Class 12

MP Board Class 12th Business Studies Important Questions Chapter 2 Principles of Management

Principles of Management Important Questions

Principles of Management Very Short Answer Type Questions

Questions 1.
What do you understand by “Mental revolution” ?
Answer:
Principle of mental revolution assumes that even if there are large physical resources and other resources in an organization but if the human resources are not satisfied, there isn’t peace and prosperity and the moral of the human resources is not high till then the resources are simply resources. They can never take the form of production and services. Hence, managers must try to attain the maximum cooperation of the workers, to remove the gaps between labour and capital and to radicate mutual enmity.

Question 2.
What do you understand by principles of management ?
Answer:
Principles of management represent the techniques and principles which are related to management tasks and to show the correct path for achievement of the main objectives of the organization.

Question 3.
Explain Fayols principle of Initiatives ?
Answer:
According to this principle Fayol has suggested that the officers and employees of an organization should be given freedom to work on the new plans and to upgrade the old plans in order to increase the productivity. The employees should take the initiative in order to give suggestions to increase the productivity of the firm.

Question 4.
What is principle of equity ?
Answer:
This principle was formulated by Henri Fayol. Equity means fair dealings, equality of treatment and cooperative attitude among the employees in the enterprise. Equity results from a combination of kindness and justice. Employees expect equity from management.

Question 5.
How is efficiency developed through division of labour ?
Answer:
According to Henri Fayol all employees is an organization or institution do not have same efficiency. So according to their efficiency and interest they should be assigned the work. Labour should be divided into various groups. By doing this workers are motivated and this helps in increasing their efficiency.
The object of division of work is to derive benefits from specialization. The various functions of management like planning, organization cannot be done alone.

Question 6.
Distinguish is between principles of unity of command and principle of unity of direction.
Answer:
Difference between Principles of unity of command and Principle of unity of direction

principal of unity command:

1. it is concerned with the functioning of personnel.
2. 2.this principal makes the relation of and officers (authority) strong.

princpals of unity of direction:

1. it is concerned with the functioning of the department and sub-department comprising the business concern.
2. the principals help the unification of plans.

Question 7.
What do you mean by decentralization ?
Answer:
When the top level management gives more role and importance to its subordinates in management and organization, it is known as decentralization. The intensity of centralization of decentralization is decided by the top level management.

Question 8.
What do you mean by standardization ?
Answer:
According to Taylor standardization is essential for high quality product with low cost of production. It involves the standardization of tools and equipment conditions of work and materials so that the best product can be produced at a minimum cost.

Question 9.
Explain the principle of unity of direction formulated by Henri Fayol.
Answer:
Principle of unity of direction means there should be one head and one plan for a group of activities having the same objectives. While unity of command is concerned with the functioning of personnel unity of direction is concerned with the functioning of the body corporate. The departments and subdepartments comprising the business concern.

Question 10.
Explain Principle of scientific selection, training and development.
Answer:
According to Taylor, selection of employees should be done very carefully, because wrong selection may create disaster. This does not mean that only well qualified persons should be employed, but that person should be employed who can do that work or is capable of -handling that work. Taylor in his book, “The principle of scientific management” has written, that no matter how good your selection procedure is, but without training, the working capacity of the employees cannot be increased. Therefore, along with scientific selection method, arrangement for training should also be made. “A person should do the work not mere know the work.”

Question 11.
What is Taylor’s Eight Boss scheme ?
Answer:
Taylor has suggested functional organizational system. This organization is called Eight Boss scheme.
They are:

1. Gang Boss
2. Speed Boss
3. Repair Boss
4. Inspector
5. Shop disciplinarian
6. Route clerk
7. Instruction card clerk
8. Time and cost clerk.

Question 12.
ExpIain the principle of ‘I’ and ‘We’ propounded by Fayol ?
Answer:
According to this principle of Henry Fayol the industrialists should use the word we instead of ‘I’ in order to get maximum cooperation of the employees. Both the aspect should work in cave and harmony.

Question. 13.
Explain Taylor’s principle of task idea.
Answer:
Principle of task idea: This pertains to the norms of productivity of individual worker. It is a technique of forecasting and viewing ahead every step in a long series of separate operations. The standard task is the amount of work which an average worker does in congenial atmosphere during a working day. Taylor called it a fair day’s work. A worker usually works much below his capacity if no standard is set for him.

Question 14.
What do you mean by motion study ?
Answer:
It is the study of a machine operator and his movements for performing the job. The purpose is to eliminate useless motion and thereby establish a standard for performance. The study is undertook in order.

Question 15.
Explain the principle of unity of direction.
Answer:
Principle of unity of direction : It means, there should be one head and one plan for a group of activities having the same objectives while unity of command is concerned with the functioning of personnel, unity of direction is concerned with the functioning of the body corporate the departments and subdepartments comprising the business concern.

Question 16.
Who formulated scientific principles of management when and where did he die ?
Answer:
Principles of scientific management was formulated by F.W. Taylor. He died in 1915 at Filadalphiya.

Question 17.
How many parts did Henri Fayol divide the works of business ?
Answer:
Fayol has classified all the works of business in six parts :

1. Technical work
2. Commercial work
3. Financial work
4. Security work
5. Administrative work
6. Managerial work.

Question 18.
What do you mean by time study ?
Answer:
lt is done to observe and study the minimum time-period required to perform any task. Through time study the precise time required for man’s work is determined. It will help in fixing the standard time required to do a particular job.

Question 19.
what do you mean by Fatigue study ?
Answer:
Fatigue study : Excessive specialization of work and poor working conditions may cause monotony resulting in both physical and mental fatigue among industrial workers. Fatigue either a physical or mental always has an adverse effect on worker’s health and his efficiency. The idea of this study is to see that workers maintain their operational efficiency without any injury to their health and happiness.

Principles of Management Short Answer Type Questions

Question 1.
Name the 14 principles of management formulated by Henri Fayol.
Answer:
AnirFayol’s Principles of Administration of Management:

1. Principle of division of work
2. Principle of authority and responsibility
3. Principle of discipline
4. Principle of unity of command
5. Principle of unity of direction
6. Principle of priority to common interest
7. Principle of remuneration of personnel
8. Principle of centralization
9. Principle of scalar chain
10. Principle of order
11. Principle of equity
12. Principle of stability
13. Principle of initiative and Esprit de corps
14. Principle of coordination of management.

Question 2.
Differentiate the time study and motion study.
Answer:
Differences between time study and motion study :

Time study:

1. Time study is done at first.
2. In time study, study of time is done of time taken by machine and time taken by labours.
3. Stopwatch is used in time study.
4. Time study is related to increase the efficiency of labourers.

Motion study:

1. Motion study is done after time study In motion study the study of work done by labours is done.
2. Camera is used in motion study.
3. The object of motion study is to minimize the movement of labours.

Question 3.
why the principles of scientific management opposed by labourers ? Give any six reasons.
Answer:
Opposition by labourers : The maximum opposition of scientific management has been done by the labourers. The reasons for opposition of scientific management are following: .

1. Increase of labourers work: To adopt the scientific management, it is necessary to increase the efficiency of a labour and so the work load on labour is increased. The increased work load creates bad effect on the health of labourer. Hence, it is opposed by the labourers.

2. Strict control: Under the scientific management, strictness has to be observed and so the labourers are kept under the strict control which they do not like.

3. Fear of unemployment: Increased use of machines create a sense of fear in the minds of the workers, thereby making them inefficient. This may further lead to retrench-ment. Hence, they are against being the standards set;

4. Exploitation of labourers : In scientific management, wages of the labourers are not increased in proportion to the increased production. So, in this system, exploitation of the labourers are maximum.

Question 4.
describe the similarities of opinions of Taylor and Fayol.
Answer:
Followings are the similarities of opinions Taylor and Fayol:

1. Both wanted to bring change in managerial conditions
2. Both were great supporters of human factor
3. Both gave much importance to forecasting and planning
4. Both were management experts of the same time, i.e., both were contemporaries
5. Both accepted management as an earned talent
6. Both were experts in their own fields
7. Both emphasised working in a technical and scientific way
8. Both formulated principles of management.

Question 5.
To increase the production efficiency in industrial organization many principles were developed by Taylor. Explain them.
Answer:
To increase the production efficiency in industrial organization following principle are adopted by Taylor :

Principle of modern machines and equipment: Modem machines and equipments yield good quality of goods. Cost incurred is also reduced, therefore under production management, best of modem machines and equipments should be used.

Principle of standardization : Taylor says that, “For production, standard material, standard method, proper procedure and good workers qualifying the standard requirements should be made available, such that good quality of goods can be produced at lesser cost.”

Principle of ideal cost accounting system : For controlling the wastage in expenditure and misuse of materials suitable system of cost accounting should be adopted. Capable and experienced cost accounts should so appointed in the organization.

Question 6.
What are the dissimilarities in study of Taylor and Fayol ?
Answer:
Dissimilarity of Opinions : Taylor and Fayol

1. Taylor’s centre of study was labour whereas Fayol’s centre of study was manager.
2. Taylor’s principles were Principles of Scientific Management whereas Fayol’s Principles were known as Principles of Administration.
3. Taylor’s experiment point was factory whereas Fayol’s experiment point was top administration.
4. Taylor justified functional organization (Eight Boss Scheme) for an enterprise whereas Fayol justified unity of command (One-man order).
5. Taylor’s study and contribution was lower level, labour to manager whereas Fayol’s_ study and contribution was manager from the beginning and he used the words order, direction, authority, etc.
6. Taylor is called as Factory Expert whereas Fayol is known as Management Expert.
7. Taylor’s principles are not applied in practice whereas Fayol’s principles have been applied in so many enterprises.

Question 7.
Mention the objectives of scientific management.
Answer:
Following are the main objectives of scientific management: .

1. Increase in the rate of production by use of standardized tools, equipments and methods
2. Increase in quality of goods by research, quality control and inspection devices
3. Reduction in the cost of production by rational planning and regulation and using cost control techniques.

Question 8.
“Principles of management are concentrated on human behaviour”. Explain.
Answer:
Management is totally related with human behaviour. The success of principles of management is affected by human behaviour. Management is related with persons. Management gives proper directions to human behaviour. Management is also done by human being and the consumer for whom the organization, is prepared is also human being. So it is said that principles of “Management are concentrated on human behaviour”.

Question 9.
The work of Taylor and Fayol are complementary.’ Explain
Answer:
F. W. Taylor is regarded as “father of scientific management and Henry Payol is regarded as “father of modern principles of industrial management”. Both were contemporaries and both were top class engineers. Both emphasized working in a technical and scientific way. Both have contributed a lot in the field of management. These existed a difference in thinking and working method of both. Economics for instance, Taylor focussed on improving efficiency of labourers while Fayol focussed on management work. The both realised that in every stage. Employees and their management will be the basis of success of an industrial Concern. To solve such problem both suggested scientific solutions and their implementation.

Despit of dissimilarities between the principles propounded by both learned economists. We can say that their work is complementary to each other.

Question 10.
Write the importance of principles of management is three points.
Or
Explain the importance of principle of management.
Answer:
Significance of Importance of Principles of Management: The principles of management are very useful as they provide useful guidelines to managerial behaviour and influence managerial practices. Actually, principles guide Them in taking and implementing decisions. Their importance can be understood by the discussion of the following points :

1. They provide useful In-sights : The principles provide the managers with useful insights into real world situations. Following these principles they increase their knowledge, ability and understanding and can face adverse situations effectively. Thus, the principles increase managerial efficiency.

2. They help in utilization to Resources of the optimum : As we know resources both human and material are limited. They have to be put to optimum use. They should be uses in such a way that they give maximum benefit with minimum cost. Principles equip the managers foresee the.cause and effect relationships of their decisions and actions. Thus, the wastage associated with a trial-and-error approach can be overcome.

3. Highlight the True Nature of Management: Lack of understanding of management principles makes it difficult to define clearly the nature and scope of the managerial job. Principles are necessary if the true nature of management is to be clearly understood. Principles act as a checklist of the meaning and contents of management. They highlight the role of managers in concrete terms and serve as a criteria to judge the appropriateness of managerial decisions and actions.

4. Based on Objective Assessment: Management principles help in thoughtful decision making. They emphasis logic rather than blind faith. Management decisions taken on the basis of principles are free from bias and prejudice. They are based on the objective assessment of the situation

5. To Guide Research : Principles of management indicate the lines along, which research should be carried out to develop new guidelines and to ascertain the validity of existing guidelines. Thus, management principles serve as focal points for useful research in management.

6. To attain Social Goals: Manager coordinate the efforts of people so that individual objectives get translated into social attainments. The standard of living of people depends upon the quality of management. If management is efficient, the resources of society are better utilized and quality of life is improved. Development of management principle by increasing, efficiency, in the use of resources, would definitely have a revolutionary impact on the cultural level of society and on human progress.

Question 11.
Distinguish between managerial principles and managerial techniques.
Answer:
Differences between Managerial principles and Managerial techniques

Managerial principles:

1. Managerial principle is flexible.
2. Managerial principles are directions for managerial work.

Managerial techniques:

1. Managerial techniques are less flexible
2. These are managerial techniques which are made fulfil the targets.

Question 12.
How are principles of management arises ?
Answer:
Principles of management arises due to following reasons :

1. Deep observation: In every business in front of management some problems arises. During this time managers observe working of employees. They see their reactions. On the basis of these observations principles are made.

2. Experiments done again and again : In this method experiments are done on various employees frequently. On the basis of results the principles are made.

Question 13.
Explain the principle of “Esprit De Corps”
Answer:
By ‘Esprit De Corps’ we mean “Power of Organization”. According to this principle each employee should consider himself/herself as the member of teach. He/she should try to achieve the target of team. Management should try to create feeling of cooperation among the workers. By following this target group target is achieved.

Question 14.
While fixing the proper wages which points are to be kept in mind ?
Answer:
While fixing the proper wages these points are kept in mind :

1. Financial situation of institution or organization.
2. Minimum wages act of government.
3. Payment of bonus and wages done by competent.

Question 15.
How principles of management make administration more effective ? Explain with example.
Answer:
Principles of management discourages partiality and prejudice. It makes administration more effective. These principles encourages the scientific principles.

For example “Principle of unity of command” and “Principle of unity of direction” lead the orderly working of the organization. Principle of direction do assimilation of effort made by workers in a specific direction. Principle of scalar chain do the flow of information in a proper way.

Question 16.
Recognize those techniques of scientific management which are analysed by the following statements :

1. When any expert supervises each labour.
2. To know the best solution of doing any work.
3. Different wages paid to labours.
4. When equality is brought in material, machines tools and working methods and conditions after proper experimentation.
5. To fix the standard time for completing the fixed work.
6. Employee is for the enterprise and enterprise is for the employee.

Answer:

1. Creative foremanship
2. Method study
3. Differential wages
4. Stan-dardization of work
5. Time study
6. Mental revolution.

Question 17.
Distinguish between principle of unity of command and functional foremanship.
Answer:
The differences between principle of unity of command and functional foremanship:
Principle of unity of command

1. This principle requires that the employees should receive orders from one superior only for any action or activity.
2. In it planning and performing work is same.
3. It means workers in a department are required to be accountable to one superior for orders.

Functional foremanship

1. Under it each worker has to takeorder for performance of any work from foreman.
2. Under it planning and performance of work is different, (excutation of function)
3. This concept was extended to the lower level.

Question 18.
Distinguish between principles of management and principles of pure science
Answer:
Principles of management:

1. Principles of management are of more flexible nature.
2. These changes in business environment according to change.
3. These principles are implemented with creativity.

Principles of puree science:

1. These principles are more strong.
2. These principles do not change according to time.
3. These are implemented with fixed or proposed way.

Question 19.
When a salesman gets order from two high authority then what bad effects may take place ?
Answer:
When a salesman gets order from two different authorities then following bad effects may be found:

1. Suspicion may take place in mind of salesman.
2. He tries to avoid the work.
3. There may be difference of opinion regarding the orders by two persons.
4. It may be difficult to maintain discipline.

Question 20.
It is seen that in organization present position is due to the non-violation of principles of management. What is this present position ?
Answer:
Due to violation of principles of management following positions may be in the organization:

1. Necessary material may not be made available at die time of requirement of it. In search of material much time will be wasted.
2. Conditions may increases.
3. Possibility of accidents may increase.

Question 21.
Write the names of those officers who work under foremanship the following works:

1. To examine the quality of produced goods. To check the standard of goods and if the goods are of not standardize, then finding the reason of it.
2. To cheek if all the labours are working their own work according to fixed rate.
3. To ensure what will be the order of any work in completing the specific work.
4. To keep the machines and tools in working conditions.
5. To give direction to workers to do work by a specific method.
6. To ensure that each work is done in proper way.

Answer:

1. Inspector
2. Speed Boss
3. Rote clerk
4. Repairs Boss
5. Card clerk
6. Displinarian.

Principles of Management Long Answer Type Questions

Question 1.
Explain some merits of scientific management.
Answer:
some merits of scientific management are as follows :

1. Division of work : It stress on specialization and division of work. It helps the producer in completing of work effectively.
2. Industrial peace: The conflicts among labours are solved scientifically under scientific management. There is no place for conflicts which helps in main taining industrial peace.
3. Maximum work: It helps in getting the maximum work done from laboures. This helps in increase in profit as well.
4. Reduction in cost of production : If reduce the cost of production, it can achieve .success. ‘
5. Increase in Income: It increases the efficiency of worker which increase the income of labours.
6. Stability in employment: Efficient and able worker provide their service regularly it will increasestability in employment.

Question .2
What do you understand by scientific management ? Write its characteristics.
Answer:
Meaning of Scientific Management: Scientific management is made of two words science and management. Science is the application of logic, reasoning and it gives a new dimension to our study approach to different problems of life. When we apply scientific approach towards management concepts this is called as scientific management.
Following are the characteristics of scientific management:

1. The objectives of every job in scientific management is fixed and predetermined,
2. There is a fixed plan to attain objectives
3. Mutual cooperation among employees is essential in scientific management
4. Scientific management works on the objective of maximum production at minimum cast
5. Motivation is considered necessary for the completion of work
6. Scientific control system is implemented under scientific management
7. Time to time evaluation of standards is performed
8. Increase in efficiency of employees helps in increase in the income of employees
9. Researches and experiments plays an important role in the field of scientific management
10. Specialisation and division of work is also given importance
11. Much stress is given on right person for the right job
12. Responsibilities are assigned or fixed for every level of work.

Question 3.
Discuss the elements of scientific management. (Imp)
Answer:
In order to implement the principles of scientific management, Taylor and his associates developed the following techniques :

1. Time Study : In refers to the technique used to measure the time that may be taken by a worker of reasonable skill and efficiency to perform various elements of a job. The aim of time study is to fix ‘Time standards’ for each operation.

2. Motion Study : It refers to have close observation of the movements of a worker’s body involved in performing a job. The purpose of motion study is to avoid wasteful motions in working.

3. Fatigue Study ; Fatigue study seeks to find out how long a person can perform the standard task without any adverse effect on this health and efficiency. It helps to determine the mount and frequency of rest intervals in completing a job.

4. Method Study : The aim of this study is to maximize efficiency in the use of materials, machinery, manpower, and capital by improving work methods.

5. Scientific Task Setting : Scientific planning of a task is the technique of forecast¬ing and viewing ahead every step in a long series of separate operations. Each step has to be taken in the right place, of the right degree, and at the right time, so that work can be done with maximum possible efficiency. Scientific task is the amount of work which an average worker working under proper working conditions can perform during a working day.

6. Standardization : Standardization is the process of fixing well thought-out and tasted norms with a view to maximise efficiency of work standardization eliminates needles variety and thereby simplifies the process of production.

7. Scientific Selection and Training : Under scientific management right men are selected for right jobs. Employees are selected according to predetermined standard in an impartial way. Workers are specifically trained for the jobs, so that they can perform their jobs effectively and efficiently.

8. Mental Revolution : It requires that there should be perfect cooperation,coordination and adjustment between the efforts of labour and management.

Question 4.
describe the principles formulated by Taylor.
Answer:
‘The principles formulated by Taylor are in fact the principles of scientific management. Sir Taylor has formulated the principles of management after studying the scientific management minutely and deeply. Taylor has presented the principles of management in L scientific way. So, it is called as principles of scientific management. The following principles have been formulated by Taylor.

1. Principle of task idea : This pertains to the norms of productivity of individual worker. It is a technique of forecasting and viewing ahead every step in a long series of separate operations. The standard task is the amount of work which an average worker does in congenial atmosphere during a working day. Taylor called it a fair day’s work. A worker usually works much below his capacity if no standard is set for him.

2. Principles of experiment: Taylor suggested that the following studies be conducted by an organization so that it has an insight in knowing the method of doing a job, movement of a machine, time required for its completion and the fatigue that is associated with performing the job.

(i) Time study : It is done to observe and study the minimum time required to perform any task. Through time study, the precise time required for man’s work is determined. It will help in fixing the standard time required to do a particular job.

(ii) Motion study : It is the study of a machine operator and his movements for performing the job. The purpose is to eliminate useless motion and thereby establish a standard for performance. This study is undertook in order to develop a less time consuming system for accomplishing the work.

(iii) Fatigue study : Excessive specialization of work and poor working conditions may cause monotony resulting in both the physical and mental fatigue among the industrial workers. Fatigue either a physical or mental always has an adverse effect on worker’s health and his efficiency. The idea of this study is to see that workers maintain their operational efficiency without any injury to their health and happiness.

3. Principles of scientific selection and training : Scientific management requires a radical change in the methods and procedures of selecting workers for the plan. Hence, it is essential to entrust the task of selection to a central personnel department. After having selected the workers, every job must be entrusted to the’best worker in the factory. This depends upon the workers aptitude, skill and intelligence.

Under scientific management, the management has to undertake the training of workmen before allotting them certain tasks. Training should be imparted from time to time to improve the efficiency of labour.

4. Principles of justified division of work : According to the principle of division of work, work should be allotted to the right worker according to his choice and efficiency. This is called as right job to the right person.

5. Principles of incentive wage system : Taylor assumes that in addition to good wages, proper incentive should be given to the employees for getting proper attention towards the work. Taylor introduced ‘Differential Wage System’ in this regard. This implies that a worker who works as per the standards set should be paid more, in comparison to the inefficient worker, who does not work as per the standards set. In simple words, a worker who works more should receive more wages than the person who doesn’t.

Question 5.
Explain the principles of Fayol.
Answer:
Principles of administration or Management of Fayol: Fayol has mentioned 14 principles in his book, ‘General and Industrial Management’. As he has accepted administration and management as being synonyms, we can say it as principles of management.

These principles are as follows :

1. Principle of division of work : The object of division of work is to derive benefits from specialization. The various functions of management like planning, organization, coordination, control, etc., cannot be performed by a single proprietor or by a group of directors and they have to be entrusted to specialists in the related fields.

2. Principle of authority and responsibility: As management consists of getting the work done through others, it implies that the manager should have the right to give orders and the power to exact obedience. Thus, the manager gives order to his subordinates to perform the job as per his direction. A manager may exercise official authority and also personal authority. An individual to whom authority is given to exercise power must also be prepared to bear responsibility to perform the work.

3. Principle of discipline : Discipline is absolutely essential for the smooth running of the business. By discipline, we mean the obedience to authority, observance of the rules of service and norms of performance, respect for agreements, sincere efforts for completing the given job, respect to the superiors, etc.

Question 6.
Explain the nature of principles of management.
Or
Describe any six characteristics which clarify the nature of the principle of management.
Answer:
Nature/Characteristics of management principles: Management includes the qualities of both science and art that’s why management is called as a science or an art. The principles of management are not rigid and fixed like the principles of physics and chemistry. The main characteristics which highlight the nature of management principles are as follows:

(1) Universality of principles: Management principles can be applied to all manage
rial situations irrespective of the size (Small or large) and nature (Business government or social organization). That’s why Henri Fayol stressed on universality of management principles. Every organization or persons tries to achieve objectives through group efforts. Management principles can be implemented by the managers or directors in any level of management.

(2) Dynamic nature of principles : Management principles are dynamic in nature and flexible. These principles are not rigid and fixed like the principles of mathematics, physics and chemistry. Management principle changes due to cause and effect relationship. Management principles which were earlier important now have changed due to social changes. Management principles are dynamic in nature because these principles are directly related with human beings.

(3) Concentrated on human behavior: Management is totally related with human behavior. Management is related with persons. The success of management principles is affected by human behavior. Principles of management are directed towards enhancing human behavior so that the persons in organization give their best performance.

(4) Relative principles : Management principles are relative not absolute. These principles are implemented according to the needs of organization. It is not compulsory to apply these principles. Every organization, enterprise or society is having different type of nature. Thus due to different circumstances, assumptions, needs etc. accordingly the principles are applied in enterprises and organizations.

(5) Equal importance : All the principles of management are having equal importance. This is true that some principles are important for one organization while other principles may be important for other organization. But these principles are having some importance in all the organization.

(6) Cause and effect relationship : Management principles are always affected by cause and effect (result). Management principles establishes relationship between cause and result. These principles indicates the effect of an important decision on the activities of enterprise.

Question 7.
Write the characterstics of scientific management.
Answer:
Meaning of Scientific Management: Scientific management is made of two words science and management. Science is the application of logic, reasoning and it gives a new dimension to our study approach to different problems of life. When we apply scientific approach towards management concepts this is called as scientific management.

Following are the characteristics of scientific management:

1. The objectives of every job in scientific management is fixed and predetermined,
2. There is a fixed plan to attain objectives
3. Mutal cooperation among employees is essential in scientific management
4. Scientific management works on the objective of maximum production at minimum cast
5. Motivation is considered necessary for the completion of work
6. Scientific control system is implemented under scientific management
7. Time to time evaluation of standards is performed
8. Increase in efficiency of employees helps in increase in the income of employees
9. Researches and experiments plays an important role in the field of scientific management
10. Specialisation and division of work is also given importance
11. Much stress is given on right person for the right job
12. Responsibilities are assigned or fixed for every level of work.

MP Board Class 12 Business Studies Important Questions

MP Board Class 12th Business Studies Important Questions Chapter 1 Nature and Significance of Management

Nature and Significance of Management Important Questions

Nature and Significance of Management Objective Type Questions

Question 1.
Write the answer in one word / sentence :

1. How many basic, elements of management are there ?
2. In how many fields of the need of management are felt ?
3. What are the levels of management ?
4. Which function of management is called basic or primary function ?
5. Write the name of that level of management in which supervision of work of employees is essential ?
6. If father is a manager and his son is also a manager. Then what is this ability called ?
7. What is systematic knowledge called ?
8. To get the work done by others is called what ?
9. With what all the activities of management are related ?
10. The directions which are given to the members for guidance and control of behavior in business is called what ?
11. To fix the policies in management is the function of what level ?
12. Need of coordination is required at what level ?
13. To achieve the results by use of personnel skill is called what ?
14. Knowledge achieved by training and experience and is used for social welfare is called ?
15. In management do all the characteristics of art are included ?

Answer:

1. Five
2. In all regions
3. There are three levels of management:
   • High level
   • Middle level
   • Low level
4. Planning
5. Lower level or supervisory level
6. It is called inborn ability
7. Science
8. Management
9. With human being
10. Code of conduct
11. At high level management
12. At all the three levels of management
13. Art
14. Profession
15. Yes, in management all the characteristics of art are included

Nature and Significance of Management Very Short Answer type Questions

Question 1.
What is management ?
Answer:
Management is both science and an art in which the functions of planning organizing, coordinating, directing, motivating and controlling of human endeavour are performed to achieve the predetermined targets and objects by adopting scientific methods.

Question 2.
What is the importance of management in business ?
Answer:
The importance of management in business is same as the role of brain in human body.

Question 3.
Write the meaning of management.
Answer:
By management, we mean the art of getting the work done by others. It includes how best quality work can be get done by workers efficiently with less cost of production.

Question 4.
What are the 5 M’s of management ?
Answer:
The 5 M’s of management are :

1. Men
2. Material
3. Machine
4. Money
5. Method

Question 5.
“Why it is said management is universal ?” Explain.
Answer:
The principles of management is applicable in all sorts of organizations whether they are trading or non-trading like schools, clubs, etc. However, the same principles of management are not applicable to all organizations, rather it depends on the type of organization and it can be modified as per the need. Hence, we can say that management is universal.

Question 6.
Write the first and last functions of management.
Answer:
Planning is the first and control is the last functions of management.

Question 7.
Which officers are included in high level of management ?
Answer:
Following officers are included in high level of management:
(a) Board of directors
(b) Managing directors
(c) Chief manager.

Question 8.
Write the definition of management.
Answer:

1. According to Henri Fayol : “The management is to forecast, to plan, to organize, to command, to coordinate and to control.”
2. Management is the art of knowing exactly by what you want your men to do and then seeing that they do it in the best and cheapest way.

Question 9.
Write any two functions of middle level management.
Answer:
Two functions are :

1. To prepare plans for the achievement of objectives.
2. To explain and interpret the policy decision taken by the management.

Question 10.
What do you mean by levels of management ?
Answer
There are three levels of management :

1. Top level
2. Middle level and
3. low level management. In the top level, board of directors and general managers see the managerial work, whereas in the middle level, different heads of department see the management work and in the lower level supervision works are done.

Though the work is done at different level still full cooperation and coordination is required at all levels of management.

Question 11.
What is effectiveness and efficiency use of resources ?
Answer
Effectiveness means completing the assigned task and achieving the predetermine goals or targets.

Efficient means not only completing the assigned task or achieving the target in the given time but also in a manner that all the available resources are utilized optimally and the task is completed in minimum cost.

Thus, effectiveness is doing assigned task within the assigned time whereas efficiency is doing the assigned task in such a manner that it earns maximum profits.

Question 12.
What do you mean by coordination ?
Answer
Coordination means bringing all the activities of various levels of management together efficiently and effectively. Proper application of management principles helps in bringing about coordination. Coordination is a grouped activity where people work together for the common goal. In common language it means “Mutual Understanding”.

Question 13.
Shyam is a manager of north block of a big construction company. He works at what level of management ?
Answer
Shyam is working at the middle level of management. His main functions are :

1. He works as a chain between the top level of management and supervisory management.
2. He explains policies to low level employees which are made by top level management.
3. He motivates people for their high efficiency and performance of their work.
4. He gives directions to his employees.

Question 14.
Why is management considered as a group activity ?
Answer:
An organization achieves, its goals with the help of diverse individuals working as a group. To make sure that each individual efforts are well coordinated and work towards the common goal an effective management provides opportunities to grow and fulfill their aspirations working as a team.

Question 15.
Why is planning considered as a primary function of management ?
Answer:
Planning gives directions for actions and lays down the framework regarding how work is to be done. All other managerial functions are performed within the framework of the plans drawn. Therefore, planning is a primary function of management which precedes all other managerial functions.

Question 16.
What is mild knowledge ? Name the person who felt management is mild.
Answer:
Mild means soft, i.e., which is not hard. In this way it means such science whose principles are not rigid. Ernest Dale felt management to be mild.

Question 17.
Why is management considered to be universal ? Explain.
Answer:
Management is found in every country, every organization and in life of every individual. Any work is not there which can be completed without management. Thus management is found in every aspect of life. Due to this reasons management is called as “Universal”.

Question 18.
“Management is an intangible force”. Why ?
Answer:
Management can be felt and experienced not seen. If a company enables its
employee to meet their targets, keeps its stake holder satisfied and eventually achieve its goal effectively. It is said to be efficiently managed. Thus, the effect of management is noticeable from the manner in which a.n enterprise function.

Nature and Significance of Management Short Answer Type Questions

Question 1.
Why is management called a multiple concept ?
(or)
Why is management considered a multifacted concept ?
Answer:
Management is considered a multifacted concept because management does not , manage a single activity rather it manages three main activities the work, the people and the operations managing work to achieve organizational goals. Three main activities are like this:

1. Managing people : Managing people is the process of getting work done with the aim to transform input into desired output.
   It means supervise the personal requirement of single employee and to look after the group of people as employees. ‘
2. Managing work: To manage work for achieving the target.
3. Managing management: To purchase raw materials and convert their problem

Question. 2.
Write any four characteristics of management.
Answer:
The characteristics of management are :

1. Management is a human activity : Management is completely related to human activity. The development of human effect is possible by management only.

2. Group efforts : The origin of management word indicates towards group efforts. Only an individual person cannot complete the work of management. Predetermined objectives can be achieved with group efforts.

3. Management is purposeful: Management is always purposeful because to attain different objectives, different junctions are performed by management.

4. Management is needed at all levels : There are three levels of management i.e. top, middle and lower. In the top level board of directors and managers see the managerial work whereas in middle level departmental heads look after the work and in lower level employees follow the instructions.

5. Management.has specific objective : Without specific objective management is not possible.

Question 3.
Explain three secondary functions of management.
Answer:
The secondary functions of management are :

1. Communication : Although communication is the secondary function of management, its importance cannot be undermined. That is why some management experts have put it in the primary function of management. Communication is an exchange of facts, ideas, opinions or emotions by two or more persons.

2. Innovation : Termed as management’s modem function, it’s basically a creative activity. It consists in doing such things, are generally not done in the ordinary course of business. It may involve various activities like introduction of a new product, new method of production, opening a new market, conquest of a new source of supply, creation of monopoly position, etc.

3. Decision making : Decision making is an important function of management. Everyday hundred of decisions are made in an enterprise. So, decision making is also a part of management activity. According to Peter F. Drucker, “Whatever a manager does, he does after making decisions.”

Question 4.
Write the functions of top level management.
Answer:
The functions of top level management of follows :

1. To determine the objective of undertaking
2. To pass of budgets
3. To investigate the achievements and results of undertaking
4. To implement the various policies like financial, marketing policies etc.
5. To set up the organizational framework
6. To discuss important matters related with organization
7. Top management assembles the resources

i. e., men, material, rrlachine and money.

Question. 5.
Write the functions of middle level management.
Answer:
The functions of middle level management are as follow :

1. To prepare plans for the achievement of objectives
2. To explain and interpret the policy decision taken by the top management
3. To motivate lower level management for efficient performance
4. To coordinate the various activities of different units
5. To act as a line between’top level and lower level
6. To develop leaders for future by training and experience
7. To assist or help in taking decisions related with administration.

Question. 6.
Write the functions of lower level management.
Answer:
The functions of lower level management are as follows :

1. To evaluate the work of employees
2. To communicate the problems and difficulties of workers to the middle management
3. To guide and direct the workers about the procedure of work.

Question. 7.
Write the objectives of management.
Answer:
Following are the objectives of management:

1. To achieve maximum with less efforts: The main objectives of management is to achieve maximum with less efforts. Mainly it is based on trade. It is tried in trade to achieve maximum profit and to produce goods with less cost.
2. Development of employee and employer : A good management always do development of employees and employers because both are the important parts of our society. Thus the main objective of management is to develop both.
3. Maintain coordination between labour and capital: The another objective of management is to maintain coordination between labour and capital.

Question 8.
Distinguish between Coordination and Cooperation.
Answer:
The difference between Coordination and Cooperation:

Question 9.
“Management is an inborn and acquired ability.” Explain.
Answer:
According to ancient concept, “Management is an inborn ability i.e., some people are so efficient and talented by birth that they lead and get success in the field of business.
Actually speaking this concept prevailed when ownership and management were not separated. According to the modem concept, managers are both bom and made. People with managerial aptitude can undoubtedly develop the ability to utilize it to better advantage through experiences which give them the opportunity to get things done through and with others.

Question 10.
Management is the development of people and not the direction of things”. Explain. ”
Answer:
According to L. Appley: Management is the development of people, employee, not for things because employee (labour) and institutions are the main parts of development of things. So management is related human elements. Under it development of employees are given importance and not on non-living things like machines.

Question 11.
Management is pervasive in nature. Explain.
(or)
Management is required at all levels. Explain.
Answer:
Management is required by all types of organizations whether big or small. Management is required at three levels top, middle and bottom. The business is managed at the top level by the board or directors, whereas at the middle level the manager’s function and e bottom level. Thus, management is required at all levels.

Question 12.
explain the different levels of managment.
Answer:
in general there are three levels of management:

1. Top level management: Top level management is having an important place compared to other levels of management in an enterprise. Generally the objectives of enterprise are established by the board of directors. According to E.F.L. Breech “The responsibility of board of directors is to form the policies and to control the entire organization”.

The actual operation of organization is performed by the managing director or general manager and they are called as chief executive. The function of chief executive is to implement the instructions given by Board of Directors and to make efforts to achieve the objectives. Top level management consists of the representatives of shareholders. Top level management includes, Chief Executives, Chairman, Managing Directors, Senior Executives and General Manager.

2. Middle level management: The officers between top level and lower level management are included in the middle level management. This level includes departmental head, sectional head, productional officers etc.

3. Supervisory or lower level management: In the levels of management, supervisory management is at the lowest level. It is also called as operative management. This level of management includes Superintendent, General foreman, Foreman, First line supervisor etc. This level is directly concerned with operative staff. In this level the function of supervisors is to explain the policies to employees, to establish standards, to motivate for standard work etc. At present supervisors are having an important place in organization since they are the friend, philosopher and guide of employees.

Question 13.
What is meant by middle management ? Write its functions.
Answer:
Middle management is mainly concerned with the task of executing the policies and plans chalked out by the top management. It consists of heads of department or sectional head of other executive officers in charge of different department. The middle level management work between top level and lower level management.

The function of middle management:

1.  To explain and interpret the policy decision taken by the top management
2. To communicate the problem and suggestion of supervisory management
3. To serve as a link between top management and supervisory management
4. To take department decision,
5. To develop leaders for future by training and experience.

Question 14.
Explain any six social responsibilities of management.
Answer:
The social responsibilities of management are as follows :

1. Towards consumer: To provide right quality of products and services to consumer at right time and in reasonable price and not to produce harmful goods for society.
2. Towards government: To pay government taxes honestly and timely and also the fallowing the rule and regulation of government in proper manner.
3. Towards environment: To promote measures for pollution control and provide help for government pollution control program, planting trees for the welfare of society.

Question. 15.
Explain the importance of coordination.
Answer:
Without coordination none of business service in market, we need coordination in all kinds of activities. The success of any team depends upon the coordination of its activities. Following points will explain the importance of coordination.

1. For unity : By coordination it makes unity in all spheres of organization and it helps in better performance.
2. Development : By coordination our organization get success in field of development. Proper development of organization is possible only by coordination among the various factors of production.
3. For the best utilization of factor of production : By coordination Organization can use best utilization of factor of production.

Question.16.
“Management is a process.” Explain. .
Answer:
Management is a distinct process in which through planning, organization, motivation, direction and control, one achieves the desired targets.
Management is always purposeful, it is concerned with achievement of objectives of an organization before head and to achieve this pre-determined goals of various management strategies are prepared.
Management is a continuous process. Management work is not for few days or months. All the management activities are carried out for the benefit of society so it is a social process.

Nature and Significance of Management Long Answer Type Questions

Question. 1.
“Management is a system.” Explain.
Answer:
Management is a system as it has various subsystems. Moreover, each subsystem (for example, purchases, sales, etc.) has a specific object and maintains coordination with other system. No subsystem can be viewed separately as a watertight compartment, but it has to maintain communications with other system. Hence, management has been rightly referred to as a system.

Another definitions of scholars :
“The management is to forecast, to plan, to organize, to command, to coordinate and to control.” —Henry Fayol

“Management is the process of planning and regulating the activities of an enterprise.”—E.F.L. Brech

“Management is simply the process of decision making and control over the actions of human being for the expressed purpose of attaining pre-determined goals.”—Stanley Vance

“Management is the art of knowing exactly by what you want your men to do and then seeing that they do it in the best and cheapest way. —F.W. Taylor

“Management is a process of releasing and directing human energies towards attaining definite goal.”

Question. 2.
“Management is a Profession”. Explain.
Answer:
Management is a profession. It has been adopted as a profession, recently. To attain the knowledge of management, persons work as a manager in different institutions as, Indian Institute of Management, International Council for Scientific Management. Before drawing any conclusion, it is necessary to know the characteristics of a profession and also of management. The characteristic of a profession are :

1. There must be an organized and systematized body of knowledge, principles and techniques.
2. Continuous acquisition of that knowledge.
3. Entrance into a profession is restricted by standards established by an association of that profession.
4. Existence of an organization to regulate the behavior of the members of the profession.
5. Presence of ethical standards to guide the activities of the members of the profession.
6. Spirit of service to the society should receive priority over economic considerations.
   From the above characteristics, we may state that there are all trends which indicate
   that management is moving in the direction of profession. So, management can be called as profession.

Question.3
What is the importance of management in India ?
Or
Explain the need of management in India.
Or
According, to five years plans explain the importance of management.
Answer:
The importance of management for a developing country like India is due to the following reasons:

1. For rapid industrialization : Exploitation of natural resources of the country is
   possible only through efficient management. ,
2. For rural development : Rural development of the country is possible when
   government policies regarding rural development are properly implemented through efficient management. ’ ’
3. Creation of employment opportunities: More jobs can be created if more industries . are established and existing ones are expanded. This is possible only if industries are properly
   managed.
4. For success of five year plans : Five year plans are essential for the progress of the country. These plans can be implemented and made successful by capable managers only.
5. For all round development of the country : Management is not confined to business activities alone. Our hospitals, schools, colleges, government offices all have to be properly managed for all round development of the country.

Question 4
Explain the limitations of management.
Answer:
Management is essentially needed in today, modern business. Though management is universal and ongoing process, it has got certain limitations. Which are :

1. Temporary : The principles of management cannot be applied permanently for all situations. It has to be changed according to the time and situation.
2. Rely on human behavior : People are motivated by management. But behavior of all people are not same. So, a particular principle cannot be applied to all blindly.
3. Not fully universal: Though principles of management are universally accepted, but it is affected as per countries, time and other reasons.
4. Depend on type of organization : Principles of management cannot be uniformly applied to all types of organizations as each has different nature, i.e, Principles, objectives, working system, etc.
5. Depend on size of organization : The principles of management depends on the size of organizations too. In small organizations, management puts an extra financial weight on the expenses of the firm. Hence, it becomes difficult to determine the shape of management . in accordance,with the size of organizations.

Question 5.
write the primary functions of management.
Answer:
Primary functions of management are as follows :

1. Planning: It is the primary function of management. Planning provides a blueprint for action. It is concerned with deciding in advance what is to be done, how it is to be done and who is to do it. It bridges the gap from where we are to where we go. It is concerned with determination of objectives to be achieved and the course of action to be followed to achieve them.

2. Organization: Organization involves bringing together the manpower and material resources for the achievement of the objectives laid down by the enterprise. Organization involves grouping the activities in a logical pattern. Organization function helps to increase the efficiency of the organization.

3. Staffing: A separate personal department is established for recruitment work in big organizations. Staffing function is a difficult managerial function because it is concerned with the selection of persons who are properly qualified and mentally well-adjusted to the situation.

4. Direction : Direction is the art and process of getting things done. Direction is concerned with activating the members of the organization to work efficiently and effectively for the attainment of organization goals.

5. Motivation : It is a complex force inspiring a person to work to use his capacities willingly for achieving stain objectives by motivating work can be done for performing any job, two important things are necessary viz,, will to work and ability to work. The important of motivation lies in converting the ability to work into will to work.

6. Control : It is concerned with seeing whether the activities have been or being performed in conformity with plans. In the age of competition, management keeps full control on his business for the attainment of predetermined targets.

7. Coordination: Due to the effect of modern contrast views and will to work freely, coordination has been an important task of management. It is cal led as mutual understanding also. Coordination aims at an orderly arrangement of group effort for the achievement of desired ends.

Question. 6
Jack Welch the CEO of JE, has been given instructions for the success of management What are they ?
Answer:
The instructions are as follows :

1. Always keep your focus on the primary
2. problem
3. Take contribution from each ‘ person and accept suggestions from all sources
4. Dream high and to fulfill to those dreams supply fuel to your organization
5. Keep focus on the social activities and responsibilities
6. Be an ideal for effective management.

Question. 7.
“Management is both science and an art”. Explain.
Answer:
What is science ? : Science is a systematised body of knowledge putting to specific field of study and contains facts that explain a phenomenon. Science has the following characteristics:

1. It has universally accepted principles
2. It is based on method of scientific enquiry.
3. It has predictable results
4. It is based on verification of principles
5. It established relationship between cause and their effects.

Management is a science: Management is a systematised body of knowledge and its principles have been evolved on the basis of observation and scruitny. Secondly management principles are developed after careful analysis scrutiny and inquires as such they can be applied to achieve the desired result: Thirdly management principles clearly establish relationship between cause and effect such as the relationship between motivation and efficiency. Fourthly, principles are practically developed after scientific inquiry and analysis. Lastly the desired results can be achieved by applying the specific principles.

What is art ?: Art refers to the application of knowledge and personal skill to achieve the desired results. The practical application of knowledge is art. The main characteristic of ‘Art’ are:

1. Practical knowledge.
2. Personal skill.
3. Concrete results.
4. Constructive objects.
5. Perfection through practice.

Management as an art : Management is also an art. Management prescribes general principles for managing various aspects of business. But the application of these principles depends on the existence and skill of manager. The manager gets perfection in the art of managing only through continuous practice.

Management is a science as well as an art: It is clear from the discussion that management combines the features of both science and art. The science of management provides certain general principles which can guide the manager in their professional effort. The art of managements consists in tackling every situation in an effective manner
Question 9
Explain the characteristics of management as a process
Answer:
In order to achieve the best possible results management functions
performed systematically and in an orderly manner. As a process it has the following characteristics ;

1. Management is a Continuous Process: Management is not confined to handling and integrating human and material resources at a particular moment. It is an ongoing process. It involves continuous handling of problems and issues. Managers are concerned with constantly identifying the problems and solving them by taking appropriate actions. .

2. Integrating Process : Management works to bring together human, physical and financial resources so that there is harmony among them.. It integrating human efforts with non-human resources like machines, materials, technology, financial assets, furniture etc. The integration is done through planning organizing coordinating and controlling.

3. A Universal Process : Management functions are not confined to business alone, they are performed in non-business organization too, management is required in developed, developing and backward countries.

4. A Social Process : It is mainly a social process because the activities performed to achieve the goals are concerned largely with relations between people. All tasks of management involve interactions with one another managers work with and through people and achieve results for the benefit of people. Human factor is the most important part of management process. As a social process’it is concerned with making interaction between people useful for achieving organizational goals.

Question 10
Explain the meaning of coordination. What is its need in our life ? Explain.
Answer:
Meaning of coordination: Coordination means bringing all the activities of various levels of management together efficiently and effectively. Proper application of management principles helps in bringing about coordination. Coordination is a grouped activity where people work together for the common goal. In common language it means “Mutual Understanding”.

Need in our life :

1. Size of the organization : Need of coordination depend on the size of the organization. When the size of organization is big coordination is required because in large-scale organization more employees are there. Each person has its own aims and need. For the efficient work coordination among various activities is essential.

2. Functional differentiation : Functions of organization is divided into various departments. Each department work separately giving weight age to its efforts. But at the same time these departments are interdependent on each other. To reduce the gap between the efforts of various departments coordination is essential.

3. Specialization : Specialization plays important role in large-scale organization. There are many specialized persons in large-scale organization. Each one of them work according to their skill and efficiency. There may be differences among them so it essential to have common direction for this need of coordination is felt.

Question. 11.
Write the characteristics of profession.
Answer:
Following are the characteristics of profession :

1. Well defined group of knowledge : Profession is an organized and systematic body of knowledge. In profession technical knowledge is gained. Knowledge related to any profession can be gained only from certain special institution.

2. Aim is to render services : Aim of profession is to render services to people. In profession, remuneration for work rendered is essential. Doctors, lowers all render some services to people.

3. Profession is related to one or the other council: All profession is related to some council. These councils make rules and regulations for the entrance in the profession. They also issue certificate to them those who have completed the course. Professional has certain ethical code which are recognized by government. For example doctors takes oath of their morality before entering their profession.

4. Through entrance examination: Each profession has some entrance examination for example lowers have to give some exams while doctors have separate entrance exam. For example if one has to become the Chartered Accountant, he has to pass the exam which is organized  Indian chartered accountant organization.

Question.12
write the difficulties which come in the way of coordination also suggest their solutions.
Answer:
Following difficulties come in the way of coordination. Some solutions are also
written here :

(i) Difference in organizational and individual objectives : It is universal truth that the aims of various workers differ form the organization where they work. The organization will always wish to earn more and more profit while the employees will wish to have more salary or wages.If the employee do the work honestly and sincerely then the aims of both can be fulfiled at a time and in establishing coordination there will be no problem.

On the other hand if the employees pay attention to achieve only their own aim then their this behaviour will be obstacle in the way of coordination. To end the gap between organizational and personal aim all employees of the organization should contribute towards the organization.

(ii) Difference in individual objectives : Employees working in organization also may have different objectives. Due to this the difference in working system may occur and negative effect on coordination is found. To overcome this problem, some objectives should be followed by all.

(iii) Problem in the evolution and measurement of work progress : Evaluation may differ in various department for the same work. It may create confusion for all workers. It has negative effect on coordination. For this organization should keep same evolution for the same type of work.

(iv) Complex organizational structure : Sometimes it is not clear who is under whom. In such case coordination becomes difficult. For this it becomes essential for high authority to make it clear the liability of each employee towards organization.

Question 13.
What challenges a global manager has to face ? Explain.
Answer:
Following are the challenges before global manager.:

(i) In the form of manager : A global manager has to establish various business relations with other organizations and other countries. For it the global manager relates himself with advocate and other officers. In U.S.A, Europe, etc. Indian technical employees are appointed. The manager keeps contacts with foreign companies for these appointments. He makes clear the importance of outsourcing.

(ii) In the form of divisional head : For global manager it becomes essential to form coordination between various divisions. He has to understand the requirements of customers. Most of the customers are of foreign countries like U.S.A, Europe etc. Global manager has to understand the requirements of these customers. He also tries to the priority of foreign customers.

(iii) In the form of leader : Global manager has to maintain awareness about the, , changing commercial situations and toward customer’s priority. He has to recognize the tendencies and opportunities getting through outsourcing. He should have inference about the expected risk.

MP Board Class 12 Business Studies Important Questions

MP Board Class 12th Accountancy Important Questions Chapter 1 Accounting for Non-profit Organization

Accounting for Non – profit Organization Important Questions

Accounting for Non – profit Organization Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Advance Subscription received during a year is a:
(a) Income
(b) Expenses
(c) Liabilities
(d) Assets.
Answer:
(c) Liabilities

Question 2.
Receipts and payment account is prepared by:
(a) Firms
(b) Sole Trader
(c) Company
(d) Non-trading Concern.
Answer:
(d) Non-trading Concern.

Question 3.
Loss on sale of machine will be recorded in:
(a) Receipts & Payment A/c
(b) Income & Expenditure A/c
(c) Balance Sheet
(d) None of these. A/c
Answer:
(b) Income & Expenditure A/c

Question 4.
Entrance fee if no specific instruction is given will be treated as:
(a) Capital item
(b) Revenue item
(c) Assets
(d) None of these.
Answer:
(b) Revenue item

Question 5.
The system of accounting used by non-trading concern and professional person:
(a) Single Entry System
(b) Double Entry System
(c) Mixed Accounting System
(d) Cash System.
Answer:
(d) Cash System.

Question 6.
Balance Sheet is a:
(a) Statement
(b) Account
(c) Account & Statement
(d) None of these.
Answer:
(a) Statement

Question 2.
Fill in the blanks:

1. Opening Balance Sheet is prepared when opening balance of is not given…………….
2. The first (debit) side of Income & Expenditure Account shows ………………
3. Non-trading concern maintain their of account by ……………
4. Receipts and payment account is a summary of ……………
5. The aim of ……………… is not to earn profit

Answer:

1. Capital fund
2. Expenditure
3. Cash System
4. Cash book
5. Non-trading Concern

Question 3.
Match the columns:

Answer:

1. (d) Opening Capital
2. (a) Non-trading Concern
3. (e) School
4. (f) Real Account.
5. (b) Nominal Accounts
6. (c) Receipts & Payment A/c

Question 4.
Write true or false:

1. Receipts and payment account is like profit and loss account.
2. Among the various fund provided by government, scholarship given to the students will be debited in Income Expenditure Account.
3. If Sport Fund is shown then all the expenses related to sports will be shown in debit side of Income and Expenditure Account.
4. Receipts and Payment Account does not show the difference between capital and receipts.
5. The Receipt and payment account is a summary of all capital receipts and payment.
6. Donation for specific purpose will be always capitalized.
7. Receipts and payment account only record the revenue matured receipts and payment.

Answer:

1. False
2. False
3. False
4. True
5. False
6. True
7. False.

Question 5.
Write the answer in one word/sentence:

1. Income & Expenditure account is prepared on the basis of which account ?
2. Which type of institution prepare Income Expenditure account ?
3. Persons who earn income utilizing their intelligence and skills are called as ?
4. Non-trading concern keep their accounts by which system ?
5. Account which is the summary of cash book is known as ?

Answer:

1. Receipts & Payment A/c
2. Non-trading Concern
3. Professional person
4. Cash system
5. Receipts & Payment A/c

Accounting for Non – profit Organization Very Short Answer Type Questions

Question 1.
Write one important difference between receipts and payment account and Income and Expenditure account
Answer:
Receipts and payment account is a real account while Income and Expenditure account is a nominal account.

Question 2.
Write any one basis of difference between trading concern and non-trading concern.
Answer:
Aim of non-trading concern is to serve the society while the aim of trading concern is to maximize profit.

Question 3.
What do you mean by non-trading concerns ?
Answer:
Non-trading concerns are those concerns, whose aim is not to do business and earn profit but to serve the public. It includes hospitals, clubs, schools and colleges, etc.

Question 4.
Write the names of incomes and expenses of a non-trading concern.
Answer:
The following are the incomes and expenses of a non-trading concerns : Income : Annual membership fee, charity, interest, rent from sublet, income from the sale of waste materials, etc.
Expenses : Stationery, wages, charity, postal charges, electricity charges, subscription to newspapers, etc.

Question 5.
What do you mean by receipt and payment account ?
Answer:
Receipt and payment account is the summary of cash book in which all receipts are debited and all the payments are credited. In receipt and payment opening balance and closing balance of cash is also shown.

Question 6.
Why non-cash transactions are not entered in receipts and payment ac¬count ?
Answer:
Non cash transactions are not entered in receipts and payment account because it is prepared on the basis of cash book of accounting.

Question 7.
What do you mean by income and expenditure account and by whom it is prepared ?
Answer:
Income and expenditure account is a type of profit and loss account, which is prepared at the end of the accounting year with the help of receipts and payments by the non-trading concerns. Income and expenditure account is prepared by non-trading institutions and professional persons to know their annual income and expenditure.

Question 8.
What is the nature of income and expenditure account ?
Answer:
Income and expenditure is a nominal account.

Question 9.
Which account is prepared by non-trading concern is place of profit and loss account ?
Answer:
Income and expenditure account is prepared in place of profit and loss account.

Question 10.
What is surplus in income and expenditure account ?
Answer:
Excess of income over expenditure refers to surplus in an income and expenditure account.

Question 11.
What is the basis of preparing income and expenditure account ?
Answer:
Income and expenditure account is prepared on accrual basis.

Question 12.
A school collected Rs. 50,000 as donation for the construction of buildings. In which side of school’s balance sheet will it be showed ?
Answer:
In the liabilities side of balance sheet.

Question 13.
Following balance appeared in the books of Modern club in the beginning of the year :

• Fixed assets Rs. 50,000,
• Bank overdraft Rs. 5,000,
• Subscribed received in advance  Rs. 2,000.
• Prepared rent Rs. 1,000.

Find the opening capital.

Answer:
Capital fund = Rs. 50,000 + Rs. 1,000 – Rs. 5,000 – Rs. 2,000
= Rs. 51,000 – Rs. 7,000 = Rs. 44,000.

Accounting for Non – profit Organization Short Answer Type Questions

Question 1.
What is life time membership fees ?
Answer:
In non-trading organization some person becomes the life time member of the firm. The fee taken from such kind of persons is called as life time membership fee. This fees is not for a particular year but for the whole period. So it is treated as capital receipts and added to capital fund. It is also shown separately in the liabilities side.

Question 2.
What do you mean by endowment fund ?
Answer:
The cash or assets received by the members of the concern from the third person is called ‘endowment fund’. It is of capital nature, hence is written in receipt and payment account and is Shown in liability side of balance sheet.

Question 3.
state the meaning of ‘Not – for – Profit’ Organizations.
Answer:
Not – for – Profit Organizations (NPO) are set up with the prime objective of providing services and not to earn profit thereby enhancing the welfare of society. Such organizations include schools, hospitals, trade unions, religious organizations, etc. The person/s or the groups of individuals who govern and manage the working of an NPO are known as trustees.

NPO’s main sources of income are donations, subscriptions, life membership fees, grants etc. As these organizations are not set up with profit motive, they do not prepare Trading and Profit and Loss Account. Instead, they maintain Receipt and Payments Ac-count, Income and Expenditure Account and Balance Sheet.

Question 4.
State the meaning of Receipt and Payment Account.
Answer:
Receipts and Payments Account is a summary of the Cash Book. All cash receipts are recorded on the Receipts side (i’. e., Debit side) and all cash payments are recorded on the Payments side (i.e., Credit side) of Receipts and Payments Account. It is prepared on the basis of cash and bank transactions recorded in the Cash Book. It begins with the opening balance of cash and bank and ends with the closing balances of cash and bank (balancing figure) at the end of the accounting period.

It records all cash and bank transactions both of capital and revenue nature. It not only records the cash and bank transactions relating to the current accounting period, but also the cash and bank receipts (or payments) received during the current accounting period that may be related to the previous or next accounting period. This account only helps us to ascertain the closing balance of the cash and bank and helps in assessing the cash position of an NPO.

Question 5.
State the meaning of Income and Expenditure Account.
Answer:
Income and Expenditure Account (I&E) is similar to the Profit and Loss Account in the sense that while the former is prepared to ascertain surplus or deficit during an accounting period, the latter is prepared to ascertain net profit or net loss incurred during an accounting period. I&E Account is a nominal account and is prepared on the accrual basis. It records all transactions of revenue nature that are related to the current accounting period (whether outstanding or prepaid) for which the books are maintained.

All expenses and losses are recorded on the debit side (Expenditure side) and all income and gains are recorded on the credit side (Income side) of I&E Account. The closing balance or the balancing figure of I&E Account is termed as surplus (or deficit), if the sum total of the Income side exceeds (is lesser than) the sum total of the Expenditure side.

Question 6.
Write the objective of receipts and payment account.
Answer:
Following are the objectives of receipts and payment account:

1. Knowledge of cash: This account is summary of cash book on the basis of which the closing cash balance for a certain period can be known.
2. Knowledge of cash receipts: By the help of this account the cash receipts of the organization at the beginning, mid of the year and at the end of the year can be easily found out.
3. Knowledge of cash payments: Receipts and payment account shows all the cash payments made by the organization at the end of the year.
4. 4. Basis of income and expenditure a/c:  With the help of receipts and payment account, income and expenditure account is prepared.

What is subscription? How is it calculated?
Answer:
subscription is the main source of income for an NPO besides entrance fees, donations, grants, etc. Subscriptions refer to the amount of money paid by the members on periodic basis for keeping their membership with the organization alive. It is paid monthly, quarterly, half yearly or annually by the members.

It is shown in the debit side of the Receipt and Payment Account with the total amount received during the year that may be related to the current period and to the previous and next accounting period. While calculating subscription for the current period, advance subscription received for the current period in the previous period and outstanding subscription for the current period are added to the subscription received during the current period.

Whereas, on the other hand, advance subscription received for the next accounting period during the current period and outstanding subscription for the preceding period are deducted from the subscription received during the current period.

Calculation of Subscription:
Subscription received during the year
Add: Subscription received (in advance) during previous year for current year
Add: Subscription outstanding at the end of the year
Less: Subscription received in advance for the next year
Less: Subscription outstanding for the previous year
## Subscription shown in Income and Expenditure Account

This subscription is related to the current accounting period and is shown in the In-come side of the Income and Expenditure Account.

Question 8.
What is Capital Fund? How is it calculated?
Answer:
Capital fund is the excess of NPOs’ assets over its liabilities. In other words, the excess of assets over the liabilities for a profit earning organization is termed as capital and the same for an NPO is termed as capital fund. Any surplus or deficit ascertained from Income and Expenditure account is added to (deducted from) the capital fund. It is also termed as Accumulated Fund.

Calculation of Capital Fund
Capital Fund at the beginning of the year
Add: Surplus from Income and Expenditure Account
Add: Subscription Amount (Capitalized amount)
Add: Life membership fee.
Less : Deficit from Income and Expenditure Account
Capital Fund at the end of the year

Question 9.
Explain the statement: “Receipt and Payment Account is a summarized version of Cash Book”.
Answer:
Receipts and Payments Account is a summary of the Cash Book. This account is prepared by those organizations which maintain their books on cash basis. All cash receipts are recorded on the Receipts side (i.e., Debit side) and all cash payments are recorded on the Payments side (i.e., Credit side) of Receipts and Payments Account. It is prepared on the basis of cash and bank transactions recorded in the Cash Book.

It begins with the opening balance of cash and bank and ends with the closing balances of cash and bank (balancing figure) at the end of the accounting period. It records all the cash and bank transactions both of capital and revenue nature. It not only records the cash and bank transformations relating to the current accounting period, but also cash and bank receipts (or payments) received during the current accounting period that may be related to the previous or next accounting period.

This account only helps us to ascertain the closing balance of the cash and bank and helps in assessing the cash position of an NPO. It also forms the basis for the preparation of Income and Expenditure Account.

Question 10.
What steps are taken to prepare Income and Expenditure Account from a Receipt and Payment Account ?
Answer:
The following steps are taken to prepare Income and Expenditure Account (I&E) from Receipts and Payment Account (R&P).

Step 1:
All the revenue expenditures paid for the current accounting period are transferred from the Payments side of R&P to the Expenditure side of I&E.

Step 2:
All the revenue receipts for the current accounting period are transferred from the Receipts side of R&P to the Income side of I&E.

Step 3:
Expenses outstanding for the current period and expenses paid in advance (prepaid expenses) for the current period in the preceding accounting periods are to be added (adjusted) to their related expenses in the Step 1.

Step 4:
Income outstanding (accrued income) for the current period and income re-ceived in advance for the current period in the preceding accounting periods are to be added (adjusted) to their related incomes in Step 2.

Step 5:
Non-cash items like depreciation, appreciation for the current accounting period are to be adjusted in the I&E.

Step 6:
After adjusting all the revenue items for the current accounting period, the Income and the Expenditure sides are totaled. If the sum total of the Income side exceeds (or is lesser than) the sum total of the Expenditure side, then the balancing figure is termed as surplus (or deficit).

Accounting for Non – profit Organization Long Answer Type Questions

Question 1.
Differentiate the receipts and payments account and cash account, (any 2)
Answer:
Difference between receipts and payments account and cash account:
Receipts and Payments A/c:

1. It is maintained by non-trading concerns.
2. It is a summary of cash account.
3. Income and expenditure account is prepared from it.
4. Date is written at the top along with the heading.

Cash A/c:

1.  It is prepared by both the trading and non-trading concerns.
2. It is main account and write daily.
3. Receipts and payments account is prepared from it.
4. Date is written along with each transact-ion.

Question 2.
“Income and Expenditure Account of a Not – for – Profit Organisation is akin to Profit and Loss Account of a business concern”. Explain the statement.
Answer:
Income and Expenditure Account (I&E) is similar to Profit and Loss Account (P&L), in the sense that the former is prepared by Not-for-profit-Organisations and the latter is prepared by profit earning organisations. Both the accounts are prepared on the accrual basis.

Similar to the P&L, all the expenses and losses pertaining to the current accounting period are recorded on the debit side (Expenditure side) and all the gains and income of the current accounting period are recorded on the credit side (Income side) of the I&E.

The balancing figure of the I&E is surplus or deficit and that of the P&L is net profit or net loss. Both the accounts record only revenue items which are related to the current accounting period. Similarities between Income and Expenditure Account and Profit and Loss Account I&E Account of an NPO is akin to the Profit and Loss Account of a profit earning business in the following manners.

1. Nature of Account: Both the concerned accounts are nominal in nature.

2. Basis of Recording: Both the accounts record only revenue expenses and revenue income related to the current accounting period. The items of capital nature are not ignored while preparing these accounts.

3. Period: Transactions related to current year are recorded in Income and Expenditure account in the same manner in which profit and loss account is prepared. Transactions related to previous year or next year are excluded.

4. Adjustments: The treatment of adjustments like, outstanding expenses, prepaid expenses, income received in advance, income due but not received, depreciation, bad debts etc. is same as that in Profit and Loss Account. Thus, both the accounts are prepared on the accrual basis.

Question 3.
Why does non-trading concerns maintain their accounts on the basis of cash system ? Given any five reasons.
Answer:
Due of the following reasons, non-trading concerns are maintaining these books of accounts according to the cash system of book-keeping.

1. Easy method: Cash system is in easy method. It is a part of cash-book and its is maintained as a cash-book is prepared.
2. Less expensive system: Under this method, no need for keeping more accounts. Even untrained and un experienced persons can do this work.
3. Limited transactions: In non-trading concerns, the number of receipts and payments are limited.
4. Cash transaction: Only cash receipts and payments are entered here as in the cash-book. If there is more credit transactions it will be entered in memorandum book.
5. Checking of correctness: Correctness can be checked by comparing with cash-book.

Question 4.
What points should be kept in mind while preparing the receipts and payments account ?
Answer:
The following points should be kept in mind while preparing receipts and payments account:

1. In this account, enter first of all the opening cash balance. This balance might be closing balance of the previous year.
2. Only cash transaction, i.e., cash receipts and payments are entered and no credit transaction.
3. All types of capital and revenue nature transactions are entered here.
4. All types of receipts and payments whether they relate to previous, current or coming year should be entered.
5. No adjustment relating to outstanding, prepaid, as earned and accrued are done here.

Question 5.
Differentiate the receipts and payments account and income and expenditure account ? (any 4 points)
Answer:
Differences between receipts and payments and income and expenditure

Question 6.
Describe in brief the sources of income of non-trading concern.
Answer:
The various sources of income of non trading concern are :

(a) Subscription:
Subscription is the main source of income for the non-trading concern. It is the amount paid by the members of the organization monthly or yearly for keeping their membership with the organization alive.

(b) Donation:
Donation refers to the amount received by the organization as gift from the members or outsiders. If the amount of donation is merge, it will be treated as revenue income. But if the amount is huge and for specific purpose, then it is treated as capital receipt.

(c) Entry fee:
Amount paid by new member at the time of joining an organization is known as entry fee. It is paid by every member once at the time of becoming a member of the organization.

(d) Interest:
For increasing its sources of income, these non-trading organization invest money outside. The interest earned on this investment is also a source of income.

(e) Income from entertainment:
These organization organize various types of cultural programmes and collected income from the common people.

(I) Rent received:
By giving the building or some other portion in rent, the rent received is also a source of income for them.

Question 7.
What are the process of preparing income and expenditure account from receipts and payments account ?
Answer:
The following is the process of preparing income and expenditure account from receipts and payment account:

(a) Opening cash and bank balance:
It is not entered in income and expenditure account. It is shown in assets side of opening balance sheets.

(b) Closing cash and bank balance:
It is not recorded in income and expenditure account. It is shown in assets side of closing balance sheet.

(c) Capital receipts:
Capital receipts are not entered in income and expenditure account. It is shown in liabilities side of closing balance sheet. Example of capital receipts are donation fund, sports fund, prize fund etc.

(d) Reserve and fund:
Reserve and funds are not recorded in income and expenditure account. It is shown in liabilities side of balance sheet and the concerned expenses are deducted from it.

(e) Sale of fixed assets:
Sale of fixed assets is not shown in income and expenditure account. It is entered in assets side of balance sheet by deducting if from concerned assets.

(f) Capital payments:
Capital payments are not entered in income and expenditure account. It is shown in assets side of closing balance sheet. Example of capital payments are purchase of assets, investment fixed deposits etc.

Question 8.
How is capital fund ascertained in non-trading organization ?
Or
How is initial capital ascertained in non-trading organizations ? Explain.
Answer:
For calculating initial capital, an opening balance sheet is prepared with the help of last year’s assets and liabilities. The total of liabilities side is deducted from the total of assets side and the difference thus gained is called initial capital.
Capital = Total assets – Total liabilities on opening date.

MP Board Class 12 Accountancy Important Questions

MP Board Class 12th Business Studies Important Questions Chapter 3 Management and Business Environment

Management and Business Environment Important Questions

Management and Business Environment Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Out of the following which is not a part of business environment:
(a) Urbanization
(b) Employee
(c) Comparative
(d) Mandatory.
Answer:
(b) Employee

Question 2.
Out of the following which is a component of business environment:
(a) Identification
(b) Improvement in performance
(c) Coping with rapid changes
(d) All the above.
Answer:
(d) All the above.

Question 3.
Which is an example of social environment:
(a) Supply of money in the economy
(b) Consumer protection ACT
(c) Structure of country
(d) Formation of family.
Answer:
(d) Formation of family.

Question 4.
Liberalization means:
(a) Integration among economies
(b) Reduced government control and restriction
(c) Policy of planned disinvestment
(d) None of these.
Answer:
(b) Reduced government control and restriction

Question 5.
Shape and distribution of population is taken as :
(a) Part of technological environment
(b) Part of legal environment
(c) Part of political environment
(d) Part of social environment.
Answer:
(d) Part of social environment.

Question 2.
Fill in the blanks :

1. The …………….. components of business are controllable.
2. The internal components of business …………….. affects the business.
3. Legal regulatory is the …………….. component of business.
4. The objectives and outlook of business is included …………….. in components or factor.
5. India formulated the policy of new economic reforms in ……………..

Answer:

1. Internal
2. Directly
3. External
4. Internal
5. 1991.

Question 3.
Write the answer in one word/sentence :

1. What do you mean by Globalization ?
2. What is the policy made for providing employment to people known as ?
3. Foreign policy is the component of which business environment ?
4. What does EPCG stands for ?
5. Name the business zone established to encourage exports.

Answer:

1. Integration of world trade economies
2. Employment policy
3. Political
4. Export promotion goods plans
5. Export promotion policy.

Question 4.
Write true or false :

1. Liberalization in India is failure in affecting industry and business.
2. Economic, Social and political situations creates business environment.
3. Globalization has affected industries and business in India.
4. External factors do not affect the business environment.
5. Changes in government policies do not affect industries and business.

Answer:

1. False
2. True
3. True
4. False
5. False.

Question 5.
Match the columns :

Answer:

1. (e)
2. (a)
3. (b)
4. (c)
5. (d)

Management and Business Environment Very Short Answer Type Questions

Question 1.
Advanced techniques lay emphasis on what factors ?
Answer:
Advanced techniques lay emphasis on the following :

1. Scientific research
2. Innovation
3. Invention
4. Improvement
5. Spreading innovation.

Question 2.
Name the change in economic surrounding of India after 1991.
Answer:
The changes in economic surroundings of India after 1991 :

1. Modernization
2. Privatization
3. Globalization,
4. Liberalization.

Question 3.
what are the reasons behind adoption of new economic policy in India ?
Answer:
The following are the reasons behind adoption of new economic policy in India:

1. Decrease in foreign exchange
2. Heavy government loss
3. Increase in cost
4. Increase in internal loan
5. Decrease in foreign business.

Question 4.
Under 1956 industrial policy how are the industries classified ?
Answer:

1. First category: Those industries whose development is completely based on its state.
2. Second category: Those industries which are run by the state and new industries in future will also be stated by them.
3. Third category: Rest all industries belong to this. Industries in this category are mainly consumer products based.

Question 5.
Write names of four economic policies :
Answer:
Four economic policies :

1. Export import policy
2. Employment policy
3. Agricultural policy
4. Industrial policy
5. Foreign investment policy
6. Tax policy.

Question 6.
Which industries according to the new economic policy compulsorily need to take licence.
Answer :
The six industries according to the new economic policy compulsorily need to take licence are :

1. Harmful/Dangerous chemical
2. Drugs and pharmaceuticals
3. Alcohol peva
4. Defense equipment
5. Industrial explosive.

Question 7.
After 1991, what were the changes in the Indian economic environment ?
Answer :
These were the important parts of new economic policy which has been modified so that objectives of liberalization, globalization and privatization can be achieved.

1. Declaration of new industrial policy : At the time, 1991 industrial policy was announced, 18 items were subject to industrial licensing but gradually their number has been reduced to 5 namely.

1. Cigars and cigarettes
2. Electronic aerospace and defence equipment
3. Indus-trial explosives
4. Hazardous chemicals
5. Certified specified drugs.

2. New trade policy :

1. Minimum restriction on trade. To increase foreign trade, import duty and tariff were slowly reduced.
2. To bring expansion in economy foreign investment and new technology was encouraged.
3. Liberal financial facilities to exporters were given.

3. Reform in capital market: SEBI was established.

4. Reforms in fiscal policy : Control on public expenditure and introduction of VAT in states.

Question 8.
Throw light on new monetary policy.
Answer:
To encourage liberalization and to enhance foreign business government decided to amend monetary policy for which Narshimham Committee was formed. The recommendations are as follows:

1. Bank should independently control the interest rates and RBI should not interfere in it.
2. To redesign the banking system.
3. To give more independence to banks.
4. To improve the accounting system of banks.
5. National banks should be given freedom to accumulate money from the market.
6. Private sector should be permitted to establish bank.
7. There should be reorientation of banks. Two or three banks should have international orientation which means that they alone should operate overseas.
8. Necessary changes should be made in accounting systems of banks.
9. There should be 8 to 10 national banks who should operate throughout the country.

Question 9.
What do you understand by business environment ?
Answer:
business environment is made of two words ‘Business’ and ‘Environment’. Business simply mean busy with an aim of profit earning. Environment means the surrounding extend agents influences or circumstances under which someone or something exits. Business environment consists of all external and internal factors that influence the complex interaction in the market production and finance.

Question 10.
Write the various components of internal environment.
Answer:
Various components of internal environment are as follows :

1. Policies, rules and plans of business.
2. Aims and objectives of business.
3. Production pattern of business. .
4. Organization of business.
5. Sources and resources of business.
6. Labour policy of business.

Question 11.
Explain India vision 2020.
Answer:
On 23rd January, 2003, the planning commission for the evaluation of economic development of next two decades issued a document or deed called “lndia Vision 2020” Through this document, planning commission has made an effort to improve the business environment.
The main features of India Vision 2020 are as follows :

1. To pay attention towards the multiplying of employment opportunities.
2. Per capital income to be increased by 4 times.
3. To lay stress on balanced environment protection.
4. To increase economic development at the rate of 9% p.a.
5. To increase the share of direct foreign investment in the total capital formation.
6. To increase the gross domestic product etc.

Management and Business Environment Short Answer Type Questions

Question 1.
What does LPG stands for in business environment ?
Answer:

• L : Liberalization
• P : Privatization
• G : Globalization.

Question 2.
Write full forms of FERA and FEMA under the new policy of India which rule has been eliminated to implement FEMA ?
Answer:
FERA: Foreign exchange regulation Act
FEMA: Foreign exchange management Act.
In India at the time of new economic policy FEMA was implemented in place FERA.

Question 3.
What do you understand by Micro and Macro environment ?
Answer:
Micro environment: It means that environment in which the included components have close relationship with business.

Question 5.
What were reasons which compelled Indian government to adopt new economic policy ?
Answer:
Following are the reasons which compelled Indian government to adopt new economic policy:

1. Fiscal threat
2. Increased international loan
3. Appreciation of currency index
4. Fall in foreign exchange accumulation
5. Depreciation in Indian rupee
6. Imbalance and negligence of payments
7. RBI and SBI Mortgaged Gold.

Question 6.
Give three examples of Indian legal environment which has affected the business.
Answer:

1. Restriction on advisement of toxic production.
2. Industries were made free from licencing.
3. Statutory warning was made compulsory on tobacco products.

Question 7.
What do you mean by liberalization ? Under liberalization what measures were taken and what was the impact of it on business and industry ?
Answer
Meaning : Liberalization refuse to removal of entry and growth restrictions on the private companies. It is an economic reform with main objective to minimize licencing requirements, remove restrictions related to scale of operations, give freedom for fixing prices and movement of goods, encourage.import and export.Measures and impact of liberalization on business and industries are as follows :

1. Ending of licensing and registration : The main objective under new economic policy is to have controlled economy. The new economic policy has benefited the private industries by providing free licensing.

2. Relief from proprietorship restrictions: The restrictions of more than 100 crores investment in proprietorship was removed.

3. Increase in investments in small scale industries : Investment limits in small scale industries have been increased to 1 crore and adopted the latest advance in technology.

4. Discount on technological imports: To promote the new economic policy emphasis was laid to promote latest technologies and discounts were given. This policy promoted industries based on computers and electronics.

Question 8.
In today’s scenario of business environment how can a businessman prove his existence ?
Answer:
The ahead points lay emphasis on what a businessman can do in today’s scenario to exist:

1. To produce goods as per customer’s need
2. To keep the employees happy with sufficient remuneration
3. To recruit trained and experienced employees
4. To produce good in such a manner which lays good impression on consumers
5. To make plans and techniques keeping in view the competitors.

Question 9.
Explain the following in short:

1. Liberalization
2. Privatization
3. Globalization.

Answer:
1. Liberalization: Means removing unnecessary trade restriction. Liberalization allows private sectors to run even those industries which are reserved for the public sectors, relaxing the restriction and regulations imposed on it.
Measures of liberalization adopted by India are :

1. Exemption from licensing
2. Concession from MRTP Act (Monopolistic and Restrictive Trade practice)
3. Increase in production capacity
4. Freedom of production,
5. Buying foreign exchange from open market.

2.Privatization: Privatization refers to transfer of ownership management and control of public sector enterprise to the entrepreneurs of private sector. It is a set of economic reforms with the main objective to allow private sector to invest in sectors earlier restricted to public sector to develop healthy competition between public and private sector and dilute government ownership in public sector enterprises. Process by which the participation of state and public sector in economic activities are reduced. The main aim of privatization to allow private sector to run and manage the industries and to minimize the reservation made for public sector.

3.Globalization : Globalization refers to integration of national economy with the world economy by removing with the objective to strengthen relationships between countries and increase imports and exports.

Question 10.
Write any two objectives of globalization.
Answer:
Two objectives of globalization are as follows :

1. Removal of inefficiency: Globalization is intended to remove inefficiency of firms and Industries.

2. Improvement in banking and financial sectors : As result of globalization, entry of foreign banks will be easy. It will improve the efficiency of banking and financial sector.

Question 11.
What do you mean by internal components of business ?
Answer:
Business environment can be classified in two categories. Internal environment and external environment. Internal environment includes those factors of business which can be controlled by business. In internal environment changes take place speedily.
Components of Internal Environment:

1. Policies,rules and plans of business.
2. Aims and objectives of business.
3. Production pattern of business.
4. Sources and resources of business.
5. Organization of business
6. Labour policy of business etc.

These are the internal factors of business which affects the business directly while it is under the control of the owners of business. Hence changes can be made according to our own will but external environment is not under the control of the owner of business.

Question 12.
Explain the social environment of business.
Answer:
Social environment: Social environment is shaped by social factors such as attitudes of people, cultural heritage, beliefs and customs of people, education system, consumers organizations, trade unions and other non-government organizations. In society, there are activities related with caste, religion, sex, family etc. which affects the business.

Question 13.
Throw light on legal regulatory of business.
Answer:
Legal Regulatory: Legal regulatory environment of business is determined by constitutional provisions, economic laws, commercial laws, industrial and labour laws, government regulations under various laws and court decisions. These help the government in the regulation of economic activities of the business enterprises. Since Indian economy is centrally planned and controlled the business enterprises are required to operate within the framework of legal regulatory environment. The important components of the legal regulatory environment include the following:

1. The constitutional framework: Directive principle fundamental rights and division of legislative powers between central and state government.
2. Commercial and economic laws and government policies under the laws relating to licensing, monopolies, foreign investment etc.
3. Government policies related to imports and exports. .
4. Government policies related to small scale industries, sick industries, consumer protection etc.
5. Government policies related to pricing and distribution of essential commodities.
6. decisions for the protection of consumers, environment and ecological balance.

Question 14.
What is meant by External Environment ? What are its dimentions ?
Answer:
External Environment which is an index of opportunities and anticipations Internal
environment inclueds. External environment indued micro environment and macro environment it is called up External Environment.

Dimention:

1. Political environment
2. Social environment
3. Legal environment
4. Demographic environment
5. Technological environment
6. Physical environment
7. International environment

Management and Business Environment Long Answer Type Questions

Question 1.
Write the main characteristics of economic environment.
Answer:
The main characteristics of economic environment are as follows :

1.Related with economic activities: All those activities of human beings are included in the economic environment which is related with the purchase sale, consumption, income- expenditure etc. of man.

2. Related with environment : Economic environment is affected by other environments. Ecological environment like land, air, forest, soil, water, mineral resources etc. also affects the economic environment. Political environment also affects the economic environment.

3. Administration by government : Economic environment is administered and directed by government. The government uses economic policies to effectively influence the economic environment of the business.

4. Related with common people : Economic environment is directly related with common people. Each person is having the will to earn money therefore he is affected by the economic environment. He cannot remain distant from economic environment.

5. Availability of capital: If capital is available in sufficient quantity then the natural and human resources are utilized properly. This leads increase in national income due to this rate Of national development also increases.

Question 2.
What are the elements components of micro environment ?
Answer:
Elements of micro environment:

1. Supplier : A person, firm, agency or organization which supplies raw materials, labour or other required things is known as supplier. More number of suppliers must be there to avoid problems of supply to firms.

2. Consumer market: A businessman can have many areas of consumer market. Consumer market can be National/International which uses the material for self or uses it to produce any other goods.

3. Competitor : The competitive business environment increases the level of productivity.

4. Public: Public has various categories, local public, press, media, etc. The behaviour of these categories effect the business unit.

5. Marketing middlemen: Marketing middlemen are those people who decrease and work as connectors between producers and consumers. The success of an industry also depends upon these marketing middlemen.

Question 3.
Name the characteristics of business environment and explain them.
Answer:
Main features/characteristics of business environment are as below :

1. Sum total of external forces.
2. Inter relatedness.
3. Dynamic Nature.
4. Uncertainly.
5. Complexity.
6. Relativity.

1. Dynamic Nature : Business environment is dynamic in nature in the sense that it keeps on changing whether in terms of technological improvement, shifts in consumer preferences of entry of new competition in the market.

2. Complexity : Environment is a complex phenomenon that is relatively easier to understand in parts but difficult to grasp in it totality.

3. Relativity: It is a relative concept since it differs from country to country and even
region to region. For e.g.,  South India may have fairly high in demand than in North India. .

4. Long-term process: business environment has a long-term effect long-term changes in affect the profitability, productivity and development.

5. Business environment influences the firms differently: It is not compulsory that all firms will be equally affected by change in business environment. Each firm is affected according to its nature.

6. Inter dependency of components of business environment: All the components of business environment are related to each other. So effect of business cannot be found out separately.

Question 4.
Explain the components of business environment.
Answer:
Different components of business environment are.

1. Economic components : It includes economic activities, Economic policies, Demand, supply, Investment, Industrial tendency/behaviour and monetary pressure, Investment flow and level, Import-Export, Fiscal and Taxation policies, Monetary policies, etc.

2.Geographical and ecological components : It includes Resources, Environment, Climate locomotion. Marine, Celestical, Geologic resource, Magnetic and solar energy.

3.Political components : It includes government monetary system and government system, Political point of view, administrative arrangements, constitutional arrangements security, etc.

4.Social and cultural components : It includes cost, beliefs, assumptions, social arrangement, materialism, religion, culture, traditions, etc.

5.Science and technological components : It includes scientific research and level, competitive development, Mechanization thermal power, Satellite transmission system, Celestical research labs, etc.

6.Legal and judical components : It includes justice system, various business, Industrial and labour control forum, administrative system, etc.

7.Others components: Population, educational level, consumer behavior, security, crisis, danger, international powers, industrial power and struggle, etc.

Question 5.
Describe the importance of business environment.
Or
What is the need of business Environment today ? Explain.
Answer:
The importance of business environment is as follows : .

1. Market conditions and situations : A business must know the structure of market and changes taking place in the market. Creation of demand, monopoly, boom periods etc. should be in the knowledge of business. These information, conditions are provided by business environment.

2. Better performance : Proper understanding of the various elements of external environment is necessary to take timely action to deal with the threats and avail opportunities for the purpose of improvement in performance of the firm.

3. Technological up gradation : Development of business also depends upon the new technology and inventions, so if a business has to grow it should be aware of new techniques of production, new products, new designs, scientific development etc., No business will survive if technology is ignored.

4. Optimum utilization of resources : Natural environment divides the nature of business, localization, stability, progress etc. Human resource decide the knowledge, techniques, production, Industrial relations, managerial efficiency etc. Management of these resources play an important role in success of business.

5. Survival and growth : A detailed study of business environment is essential for the survival and growth of business enterprise. Business has limited capacity to influence the environment. Therefore adjustment of business, policies and strategies to be changed environment is essential for business survival.

6. Reputation of the company: The management of a company is interested not only in building up its business but also to build an image and reputation of the company in the public mind. Therefore study of environment will help the company to be friendly and avoid such policies which will affect the public.

Question 6.
Explain the political factors affecting business environment.
Answer:
The political conditions and situations of any country affects the business environment. The main factors of political environment are as follows :

1. Ideology of government: The three forms of government are capitalism, socialism and mixed economy. These governments have different ideologies and thinking. These political ideologies and thinking affects the economic and business environment.

2. Policies of government: The policies of government are related with their ideology and thinking. Their policies directly affects the business, trade and industries.

3. Political stability : Stability of government is also an important factor whenever there is political instability in the country, business is affected adversely. The stable government and peaceful atmosphere of the society encourages economic growth.

4. Foreign policy: The foreign policy of any country affects its business environment. The foreign policies helps in maintaining peaceful atmosphere which develops the international business.

5. National security : The business of any country is also affected due to its national
security. Terrorist activities, civil wars disrupt. The business activities while in peaceful atmosphere of the country economic growth takes place.

6. Constitution : The laws and provisions made in constitution is not changeable for a long-term if political situation is stable and changes are not made in constitution then it is not getting to affect the business.

Question 7.
Explain the impact of government policy changes on business and industry.
Answer:
Following are the impacts on business and Industry due to changes in government policies:

1. Increase in competition: Due to government polices of liberalization, privatization and globalization has increase the competition in day to day world. Increasing competition has adver self affected many industries.

2. Increase in demands: Due to economic reforms.speedy industrialization has taken place and various high quality products are available for customers. The customers are demanding these products for improvement in standard of living.

3. Technological environment: Due to rapid changes in technological environment competition is increasing. Only those companies will survive who are going to follow the changes in technology. For smaller firms it is becoming difficult to face the competition.

4. Need of efficient human resource : To achieve success it is necessary to have efficient human resource and it should be properly utilized. That’s why in all industries the demand for efficient human resources and their salary have also increased.

5. Decline of Public Sector : Due to government policies and economic reforms importance of public sector has declined. Financial assistance is also not provided to the public sectors property.

6. Decline of small-scale and cottage industries : Economic reforms and policies had played a major role decline of small-scale and cottage industries. It is becoming difficult for small-scale and cottage industries to face the competition with big companies with limited resources.

7. Open economy : The policies of the government has made our economy as open economy. Due to minimisation of restrictions our country is now developing into as international market.

Question 8.
Explain the managerial response of changing pattern in business environment.
Answer:

1. Large-scale production : Big companies are increasing their production to reduce the cost of production and large-scale production makes it possible to get subsidies.

2. New technology: Various companies are using the new technology with the help of foreign countries to increase their share in market.

3. Direct distribution : By eliminating the middle men companies are trying to sell their products directly to the the consumers. For e.g.Amway is distributing its products directly to consumers.

4. Variety of line of goods and services : Various types of products and services are provided to the consumers at reasonable prices. For e.g., Tata’s are dealing in Iron products, Tea, cars etc.

5. Capital Formation : To expand the business to increase production, to open new branches in national and international market big companies are changing capital structure and forming huge capital.

Question 9.
Explain the economic policy of 2007.
Or
Explain new Exim policy 2002-07.
Answer:
The export-import policy for the period 2002-07 was announced on 31st March, 2002 with the objective to consolidate. The gains of the previous export-import policies and to set for itself a challenging target to achieve one per cent share in global trade by 2007 but this policy was replaced by Foreign Trade Policy by UPA Govt, for the period of2004-2009. This policy aimed to have exports of US 80-48 billion $ per year. The salient features of this policy are: ‘

1. Increase in export: The target of 11 -9% per annum average growth by increase annual exports up to 80 Billion $.

2. Abolishing Quantitative restrictions : Exim policy removed all quantitative re-strictions on all agricultural products except a few sensitive items like jute and onions.

3. Export promotion schemes: Value cap exemption granted on 429 items to continue.

4. More facilities to SEZs : A scheme for setting up special economic zones (SEZs) in the country to promote exports was announced by the Govt, in the export and import policy on 31st March, 2000. The SEZs are to provide an internationally competitive and hassle free environment for exports and exported to give a further boost to the country’s exports. 13 new SEZs were given the permission to start their work by Govt. Units set up in SEZ would be eligible for income tax concessions, exemptions from customs and excise duties.

5. Transport subsidies : Processed fruits and vegetables, poultry, dairy products, wheat and rice etc., Their cost of transportation is more than 80S per ton. This leads to increase in prices of products. To increase exports in the international market Govt, provided a transport subsidy of Rs. 100 crores in the year 2002-03.

6. Overseas banking : Permission have been given by the Govt, to start overseas banking branches in SEZs. In these branches the rules like CRR and SLR of Reserve Bank of India will not be followed.

7. Export to Russia: To increase the export of butter, wheat, groundnut, oil, cashewnuts and products of wheat to Russia, packaging and registration have been stopped.

8. Expansion of market: The countries separated from Africa and Russia and some Latin American countries are the new markets of India. Hence, the exports have been in-creased by 40%. The policy has also made provision for expansion of export market. In first phase Nigeria,.South Africa, Mauritius, Kenya, Ethiopia, Tanzania and Ghana were selected which helped in expansion of market.

By increasing the exports in international market it helped India in increasing foreign trade. This happened due to the efforts and policy of 1991. India’s current share in global merchandise trade is about 0-8 percent ($ 55 billion out of $ 7274 billion in 2003).

MP Board Class 12 Business Studies Important Questions

MP Board Class 12th Chemistry Solutions Chapter 9 Coordination Compounds

Coordination Compounds NCERT Intext Exercises

Question 1.
Write the formulas for the following coordination compounds:

1. Tetraamminediaquacobalt (III) chloride
2. Potassium tetracyanonickelate (II)
3. tris (ethane-1,2-diamine) chromium (III) chloride
4. Amminebromidochloridonitrito-N- platinate(II)
5. Dichlorido 6/s(ethane-1,2-diamine) platinum (IV)nitrate
6. Iron(III) hexacyanoferrate(II).

Answer:

1. [CO(NH₃)₄(H₂O)2]Cl₃
2. K₂[Ni(CN)₄]
3. [Cr(en)₃]Cl₃
4. [Pt(NH₃)BrCl NO₂]^–
5. [PtCl₂(en)₂](NO₃)₂
6. Fe₄[Fe(CN)₆]₃.

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

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

Answer:

1. Hexa ammine cobalt(III) chloride
2. Pentaamminechloridocobalt(IlI) chloride
3. Potassium hexacyanoferrate(III)
4. Potassium trioxalatoferrate(III)
5. Potassium tetrachloridopalladate(II)
6. Diamminechlorido(methanamine)platinum(II) chloride.

Question 3.
Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers :
(i) K[Cr(H₂O)₂(C₂O₄)₂]
(ii) [Co(en)₃]Cl₃
(iii) CO(NH₃)5(NO₂)](NO₃)₂]
(iv) [Pt(NH₃)(H₂O)Cl₂].
Answer:
(i) K[Cr(H₂O)₂(C₂O₄)₂] shows geometrical isomers :

cls-isomers exhibits optical isomerism :

(ii) [Co(en)₃Cl₃] exhibits optical isomerism :

(iii) [CO(NH₃)₅NO₂](NO₃)₂ shows ionisation isomerism [Co(NH₃)₅NO₂](NO₃)₂ and [Co(NH₃)_5-NO₃](NO₂)(NO₃). This also shows linkage isomerism.
[CO(NH₃)₅NO₂](NO₃)₂ and [Co(NH₃)₅(ONO)] NO₃.

(iv) [Pt(NH₃)(H₂O)Cl₂] shows geometrical isomerism.

Question 4.
Give evidence that [CO(NH₃)5Cl]SO₄ and [CO(NH₃)₅SO₄]Cl are ionization isomers.
Answer:
The ionisation isomers dissolve in water to yield different ions and thus react differently to various reagents:
[CO(NH₃)₅Br]SO₄ + Ba²⁺ → BaSO_4(s) (White precipitate)
[CO(NH₃)₅SO₄]Br + Ba²⁺ → No reaction
[CO(NH₃)₅Br]SO₄ + Ag⁺ → No reaction
[CO(NH₃)₅SO₄]Br + Ag⁺ → AgBr_(s) (Pale yellow precipitate)

Question 5.
Explain on the basis of valence bond theory that [Ni(CN)₄]^2- ion with square planar structure is diamagnetic and the [NiCl₄]^2- ion with tetrahedral geometry is paramagnetic.
Answer:
[Ni(CN)₄]^2-: The oxidation state of Ni is +2 in the complex. Its electronic configuration is [Ar]3d⁸. It undergoes dsp² hybridization. Because CN^– being a strong ligand causes the pairing of electrons.

[NiCl₄]^2-: The oxidation state of Ni is +2 in the complex. Its electronic configuration is (Ar)3d⁸. It undergoes sp³ hybridization. Because Cl being a weak field ligand does not cause pairing of the electron.

Two unpaired electrons are present in the complexion, hence it is paramagnetic in nature.

Question 6.
[NiCl₄]^2- is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Why?
Answer:
In Ni(CO)₄, Ni is in zero oxidation state whereas in NiCl^2-₄, it is in +2 oxidation state. In the presence of CO ligand, the unpaired d electrons of Ni pair up but Cl^– being a weak ligand is unable to pair up the unpaired electrons.

Question 7.
[Fe(H₂O)₆]³⁺ is strongly paramagnetic whereas [Fe(CN)₆]^3- is weakly para-magnetic. Explain.
Answer:
In both the complexes, Fe is in a +3 oxidation state with the configuration 3d⁵. CN^– is a strong ligand. In its presence, 3d-electrons pair up leaving only one unpaired electron. The hybridization is d²sp³ forming an inner orbital complex. H₂O is a weak ligand. In its presence, 3d-electrons do not pair up. The hybridization is sp³d² forming an outer orbital complex containing five unpaired electrons. Hence, it is strongly paramagnetic.

Question 8.
Explain [Co(NH₃)₆]³⁺ is an inner orbital complex whereas [Ni(NH₃)₆]²⁺ is an outer orbital complex.
Answer:
In the presence of NH₃, the 3d electrons pair up leaving two d orbitals empty to be involved in d²sp³ hybridisation forming inner orbital complex in case of [CO(NH₃)₆)²⁺. In Ni(NH₃)₆²⁺, Ni is in a +2 oxidation state and has a ds configuration, the hybridisation involved is sp³d² forming an outer orbital complex.

Question 9.
Predict the number of unpaired electrons in the square planar [Pt(CN)₄]^2- ion.
Answer:
The platinum atom has the ground state electronic configuration 5d⁴ 6s¹. In the complex, Ft is in a +2 oxidation state and has the electronic configuration 5d⁸. To have square planar geometry, one of the 5d-orbitals is vacated and all the other four orbitals get a pair of electrons. Therefore, it involves dsp² hybridization and is diamagnetic.

Question 10.
The hexaaquomanganese(II) ion contains five unpaired electrons, while the hexacyano ion contains only one unpaired electron. Explain using Crystal Field Theory.
Answer:
Mn in +2 oxidation state has the electronic configuration 3d⁵. H₂O is a weak ligand. In presence of H₂O molecules, the distribution of electrons is \(t^{3}_{2 g} e_{g 2}\), all the electrons are unpaired.

CN^– is a strong ligand. In its presence, the distribution of electrons is \(t_{2 g}^{5}\) i e one unpaired electron is present.

Question 11.
Calculate the overall complex dissociation equilibrium constant for the Cu(NH₃)²⁺₄ ion, given that β₄ for this complex is 2.1 × 10¹³.
Solution:
Overall complex dissociation equilibrium constant is the reciprocal of the overall stability constant.

Coordination Compounds NCERT Textbook Exercises

Question 1.
Explain the bonding in coordination compounds in terms of Werner’s postulates.
Answer:
Werner’s postulates explain the bonding in coordination compounds as follows:
(i) A metal exhibits two types of valencies namely, primary and secondary valencies.
Primary valencies are satisfied by negative ions while secondary valences are satisfied by both negative and neutral ions.
(In modem terminology the primary valency corresponds to the oxidation number of the metal ion, whereas the secondary valency refers to the coordination number of the metal ion)

(ii) A metal ion has a definite number of secondary valences around the central atom. Also, these valences project in a specific direction in the space assigned to the definite geometry of the coordination compound.

(iii) Primary valences are usually ionisable, while secondary valences are non ionisable.

Question 2.
FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1:1 molar ratio gives the test of Fe²⁺ ion but CuSO₄ solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu²⁺ ion. Explain, why?
Answer:
FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1 : 1 molar ratio given double salt, FeSO₄(NH₄)₂.SO₄.6H₂O (Mohr’s salt). Being double salt, it ionizes in the solution to give Fe²⁺ ions. Hence, it gives the test of Fe²⁺ ions. CuSO₄ solution mixed with aqueous ammonia in 1 : 4 molar ratio gives a complex compound.
CuSO₄ + 4NH₃ → [CU(NH₃)₄]SO₄.
The complexion [Cu(NH₃)₄]²⁺ does not ionise to give Cu²⁺ ions. Hence, it doesn’t give the test of Cu²⁺ ions.

Question 3.
Explain with two examples each of the following: co-ordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.
Answer:
Co-ordination entity: This entity usually constitutes a central metal atom or ion, to which are attached a fixed number of other atoms or ions or groups by co-ordinate bonds.
e.g. [Ni(CO₄)], [CoCl₃(NH₃)₃], etc.
Ligand: It is an ion having at least one lone pair of electrons and capable of forming a coordinate bond with the central atom/ion in the coordination entity.
e.g. Cl^–, (OH)^–, (CN)^–, etc.
Co-ordination number: The total number of co-ordination bonds with which the central atom/ion is linked to ligands in the co-ordination entity is called the co-ordination number of central atom/ion.
Co-ordination polyhedron: The spatial arrangement of the ligands which are directly attached to the central atom/ion defines a co-ordination polyhedron about the central atom.
e.g. [CO(NH₃)₆]³⁺ is octahedral, [Ni(CO)₄] is tetrahedral.
Homoleptic and heteroleptic: Complexes in which a metal is bound to only one kind of donor groups are known as homoleptic.
eg. [CO(NH₃)₆]³⁺.
Complex in which a metal is bound to more than one kind of donor groups are called heteroleptic. e.g. [CO(NH₃)₄Cl₂]⁺

Question 4.
What is meant by unidentate, bidentate and ambidentate ligands? Give two examples for each.
Answer:
Unidentate ligand: When a ligand is bound to a metal ion through a single donor atom as with Cl^–, H₂O. The ligand is said to be unidentate.
Didentate ligand : When a ligand is bind through two donor atoms the ligand is said to be didentate.
Example: C₂O^2-₄ (Oxalate)
H₂NCH₂ – CH₂NH₂ (ethane 2 diammine)
Ambidentate ligand : Ligand which can ligate through two different atoms is called ambidentate ligand.
Example : -NO₂– and SCN^–.

Question 5.
Specify the oxidation numbers of the metals in the following co-ordination entities:
(i) [Co(H₂O)(CN)(en)₂]²⁺
(ii) [CoBr₂(en)₂]⁺
(iii) [PtCl₄]^2-
(iv) K₃[Fe(CN)₆]
(v) [Cr(NH₃)₃Cl₃].
Answer:
(a) x + (-1) = +2, x = +3
(b) x + 4(-1) = -2,x = +2
(c) x + 3 (-1) = 0, x = +3
(d) x + 2 (-1) = +1, x = +3
(e) x + 6 (-1) = -3, x = +3.

Question 6.
Using IUPAC norms write the formulas for the following :

1. Tetrahydroxozincate(II)
2. Potassium tetrachloridopalladate(II)
3. Diamminedichloridoplatinum(II)
4. Potassium tetracyanonickelate(II)
5. Pentaamminenitrito-o-cobalt(III)
6. Hexaamminecobalt(III) sulphate
7. Potassium tri-(oxalato)chromate(III)
8. Hexaammineplatinum(IV)
9. Tetrabromidocuprate(II)
10. Pentaamminenitrito-N-cobalt(III).

Answer:

1. [Zn(OH]^2-
2. K₂[PdCl₄]
3. [Pt(NH₃)₂Cl₂]
4. K₂[Ni(CN)₄]
5. [CO(NH₃)₅(ONO)]²⁺
6. [CO(NH₃)₆]₂ (SO₄)₃
7. K₃[Cr(C₂O₄)₃]
8. [Pt(NH₃)₆]⁴⁺
9. [Cu(Br)₄]^2-
10. [CO(NH₃)₅(NO₂)]²⁺.

Question 7.
Using IUPAC norms write the systematic names of the following :

1. [CO(NH₃)₆]CI₃
2. [Pt(NH₃)₂Cl(NH₂CH₃)]Cl
3. [Ti(H₂O)₆]³⁺
4. [CO(NH₃)₄Cl(NO₂)]Cl
5. [Mn(H₂O)₆]²⁺
6. [NiCl₄]^2-
7. [Ni(NH₃)₆]Cl₂
8. [Co(en)₃]³⁺
9. [Ni(CO)₄].

Answer:

1. Hexaamminecobalt(III) chloride
2. Diammine chlorido (methylamine) platinum (II) chloride
3. Hexaquatitanium(III) ion
4. Tetraamminechloridonitrito-N-Cobalt(III) chloride
5. Hexaquamanganese(II) ion
6. Tetrachloridonickelate(II) ion
7. Hexaamminenickel(II) chloride
8. tris (ethane-1, 2-diammine) cobalt(III) ion
9. Tetracarbonylnickel (0).

Question 8.
List various types of isomerism possible for co-ordination compounds, giving an example of each.
Answer:
Isomerism in Coordination Compounds: Two or more compounds having the same molecular formula but a different arrangement of atoms are-called isomers.
Isomerism in Coordination Compounds is given below:

1. Structural Isomerism:
This type of isomerism arises due to the difference in structures of coordination compounds. It is further subdivided into four different types as explained below :

(a) Ionization isomerism: This type of isomerism is shown by such compounds which have the same composition but liberate different ions in solution.
Example : [Co(NH₃)₅Br]SO₄ (dark violet). It liberates SO^2-₄ ions in solution, whereas [Co(NH₃)₅ SO₄] (red) liberates Br- ions in solution. Both of these are ionization isomers.

(b) Co-ordination isomerism: This type of isomerism is shown by such a compound which contains both cationic and anionic species and it arises from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex.
For example, [Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆]
hexaamminecobalt (III) hexacyano chromate (III) and hexaamminechromium (III) hexacyano- cobalt (III).

(C) Linkage isomerism: Linkage isomerism occurs when different atoms of the ligand are attached to the central metal ion. The structure obtained are called linkage isomers. Such type of ligands are called ambidentate ligands.

In structure, I Ni²⁺ is linked by thiocyanate sulphur and in structure II it is linked by a nitrogen atom.
Ionisation Isomerism: This type of isomerism is shown by such compounds which have the same composition but liberate different ions in solution.
[Co (NH₃)₅ Br]SO₄ – It liberates SO^2-₄ ions.
[Co(NH₃)₅SO₄]Br – It liberates Br^– ions.

(d) Hydrate isomerism: This type of isomerism is shown by such compounds which differ in the number of molecules of water inside the coordination sphere.

Example: The three hydrate isomers are as follows:

1. [Cr(H₂O)₆]Cl₃ (violet)
2. [Cr(H₂O)5Cl]Cl₂.H₂O (green)
3. [Cr(H₂O)₄Cl₂]Cl.2H₂O (darkgreen).

2. Stereoisomerism
Two compounds are called stereoisomers when they contain the same ligand in their co-ordination sphere but differ in their spatial arrangement. Stereoisomerism is further classified as geometrical and optical isomerism.

(a) Geometrical isomerism: When the ligands are situated at a different position around the metal it gives rise to geometrical isomerism. The isomer in which similar groups are in adjacent position (making an angle 90° with metal ion) is called c/s-isomer. The isomer in which the similar group occupies the opposite position (making an angle of 180° with metal ion) is called trans-isomer.

This type of isomerism is observed in square planar (CN = 4) and octahedral (CN = 6) types of complexes.

(b) Optical Isomerism: Optical isomers are sensitive to plane polarised light. When light passed through the solution of optically active substances these rotate the light-plane in a clockwise or anticlockwise direction.

This type of isomerism is exhibited by the substances which are a mirror images of each other but cannot be superimposed. Optical isomerism is shown by the substances which have a minimum one asymmetric carbon atom.

The compound which rotates plane-polarized light in left side are called l-isomers and which rotates in right side called d-isomers. These types of isomerism are generally found in the tetrahedral complex [-CN = 4] and octahedral complex [-CN = 6].

(i) Tetrahedral complex : [As (CH₃) (C₂H₅) (S) (C₆H₄COO)]²⁺

(ii) Octahedral complex : [CO(en)₂]³⁺

Question 9.
How many geometrical isomers are possible in the following co-ordination entities :

1. [Cr(C₂O₄)₃]^3-
2. [CO(NH₃)₃Cl₃].

Answer:

1. Nil
2. Two (fac and mer).

Question 10.
Draw the structures of optical isomers of:
(i) [Cr(C₂O₄)₃]^3-
(ii) [PtCl₂(en)₂]²⁺
(iii) [Cr(NH₃)₂Cl₂(en)]⁺.
Answer:

Question 11.
Draw all the isomers (geometrical and optical) of:
(i) [CoCl₂(en)₂]⁺
(ii) [Co(NH₃)Cl(en)₂]²⁺
(iii) [Co(NH₃)₂Cl₂(en)]⁺.
Answer:

cis-isomer will show optical isomer [see question (ii)]



cis isomer shows optical isomers

Question 12.
Write all the geometrical isomers of [Pt(NH3)(Br) (Cl)(py)] and how many of these will exhibit optical isomers?
Answer:
Three isomers:

These type of isomer do not show optical isomerism. Optical isomerism in square planar complexes occurs only with unsymmetrical chelating ligand.

Question 13.
Aqueous copper sulphate solution (blue in colour) gives:
(i) A green precipitate with aqueous potassium fluoride, and
(ii) A bright green solution with aqueous potassium chloride. Explain these experimental results.
Answer:
Aqueous copper sulphate exists as [Cu(H₂O)₄]SO₄ which has blue colour due to [CU(H₂O)₄]²⁺ ions.

• When KF is added, the weak water ligand are replaced by F^– ligand and forms [CUF₄]²⁺ ions which is green precipitate.
  [CU(H₂O)₄]SO₄ + 4F^– → [CUF₄]^2- + 4H₂O.
• When KCl is added, the weak water ligand are replaced by CT ion forming [CuCl₄]^2- which has bright green colour.
  [CU(H₂O)₄C1] SO₄ + 4C1^– → [CUC1₄]^2- + 4H₂O.

Question 14.
What is the coordination entity formed when an excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H₂S_(g) is passed through this solution?
Answer:
CuSO_4(aq) + 4KCN _(aq) → K₂Cu (CN)₄]_(aq) + K₂SO₄
i.e. [Cu (H₂O)₄]²⁺ + 4CN^– → [Cu (CN)₄]^2- + 4H₂O
Thus, the coordination entity formed in the process is K₂ [Cu (CN)₄]. K₂ [Cu (CN)₄] is a very stable complex which does not ionize to give Cu²⁺ ions when added to water. Hence Cu²⁺ ions are not precipitated when H₂S_(g) is passed through the solution.

Question 15.
Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory :

1. [Fe(CN)₆]^4-
2. [FeF₆]^3-
3. [Co(C₂O₄)₃]^3-
4. [CoF₆]^3-

Answer:

1. [Fe(CN)₆]^4- : d₂sp³, octahedral, diamagnetic (Refer NCERT Text-book).
2. [FeF₆]^3- : sp³d², octahedral, paramagnetic (Refer NCERT Text-book) Similar to [CoF₆],
3. [Co(C₂O)₄]^3- : d²sp³, octahedral, paramagnetic (similar to [Fe(CN)₆]^4-).
4. [CoF₆]^3- : sp³ d², octahedral, paramagnetic (Refer NCERT Text-book).

Question 16.
Draw a figure to show the splitting of 4-orbitals in an octahedral crystal field.
Answer:
CFSE can be calculated as :
CFSE = [-0.4x + 0.6y]∆₀
Where, ∆₀ = CFSE in octahedral complex
x = Number of electrons in t_2g orbitals
y = Number of unpaired electrons in e_g orbitals.

Question 17.
What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Answer:
A spectrochemical series is the arrangement of common ligands in the increasing order of their crystal field splitting energy (CSFE) values the ligands present on the RHS of the series are strong fields ligands while those on the LHS are weak field ligands. Also, strong-field ligands cause higher splitting in the d orbitals than weak field ligands.
I^– < Br^– < S^2- < SCN^–< Cl^– < N₃ < F^– < OH^– < C₂O₄^2- ~ H₂O < NCS^– ~ H^– < CN^– < NH₃, < en ~ SO₃^2-2- < NO₂^– < phen < CO

Question 18.
What is crystal field splitting energy (CFSE)? How does the magnitude of ∆0 decide the actual configuration of d-orbitals in a co-ordination entity?
Answer:
When ligands approach a transition metal ion, the 4-orbital split into two sets, with lower energy and higher energy. The difference of energy between two sets of orbitals is called crystal field splitting energy (CFSE) i.e. ∆₀ for the octahedral field.
For example, the d⁴ system has the following configuration depending upon ∆₀.

• If ∆₀ < P (pairing energy), the 4^th e^– entres one of the e_g orbital giving configuration. \(t_{2 \mathrm{g}}^{4} e_{g 0}\)
• If ∆₀ > P, the 4^th e^– paired up in one of the e_g orbital giving \(t_{2 \mathrm{g}}^{3} e_{g 1}\) configuration.

Question 19.
[Cr(NH₃)₆]³⁺ is paramagnetic while |Ni(CN)₄]^2- is diamagnetic. Explain, why?
Answer:
[Cr(NH₃)₆]³⁺ ion is paramagnetic whereas [Ni(CN₄)]²⁺ is diamagnetic.

Six NH₃ ligands donate electrons and form d² sp³ hybrid orbitals. Since there are three unpaired electrons in 3d subshell. Thus, [Cr(NH₃)₆]³⁺ ion is paramagnetic.

CN^– ion is a strong ligand.
Therefore, it forces the electrons to pair up in 3d subshell and vacant them which participate in hybridization. One 4.9, two 4p, and one 4d subshell form dsp² hybrid orbital.

Question 20.
A solution of [Ni(H₂O)₆]²⁺ is green but a solution of [Ni(CN)₄]^2- is colourless. Explain.
Answer:
H₂O is a weak ligand [Ni(H₂O)₆]^2- is an outer orbital complex. The complex has two unpaired electrons. The d-d transition is possible. It absorbs red light and complementary green light is emitted.

CN^– is a strong ligand. The unpaired electrons are paired up. The central atoms undergo dsp² hybridization. The square planar complex is formed. No unpaired electrons are present i. e. d-d transition is not possible, hence the complex is colourless.

Question 21.
[Fe(CN)₆]^4- and [Fe(H₂O)₆]²⁺ are of different colours in dilute solutions, why?
Answer:
The colour of a particular coordination compound depends on the magnitude of crystal field splitting energy ∆. This CFSE in turn depends on the nature of the ligand. In case of [Fe (CN)₆)^4- and [Fe (H₂O)₆]^2- the colour differs because there is a difference in the CFSE. Now, CN^– is a strong field ligand having a higher CFSE value as compared to the CFSE value of water. This means that the absorption of energy for the extra d- d transition also differed. Hence, the transmitted colour also different.

Question 22.
Discuss the nature of bonding in metal carbonyls.
Answer:
The metal-carbon bonds in metal carbonyls have both s and p characters. M-C σ-bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal. M-C π-bond is formed by the donation of a pair of electrons from the filled metal d orbital into the vacant anti-bonding π^* orbital of carbon monoxide. This is also known as back bonding of the carbonyl group. The metal to ligand bonding creates a synergic effect which strengthens the bond between CO and the metal. This synergic effect strengthens the bond between CO and the metal.

Question 23.
Give the oxidation state, rf-orbital occupation and co-ordination number of the central metal ion in the following complexes :
(i) K₃[CO(C₂O₄)₃]
(ii) cis-[Cr(en)₂Cl₂]Cl
(iii) (NH₄)₂[CoF₄]
(iv) [Mn(H₂O)₆]SO₄.
Answer:

Question 24.
Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration, and coordination number. Also, give stereochemistry and magnetic moment of the complex:

1. K[Cr(H₂O)₂(C₂O₄)₂].3H₂O
2. [Co(NH₃)₅Cl]Cl₂
3. CrCl₃(py)₃
4. Cs[FeCl₄]
5. K₄[Mn(CN)₆].

Answer:

1. Potassium diaquadioxalatochromate (III) hydrate.
2. Pentaamminechloridocobalt(III) chloride
3. Trichloridotripyridinechromium (III)
4. Caesium tetrachloroferrate (III)
5. Potassium hexacyanomanganate(II).

Question 25.
What is meant by the stability of a coordination compound in solution? State the factors which govern the stability of complexes.
Answer:
The stability of a complex in a solution refers to the degree of association between the two species involved in a state of equilibrium. Stability can be expressed quantitatively in terms of stability constant or formation constant.

For this reaction, the greater the value of the stability’ constant, the greater is the proportion of ML₃ in the solution. Stability can be of two types:

(a) Thermodynamic stability : The extent to which the complex will be formed or will be transformed into another species at the point of equilibrium is determined by thermodynamic stability.

(b) Kinetic stability: This helps in determining the speed with which the transformation will occur to attain the state of equilibrium.

Factors affect complex compounds :

1. Nature of metal ion: The term nature means change density on the central metal ion, the greater the charge density more will be the stability of the complex.
2. Size of metal ion:
   
3. Nature of ligand (a): If the ligand is small in size and its charge is more, the complex will be more stable.
4. Greater the basicity of ligand, greater the stability of the complex.
5. Chelate ligand: Complex formed by chelate ligand is ten times more stable than ordinary ligand.
6. Nature of solvent: If metal and ligand are made to react in a solvent of low dielectric constant and low dipole moment, the stability of the complex increases.

Question 26.
What is meant by the chelate effect? Give an example.
Answer:
When a ligand attaches to the metal ion in a manner that forms a ring, then the metal-ligand association is found to be more stable. In other words, we can say that com-plexes containing chelate rings are more stable than complexes without rings. This is known as the chelate effect for example :

Question 27.
Discuss briefly giving an example in each case the role of coordination compounds in:
1. Biological system
2. Medicinal chemistry
3. Analytical chemistry
4. Extraction/metallurgy of metals.
Answer:
1. Biological system: Zinc and iron deficiency of plants is very common. Plants require iron and zinc for healthy growth. When applied to the soil in the form of their soluble salts, these increase metals get precipitated in the form of their hydroxides at pH 7-8 the soil.

To prevent it and their rapid and absorption by the plants, their soluble complexes like iron-EDTA and zinc-EDTA are used.

2. Medicinal chemistry: Many coordination complexes are used in medicines:
(i) Vitamin B₁₂ which is used to prevent anemia is a complex compound of cobalt (I), an unusual oxidation state of cobalt.
(ii) The platinum complex cis-[Pt(NH₃)₂Cl₂] known as cisplatin is used in the treatment of cancer. It also inhibits the growth of fiimours. This is known as Chemotherapy.
(iii) The excess of copper and iron are removed by the chelating ligands D-penicillamine and desferrioxamine-B via the formation of coordination compounds.
(iv) The chelate complex of calcium and EDTA is used in the treatment of lead poi¬soning. Inside the body calcium in the complex is replaced by lead. The more soluble lead- EDTA complex is eliminated in urine.

3. Analytical chemistry: Co-ordination compounds find their applications in both qualitative and quantitative methods of analysis.
(a) In qualitative analysis: The formation of a complex substance by using suitable reagents is very effectively used in the separation and detection of cations in qualitative analysis. For example :

(i) The separation of Ag⁺ from \(\mathrm{Hg}_{2}^{+2}\) in the first group of analysis is based on the fact that, while silver chloride is soluble in aqueous ammonia and Hg₂Cl₂ forms back. The insoluble complexes can be separated by filtration.

(ii) Similarly, separation of Cd⁺² from Cu⁺² could also be achieved by treating the solution of Cu⁺² and Cd⁺² with excess of CN^– reaction takes place.

Now, when H₂S is passed in the solution, ionic product does not exceed the K_sp of Cl₂S due to the formation of a highly stable complex of copper and hence only CdS gets precipitated.

(b) In quantitive analysis : (i) Gravimetric estimation of Ni⁺² is carried out by precipitating Ni+2 as red nickel-dimethyl oxime complex in thr presence of ammonia.

(ii) EDTA is a versatile complexing agent. It is used in the complexometric determina¬tion of several metal ions such as Ca⁺², Mg⁺², Zn⁺², Mn⁺², Fe⁺², Co⁺², Ni⁺² etc.

4. Extraction/metallurgy of metals: The extraction of silver and gold is done by the formation of cyanide complexes. The crushed ore is treated with an aqueous cyanide solution in the presence of air to dissolve gold by forming the soluble complex ion [Au(CN)₂]^–.

The complex is separated and then treated with an electropositive metal, such as zinc to recover gold.

The extraction of nickel by means of Mond’s process is based on the preparation of volatile nickel carbonyl. Carbon monoxide gas is passed over impure nickel metal at about 70 °C to form the volatile nickel carbonyl. Distillation of this at 200°C yields nickel.

Question 28.
How many ions are produced from the complex [Co(NH₃)₆CI₂] in solution :
(i) 6
(ii) 4
(iii) 3
(iv) 2.
Answer:

Question 29.
Amongst the following ions which one has the highest magnetic moment value:
(i) [Cr(H₂O)₆]³⁺
(ii) [Fe(H₂O)₆]²⁺
(iii) [Zn(H₂O)₆]²⁺
Answer:
(ii) [Fe(H₂O)₆]²⁺ has the highest magnetic moment because the Fe²⁺ ion has maximum Cl unpaired electrons.

Question 30.
The oxidation number of cobalt in K[Co(CO)₄] is :
(i) +1
(ii) +3
(iii) -1
(iv) -3.
Answer:
(c) K[Co(CO)₄]
1 + x + 0 = 0
x = -1.

Question 31.
Amongst the following, the most stable complex is :
(i) [Fe(H₂O)₆]³⁺
(ii) [Fe(NH₃)₆]³⁺
(iii) [Fe(C₂O₄)₃]^3-
(iv) [FeCl₆]^3-.
Answer:
(iii) [Fe(C₂O₄)₃]^3- is most stable because it is chelate ligand complex.

Question 32.
What will be the correct order for the wavelengths of absorption in the visible region for the following:
[Ni(NO₂)₆]^4-, [Ni(NH₃)₆]²⁺, [Ni(H₂O)₆]²⁺.
Answer:
In all the three complexes, the central metal ion is the same hence, the increasing field strengths of the ligand from the spectrochemical series are in the order :
H₂O < NH₃ < NO^–₂
Thus, the energies absorbed for excitation will be in the order :
[Ni(H₂O)₆]²⁺ < [Ni(NH₃)₆]²⁺ < [Ni(NO₂)₆]^4-.
As E = \(\frac{h c}{\lambda}\) , the wavelength absorbed will be in opposite order.

Coordination Compounds Other important Questions and Answers

Coordination Compounds Objective Type Questions

Choose the correct answer:

Question 1.
The correct structural formula of Zeise’s salt is :
(a) K⁺ [PtCl₃(C₂H₄)]^–
(b) K⁺ [PtCl₂(n² – C₂H₄]Cl^–
(c) K₂ [PtCl₃(n – C₂H₄)]
(d) K⁺ [PtCl₃(n² – C₂H₄)]^–
Answer:
(d) K⁺ [PtCl₃(n² – C₂H₄)]^–

Question 2.
Which of the following is not an olefinic organometallic :
(a) C₄H₄Fe(CO)₃
(b) (C₂H₄PtCl₃)₂
(c) Be(CH₂)₂
(d) K[C₂H₄PtCl₃].3H₂O.
Answer:
(c) Be(CH₂)₂

Question 3.
IUPAC name of K₃[Al(C₂O₄)₃] is :
(a) Potassium alumino oxalate
(b) Potassium trioxalato aluminate (III)
(c) Potassium aluminium (III) oxalate
(d) Potassium trioxalato aluminate (IV).
Answer:
(b) Potassium trioxalato aluminate (III)

Question 4.
AgCl is soluble in aqueous ammonia due to formation of:
(a) [Ag(NH₃)²⁺
(b) [Ag(NH₄)₂]⁺
(c) [Ag(NH₃ )₄ ]⁺
(d) [Ag(NH₃)₂]⁺.
Answer:
(d) [Ag(NH₃)₂]⁺.

Question 5.
Which of the following gives a white precipitate in aqueous solution with silver nitrate :
(a) [Cr(NH₃)₅Cl](NO₂)₂
(b) [Pt(NH₃)₂Cl₂]
(c) [Pt(CN)₂Cl₂]
(d) [Pt(NH₃)₂]Cl₂.
Answer:
(d) [Pt(NH₃)₂]Cl₂.

Question 6.
Correct nomenclature of Fe₄[Fe(CN)₆]₃ is :
(a) Ferroso ferric cyanide
(b) Ferric ferrous hexacyanate
(c) Iron (III) hexacyanoferrate (III)
(d) Hexa cyano ferrate (III-II).
Answer:
(c) Iron (III) hexacyanoferrate (III)

Question 7.
Number of geometrical isomers of [Pt(NH₃)₂Cl₂] will be :
(a) Two
(b) One
(c) Three
(d) Four.
Answer:
(a) Two

Question 8.
In co-ordination compounds, coordination number of metal is :
(a) Equal to primary valency
(b) Sum of primary and secondary valency
(c) Equal to secondary valency
(d) None of these.
Answer:
(c) Equal to secondary valency

Question 9.
In which of the following complex, oxidation state of metal is zero :
(a) [Pt(NH₃)₂Cl₂]
(b) [Cr(CO)₆]
(c) [Cr(NH₃)₃Cl₃]
(d) [Cr(CN)₂Cl₂].
Answer:
(b) [Cr(CO)₆]

Question 10.
Which of the following is not an organometallic compound :
(a) Ethyl magnesium bromide
(b) Tetra-ethyl lead
(c) Sodium ethoxide
(d) Tetra methyl aluminium.
Answer:
(c) Sodium ethoxide

Question 11.
Co-ordination number of Fe in [Fe(CN)₆]^4-, (Fe(CN)₆]^3- and [Fe(Cl)₄]^– is:
(a) 2, 2, 3
(b) 6, 6, 4
(c) 6, 3, 3
(d) 6, 4, 6.
Answer:
(b) 6, 6, 4

Question 12.
Example of dsp² hybridization is:
(a) [Fe(CN)₆]^3-
(b) [Ni(CN)₄]^2-
(c) [Zn(NH₃)₄]²⁺
(d) [FeF₆]^3-
Answer:
(b) [Ni(CN)₄]^2-

Question 13.
Which of the following complex is used as an anti-cancer agent:
(a) trans[Co(NH₃)₃Cl₃]
(b) cis[Pt(NH₃)₂Cl₂]
(c) cis-K₂[PtCl₂Br₂]
(d) Na₂CO₃.
Answer:
(b) cis[Pt(NH₃)₂Cl₂]

Question 14.
Correct order of hybridization of central atoms in NH₃.[PtCl₄]^2- PCl₅ and BCl₃ is:
(a) dsp², dsp³, sp² and sp³
(b) sp³, sp³, sp² and sp²
(c) dsp², sp², sp³ and dsp³
(d) dsp², sp³, sp² and dsp³
Answer:
(b) sp³, sp³, sp² and sp²

Question 15.
Oxidation state of Fe in Fe(CO)₅ complex is:
(a) -1
(b) +2
(c) +4
(d) Zero.
Answer:
(d) Zero.

Question 16.
Grignard reagent is:
(a) Organometallic compound
(b) Complex compound
(c) Double salt
(d) Neutral compound.
Answer:
(a) Organometallic compound

Question 17.
Structure of complex salts was proposed by:
(a) Berzelius
(b) Werner
(c) Raoult
(d) Faraday.
Answer:
(b) Werner

Question 18.
Mohrs salt is :
(a) Double salt
(b) Complex salt
(c) Neutral salt
(d) Reagent.
Answer:
(a) Double salt

Question 19.
The formula of sodium nitro prusside is :
(a) Na₄ [Fe (CN)₅ NO₅]
(b) Na₂ [Fe (CN)₅ NO]
(c) NaFe [Fe (CN)₆]
(d) Na₂ [Fe (CN)₆ NO₂].
Answer:
(b) Na₂ [Fe (CN)₅ NO]

Question 20.
Which of the following compound is different:
(a) Potassium ferrocyanide
(b) Ferrous ammonium sulphate
(c) Potassium ferricyanide
(d) Tetrammine copper (II) sulphate.
Answer:
(b) Ferrous ammonium sulphate

Question 2.
Fill in the blanks :

1. cis [Pt(NH₃)2Cl₂] complex is used as an …………………… agent.
2. Haemoglobin is a …………………… compound of iron.
3. Geometrical isomerism is found in both …………………… and …………………… complexes.
4. Oxidation state of Ni in Ni(CO)₄ is ……………………
5. Diethyl zinc is a …………………… compound.
6. The correct I.U.P.A.C. name of K₄[Fe(CN)₆] is ……………………
7. Oxidation state of Co in [Co (E.D.T.A)] is ……………………
8. The formula of dibromo chlorotriaquo chromium (III) is ……………………
9. [COF₆]^-3 is a spin …………………… complex.
10. The formula of antiknock organo-metallic substance is ……………………
11. E. D. T. A is …………………… ligand.

Answers:

1. Anti-cancer
2. Complex
3. Tetrahedral, Octahedral
4. Zero
5. Organo-metallic compound
6. Pottassium hexacyano ferrate (II)
7. +3
8. [Cr(H₂O)₃Cl Br₂]
9. High
10. Tetra ethyl lead (C₂H₅)₄Pb
11. Hexadentate.

Question 3.
Match the following :

Answer:

1. (h)
2. (g)
3. (f)
4. (e)
5. (b)
6. (d)
7. (a)
8. (c)
9. (i).

Question 4.
Answer in one word/sentence :

1. Which type of isomerism is found in [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br ?
2. Name the organo-metallic compound which is used as an antiknock compound in petrol.
3. Complexes of E.D.T.A. formed with calcium is used to remove the poisoning caused due to which metal ?
4. How is the structure of dibenzene ?
5. What type of hybridization is found in [Ni(CO)₄] ?
6. Which type of isomerism is represented by [Cr(H₂O)₅SCN]²⁺ and [Cr(H₂O)₅NCS]²⁺ ?

Answers:

1. Ionization isomerism
2. Tetra ethyl lead
3. Lead
4. Sandwich
5. sp³
6. Linkage isomerism.

Coordination Compounds Short Answer Type Questions

Question 1.
Explain double salt and complex salt. Give one-one example of each.
Answer:
Double salt: Double salts are additive compounds which are stable in the crys-tal lattice but when dissolved in water break into different compounds.
Example : Ferrous ammonium sulphate is a double salt which ionise in water as

Complex salt : The compounds in which ligand with lone-pair electron are linked with any metal atom or metal ion by coordinate bonds, are called coordination compounds. In these compounds metal ion and ligand in combined state act as complex ion and thus these compounds are also known as complex compounds.
Example : K₄[Fe(CN)₆].

Question 2.
Explain Ambidentate ligand.
Answer:
Ligands which contain more than one atom to donate electron pair to central metal atom but it donates through one atom at a time are known as ambidentate ligand.
Example : NO₂^– ion ligand can co-ordinate to central metal atom either through N or by O.

Question 3.
What do you understand by ligand ? Explain giving example.
Answer:
Ligand : Any atom, ion or molecule which can donate electron pair to central ion and forms co-ordinate bond are called ligand. Example : In K₄[Fe(CN)₆], CN is ligand.

In ligand the specific atom which donates the electron pair is known as donor atom. On the basis of number of donor atoms in a ligand, they are classified as monodentate, bidentate, tridentate, polydentate ligands. Some such ligands are given below :

Question 4.
What is complex ion ?
Answer:
Complex Ion : A complex ion is an electrically charged redical or a species, carrying either positive or negative charge, in which the central metal ion is surrounded through co-ordinate bond by a suitable number of ligands (neutral molecules or negative ions).

Example : Complex ferrocyanide ion[Fe(CN)₆]^4- is formed by the union of six CN^– ions with one Fe²⁺ ion. While writing the formula of a complex ion, the co-ordinating groups are written inside the bracket ( ), and the whole of the complex ion in a square bracket [ ]. The net charges is written on right hand top comer of the square bracket.

The square bracket is known as co-ordination sphere.

Question 5.
Give example of bidentate and hexadentate ligand.
Answer:
Bidentate ligand :

Hexadentate ligand:

Question 6.
The following are two coordination complexes.

1. [CO(NH₃)₆]Cl₃
2. Na₃ [Cr(CN)₆]

Read the following statements and identify the wrong one/s, then correct them.
(a) The IUPAC names of both end in ‘ate’.
(b) Secondary valency of both is equal to three.
(c) Both can show ionization isomerism.
(d) Oxidation number of central metal in both is +3.
(e) Both can exhibit optical isomerism.
Answer:
(a) Wrong. IUPAC name of (ii) only ends in ‘ate’.
(b) Wrong. The secondary valency of both is 6.
(c) Wrong. Both cannot exhibit optical isomerism.

Question 7.
What are organometallic compounds ? Write its two applications.
Answer:
Those compounds in which the carbon atom of organic groups are directly bonded to metal atoms are called organometallic compound. The compound of element such as boron, phosphorus, silicon, germanium and antimony with organic groups are also included in the organometallics.

Applications of organometallic compounds :

1. Tetra ethyl lead (C₂H₅)₄ Pb is used as on antiknock compound.
2. Ziegler-Natta catalysis is used in the polymerization of ethylene or other alkenes.
3. Ethyl mercuric chloride (C₂H₅HgCl) is used in agriculture as an insecticide.
4. Willkinson catalyst is used in the hydrogenation of some alkenes.

Question 8.
Explain geometrical isomerism with an example.
Answer:
Geometrical isomerism : When the ligands are situated at different position around the metal it gives rise to geometrical isomerism. The isomer in which similar groups are in adjacent position (making an angle 90° with metal ion) is called c/s-isomer. The iso-mer in which the similar group occupy the opposite position (making an angle of 180° with metal ion) is called trans-isomer.

This type of isomerism is observed in square planar (CN = 4) and octahedral (CN = 6) type of complexes.

Question 9.
Explain the principle of Werner with the example of K₄[Fe(CN)₆] complex.
Answer:
The various terms used for a co-ordination compound are shown in the following example:

(i) When it is dissolved in water it does not break into its constitutent ions K⁺, Fe²⁺, CN^– but it forms K⁺ and complex ion [Fe(CN)₆]^4-.
(ii) IUPAC name of the compound is Potassium hexacyanoferrate (II).
(iii) Ionisation of K₄Fe(CN)₆ in aqueous solution.

According to Werner’s theory : Central metal atom in the compound is Fe whose oxidation number (Primary valency) is 4 and co-ordination number (Secondary valency) is 6.

Question 10.
Explain structure of [Ni(CN)₄]^2- on the basis of valence bond theory.
Answer:
In [Ni(CN)₄]^2- ion, Nickel is present as Ni²⁺ ion and its valency shell configuration is

Co-ordination number of Ni²⁺ in this complex ion is 4 and experimentally it is found to be diamagnetic. This is possible if it does not have unpaired electron i.e., \(3 d_{z^{2}}\) gives its electron to 3d_x²-y² Now \(3 d_{z^{2}}\), 4s, 4p_x and 4p_y orbital hybridizes forming four dsp² hybrid orbital oriented along the comer of a planar square.

These hybrids accept lone pair of electron from four CN^– ion forming σ – bonds. Now there is no unpaired electron in it. Therefore, [Ni(CN)₄]^2- is diamagnetic.

Question 11.
What is effective atomic number (EAN)?
Answer:
In co-ordination compounds, sum of electrons present in central metal ion and electrons received in bond formation is called effective atomic number.
EAN = Atomic number – Electrons lost in ion formation + Electrons obtained in bond formation
In K₄[Fe(CN)₆]EAN for Fe = 26 – 2 + 12 = 36.

Question 12.
Write the I.U.P.A.C. names of the following compounds:
(i) (a) [Hgl₄]^2-
(b) [Ag (CN)₂]^–
(c) [Fe (C₅H₅)2l
(d) K [Ag (CN)₂
(ii) What is Zeise’s Salt and Ferrocene ? Explain with structure.
Answer:
(i) (a) Tetraiodomercurate (II) ion.
(b) Dicyanoargentate (I) ion.
(c) Bis (cyclopentadienyl) iron (II).
(d) Potassium dicyano argentate (I).
(ii) (a) Zeise’s salt K [PtCl₃ – n² (C₂H₄):

This salt was prepared by Danish pharmacist Zeise in 1830. it is one of the compound of transition metals which was prepared earlier. The plane of ethylene molecule and the C = C axis are perpendicular to the expected bond direction of the central atom.

(b) Ferrocene Fe (n⁵ – C₅H₅)₂: It is an orange-yellow coloured compound. Kealy and Pauson reported it in 1951. It has sandwich structure in which iron atom is in between two cyclopentadienyl rings. The planes of the rings are parallel so that all the carbon atoms are equidistant from iron atom.

Question 13.
What are chelates? Give example. Write importance of Chelate.
Answer:
Chelates is a multidentate ligand which co-ordinates with central metal atom to form ring compound, the process is called chelation.

Importance of chelate ligands :

1. In qualitative analysis, for the detection of some elements.
2. They are used in the separation of lanthanides and actinides.
3. In softening of hard water.

Question 14.
What is the difference between primary and secondary valency? Give example also.
Answer:
Primary valency is ionisable while secondary valency does not ionise. Primary valency is represented in figure by solid lines and secondary valency by broken or dotted lines.
Example : [Co (NH₃)₆] Cl₃, primary valency is 3 and secondary valency is 6.

Question 15.
What is the importance of coordination compound in extraction of metals?
Answer:
In metallurgy, extraction of metals like gold and silver is down by forming complexes. Metals are converted into soluble cyanide complexes by treating with dilute cyanide solution. These soluble complexes liberate metals when treated with electropositive metals like zinc.

Question 16.
Write I.U.P.A.C. name of the following :
(a) [CU(NH₃)₄]SO₄
(b) [Cr(H₂O)₆ ]Cl₃
(c) [Pt(NH₃)₆]Cl₄
(d) K₂[HgI₄].
Answer:
(a) Tetraammine copper (II) sulphate
(b) Hexaaquo chromium (III) chloride
(c) Hexaammine platinum (IV) chloride
(d) Potassium tetraiodomercurate (II).

Question 17.
Explain differences between Double salt and Complex compound.
Answer:
Differences between Double salt and Complex compound:

Double salt

1. Addition compound in solid-state but ionize in components when dissolved in the water.
2. These dissociate into constituent ions in an aqueous solution.
3. The physical and chemical properties are similar to the properties of constituents.
4. The nature of bond is ionic or electro-valent.

Complex compound

1. Stable in solid or in solution may be partialy ionize in to complex ion.
2. When dissolved in water form a complex ion.
3. Physical and chemical properties of complex compounds are quite different from constituents.
4. In complex compound, the bond present between the central metal ion and ligand is always co-ordinate.

Question 18.
Write the chemical formulae of the following :
(a) Trinitritotriamminecobalt (III)
(b) Tris (ethylenediamine) Chromium (III) chloride.
(c) Pentacarbonyliron (0)
(d) Tetrachloroplatinate (II) ion.
Answer:
(a) [Co(NH₃)₃(ONO)₃]
(b) [Cr (en)₃]Cl₃
(c) [Fe (CO)₅]
(d) [Pt Cl₄]^2-

Question 19.
Write IUPAC name of the following :

1. K₄[Ni(CN)₄]
2. H₂[CUCl₄]
3. [Ag(NH₃)₂]Cl
4. [Ni(CO)₄].

Answer:

1. Potassium tetracyanonickelate (0)
2. Hydrogen tetrachloro cuprate (II)
3. Diammine silver (I) chloride.
4. Tetra carbonyl nickel (0).

Question 20.
Explain Linkage and Ionisation isomerism with example.
Answer:
Linkage isomerism : Linkage isomerism occurs when different atoms of the ligand are attached to the central metal ion. The structure obtained are called linkage isomers. Such type of ligands are called ambidentate ligand.

In structure I Ni²⁺ is linked by thiocyanate sulphur and in structure II it is linked by nitrogen atom.
Ionisation Isomerism : This type of isomerism is shown by such compounds which have same composition but liberate different ions in solution.
[Co (NH₃)₅ Br]SO₄ – It liberates SO^2-₄ ions.
[Co(NH₃)₅SO₄]Br – It liberates Br^– ions.

Question 21.
(i) Write I.U.P.A.C. name of

(ii) Write the ligands and co-ordination number of the complex [Cr (NH₃)₄(ONO) Cl]NO₃.
(iii) Write chemical formula of carbonate pentaammine cobalt (III) chloride.
Answer:
(i) µ -amido- µ -hydroxobis-(tetraammine cobalt) (III) ion.
(ii) Ligands (a) NH₃, (b) ONO, (c) Cl. Thus, number of ligands are 3 and co-ordination number is 6.
(iii) [CO(NH₃)₅CO₃]Cl.

Question 22.
Write IUPAC names of the following complex compounds.

1. NH₄[Cr(NH₃)₂ (NCS)₄]
2. K₂[Pt Cl₆]
3. [CoCI(en)₂NH₃]⁺⁺,
4. K[Pt(NH₃)Cl₅]
5. [Fe(CO)₅].

Answer:

1. Ammonium tetraisothiocyanatodiamminechromate (III)
2. Potassium hexachloroplatinate (IV)
3. Chlorobis-(ethylenediamine) amminecobalt (III).
4. Potassium pentachioroammineplatinate (IV).
5. Pentacarbonyl iron.

Question 23.
Write IUPAC name of the following :

1. Li(AlH₄)
2. [Cr(NH₃)₆](NO₃)₃
3. (Cr(H₂O)₆]Cl₃
4. K₄[Fe(CN)₆]

Answer:

1. Lithium tetrahydridoaluminate (III)
2. Hexaammine chromium (III) nitrate
3. Hexaaquoch’romium (III) chloride
4. Potassium hexacyanoferrate (II).

Question 24.
Explain optical isomerism in co-ordination compounds.
Answer:
Optical isomerism : This type of isomerism is observed in such similar compounds which are the mirror images of each other and are non-superimposable. They rotate the path of plane polarized light to the left (l) or to the right (d).

Question 25.
Describe important applications of organometallic compounds.
Answer:
Important applications of organometallic compounds are :

1. In medicine : Antiseptics like mercurochrome, mercury hydrix tincture are organometallic compounds of mercury.
2. In agriculture: Organometallic compounds of mercury like ethyl mercury chloride is used for treatment of seeds.
3. Antiknock compound : Tetraethyl lead (TEL) is an important antiknock compound.
4. In industries : Organolithium and organoaluminium compounds are used as catalyst for the manufacture of dyes and chemicals.
5. In chemical synthesis: Grignards reagent and organolithium compounds are useful in synthesis of organic compound.
6. Catalyst: Organometallic compounds of transition metals acts as heterogeneous catalyst, e.g., Wilkinson’s catalyst -(Ph₃P)₃ RhCl is used in hydrogenation of double bonds.

Question 26.
Write IUPAC names of the following :

1. [Cr(H₂O)₆]Cl₃
2. [Ag (NH₃)₂]CI
3. H₂[CuCl₄]
4. [CO(NH₃)₆]Cl₃
5. K₂[PtCl₆]
6. [Pt Cl₄(NH₃)₂]

Answer:

1. Hexaaquachromium (III) chloride.
2. Diammine silver (I) chloride.
3. Hydrogen tetrachlorocuprate (II).
4. Hexammine cobalt (III) chloride.
5. Potassium hexa cloroplatinum (IV).
6. Diammine tetrachloroplatinum (IV).

Question 27.
Write IUPAC name of the following co-ordination compounds :

1. K[Ag(CN)₂]
2. K₄[Fe(CN)₆]
3. [Ag(NH₃)₂]Cl
4. [Cr(NH₃)₆]Cl₃.

Answer:

1. Potassium dicyanoargentate (I).
2. Potassium hexacyanoferrate (II).
3. Diamminesilver (I) chloride.
4. Hexamminechromium (III) chloride.

Question 28.
Explain ionization isomerism and hydrate isomerism in coordination complexes giving suitable example.
Answer:
Ionization isomerism : This type of isomerism is shown by such compounds which have same composition but liberate different ions in solution.
Example : [Co(NH₃)₅Br]SO₄ (dark violet). It liberates SO^2-₄ ions in solution, whereas [Co(NH₃)₅ SO₄] (red) liberates Br^– ions in solution. Both of these are ionization isomers.
Hydrate isomerism : This type of isomerism is shown by such compounds which differ in the number of molecules of water inside the coordination sphere.

Example : The three hydrate isomers of are as follows :

1. [Cr(H₂O)₆]Cl₃ (violet)
2. [Cr(H₂O)5Cl]Cl₂.H₂O (green)
3. [Cr(H₂O)₄Cl₂]Cl.2H₂O (darkgreen).

Question 29.
Write chemical formulae of the following co-ordination compounds :
(a) Hexammine cobalt (III) chloride
(b) Tetrammine copper (II) sulphate
(c) Tetrammine platinum (II) chloride
(d) Potassiumhexacyanoferrate (II).
Answer:
(a) [CO(NH₃)₆]Cl₃
(b) [Cu(NH₃)₄]SO₄
(c) [Pt(NH₃)₄]Cl₂
(d) K₄[Fe^II(CN)₆].

Question 30.
Write chemical formulae of the following co-ordination compounds :
(a) Chloropentammine cobalt (II) chloride
(b) Potassiumdicyanoargentate (I)
(c) Hexaquachromium (III) chloride
(d) Tetracyano nickelate (I)
Answer:
(a) [CO(NH₃)5Cl]Cl₂
(b) K[Ag(CN)₂]
(c) [Cr(H₂O)₆]Cl₃
(d) [Ni(CN)₄]^2-.

Question 31.
Write chemical formulae of co-ordination compounds :
(a) Hexammineplatinum (IV) chloride
(b) Potassiumhexacynoferrate (III)
(c) Dichlorodiammine platinum (II)
(d) Sodium pentacyanonitrosyl ferrate (III)
Answer:
(a) [Pt(NH₃)₆]Cl₄
(b) K₃[Fe(CN)₆]
(c) [Pt(NH₃)2Cl₂]
(d) Na₂[Fe(CN)₅NO]

Question 32.
Write IUPAC name of the following :
(a) [Pt(NH₃)₂Cl₂]
(b) K₃[Fe(CN)₆]
(c) [CO(NH₃)₆]Cl₃
(d) Pt[(NH₃)₆]Cl₄
(e) CuCl^2-₄.
Answer:
(a) Diachlorodiammine platinum (II)
(b) Potassium hexacyanoferrate (III)
(c) Hexammine cobalt (III) chloride
(d) Hexammine platinum (IV) chloride
(e) Tetrachlorocuprate (II).

Question 33.
Write IUPAC name of the following :
(a) K₄[Fe(CN)₆]
(b) [Cr(NH₃)₆(NO₂)₃]
(c) [Ni(CN)₃]CI₃
(d) K₂[Pt(Cl)₆]
(e) Ni (CO)₄ U.
Answer:
(a) Potassium hexacyanoferrate (II)
(b) Trinitrohexammine chromium (III)
(c) Tricyanonickel (III) chloride
(d) Potassiumhexachloro platinate (IV)
(e) Tetracarbonylnickel (0).

Question 34.
[NiCI₄]^-2 is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Why ?
Answer:
In [NiCl₄]^-2, Ni is +2 oxidation state with the configuration 4s⁰3d⁸. Cl^– is weak ligand. It cannot pair up electrons in 3d- orbitals. Hence, it is paramagnetic. In [Ni(CO)₄], Ni is in zero oxidation state with the configuration 4s² 3d⁸. In the presence of CO ligand the 45 electron shift to 3d to pair up 3d electrons. Thus, there is no unpaired electron present. Hence, it is diamagnetic.

Question 35.
How will you differentiate [Co(NH₃)₅Br]SO₄ between and [Co(NH₃)₅SO₄]Br ?
Answer:
Both these complexes are ionisation isomers. First will give white precipitate (BaSO₄) with BaCl₂, second will give no precipitate with BaCl₂. Similarly, second complex [Co(NH₃)₅SO₄]Br will give yellow precipitate with AgNO₃ whereas first will not give any precipitate.

Coordination Compounds Long Answer Type Questions

Question 1.
Find the oxidation no. of central metal in the following :
(a) [Pt(NH₃)₃Cl₃]^–
(b) [Zn(H₂O)₃OH]⁺
(c) Na₄[Ni(CN)₄]
(d) K₂[Zn(OH)₄]
Answer:

Question 2.
Give methods of preparation of following (one method for each):
(i) Tetrabutyl tin
(ii) Tetraethyl lead
(iii) n-butyl lithium
(iv) Ferrocene
(v) Nickel tetracarbonyl
(vi) Zeise’s salt.
Answer:
(i) Tetra butyl Tin:

(ii) Tetraethyl lead:

(iii) n-butyl lithium:

(iv) Ferrocene:

(v) Nickel tetracarbonyl:

(vi) Zeise’s salt:

Question 3.
Explain the structure of [Ni(CO)₄] on the basis of valence bond theory.
Answer:
Structure of [Ni(CO)₄]: In nickel tetracarbonyl oxidation state of nickel atom is zero (0). Electronic configuration of Ni is 4s² 3d⁸ or 3d¹⁰. As a result of sp³ hybridization four sp³ orbitals are formed in tetrahedral form which are vacant. Four CO molecule join with them, as a result of which tetrahedral nickel tetracarbonyl molecule is formed.

Question 4.
Explain the structure of [Zn(NH₃)₄]²⁺ on the basis of valence bond theory.
Answer:
Structure of [Zn(NH₃)₄]²⁺ Tetraammine zinc (II) ion : Outer electronic configuration of zinc (Z = 30) in ground state is 3d¹⁰ 4s². In this complex zinc is in + 2 oxidation state with the outer electronic configuration of 3d¹⁰.
The 3d orbital being completely filled does not take part in hybridization. The vacant 4s and 4p orbitals hybridize and form four hybridized sp3 rorbitais directed

towards the four comers of a tetrahedron. The four lone pairs from four NH₃ overlap with these sp³ orbitals and forms four cr-bonds. The compound is diamagnetic as it does not contain unpaired electrons.

Question 5.
Explain optical isomerism with the example of tetrahedral and octahedral co-ordinate compounds.
Answer:
Optical Isomerism : Optical isomers are sensitive to plane polarised light. When light passed through the solution of optically active substances these rotates the light-plane in clockwise or anticlockwise direction.

This type of isomerism is exhibited by the substances which are mirror image of each other but cannot be superimposed. Optical isomerism is shown by the substances which have minimum one assymmetric carbon atom.

The compound which rotate plane polarized light in left side are called l-isomers and which rotate in right side called d-isomers. These types of isomerism are generally found in tetrahedral complex [-CN = 4] and octahedral complex [-CN = 6].

(i) Tetrahedral complex : [As (CH₃) (C₂H₅) (S) (C₆H₄COO)]²⁺

(ii) Octahedral complex : [CO(en)₂]³⁺

Question 6.
Write the application of complex compounds in the extraction of metals and detection of metal ions.
Answers:
(i) In extraction of metals: Silver and gold ores are treated with NaCN solution and form soluble complex.
(a) Silver glance (Ag₂S) dissolves in NaCN solution and forms soluble complex Na- [Ag(CN)₂] from which Ag is precipitated by adding Zn dust.
Ag₂S + 4NaCN → 2Na[Ag(CN)₂] + Na₂S
2Na [Ag(CN)₂ ] + Zn → Na₂ [Zn (CN)₄] + 2 Ag ↓

(b) Gold is found in native state. Its powdered ore is kept in NaCN or KCN solution for 12 to 24 hours. Au dissolves by forming soluble complex K[Au(CN)₂], Now zinc sharings are added to precipitate Au.
4AU + 8KCN + 2H₂O + O₂ (from air) → 4K[Au(CN)₂] + 4KOH
2K[AU(CN)₂] + Zn → K₂ [Zn(CN)₄] + 2Au ↓

(ii) In the detection of metal ion : In the qualitative analysis of Ni²⁺ a scarlet pink colour complex is obtained with (D.M.G.) dimethyl glyoxine

Question 7.
Explain crystal field theory.
Answer:
This theory was proposed by Bethe and Von Black. According to this theory bonding between central metal ion and ligand is due to pure electrostatic attraction. If ligand is anion then attraction towards cation is just like the attraction between two oppositely charged particles. If ligand is neutral molecule then the anionic end of this dipole is attracted to central positive ion. Thus, bonding between them is due to ion-ion attraction or ion dipole attraction.

In crystal field theory due to approaching ligands the d-orbitals split into different energy levels on the basis of extent of splitting (depend on metal ion and nature of ligand) structure and properties of complex can be discussed. Colour of transition metal complex is due to absorption of visible light which leads to excitation of electron from one d-orbital to the other d-orbital (d-d transition). This way this theory is easy and successfully explains maximum properties of the complexes.

MP Board Class 12th Chemistry Solutions

MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

Haloalkanes and Haloarenes NCERT Intext Exercises

Question 1.
Write structures of the following compounds :
(i) 2-Chloro-3-methylpentane
(ii) 1-Chloro-4-ethylcyclohexane
(iii) 4-tert. ButyI-3-iodoheptane
(iv) 1,4-Dibromobut-2-ene
(v) 1-Bromo-4-sec. buty1-2-methylbenzene.
Answer:

Question 2.
Why is sulphuric acid not used during the reaction of alcohols with KI?
Answer:
H₂SO₄ cannot be used along with KI in the conversion of an alcohol to an alkyl iodide as it converts KI to corresponding HI and then oxidises it to I₂.

Question 3.
Write structures of different dihalogen derivatives of propane.
Answer:

Question 4.
Among the isomeric alkanes of molecular formula C₅H₁₂, identify the one that on photochemical chlorination yields:
(i) A single monochloride
(ii) Three isomeric monochlorides
(iii) Four isomeric monochlorides.
Answer:

All the hydrogen are the same i.e. 1°. Therefore, the replacement of only one of them will give the same product.

Replacement of a, b and c hydrogen atoms give three isomeric monochlorides.

Replacement of a, b, c and d hydrogen atoms give four isomeric monochlorides.

Question 5.
Draw the structures of major monohalo products in each of the following reactions:

Answer:

Question 6.
Arrange each set of compounds in order of increasing boiling points.
(i) Bromomethane, Bromoform, Chloromethane, Dibromomethane.
(ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.
Answer:
(i) Chloromethane, Bromomethane, Dibromomethane, Bromoform (Boiling point increases with increase in molecular mass).
(ii) Isopropylchloride, 1-Chloropropane, 1-Chlorobutane. (Isopropylchloride being branched has lower b.p. than 1-Chloropropane).

Question 7.
Which alkyl halide from the following pairs would you expect to react more rapidly by an SN₂ mechanism? Explain your answer.

Answer:
If the leaving group is the same in different isomers of a particular formula, the reactivity of the isomers towards the S_N2 mechanism decreases with the steric hindrance.

which is 2° alkyl halide having some steric hindrance.

which is 3° alkyl halide having much more steric hindrance (than 2°).

are 2° alkyl halides. But in (II) the CH₃ group is at the C₂ atom which is closer to Br (exerts more steric hindrance) to the attacking nucleophile at the C₁ atom as compared to (I).

Question 8.
In the following pairs of halogen compounds, which compound undergoes faster S_N1 reaction:

Answer:
The reactivity of alkyl halide towards S_N1 reaction depends upon the stability of the intermediate carbocation formed as 3° > 2° > 1°

Question 9.
Identify A, B, C, D, E, R and R’ in the following :


Answer:

Since, D of D₂O gets attached to same C-atom on which – MgBr or Br was present so that

tert-Alkyl halides do not undergo Wurtz reaction. Therefore, the question is not correct. They undergo dehydrohalogenation to give alkenes. Hence,

Haloalkanes and Haloarenes NCERT Textbook Exercises

Question 1.
Name of the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl, or aryl halides:

1. (CH₃)₂CHCH(Cl)CH₃
2. CH₃CH₂CH(CH₃)CH(C₂H₅)Cl
3. CH₃CH₂C(CH₃)₂CH₂I
4. (CH₃)₃CCH₂CH(Br)C₆H₅
5. CH₃CH(CH₃)CH(Br)CH₃
6. CH₃C(C₂H₅)₂CH₂Br
7. CH₃C(Cl)(C₂H₅)CH₂CH₃
8. CH₃CH=C(Cl)CH₂CH(CH₃)₂
9. CH₃CH=CHC(Br)(CH₃)₂
10. p-ClC₆H₄CH₂CH(CH₃)₂
11. m-ClCH₂C₆H₄CH₂C(CH₃)₃
12. o-Br-C₆H₄CH(CH₃)CH₂CH₃.

Answer:

1. 2-Chloro-3-methylbutane (2° alkyl)
2. 3-Chloro-4-methylhexane (2° alkyl)
3. l-Iodo-2,2-dimethylbutane (1° alkyl)
4. l-Bromo-3,3-dimethyl-l-phenylbutane (2° benzylic)
5. 2-Bromo-3-methylbutane (2° alkyl)
6. 3-Bromomethyl-3-methylpentane (1° alkyl)
7. 3-Chloro-3-methylpentane (3° alkyl)
8. 3-Chloro-5-methylhex-2-ene (vinyl)
9. 4-Bromo-4-methylpent-2-ene (allylic)
10. 1-Chloro-4-(2′-methylpropyl) benzene (aryl) or p-Chloro isobutyl benzene
11. 1-Chloromethyl-3-(2’2′-diethylpropyl) benzene (benzylic) or /n-Neopentyl ben-zyl chloride
12. 1-Bromo-2-(l’-methylpropyl) benzene (aryl).

Question 2.
Give the IUPAC names of the following compounds :

1. CH₃CH(Cl)CH(Br)CH₃
2. CHF₂CBrCIF
3. ClCH₂C = CCH₂Br
4. (CCl₃)₃CCl
5. CH₃C(p-CIC₆H₄)2CH(Br)CH₃
6. (CH₃)₃CCH=ClC₆H₄I-p.

Answer:

1. 2-Bromo-3-chlorobutane
2. 1 -Bromo-1 -chloro-1,2,2-trifluoroethane
3. l-Bromo-4-chlorobut-2-yne
4. 1,1,1,2,3,3,3-heptachloro-2-(trichloromethyl) propane
5. 3-Bromo-2,2,-bis (4′-chlorophenyl) butane
6. 1-Chloro-l-(4′-iodophenyl)-3,3-dimethyl-but-1-ene.

Question 3.
Write the structures of the following organic halogen compounds :
(i) 2-Chloro-3-methylpentane
(ii) p-Bromochlorobenzene
(iii) 1-Chloro-4-ethylcyclohexane
(iv) 2-(2-Chlorophenyl)-1-iodooctane
(v) Perfluorobenzene
(vi) 4-tert-Butyl-3-iodoheptane
(vii) 1-Bromo-4-sec-butyl-2-methylbenzene
(viii) 1,4-Dibromobut-2-ene.
Answer:

Question 4.
Which one of the following has the highest dipole moment:
(i) CH₂Cl₂
(ii) CHCl₃
(iii) CCl₄.
Answer:

CCl₄ (iii) is a symmetrical and has a resultant zero dipole moment. In CHCl₃ (ii), the resultant of two C – Cl dipoles is opposed by the resultant of C – H and C – Cl bonds. Since the latter result is expected to be smaller than the former, therefore, CHCl₃ has a finite dipole (1.03 D) moment.
In CH₂Cl₂ (i), the resultant of two C-Cl dipole moments is reinforced by the resultant of two C-H dipoles, therefore, CH₂Cl₂ (1.62 D) has a dipole moment higher than that of CHCl₃.
Thus, CH₂Cl₂ has the highest dipole moment.

Question 5.
A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monochloro compound C₅H₉Cl in bright sunlight Identify the hydrocarbon.
Answer:

1. The hydrocarbon with molecular formula C₅H₁₀ can be either a cycloalkane or alkene.
2. Since the hydrocarbon does not react with Cl₂ in the dark, it cannot be an alkene but it must be cyclohexane.
3. Since the compound can give a single monochloro derivative, the cycloalkane is cyclopentane.

Question 6.
Write the isomers of the compound having the formula C₄H₉Br.
Answer:
The isomers of C₄H₉Br along with their common names are given below :

Question 7.
Write the equations for the preparation of 1-iodobutane from :
(i) 1-butanol
(ii) 1-chlorobutane
(iii) but-1-ene.
Answer:

Question 8.
What are ambident nucleophiles? Explain with an example.
Answer:
The nucleophiles with two nucleophilic centres are called ambident nucleophiles. Depending on the reagent and the reaction conditions, the reaction may take place predominantly at one of these centres.
Example: CN^– has two nucleophilic centres, one each at carbon and nitrogen and forms cyanides and isocyanides respectively.

Question 9.
Which compound in each of the following pairs will react faster in S_N2 reaction with OH-:
(i) CH₃Br or CH₃I
(ii) (CH₃)₃CCl or CH₃Cl.
Answer:
(i) CH₃I reacts faster than CH₃Br in S_N2 reaction with OH^– because I^– ion is a better leaving group than Br^– ion because of its large size.
(ii) CH₃Cl reacts faster than (CH₃)₃CCl because of steric hindrance in (CH₃)₃CCl.

Question 10.
Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene :
(i) 1-Bromo-1-methylcyclohexane
(ii) 2-Chloro-2-methylbutane
(iii) 2,2,3-Trimethyl-3-bromopentane.
Answer:
(i) In 1-Bromo-l-methylcyclohexane, the β hydrogens on either side of the Br atom are equivalent, therefore, only 1-alkene is formed.


Since, the alkene (A) is more substituted according to SaytzefTs rule, it is more stable and will be the major product.

Question 11.
How will you bring about the following conversions :
(i) Ethanol to but-1-yne
(ii) Ethane to bromoethene
(iii) Propene to 1-nitropropane
(iv) Toluene to benzyl alcohol
(v) Propene to propyne
(vi) Ethanol to ethyl fluoride
(vii) Bromomethane to propanone
(viii) But-l-ene to but-2-ene
(ix) 1-Chlorobutane to n-octane
(x) Benzene to biphenyl.
Answer:

Question 12.
Explain why:
(i) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride ?
(ii) Alkyl halides, though polar, are immiscible with water ?
(iii) Grignard reagents should be prepared under anhydrous conditions ?
Answer:

Due to sp² hybridization of C-atom in chlorobenzene, C-atom is more electronegative (greater s-character) whereas, cyclohexyl chloride, C-atom is sp3 hybridized i.e., less electronegative (lesser 5-character). So, polarity of C—Cl bond in chlorobenzene is less than C—Cl bond in cyclohexyl chloride. Further due to delocalisation of lone pair of electrons of Cl-atom over the benzene ring, C—Cl bond in chlorobenzene acquires partial double bond character while C—Cl bond in cyclohexyl chloride is a pure single bond. Thus, C—Cl bond in chlorobenzene is shorter than in cyclohexyl chloride. As dipole moment is a product of charge and distance, therefore, the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.

(ii) Water molecules have enough strong intermolecular H-bonding which is difficult to be broken by alkyl halides though they are polar in nature as well. Therefore, alkyl halide do not dissolve in water and form separate layers.

(iii) Grignard reagents (R-Mg-X) are readily decomposed by water to produce alkanes. That is why, they should be prepared under anhydrous conditions. Instead, ether is used as a solvent during the preparation of the Grignard reagent.

Question 13.
Give the uses of freon-12, D.D.T., Carbon tetrachloride and Iodoform.
Answer:
Uses of freon – 12:
Freon -12 (dichlorodifluoromethane, CF₂Cl₂) is commonly known as CFC. It is used as a refrigerant in refrigerators and air conditioners. It is also used in aerosol spray propellants such as body sprays, hair sprays etc. However, it damages the ozone layer. Hence its manufacture is banned.

Uses of DDT:
DDT (p, p – dichlorodiphenyltrichloroethane) is one of the best-known pesticides. It is very effective against mosquitoes and lice. However, due to its harmful effects, its manufacture is banned;
Uses of carbon tetrachlorides (CCl₄):

• It is used for manufacturing refrigerants and propellants for aerosol cans.
• It is used as feedstock in the synthesis of CFCs and other chemicals.
• It is used as a solvent in the manufacture of pharmaceutical products.
• CCl₄ was once widely used as a cleaning fluid, a degreasing agent in industries, a spot remover homes, and a fire extinguisher.

Uses of iodoform (CHI₃):
Iodoform was used as an antiseptic, but now it has been replaced by other formulations containing iodine-due to its objectionable smell. The antiseptic property of CHI₃ is only due to the liberation of free iodine when it comes in contact with the skin.

Question 14.
Write the structure of the major organic product in each of the following reactions :


Answer:

Question 15.
Write the mechanism of the following reaction :

Answer:
KCN gives CN- ion as a nucleophile in an aqueous medium which is resonance hybrid of the following :

Thus, cyanide ions is an ambident nucleophile. Therefore, it can attack the C-atom of C—Br bond in «-BuBr either through C-atom or through N-atom. Thus, two possible products are cyanides and isocyanides respectively.

But C—C bond is more stable than C—N bond, so attack occurs through C-atom and hence, cyanide is predominantly formed.

Question 16.
Arrange the compounds of each set in order of reactivity towards SN2 displacement:
(i) 2-Bromo-2-methylbutane, 1 Bromopentane, 2-Bromopentane
(ii) 1-Bromo-3-methyIbutane, 2-Bromo-2-methylbutane, 3-Bromo-2- methylbutane
(iii) 1-Bromobutane, l-Bromo-2,2-dimethyl-propane, l-Bromo-2-methylbutane, l-Bromo-3-methylbutane.
Answer:
The reactivity of S_N2 reaction depends upon steric hindrance. More the steric hindrance lesser will be the reactivity. Therefore, the reactivity of different alkyl halides towards S_N2 reaction is 1° > 2° > 3°.

Question 17.
Out of C₆H₅CH₂Cl and C₆H₅CHClC₆H₅, which is more easily hydrolysed by aqueous KOH ?
Answer:

However, under S_N2 mechanism, the reactivity depends on steric hindrance, therefore, C₆H₅CH₂Cl gets hydrolysed more easily than C₆H₅CHClC₆H₅ under S_N2 conditions.

Question 18.
p-Dichlorobenzene has higher m.p. and solubility than those of o- and m- isomers. Discuss.
Answer:
p-Dichlorobenzene is more symmetrical than o- and m- isomers. For this season, it fits more closely than o- and m-isomers in the crystal lattice. Therefore, more energy is required to break the crystal lattice of p-dichlorobenzene. As a result, p-dichlorobenzene has higher mp and lower solubility than o- and m- isomers.

Question 19.
How the following conversions can be carried out:
(i) Propene to propan-1-ol
(ii) Ethanol to but-l-yne
(iii) 1-Bromopropane to 2-bromopropane
(iv) Toluene to benzyl alcohol
(v) Benzene to 4-bromonitrobenzene
(vi) Benzyl alcohol to 2-phenylethanoic acid
(vii) Ethanol to propanenitrile
(viii) Aniline to chlorobenzene
(ix) 2-Chlorobutane to 3,4-dimethyIhexane
(x) 2-Methyl-1-propene to 2-chloro-2-methyl-propane
(xi) Ethyl chloride to propanoic acid
(xii) But-1-ene to n-butyliodide
(xiii) 2-Chloropropane to 1-propanol
(xiv) Isopropyl alcohol to iodoform
(xv) Chlorobenzene to p-nitrophenol
(xvi) 2-Bromopropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to diphenyl
(xix) ttert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenyl isocyanide.
Answer:

Question 20.
The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.
Answer:
In the presence of water, KOH dissociates completely into OH^– ions which being a strong nucleophile brings about substitution on alkyl halides and produce alcohols from alkyl halide. Further in an aqueous solution, OH^– ions are highly solvated (hydrated). This solvation reduces the basic character of OH^– ions which, therefore, fails to abstract a proton from β-carbon of alkyl chloride to form alkenes. In an alcoholic medium, (less polar than H₂O) OH- is less highly hydrated, therefore, acts as the strong base and abstracts the proton from β-carbon giving alkene as a major product (dehydrohalogenation). Moreover, the alcoholic solution contains C₂H₅O- ethoxide ion in addition to OH^– ions. Being a stronger base than OH^–, the abstract proton gives alkene.

Question 21.
Primary alkyl halide C₄H₉Br (a) is reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d). C₈H₁₈ is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.
Answer:
Two possible isomers of given 1° alkyl halide C₄H₉Br are :

According to the question, compound (a) on reaction with sodium does not give the same product produced by n-butyl bromide. So (a) cannot be (I).

Question 22.
What happens when:
(i) n-butyl chloride is treated with alcoholic KOH.
(ii) bromobenzene is treated with Mg in the presence of dry ether.
(iii) chlorobenzene is subjected to hydrolysis.
(iv) ethyl chloride is treated with aqueous KOH.
(v) methyl bromide is treated with sodium in the presence of dry ether.
(vi) methyl chloride is treated with KCN?
Answer:
(i) CH₃CH₂CH₂CH₂Cl + KOH (alc.)

Haloalkanes and Haloarenes Other Important Questions and Answers

Haloalkanes and Haloarenes Objective Type Questions

Choose the correct answer:

Question 1.
Which of the following compound gives a yellow precipitate with AgNO3 solution:
(a) KIO₃
(b) CHI₃
(c) KI
(d) CH₂I₂.
Answer:
(c) KI

Question 2.
What is formed by the reaction of Ethyl bromide with lead sodium alloy :
(a) Tetraethyl lead
(b) Tetraethyl bromide
(c) Both (a) and (b)
(d) None of these.
Answer:
(a) Tetraethyl lead

Question 3.
Reaction CH₃Br + OH → CH₃ – OH + Br- is:
(a) Electrophilic substitution
(b) Electrophilic addition
(c) Nucleophilic addition
(d) Nucleophilic substitution.
Answer:
(d) Nucleophilic substitution.

Question 4.
When acetylene added to HC1, the product formed is :
(a) CH₂ = CHCl
(b) CH₃ – CH – Cl₂
(c) Cl – CH = CHCl
(d) None of these.
Answer:
(b) CH₃ – CH – Cl₂

Question 5.
In Aryl halide, the carbon linked to the halogen atom is :
(a) sp hybridized
(b) sp hybridized
(c) sp hybridized
(d) sp³d hybridized.
Answer:
(b) sp hybridized

Question 6.
In S_N1 reaction in the first step is formed :
(a) Free radical
(b) Carbanion
(c) Carbocation
(d) Final product.
Answer:
(c) Carbocation

Question 7.
Chlorobenzene reacts with chloral and concentrated H₂SO₄ to form :
(a) PVC
(b) TNT
(c) B.H.C.
(d) DDT.
Answer:
(d) DDT.

Question 8.
Reaction CH₃OH + OH^– → CH₃OH + Br^– is :
(a) S_N1
(b) S_N2
(C) S_E-1
(d)S_E-2.
Answer:
(b) S_N2

Question 9.
The following compound reacts with silver powder to form acetylene :
(a) CH₂I₂
(b) CH₃I
(c) CHI₃
(d) Cl₄.
Answer:
(c) CHI₃

Question 10.
Pyrene is used with any one of the following for extinguishing fire :
(a) CO₂
(b) CH₂Cl₂
(c) CCl₄
(d) CH₂ = CHCl.
Answer:
(c) CCl₄

Question 11.
Which of the following compound is known by the name freon :
(a) CHCl₃
(b) CCl₄
(c) CCl₂F₂
(d) CF₄.
Answer:
(c) CCl₂F₂

Question 12.
On heating ethyl iodide with alcoholic KOH the product formed is :
(a) Ethanol
(b) Ethane
(c) Acetylene
(d) Ethylene.
Answer:
(d) Ethylene.

Question 13.
On heating C₂H₅OH with iodine and a base the product formed is :
(a) CH₃I
(b) CHI₃
(c) CH₃CHO
(d) CHCI₃.
Answer:
(b) CHI₃

Question 14.
Which of the chemical formula is of chloropicrin :
(a) CCl₃ – CHO
(b) C(NO₂)Cl₃
(c) CH₃ – C(NO₂)Cl₂
(d) CCl₃ – NH₂.
Answer:
(b) C(NO₂)Cl₃

Question 15.
Order of polarity of CH₃I, CH₃Br and CH₃Cl molecules is :
(a) CH₃Br > CH₃Cl > CH₃I
(b) CH₃I > CH₃Br > CH₃Cl
(c) CH₃CI > CH₃Br > CH₃I
(d) CH₃CI > CH₃I > CH₃Br.
Answer:
(c) CH₃CI > CH₃Br > CH₃I

Question 16.
Correct order of reactivity of Alkyl halides is :
(a) Iodide > Bromide > Chloride
(b) Iodide < Bromide < Chloride
(c) Bromide > Iodide > Chloride
(d) Bromide < Chloride < Iodide.
Answer:
(a) Iodide > Bromide > Chloride

Question 17.
What is formed on heating iodoform with silver powder :
(a) Alkane
(b) Ethylene
(c) Acetylene
(d) Isocyanide.
Answer:
(c) Acetylene

Question 18.
Raschig method is used for the manufacture of which of the following :
(a) Chlorobenzene
(b) Benzene
(c) Toluene
(d) Nitro-benzene.
Answer:
(a) Chlorobenzene

Question 2.
Fill in the blanks :

1. The formula of the harmful product formed on keeping chloroform open is ………………
2. General formula of alkyl halide is ………………
3. On heating aromatic primary amine with chloroform and alcoholic caustic potash a bad smelling gas ……………… is formed.
4. B.H.C. is an insecticide whose commercial name is ………………
5. Chloretone is a high grade ………………
6. SN1 reaction occur in ……………… steps.
7. In haloarene substitution reactions are mainly ………………
8. Formula of refrigerant Freon is ………………

Answer:

1. 1. COCl₂

1. C_nH_2n+1X
2. Phenyl isocyanide
3. Gammaxene or lindane
4. Hypnotic medicine
5. Two
6. Electrophilic
7. CCl₂F₂.

Question 3.
Match the following :

Answer:

1. (d)
2. (e)
3. (f)
4. (b)
5. (c)
6. (g)
7. (a).

Question 4.
Answer in one word/sentence :

1. On reacting Benzene with methyl chloride in presence of AlCl₃, Toluene is formed, what is the name of the reaction ?
2. Alkyl halide is polar in nature still it is insoluble in water.
3. On heating alkyl halide with sodium metal, product formed is.
4. On heating Benzene diazonium salt with cuprous halide and its corresponding acid haloarene is formed. Write name of the reaction.
5. On heating iodobenzene with copper powder at 200°C, product obtained is.
6. What is formed on treating benzene with Cl₂ in the presence of sunlight ?
7. Write laboratory method of preparation of chlorobenzene.
8. Write the name of nucleophilic substitution reaction in primary alkyl halide.

Answer:

1. Friedel-Crafts reaction
2. Due to inability to form hydrogen bond
3. Alkane
4. Sandmeyer reaction
5. Diphenyl
6. B.H.C.
7. Raschig method
8. Bimolecular nucleophilic substitution reaction.

Haloalkanes and Haloarenes Short Answer Type Questions

Question 1.
(i) Write Iodoform reaction.
(ii) Iodoform gives yellow ppt with AgNO₃ solution but chloroform doesn’t Why ?
(iii) What happens when ethyl bromide is heated with alcoholic KOH ?
Answer:
(i) Iodoform reaction : When ethyl alcohol or acetone is heated with iodine and NaOH, yellow crystals of iodoform are formed.
C₂H₅OH +4I₂ + 6NaOH → 5NaI + HCOONa + 5 H₂O + CHI₃

(ii) When iodoform is heated with AgNO₃ solution a yellow ppt. (Agl) is obtained but chloroform doesn’t give this reaction because in chloroform C-Cl bond is more stable than C-I bond in iodoform.

(iii) On boiling Ethyl bromide with alcoholic KOH ethylene is formed.

Question 2.
Explain the Sandmeyer reaction with example.
Answer:
Decomposition of diazonium salts (Sandmeyer reaction): When a diazonium salt solution is added to a solution of cuprous halide dissolved in the corresponding halogen acid, the diazo group is replaced by a halogen atom.

Question 3.
Give chemical reaction between chlorobenzene and chloral in presence of cone. H₂SO₄.
Or,
How is D.D.T. formed ? Write its one application.
Answer:
DDT (Dichlorodiphenyl trichloroethane) is formed by the condensation of one molecule of chloral with two molecules of chlorobenzene in presence of cone. H₂SO₄.

Application : It is an important pesticide.

Question 4.
What are Gem-dihalide and Vicinal-dihalide ?
Answer:
When both halogen atoms are linked to one carbon atom of hydrocarbon then it is known as Gem-dihalide. Gem means geminal i.e., same position.

When both halogen atoms are connected to two different neighbouring carbon atoms, then it is known as vicinal dihalide. Vic means vicinal which means adjacent position.

Question 5.
What is Lucas reagent? Give its application.
Answer:
Solution of ZnCl₂ in conc. HCl is known as Lucas reagent.
Application: It is used to differentiate primary, secondary and tertiary alcohols. On adding the alcohol to Lucas reagent, a tertiary alcohol reacts immediately forming a ppt. of alkyl chloride. If the ppt. appears after few minutes, then the alcohol is secondary. If no ppt. is obtained in cold the alcohol is primary.

Question 6.
Explain, Carbylamine reaction and give one application of this reaction.
Answer:
Carbylamine reaction: On heating chloroform with primary amine (e.g., aniline) and alcoholic KOH solution, phenyl isocyanide or carbylamine is formed which has a very bad smell and is poisonous.

Application: Chloroform and primary amine can be tested by this reaction.

Question 7.
What is 666 (lindane)? Explain its preparation and use in agriculture.
Answer:
It is 1, 2, 3, 4, 5, 6-Hexachlorocyclohexane. It is obtained by heating benzene with chlorine in presence of sunlight.
Preparation: It is prepared by the chlorination of benzene in the presence of ultra-violet light.

1, 2, 3, 4, 5, 6-Hexachlorocyclohexane (B.H.C.)
Uses : Benzene hexachloride is an addition compound and its 7 -isomer is called gammexane. It is an important pesticide used in agriculture. It is also called lindane or 666.

Question 8.
Explain the following reaction of chlorobenzene :
(i) Reaction with chlorine in the presence of FeCI₃ in dark
(ii) Fittig reaction.
Answer:
(i) When Chlorobenzene reacts with Cl₂ in the presence of FeCl₃ in dark o – dichlorobenzene and P – dichlorobenzene is obtained.

(ii) Fittig reaction : When Chlorobenzene is heated at 200°C with Cu powder in a sealed tube Diphenyl is formed.

When two molecules of aryl halide reacts with sodium metal in presence of dry ether, then diphenyl is formed. This reaction is known as Fittig reaction.

Question 9.
How can you obtained following compounds from chloroform ? Give equations :
(a) Methane
(b) Acetylene
(c) Carbon tetrachloride.
Answer:
(a) Chloroform reduces into methane by Zn and H₂O.

(b) When CHCl₃ heated with Ag powder it gives C₂H₂.

(c) By the chlorination of CHCl₃ in presence of sunlight CCl4 is formed.

Question 10.
Write short notes on :
(a) Hunsdiecker method and
(b) Raschig process.
Ans.
(a) Hunsdiecker method : When silver salt of a carboxylic acid is heated with bromine, in the presence of an inert solvent like CCl₄, aryl bromide is formed.

This method is called Hunsdiecker method.

(b) Raschig process : When benzene vapours mixed with air and HCl gas is passed over CuCl₂(catalyst) at 230°C, chlorobenzene is formed.

Question 11.
Write method of preparation, properties and uses of Freon.
Answer:
Freon : Dichloro, Difluoro methane.
It is formed by the action of SbF₃ with CCl₄ in presence of SbCl₅.

It has very low boiling point due to which by increasing the pressure at room temperature it can be liquefied.
It is a non-poisonous, non-combustible and inactive substance which is used as a cooling agent in the refrigerator. It is used in aerosol and foam.

Question 12.
(i) b.p. of ethyl iodide is higher than b.p. of ethyl bromide. Give reason
(ii) Explain why the m.p. of para dichlorobenzene is higher than its ortho and meta derivatives.
Answer:
(i) In alkyl halides containing same alkyl group boiling point increases with increases in atomic weights of halogen atoms. Molecular weight of ethyl iodide is more than ethyl bromide and therefore boiling point of ethyl iodide is also high.
(ii) Para derivatives of dichlorobenzene is more symmetrical than its ortho and meta derivatives therefore its m.p. is higher.

Question 13.
Give main nucleophilic substitution reaction of alkyl halides.
Answer:
Nucleophilic substitution Reactions :

1. Substitution by —OH group (Hydrolysis): Alkyl halide on reacting with water or aqueous KOH hydrolyse to form alcohol.
C₂H₅Br + KOH → C₂H₅ — OH + KBr

2. Substitution by —OR group : Alkyl halide reacts with sodium alkoxide or Ag₂O to form ether by substitution of halogen atom by —OR group.
C₂H₅Br + NaOC₂H₅ → C₂H₅ — OC₂H₅ + NaBr
2C₂ H₅I + Ag₂O → (C₂H₅)₂ O + 2AgI

3. Substitution by —CN group: Alkyl halide reacts with aqueous or alcoholic KCN to form alkyl cyanide.
C₂H₅Cl + KCN → C₂H₅CN + KCl

4. Substitution by ammonia (Hofmann method) : On heating alkyl halide with aqueous or alcoholic solution of NH₃ in a sealed tube at 100°C, a mixture of various amine is obtained.

Question 14.
Give the laboratory method for the preparation of chlorobenzene and explain its nitration and sulphonation reactions.
Answer:
Laboratory method: Chlorobenzene is prepared by direct halogenation of arene. Aryl chloride may be prepared from arenes by the action of chlorine or presence of halogen carrier like FeCl₃, FeBr₃ and AlCl₃. Iodine and iron filings can also be used as halogen carrier.

Nitration : Haloarenes react with nitrating mixture to form o-nitro and p-nitro sub-stituted haloarenes.

Sulphonation : On heating with cone. H₂SO₄, 2-Chlorobenzene sulphonic acid and 4-Chlorobenzene sulphonic acid are formed.

Question 15.
Give the chemical reactions when ethyl halide reacts with the following :
(i) Alloy of Pb-Na
(ii) Mg metal
(iii) AgNO₂
(iv) Na metal.
Answer:
(i) Reaction with lead-sodium alloy: Alkyl halides, form alkyl lead when treated with lead-sodium alloy.

Tetraethyl lead (TEL) is an antiknock compound which is added in petrol.

(ii) Reaction with magnesium : Alkyl halide forms alkyl magnesium halide i.e., Grignard reagent with magnesium in presence of dry ether as a solvent.

(iii) Reaction with AgNO₂: Mainly nitroethane is formed.
C₂H₅I + AgNO₂ → C₂H₅ NO₂ + Agl

(iv) Reaction with sodium (Wurtz reaction) : When alkyl halide is heated with sodium, in the presence of dry ether, alkane is formed.

Question 16.
Give the preparation method, properties and applications of dichloroethane. Ans. Preparation of dichloroethane: It can be prepared by replacing two hydrogen of ethane by two chlorine atoms.
(i) By the passage of Cl₂ into ethene :

(ii) By the heating of mix of ethane diol and HCl in presence of anhydrous ZnCl₂.

Properties : (a) Reaction with aqueous KOH :

(b) Reaction with alcoholic KOH :

Vinyl chloride reacts with ale. KOH and form vinyl ethyl other.
CH₂ = CHCl + HOC₂H₅ + KOH → CH₂ = CH – OC₂H₅ + KCl + H₂O
(c) Reaction with KCN :

(d) Reaction with Zn powder and methanol:

Use : (i) As a solvent, (ii) As an antiknocking fuel, (iii) Removing paints.

Question 17.
Write Frankland reaction.
Answer:
Reaction with zinc (Frankland’s reaction) : This reaction is similar to the Wurtz reaction but zinc is used in place of sodium.

Question 18.
Explain Friedel-Craft’s reaction with chemical equation.
Answer:
Friedel-Craft’s reaction (alkylation) : Alkyl halides react with benzene in presence of anhydrous aluminium chloride to give alkyl benzene.

Acetylation : Acetyl chloride reacts with benzene in presence of anhydrous aluminium chloride to give acetophenone.

Question 19.
Identify ‘A’, ‘B’ ‘C’ and ‘D’

Answer:

Question 20.
An alcohol ‘A’ on reaction with cone. H₂SO₄ gives an alkene ‘B’. ‘B’ after bromination with sodamide gives dehydrogenated compound ‘C’. ‘C’ on reaction of H₂SO₄ in presence of HgSO₄ gives ‘D’ Identify ‘A’, ‘B% ‘C% and ‘D’.
Answer:

Haloalkanes and Haloarenes Long Answer Type Questions

Question 1.
Write short notes on :
(a) Hunsdiecker method
(b) Raschig process
(c) Wurtz’s reaction
(d) Westron
(e) Frankland reaction
(f) Carbylamine reaction
(g) Iodoform reaction
(h) Fittig reaction.
Answer:
(a) Hunsdiecker method : When silver salt of a carboxylic acid is heated with bromine, in the presence of an inert solvent like CCl₄, aryl bromide is formed.

This method is called Hunsdiecker method.
(b) Raschig process : When benzene vapours mixed with air and HCl gas is passed over CuCl₂(catalyst) at 230°C, chlorobenzene is formed.

(c) Wurtz’s reaction : When alkyl halide is heated with sodium, in the presence of dry ether, alkane is formed.

(d) Westron : Symmetrical tetrachloromethane or acetylene tetrachloride are known as Westron. It can be prepared by the chlorination of acetylene.
Westron, it can be prepared by the chlorination of acetylene.

It is poisonous, non- flammable liquid, which gives westrol when it is boiled with lime.
(e) Frankland reaction : When alkyl halide is heated with zinc dust, alkane is formed.

(f) Carbylamine reaction : On heating chloroform with primary amine (e.g., aniline) and alcoholic KOH solution, phenylisocyanide or carbylamine is formed which has bad smell and is poisonous.

(g) Iodoform reaction : When ethyl alcohol or acetone is heated with iodine and NaOH, yellow crystals of iodoform are formed.
C₂H₅OH +4I₂ + 6NaOH → 5NaI + HCOONa + 5 H₂O + CHI₃

(h) Fittig reaction : When Chlorobenzene is heated at 200°C with Cu powder in a sealed tube Diphenyl is formed.

When two molecules of aryl halide reacts with sodium metal in presence of dry ether, then diphenyl is formed. This reaction is known as Fittig reaction.

Question 2.
Give the laboratory method for the preparation of chloroform. Describe the formation of chloroform by ethanol with labelled diagram, equation and principle.
Answer:
Laboratory method : Chloroform is prepared in the laboratory by the action of water and bleaching powder on ethyl alcohol or acetone.

Method : About 100 gm of bleaching powder made into a paste by adding about 200 ml of water and taken in a flask fitted with a condenser. Now, 25 ml of alcohol or acetone is added and the mixture is distilled, chloroform collects as a heavy liquid under water.

It is washed with dilute NaOH solution then with water, dried over fused calcium chloride and redistilled.

The available chlorine of bleaching powder acts as oxidising as well as chlorinating agent during the preparation of chloroform from alcohol and acetone.

CaOCl₂ + H₂O →(OH)₂ + Cl₂

The chemistry involved in the conversion of alcohol and acetone into chloroform is as shown below:

(A) From alcohol: The steps involved are :
(i) Ethyl alcohol is oxidized by chlorine to acetaldehyde.

(ii) Acetaldehyde reacts with chlorine to give chloral, i.e. trichloro acetaldehyde.

(iii) Two moles of chloral react with one mole of calcium hydroxide to produce chloroform.

Question 3.
What product is formed by the reduction of chloroform ? Give the chemical equation when it reacts to nitric acid and acetone.
Answer:
Reduction: (i) On heating with Zn and HCl, it reduces to form methylene dichloride.

(ii) On heating with zinc dust and water, methane is formed.

Nitration : On treating chloroform with concentrated nitric acid, the hydrogen atom of chloroform is replaced by nitro group and nitro chloroform (or chloropicrin) is formed. It is a liquid (b.p. 112°C) which is used in war as a poisonous gas.

Condensation : Chloroform condenses with acetone in presence of sodium hydroxide to form chloretone which is a hypnotic (sleep inducing drug) of high grade.

Question 4.
Give the Chemical reaction when chloroform reacts with following :
(a) Oxidation, (b) Carbylamine reaction, (c) Ag powder, (d) Nitration, (e) Reimer- Tiemann reaction.
Or,
How will you obtain the following from chloroform : (i) Carbonyl chloride, (ii) Acetylene, (iii) Chloropicrin, (iv) Phenyl isocyanide, (v) Chloretone, (vi) Salicylal- dehyde.
Or,
How trichloro methane reacts with : (a) Atmospheric air, (b) Aniline and ale. KOH, (c) Ag powder, (d) Cone. HNO₃, (e) Phenol.
Answer:
(a) Action of air and light (Oxidation): Chloroform oxidizes in presence of sunlight and air and forms a poisonous gas, phosgene (carbonyl chloride).

Ordinary chloroform contains phosgene gas and is used as a solvent. Pure chloroform which is used as an anaesthetic does not contain even traces of phosgene. While preserving chloroform which is to be used as an anaesthetic, the following precautions are taken :

(i) The chloroform is filled in blue or brown coloured bottle up to the neck. After putting a stopper, the bottle is kept in dark. As there is no empty space in the bottle, it is also free from air.
(ii) One percent ethyl alcohol is added in the bottle. If phosgene gas is formed, alcohol reacts with it to form diethyl carbonate, a non-toxic substance, i.e. (C₂H₅)₂CO₃,

(b) Carbylamine reaction : On heating chloroform with primary amine (e.g., aniline) and alcoholic KOH solution, phenylisocyanide or carbylamine is formed which has a very bad smell and is poisonous.

(c) Reaction with Ag powder or Dehalogenation : On heating chloroform with sil-ver powder, pure acetylene gas is formed.
CHCl₃ + 6 Ag + Cl₃CH → HC ≡ CH + 6 AgCl

(d) Nitration : On treating chloroform with concentrated nitric acid, the hydrogen atom of chloroform is replaced by nitro group and nitro chloroform (or chloropicrin) is formed. It is a liquid (b.p. 112°C) which is used in war as a poisonous gas.

(e) Reimer-Tiemann reaction : On heating chloroform with concentrated alkali and phenol at 60-70°C, o-hydroxy benzaldehyde (salicylaldehyde) is formed. Traces of p-hydro- xybenzaldehyde are also formed.

Question 5.
Explain nucleophilic substitution reaction of chlorobenzene (Give only equations).
Answer:
Nucleophilic substitution reaction of Chlorobenzene : Halogen atom in haloarenes is very strongly linked directly to benzene ring due to which it cannot be easily substituted by nucleophilic reagents like -OH,-OR,-NH₂,-CN etc. but at high pressure, temperature and in presence of suitable catalyst halogen can be substituted by these groups.

(i) Substitution by -OH group : On heating chlorobenzene with NaOH at 200 at-mospheric pressure and 300°C temperature phenol is formed.

(ii) Substitution by alkoxy (-OR) group :

With sodium alkoxide mixed ether is formed.

(iii) Substitution by Amino group : On heating with aqueous ammonia in presence of Cu₂O at 60°C atmospheric pressure and 200°C temperature aromatic amine is formed.

(iv) Substitution by Cyano group:

Question 6.
Explain the nucleophilic substitution reaction in alkyl halide by S_N1 and S_N2 mechanism.
Answer:
Nucleophilic substitution reaction : In the carbon halogen bond of haloalkane, halogen atom is more electronegative as compared to the carbon atom hence, the shared pair of electrons between carbon and halogen is more attracted by the halogen atom. As a result a small negative charge and an equivalent positive charge develops on halogen atom and carbon atom respectively.

Nucleophile attacks the electron deficient carbon due to the presence of partial posi-tive charge on it and replaces the weaker nucleophilic ion i.e. the halide ion. Thus, the reaction is known as nucleophilic substitution reaction.

The order of reactivity of different alkyl halide towards nucleophilic substitution reaction is:
RI > RBr > RCl > RF
Mechanism of Nucleophilic substitution reactions :
Nucleophilic substitution reaction occurs through two different mechanism :

(1) S_N1 Mechanism (Unimolecular nucleophilic substitution) : In this mechanism following steps are involved :
(a) Formation of carbocation by dissociation of substrate i.e., reactant molecule.

(b) Attack of nucleophile on carbocation forming the product.

(2) S_N2 mechanism (Bimolecular nucleophilic substitution): Reactions of this type occur in one step i.e. they are concerted reactions. These reaction nucleophilic attack results in a transition state in which both the reactant molecules are partially bonded to each other and then the halide ion escapes out forming the product.

Rate of reaction = K[(RX)OH^–]
Order of reactivity of alkyl halide is : Primaiy > Secondary > Tertiary.

Question 7.
Draw labelled diagram of laboratory method for preparation at iodoform from alcohol. Write related chemical equation.
Answer:

Chemical reaction :

Question 8.
Explain the following reactions of chlorobenzene :
(a) Reaction with chlorine in the absence of FeCl₃ in dark.
(b) Ullmann reaction.
Answer:
(a) Chlorine reacts with chlorine in dark, in the presence of FeCl3 to form ortho -dichlorobenzene and p-dichlorobenzene.

(b) On heating bromo or iodobenzene at 200°C temperature with Cu in a sealed tube, diphenyl is formed. This reaction is known as Ullmann reaction.

Question 9.
Write the equation of following reactions of chlorobenzene :
(i) Halogenation
(ii) Nitration
(iii) Sulphonation
(iv) Alkylation.
Answer:
(i) Halogenation : Haloarene reacts with halogen in presence of halogen carrier like FeCl₃ to form ortho and para substituted dihaloarene.

(ii) Nitration : Haloarenes react with nitrating mixture to form o-nitro and p-nitro substituted haloarenes.

(iii) Sulphonation : On heating with cone. H₂SO₄, 2-Chlorobenzene sulphonic acid and 4-Chlorobenzene sulphonic acid are formed.

(iv) Alkylation: Alkylation takes place with alkyl halide in presence of anhydrous AlCl₃.

Question 10.
Haloalkanes are more reactive than haloarenes. Give reason.
Or,
Why, aryl halides are less reactive than alkyl halides ?
Answer:
In aryl halides, halogen atom is attached more strongly to the nucleus therefore the nucleophilic substitution takes slowly than alkyl halides. There is two reasons for the less reactivity of aryl halides.

(i) In aryl halides sp3 hybridization takes place whereas in alkyl halides sp2 hybridization is present due to sp2 hybridization in haloarenes the halogen are attached to nucleus more strongly.

(ii) Due to the presence of resonance in aryl halides there is some double bond character in C—Cl bond. Thus, the bond length of C—Cl bond is lesser than C—Cl bond in haloalkanes. Therefore, it is difficult to replace the halogen of haloarenes.

MP Board Class 12th Chemistry Solutions

MP Board Class 12th General English Supplementary Materials (Based on ‘Reading Time’ Section of Work Book)

Supplementary Materials Long Answer Type Questions:

Question 1.
Summarize the poem ‘Words’ in about 80 words. [2018]
Answer:
The poem classifies words into various categories. The poet says that some words are happy others are sad. There are bitter and sweet words. Some words are honest while others cheat. Some words are remembered others are forgotten. The classification represents not various categories of words but of people because words give expression to thoughts of people. Their thoughts, in turn, represent their personality and inner self. People appear to be bitter and sweet; open and sly; honest and dishonest; good and bad on account of words only.

Question 2.
Write the central idea of the poem “We are the Future”.
Answer:
The poem exhorts Indians to come together as a team and work together. The real power and strength lies in unity. As a team, they can accept every challenge and achieve every goal. The future is unknown and holds many difficulties but it is to be made, bright by Indians together. Thus, people should come from various parts of the country and build the nation.

Question 3.
Write a speech on the importance of English in everyday life. [2009]
Answer:
“The English language is an advanced and flexible language”, said Dr. Rajendra Prasad, our first President. Though it is irregular in grammar, spelling and pronunciation, it is one of the most effective means of expression. English has the richest vocabulary in the world. Every year almost ten thousand words are added to the English dictionary. Besides this, it is used in fields of study other than arts. Another thing about English is that it is used frequently in social intercourse. It is a key to scientific knowledge and learning.

Question 4.
Describe the importance of ‘The Gita’.
Answer:
The Gita is an aphoristic pertaining to a short clever saying which is intended to express a general truth] work. It is great religious poem. Gita is not a collection of do and don’ts. It has sung the praises of knowledge, but it is beyond mere intellect. Gita is considered to be one of the greatest books. So many wise men have written their interpretations of Gita. It has been translated in almost all languages of the world. It is beyond time.

Question 5.
How can one attain mastery in the art of meditation? [2012]
Answer:
One can attain mastery in the art of meditation by focussing on breathing. Since our breath is always with us, it can be done any time, any place. But the best is to settle down in a cool place that has dim light and relatively quiet. You must focus attention on your breathing. Observe your breath as it comes in and goes out. In this way, we can attain mastery in the art of meditation.

Question 6.
How is the Angel Canyon Falls compared to the Niagara Falls? [2013]
Answer:
The world’s highest waterfall is situated in Eastern Venezuela in South Africa. It is known as the Angel Falls. It hurtles down from a height of 960 m which is sixteen times more than the height of the famous Niagara Falls and provides r. truly breathtaking view.

Question 7.
How can you say that nature’s bounty is boundless? [2009, 10]
Answer:
We can say that nature’s bounty is boundless because there are millions of things that nature possesses. Even the simplest things which we take for granted are not really simple. Colorful, pretty flowers and various kinds of fruits are an example. There are thousands of other things in nature’s treasure trove. There are an entire range of high hills smoking fumes in the air for years together. An elephant’s trunk comprises of more than forty thousand muscles. The human body is a wonderful collection of so many organic and inorganic matters.

Question 8.
Summarize the poem ‘A Psalm of Life’. [2016]
Answer:
The poem ‘A Psalm of Life’ invokes people to treat life as real. The poet asks people not to waste life though it is short and will soon end up. But the soul is immortal. Life should be lived in a better way. People should make good use of time and fight against all odds. They should not follow others blindly. They should lead a noble and meaningftiTlife. They should work hard and wait patiently for God to reward.

Question 9.
Why does the author say, ‘Your heart is just as smart as the U. S. Army’? [2011,15]
Answer:
The U. S. Army is very smart. Our heart is just as smart as the U. S. Army. Our heart pumps enough blood through our body every day sufficient to fill a railway tank car. It exerts enough energy every twenty-four hours to shovel twenty tons of coal on to a platform three feet height. It does this incredible amount of work for fifty, seventy or maybe ninety years. So we say our heart is very smart.

Question 10.
What will help you to improve your English at the beginning? [2012]
Or
On what does the improvement of your English depend? [2010,11]
Answer:
In the beginning, to read a lot is essential. Read as many books you find, with the idea of listing and learning as many words as possible. Choose what is likely to interest you and be sure in advance that it is not too hard. You should not have to be constantly looking up new words here and there, but as a general policy, try to push ahead, guessing what words means from the context. It is extensive and not intensive reading that normally helps you to get interested in extra reading and thereby improve your English.

Question 11.
Write the central idea of the poem ‘Risks’. [2013, 16, 17]
Answer:
Man has various types of needs. Hard work is required to fulfill them. There are many obstacles in the path of fulfilling them. Nature also puts hinderances before man. However, these obstacles develop the ability in a man. Life is full of all kinds of risks. One will have to take risks to achieve them. One who cannot take risk, cannot gain anything. That is why it is rightly said, No risk, no gain.

Question 12.
In what categories does the author divide the books? [2014,18]
Answer:
The author divides the books into various categories. Some books are to be read in parts only, it is not essential to read them thoroughly. Some books are to be read without paying much attention they can be read, but not cursorily. There is still author type of books that are supposed to be read thoroughly or cover to cover. They demand due attention and concentration. They are to be fully absorbed. Lastly, there are some books which a person needs not read himself. He can either depute some other person to read them or rely on their extracts prepared by others. But this can be done only in the case of some less important books.

Question 13.
How can one reach Chitrakoot? Mention all the three ways? [2015]
Answer:
The following are the ways to reach Chitrakoot:

1. By Air, The nearest airport is at Khajuraho, connected with Delhi, Agra, and Varanasi.
2. By Rail: The nearest railhead is at Chitrakoot Dham on the Jhansi Manikpur mainline.
3. By Road: Regular bus services connect Chitrakoot with Jhansi, Mahoba, Chitrakoot Dham, Harpalpur, Satna and Chhatarpur.

Supplementary Materials Short Answer Type Questions

Question 1.
What makes ‘a team’ a whole? [2009, 16]
Answer:
When a team works together with co-operation it makes a team a whole. –

Question 2.
Which is the world’s highest waterfall? Where is it? [2009, 18]
Answer:
The world’s highest waterfall is Angel Falls. It is situated in Eastern Venezuela in South America.

Question 3.
What accounts for beauty in yourself? [2009, 13, 18]
Answer:
Good deeds and happy thoughts account for beauty in ourselves.

Question 4.
What are the things education is linked with? [2014]
Answer:
Education is linked with the economic, political, religious and other sub-systems.

Question 5.
What was Upcourt’s challenge to Tatas?
Answer:
Upcourt’s challenge was “If Tatas make steel rails to British specification, I will undertake to eat every pound of rail they make.”

Question 6.
What is the saying about values? [2017]
Answer:
It is said about values that they cannot be caught, they are taught.

Question 7.
How much does the heart work in a day?
Answer:
When beating at a moderate rate of seventy pulses per minute, the heart is actually working only nine hours out of the twenty-four hours.

Question 8.
What does the poet say about the Past and the Future? [2017]
Answer:
The poet says that one should not trust the Future however pleasant it is and bury the ugly past.

Question 9.
What did the truck driver say after checking the car? [2014]
Answer:
After checking the car, the truck driver said that he happened to be driving down the highway a few hours ago and saw the author helping a lady with her car troubles. He also added that what goes around comes around.

Question 10.
What is ‘Angel Canyon’? And what is the ‘Charan River’? [2009]
Answer:
The Angel Canyon is a large pool at the cliffs foot. Charan river is a tributary of the Caroni River.

Question 11.
What are the distilled books like? [2009, 16]
Answer:
Distilled books are like common distilled waters, flashy things.

Question 12.
Who will certainly preach Lord Krishna? [2017]
Answer:
People who extend the precious treasure of knowledge to his devotees will certainly reach Lord Krishna.

Question 13.
What did wooden always focus on? [2009]
Answer:
Wooden always focused on today.

Question 14.
What happens when you dream big? [2017]
Answer:
When you dream big, you pursue your dreams and change them into reality. ‘

Question 15.
What is the simplest form of meditation? [2009, 10, 12, 17]
Answer:
The simplest form of meditation is focussing on breathing.

Question 16.
What things in nature arouse your curiosity?
Answer:
The Etna Volcano Italy, the Bermuda Triangle, arouses our curiosity.

Question 17.
How does the author compare the crafty, simple and wise men? [2018]
Answer:
The author compares by saying that crafty, men condemn studies, simple men admire them and wise men use them.

Question 18.
Why is the Gita beyond the mere intellect?
Answer:
The Gita is beyond the mere intellect as it is essentially addressed to the heart and capable of being understood only by the heart.

Question 19.
Which are the two types of reading?
Answer:
They are extensive and intensive reading.

Question 20.
What do people do for their harvest?
Answer:
They dance for their harvest.

Question 21.
Why was Jamshedji prohibited from entering the Majestic hotel? [2010]
Answer:
Jamshedji was prohibited because he was an Indian.

Question 22.
Who took the kids into the restaurant? [2013]
Answer:
The truck driver took the kids into the restaurant.

Question 23.
What are the benefits of meditation? [2013, 14, 16]
Answer:
Calm mind, better retention, better health, focus, and concentration are the obvious benefits.

Question 24.
Why does the author say that the Gita is not for those who have no faith? [2013, 14]
Answer:
The author says this because the Gita is beyond the mere intellect. It is essentially addressed to the heart and capable of being understood by the heart.

Question 25.
What do you understand by do’s and don’ts? [2014]
Answer:
Do’s and don’ts are nothing but things permissible or unpermissible at a time or place; things lawful or unlawful for a person.

Question 26.
From what did wooden create the major portion of his coaching philosophy? [2014]
Answer:
Wooden created a major portion of his coaching and living philosophy from one thought [passed on to him by his father] : “Make each day your masterpiece.”

Question 27.
Which is considered the best place for meditation?[2015, 18]
Answer:
The best place for meditation is to settle down in a cool place that has dim light and is relatively quiet.

Question 28.
Does the heart work all the time? [2015]
Answer:
Most people have the idea that the heart is working all the time. As a matter of fact, there is a definite rest period after each contraction. In the aggregate, its rest periods total a full fifteen hours a day.

Question 29.
Why do we want to live in future? [2015]
Answer:
We want to live in future because we believe that the future is where our happiness lies.

Question 30.
What is idleness according to author? [2015]
Answer:
According to author to spend too much time in studies is idleness. It indicates that you are checking work.

Question 31.
Where is Chitrakoot situated? [2015,16,18]
Answer:
Chitrakoot nestles peacefully in the northern spurs of the Vindhyas, a place of tranquil forest glades and quiet rivers, and streams where calm and repose are all-pervading.

Question 32.
Why we must take risk? [2016]
Answer:
We must take risk to overcome obstacles. Obstacles develop ability in a man. Taking risk means achieving victory over them. One who cannot take risk, cannot gain anything. That is why it is rightly said—No risk, no gain.

Question 33.
What do understand by‘spiritual Guru’? [2016]
Answer:
A spiritual Guru is one who directs and helps others in making their minds calm.

MP Board Class 12th English Solutions

MP Board Class 12th English Important Extracts from the Poems

निम्नलिखित पंक्तियों को ध्यानपूर्वक पढ़िए एवं उन पर आधारित प्रश्नों के उत्तर दीजिए:

1. (Poem 1)

Teach me to listen, Lord
To those far from me The
The plea of the forgotten
The cry of the anguished. [2009, 10, 12, 16, 17]

Questions:
(i) Find out a word from the lines given a word having a similar meaning to ‘severe pain.’ [2009, 10]
(ii) The poet wants us to listen to those [2009, 10]
(iii) What does the poet request? [2009, 10]
(iv) A word similar in meaning to the word ‘hear’ : [2012, 16, 17]
(a) pear (b) listen (c) peep.
(v) The opposite word ‘near’ is : [2012, 16, 17]
(a) far (b) clear (c) further (d) fear.
(vi) Give the name of the poem from which these lines have been taken. [2012, 16, 17]
Answer:
(i) anguish.
(ii) Who are far from him and who are hopeless, forgotten and anguished.
(iii) The poet requests God to teach him to listen to the hopeless and the anguished.
(iv) (b) listen.
(v) (a) far.
(vi) Teach Me to Listen, Lord.

2. (Poem 4)

Ah, you are so great, and I am so small,
I tremble to think of you, World, at all;
And yet, when I said my prayers today,
A whisper inside me seemed to say,
“You are more than the Earth, though you are such a dot.
You can love and think, and the Earth cannot !” [2009]

Questions:
(i) A word opposite in meaning to ‘inside’:
(a) great (b) tremble (c) outside (d) None of these.

(ii) A word meaning ‘shake’ in the, above lines is :
(a) whisper, (b) tremble (c) Vet (d) great.

(iii) Write the name of the poet who wrote this poem.
Answer:
(i) (c) outside.
(ii) (b) tremble.
(iii) William Brighty Rands.

3. [Poem 16]

Some people say that you’re a dear
Yet dear is far from cheap,
A jumper is a thing you wear,
Yet a jumper has to leap,
It’s very clear, it’s very queer,
And pray who is to blame
For different meanings to some words
Pronounced and spelt the same? [2018]

Questions:
(i) What is the opposite of the word dear?
(a) costly (b) cheap (c) precious (d) beloved.

(ii) The word ‘jumper’ has two meanings. It’s one meaning is ‘one who jumps’. What is its second meaning?
(a) a woman’s loose outer garment (b) strange (c) different (d) blame.

(iii) Name the poem from which this extract has been taken?
Answer:
(i) (b) cheap.
(ii) (a) a woman’s loose outer garment
(iii) The English Language.

4. [Poem 10]

After unnumbered steps of a hill-stair
I saw upon earth’s head brilliant with sun
The immobile Goddess in her house of stone
In the loneliness of meditating air. [2013]

Questions:
(i) A word similar in meaning to ‘unnumbered’ is :
(a) numberless (b) few (c) some (d) none of these.
(ii) The opposite of brilliant’ is :
(a) bright (b)dull (c) intelligent (d) sharp.
(iii) Give the name of the poem and poet.
Answer:
(i) (a) numberless.
(ii) (b) dull.
(iii) Name of the poem – The Hill-Top Temple
Name of the poet – Sri Aurobindo.

5. [Poem 7]

Can I admire the statue great,
When living men starve at its feet.
Can I admire the parks green tree,
A roof for homeless misery? [2016]

Questions:
(i) Name the poet :
(a) Sri Aurobindo (b) W.H. Davies (c) William Rands (d) Anonymous.
(ii) Give the meaning of the word ‘Starve’ :
(a) die of hunger (b) labour hard (c) work comfortably (d) live happily.
(iii) Why does the poet say that he cannot admire the great statue?
Or
Give noun form of admiring.
Answer:
(i) (b) W. H. Davies.
(ii) (a) die of hunger.
(iii) The poet cannot admire the great statue because the living men are starving at its feet.
Or
admiration.

6. [Poem 16]

Some words have different meanings,
and yet they’re spelt the same.
A cricket is an insect,
to play it it’s a game.
On every hand, in every land,
it’s thoroughly agreed,
the English language to explain,
is very hard indeed. [2010, 14, 17]

Questions:
(i) The above lines have been taken from the poem [2010]
(ii) What is the name of the insect mentioned in the stanza? [2010]
(iii) Give the opposite of the word ‘easy’ from the above lines. [2010]
(iv) The opposite of the word ‘same’ in the stanza is : [2014,17]
(a) difficult (b) different (c)hard (d) every.

(v) The meaning of the word ‘hard’ is : [2014]
(a) difficult (b) easy (c) rough (d) plain.

(vi) What are the two meanings of cricket? [2014]
(vii) An insect and a game both can be called : [2017]
(a) chess (b) cricket (c) fly (d) similar.

(viii) What is thoroughly agreed everywhere? [2017]
Answer:
(i) The English Language.
(ii) pricket.
(iii) mrd.
(iv) (b) different.
(v) (a) difficult.
(vi) The two meanings of cricket’ are ‘insect’ and ‘a game’.
(vii) (b) cricket.
(viii) It is thoroughly agreed everywhere that English is a difficult language.

7. (Poem 1)

Teach me to listen, Lord
To myself
Help me to be less afraid
To trust the voice inside
In the deepest part of me.

Questions:
(i) A word having the same meaning as the word ‘belief is :
(a) afraid (b) deepest (c) trust (d) inside.
(ii) A word from the stanza which is the opposite of ‘more’ is :
(a) part (b) less (c) lord (d) voice.
(iii) What does the poet mean by “in the deepest part of me”?
Answer:
(i) (c) trust.
(ii) (b) less.
(iii) The poet means the deepest corner of his heart.

8. [Poem 10]

Our body is an epitome of some Vast
That masks its presence by our humanness.
In us, the secret Spirit can indite.
A page and summary of the Infinite.
A nodus of Eternity expressed
Live in an image and a sculptured face. [2015]

Questions:
(i) The poem is :
(a) In-country (b) Forest and River (c) The Hill-Top Temple (d) Wonderful world

(ii) What is the meaning of “infinite”.
(a) that means no end (b) that means for some time (c) that means begin (d) that means go on.

(iii) Write the correct transcription of the word ‘epitome’ to speak it accurately.
(iv) Write the adjective form of the word ‘eternity’.
Answer:
(i) (c) The Hill Top Temple
(ii) (a) that means no end.
(iii) i-pit’ ∂-m
(iv) eternal.

9. (Poem 1)

To those nearest me
My family, my friends, my co-workers
Help me to be aware that
No matter what words I hear
The message is
‘Accept the person I am, listen to me.’ [2011, 15]

Questions:
(i) A word similar in meaning to the word ‘attentive’ :
(a) preventive (b) active (c) aware (d) co-worker.

(ii) The word opposite in meaning of ‘aware’ :
(a) unaware (b) inaware (c) unawares (d) disaware.

(iii) The poet wants us to listen to those [nearest/farthest]
(iv) Write the name of the poem from which these lines have been taken.
Answer:
(i) (c) aware
(ii) (a) unaware
(iii) nearest
(iv) Teach Me to Listen, Lord.

10. (Poem 13)

But what am I?
Only a captive,
Chained to the earth.
In silence I grow old
In silence I wither and die. [2013]

Questions:
(i) Who is T in these lines?
(a) River (b) Forest (c) Sky

(ii) From These lines opposite of the word ‘free’.
(a) captive (b) old (c) silent (d) with.

(iii) Give the name of the poet.
Answer:
(i) (b) Forest.
(ii) (a) captive.
(iii) Jalaluddin Rumi.

11. (Poem 13)

On every hand, in every land
it’s thoroughly agreed
the English language to explain
is very hard indeed. [2012]

Questions:
(i) The same meaning of ‘completely’ is
(a) purely (b) wrongly (c) thoroughly (d) explain.

(ii) The opposite word of ‘easy’ is
(a) cused (b) hard (c) soft (d) comfort.

(iii) Give the name of the poem from which these lines have been taken.
Answer:
(i) (c) thoroughly.
(ii) (b) hard.
(iii) The English Language.

12. (Poem 4)

The wonderful air is over me,
And the wonderful wind is shaking the tree,
It walks on the water, and whirls the mills,
And talks to itself on the tops of the hills. [2014,18]

Questions:
(i) The opposite of the word ‘bottom’ in the stanza is : [2014, 18]
(a) wonderful (b) shake (c)top (d) whirl.

(ii) The word that rhymes with ‘tree’ is : [2014,18]
(a) hill (b)me (c)top (d)mill.

(iii) What is shaking the tree? [2014]
(iv) Give the name of the poem from which this extract has been taken. [2018]
Answer:
(i) (c) top.
(ii) (b)me.
(iii) Wonderful wind is shaking the tree.
(iv) The poem is wonderful World’.

MP Board Class 12th English Solutions

MP Board Class 12th Biology Solutions Chapter 15 जैव-विविधता एवं संरक्षण

जैव-विविधता एवं संरक्षण NCERT प्रश्नोत्तर

प्रश्न 1.
जैव विविधता के तीन आवश्यक घटकों (Component) के नाम बताइए।
उत्तर
जैव विविधता (Bio diversity) के तीन आवश्यक घटक निम्नलिखित हैं

• आनुवंशिक विविधता (Genetic diversity)
• जातीय विविधता (Species diversity)
• पारिस्थितिकीय विविधता (Ecological diversity)।

प्रश्न 2.
पारिस्थितिकीविद् किस प्रकार विश्व की कुल जातियों का आंकलन करते हैं ?
उत्तर
वैश्विक विविधता के निर्धारण के लिए UNEP (United Nations Environmental Programme) के अन्तर्गत चलाई गई एक योजना के अनुसार पृथ्वी पर जीवों की 13-14 मिलियन जाति का अनुमान लगाया गया है। लेकिन इनमें से केवल 1.75 मिलियन जीव-जातियों का ही वर्णन किया गया है। यह कुल जीवधारियों का केवल 15% ही है।

ज्ञात जातियों में से लगभग 61% (10,2,500) कीटों की जातियाँ है। स्तनधारियों की जातियाँ सापेक्षिक रूप से बहुत कम (4650 जातियाँ) है। इनमें भी शैवाल एवं जीवाणुओं की संख्या कम है। अनेक वर्गिकीविज्ञों का विश्वास है कि अभी भी असंख्य जीव जातियों की खोज नहीं हुई है। विषाणुओं, जीवाणुओं, प्रोटिस्टा के बारे में अभी भी ज्ञान अधूरा है। उपलब्ध रिकॉर्ड से ज्ञात होता है कि विषाणुओं की 1,550, जीवाणुओं की 40, 000 जातियाँ ज्ञात हैं।

प्रश्न 3.
उष्ण कटिबन्ध क्षेत्रों में सबसे अधिक स्तर की जाति-समृद्धि क्यों मिलती है ? इसकी तीन · परिकल्पनाएँ दीजिए।
उत्तर
उष्ण कटिबन्ध क्षेत्रों में सबसे अधिक स्तर की जाति समृद्धि पायी जाती है, इसको समझाने के लिए पारिस्थितिक तथा जैव विकास विदों ने अनेक परिकल्पनाएँ प्रस्तुत की हैं जिनमें से प्रमुख निम्नानुसार हैं

• जाति उद्भवन (Speciation) समयानुसार होता है। शीतोष्ण क्षेत्र में प्राचीन काल से ही बार-बार हिमनद (Glaciation) होता रहता है, जबकि उष्ण कटिबन्ध क्षेत्र लाखों वर्षों से बाधा से मुक्त रहता है। इसी कारण जाति विकास तथा विविधता के लिए लंबा समय मिलता है।
• उष्ण कटिबन्धीय पर्यावरण शीतोष्ण पर्यावरण (Temperate environment) से भिन्न तथा कम मौसमी परिवर्तन को दर्शाता है। यह स्थिर पर्यावरण निकेत (Niches) विशिष्ट करण को प्रोत्साहित करता रहता है जिसके कारण अधिकाधिक जाति विविधता उत्पन्न हुई ।
• उष्ण कटिबन्धीय क्षेत्रों में अधिक सौर ऊर्जा उपलब्ध है जिससे उत्पादन अधिक होता है जिससे परोक्ष रूप से अधिक जैव विविधता उत्पन्न हुई है।

प्रश्न 4.
जातीय क्षेत्र संबंध में समाश्रयण (रिग्रेशन) की ढलान का क्या महत्व है ?
उत्तर
जर्मनी के महान् प्रकृतिविद् व भूगोलशास्त्री अलेक्जेंडर वॉन हम्बोल्ट ने दक्षिणी अमेरिका के जंगलों के गहन अन्वेषण के समय दर्शाया कि कुछ सीमा तक किसी क्षेत्र की जातीय समृद्धि अन्वेषण क्षेत्र की सीमा बढ़ाने के साथ बढ़ती है। वास्तव में जाति समृद्धि और वर्गकों (अनावृत्तबीजी पादप, पक्षी, चमगादड़, अलवणजलीय मछलियाँ) की व्यापक किस्मों के क्षेत्र के बीच संबंध आयताकार अतिपरवलय (रेक्टंगुलर हाइपरबोल) होता है (चित्र)। लघुगणक पैमाने पर यह संबंध एक सीधी रेखा दर्शाता है जो कि निम्न समीकरण द्वारा प्रदर्शित है |

log S = log C+Z log A
जहाँ पर S = जातीय समृद्धि, A = क्षेत्र, Z = रेखीय ढाल (समाश्रयण गुणांक रिग्रेशन कोएफिशिएट), S = Y अंत: खंड (इंटरसेप्ट)

पारिस्थितिक वैज्ञानिकों ने बताया कि Z का मान 0.1, से 0-2 परास में होता है भले ही वर्गिकी समूह अथवा क्षेत्र (जैसे कि ब्रिटेन के पादप, कैलिफोर्निया के पक्षी या न्यूयार्क के मोलस्क) कुछ भी हो। समाश्रयण रेखा (रिग्रेसन लाइन) की ढलान आश्चर्यजनक रूप से एक जैसी होती हैं। लेकिन यदि हम किसी बड़े समूह, जैसे संपूर्ण महाद्वीप के जातीय क्षेत्र संबंध का विश्लेषण करते हैं तब ज्ञात होता है कि समाश्रयण रेखा की ढलान पैमाने पर संबंध रेखीय हो जाते हैं तीव्र रूप से तिरछी खड़ी है (Z का मान की परास 0.6 से 1.2 है)।
उदाहरणार्थ – विभिन्न महाद्वीपों के उष्ण कटिबंध वनों के फलाहारी पक्षी तथा स्तनधारियों की रेखा की ढलान 1.15 है।

प्रश्न 5.
किसी भौगोलिक क्षेत्र में जाति क्षति के मुख्य कारण क्या हैं ?
उत्तर
वन्य जातियों के या जाति क्षति के लिए जिम्मेदार कारण निम्नलिखित हैं

(1) मानव सभ्यता के विकास के साथ मानव आवश्यकताएँ बढ़ती गयी हैं। इन आवश्यकताओं की पूर्ति के लिए मनुष्य ने लगभग आधे से अधिक जंगलों को नष्ट कर दिया है। भारतवर्ष में 18% से भी कम स्थान में आज वन रह गये हैं। इस कारण जलवायु परिवर्तित हो गयी है, जिसके कारण वन्य जीव विलुप्त हो रहे हैं, क्योंकि इनका प्राकृतिक आवास नष्ट एवं परिवर्तित हो गया है ।

(2) वन्य जीवों की संख्या उनके शिकार के कारण भी कम हो रही है, क्योंकि मनुष्य मनोरंजन तथा विभिन्न आवश्यकताओं की पूर्ति के लिए जीवों का शिकार करता है, जैसेकस्तूरी मृग को कस्तूरी के लिए, हिरन, सांभर, तेंदुआ, शेर, चीता, खरगोश को खाल के लिए तथा हाथी को दाँत के लिए।

(3) वन्य जीवों के संरक्षण के लिए सरकार की तरफ से किसी भी प्रकार का प्रतिबन्ध न होना भी वन्य जीव विलुप्तीकरण का एक प्रमुख कारण है।

(4) प्रदूषण (जैसे-शोर इत्यादि) भी वन्य जीवों को प्रभावित करता है, जिसके कारण ये आज विलुप्त हो रहे हैं।

(5) वनों की कटाई, उद्योगों की स्थापना तथा विभिन्न रसायनों व प्रदूषकों के उपयोग इत्यादि के कारण वातावरण में परिवर्तन आया है, जिससे कई जीव विलुप्त हो गये हैं या हो रहे हैं ।

(6) मनुष्य सुरक्षात्मक कारणों से भी कुछ वन्य जीवों को मार देता है, जिसके कारण ये विलुप्त हो रहे हैं। जैसे-चीता, शेर इत्यादि से मनुष्य डरता है । इस कारण इन्हें मार देता है।

(7) कुछ जन्तुओं तथा उनके उत्पादों जैसे-हाथी दाँत, चमड़ा, सींग, मोती आदि की यूरोपीय देशों में बहुत अधिक माँग है और ये वहाँ पर उच्च दामों में बिकते हैं। भारत तथा दूसरे विकासशील राष्ट्रों में अवैध शिकार हो रहा है जो एक प्रमुख कारण है।

प्रश्न 6.
पारितंत्र के कार्यों के लिये जैव-विविधता कैसे उपयोगी है ?
उत्तर
जैव विविधता की पारितंत्र के कार्यों के लिए उपयोगिता (Utility of Biodiversity for ecosystem functioning)-समृद्ध जैव-विविधता अच्छे पारितंत्र के लिए आवश्यक है। प्रकृति द्वारा प्रदान की गई जैव विविधता को अनेक पारितंत्र सेवाओं में मुख्य भूमिका है। तीव्र गति से नष्ट हो रहा अमेजन वन पृथ्वी के वायुमण्डल को लगभग 20% ऑक्सीजन, प्रकाश संश्लेषण द्वारा प्रदान करता है। अन्य उपयोग हैं परागण क्रिया जिसके बिना पौधे फल तथा बीज नहीं दे सकते जो परागणकर्ता, जैसे-मधुमक्खी, पक्षी, चमगादड़ आदि द्वारा सम्पन्न होते हैं।

पादप एवं जंतुओं को हजारों खाने योग्य प्रजातियाँ ज्ञात हैं, फिर भी विश्व में 85% खाद्य उत्पादन 20 पादप जातियों से ही प्राप्त होता है। जैव विविधता उन्नत प्रजातियों के विकास के लिये प्रजनन पदार्थ ‘भी उपलब्ध कराती है। अनेक पदार्थों में उपचारात्मक गुण होते हैं जिन्हें अनेक पौधों और जंतुओं से प्राप्त किया जाता है। राष्ट्रीय उद्यानों तथा अभयारण्य में घूमना आनन्ददायक तथा रोमांचक होता है। सभी जीवधारी खाद्य शृंखलाओं में व्यवस्थित होते हैं। ये सभी अपने जैविक पर्यावरण से संबंधित होकर पारिस्थितिक संतुलन बनाते हैं। प्रकृति में अनेक चक्र लगातार चलते रहते हैं जिनमें जीवधारी एवं अजैविक वातवरण अपनी अपनी भूमिका पूर्णतः निभाते हैं।

प्रश्न 7.
पवित्र उपवन क्या है ? उनकी संरक्षण में क्या भूमिका है ?
उत्तर
पवित्र उपवन (Sacred groves)- भारत में सांस्कृतिक व धार्मिक परम्परा का इतिहास जो प्रकृति की रक्षा करने पर जोर देता है। बहुत-सी संस्कृतियों में वनों के लिए अलग भू-भाग छोड़े जाते थे और उनमें सभी पौधों तथा वन्य जीवों की पूजा की जाती थी। पवित्र उपवन पूजा स्थलों के चारों ओर पाया जाने वाला वनखण्ड है। ये जातीय समुदायों/राज्य या केन्द्र सरकार द्वारा स्थापित किये गये हैं। ये उपवन भारत के कई राज्यों, मेघालय, महाराष्ट्र, कर्नाटक, केरल आदि में है।

जातीय समुदाय द्वारा निर्मित मंदिर के आस-पास देवदार के वृक्ष लगाये गये हैं, जैसे-कुमाऊं क्षेत्र । इसी प्रकार राजस्थान में विश्नोई समुदाय के लोगों ने प्रोस्पिस व ब्लैक बक को धार्मिक रूप से बचाया है। पवित्र उपवन में किसी भी पौधे को तोड़ने की अनुमति नहीं होती है। अतः इनमें सभी स्थानिक (Endemic) प्रजातियाँ भली प्रकार से वृद्धि करती हैं और संरक्षित रहती हैं।

प्रश्न 8.
पारितंत्र सेवा के अन्तर्गत बाढ़ व भू-अपरदन (Soil Erosion) नियंत्रण आते हैं। यह किस प्रकार पारितंत्र के जीवीय घटकों (बायोटिक कम्पोनेंट) द्वारा पूर्ण होते हैं ?
उत्तर
वृक्ष तथा पौधे बाढ़ व भू-अपरदन को नियंत्रण करने में सहायक सिद्ध होते हैं । वृक्ष तथा पौधे भूअपरदन को विभिन्न तरीकों के द्वारा रोक सकते हैं

• पौधे तथा वृक्षों की जड़ें मृदा या भूमि को मजबूती से जकड़े रहती है जिससे जल तथा वायु प्रवाह में अवरोध उत्पन्न होते हैं।
• वृक्ष वायु गति की तीव्रता को कम करने में सहायक होते हैं। जिससे अपरदन की दर कम हो जाती है।
• वृक्ष मृदा को छाया उपलब्ध कराते हैं। जिससे ग्रीष्म के दौरान मृदा के शुष्क होने से बचाव होता है।
• वृक्षों की गिरी हुई पत्तियाँ वर्षा की बूंदों की तीव्रता से होने वाली मृदा को हानि से बचाव करती है।
• वृक्षारोपण बाढ़ नियंत्रण में प्रमुख भूमिका निभाता है।
• वृक्ष मरुस्थलों में वायवीय अपरदन (Wind erosion) को रोकने में उपयोगी होते हैं।

प्रश्न 9.
पादपों की जाति विविधता ( 22 प्रतिशत), जंतुओं ( 72 प्रतिशत) की अपेक्षा बहुत कम है। क्या कारण है कि जन्तुओं में अधिक विविधता मिलती है ?
उत्तर
किसी भी पारितंत्र में जंतुओं में पौधों की तुलना में अधिक जैव विविधता पायी जाती है, इसके निम्नलिखित कारण हैं

• जंतुओं में अनुकूलन की क्षमता पौधों की अपेक्षा बहुत अधिक होती है। जंतुओं में तंत्रिका तंत्र तथा अन्त:स्रावी तंत्र पाये जाने के कारण वे स्वयं को वातावरण के प्रति अनुकूलित कर लेते हैं।
• प्राणियों में प्रचलन का गुण पाया जाता है जिसके कारण वे विपरीत परिस्थितियाँ होने पर स्थान परिवर्तन कर लेते हैं और स्वयं को बचाये रखते हैं। जबकि पौधे स्थिर होने के कारण वे विपरीत परिस्थितियों का सामना करने के लिये विवश रहते हैं।

जैव-विविधता एवं संरक्षण अन्य महत्वपूर्ण प्रश्नोत्तर

जैव-विविधता एवं संरक्षण वस्तुनिष्ठ प्रश्न

1. सही विकल्प चुनिए

प्रश्न 1.
हमारे देश में वन्य जीवन संरक्षण अधिनियम कब पारित किया गया था
(a) 1883
(b) 1972
(c) 1973
(d)1982
उत्तर
(b) 1972

प्रश्न 2.
इण्डियन बोर्ड ऑफ वाइल्ड लाइफ (IBWL) की स्थापना कब हुई थी
(a) 1952
(b) 1981
(c) 1971
(d) 1972.
उत्तर
(a) 1952

प्रश्न 3.
NBPGR कहाँ स्थित है
(a) दिल्ली
(b) कोलकाता
(c) लखनऊ
(d) मुम्बई।
उत्तर
(a) दिल्ली

प्रश्न 4.
हमारे देश में बायोस्फियर रिजर्व की संख्या है
(a)73
(b) 7
(c)416
(d) 23
उत्तर
(b) 7

प्रश्न 5.
कौन-सा राष्ट्रीय उद्यान सफेद शेर से सम्बन्धित है
(a) कांकेर
(b) सतपुड़ा
(c) बाँधवगढ़
(d) कान्हा
उत्तर
(c) बाँधवगढ़

प्रश्न 6.
म. प्र. का प्रथम राष्ट्रीय उद्यान है
(a) शिवपुरी
(b) बाँधवगढ़
(c) कान्हा
(d) कांकेर।
उत्तर
(c) कान्हा

प्रश्न 7.
पादप जीवाश्म राष्ट्रीय उद्यान कहाँ पर स्थित है
(a) शिवपुरी
(b) मण्डला
(c) कान्हा
(d) कांकेर।
उत्तर
(b) मण्डला

प्रश्न 8.
म. प्र. का राष्ट्रीय उद्यान जिसे बायोस्फियर रिजर्व के रूप में चिह्नित किया गया है, वह है
(a) कान्हा
(b) शिवपुरी
(c) बाँधवगढ़
(d) सतपुड़ा।
उत्तर
(a) कान्हा

प्रश्न 9.
बांदीपुर (कर्नाटक) का राष्ट्रीय उद्यान किसके संरक्षण से संबंधित है
(a) मोर
(b) हिरण
(c) शेर
(d) हाथी।
उत्तर
(d) हाथी।

प्रश्न 10.
भारत से विलुप्त हुआ जन्तु है
(a) हिप्पोपोटामस
(b) स्नो लेपर्ड
(c) चीता
(d) भेड़िया।
उत्तर
(c) चीता

प्रश्न 11.
निम्नलिखित में से कौन-सी संरक्षण की स्व-स्थाने (In-situ) विधि है- .
(a) वानस्पतिक उद्यान
(b) राष्ट्रीय उद्यान
(c) ऊतक संवर्धन
(d) क्रायो-परिरक्षण।
उत्तर
(b) राष्ट्रीय उद्यान

प्रश्न 12.
वन्य जीव अभयारण्य में निम्न में से क्या नहीं होता है
(a) फ्लोरा का संरक्षण
(b) फोना का संरक्षण
(c) मृदा एवं फ्लोरा का उपयोग
(d) शिकार का निषेध।
उत्तर
(c) मृदा एवं फ्लोरा का उपयोग

प्रश्न 13.
भारत में अधिकतर क्षेत्र वनों से आच्छादित है
(a) उड़ीसा
(b) अरूणाचल-प्रदेश
(c) मध्यप्रदेश
(d) केरल।
उत्तर
(b) अरूणाचल-प्रदेश

प्रश्न 14.
वनों का विनाश
(a) प्राकृतिक स्रोतों का विनाश
(b) पर्यावरण का प्रदूषण
(c) आनुवंशिक विकृति
(d) उपर्युक्त सभी।
उत्तर
(b) पर्यावरण का प्रदूषण

प्रश्न 15.
वन्य अनुसंधान केन्द्र स्थित है
(a) शिमला
(b) चेन्नई
(c) देहरादून
(d) कलकत्ता।
उत्तर
(c) देहरादून

प्रश्न 16.
विश्व पर्यावरण दिवस मनाया जाता है
(a)4 मई
(b)5 जून
(c) 15 मार्च
(d) 15 अप्रैल।
उत्तर
(b) 5 जून

प्रश्न 17.
कान्हा राष्ट्रीय उद्यान किस राज्य में स्थित है
(a) उत्तर प्रदेश
(b) राजस्थान
(c) गुजरात
(d) मध्यप्रदेश।
उत्तर
(d) मध्यप्रदेश।

प्रश्न 18.
वे जातियाँ लगातार विलुप्त होती जा रही हैं
(a) दीर्घ जीवी जीव
(b) खतरनाक जीव
(c) साधारण जीव
(d) इनमें से कोई नहीं ।
उत्तर
(c) साधारण जीव

प्रश्न 19.
काजीरंगा राष्ट्रीय उद्यान स्थित है
(a) पश्चिम बंगाल
(b) केरल
(c) असम
(d) गुजरात।
उत्तर
(c) असम

प्रश्न 20.
बान्धवगढ़ नेशनल पार्क किस जिले में है
(a) सतना
(b) शिवपुरी
(c) मंडला
(d) उमरिया।
उत्तर
(d) उमरिया।

2. रिक्त स्थानों की पूर्ति कीजिए

1. FR… …………… में स्थित है।
2. काजीरंगा अभयारण्य …………… के संरक्षण से संबंधित है।
3. रेड डाटा बुक …………… के संरक्षण से संबंधित है।
4. …………… भारतवर्ष का प्रथम बायोस्फियर रिजर्व है।
5. भारत में पक्षियों की …………… प्रजातियाँ पायी जाती हैं।
6. ………….ऊर्जा का अक्षय साधन है।
7. मध्यप्रदेश में …………… के संरक्षण हेतु नेशनल पार्क बनाए गए हैं।
8. वन अपरोपण का मुख्य कारण …………. है।
9. भारतीय शेर व चीता का नाम …………….पशु में आता है।
10. बंजर व खाली (परती) भूमि पर वनों को विकसित करना ………….. कहलाता है।
11. विश्व पर्यावरण दिवस …………… को मनाया जाता है।
उत्तर

1. देहरादून
2. गेंडा
3. विलुप्त प्रायः प्रजाति
4. नीलगिरी
5. 1200
6. सूर्य
7. विलुप्तप्राय जातियों
8. बढ़ती जनसंख्या
9. दुर्लभ
10. वनीकरण
11. 5 जून।

3. सही जोड़ी बनाइए

I. ‘A’ – ‘B’

1. MAB – (a). जन्तु उत्पाद
2. वन संरक्षण अधिनियम – (b) कान्हा
3. पादप जीवाश्म उद्यान – (c) सन् 1973
4. टाइगर प्रोजेक्ट – (d) सन् 1980
5. लाख – (e) मण्डला।
उत्तर
1.(c), 2.(d), 3.(e), 4. (b), 5. (a).

II. ‘A’ – ‘B’

1. संकटमयी जातियाँ – (a) R
2. भेद्य जातियाँ – (b) T
3. दुर्लभ जातियाँ – (c) E
4. आशंकित जातियाँ – (d) V
5. विलुप्त जातियाँ – (e) EVR
उत्तर
1. (e), 2.(d), 3. (a), 4. (b), 5.(c)

III. ‘A’ – ‘B’

1. उत्प्रेरक परिवर्तक – (a) कणकीय पदार्थ
2 स्थिर वैद्युत् अपक्षेपित्र – (b) कार्बन मोनोऑक्साइड और नाइट्रोजन ऑक्साइड (इलेक्ट्रोस्टैटिक प्रेसिपिटेटर)
3. कर्ण सफ (इयर मफ्स) – (c) उच्च शोर स्तर
4. लैण्ड: फेल – (d) ठोस अपशिष्ट।
उत्तर
1. (b), 2. (a), 3. (c), 4. (d).

4. एक शब्द में उत्तर दीजिए

1. विश्व पर्यावरण दिवस कब मनाया जाता है ?
2. म.प्र. में प्रोजेक्ट टाइगर कहाँ स्थापित है ?
3. सिंह भारत में कहाँ पाये जाते हैं ?
4. जंगली गधा कहाँ पाया जाता है ?
5. सोनचिड़िया भारत में कहाँ पायी जाती है ?
6. घाना पक्षी अभयारण्य यह किस राज्य में स्थित है ?
7. भारत के राष्ट्रीय पशु एवं राष्ट्रीय पक्षी का नाम बताइए।
8. चिपको आन्दोलन किस व्यक्ति से संबंधित है।
9. IUCN का मुख्यालय कहाँ है ?
10. भारत का पहला राष्ट्रीय उद्यान कौन-सा है ?
11. जैव विविधता का अन्य नाम लिखिए।
12. विश्व के किस भाग में न्यूनतम जैव विविधता पायी जाती है ?
13. जैव मण्डल की जैविक विविधता का मूलभूत आधार क्या है ?
14. किन्हीं दो प्रान्तों में पूजे जाने वाले पौधों के नाम लिखिए।
15. विश्व में सर्वाधिक जैव विविधता कहाँ होती है ?
16. IUCN का पूर्ण शब्द विस्तार लिखिए।
17. प्रकृति संतुलन में महत्वपूर्ण योगदान कौन करते हैं ?
18. वन्य जीव संरक्षण का अध्ययन विज्ञान की किस शाखा के अंतर्गत किया जाता है ?
19. सामाजिक वानिकी कार्यक्रम कब प्रारम्भ हुआ?
20. अभयारण्य का आधुनिक नाम लिखिए।
उत्तर

1. 5 जून
2. कान्हा शरणस्थल
3. गिर-वन (गुजरात)
4. कच्छ के रन में
5. राजस्थान में
6. राजस्थान में
7. टाइगर, मोर
8. सुन्दरलाल बहुगुणा
9. मॉर्गेस में
10. जिम कार्बेट
11. जैविक विविधता
12. ध्रुवों पर न्यूनतम जैव विविधता पायी जाती है
13. जीन
14. (i) राजस्थान में कदम्ब
    (ii) उड़ीसा में आम,
15. ब्राजील में
16. International Union for Conservation of Nature and Natural Resources
17. वन्य जीव
18. वानिकी
19. 1976
20. शरण-स्थल।

जैव-विविधता एवं संरक्षण लघु उत्तरीय प्रश्न

प्रश्न 1.
जैविक विविधता के संरक्षण को समझाइये।
उत्तर
कोई जीव कभी-भी अकेला नहीं रहता, बल्कि सभी जीव समूहों में दूसरे जीवों के साथ रहने का प्रयास करते हैं। इसी कारण किसी स्थान विशेष में विविध प्रकार के जीव पाये जाते हैं । जैविक विविधता का सबसे बड़ा कारण यह है कि सभी जीव भोजन, आवास या पर्यावरण तथा दूसरी उपयोगी वस्तुओं के लिए एक-दूसरे पर निर्भर रहते हैं। चूँकि सभी एक-दूसरे पर निर्भर हैं इस कारण जैविक विविधता को बनाये रखना आवश्यक है। जैविक विविधता को बनाये रखने वाले उपायों को ही जैविक विविधता का संरक्षण कहते हैं।

प्रश्न 2.
सामुदायिक वानिकी के चार उद्देश्य लिखिए।
उत्तर
सामुदायिक वानिकी सन् 1976 में शुरू की गई एक परियोजना है, जिसके द्वारा वनों का विकास तथा संरक्षण किया जाता है। यह परियोजना भारत सरकार द्वारा चलायी जा रही है। इसके प्रमुख उद्देश्य निम्न हैं|

• वनों में उपयोगी वृक्षों को रोपना।
• व्यक्तिगत क्षेत्रों में सहकारी सहयोग से वनों का विकास।
• प्रदूषण से उत्पन्न खतरों को कृत्रिम वनों के विकास द्वारा दूर करना।
• विलुप्त हो रही वन्य जातियों को विशेष संरक्षण प्रदान करना।

प्रश्न 3.
वनों का महत्व लिखिए।
उत्तर
वन हमारे लिए बहुत अधिक महत्व रखते हैं। इनके महत्व निम्नलिखित हैं

• ये वातावरण को सन्तुलित रखते हैं।
• ये वर्षा को नियन्त्रित करते हैं।
• ये भूमि कटाव तथा बाढ़ पर नियन्त्रण रखते हैं।
• ये प्रदूषण को नियन्त्रित करते हैं ।
• वनों से हमें कई पादप उत्पाद एवं औषधियाँ प्राप्त होती हैं ।
• इनसे कई जन्तु उत्पाद लाख, रेशम, मोम आदि प्राप्त होते हैं ।
• वैज्ञानिक प्रयोगों के लिए जीन कोष का

प्रश्न 5.
वन्य प्राणियों के नष्ट होने के कोई पाँच कारण लिखिए।
अथवा
वन्य जीवों के विलुप्तीकरण के पाँच कारणों को लिखिए।
उत्तर
वन्य प्राणियों के नष्ट होने के पाँच कारण निम्न हैं-

• वन्य प्राणियों के आवासों का प्रतिकूल परिवर्तन-वन कटने के कारण वन्य प्राणियों का आवास परिवर्तित हो गया है।
• शिकार-पैसा कमाने तथा मनोरंजन के लिए वन्य प्राणियों का अन्धाधुन्ध शिकार किया गया है जिसके कारण ये नष्ट हो रहे हैं।
• वैधानिक नियमों की कमी-हमारे देश में शक्तिशाली वैधानिक नियम न होने के कारण वन्य प्राणी नष्ट हुए हैं।
• प्राकृतिक विपदाएँ-प्राकृतिक सम्पदा के अनियन्त्रित दोहन से आयी विपदाओं के कारण भी वन्य प्राणी घटे हैं।
• प्रदूषण-वन्य प्राणियों के आवासों में प्रदूषण के कारण भी वन्य प्राणी नष्ट हुए हैं।

प्रश्न 6.
वन्य प्राणियों के संरक्षण की आवश्यकता किन कारणों से है ? संक्षेप में लिखिए।
उत्तर
वन्य प्राणियों के संरक्षण की आवश्यकता-वन्य प्राणी हमारे लिए अत्यन्त महत्वपूर्ण होते हैं। नीचे कुछ ऐसे लाभ या कारण दिये गये हैं जिसके कारण इनका संरक्षण आवश्यक है

• प्राकृतिक सन्तुलन-ये प्राकृतिक सन्तुलन स्थापित करते हैं।
• आर्थिक महत्व-इनसे कई आर्थिक महत्व के उत्पाद जैसे-दाँत, त्वचा, सींग आदि प्राप्त होते हैं, जिनसे अनेक उपयोगी वस्तुएँ बनाई जाती हैं।
• वैज्ञानिक शोधकार्य-इनका उपयोग वैज्ञानिक शोध कार्यों में किया जाता है।
• मनोरंजन-वन्य प्राणियों को उनके प्राकृतिक आवास में देखने से आनन्द की प्राप्ति होती है।
• धार्मिक तथा सांस्कृतिक मूल्य-इनके साथ हमारी धार्मिक भावनाएँ तथा सांस्कृतिक भावनाएँ भी जुड़ी हैं।

जैव-विविधता एवं संरक्षण दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
संकटग्रस्त प्रजातियाँ क्या हैं ? इनके विभिन्न प्रकारों का वर्णन कीजिये।
उत्तर
पौधों की लगभग 20,000 से 25,000 प्रजातियाँ विनाश के कगार पर हैं (एक अनुमान)। इनका संरक्षण आवश्यक है। ऐसी जातियों के पौधे व जीव जन्तु को ही संकटापन्न या लुप्तप्राय जातियाँ कहते हैं । इनका वर्णन रेड डाटा बुक में किया गया है। द इण्डियन प्लाण्ट्स रेड डाटा बुक में संकटापन्न जातियों को निम्नलिखित समूहों में वर्गीकृत किया गया है

(1) विलुप्त (Extinct)—ऐसे पौधे जो कि पूर्व में किसी स्थान विशेष में पाये जाते थे, लेकिन वर्तमान में वे अपने प्राकृतिक स्थानों से लुप्त हो गये हैं, उन्हें ही विलुप्त प्रजातियाँ कहते हैं। ऐसे पौधे जब उनके प्राकृतिक आवास स्थानों पर उपलब्ध नहीं होते तब इन्हें विलुप्त प्रजाति माना जाता है, अत: इनका संरक्षण असम्भव होता है।

(2) लुप्तप्राय (Endangered)-ऐसी पादप प्रजातियाँ जो कि लुप्त होने की स्थिति में हों एवं यदि वही पारिस्थितिक परिस्थितियाँ बनी रहें तब उन्हें विलुप्त होने से नहीं बचाया जा सकता है, अर्थात् जिन प्रजातियों के लुप्त होने का खतरा बना रहता है उन्हें लुप्तप्राय (Indangered) प्रजाति कहते हैं। ऐसी प्रजातियों की संख्या धीरेधीरे इतनी कम हो जाती है कि उनमें प्रजनन की सम्भावनाएँ लगभग समाप्त हो जाती हैं जिसके कारण धीरे-धीरे ये विलुप्त होने लगती हैं।

(3) वल्नेरेबिल या चपेट में (Vulnerable)—ऐसी पादप प्रजातियाँ जो कि कुछ ही समय में लुप्तप्राय स्थिति में पहुँचने वाली हों उन्हें ही वल्नरेबिल प्रजातियाँ कहते हैं। यदि इन्हें लगातार उन्हीं पारिस्थितिक स्थितियों का सामना करना पड़ता है तब ये प्रजातियाँ भी लुप्त होने लगती हैं।

(4) दुर्लभ (Rare)–ऐसी पादप प्रजातियाँ जो कि संसार में कहीं-कहीं पर और बहुत कम संख्या में उपलब्ध हों उन्हें ही दुर्लभ प्रजातियाँ कहते हैं । ऐसी प्रजातियाँ प्रारम्भ से ही लुप्तप्राय या वल्नेरेबिल नहीं होती, लेकिन धीरे-धीरे लुप्तप्राय स्थिति में आ जाती हैं और अन्त में समाप्त हो जाती हैं।

(5) अपर्याप्त जानकारी (Insufficient knowledge)—ऐसी पादप प्रजातियाँ जिनके सम्बन्ध में यह नहीं कहा जा सकता है कि वे किस समूह (1 से 4) के अन्तर्गत आती हैं, ऐसी प्रजातियों के बारे में हमें सही जानकारी नहीं मिल पाती, लेकिन धीरे-धीरे ये लुप्तप्राय स्थिति में आ जाती हैं।

(6) खतरे से बाहर (Out of danger)—ऐसी समस्त पादप प्रजातियाँ जो कि उपर्युक्त श्रेणियों (1 से 5) के अन्तर्गत पहुँच चुकी होती हैं, लेकिन उनके संरक्षण के पश्चात् पूर्व के वास स्थान पर स्थापित हो जाती हैं उन्हें इस समूह में रखा गया है।

प्रश्न 2.
राष्ट्रीय उद्यान क्या है ? भारतवर्ष के किन्हीं 5 राष्ट्रीय उद्यानों का वर्णन कीजिये।
उत्तर
वह क्षेत्र जो भारत सरकार द्वारा वन्य जीवन के विकास के लिये घोषित किया गया हो, राष्ट्रीय उद्यान कहलाता है। ऐसे क्षेत्रों में वनों के काटने, पशु चारागाह या पशुचारण, खेती इत्यादि की अनुमति नहीं दी जाती। हमारे देश में कुल 66 राष्ट्रीय उद्यान हैं, जिनका कुल क्षेत्रफल 33,988.14 वर्ग किलोमीटर है।
हमारे देश के कुछ प्रमुख राष्ट्रीय उद्यान निम्नलिखित हैं

• शिवपुरी पार्क (Shivpuri Park)—यह मध्य प्रदेश में ग्वालियर के पास शिवपुरी में एक झील के किनारे स्थित है। इसमें चीतल, साँभर, बाघ प्रमुख रूप से पाये जाते हैं।
• गुण्डी डीयर पार्क (Gundy Deer Park)—यह काले चीतल तथा ऐल्विनों हिरणों के लिए स्थापित किया गया है। यह चेन्नई के पास तमिलनाडु में स्थित है।
• जिम कार्बेट पार्क (Jim Corbett Park)—यह उत्तराखण्ड में नैनीताल के पास बनाया गया है। यहाँ शेरों को संरक्षित किया गया है।
• बेतला राष्ट्रीय उद्यान (Betla National Park)—यहाँ बाघों व हाथियों का संरक्षण किया गया है। बिहार राज्य के पलामू जिले में है।
• डचिगम राष्ट्रीय उद्यान (Dachigam National Park)—चीता, काले भालू, कस्तूरी मृग, एण्टिलोप, हिमालय टहर, जंगली बकरी तथा कश्मीरी बारहसिंगों का संरक्षण कश्मीर में किया गया है।

प्रश्न 3.
भारत में वन्य जीवन की विलुप्ति के कारणों की विस्तृत व्याख्या कीजिये।
उत्तर
आदिकाल से ही जन्तुओं की विभिन्न जातियाँ प्राकृतिक कारणों से विलुप्त होती जा रही हैं, जैसे-एमोनाइट्स (Ammonites), दैत्याकार सिफैलोपॉड्स (Cephalopodes), ब्रैकियोपॉड्स (Brachiopods) तथा डाइनोसॉर्स (Dinosaurs) मानव के आगमन के पूर्व ही मध्यजीवी कल्प (Mesozoic era) के समाप्त होतेहोते विलुप्त हो गये। बढ़ती मानव जनसंख्या ने वन्य जीवन एवं उनके प्राकृतिक आवासों का दोहन किया है। साथ ही बढ़ते शहरीकरण, औद्योगीकरण एवं प्रदूषण के परिणामस्वरूप जन्तुओं के लुप्त होने की गति बढ़ती जा रही है। 17वीं, 18वीं एवं 19वीं सदी में जन्तुओं की क्रमशः 7, 11 एवं 27 जातियाँ विलुप्त हुई जबकि बीसवीं सदी में जन्तुओं की 67 जातियाँ विलुप्त हुईं। मानव द्वारा वन्य जीवन के विनाश के विभिन्न कारण निम्नलिखित दो श्रेणियों में आते हैं

(1) प्रत्यक्ष विनाश (Direct destruction)—सुरक्षा, क्रीड़ा, मनोरंजन, मांस, गौरव, उपहार आदि के लिए मानव द्वारा वन्य जन्तुओं के शिकार को हमेशा से प्रोत्साहित किया गया है। शेर, बाघ, तेंदुआ, भेड़िया, आदि महत्वपूर्ण वन्य जन्तुओं का शिकार राजा-महाराजाओं द्वारा किया जाता रहा है। सुरक्षा की दृष्टि से अथवा पशुधन को सुरक्षित रखने के लिए भी इनका वध किया जाता है ।

सौन्दर्य प्रसाधनों, सुगन्ध द्रव्यों, साज-सज्जा आदि के लिए भी इनका शिकार किया गया। ढेलों को प्रसाधनों एवं साबुन उद्योगों में प्रयुक्त होने वाले वसा के लिए हजारों की संख्या में प्रतिवर्ष मारा जाता है। इसी प्रकार हाथी दाँत के लिए हाथियों का, कामोत्तेजक औषधियों (Aphrodisiac) के संश्लेषण में प्रयुक्त सींग के लिए गेंडे (Rhinoceros) का, कस्तूरी के लिए कस्तूरी मृगों (Musk deers) का, फर या समूर के लिए हिमालयी हिमचीते (Himalayan snow-leopard) आदि का वध होता चला आ रहा है जिसके कारण आज यह संकटग्रस्त अवस्था में पहुँच गये हैं।

(2) अप्रत्यक्ष विनाश (Indirect destruction)-वन्य जीवन के अप्रत्यक्ष रूप से विनाश के भी अनेक कारण हैं। इनमें सर्वाधिक प्रमुख कारण है–मनुष्य की निरन्तर बढ़ रही जनसंख्या की विभिन्न आवश्यकताओं की पूर्ति के लिए होने वाले भूमि अधिग्रहण । जैसे-जैसे मानव जनसंख्या में वृद्धि होती गयी, आवास, कृषि, ईंधन एवं औद्योगीकरण आदि की आवश्यकताओं में भी वृद्धि होती गयी।

परिणामस्वरूप वकेन्मूलन (Deforestation), आवासों का विनाश, मरुस्थलों का प्रसार आदि से मानव आवश्यकताओं की पूर्ति करता चला गया, जिसका प्रत्यक्ष प्रभाव वन्य जीव-जन्तुओं पर पड़ा और उनकी संख्या में लगातार कमी आती चली गई। कीटनाशकों के प्रयोग एवं पर्यावरणीय प्रदूषण ने भी जीवों का विनाश किया है।

प्रश्न 4.
वन संरक्षण के राष्ट्रीय तथा अन्तर्राष्ट्रीय प्रयासों का उल्लेख कीजिए।
उत्तर
वन संरक्षण के लिए राष्ट्रीय स्तर पर प्रयास अंग्रेजी शासन के समय ही प्रारम्भ हो गया था। सन् 1856 में लॉर्ड डलहौजी ने बर्मा के वनों के संरक्षण के लिए एक नीति बनायी थी, जो बाद में पूरे देश में लागू की गयी। सन् 1894 में भारत सरकार ने वनों के संरक्षण के लिए एक राष्ट्रीय स्तर पर नीति बनायी। इस नीति में निम्नलिखित बातों पर अधिक ध्यान दिया गया

• वन प्रबन्धन
• वन भूमि का उचित उपयोग
• सुरक्षित वनों की नीति
• वन उत्पादों की बढ़ोतरी।

इस नीति के तहत् वनों तथा जन्तुओं के संरक्षण के लिए भारत सरकार ने राष्ट्रीय अभ्यारण्य एवं प्राणी उद्यानों की स्थापना की है।
अन्तर्राष्ट्रीय स्तर पर भी वनों के संरक्षण के लिए संयुक्त राष्ट्र संघ की एफ. ए. ओ. संस्था कार्य कर रही है। यह संस्था वनों के संरक्षण के लिए आर्थिक मदद भी करती है। इस संस्था के द्वारा समय-समय पर सेमिनार तथा वर्क शॉप आयोजित किये जाते हैं। इसी संस्था के द्वारा बस्तर क्षेत्र का सर्वेक्षण कराया गया है और वहाँ पर देवदार ( पाइन) एवं बाँस लगाने की योजना बनायी गयी है । सन् 1952 में भारत सरकार ने इसी संस्था के निर्देश पर “India’s New National Forest Policy” नामक एक राष्ट्रीय वन नीति बनायी है। इस नीति में निम्नलिखित बातों पर विशेष ध्यान दिया गया है

• पहाड़ी क्षेत्रों में वृक्षों को कटने से बचाना ।
• नष्ट हुए वनों के पूरक वनों को विकसित करना।
• अपरदन रोकने के लिए वृक्षों को लगाना ।
• चारागाहों का विकास करना जिससे वनों पर कम भार पड़े।
• औद्योगिक दृष्टि से उपयोगी वनों को लगाना।
• वनों से होने वाली सरकारी आय को बढ़ाना।

आजकल वनों के संरक्षण के लिए सामाजिक संस्थाओं के साथ आम जनता भी बहुत जागरूक हो गयी है और इसने कई आन्दोलन प्रारम्भ कर दिये हैं।  पं. सुन्दर लाल बहुगुणा द्वारा उत्तर प्रदेश के टेहरी गढ़वाल क्षेत्र में चलाया जाने वाला चिपको आन्दोलन दिशा में एक महत्वपूर्ण उदाहरण है।

प्रश्न 5.
भारत के प्रमुख वन्य प्राणियों पर एक लेख लिखिए।
उत्तर
भारत के प्रमुख वन्य प्राणी-भारत की जलवायु में बहुत अधिक विविधता पायी जाती है। एक तरफ हिमालयी क्षेत्र का तापमान 0°C होता है, दूसरी तरफ राजस्थान का तापमान 49°C रहता है। जलवायु की विविधता के कारण भारत के वनों एवं प्राणियों में भी बहुत अधिक विविधता पायी जाती है। भारत के कुछ प्रमुख वन्य प्राणी निम्नानुसार हैं
(1) पशु-भारत के वनों में पाये जाने वाले प्रमुख पशु निम्नलिखित हैं

1. मृग या हिरण-हमारे देश में इनकी कई जातियाँ, जैसे-कस्तूरी मृग, बार्किग हिरण, सांभर, चीतल आदि पायी जाती हैं।
2. एण्टीलोप-ये मृग के ही समान होते हैं, जैसे-नीलगाय, बारहसिंघा, चौसिंघा, भारतीय गजेले (Gazelle) आदि।
3. हाथी-यह वर्तमान में पृथ्वी का सबसे बड़ा चौपाया है, जो अधिकतर केरल तथा उत्तर के तराई भागों में पाया जाता है।
4. गैण्डा-यह हिमालय क्षेत्र, बंगाल एवं असम के जंगलों में पाया जाता है। सींग के कारण इनका इतना शिकार हुआ है कि यह विलुप्त होने के कगार पर है।
5. जंगली गधा-संसार में यह और कहीं नहीं पाया जाता है । अब भारत में भी यह केवल कच्छ के रन क्षेत्र तक ही सीमित रह गया है।
6. मांसाहारी पशु-कुछ भारतीय वन्य मांसाहारी पशु निम्नलिखित हैं-

• भारतीय सिंह-अब गिर के वनों तक ही सीमित हैं।
• चीता-यह विलुप्त होने के कगार पर है।
• शेर-भारत का राष्ट्रीय पशु है, हमारे देश में वर्तमान में इनकी संख्या 3000 से ज्यादा है।
• तेंदुआ-ये चीते के समान, लेकिन छोटे होते हैं।

(2) पक्षी-हमारे देश के वनों में मोर, जंगली मुर्गा, कई प्रकार के बत्तखें, बगुले, कबूतर, तीतर, बटेर, गरुढ़, गिद्ध, सारस, उल्लू, बाज, दूधराज आदि पक्षी पाये जाते हैं।

(3) सरीसृप-हमारे देश के वनों में मगर, घड़ियाल, कछुए, छिपकलियाँ, सर्प आदि वन्य प्राणी सरीसृप वर्ग के पाये जाते हैं। इसके अलावा भी भारत के वनों में कई कशेरुकी तथा अकशेरुकी प्राणी पाये जाते हैं।

प्रश्न 6.
राष्ट्रीय वन नीति के अन्तर्गत किन प्रमुख बातों पर ध्यान दिया गया है?
उत्तर
राष्ट्रीय वन नीति के अन्तर्गत निम्नलिखित प्रमुख बातों पर ध्यान दिया गया है
(1) वन प्रबन्ध-इसके अन्तर्गत सरकार ने उपयोगी वनों के रख-रखाव व प्रबन्ध की व्यवस्था की है।
(2) वन भूमि का उचित उपयोग-वन में पड़ी फालत भूमि के विवेकपूर्ण उपयोग करने हेत नीति का निर्धारण किया गया है।
(3) सुरक्षित वनों की नीति-सरकार ने कुछ वनों को सुरक्षित वन घोषित किया है। इसमें वृक्षों की कटाई पर पूरी तरह से प्रतिबन्ध लगायी गयी है।
(4) बन उत्पाद की बढ़ोत्तरी-इसके अन्तर्गत वनों के विभिन्न उत्पादों को बढ़ाने हेतु प्रयास एवं नयी जानकारियों को खोजा गया है। उपर्युक्त बातों के अलावा इस नीति में निम्नलिखित बातों पर विशेष रूप से ध्यान दिया गया है

• पहाड़ी क्षेत्रों के वृक्षों को काटने से रोकना
• नष्ट किये वनों के स्थान पर पूरक वनों को रोपना
• भूमि अपरदन को रोकने के लिए वृक्षारोपण करना।
• वनों के बीच चारागाहों का विकास करना, जिससे पशुओं के द्वारा वनों को नष्ट होने से बचाया जा सके।
• औद्योगिक दृष्टि से उपयोगी वनों को लगाना ।
• वनों से होने वाली आय को बढ़ाने का प्रयास करना।

प्रश्न 7.
वन तथा वन्य प्राणी एवं उनके संरक्षण के अन्तर्सबन्धों पर एक लेख लिखिए।
उत्तर
बन तथा वन्य प्राणी-मानव के आवासीय क्षेत्र से बाहर वृक्षों, झाड़ियो तथा घासों से आच्छादित क्षेत्र जिसमें वृक्ष प्रभावी रूप में हो वन कहलाता है, जबकि वह वन जिसमें किसी भी प्रकार का मानव हस्तक्षेप न हो जंगल कहलाता है। वनों में निवास करने वाले प्राणियों को वन्य प्राणी कहते हैं। भारत की जलवायु में विविधता के कारण यहाँ के वनों तथा वन्य प्राणियों में बहुत अधिक विविधता पायी जाती है।

भारतीय वन वन्य प्राणियों के मामले मे समृद्ध हैं। भारत में कुछ 500 प्रकार के स्तनी, 1200 प्रकार के पक्षियाँ, 220 प्रकार के सर्प, 150 प्रकार की छिपकलियाँ, 30 प्रकार के कहुए, 30 प्रकार के मगर तथा घड़ियाल पायी जाती हैं। इनमें से हिरण, ऐण्टिीलोप, जंगली भैंसा, बाइसन, हाथी, गेण्डा, जंगली गधा, सिंह, चीता, शेर, तेंदुआ आदि जातियों के पशु एवं मोर, जंगली मुर्गा, बत्तख, तीतर, बटेर, कबूतर, सारस, गिद्ध, सोनचिड़िया आदि प्रमख पक्षियों एवं मगर, घडियाल तथा सर्प प्रमुख रूप से पाये जाने वाले प्राणी हैं।

वन तथा वन्य प्राणी के संरक्षण के अंतर्सम्बन्ध-चूँकि वन्य प्राणी वनों में पाये जाने वाले जन्तु हैं। इस कारण वनों के बिना वन्य प्राणियों की कल्पना ही नहीं की जा सकती। अगर हमें वन्य प्राणियों को संरक्षित रखना है, तो उनके प्राकृतिक आवासों अर्थात् वनों को संरक्षित रखना अत्यन्त आवश्यक है। हमारे कई उपयोगी वन्य प्राणियों के विलुप्त होने का एकमात्र कारण यह है कि उनका प्राकृतिक आवास अर्थात् वन, दिन-प्रतिदिन कम होता जा रहा है।

वनों में पाये जाने वाले वन्य प्राणी भी वनों को संरक्षित करते हैं, क्योंकि इनके भय से वन को हानि पहुँचाने वाले जीव वनों में प्रवेश नहीं करते। इस प्रकार वन तथा वन्य प्राणी एक-दूसरे से घनिष्ट रूप से जुड़े हैं था एक-दूसरे का प्राकृतिक रूप से संरक्षण एवं पोषण करते हैं। इस कारण दोनों के अस्तित्व को बनाये रखने के लिए दोनों का ही संरक्षित रहना आवश्यक है।

MP Board Class 12th Biology Solutions

MP Board Class 12th Chemistry Solutions Chapter 14 जैव-अणु

जैव-अणु NCERT पाठ्यनिहित प्रश्नोत्तर

प्रश्न 1.
ग्लूकोज तथा सुक्रोज जल में विलेय है जबकि साइक्लो-हेक्सेन अथवा बेंजीन (सामान्यतः छः सदस्यीय वलय युक्त यौगिक) जल में अविलेय होते हैं, समझाइए।
उत्तर
बेंजीन व साइक्लोहेक्सेन में न तो यह ध्रुवीय और न ही इनमें -OH समूह होता है। अत: यह जल के अणुओं के साथ हाइड्रोजन बंध बनाने के योग्य नहीं होते हैं, अतः जल में अघुलनशील होते हैं।

ग्लूकोज व सुक्रोज ध्रुवीय अणु होते हैं तथा इनमें बड़ी संख्या में -OH समूह उपस्थित होते हैं (ग्लूकोज में पाँच तथा सुक्रोज में आठ) हैं ये व्यापक (विस्तृत) रूप में जल के साथ हाइड्रोजन बंध बनाते हैं, अत: ये जल में घुलनशील होते हैं।

प्रश्न 2.
लैक्टोज के जल-अपघटन से किन उत्पादों के बनने की अपेक्षा करते हैं ?
उत्तर
गैलैक्टोज तथा ग्लूकोज, लैक्टोज के जल-अपघटन पर बनने वाले उत्पाद है।

प्रश्न 3.
D-ग्लूकोज के पेन्टाऐसीटेट में आप ऐल्डिहाइड समूह की अनुपस्थिति को कैसे समझाएँगे?
उत्तर
जब ग्लूकोज ( α या β ) की क्रिया एसीटिक ऐनहाइड्राइड से की जाती है, तो यह एक पेन्टाएसीटिल व्युत्पन्न बनाता है, जिसमें C-1 पर मुक्त -OH समूह नहीं होता है। ये जलीय विलयन में जलअपघटित होकर खुली-श्रृंखला वाला ऐल्डिहाइडिक रूप नहीं बनाता है, अत: ग्लूकोज पेन्टाऐसीटेट NH₂OH के साथ क्रिया कर ग्लूकोज ऑक्सिम नहीं बनाता है।

प्रश्न 4.
एमीनो अम्लों के गलनांक व जल में विलेयता सामान्यतः संगत हैलो अम्लों की तुलना में अधिक होती है, समझाइए।
उत्तर
एमीनो अम्ल ज्विटर आयन के रूप में प्रदर्शित होते हैं  इस द्विध्रुव के लवण के समान गुण होने के कारण इनमें प्रबल द्विध्रुव-द्विध्रुव आकर्षण या स्थिर विद्युतीय आकर्षण पाया जाता है। अतः इनके गलनांक हैलो अम्लों से उच्च होते हैं जिनमें लवण के समान लक्षण नहीं होते हैं तथा लवण के समान गुण होने के कारण इनका जल के साथ प्रबलतम आकर्षण होता है । अतः एमीनो अम्ल की जल में घुलनशीलता उनके संगत हैलो अम्लों जिनमें लवण के समान लक्षण नहीं होते हैं से ज्यादा होती है।

प्रश्न 5.
अण्डे को उबालने पर उसमें उपस्थित जल कहाँ चला जाता है ?
उत्तर
जब अण्डे को उबाला जाता है तब प्रोटीन के विकृतिकरण तथा स्कंदन संभवत: H- बंध के द्वारा होता है । अण्डे में उपस्थित जल अवशोषित हो जाता है या विकृतिकरण के दौरान अवशोषित या विलुप्त हो जाता है। इस क्रिया में गोलिकाकार प्रोटीन अघुलनशील रेशेदार प्रोटीन में परिवर्तित हो जाते हैं।

प्रश्न 6.
हमारे शरीर में विटामिन C संचित क्यों नहीं होता है ?
उत्तर-
विटामिन-C जल में घुलनशील होता है। यह हमारे शरीर में संचित नहीं होता है क्योंकि ये मूत्र के द्वारा आसानी से निष्कासित हो जाता है।

प्रश्न 7.
यदि DNA के थायमीन युक्त न्यूक्लियोटाइड का जल-अपघटन किया जाये, तो कौनकौन से उत्पाद बनेंगे?
उत्तर
जल-अपघटन उत्पाद-2-डिऑक्सी-D-राइबोज (शर्करा) + थायमीन (क्षार) + फॉस्फोरिक अम्ल।

प्रश्न 8.
यदि RNA का जल-अपघटन किया जाता है, तो प्राप्त क्षारकों की मात्राओं के मध्य कोई संबंध नहीं होता। यह तथ्य RNA की संरचना के विषय में क्या संकेत देता है ?
उत्तर
जब RNA जल-अपघटित होता है, तब चारों क्षारों की मात्राओं में कोई संबंध नहीं होता है अर्थात् साइटोसीन (C), ग्वानीन (G), एडेनीन (A) तथा यूरेसिल (U) प्राप्त होते हैं। अतः क्षारयुग्मन सिद्धान्त अर्थात् (A) का (T) के साथ युग्म तथा (C) का (G) के साथ युग्मन नहीं होता (जैसा कि DNA में होता है)। अतः RNA एक एकल रज्जुक (तन्तु) अणु है।

जैव-अणु NCERT पाठ्य-पुस्तक प्रश्नोत्तर

प्रश्न 1.
मोनोसैकेराइड क्या होते हैं ?
उत्तर
मोनोसैकेराइड कार्बोहाइड्रेट होते हैं, जो पुनः जल-अपघटित होकर पॉलीहाइड्रॉक्सी ऐल्डिहाइड व कीटोन की सरलतम इकाई नहीं देते हैं।

प्रश्न 2.
अपचायी शर्करा क्या होती है ?
उत्तर
अपचायी शर्करा कार्बोहाइड्रेट है, जो फेहलिंग विलयन को अपचयित करके Cu₂O का लाल अवक्षेप बनाते हैं तथा टॉलेन्स अभिकर्मक से चमकदार धात्विक सिल्वर बनाती है। सभी मोनोसैकेराइड (एल्डोज तथा कीटोज दोनों) तथा डाइसैकेराइड सुक्रोज को छोड़कर अपचायी शर्करा होते हैं। अत: D(+) ग्लूकोज, D(+) गैलैक्टोज, D(-) फ्रक्टोज, D(+) माल्टोज तथा D(+) लैक्टोज अपचायी शर्करा है।

प्रश्न 3.
पौधों में कार्बोहाइड्रेटों के दो मुख्य कार्यों को लिखिए।
उत्तर

• सेल्युलोज प्रमुखतः पादप-कोशिकाओं की कोशिका भित्ति (Cell wall) बनाते हैं।
• स्टार्च पौधों में संग्रहित मुख्य पॉलीसैकेराइड है। स्टार्च बीजों में जमा होता है जब तक कि वे अपना भोजन प्रकाश-संश्लेषण द्वारा स्वतः नहीं बना लेते हैं तब तक उन्हें खाद्य पदार्थ प्रदान करते हैं।

प्रश्न 4.
निम्नलिखित को मोनोसैकैराइड तथा डाइसैकेराइड में वर्गीकृत कीजिए-राइबोज, 2-डि-ऑक्सीराइबोज, माल्टोज, गैलैक्टोज, फ्रक्टोज तथा लैक्टोज।
उत्तर
मोनोसैकेराइड- राइबोस, 2-डि-ऑक्सीराइबोज, गैलैक्टोज तथा फ्रक्टोज।
डाइसैकेराइड- माल्टोज तथा लैक्टोज।

प्रश्न 5.
ग्लाइकोसाइडी बंध से आप क्या समझते हैं ?
उत्तर
ऑक्सीजन लिंकेज जिसके द्वारा दो मोनोसैकेराइड इकाई जल के एक अणु को निष्कासित कर डाइसैकेराइड का एक अणु बनाते हैं । यह ग्लाइकोसिडिक लिंकेज या ग्लाइकोसाइडी बंध कहलाते हैं। उदाहरण के लिये-सुक्रोज (एक डाइसैकेराइड) a-ग्लूकोज के C₁ तथा 8-फ्रक्टोज के C₂ में ग्लाइकोसाइडी बंध द्वारा बनते हैं।

प्रश्न 6.
ग्लाइकोजेन क्या होता है तथा ये स्टार्च से किस प्रकार भिन्न है ?
उत्तर
1. कार्बोहाइड्रेट जन्तुओं के शरीर में ग्लाइकोजेन के रूप में संगृहित होते हैं । ये यकृत, माँसपेशियों तथा मस्तिष्क में उपस्थित होते हैं । एन्जाइम, ग्लाइकोजेन को ग्लूकोज में तोड़ता है जब शरीर को ग्लूकोज की आवश्यकता होती है।

2. ग्लाइकोजेन एमाइलोपेक्टीन (स्टार्च) की तुलना में ज्यादा शाखित होता है। ग्लाइकोजेन श्रृंखला 10-14 ग्लूकोज इकाइयों से बनी होती है, जबकि एमाइलोपेक्टीन 20-25 ग्लूकोज इकाइयों से बना होता है।

प्रश्न 7.
(अ) सुक्रोज तथा (ब)लैक्टोज के जल-अपघटन से कौन-से उत्पाद प्राप्त होते हैं ?
उत्तर
(अ) ग्लूकोज तथा फ्रक्टोज।

(ब) D-ग्लूकोज तथा D-गैलैक्टोज।

प्रश्न 8.
स्टार्च तथा सेल्युलोज में मुख्य संरचनात्मक अंतर क्या है ?
उत्तर
स्टार्च दो यौगिकों का बना होता है-एमाइलोज तथा एमाइलोपेक्टीन । एमाइलोज 200-1000 α-D(+) ग्लूकोज इकाई का लम्बा रेखीय बहुलक है, जो C1-C₄ ग्लाइ-कोसिडिक बंध द्वारा संघटित होता है। ये पानी में घुलनशील होता है। एमाइलोपेक्टीन α-D(+) ग्लूकोज बंध का शाखित शृंखलित बहुलक होता है, जो C₁-C₆ ग्लाइकोसिडिक बंध द्वारा शाखित होता है। ये पानी में अघुलनशील होता है।

दूसरी तरफ सेल्युलोज सीधी शृंखला वाला पॉलीसैकेराइड है, जो β-D(+) ग्लूकोज इकाई द्वारा संघटित होता है जिसमें एक ग्लूकोज इकाई के C₁ तथा दूसरी ग्लूकोज इकाई के C₄ के बीच ग्लाइकोसाइड बंध बनता है।

प्रश्न 9.
क्या होता है, जब D-ग्लूकोज की अभिक्रिया निम्नलिखित अभिकर्मकों से करते हैं

1. HI
2. ब्रोमीन जल
3. HNO₃

उत्तर
1. जब ग्लूकोज की क्रिया HI से करायी जाती है, तो यह n-हेक्सेन बनाता है, जो पुष्टि करता है, कि सभी छ: कार्बन सीधी श्रृंखला में बँधे होते हैं।

2. ग्लूकोज को ब्रोमीन जल के साथ गरम करने पर यह छ: कार्बन वाले कार्बोक्सिलिक अम्ल ग्लूकोनिक अम्ल में ऑक्सीकृत हो जाता है।

3. ग्लूकोज को नाइट्रिक अम्ल के साथ क्रिया करने पर यह डाइकार्बोक्सिलिक अम्ल सैकेरिक अम्ल देता है

प्रश्न 10.
ग्लूकोज की उन अभिक्रियाओं का वर्णन कीजिए जो इसकी विवृत्त श्रृंखला संरचना के द्वारा नहीं समझायी जा सकती है।
उत्तर
निम्न अभिक्रियाएँ ग्लूकोज की विवृत्त श्रृंखला संरचना के द्वारा नहीं समझायी जा सकती है

• ग्लूकोज का पेण्टा-ऐसीटेट हाइड्रॉक्सिल एमीन से क्रिया नहीं करता है। इसमें मुक्त –CHO समूह की अनुपस्थिति प्रदर्शित करता है।
• ऐल्डिहाइडिक समूह के बावजूद –
  • ग्लूकोज NaHSO₃ के साथ हाइड्रोजन सल्फाइड योगा-त्मक उत्पाद नहीं बनाता है।
  • ग्लूकोज शिफ परीक्षण नहीं देता है।

प्रश्न 11.
आवश्यक तथा अनावश्यक ऐमीनो अम्ल क्या होते हैं ? प्रत्येक प्रकार के दो उदाहरण दीजिए।
उत्तर
1. आवश्यक एमीनो अम्ल-एमीनो अम्ल जिन्हें हमारा शरीर नहीं बनाता है, ये आहार से प्राप्त होते हैं । उदाहरण-वैलीन, आइसोल्यूसीन, आर्जिनीन, ल्यूसीन, थ्रिऑनीन आदि।

2. अनावश्यक एमीनो अम्ल-ये वे एमीनो अम्ल हैं जिन्हें हमारा शरीर बनाता है। उदाहरणग्लूसीन, ऐलानिन, ग्लूटेमिक अम्ल, ऐस्पार्टिक अम्ल, ग्लूटेमिन, सेरीन इत्यादि।

प्रश्न 12.
प्रोटीन के संदर्भ में निम्नलिखित को परिभाषित कीजिए

1. पेप्टाइड बंध,
2. प्राथमिक संरचना,
3. विकृतिकरण।

उत्तर
1. पेप्टाइड बंध-पेप्टाइड बंध एक एमाइड बंध है, जो -COOH समूह एक -एमीनो अम्ल के तथा दूसरे -एमीनो अम्ल के -NH₂ समूह के बीच जल के एक अणु के निष्कासन द्वारा बनता है। . ये दो एमीनो अम्ल की इकाइयों को एक पेप्टाइड अणु में जोड़ देती है।

2. प्रोटीन – एमीनो अम्लों का पॉलीमर है-यह पॉलीमर (पॉलीपेप्टाइड के नाम से भी जाने जाते हैं), एमीनो अम्लों से जो एक-दूसरे से विशिष्ट क्रम के साथ बंधा होता है के द्वारा संघटित होता है। एमीनो अम्लों का यह क्रम प्रोटीन की प्राथमिक संरचना कहलाता है। एमीनो अम्लों के इस क्रम में कोई भी परिवर्तन (अर्थात् प्राथमिक संरचना) एक अलग प्रोटीन का निर्माण करता है।

3. विकृतिकरण- वह अभिक्रिया जो प्रोटीन की भौतिक तथा जैविक गुणों को बिना प्रोटीन की रासायनिक संघटन में प्रभाव डाले परिवर्तित कर देते हैं, विकृतिकरण कहलाते हैं। विकृतिकरण निश्चित भौतिक व रासायनिक उपचार है, जैसे-pH में परिवर्तन, ताप, कुछ लवणों की उपस्थिति या निश्चित रासायनिक कारकों के कारण होता है।

प्रश्न 13.
प्रोटीन की द्वितीयक संरचना के सामान्य प्रकार क्या हैं ?
उत्तर
पॉलीपेप्टाइड श्रृंखला का संरूपण जिसकी कल्पना हाइड्रोजन बंध के फलस्वरूप की जाती है। प्रोटीन की द्वितीयक संरचना कहलाती है। इनके दो प्रकार की द्वितीयक संरचना है-

• a-हेलिक्स तथा
• P-प्लेटेडेड शीट संरचना (विस्तार के लिए NCERT पाठ्य-पुस्तक में देखें)।

प्रश्न 14.
प्रोटीन की -हेलिक्स संरचना के स्थायीकरण में कौन-से आबंध सहायक होते हैं?
उत्तर
6,6 पेप्टाइड बंधों से जो -NH तथा -C=0 समूह के बीच हाइड्रोजन बंध से बनता है। -हेलिक्स संरचना में स्थायित्व देता है (विस्तार के लिये NCERT पाठ्य-पुस्तक में देखें)।

प्रश्न 15.
रेशेदार तथा गोलिकाकार (Globular) प्रोटीन को विभेदित कीजिए।
उत्तर
रेशेदार तथा गोलिकाकार प्रोटीन में अन्तर

प्रश्न 16.
ऐमीनो अम्लों की उभयधर्मी प्रकृति को आप कैसे समझायेंगे?
उत्तर
द्विध्रुवीय या ज्विटर आयन संरचना के कारण एमीनो अम्ल उभयधर्मी स्वभाव के होते हैं । एमीनो अम्लों का अम्लीय स्वभाव  समूह के कारण तथा क्षारीय गुण – COOH समूह के कारण जैसा कि निम्न में दिखाया गया है, होता है

प्रश्न 17.
एन्जाइम क्या होते हैं ?
उत्तर
एन्जाइम गोलिकाकार प्रोटीन होते हैं जो जैव-उत्प्रेरक की तरह कार्य करते हैं। ये बहुत विशिष्ट तथा अपनी क्रियाओं में दक्ष होते हैं । लगभग सभी क्रियाएँ जो हमारे शरीर में होती है, एन्जाइम द्वारा उत्प्रेरित होती
है।

प्रश्न 18.
प्रोटीन की संरचना पर विकृतिकरण का क्या प्रभाव होता है ?
उत्तर
वकृतिकरण के दौरान प्रोटीन की 2° तथा 3° संरचना नष्ट हो जाती है, प्राथमिक संरचना वैसी ही रहती है। विकृतिकरण के परिणामस्वरूप गोलिकाकार प्रोटीन (H₂O में घुलनशील) रेशेदार प्रोटीन में (H₂O में अघुलनशील) परिवर्तित हो जाते हैं तथा उनकी जैविक क्रियाशीलता नष्ट हो जाती है। अण्डे की सफेदी को उबालने पर स्कंदन विकृतिकरण का सामान्य उदाहरण है।

प्रश्न 19.
विटामिनों को किस प्रकार वर्गीकृत किया गया है ? रक्त के थक्के जमने के लिए जिम्मेदार विटामिन का नाम दीजिए।
उत्तर
वसा या जल में घुलनशीलता के आधार पर विटामिन सामान्यतः निम्न दो प्रकारों में वर्गीकृत होते है –

1. जल में विलेय विटामिन-इसमें विटामिन B-संकुल (B₁,B₂, B₃,B₄, B₆, B₁₂ तथा निकोटिनिक अम्ल इत्यादि) तथा विटामिन C शामिल है।

2. वसा में विलेय विटामिन-इसमें विटामिन A,D,E तथा K शामिल हैं। यकृत कोशिका में वसा में विलेय विटामिन बहुतायत में पाये जाते हैं। विटामिन K रक्त के स्कंदन के लिये उत्तरदायी होता है।

प्रश्न 20.
विटामिन A व C हमारे लिए आवश्यक क्यों हैं ? उनके महत्वपूर्ण स्रोत दीजिए।
उत्तर
विटामिन A -ये हमारे लिये आवश्यक होते हैं क्योंकि इनकी कमी से रतौंधी तथा जीरोफ्थैल्मिया (आँख की कॉर्निया का कठोरीपन) का कारण बनती है।
स्रोत- गाजर, दूध, मक्खन, मछली के यकृत का तेल, अण्डे का योक, पीली व हरी सब्जियाँ।

विटामिन C- ये हमारे लिये आवश्यक होते हैं क्योंकि इनकी कमी के कारण स्कर्वी (मसूड़ों में रक्त बहाव), दाँतों का टूटना, पाइरिया इत्यादि होता है।
स्रोत- नींबू, संतरा (रसीले फल), आँवला, टमाटर, आलू तथा हरी पत्तेदार सब्जियाँ।

प्रश्न 21.
न्यूक्लिक अम्ल क्या होते हैं ? इनके दो महत्वपूर्ण कार्य लिखिए।
उत्तर
न्यूक्लिक अम्ल न्यूक्लियोटाइड की लंबी शृंखलित बहुलक होते हैं। इन्हें पॉलीन्यूक्लियोटाइड भी कहते हैं। न्यूक्लिक अम्ल मुख्यतः दो प्रकार के होते हैं,

• डि-ऑक्सी-राइबोन्यूक्लिक अम्ल (DNA) तथा
• राइबोन्यूक्लिक अम्ल (RNA) के होते हैं।

कार्य-
1. DNA एक पीढ़ी से दूसरी पीढ़ी तक वंशागत प्रभाव को स्थानान्तरित करते हैं। ये कोशिका विभाजन के दौरान प्रतिकरण (Replication) के विशिष्ट गुण तथा दो DNA रज्जुक (Strand) के पुत्री कोशिका में स्थानान्तरित होने के कारण होता है।

2. DNA तथा RNA सभी प्रोटीन के संश्लेषण के लिए उत्तरदायी होते हैं तथा हमारे शरीर के विकास तथा संरक्षण के लिए आवश्यक होते हैं। वास्तविकता में प्रोटीन का संश्लेषण कोशिका में विभिन्न RNA अणु (m-RNA & t-RNA इत्यादि) के द्वारा होता है। वस्तु विशेष प्रोटीन के संश्लेषण की सुचना DNA में रहती है।

प्रश्न 22.
न्यूक्लियोसाइड तथा न्यूक्लियोटाइड में क्या अन्तर होता है ?
उत्तर
एक न्यूक्लियोसाइड श्रृंखला केवल न्यूक्लिक अम्ल का आधार घटक है, जिसका नाम पेन्टोज शर्करा तथा एक नाइट्रोजनीय क्षार की बनी होती है। एक न्यूक्लियोसाइड में न्यूक्लिक अम्ल के सभी तीन घटक जिनका नाम एक फॉस्फोरिक अम्ल समूह, एक पेन्टोज शर्करा तथा एक नाइट्रोजनीय क्षार होता है।

प्रश्न 23.
DNA के दो रज्जुक समान नहीं होते, अपितु एक-दूसरे के पूरक होते हैं। समझाइए।
उत्तर
DNA अणु में दो रज्जुक (तन्तु) एक-दूसरे से एक रज्जुक के प्यूरीन क्षार तथा दूसरे के पिरिमिडीन क्षार के बीच हाइड्रोजन बंध द्वारा बँधे रहते हैं । क्षारों के विभिन्न आकार तथा ज्यामिति के कारण, DNA में संभावित युग्मन है-ग्वानीन (G) तथा साइटोसीन (C) तीन हाइड्रोजन बंध द्वारा अर्थात् (C=G) तथा एडेनीन A तथा थायमीन T दो हाइड्रोजन बंधों द्वारा (अर्थात् A = T) (चित्र के लिये पाठ्य-पुस्तक देखिए)। इस क्षार युग्मन सिद्धान्तानुसार एक रज्जुक में क्षारों का क्रम स्वतः दूसरे रज्जुक में क्षारों के क्रम को स्थिर करता है। अतः दो रज्जुक एक-दूसरे के पूरक तथा असमान होते हैं।

प्रश्न 24.
DNA तथा RNA में महत्वपूर्ण संरचनात्मक एवं क्रियात्मक अंतर लिखिए।
उत्तर
DNA तथा RNA में अंतर

प्रश्न 25.
कोशिका में पाए जाने वाले विभिन्न प्रकार के RNA कौन-से हैं ?
उत्तर
RNA निम्न तीन प्रकार के होते हैं

• राइबोसोमल RNA(r-RNA),
• संदेशवाहक RNA(m-RNA),
• अंतरण (स्थानान्तरण) RNA(t-RNA)

जैव-अणु अन्य महत्वपूर्ण प्रश्नोत्तर

जैव-अणु वस्तुनिष्ठ प्रश्न

1. सही विकल्प चुनकर लिखिए

प्रश्न 1.
कौन-सा प्रोटीन रक्त प्रवाह द्वारा 0₂ का अभिगमन करता है
(a) मायोग्लोबिन
(b) इन्सुलिन
(c) ऐल्बुमिन
(d) हीमोग्लोबिन।
उत्तर
(d) हीमोग्लोबिन।

प्रश्न 2.
बेरी-बेरी रोग किस विटामिन की कमी से होता है
(a) विटामिन-A
(b) विटामिन-C
(c) विटामिन-B
(d) विटामिन-D.
उत्तर
(c) विटामिन-B

प्रश्न 3.
एन्जाइम जो ग्लूकोज के एथेनॉल में रूपान्तरण को उत्प्रेरित करता है
(a) जाइमेज
(b) इन्वर्टेस
(c) माल्टेस .
(d) डायस्टेज।
उत्तर
(a) जाइमेज

प्रश्न 4.
मानव शरीर में कार्बोहाइड्रेट का संचयन होता है
(a) ग्लूकोज के रूप में
(b) ग्लाइकोजन के रूप में
(c) स्टार्च के रूप में
(d) फ्रक्टोस के रूप में।
उत्तर
(b) ग्लाइकोजन के रूप में

प्रश्न 5.
शर्करा के ताजे विलयन का प्रकाशीय घूर्णन कुछ समय बाद परिवर्तन होना कहलाता है
(a) घूर्णन गति
(b) इन्वर्सन
(c) विशिष्ट घूर्णन
(d) म्यूटारोटेशन।
उत्तर
(b) इन्वर्सन

प्रश्न 6.
बहुधा प्रायोजित डाइसैकेराइड अणु का सूत्र है
(a) C_1oH₁₈O₉
(b) C_1oH₂₀O_1o
(c) C₁₈H₂₂O₁₁
(d) C₁₂H₂₂O₁₁
उत्तर
(d) C₁₂H₂₂O₁₁

प्रश्न 7.
राइबोज के संबंध में निम्न कथन असत्य है
(a) यह पॉलीहाइड्रॉक्सी यौगिक
(b) यह ऐल्डिहाइड शर्करा है
(c) इसमें छ: कार्बन परमाणु हैं
(d) इसमें ध्रुवण घूर्णकता है।
उत्तर
(c) इसमें छ: कार्बन परमाणु हैं

प्रश्न 8.
कार्बोहाइड्रेट को बनाने में आवश्यक होते हैं
(a) 2 कार्बन
(b) 3 कार्बन
(c) 4 कार्बन
(d) 6 कार्बन।
उत्तर
(d) 6 कार्बन।

प्रश्न 9.
हीमोग्लोबिन है
(a) एन्जाइम
(b) ग्लोब्यूलर प्रोटीन
(c) विटामिन
(d) कार्बोहाइड्रेट।
उत्तर
(d) कार्बोहाइड्रेट।

प्रश्न 10.
कौन-सा कार्बोहाइड्रेट पौधों की कोशिकाओं का महत्वपूर्ण अवयव है
(a) सेल्युलोज
(b) स्टार्च
(c) इक्षु शर्करा
(d) विटामिन।
उत्तर
(b) स्टार्च

प्रश्न 11.
हीमोग्लोबिन में कितनी उप-इकाइयाँ उपस्थित होती हैं
(a) 2
(b) 3
(c) 4
(d) 5.
उत्तर
(b) 3

प्रश्न 12.
स्टार्च किसका बहुलक है
(a) ग्लूकोज
(b) सुक्रोज
(c) (a) तथा (b) दोनों का
(d) इसमें से कोई नहीं।
उत्तर
(a) ग्लूकोज

प्रश्न 13.
मानव रक्त में कौन-सी शर्करा अधिकतम विद्यमान है
(a) d-फ्रक्टोज
(b) d-ग्लूकोज
(c) सुक्रोज
(d) लैक्टोज।
उत्तर
(b) d-ग्लूकोज

प्रश्न 14.
विटामिन B₁₂ में धातु होता है
(a) Pb
(b) Zn
(c) Fe
(d) Co.
उत्तर
(d) Co.

प्रश्न 15.
रक्त में ग्लूकोज का मात्रात्मक निर्धारण किया जाता है
(a) टॉलेन अभिकर्मक
(b) बेनेडिक्ट विलयन
(c) क्षारीय आयोडिन विलय
(d) ब्रोमीन जल,
उत्तर
(b) बेनेडिक्ट विलयन

प्रश्न 16.
रिकेट्स किस विटामिन की कमी से होता है
(a) विटामिन-C
(b) विटामिन-B
(c) विटामिन-A
(d) विटामिन-D.
उत्तर
(d) विटामिन-D.

प्रश्न 17.
उपापचयी विधियों में निम्नलिखित में से कौन-सा सर्वाधिक ऊर्जा प्रदान करता है
(a) प्रोटीन
(b) विटामिन
(c) लिपिड .
(d) कार्बोहाइड्रेट।
उत्तर
(d) कार्बोहाइड्रेट।

प्रश्न 18.
विटामिन B, है
(a) राइबोफ्लेविन
(b) कोबालामीन
(c) थायमिन
(d) पिरीमिडीन।
उत्तर
(a) राइबोफ्लेविन

प्रश्न 19.
विटामिन C की कमी से होता है
(a) स्कर्वी
(b) रिकेट्स
(c) पायरिया
(d) रक्ताल्पता।
उत्तर
(a) स्कर्वी

प्रश्न 20.
सभी जीवित कोशिकाओं के अधिकतम प्रभावशाली ऊर्जा वाहक हैं
(a) A.M.P. .
(b) A.T.P.
(c) A.D.P.
(d) U.D.P.
उत्तर
(b) A.T.P.

प्रश्न 21.
दूध में उपस्थित डाइसकेरॉइड है
(a) सुक्रोस
(b) लैक्टोस
(c) माल्टोस
(d) सेलुलोस।
उत्तर
(b) लैक्टोस

प्रश्न 22.
कौन ग्लिसराइड नहीं है
(a) वसा
(b) तेल
(c) फॉस्फोलिपिड
(d) साबुन।
उत्तर
(d) साबुन।

प्रश्न 23.
RNA में नहीं पाया जाता है
(a) थायमीन
(b) यूरेसिल .
(c) ऐडिनीन
(d) ग्वानीन।
उत्तर
(a) थायमीन

प्रश्न 24.
एन्जाइम होते हैं
(a) नाइट्रोजन युक्त जटिल यौगिक
(b) कार्बोहाइड्रेट
(c) उपसहसंयोजी यौगिक
(d) धात्विक यौगिक।
उत्तर
(a) नाइट्रोजन युक्त जटिल यौगिक

प्रश्न 25.
विटामिन C का रासायनिक नाम है
(a) सायनो कोबाल्ट ऐमीन
(b) एस्कार्बिक अम्ल
(c) टोकोफेरॉल
(d) बायोटिन।
उत्तर
(b) एस्कार्बिक अम्ल

26. हीमोग्लोबिन आयरन का ……. यौगिक है।
उत्तर
संकुल।

2. एक शब्द / वाक्य में उत्तर दीजिए

1. विटामिन-C का रासायनिक नाम लिखिए।
2. विटामिन-K का स्रोत बताइए।
3. खून का थक्का न जमने के लिए उत्तरदायी है।
4. अमीनो अम्लों को आपस में कौन-सा बंध जोड़ता है ?
5. मनुष्य के शरीर के द्वारा कितने अमीनो अम्ल संश्लेषित होते हैं ?
6. सेल्यूलोस किस ग्लूकोज का रेखीय बहुलक है ?
7. RNA अणु में थायमिन के स्थान पर कौन-सा पिरामिडीन होता है ?
8. लैक्टोज जल-अपघटन पर देता है।
9. ग्लूकोस में पाइरेनोज वलय होता है, जबकि फ्रक्टोज में।
10. पॉलीसैकेराइडों में मोनोसैकेराइड की इकाइयाँ आपस में एक-दूसरे से किस बन्ध के द्वारा जुड़ी रहती हैं ?
11. रक्त का थक्का बनाने में सहायक प्रोटीन क्या कहलाता है?
12. मोनोसैकेराइड कार्बोहाइड्रेट का एक उदाहरण लिखिए।
13. दूध में उपस्थित डाईसैकेराइड शर्करा क्या कहलाती है ?

उत्तर

1.  ऐस्कार्बिक अम्ल
2. हरे पत्तेदार सब्जियाँ
3. विटामिन-K (फाइलो क्वीनोन)
4. पेप्टाइड बंध
5. दस
6. B-ग्लूकोज
7. यूरेसिल
8. ग्लूकोज और लैक्टोज
9. फ्यूरेनोज वलय
10. ग्लाइकोसाइडिक
11. फाइब्रिनोजेन
12. ग्लूकोज या फ्रक्टोज
13. लैक्टोस।

3. रिक्त स्थानों की पूर्ति कीजिए

1. ग्लूकोज के ऑक्सीकरण में ATP के ………. अणु उत्पन्न होते हैं।
2. जीवों में जटिल अणुओं का टूटना …………. कहलाता है।
3. हाइपरग्लाइसेमिया में रक्त में ……….. की मात्रा बढ़ जाती है।
4. ………….. की कमी से आँखों का रोग होता है।
5. आयोडीन की कमी से …………… रोग होता है।
6. रक्त सम्पूर्ण शरीर के ताप को ………. बनाये रखता है।
7. …………. हॉर्मोन रक्त में शर्करा की मात्रा को संतुलित रखता है।
8. ………. रक्त थक्का बनने के लिए उत्तरदायी है।
9. विकृतिकरण प्रोटीन की ………… संरचना को प्रभावित नहीं करता।
10. प्रोटीन ………… का बहुलक है।
11. ………… प्रोटीन की मौलिक इकाई है।
12. ………… DNA में नहीं पाया जाता है।
13. हीमोग्लोबिन आयरन का ………….. यौगिक है।
14. जन्तुओं एवं पौधों से प्राप्त तेल व वसा ………….. कहलाते हैं।

उत्तर

1. 38,
2. कैटाबोलिज्म
3. शर्करा
4. विटामिन-A
5. पेंघा
6. एकसमान
7. इन्सुलिन
8. विटामिन-K
9. प्राथमिक
10. एमीनो अम्लों
11. एमीनो अम्ल
12. यूरेसिल
13. संकुल
14. लिपिड।

4. उचित संबंध जोड़िए

उत्तर

1. (1)
2. (c)
3. (b)
4. (e)
5. (d)
6. (a)
7. (h)
8. (g).

जैव-अणु लघु उत्तरीय प्रश्न

प्रश्न 1.
प्रोटीन की कमी से कौन-सा रोग होता है व इसका शरीर पर क्या प्रभाव पड़ता है ?
उत्तर
प्रोटीन की कमी से होने वाले रोग –

1. ऐनीमिया-मनुष्य में हीमोग्लोबिन नामक प्रोटीन की कमी से यह रोग होता है। इसमें रोगी के शरीर में रक्त की कमी होने से चक्कर आना, शरीर पर झुर्रियाँ पड़ जाना आदि परिलक्षित होते हैं।

2. क्वाशियोरकर-यह रोग मुख्यतः बच्चों में पाया जाता है, इसमें रोगी का शरीर सूजकर बेडौल हो जाता है।

प्रश्न 2.
कार्बोहाइड्रेट को परिभाषित कीजिए।
उत्तर
वे पदार्थ जो जल, हाइड्रॉक्सी ऐल्डिहाइड या पॉलीहाइड्रॉक्सी कीटोन हैं अथवा वे पदार्थ जो जल-अपघटित होने पर ये यौगिक देते हैं, कार्बोहाइड्रेट कहलाते हैं।

प्रश्न 3.
विटामिन-C का रासायनिक नाम, स्रोत, सूत्र तथा इसकी कमी से उत्पन्न रोगका नाम लिखिए।
उत्तर
विटामिन-C का रासायनिक नाम-ऐस्कार्बिक अम्ल हैं।
विटामिन-C के स्रोत–सन्तरा, नीबू, आँवला, टमाटर।
विटामिन-C की कमी से होने वाले रोग-स्कर्वी, पायरिया, हड्डियों का कमजोर होना।
विटामिन-C का सूत्र- C₆H₈O₆.

प्रश्न 4.
निम्नलिखित विटामिन की कमी से होने वाले रोग लिखिए
(a) विटामिन-A
(b) विटामिन-B
(c) विटामिन-D
(d) विटामिन-E
उत्तर
उपर्युक्त विटामिनों की कमी से होने वाले रोग निम्नलिखित हैं
(a) विटामिन-A की कमी-रतौंधी (Night blindness)।
(b) विटामिन-B की कमी- भूख न लगना, पैरों में दर्द व सूजन (बेरी-बेरी)।
(c) विटामिन-D की कमी-बच्चों में सूखा रोग (रिकेट्स)।
(d) विटामिन-E की कमी-बन्ध्यता रोग (बाँझपन)।

अथवा निम्नलिखित विटामिनों के कार्य लिखिए
(a) विटामिन-A
(b) विटामिन-D
(c) विटामिन-E
(d) विटामिन-K.
उत्तर
उपर्युक्त विटामिनों के कार्य
(a) विटामिन-A के कार्य-दृष्टि, वृद्धि व प्रतिरोधात्मक शक्ति प्रदान करना।
(b) विटामिन-D के कार्य-सुदृढ़ अस्थि, फॉस्फोरस तथा कैल्सियम चयापचय का नियंत्रण करना।
(c) विटामिन-E के कार्य-नर-प्रजनन क्षमता बढ़ाना।
(d) विटामिन-K के कार्य-रक्त का जमना।

प्रश्न 5.
किन्हीं चार प्रोटीनों के नाम देते हुए उनके द्वारा मनुष्य के शरीर में किये जाने वाले कार्य लिखिए।
उत्तर
प्रोटीन और उनके कार्य –

1. हीमोग्लोबिन- यह रक्त में पाया जाता है तथा श्वसन क्रिया में ऑक्सीजन के साथ मिलकर ऑक्सीहीमोग्लोबिन नामक अस्थायी यौगिक बनाता है, जो फेफड़ों से विभिन्न ऊतकों तथा ऑक्सीजन का संवहन करता है।

2. मायोसिन-यह मांसपेशियों में पाया जाता है तथा यह मांसपेशियों के संचालन में सहायक होता है।

3. पेप्सिन-यह शरीर के आहार नाल के आमाशय (Stomach) में पाया जाता है तथा यह भोजन के पाचन में सहायक होता है।

4. फाइब्रिनोजेन-यह रक्त में पाया जाता है तथा यह रक्त का थक्का बनाने में सहायक होता है।

प्रश्न 6.
एन्जाइम क्या है ? उद्योगों में इनके चार अनुप्रयोग लिखिए।।
उत्तर
एन्जाइम (Enzyme)—एन्जाइम उच्च अणुभार के नाइट्रोजनयुक्त जटिल कार्बनिक यौगिक हैं, जो जीवित कोशिकाओं में उत्पन्न होते हैं। ये मुख्य रूप से रासायनिक क्रियाओं को उत्प्रेरित करते हैं, अतः एन्जाइम जीव-उत्प्रेरक कहलाते हैं।
उदाहरण- इनवर्टेस एन्जाइम सुक्रोस के जल-अपघटन को उत्प्रेरित करता है।

ग्लूकोस फ्रक्टोस एन्जाइम की क्रियाशीलता को प्रभावित करने वाले कारक-

• ताप
• pH
• एन्जाइम की सान्द्रता
• पदार्थ की सान्द्रता
• बनने वाले उत्पाद की सान्द्रता।

अनुप्रयोग-

• कार्बोहाइड्रेट के किण्वन से बीयर, विस्की, मादक पेय द्रव्य निर्माण में।
• खाद्य उपयोग में मक्का के स्टार्च से शरबत बनाने में।
• पनीर बनाने में।
• किण्वन क्रिया द्वारा शीरा से ऐल्कोहॉल निर्माण में।

प्रश्न 7.
मोनोसैकेराइड किसे कहते हैं ? उदाहरण सहित समझाइए।
उत्तर
मोनोसैकेराइड- ये सबसे सरल कार्बोहाइड्रेट हैं और इन्हें जल-अपघटन द्वारा अधिक सरल (लघु अणुओं में) कार्बोहाइड्रेटों में परिवर्तित नहीं किया जा सकता। इनका सामान्य सूत्र (C_nH_2nO_n) जिसके कुछ अपवाद भी हैं । जहाँ n का मान 2 से 10 तक हो सकता है । ये ऐल्डोस और कीटोस प्रकार के हो सकते हैं। जैसे

ऐल्डोपेण्टोसेस- एरेबिनोस, जाइलोस, राइबोस आदि (C₅H₁₀O₅) ।
ऐल्डोहेक्सोसेस- ग्लूकोस, गैलेक्टोस, मैनोस आदि (C₆H₁₂O₆)।
कीटोहेक्सोसेस- फ्रक्टोस, सारबोस आदि (C₆H₁₂O₆)।

प्रश्न 8.
प्रोटीन क्या होते हैं ?
उत्तर
प्रोटीन शब्द की उत्पत्ति ग्रीक शब्द प्रोटियोस (Protios = To take the first) से हुई अर्थात् प्रथम या अतिआवश्यक है। प्रोटीन उच्च अणु भार के नाइट्रोजन युक्त जटिल कार्बनिक यौगिक हैं, जो सभी जन्तु तथा पादप के प्रोटोप्लाज्म में पाये जाते हैं । इसमें हाइड्रोजन, ऑक्सीजन, कार्बन, नाइट्रोजन तथा अल्प मात्रा में सल्फर भी पाया जाता है।
रासायनिक रूप से प्रोटीन अल्फा अमीनो अम्ल के संघनन बहुलक हैं।

प्रश्न 9.
कार्बोहाइड्रेट क्या है ? कार्बोहाइड्रेट की कौन-सी इकाई मानव शरीर को ऊर्जा प्रदान करती है ?
उत्तर
कार्बोहाइड्रेट, कार्बन, हाइड्रोजन तथा ऑक्सीजन से बने यौगिक हैं। इनका सामान्य सूत्र C_x(H₂O)_y होता है, जहाँ x और y गुणांक हैं। इन कार्बनिक यौगिकों में हाइड्रोजन तथा ऑक्सीजन का अनुपात जल (H₂O) के समान 2 : 1 होने के कारण इन्हें कार्बन का हाइड्रोजन माना गया है तथा इन यौगिकों का नाम कार्बोहाइड्रेट रखा गया है। कार्बोहाइड्रेट श्रेणी के यौगिकों के अन्तर्गत ग्लूकोस (C₆H₁₂O₆), फ्रक्टोस(C₆H₁₂O₆), सुक्रोस (C₁₂H₂₂O₁₁), स्टार्च आदि आते हैं।

ग्लूकोस, कार्बोहाइड्रेट्स की वह इकाई है, जो शरीर में उपस्थित एन्जाइम की सहायता से ऑक्सीकरण द्वारा धीरे-धीरे CO₂ तथा जल में अपघटित हो जाती है तथा शरीर को ऊर्जा प्रदान करती है।

प्रश्न 10.
प्रोटीन की संरचना में पेप्टाइड लिंक का बनना स्पष्ट कीजिए। अथवा ऐमाइड एवं पेप्टाइड बन्ध क्या हैं ?
उत्तर
ऐमाइड बन्ध- प्रोटीन एक जटिल कार्बनिक पदार्थ है, जो विभिन्न अमीनो अम्लों के आपस में संयुक्त होने से बनता है। एक अमीनो अम्ल का कार्बोक्सिलिक समूह दूसरे अमीनो अम्ल के अमीनो समूह से संयोग करके ऐमाइड बन्ध बनाता है।

पेप्टाइड बन्ध- अमीनो अम्ल प्रोटीन के निर्माण की इकाई होती हैं। प्रोटीन में अमीनो अम्ल पेप्टाइड बन्धों द्वारा अर्थात् -CONH- समूह द्वारा एक-दूसरे से जुड़े होते हैं । पेप्टाइड बन्ध का निर्माण α -अमीनो समूह और दूसरे – अमीनो समूह के कार्बोक्सिलिक समूह की परस्पर क्रिया के फलस्वरूप होता है। पार्श्व श्रृंखला के R में उपस्थित कोई भी समूह पेप्टाइड बन्ध के निर्माण में भाग नहीं लेता। R R
i

प्रश्न 11.
पॉलिसैकेराइड क्या हैं ? इनके दो उदाहरण दीजिए।
उत्तर
पॉलिसैकेराइड प्राकृतिक बहुलक हैं, जिनका आण्विक द्रव्यमान कुछ हजार से कई लाख तक होता है। इनका सामान्य सूत्र (C₆H₁₀O₅)_n है जिसमें n का मान 12 से कई हजार तक होता है। ये अत्यन्त जटिल पदार्थ हैं। ये मोनोसैकेराइडों के संघनन बहुलीकरण द्वारा बनते हैं। इनमें ग्लाइकोसाइडिक बन्ध पाया जाता है। पॉलिसैकेराइडों के दो प्रमुख उदाहरण – स्टार्च एवं सेल्युलोस हैं।

प्रश्न 12.
इन्वर्ट शर्करा किसे कहते हैं ?
उत्तर
शर्करा दक्षिण ध्रुवण घूर्णक [D या +] होता है परन्तु जल-अपघटन पर दो मोनोसैकेराइडों का एक सम-अणुक मिश्रण प्राप्त होता है; जो वाम ध्रुवण घूर्णक [L या –] हो जाता है। अतः प्राप्त ग्लूकोस और फ्रक्टोस का मिश्रण इन्वर्ट शर्करा कहलाता है।

प्रश्न 13.
डाइसैकेराइड क्या हैं ? किसी सामान्य डाइसैकेराइड का अणुसूत्र लिखिए ।
उत्तर
डाइसैकेराइड वे शर्करा हैं, जो मोनोसैकेराइड के दो अणुओं के संयुक्त होने पर जल के एक अणु के निष्कर्षण द्वारा बनते हैं। दोनों मोनोसैकेराइड प्रायः हेक्सोस होते हैं तथा उनमें से एक ग्लूकोस होता है। इस प्रकार ऐल्डोस-ऐल्डोस तथा ऐल्डोस-कीटोस प्रकार के डाइसैकेराइड पाये जाते हैं। इन डाइसैकेराइडों का अणुसूत्र C₁₂H₂₂O₁₁ होता है। उदाहरणार्थ सुक्रोस, माल्टोस ,लैक्टोस आदि।

प्रश्न 14.
सुक्रोज और माल्टोज के पाइरानोस संरचना दीजिए।
उत्तर
सुक्रोज-सुक्रोज में दो मोनोसैकेराइड्स अणु (ग्लूकोज और फ्रक्टोस) परस्पर ग्लाइकोसाइडिक बन्ध के द्वारा जुड़े होते हैं । जो α – ग्लूकोस के C₁ तथा β- फ्रक्टोस के C₂ के मध्य जुड़ा होता है।

प्रश्न 15.
प्रोटीन की संरचना स्पष्ट कीजिए।
उत्तर
प्रोटीन के अणु का निर्माण ऐमीनो अम्लों से होता है। वस्तुतः प्रोटीन अणु ऐमीनो अम्लों के रैखिक बहुलक (linear polymers) होते हैं। इनकी सम्पूर्ण संरचना चार पदों में निर्धारित की जाती है।
1. प्राथमिक संरचना- इसमें प्रोटीन की पॉलिपेप्टाइड श्रृंखला में विभिन्न ऐमीनो अम्लों के परस्पर जुड़ने के क्रम का ज्ञान होता है।

2. द्वितीयक संरचना- यह प्रोटीन की पेप्टाइड शृंखलाओं के संरूपण (Conformations) का ज्ञान कराती है।

3. तृतीयक संरचना- इससे यह ज्ञान होता है कि प्रोटीन अणु किस प्रकार मुड़कर एक विशिष्ट आकृति प्राप्त कर लेता है।

4. चतुर्थक संरचना- इससे यह पता चलता है कि दो पॉलिपेप्टाइड शृंखलाएँ एक-दूसरे के सापेक्ष किस प्रकार व्यवस्थित हैं।

प्रश्न 16.
न्यूक्लिओसाइड तथा न्यूक्लिओटाइड से आप क्या समझते हैं ?
उत्तर
न्यूक्लिओसाइड- जब कोई प्यूरीन या पाइरामिडीन बेस, पेण्टोस शुगर अणु के साथ जुड़ जाता है, तो इसे न्यूक्लिओसाइड कहते हैं।
Base + Sugar → Nucleoside
न्यूक्लिओटाइड- यह न्यूक्लिओसाइड का फॉस्फेट एस्टर है।

प्रश्न 17.
कार्बोहाइड्रेट क्या है ? मोनो, डाइ तथा पॉलीसैकेराइडों को उदाहरण सहित समझाइए।
उत्तर
कार्बोहाइड्रेट- वे पदार्थ जो पॉलीहाइड्रॉक्सी ऐल्डीहाइड या पॉलीहाइड्रॉक्सी कीटोन है। अथवा वे पदार्थ जो जल-अपघटन के पश्चात् ये यौगिक देते हैं कार्बोहाइड्रेट्स कहलाते हैं। कार्बोहाइड्रेट्स को जलअपघटन के आधार पर तीन भागों में वर्गीकृत किया गया है।

1. मोनोसैकेराइड- ये सबसे सरल कार्बोहाइड्रेट्स हैं। इन्हें जल-अपघटन द्वारा अधिक सरल कार्बोहाइड्रेटों में परिवर्तित नहीं किया जा सकता है। इनका सामान्य सूत्र C_nH_2nO_n है, जहाँ n का मान 1 से 10 तक हो सकता है। ये क्रिस्टलीय ठोस हैं। जल में घुलनशील और स्वाद में मीठे होते हैं। ये दो प्रकार के होते हैं

• ऐल्डोस,
• कीटोस।

2. डाइसैकेराइड- ये वे शर्करा हैं, जो मोनोसैकेराइड के दो अणुओं के संयुक्त होने या जल के एक अणु के निष्कर्षण द्वारा बनता है। दोनों मोनोसैकेराइड प्रायः हेक्सोस होते हैं तथा उनमें से एक अणु ग्लूकोस होता है। इन डाइसैकेराइड का अणुसूत्र C₁₂H₂₂O₁₁ है।
उदाहरण- सुक्रोस, माल्टोस।

3. पॉलीसैकेराइड- ये वे सैकेराइड हैं, जो जल-अपघटन के पश्चात् मोनोसैकेराइड के n अणु देते हैं। ये रंगहीन तथा स्वादहीन होते हैं। इनका अणुसूत्र C_n(H₁₀0₅)_n होता है।
उदाहरण- स्टार्च, सेल्यूलोज।

प्रश्न 18.
क्या होता है जब, प्रोटीन का विकृतिकरण होता है ?
उत्तर
प्रोटीन का विकृतिकरण-प्रोटीन, ऊष्मा तथा रसायनों से प्रभावित होते हैं। प्रोटीन को गर्म करने पर अथवा रासायनिक यौगिकों से क्रिया कराने पर इसकी जैविक क्रियाशीलता नष्ट हो जाती है ये विकृत और स्कन्दित होकर अविलेय हो जाते हैं। इस क्रिया को प्रोटीन का विकृतिकरण कहते हैं।

विकृतिकरण से प्रोटीन की प्राथमिक संरचना अपरिवर्तित रहती है, किन्तु द्वितीयक एवं तृतीयक संरचना में परिवर्तन हो जाता है। जैसे-जब अण्डे को उबलते हुए पानी में कुछ समय के लिए रखा जाता है, तो अण्डे की प्रोटीन अविलेय रेशेदार प्रोटीन में परिवर्तित हो जाती है, जिससे प्रोटीन स्कन्दित हो जाता है अर्थात् प्रोटीन का विकृतिकरण हो जाता है।

प्रश्न 19.
विटामिन-B के कार्य तथा विटामिन-B के अभाव में होने वाले दो रोगों के नाम लिखिए।
उत्तर
विटामिन-B₁ बहुत से विटामिन के समूह को कहते हैं, जो निम्नलिखित हैं –

विटामिन-B₁(थायमीन)
विटामिन-B₂ (राइबोफ्लेविन)
विटामिन-B₃(पेन्टोथेनिक ऐसिड)
विटामिन-B₆ (पायरीडॉक्सीन)
विटामिन-B₁₂ (सायनोबलेमीन)।

विटामिन- B₁(थायमीन)-स्रोत-बिना पॉलिश किया चावल, हरी सब्जी, अण्डा।
कार्य- तंत्रिका तन्त्र की क्रियाशीलता बनाये रखना।
अभाव से रोग- (i) बेरी-बेरी (हाथ पैर में सूजन), (ii) गैस्ट्रिक (पाचन क्रिया का अनियमित होना)।
विटामिन- B₂ (राइबोफ्लेविन) स्रोत-खमीर, अण्डा का पीला भाग आदि।
कार्य- शारीरिक वृद्धि हेतु।
अभाव से रोग- होठों का फटना, नेत्र दृष्टि कम होना।

प्रश्न 20.
विटामिन्स क्या हैं ? उन विटामिन्स के नाम लिखिये जिनकी कमी से निम्नलिखित बीमीरियाँ उत्पन्न होती हैं –

• खून का थक्का न जमना
• रतौंधी
• रक्त अल्पता
• सूखा रोग
• पायरिया
• बन्ध्यता
• अरक्तता।

उत्तर
विटामिन जटिल कार्बनिक यौगिक हैं, जो शरीर के लिए आवश्यक पोषक तत्व के समान कार्य करते हैं, यद्यपि ये हमारे शरीर में बनते नहीं, परन्तु इनके अभाव से अनेक रोग उत्पन्न हो जाते हैं। विटामिनों की थोड़ी-सी मात्रा भी शरीर के सुचारु रूप से कार्य करने और वृद्धि के लिए आवश्यक हैं। विटामिन कई प्रकार के होते हैं। जैसे-A, B, C, D, E व K. .

• खून का थक्का न जमना-विटामिन-K(फाइलोक्विनोन)
• रतौंधी-विटामिन-A (रेटिनॉल) एक्सेरोफाइटॉल
• रक्त अल्पता-विटामिन-B₁₂ (सायनोबलेमीन)
• सूखा रोग-विटामिन-D (कैल्सिफेरॉल)
• पायरिया-विटामिन-C (ऐस्कार्बिक एसिड)
• बन्ध्यता-विटामिन-E (aटोकॉफेरॉल)
• अरक्तता- विटामिन-B₆ (पिरिडॉक्सीन)।

प्रश्न 21.
निम्नलिखित हॉर्मोन्स के बारे में लिखिए

• टेस्टोस्टेरॉन
• थायरॉक्सिन
• इन्सुलिन
• कार्टिसोन।

उत्तर

• टेस्टोस्टेरॉन- टेस्टोस्टेरॉन हॉर्मोन को सावित करने वाली ग्रंथि वृषण है तथा इसका कार्य पुरुषों में जनन अंगों का नियंत्रण है।
• थायरॉक्सिन- इस हॉर्मोन को स्रावित करने वाली ग्रंथि का नाम, थायरॉयड है। इसका कार्य उपापचय क्रियाओं एवं वृद्धि का नियंत्रण है।
• इन्सुलिन- इन्सुलिन हॉर्मोन अग्न्याशय द्वारा स्रावित होता है तथा इसका कार्य रक्त में ग्लूकोज की मात्रा का नियंत्रण करना है।
• कार्टिसोन- कार्टिसोन ऐड्रीनल कॉर्टेक्स द्वारा स्रावित होता है तथा इसका कार्य वसा, प्रोटीन तथा जल के उपापचय का नियंत्रण करना है।

प्रश्न 22.
रेटिनॉल, थायमीन, ऐस्कार्बिक एसिड व राइबोफ्लेवीन की कमी से होने वाले दो-दो रोगों के नाम लिखिए।
उत्तर
रेटिनॉल- रतौंधी, अतिसार । थायमीन- बेरी-बेरी, शरीर की वृद्धि रुक जाती है।
ऐस्कार्बिक एसिड- स्कर्वी, दंतक्षय। राइबोफ्लेवीन-नेत्र दृष्टि कम होना, त्वचा फटना।

प्रश्न 23.
विटामिन-A विटामिन-C के दो-दो स्रोत लिखिए। इनकी कमी से होने वाले एक-एक रोग बताइये।
उत्तर

प्रश्न 24.
विटामिन A, D, E एवं K के कार्य लिखिए।
उत्तर
विटामिन A – यह रोडोप्सिन एवं आयोडोप्सिन नामक दृश्य पिगमेन्ट के निर्माण में भाग लेता
विटामिन D – यह हड्डियों के निर्माण में उपयोगी होता है।
विटामिन E – यह RBC के टूटने की क्रिया को रोकता है।
विटामिन K – यह रक्त को जमने में सहायता करता है।

प्रश्न 25.
निम्नलिखित जैव-अणुओं/तत्वों के कार्य व प्राप्ति के स्रोत लिखिए

1. प्रोटीन
2. कार्बोहाइड्रेट
3. वसा
4. कैल्सियम।

उत्तर
1. प्रोटीन – शरीर के अंगों का निर्माण करना।
प्राप्ति – दूध, पनीर, अंडा, मछली आदि।

2. कार्बोहाइड्रेट – ऊर्जा प्रदान करना ।
प्राप्ति – अनाज, चावल, फल, आलू, शक्कर आदि।

3. वसा – ऊर्जा प्रदान करना।
प्राप्ति – घी, तेल, मेवे, दूध, अंडा।

4. कैल्सियम – दाँत व हड्डी की वृद्धि।
प्राप्ति – पत्तेदार सब्जियाँ, साबुत अनाज, दूध।

प्रश्न 26.
α -ऐमीनो अम्ल तथा प्रोटीन में दो-दो अन्तर लिखिए।
उत्तर
α -ऐमीनो अम्ल तथा प्रोटीन में अन्तर

प्रश्न 27.
संक्षेप में समझाइये
(a) दो ऐन्जाइमों के नाम तथा उनके कार्य।
(b) जल में घुलनशील दो विटामिनों के नाम एवं इनके अभाव से होने वाले रोग।
उत्तर
(a) दो ऐन्जाइमों के नाम तथा उनके कार्य
नाम-
1. ऐमाइलेज (टायलिन)
कार्य- यह स्टार्च को ग्लूकोज में बदल देता है।

2. पेप्सिन –
कार्य- यह प्रोटीन को ऐमीनो अम्ल में बदल देता है।

(b) जल में घुलनशील विटामिन –

1. विटामिन B₁– थायमीन –
अभाव में रोग – बेरी-बेरी

2. विटामिन C- ऐस्कार्बिक अम्ल
अभाव में रोग – स्कर्वी
जल में घुलनशील विटामिनों के अन्य उदाहरण- विटामिन B₂, B₆, B₁₂ तथा K.

जैव-अणु दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
न्यूक्लिक अम्ल पर संक्षिप्त टिप्पणी लिखिए।
उत्तर
न्यूक्लिक अम्ल-यह जीव कोशिका के केन्द्रक में पाया जाता है। इसमें फॉस्फोरस की मात्रा अधिक होती है। न्यूक्लिक अम्ल पॉली न्यूक्लिओटाइड होते हैं, जो अनेक न्यूक्लिओटाइड की इकाइयों के , मिलने से बनती है।
प्रत्येक न्यूक्लिओटाइड तीन रासायनिक घटकों का बना होता है –

• फॉस्फेट समूह,
• पेण्टोज राइबोज शर्करा या डी-ऑक्सीराइबोज,
• विषमचक्रीय बेस, जैसे-पायरीमिडीन के व्युत्पन्न (थाइमीन, यूरेसिल, साइटोसीन) एवं प्यूरीन के व्युत्पन्न (ऐडीनीन एवं ग्वानीन)।

न्यूक्लिक अम्ल दो प्रकार के होते हैं
(A) DNA-डी-ऑक्सीराइबो न्यूक्लिक अम्ल

(B) RNA- राइबोन्यूक्लिक अम्ल ।
DNA के घटक
(a) डी-ऑक्सीराइबोस शर्करा अणु,
(b) फॉस्फोरिक अम्ल के अणु,

(c) नाइट्रोजन बेस। ये दो तरह के होते हैं

• पिरीमिडीन बेस-इसके अन्तर्गत साइटोसीन (C) और थायमीन (T) आते हैं।
• प्यूरीन बेस- इसके अन्तर्गत एडीनीन (A) और ग्वानीन (G) आते हैं।

RNA के घटक-RNA में राइबोज तथा नाइट्रोजन बेस, जैसे-ऐडीनीन (A), ग्वानीन (G), यूरेसिल (U), और साइटोसीन (C) होते हैं।

प्रश्न 2.
कार्बोहाइड्रेट क्या होते हैं ? इनका वर्गीकरण करके चार प्रमुख कार्य लिखिए।
उत्तर
परिभाषा-प्रकाश सक्रिय पॉलीहाइड्रॉक्सी ऐल्डिहाइड या कीटोन या वे पदार्थ जो जल-अपघटित होकर इनका निर्माण करते हैं, कार्बोहाइड्रेट कहलाते हैं।
उदाहरण- ग्लूकोस, स्टार्च, सेल्युलोस, सुक्रोस आदि।
कार्बोहाइड्रेट का वर्गीकरण –

कार्बोहाइड्रेट के कार्य –
1.यह कोशिका का प्रमुख संरचनात्मक घटक है।

2. यह जैव-ईंधन की तरह कार्य करता है और जीवधारियों को कार्य करने के लिए ऊर्जा प्रदान करता है।

3. लीवर (Liver) में कार्बोहाइड्रेट ग्लाइकोजन के रूप में आरक्षित रहते हैं, जो जल-अपघटित होकर आवश्यक ऊर्जा प्रदान करते हैं।

4. सेल्युलोस घास और पौधों में पाया जाता है जो घास चरने वाले जानवरों को ऊर्जा प्रदान करता है क्योंकि जानवरों के शरीर में सेल्युलोस को ग्लूकोस में जल-अपघटित करने वाले विशिष्ट एन्जाइम पाये जाते हैं।

प्रश्न 3.
ऐस्कार्बिक अम्ल, थायमिन, रेटिनॉल एवं निकोटिनिक अम्ल की कमी से होने वाले बीमारियों के नाम लिखिए। (प्रत्येक के दो-दो नाम दीजिये)
अथवाविटामिन A,B,C और D की कमी से कौन-कौन से रोग होते हैं ? इनके नाम व एक-एक स्रोत लिखिये।
अथवा विटामिन A, C, D एवं E की कमी से होने वाले रोग एवं प्राप्ति के स्रोत बताइये।
उत्तर

प्रश्न 4.
मोनोसैकेराइड, डाइसैकेराइड और पॉलिसैकेराइड में अन्तर लिखिए।
उत्तर
मोनोसैकेराइड, डाइसैकेराइड और पॉलिसैकेराइड में अन्तर –

MP Board Class 12th Chemistry Solutions