crystal
(c) ionic crystal
(d) molecular crystal
Answer:
(b) covalent crystal

6. Anisotropy is observed in
(a) Pyrex glass
(b) plastic
(c) K2SO4
(d) fullerene
Answer:
(c) K2SO4

7. The number of crystal systems (or types) is
(a) 4
(b) 7
(c) 8
(d) 12
Answer:
(b) 7

8. The number of Bravais lattices are
(a) 6
(b) 8
(c) 12
(d) 14
Answer:
(d) 14

9. Face centred cubic crystal is
(a) Cubic lattice of Bravais system
(b) Bravais lattice of HCP
(c) Bravais lattice of cubic system
(d) Cubic lattice with 5 atoms
Answer:
(c) Bravais lattice of cubic system

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

10. The number of tetrahedral sites per sphere in ccp structure is,
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

11. The packing fraction for a body centred cubic structure is
(a) 0.42
(b) 0.53
(c) 0.68
(d) 0.82
Answer:
(c) 0.68

12. If r is the radius of an atom and a is an edge length of fcc unit cell, then
(a) r = \(\frac{\sqrt{3}}{4} a\)
(b) r = \(\frac{a}{2 \sqrt{2}}\)
(c) r = \(\frac{a}{2}\)
(d) r = \(2 \sqrt{2} a\)
Answer:
(b) r = \(\frac{a}{2 \sqrt{2}}\)

13. The ratio of packing efficiency in see, bcc and fee crystalline structures is
(a) 1 : 1.2 : 1.3
(b) 1 : 1.12 : 1.23
(c) 1 : 1.3 : 1.4
(d) 1 : 1.25 : 1.38
Answer:
(c) 1 : 1.3 : 1.4

14. The correct sequence of the atomic layers in cubic close packing is
(a) ABABA
(b) ABACABAC
(c) ABCABC
(d) AABBAABB
Answer:
(c) ABCABC

15. The major binding force in diamond is
(a) Covalent bond
(b) Ionic bond
(c) Metallic bond
(d) Coordinate covalent bond
Answer:
(a) Covalent bond

16. The ratio of close packed atoms to octahedral holes in cubic packing is
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 3
Answer:
(a) 1 : 1

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

17. A defect present in AgCl is
(a) Frenkel defect
(b) Schottky defect
(c) point defect
(d) interstitial impurity defect
Answer:
(a) Frenkel defect

18. An ionic solid crystallises in bcc structure. If the ionic radii of cation and anion are 0.84 Å and 1.07 Å, the length of the body diagonal is
(a) 1.91 Å
(b) 2.75 Å
(c) 3.82 Å
(d) 2.32 Å
Answer:
(c) 3.82 Å

19. The type of defect in NaCl crystal will be
(a) point defect
(b) interstitial defect
(c) vacancy defect
(d) impurity defect
Answer:
(c) vacancy defect

20. Schottky defects are observed in which solid among the following ?
(a) Brass
(b) Cesium chloride
(c) Zinc sulphide
(d) Stainless steel
Answer:
(b) Cesium chloride

21. An ionic compound crystallises in FCC type structure with ‘A’ ions at the centre of each face and ‘B’ ions occupying comers of the cube. The formula of compound is
(a) AB4
(b) A3B
(c) AB
(d) AB3
Answer:
(b) A3B

22. Total number of different primitive unit cells are
(a) 6
(b) 7
(c) 12
(d) 14.
Answer:
(d) 14

23. The volume of atoms present in body centred cubic unit cell of a metal of atomic radius r is,
(a) \(\frac{16}{3} \pi r^{3}\)
(b) \(\frac{8}{3} \pi r^{3}\)
(c) \(\frac{12}{3} \pi r^{3}\)
(d) \(\frac{24}{3} \pi r^{3}\)
Answer:
(b) \(\frac{8}{3} \pi r^{3}\)

24. The substances which can be permanently magnetised are
(a) diamagnetic
(b) paramagnetic
(c) ferromagnetic
(d) non-magnetic
Answer:
(c) ferromagnetic

25. CrO2 is
(a) diamagnetic
(b) paramagnetic
(c) metalic
(d) ferromagnetic
Answer:
(d) ferromagnetic

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

26. A metallic element crystallises in face centred cubic structure. If the radius of metal ion is 0.92 Å, the edge length of the unit cell of the crystal is
(a) 0.8464 Å
(b) 1.252 Å
(c) 5.187 Å
(d) 2.6 Å
Answer:
(d) 2.6 Å

27. The volume of unit cell of a metallic crystal of bcc type is 8.4 × 10-23 cm3. The volume occupied by 10 atoms in the crystalline structure is
(a) 4.2 × 10-22 cm3
(b) 3.12 × 10-23 cm3
(c) 1.74 × 10-23 cm3
(d) 2.856 × 10-22 cm3
Answer:
(d) 2.856 × 10-22 cm3

28. Copper crystallises in face centred cubic structure. If the unit cell length is 360 pm, the radius of copper atom is
(a) 180 pm
(b) 156 pm
(c) 127 pm
(d) 110 pm
Answer:
(c) 127 pm

29. If all the lattice points in ccp structure namely comers, face and edge centres and body centre are occupied by atoms then the total number of atoms in the unit cell will be
(a) 8
(b) 12
(c) 14
(d) 16
Answer:
(a) 8

30. Gold crystallises in face centred cubic structure. If atomic mass of gold is 197 g mol-1, the mass of the unit cell of gold will be
(a) 3.25 × 10-23 kg
(b) 6.5 × 10-23 kg
(c) 3.9 × 10-24 kg
(d) 1.3 × 10-24 kg
Answer:
(d) 1.3 × 10-24 kg

31. The mass of a unit cell of a body centred cubic crystal of a metal is 72.2 × 10-23 g. The atomic mass of the metal is
(a) 128.6 gmol-1
(b) 108.7 gmol-1
(c) 217.3 gmol-1
(d) 57.86 gmol-1
Answer:
(c) 217.3 gmol-1

32. An element crystallises in fee structure. If the atomic mass of the element is 72.7 U, the mass of one unit cell of it will be
(a) 2.9 × 10-24 g
(b) 4.83 × 10-25 kg
(c) 1.2 × 10-22 g
(d) 2.41 × 10-24 kg
Answer:
(b) 4.83 × 10-25 kg

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

33. Edge length of a cubic unit cell is 354 pm. The distance between two atoms diagonally opposite on the face is
(a) 500 pm
(b) 354 pm
(c) 708 pm
(d) 627 pm
Answer:
(a) 500 pm

34. The unit cell has an edge length 403 pm. The distance between two atoms placed opposite ends of body diagonal will be
(a) 806 pm
(b) 201.5 pm
(c) 698 pm
(d) 578 pm
Answer:
(c) 698 pm

35. An element having atomic mass 115 u, crystallises in bcc structure. The number of unit cells in 1 g of the element will be
(a) 2.6 × 1021
(b) 3.8 × 1023
(c) 8.7 × 10-3
(d) 6.17 × 1020
Answer:
(a) 2.6 × 1021

36. The edge length of a bcc unit cell of a metallic crystal is 2.9 Å. Hence the diameter of an atom is
(a) 1.025 Å
(b) 2.512 Å
(c) 1.45 Å
(d) 1.31 Å
Answer:
(b) 2.512 Å

37. An element crystallises in fee structure. If the atomic radius is 130 pm, the edge length of unit cell is
(a) 332.5 pm
(b) 410 pm
(c) 390 pm
(d) 367.6 pm
Answer:
(d) 367.6 pm

38. The arrangement of layers in hexagonal close packing is
(a) ABCABC
(b) ABAB
(c) ABBABBA
(d) ABBCABBC
Answer:
(b) ABAB

39. For square close packing, the planar arrangement is
(a) AAAA
(b) ABAB
(c) ABCABC
(d) AABBAA
Answer:
(a) AAAA

Maharashtra Board Class 12 Chemistry Important Questions Chapter 1 Solid State

40. Semiconductors are manufactured by addition of impurities of
(a) p-block elements
(b) actinoids
(c) Lanthanoids
(d) s-block elements
Answer:
(a) p-block elements

41. p-type semiconductor is formed when trace amount of impurity is added to silicon. The number of valence electrons in the impurity atom must be
(a) 3
(b) 5
(c) 1
(d) 2
Answer:
(a) 3

42. n-type semiconductor is formed when trace amount of impurity is added to silicon. The number of valence electrons in the impurity atom must be
(a) 3
(b) 5
(c) 1
(d) 2
Answer:
(b) 5