Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Commerce Guide Pdf Chapter 1 Principles of Management Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Commerce Solutions Chapter 1 Principles of Management

12th Commerce Guide Principles of Management Text Book Back Questions and Answers

I. Choose the correct answer.

Question 1.
Management is what a……………..does?
a) Manager
b) Subordinate
c) Supervisor
d) Superior
Answer:
a) Manager

Question 2.
Management is an …………………
a) Art
b) Science
c) Art and Science
d) Art or Science
Answer:
c) Art and Science

Question 3.
Scientific management is developed by
a) Fayol
b) Taylor
c) Mayo
d) Jcob
Answer:
b) Taylor

Question 4.
Dividing the work into small tasks is known as
a) Discipline
b) Unity
c) Division of work
d) Equity
Answer:
c) Division of work

Question 5.
With a wider span, there will be …………… hierarchical levels.
a) More
b) Less
c) Multiple
d) Additional
Answer:
b) Less

II. Very short answer questions.

Question 1.
What is Management?
Answer:
Management is part and parcel of our day to day life. So management is goal oriented and it is an art of getting things done with and through others.

Question 2.
List out the management tools. (BASED)
Answer:

1. Business Law
2. Accounting
3. Statistics
4. Economics
5. Data processing

Question 3.
Who is a manager?
Answer:
A manager is a dynamic and life giving element in every business. Without efficient management it cannot be possible to secure the best allocation and utilisation of human, material and financial resources.

Question 4.
State the meaning of Authority.
Answer:
“Authority” means the right of a superior given to his subordinate to get work from him.

Question 5.
What do you mean by the span of management?
Answer:
The Span of Management refers to the number of subordinates who can be managed efficiently by a superior. Simply, the manager having the group of subordinates who report him directly is called as the span of management.

III. Short Answer Questions.

Question 1.
Define the term management.
Answer:
“Management is a multi-purpose organ that manages a business, and manages manager and manages workers and work”. – PETER F. DRUCKER.

Question 2.
Is management an Art or Science?
Answer:
Management is neither a science nor an art, but a combination of both requiring people holding managerial positions to apply the scientific management principles and displaying popular managerial skills to accomplish the organizational goals as efficiently and as quickly as possible so as to be competitive in the globalized environment of business.

Question 3.
Differentiate management from Administration. MA CARD
Answer:

Question 4.
What are the principles of Taylor?
Answer:

• Science, not a Rule of Thumb.
• Harmony, not Discord.
• Mental Revolution
• Cooperation, not Individualism
• Development of each and every person to his or her greatest efficiency and prosperity.

Question 5.
What determines the span of management?
Answer:
The Span of Management has two implications:

1. Influences the complexities of the individual manager’s job.
2. Determine the shape or configuration of the organization.
3. There is a wide and a narrow span of management.

IV. Long answer questions

Question 1.
Explain the concept of management.
Answer:

The substance of management should be identified as a process. A process is something that what a person does in the context of his individual duties and responsibilities assigned by his or her immediate higher authority.
There are twin purposes of the management process:

1. Maximum productivity or profitability
2. Maximum human welfare and satisfaction.

There are five parts of management as a process:

1. Co-ordination of resources: The manager of an enterprise must effectively coordinate all activities and resources of the organization, namely, men, machines, materials, and money, the four M’s of management.
2. Management is a Process: The manager achieves proper coordination of resources by means of the managerial functions of planning, organizing, staffing, directing (or leading and motivating), and controlling.
3. Management is a Purposive Process: It is directed toward the achievement of predetermined goals or objectives. Without an objective, we have no destination to reach or a path to follow to arrive at our destination, i.e., a goal, both management, and organization must be purposive or goal-oriented.
4. Management is a Social Process: It is the art of getting things done through other people.
5. Management is a Cyclical Process: It represents a planning-action-control-replanning cycle, i.e., an ongoing process to attain the planned goals.

Question 2.
Explain the management process in detail.

Question 3.
Describe the principles of scientific management. (S) (HM) (CD)
Answer:
Science, not Rule of Thumb:

• Rule of thumb means decisions taken by managers as per their personal judgements.
• Taylor says, even a small production activity like loading – iron sheets into box cars can be scientifically planned.
• It saves time and energy of human.

Harmony, not Discard:

• Taylor emphasized that there should be complete harmony between the workers and the management.
• If there is any conflict between the two, it will not be beneficial either for workers or for management.

Mental Revolution:

• Taylor suggested complete mental revolution on the part of both workers and Management.
• It means that there should be complete change in the attitude and out look of both workers and management.
• It becomes possible – Teamwork – Sharing profits – Division of work etc.

Co-operation, not Individualism:

• It is an extension of principle of “Harmony, not discord”.
• It lays stress on mutual co-operation between workers and the management.
• Both workers and management should realize the importance of each other.

Development of Each and Every person:

• Efficiency of any organisation also depends on the skills and capabilities of its employees to a great extent.
• Thus, providing to the workers was considered essential in order to learn the best method.
• It helps to attain the efficiency and property for both workers and management.

Question 4.
Explain the principles of modern management.
Answer:
The Father of Modem Management is Mr.Henry Fayol, and according to him, there are 14 major principles of management which every manager has to practice for the success of the organization.

1. Division of Work: According to this principle the whole work is divided into small tasks. This leads to specialization which increases the efficiency of labour.
2. Authority and Responsibility: This is the issue of commands followed by responsibility for their consequences.
3. Discipline: It is obedience, proper conduct in relation to others, respect of authority, etc. It is essential for the smooth functioning of all organizations.
4. Unity of Command: This principle states that each subordinate should receive orders and be accountable to one and only one superior.
5. Unity of Direction: All related activities should be put under one group, there should be one plan of action for them, and they should be under the control of one manager.
6. Subordination of Individual Interest to Mutual Interest: The management must put aside personal considerations and put company objectives firstly.
7. Remuneration: Workers must be paid sufficiently as this is a chief motivation of employees and therefore greatly influences productivity.
8. The Degree of Centralization: The amount of power wielded with the central management depends on company size.
9. Line of Authority/Scalar Chain: This refers to the chain of superiors ranging from top management to the lowest rank.
10. Order: Social order ensures the fluid operation of a company through the authoritative procedure.
11. Equity: Employees must be treated kindly, and justice must be enacted to ensure a just workplace.
12. Stability of Tenure of Personnel: Stability of tenure of personnel is a principle stating that in order for an organization to run smoothly, personnel (especially managerial personnel) must not frequently enter and exit the organization.
13. Initiative: Using the initiative of employees can add strength and new ideas to an organization.
14. Esprit de Corps/Team Spirit: This refers to the need of managers to ensure and develop morale in the Workplace; individually and communally.

12th Commerce Guide Principles of Management Additional Important Questions and Answers

I. Choose the correct answer.

Question 1
_______ is a global and universal concept.
(a) Management
(b) Process
(c) Art
(d) Science
Answer:
(a) Management

Question 2.
To manage is to forecast, to plan, to organize, to command, to co-ordinate and to control said by ……………………
a) Henry Fayol
b) Peter F. Drucker
c) Taylor
d) NOTA
Answer:
a) Henry Fayol

Question 3.
“To manage is to forecast, to plan, to organise, to command, to co-ordinate and to control.” said by _______
(a) Henry Fayol
(b) Peter F. Drucker
(c) Walker
(d) Carter
Answer:
(a) Henry Fayol

Question 4.
What should be done? and when is should be done decided by ……………
a) Management
b) Supervisor
c) Director
d) Administrator
Answer:
d) Administrator

Question 5.
Executive and governing functions are done by ……………..
a) Manager
b) Supervisor
c) Worker
d) Administrator
Answer:
a) Manager

Question 6.
The father of Modem Management is _______
(a) F.W. Taylor
(b) Henry Fayol
(c) Peter F. Drucker
(d) Louis A. Allen
Answer:
(b) Henry Fayol

Question 7.
The Authority of Top-level is ………………
a) Management
b) Administration
c) Both
d) NOTA
Answer:
b) Administration

Question 8.
One employee should receive an order from only one boss is known as ………………
a) Unity of command
b) Unity of objective
c) Unity of direction
d) Unity of strength
Answer:
a) Unity of command

Question 9.
Superiors ranging from top management to the lowest rank is called …………….
a) Responsibility
b) Authority
c) Remuneration
d) Scalar chain
Answer:
d) Scalar chain

Question 10.
With the narrow span, the hierarchical levels……………..
a) Decreases
b) Increases
c) Ups and downs
d) Stagnant
Answer:
b) Increases

Question 11.
Find the odd one out:
a) Harmony, Not Discord
b) Mental Revaluation
c) Cooperation, not individualization
d) Order
Answer:
d) Order

Question 12.
Find the odd one out:
a) Division of work
b) Scalar chain
c) Equity
d) Science not thum of Rule
Answer:
d) Science not thum of Rule

Question 13.
Which one of the following is not correctly matched?
a) Business law – Management Tool
b) Specialisation – Highly qualified
c) Code of conduct – Social conscience
d) Behavioural school – Process of planning
Answer:
d) Behavioural school – Process of planning

Question 14.
Which one of the following is correctly matched?
a) Cyclical process – Ongoing process
b) Purposive process – Co-ordinate
c) Social process – Goal-oriented
d) Not Individualisation – Separation
Answer:
a) Cyclical process – Ongoing process

Question 15.
Which is the correct statement?
i) Division of work is whole work divided into small tasks.
ii) Discipline is obedience and respect for authority.
iii) Unity of direction is related activities put under many groups.
a) (i) is correct
b) (i) and (ii) are correct
c) All are correct
d) All are wrong
Answer:
b) (i) and (ii) are correct

II. Match the following.

Question 1.
Match List I with List II

a) iv-1, iii-2, ii-3, i-4
b) iv-2, iii-1, ii-3, i-4
c) iv-3, iii-4, ii-1, i-1
d) iv-4, iii-3, ii-2, i-1
Answer:
a) iv-1, iii-2, ii-3, i-4

Question 2.
Match List I with List II

a) i-1, ii-3, iii-2, iv-4
b) i-1, ii-2, iii-4, iv-3
c) i-1, ii-4, iii-3, iv-2
d) i-1, ii-2, iii-3, iv-4
Answer:
a) i-1, ii-3, iii-2, iv-4

III. Assertion and Reason

Question 1.
Assertion (A): line employees should recei\e orders from one superior.
Reason (R): If the receives orders from more superiors, it creates confusion and conflict,
a) both (A) and (R) are true. (R) is the correct explanation of (A)
b) both (A) and (R) are false. (R) is the correct explanation of (A)
c) Roth (A) and (R) are I rue. (R) is not the correct explanation of (A)
d) (A) is true (R) is False
Answer:
a) both (A) and (R) are true. (R) is the correct explanation of (A)

Question 2.
Assertion (A) : All related activities should be put under one group.
Reason (R) : There should be under the control of one manager.
a) (A) is true (R) is False
b) (A) is False (R) is True
s) Both tA) and (R) are True
d) Moth (A) and iR) are False
Answer:
c) Both (A) and (R) arc True.

Question 3.
Assertion (A): Equity must be treated kindly and justice.
Reason (R): Managers should be fair and partial.
a) (A) is true (R) is False
b) (A) is False (R) is True
c) t A) and (R) are I rue
d) (A) and (R’ are False
Answer:
a) (A) is True (R) is False

IV. Very Short Answer Questions:

Question 1.
What is a Scalar chain or Line of Authority?
Answer:
This refers to the chain of superiors ranging from top management to the lowest rank. The principle suggests that there should be a clear line of authority from top to bottom linking all managers at all levels.

V. Short Answer Questions:

Question 1.
Who is a Professional Manager?
Answer:
“A professional manager is one who specializes in the work of planning, organising, leading and controlling the efforts of others and does so through the systematic use of classified knowledge, a common vocabulary, and principles and who subscribes to the standards of practice and code of ethics established by a recognized body.” – Louis A. Allen.

VI. Long Answer Questions:

Question 1.
Formulate a new policy relating to the timing of employees.
Answer:
It has always been a struggle for people to maintain a work-life balance. Working excessive hours poses a danger to workers’ health and to their families. Earlier all the work was done manually. However, with the help of machinery and computers, in modem times certain industries don’t require as many working hours anymore. Technology has introduced new ways to increase productivity. More recently, the global trend leans toward a four-day workweek (counting one workday as eight hours).

Even the ILO standards on working time provide the framework for regulated hours of work, daily and weekly rest periods, and annual holidays. These instruments ensure high productivity while safeguarding workers’ physical and mental health. If the workers earn enough to pay for their necessities, they may opt to spend more time at home or in leisure. Countries around the globe are already experimenting and implementing shorter workweeks in varying degrees. So, in my opinion, the new policy relating to the timing of employees should be a four-day workweek.

We hope the information prevailed in this article is helpful for all the students of Class 12th. The Tamilnadu State Board Solutions for Class 12th Chapter 1 Principles of Management Questions and Answers And Pdf enhance your skills and score good marks in the exams. Stay tuned to get the latest information about the Samacheer Kalvi Class 12th Chapter 1 Principles of Management Solutions.

Question 2.
Discuss the implications of the span of management.
Answer:
Can a superior effectively manage, supervise and control how many subordinates.
It is known as “Span of Control” or “Span of Management” or “Span of Supervision”.

The span of management has two implications:

1. Influences the complexities of the individual Manager’s job.
2. Determine the shape or configuration of the organization.

It has two horizontal levels of span of management.

1. Wider-Span
2. Narrow Span

Wider-Span:

• Wider-Span-Less hierarchical levels – More subordinates – Less expensive.
• It imposes more challenges.
• It is very difficult for a superior to manage a large number of subordinates.
• In this structure, managers get reduced and remuneration saved.

Narrow Span:

• Narrow Span- more hierarchical levels – fewer subordinates – more expensive.
• With more levels of hierarchy the communication sutlers drastically.
• It takes a lot of time to reach the appropriate points and hence the actions delayed.
• In this structure, less number of subordinates under one superior requires more managers to be employed

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 5 Coordination Chemistry Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 5 Coordination Chemistry

12th Chemistry Guide Coordination Chemistry Text Book Questions and Answers

Part – I Text Book Evaluation

I. Choose the correct answer

1. The sum of primary valence and secondary valance of the metal M in the complex [M(en))₂ (Ox)] Cl is L
a) 3
b) 6
c) -3
d) 9
Answer:
d) 9

2. An excess of silver nitrate is added to 100ml of aO.OlM solution of pentaaquachlori- dochromium (III) chloride. The number of moles of AgCl precipitated would be
a) 0.02
b) 0.002
c) 0.01
d) 0.2
Answer:
b) 0.002

3. A complex has a molecular formula MSO₄Cl.6H₂O. The aqueous solution of it gives white precipitate with Barium chloride solution and no precipitate is obtained when it is treated with silver nitrate solution. If the secondary valence of the metal is six, which one of the following correctly represents the complex?
a) [M(H₂O)₄Cl] SO₄.2H₂O
b) [M(H₂O)₆] SO₄
c) [M(H₂O)₅Cl]SO₄.H₂O
d) [M (H₂O)₃Cl] SO₄.3H₂O
Answer:
c) [M(H₂O)₅Cl]SO₄.H₂O

4. Oxidation state of Iron and the charge on the ligand NO in [Fe (H₂O)₅ NO] SO₄ are
a) +2 and 0 respectively
b) +3 and 0 respectively
c) +3 and -1 respectively
d) +1 and +1 respectively
Answer:
d) +1 and +1 respectively

5. As per IUPAC guidelines, the name of the complex [Co(en)2 (ONO)Cl] Cl is
a) chlorobisethylenediaminenitritocobalt(III) chloride
b) chi or id obis (e thane-1, 2-diamine) nitrito K-Ocobaltate(III) chloride
c) chloridobis (ethane-1, 2-diammine) nitrito K -Ocobalt(II) chloride
d) chloridobis (ethane-1, 2-diamine) nitrito K -Ocobalt(III) chloride
Answer:
d) chloridobis (ethane-1,2-diamine) nitro K -Ocobalt(III) chloride

6. IUPAC name of the complex K₃[Al(C₂O₄)₃] is
a) potassiumtrioxalatoaluminium(III)
b) potassiumtrioxalatoaluminate(II)
c) potassiumtrisoxalatoaluminate(III)
d) potassiumtrioxalatoaluminate(III)
Answer:
d) potassiumtrioxalatoaluminate(III)

7. A magnetic moment of 1.73BM will be shown by one among the following (NEET)
a) TiCl₄
b) [CoCl₆]^4-
c) [CU(NH₃)₄]²⁺
d) [Ni(CN)₄]^2-
Answer:
c) [CU(NH₃)₄]²⁺

8. Crystal field stabilization energy for high spin d⁵ octahedral complex is
a) -0.6∆₀
b) 0
c) 2(P – ∆₀)
d) 2(P + ∆₀)
Answer:
b) 0

9. In which of the following coordination entities the magnitude of ∆₀ will be maximum?
a) [Co(CN)₆]^3-
b) [Co(C₂O₄)₃]^3-
c) [Co(H₂O)₆]³⁺
d) [Co(NH₃)₆]³⁺
Answer:
a) [Co(CN)₆]^3-

10. Which one of the following will give a pair of enantiomorphs?
a) [Cr(NH₃)₆][Co(CN))₆]
b) [Co(en)₂Cl₂]Cl
c) [Pt(NH₃)₄][FtCl₄]
d) [CO(NH₃)₄Cl₂]NO₂
Answer:
b) [Co(en)₂Cl₂]Cl

11. Which type of isomerism is exhibited by [Pt(NH₃)₂Cl₂]?
a) Coordination isomerism
b) Linkage isomerism
c) Optical isomerism
d) Geometrical isomerism
Answer:
d) Geometrical isomerism

12. How many geometrical isomers are possible for [Pt(Py)(NH₃)(Br)(Cl)]?
a) 3
b) 4
c) 0
d) 15
Answer:
a) 3

13. Which one of the following pairs represents linkage isomers?
a) [Cu(NH₃)₄] [PtCl₄] and [Pt (NH₃)₄] [CuCl₄]
b) [Co(NH₃)₅ (NO₃)] SO₄ and [CO(NH₃)₅ (ONO)]
c) [Co(NH₃)₄ (NCS)₂] Cl and [Co(NH₃)₄ (SCN)₂]Cl
d) both (b) and (c)
Ans:
c) [Co(NH₃)₄ (NCS)₂] Cl and [Co(NH₃)₄ (SCN)₂]Cl

14. Which kind of isomerism is possible for a complex [Co(NH₃)₄Br₂]Cl? (PTA -3)
a) geometrical and ionization
b) geometrical and optical
c) optical and ionization
d) geometrical only
Answer:
a) geometrical and ionization

15. Which one of the following complexes is not expected to exhibit isomerism?
a) [Ni(NH₃)₄ (H₂O)₂]²⁺
b) [Pt(NH₃)₂Cl₂]
c) [Co(NH₃)₅SO₄]Cl
d) [FeCl₆ ]^3-
Answer:
d) [FeCl₆]^3-

16. A complex in which the oxidation number of the metal is zero is
a) K4 [Fe(CN)₆]
b) [Fe(CN)₃ (NH₃)₃]
c) [Fe(Co)₅]
d) both (b) and (c)
Answer:
c) Fe(CO)s]

17. Formula of tris(ethane-l,2-diamine) iron (II) phosphate
a) [Fe(CH₃ – CH(NH₂)₂)₃](PO₄)₃
b) [Fe(H₂N – CH₂ – CH₂ – NH₂)₃] (PO₄)
c) [Fe(H₂N-CH₂-CH₂-NH₂)₃](PO₄)₂
d) [Fe(H₂N-CH₂-CH₂-NH₂)₃]₃(PO₄)₂
Answer:
d) [Fe(H₂N-CH₂-CH₂-NH₂)₃]₃(PO₄)₂

18. Which of the following is paramagneticin nature? (PTA -5)
a) [Zn(NH₃)₄]²⁺
b) [CO (NH₃)₆]³⁺
c) [Ni(H₂O)₆]²⁺
d) [Ni (CN)₄]^2-
Answer:
c) [Ni(H₂O)₆]²⁺

19. Fac-mer isomerism is shown by
a) [Co (en)₃]³⁺
b) [Co(NH₃)₄(Cl)₂]⁺
c) [Co (NH₃)₃ (Cl)₃]
d) [Co (NH₃)₅ Cl]SO₄
Answer:
c) [Co (NH₃)₃ (Cl)₃]

20. Choose the correct statement.
a) Square planar complexes are more stable than octahedral complexes
b) The spin only magnetic moment of [Cu (Cl)₄]₄^2- is 1.732 BM and it has square planar structure.
c) Crystal field splitting energy (∆_o) of [FeF₆]^4- is higher than the (Ao) of [Fe (CN)6]4
d) Crystal field stabilization energy of [V(H₂O)f₆]²⁺ is higher than the crystal field stabilization of [Ti(H₂O)₆]²⁺
Answer:
d) Crystal field stabilization energy of [V(H₂O)f₆]²⁺ is higher than the crystal field stabilization of [Ti(H₂O)₆]²⁺

II. Answer the following questions

Question 1.
Write the IUPAC names for the following complexes. (PTA -3)

1. Na₂ [Ni (EDTA)]
2. [Ag(CN)₂]^–
3. [Co(en)₃]₂ (SO₄)₃
4. [Co (ONO) (NH₃)₅]²⁺
5. [Pt(NH₃)₂ Cl(NO₂)]

Answer:

1. Sodium 2, 2¹, 2¹¹, 2¹¹¹-(ethane-1, 2-diyldinitrilo) tetra acetato nickelate (II)
2. Dicyanido k-c argentate (I) ion
3. Tris (ethane -1, 2 – diamine) cobalt (III) sulphate
4. Pentaammine nitrito _ kO cobalt (III) ion
5. Diammine chlorido nitrito – kN Platinum (II)

Question 2.
Write the formula for the following coordination compounds.

1. Potassiumhexacyanidoferrate(II)
2. pentacarbonyl iron(O)
3. pentaamminenitrito -k -Ncobalt(III)ion
4. hexaamminecobalt(III) sulphate
5. sodium tetra fluoridodi hydroxidochromate (III)

Answer:

1. K₄[Fe(CN)₆]
2. [Fe(CO)₅]
3. [CO(NH₃)₅ (NO₂)]²⁺
4. [CO[NH₃)₆]₂ (SO₄)₃
5. Na₃[CrF₄(OH)₂]

Question 3.
Arrange the following in order of increasing molar conductivity
i) Mg[Cr(NH₃)(Cl)₅]
ii) [Cr(NH₃)₅Cl]₃[CoF₆]₂
iii) [Cr(NH₃)₃Cl₃]
Answer:


iii) [Cr(NH₃)₃Cl₃] – Neutral complex – No ions Molar conductivity increases as the number of ions increase
∴ Molar conductivity is in the increasing order. [Cr(NH₃)₃Cl₃] < Mg (Cr(NH₃)Cl₅] < [Cr(NH₃)₅Cl]₃[CoF₆]₂

Question 4.
Give an example of a coordination compound used in medicine and two examples of biologically important coordination compounds.
Answer:
Medical uses of coordination compounds:

1. Ca-EDTA chelate is used in the treatment of lead and radioactive poisoning. That is for removing lead and radioactive metal ions from the body.
2. Cis-platin is used as an antitumor drug in cancer treatment.

Biological importance of coordination compounds:

1. A red blood corpuscles (RBC) is composed of heme group, which is Fe²⁺ Porphyrin complex.it plays an important role in carrying oxygen from lungs to tissues and carbon dioxide from tissues to lungs.

2. Chlorophyll, a green pigment present in green plants and algae, is a coordination complex containing Mg²⁺ as a central metal ion surrounded by a modified Porphyrin ligand called corrin ring. It plays an important role in photosynthesis, by which plants convert CO, and water into carbohydrates and oxygen.

3. Vitamin B₁₂ (cyanocobalamin) is the only vitamin consist of a metal ion. it is a coordination complex in which the central metal ion is CO⁺ surrounded by Porphyrin ligand.

4. Many enzymes are known to be metal complexes, they regulate biological processes. For example, Carboxypeptidase is a protease enzyme that hydrolytic enzyme important in digestion contains a zinc ion coordinated to the protein.

Question 5.
Based on VB theory explain why [Cr(NH₃)₆]³⁺ is paramagnetic, while [Ni(CN)₄]^2- is diamagnetic.
Answer:

Question 6.
Draw all possible geometrical isomers of the complex [Co(en)₂Cl₂]⁺ and identify the optically active isomer.
Answer:

Question 7.
[Ti (H₂O)₆]³⁺ is coloured, while [Sc (H₂O)₆]³⁺ is colourless – Explain. (PAT -4; MARCH 2020)
Answer:

• In [Ti (H₂O)₆]³⁺ the – outer electronic configuration of metal ion Ti³⁺ is 3d¹.
• This single electron present in lower energy t_2g orbitals in the octahedral aqua ligand field absorbs light and excited to higher energy eg level.
• This is known as d-d transition.
• For this excitation absorption maximum is at 20000 cm^-1 corresponding to the CFSE ∆₀ 239.7 KJ mol^-1
• The transmitted colour associated with this absorption in purple and hence the complex appears purple in colour. [Ti(H20)6]³⁺
• But in [Sc(H₂O)₆]³⁺, the outer electronic configuration of metal ion Sc³⁺ is 3d°
• Since there is no electron in d orbital, d-d transition is not possible.
• Hence [Sc(H₂O)₆]³⁺ is colourless.

Question 8.
Give an example for complex of the type [Ma₂b₂C₂] where a, b, c are monodentate ligands and give the possible isomers.
Answer:

• Example for the complex of the type [Ma₂b₂c₂] is [Cr(NH₃)₂ (H₂O)₂Br₂]⁺
• There are totally 6 isomers possible Geometrical isomers – 5

Question 9.
Give one test to differentiate [Co(NH₃)₅Cl]SO₄ and [Co(NH₃)₅ SO₄]Cl
Answer:

• [CO(NH₃)₅Cl]SO₄ → [CO(NH₃)₅Cl]⁺² + SO₄^2-
• [CO(NH₃)₅SO₄] Cl → [CO(NH₃)₅Cl] SO₄

1. Aqueous solution of (a) gives sulphate ion. When the addition of BaCL, solution (a) gives a white precipitate of BaS04. But (b) does not give any precipitate.

2. Aqueous solution of (b) gives chloride ion. When the addition of AgNO₃ solution (b) gives curdy white precipitate of AgCl. But (a) does not give any precipitate.

Question 10.
In an octahedral crystal field, draw the figure to show the splitting of d orbitals.
Answer:

Question 11.
What is linkage isomerism? Explain with an example.
Answer:

Linkage isomerism arises when an ambidentate ligand is bonded to the central metal atom/ion through either of its two different donor atoms.
(eg) [Co(NH₃)₅NO₂]²⁺
[Co(NH₃)₅NO₂]²⁺
Pentaammine nitrito – kN Cobalt (III) ion – Nattached [CO(NH₃)₅ONO]²⁺
Pentaammine nitrito – kO Cobalt (III) ion – O attached

Question 12.
Classify the following ligand based on the number of donor atoms,
a) NH₃
b) en
c) ox^2-
d) triaminotriethylamine
e) pyridine
Answer:

Question 13.
Give the difference between double salts and coordination compounds.
Answer:

Question 14.
Write the postulates of Werner’s theory?
Answer:
In a complex, every metal atom possesses two types of valency.
(i) Primary Valency
(ii) Secondary Valency

15. Why tetrahedral complexes do not exhibit geometrical isomerism.
Answer:
Cis-trans isomerism is not possible in tetrahedral complexes because all the four ligands are adjacent to one another [Ma₂b₂]

Question 16.
Explain optical isomerism in coordination compounds with an example.
Answer:

• Coordination compounds which possess chirality exhibit optical isomerism similar to organic compounds.
• The pair of two optically active isomers which are mirror images of each other are called enantiomers.
• Their solutions rotate the plane of the plane polarised light.
• If the rotation is in the clockwise direction the isomer is called dextro rotatory ‘d’ form.
• If the rotation is in the anti-clockwise direction the isomer is called laevo rotatory T form.
• The octahedral complexes exhibit optical isomerism.
• Examples: [CoCl₂(en)₂]⁺

Two Cis isomers are optically active. One trans isomer is optically inactive

Question 17.
What are hydrate isomers? Explain with an example. (PTA- 6; March 2020)
Answer:

• The exchange of free solvent molecules such as water, ammonia, alcohol etc., in the crystal lattice with a ligand in the coordination entity will give Solvate isomers.
• If the solvent molecule is water, then these isomers are called hydrate isomers.
• Example: CrCl₃.6H₂O has three hydrate isomers

Question 18.
What is crystal field splitting energy?
Answer:
The orbitals lying along the axes dx²-y² and dz² orbitals will experience strong repulsion and raise in energy to a greater extent than the orbitals with lobes directed between the axes (d_xy, d_yz and d_zx). Thus the degenerated orbitals now split into two sets and the process is called crystal field splitting.

Question 19.
What is crystal field stabilization energy? (CFSE)? (PTA -1)
Answer:
Crystal field stabilisation energy (CFSE) is defined as the energy difference of electronic configurations in the ligand field (E_LF) and the isotropic field/barycentre (E_iso).
CFSE (∆Eo) = {FLF}-{Eiso} = {[nt2g(-0.4)+neg(0.6)] ∆o + npP} – {n’p P}
Where
nt_2g = No. f electrons in t_2g orbitals
n_eg = No. of electrons in e_g orbitals
n_p = No. of electron in the ligand field
n¹_p = No. of electron pairs in the isotropic field (barycentre)
P = pairing energy

Question 20.
A solution of [Ni(H₂O)₆]²⁺ is green, whereas a solution of [Ni(CN)₄]^2- is colorless – Explain.
Answer:

• In [Ni(H₂O)₆]²⁺, the Ni²⁺ ion has two unpaired electrons that do not pair up in presence of weak ligand H₂O.
• Hence d-d-the transition is possible.
• Red colour is absorbed and complementary colour green is emitted. So it is green in colour.
• In [Ni(CN)₄]^2-, the Ni²⁺ ion has no unpaired electrons because of the strong ligand CN^– pairs up the electrons.
• Hence no d-d transition is possible.
• So [Ni(CN)₄]^2- is colourless

Question 21.
Discuss briefly the nature of bonding in metal carbonyls. (PTA – 2)
Answer:

• In metal carbonyls, the bond between metal atom and the carbonyl ligand consists of two components.
• The first component is an electron pair donation from the carbon atom of carbonyl ligand into a vacant d-orbital of central metal atom.
• This electron pair donation forms M←CO sigma bond.
• This sigma bond formation increases the electron density in metal d-orbitals and makes the metal electron-rich.
• In order to compensate for this increased electron density, a filled metal d-orbital interacts with the empty π* orbital on the carbonyl ligand and transfers the added electron density back to the ligand.
• This second component is called n-back bonding.
• Thus in metal carbonyls, electron density moves from ligand to metal through sigma bonding and from metal to ligand through pi bonding.
• This synergic effect accounts for a strong M←CO bond in metal carbonyls.
• This is shown diagrammatically as follows.

Question 22.
What is the coordination entity formed when an excess of liquid ammonia is added to an aqueous solution of copper sulphate?
Answer:
When excess liquid ammonia is added to an aqueous solution of copper sulphate tetraammine copper (II) sulphate is formed.
CuSO₄ + 4NH₃ → [Cu(NH₃)₄]SO₄. The Coordination entity is [Cu(NH₃)₄]²⁺.

Question 23.
On the basis of VB theory explain the nature of bonding in [Co(C₂O₄)₃]^3-
Answer:

Question 24.
What are the limitations of VB theory?
Answer:
Limitations of VB – Theory:
1. It does not explain the colour of the complex

2. It considers only the spin only magnetic moments and does not consider the other components of magnetic moments.

3. It does not provide a quantitative explanation as to why certain complexes are inner orbital complexes and the others are outer orbital complexes for the same metal. For example, [Fe(CN)₆]^4- is diamagnetic (low spin) whereas [Fe(CN)₆]^4- is paramagnetic (high spin).

Question 25.
Write the oxidation state, coordination number, nature of ligand, magnetic property and electronic configuration in octahedral crystal field for the complex K₄[Mn(CN)₆].
Answer:

III. Evaluate yourself

Question 1.
When a coordination compound CrCl₃.4H₂O is mixed with silver nitrate solution, one mole of silver chloride is precipitated per mole of the compound. There are no free solvent molecules in that compound. Assign the secondary valence to the metal and write the structural formula of the compound.
Answer:
Since 1 mole of silver chloride is precipitated only one Cl^– ion is present outside the coordination sphere.
∴ The formula is [Cr(H₂O)₄Cl₂]Cl
Secondary valence of the metal is 6

Question 2.
In the complex, [Pt(NO₂)(H₂O)(NH₃)₂]Br, identify the following
i. Central metal atom/ion
ii. Ligand(s) and their types
iii. Coordination entity
iv. Oxidation number of the central metal ion
v. Coordination number
Answer:
[Pt(NO₂)(H₂O)(NH₃)₂]Br
Central metal atom/ ion : Pt (II)
Ligands and their types :
NO₂^– nitrito-k N – Anionic ligand
H₂O aqua – Neutral ligand
NH₃ ammine – Neutral ligand
Coordination entity: [Pt(NO₂)(H₂O)(NH₃)₂]⁺
Oxidation number of the central metal ion x + (-1) + 0 + 2 (0) = +1; x = +2
Coordination number = 4

Question 3.
Write the IUPAC name for the following compounds.
(i) K₂[Fe(CN)₃(Cl)₂(NH₃)]
(ii) [Cr(CN)₂(H₂O)₄][Co(OX)₂(en)]
(iii) [Cu(NH₃)₂Cl₂]
(iv) [Cr(NH₃)₃(NC)₂(H₂O)]⁺
(v) [Fe(CN)₆]^4-
Answer:
(i) K₂[Fe(CN)₃(Cl)₂(NH₃)]
Potassium amminedichloridotricyanido- kC ferrate (III)
(ii) [Cr(CN)₂(H₂O)₄][Co(OX)₂(en)]
Tetra aqua dicyanido kc chromium (III) (ethane -1,2-diamine) dioxalato cobaltate (III)
(iii) [Cu(NH₃)₂Cl₂]
Diammine dichlorido copper (II)
(iv) [Cr(NH₃)₃(NC)₂(H₂O)]⁺
Triammine aqua dicyanido -kN chromium (III) ion
(v) [Fe(CN)₆]^4- Hexacyanido-kc ferrate (II) ion

Question 4.
Give the structure for the following compounds.
Answer:
(i) diamminesilver(I) dicyanidoargentate(I)
(ii) Pentaammine nitrito-KNcobalt (III) ion
(iii) hexafluorido cobaltate (III) ion
(iv) dichloridobis (ethylenediamine)
Cobalt (III) sulphate
(v) Tetracarbonylnickel (0)
Answer:
(i) diamminesilver(I) dicyanidoargentate(I) : [Ag(NH₃)₂] [Ag(CN)₂]
(ii) Pentaammine nitrito-KNcobalt (III) ion : (CO(NH₃)₅ (NO)₂]²⁺
(iii) hexafluorido cobaltate (III) ion : [CoF₆]^3-
(iv) dichloridobis (ethylenediamine) Cobalt (III) sulphate : [Co(en)₂Cl₂]₂SO₄
(v) Tetracarbonylnickel (0): [Ni(CO)₄]

Question 5.
A solution of [Co(NH₃)₄I₂]Cl, when treated with AgNO3, gives a white precipitate. What should be the formula of isomer of the dissolved complex that gives yellow precipitate with AgNO₃. What are the above isomers called?
Answer:

• Since yellow precipitate is obtained with AgN03 iodide ion is present outside the coordination sphere.
• Hence the formula is [Co(NH₃)₄I Cl]I
• This complex is an ionisation isomer of the complex [Co(NH₃)₄I₂]Cl

Question 6.
Three compounds A, B and C have empirical formula CrCl₃.6H₂O. They are kept in a container with a dehydrating agent and they lost water and attaining constant weight as shown below.
Answer:

Question 7.
Indicate the possible type of isomerism for the following complexes and draw their isomers,
i) [Co(en)₃][Cr(CN)₆]
ii) [Co(NH₃)₅(NO₂)]²⁺
iii) [Pt(NH₃)₃(NO₂)]Cl
Answer:
i) [Co(en)₃][Cr(CN)₆] and [Cr(en)₃][Co(CN)₆] are coordination isomers
ii) [CO(NH₃)₅(NO₂)]²⁺ and [Co(NH₃)₅ONO]²⁺ are linkage isomers
iii) [Pt(NH₃)₃(NO₂)]Cl and [Pt(NH3)3Cl] N02 are ionisation isomers
iv) [Pt(NH₃)₃(NO₂)] and [Pt(NH₃)₃ONO)]Cl are linkage isomers.

Evaluate yourself: 8
Question 1.
Draw all possible stereoisomers of a complex Ca[Co(NH₃)Cl(Ox)₂]
Answer:

Evaluate yourself: 9
Question 1.
The spin-only magnetic moment of the Tetrachloridomanganate(II)ion is 5.9 BM. On the basis of VBT, predict the type of hybridisation and geometry of the compound.
Answer:

Question 2.
Predict the number of unpaired electrons in [C0Cl₄]^2- ion on the basis of VBT. [C0Cl₄]^2-
Answer:

Question 3.
A metal complex having composition Co(en)₂Cl₂Br has been isolated in two forms A and B. (B) reacted with silver nitrate to give a white precipitate readily soluble in ammonium hydroxide. Whereas A gives a pale yellow precipitate. Write the formula of A and B. state the hybridization of Co in each and calculate their spin only magnetic moment. [Co(en)₂Cl₂Br.
Answer:
Since B gives white precipitate with AgNO₃, it contains free Cl^– ion outside the Coordination sphere. Hence B is [Co(en)₂Cl Br]Cl

Since A gives pale yellow precipitate with AgNO₃ it contains free Br^– ion outside the coordination sphere. Hence A is [Co(en)₂Cl₂]Br

Evaluate yourself: 10
Question 1.
The mean pairing energy and octahedral field splitting energy of [Mn(CN)₆]^3- are 28,800 cm^-1 and 38500 cm^-1 respectively. Whether this complex is stable in low spin or high spin?
Answer:
High spin Mn³⁺ complex

Question 2.
Draw energy level diagram and indicate the number of electrons in each level for the complex [Cu(H₂O)₆]²⁺. Whether the complex is paramagnetic or diamagnetic?
Answer:
[Cu(H₂O)₆]²⁺

Since one unpaired electron is present, the complex is paramagnetic and C.N is 6 the geometry is Octahedral.

Question 3.
For the [CoF₆]^3- ion the mean pairing energy is found to be 21000 cm^-1. The magnitude of ∆₀ is 13000cm^-1. Calculate the crystal field stabilization energy for this complex ion corresponding to low spin and high spin states. [CoF₆]^3-
Answer:

Since the high spin complex possesses a large nature CFSE value. It is more favoured.

12th Chemistry Guide Coordination Chemistry Additional Questions and Answers

Part II – Additional Questions

I. Choose the best answer.

1. A theory of co-ordination compounds was proposed by …………….
a) Rutherford
b) J.J.Thomson
c) Alfred Werner
d) Neils Bohr
Answer:
c) Alfred Werner

2. The primary valence of a metal ion refers to …………….
a) atomic number
b) oxidation state
c) atomic mass
d) Co-ordination number
Answer:
b) oxidation state

3. The secondary valence of a metal ion refers to …………….
a) atomic number
b) oxidation state
c) atomic mass
d) Co-ordination number
Answer:
d) Co-ordination number

4. Primary valence of a metal ion is always satisfied by …………….
a) Positive ions
b) negative ions
c) neutral molecules
d) all of the above
Answer:
b) negative ions

5. Among the following complexes, which one shows Zero Crystal field stability: energy (CFSE) is (PTA- 6)
a) [Mn(H₂O)₃]³⁺
b) [Fe(H₂O)₆]3³⁺
c) [Co(H₂O)₆]²⁺
d) [CO(H₂O)₆]³⁺
Answer:
b) [Fe(H₂O)₆]3³⁺

6. The sphere in which the metal ion and ligands are firmly attached is called
a) outer sphere
b) ionization sphere
c) Coordination sphere
d) none of the above
Answer:
c) Coordination sphere

7. The sphere in which the metal ion and other ions are loosely bound is called
a) Inner sphere
b) Ionization sphere
c) Coordination sphere
d) none of the above
Answer:
b) ionization sphere

8. The groups which satisfy the secondary valence are called
a) Central metal ion
b) ligands
c) Complexion
d) Complex
Answer:
b) ligands

9. In a complex a central metal ion acts as a
a) Lewis acid
b) Lewis base
c) Bronsted acid
d) Bronsted base
Answer:
a) Lewis acid

10. In a complex a ligand acts as a
a) Lewis acid
b) Lewis base
c) Bronsted acid
d) Bronsted base
Answer:
b) Lewis base

11. In a complex primary valence of a metal ion is
a) ionizable
b) non-ionisable
c) directional
d) Coordination number
Answer:
a) ionizable

12. In a complex secondary valence of a metal ion is
a) ionizable
b) non-ionisable
c) non-directional
d) Oxidation state
Answer:
b) non – ionisable

13. In a complex primary valence of a metal ion is
a) non-ionizable
b) directional
c) non-directional
d) Co-ordination number
Answer:
c) non-directional

14. According to spectrochemical series which of the following ligand produces strongest field and cause maximum splitting? (PTA -1)
a) F^–
b) CO
c) H₂O
d) C^–
Answer:
b) CO

15. In the complex K₄[Fe(CN)₆] the central metal ion is
a) K⁺
b) Fe²⁺
c) Fe³⁺
d) CN^–
Answer:
b) Fe²⁺

16. In the complex K₄[Fe(CN)₆] the ligand is
a) K⁺
b) Fe²⁺
c) Fe³⁺
d) CN^–
Answer:
d) CN^–

17. In the complex K₄[Fe(CN)₆] the primary valence of the central metal ion is
a) +2
b) +3
c) +4
d) +6
Answer:
a) +2

18. In the complex K₄[Fe(CN)₆] the secondary valence of the central metal ion is
a) 2
b) 3
c) 4
d) 6
Answer:
d) 6

19. The geometry of the complex K₄[Fe(CN)₆] is
a) Square planar
b) tetrahedral
c) Octahedral
d) trigonal bipyramidal
Answer:
c) Octahedral

20. K [PtCl₃(C₂H₄)] was called as
a) Wilkinson Catalyst
b) Zeiglar Natta catalyst
c) Magnus’s green salt
d) Zeise’s salt
Answer:
d) Zeise’s salt

21. Which among the following is not a neutral ligand?
a) aqua
b) ammine
c) Oxalato
d) Pyridine
Answer:
c) Oxalato

22. Which is a monodentate ligand?
a) Carbonato
b) Oxalato
c) Cyanido
d) en
Ans :
c) Cyanido

23. [Co (NH₃)₅ NO₂]²⁺ ion exhibits
a) Ionisation isomerism
b) Linkage isomerism
c) Coordination isomerism
d) solvate isomerism
Answer:
b) Linkage isomerism

24. [Pt (NH₃)₄] [Pd(Cl)₄] complex exhibits
a) Ionisation isomerism
b) Linkage isomerism
c) Coordination isomerism
d) solvate isomerism
Answer:
c) Coordination isomerism

25. [Cr(NH₃)₄ClBr] NO₂ and [Cr(NH₃)₄ClNO₂] Br are
a) Linkage isomers
b) Ionisation isomers
c) Coordination isomers
d) solvate isomers
Answer:
b) Ionisation isomers

26. Which among the following gives a curdy white precipitate with silver nitrate solution?
a) [Co (NH₃)₄ Cl Br] Br
b) [Co (NH₃)₄ Br₂] Cl
c) [Cr (NH₃)₄ Cl Br] NO₂
d) [Cr (NH₃)₄ ClNO₂] Br
Answer:
b) [Co (NH₃)₄ Br₂] Cl

27. Which among the following gives three moles of silver chloride with silver nitrate solution?
a) [Cr(H₂O)₄ Cl₂]Cl. 2H₂O
b) [Cr(H₂O)₅ Cl]Cl₂. 2H₂O
c) [Cr(H₂O)₆]Cl₃
d) [Cr(H₂O)₃ Cl₃]. 3H₂O
Answer:
c) [Cr(H₂O)₆]Cl₃

28. Which among the following gives a white precipitate with barium chloride solution?
a) [Co (NH₃)₅ Cl] Cl₂
b) [Cu(NH₃)₄] SO₄
c) [Ag (NH₃)₂] Br
d) [Co (NO₂)₃ (NH₃)₃]
Answer:
b) [Cu (NH₃)₄] SO₄

29. According to VBT, in a Coordination complex the bonding between central metal ion and ligands is
a) ionic
b) covalent
c) metallic
d) vander waals force
Answer:
b) covalent

30. In which of the following complexes the central metal ion undergoes SP3 hybridisation
a) [Ni(CN)₄]^2-
b) [Pt(NH₃)₄]²⁺
c) [NiCl₄]^2-
d) [Cu (NH₃)₄]²⁺
Ans :
c) [[NiCl₄]^2-

31. If the (n-1) d orbitals are involved in hybridisation, the complex is called as
a) Outer orbital complex
b) Spin paired complex
c) High spin complex
d) Spin free complex
Answer:
b) Spin paired complex

32. If the n d orbitals are involved in hybridisation, the complex is called as
a) Inner orbital complex
b) low spin complex
c) high spin complex
d) spin paired complex
Answer:
c) high spin complex

33. Which statement is incorrect?
a) [Ni(CO)₄] – Tetrahedral, Paramagnetic
b) [Ni(CN)₄]^2- – Square planar, diamagnetic
c) [Ni(CO)₄] – Tetrahedral, diamagnetic
d) [Ni(Cl)₄]^2- – Tetrahedral, Paramagnetic
Answer:
a) [Ni(CO)₄] – Tetrahedral, Paramagnetic

34. In [Ni(CO)₄], the hybridisation of central metal ion is
a) dsp²
b) sp³
c) d²sp³
d) dsp³
Answer:
b) Sp³

35. Co-ordination number of Ni in [Ni(C₂O₄)₃]^4- is (PTA – 4)
a) 3
b) 6
c) 4
d) 2
Answer:
b) 6

36. According to CFT, the bonding between central metal ion and ligands is
a) ionic
b) covalent
c) metallic
d) Vanderwaal’s force
Answer:
a) ionic

37. Which is a strong field ligand?
a) I^–
b) CN^–
c) Cl^–
d) S^2-
Answer:
b) CN^–

38. Which is a weak field ligand?
a) NO^–₂
b) NH³
c) CO
d) SCN^–
Answer:
d) SCN^–

39. The observed colour of a coordination compound can be explained using
a) Valence bond theory
b) Werner’s theory
c) Crystal field theory
d) Molecular orbital theory
Answer:
c) Crystal Field theory

40. Which among the following is a mononuclear carbonyl?
a) [C0₂(C0)₈]
b) [Fe₃(CO)₁₂]
c) [Fe(CO)₅]
d) [Fe₂(CO)₉]
Answer:
c) [Fe (CO)₅]

41. If the instability constant value of [Cu(NH₃)₄]²⁺ is 1.0 × 10^-12, its stability constant value is:
a) 1.0 × 10^-12
b) 1.0 × 10¹²
c) 12
d) -12
Answer:
b) 1.0 × 10¹²

42.

From the above table which of the following complex is most stable?
a) [Cu (NH₃)₄]²⁺
b) [Fe (SCN)]²⁺
c) [CO(NH₃)₆]³⁺
d) [Ag(CN)₂]^–
Answer:
c) [CO(NH₃)₆]³⁺

43. Phthalo blue – a bright blue pigment is a complex of
a) Ni²⁺
b) Co³⁺
c) Cu²⁺
d) Ag⁺
Ans:
c) Cu²⁺

44. For removing lead poisoning, the chelating ligand used is
a) DMG
b) EDTA
c) en
d)CO
Answer:
b) EDTA

45. The process in which coordination complexes are used in the extraction of silver and gold from their ores is
a) Mond’s process
b) Alumino thermic process
c) Mac – Arthur – Forrest cyanide process
d) Bessimerisation
Answer:
c) Mac- Arthur – Forrest cyanide process

46. Ni ions present in nickel chloride solution is estimated accurately using the ligand
a) DMG
b) EDTA
c) en
d) CO
Ans:
a) DMG

47. Catalyst used for hydrogenation of alkenes is
a) Ziegler – Natta Catalyst
b) Zeise’s Salt
c) Wilkinson’s catalyst
d) Magnus’s green Salt
Answer:
c) Wilkinson’s Catalyst

48. The catalyst used in the polymerization of ethene is
a) Ziegler – Natta Catalyst
b) Zeise’s Salt
c) Wilkinson’s catalyst
d) Magnus’s green Salt
Answer:
a. Ziegler – Natta Catalyst

49. The complex used as an antitumor drug in cancer treatment is
a) Ca – EDTA Chelate
b) Cisplatin
c) Trans – Platin
d) Cyano cobalamine
Answer:
b) Cis – Platin

50. RBC is composed of
a) Mg²⁺
b) Fe²⁺
c) Fe³⁺
d) Co⁺
Answer:
b) Fe²⁺

51. In RBC and chlorophyll, the ligand is
a) EDTA
b) DMG
c) Porphyrin
d) en
Answer:
c) Porphyrin

II. Match the following
Question 1.

Answer:

Question 2.

Answer:

Question 3.

Answer:

Question 4.

Answer:

Question 5.

Answer:

III. Pick the odd man out

1. Pick the odd man out w.r.t complexion
a) [CO(NH₃)₆] Cl₃
b) [Fe (H₂O)₆] Cl₂
c) [Cu(NH₃)₄]SO₄
d) K₄ [Fe(CN)₆]
Answer:
K₄ [Fe(CN)₆] – Anionic complex others are cationic complexes

2. Pick the odd man out w.r.t geometry
a) [CO(NH₃)₆] Cl₃
b) [Fe (H₂O)₆] Cl₂
c) [Cu(NH₃)₄]SO₄
d) K₄ [Fe(CN)₆]
Answer:
c) [Cu(NH₃)₄]SO₄

3. Pick the add man out w.r.t primary valence of the metal ion
a) [CO(NH₃)₆]Cl₃
b) K₃[Fe(CN)₆]
c) [Fe(CO)₅]
d) [CO(NH₃)₃ Cl₃]
Answer:
c) [Fe(CO)₅] – Primary valence is zero, in other complexes the primary valence is +3

4. Pick the odd man out w.r.t the nature of the ligand
a) aqua
b) carbonyl
c) nitrosyl
d) nitrato
Answer:
d) nitrato – negative ligand others are neutral ligand

IV. Pick the wrong statement

Question 1.
i) The secondary valence of a metal ion is satisfied by ligands.
ii) The secondary valence determines the geometry of the complex.
iii) The secondary valence is oxidation state of the metal ion.
iv) If the secondary valence is six, the geometry is tetrahedral.
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
d) (iii) & (iv)
Correct statement : (iii) The secondary valence is the coordination number of the metal ion (iv) If the secondary valence is six, the geometry is octahedral.

Question 2.
i) Central metal ion and ligands are bonded through ionic bond.
ii) Central metal ion accepts electron pairs donated by ligands.
iii) Ligands act as lewis acids
iv) Charge on the coordination sphere is the net change inside the coordination sphere.
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
b) (i) & (iii)
Correct statement: (i) Central metal ions and ligands are bonded through a coordinate covalent bond, (iii) Ligands act as Lewis bases.

Question 3.
i) Trans [Co Cl₂ (en)₂]⁺ is optically active.
ii) In octahedral complexes if nd orbitals are involved in hybridisation, they are called as low spin complexes.
iii) In a complex if all the electrons are paired, it is diamagnetic
iv) Ligands which cause the pairing of electrons are called strong field ligands.
a) (i) &(ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
a) (i) & (ii)
Correct statement:
i) Trans [Co Cl₂ (en)₂]⁺ is optically inactive.
ii) In octahedral complexes if nd orbitals are involved in hybridisation, they are called high spin complexes.

Question 4.
i) The ligands present on the right side of the spectrochemical series are called strong field ligands.
ii) The ligands present on the left side of the spectrochemical series are called weak field ligands.
iii) CFSE (∆E₀ = E_iso – E_LF
iv) \(\mu_{S}=\sqrt{n(n-2)}\)
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
c) (iii) & (iv)
Correct statement: (iii) CFSE (∆E₀) = E_LF – E_iso
iv) \(\mu_{s}=\sqrt{n(n+2)}\)

V. Pick the Correct statement

Question 1.
i) The primary valence of a metal ion is positive in most cases and zero in certain cases.
ii) The primary valences are directional
iii) The primary valence is satisfied by ligands in certain cases.
iv) The primary valence is co-ordination number.
a) (i) &(ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
b) (i) & (iii)
Correct statement: (ii) The primary valences are non – directional
(iv) The primary valence is oxidation state.

Question 2.
i) Coordination entity consists of central metal ion and ligands.
ii) In K₄[Fe(CN)₆], the coordination entity is [Fe(CN)₆]^4-
iii) The groups present inside the coordination entity can be ionised
iv) The coordination entity is not responsible for the geometry of the complex.
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
a) (i) & (ii)
Correct statement : (iii) The groups present inside the coordination entity can not be ionised.
(iv) The coordination entity is responsible for the geometry of the complex.

Question 3.
i) In thiocyanate ligand if sulphur forms a coordination bond with metal, the ligand is named thiocyanato-KN
ii) The ligand NH₃ is named amine.
iii) Ethylenediamine is a neutral ligand.
iv) The name of cationic ligand ends with – ium
a) (i) &(ii)
b) (i) & (iii)
c) (iii) & (iv)
d) (i) & (iii)
Answer:
c) (iii) & (iv)
Correct statement : (i) In thiocyanate ligand if sulphur forms a coordination bond with metal, the ligand is named thiocyanato-kS.
(ii) The ligand NH₃ is named as ammine.

Question 4.
i) Linkage isomer is not possible with nitrite ligand.
ii) Coordination isomerism is found in coordination compounds in which both cation and anion are complexions.
iii) Ionisation isomer arises when an ionisable counter ion itself can act as a ligand.
iv) In solvate isomer if the solvent molecule is alcohol it is called a hydrate isomer.
a) (i) &(ii)
b) (i) & (iii)
c) (iii) & (iv)
d) (i) & (iii)
Answer:
b) (ii) & (iii)
Correct statement : (i)Lingake isomer is possible with nitrite ligand (ii) In solvate isomer if the solvent molecule is water it is called as hydrate isomer.

VI. Assertion and Reason

i) Both A and R are correct, R explains A.
ii) A is correct, R is wrong
iii) A is wrong, R is correct
iv) Both A and R are correct, but R does not explain A.

Question 1.
Assertion (A) : [Co(NH₃)₄ Br₂] Cl and [CO(NH₃)₄ Cl Br] Br are ionisation isomers.
Reason (R) : In solution, they give free Cl^– and Br^– ions respectively
Answer:
i) Both A and R are correct, R explains A.

Question 2.
Assertion (A): Geometrical isomerism exists in homoleptic complexes.
Reason (R): Geometrical isomerism is due to different possible three-dimensional spatial arrangements of the ligands around the central metal atom.
Answer:
iii) A is wrong, R is correct.
Correct Assertion (A): Geometrical isomerism exists in heteroleptic complexes.

Question 3.
Assertion (A) : According to VBT the ligand → metal bond in a coordination complex, is covalent in nature.
Reason (R) : Mutually shared electrons are provided equally by the central metal atom and the ligand
Answer:
ii) A is correct, R is wrong Correct Reason: Mutually shared electrons are provided by ligands to the central metal atom.

VII. Two Mark Questions

Question 1.
Write any two medicinal uses of coordination compounds? (PTA – 6)
Answer:
Ca-EDTA chelate is used in the treatment of lead and radioactive poisoning. That is for removing lead and radioactive metal ions from the body.
cisplatin is used as an antitumor drug in cancer treatment.

Question 2.
What is a central metal ion?
Answer:
The metal ion which accepts electron pair donated by ligands is called the central metal ion of a complex.
Central metal ions are electron-pair acceptors.
Central metal ions are Lewis acids.

Question 3.
What are ligands?
Answer:
In a complex, the negative ions, neutral molecules positive ions which donate pair of electrons to the central metal ion are called ligands.
Ligands are electron-pair donors.
Ligands are Lewis bases.

Question 4.
[Fe(CN₆)]^4- and [Fe(H₂O)₆]²⁺ are of different colours in dilute solutions, why? (PTA – 2)
Answer:

• [Fe(CN₆)]^4- and [Fe(H₂O)₆]²⁺ show different colours in dilute solution because
• CN^– is a strong field ligand and H₂O is a weak ligand hence magnitude of CFSE is different
• both CN^– and H₂O absorb same wavelength of energy.
• complexes of weak field ligands are generally colourless
• the sizes of CN^– and H₂O are different hence their colours are also different.

Question 5.
Define coordination number.
Answer:
The number of ligand donor atoms bonded to a central metal ion in a complex is called the coordination number of the metal.

Question 6.
What is the coordination sphere?
Answer:
The complexion containing the central metal atom/ion and the ligands attached to it is collectively called a coordination sphere or inner sphere.

Question 7.
What are inert and labile complexes? (PTA – 4)
Answer:
In some cases, complexes can undergo rapid ligand substitution, such complexes are called labile complexes. However, some complexes undergo ligand substitution very slowly (or sometimes no substitution), such complexes are called inert complexes.

Question 8.
What is d-d transition?
Answer:
The excitation of d – electrons of central metal ion from the lower energy t_2g level to the higher energy eg level by the absorption of light is known as d-d transition.

Question 9.
What are metallic carbonyls?
Answer:
Metal carbonyls are the transition metal complexes of carbon monoxide containing metal-carbon bond.

Question 10.
What is stability constant? (PTA – 5)
Answer:
The reciprocal of dissociation equilibrium constant or instability constant (∝) of a complex is called as the formation equilibrium constant or stability constant (β)

Question 11.
Calculate the magnetic moment and magnetic property of [CoF₆]^3- (MARCH 2020)
Answer:

VIII. Three Mark questions

Question 1.
Explain geometrical isomerism in square planar complexes.
Answer:

• Geometrical isomerism exists in heteroleptic complexes due to different possible three-dimensional spatial arrangements of the ligands around the central metal atom.
• In square planar complexes of the form [MA₂B₂]^n± and [MA₂BC]^n±, if similar groups present on the same side of the metal atom it is called Cis – isomer and if they are on the opposite sides of the metal atom it is called trans isomer.
• Square planar complex of the type [M(xy)₂]^n± where XY is a bidentate ligand with two different coordinating atoms and of the type [MABCD]^n± by considering anyone ligands as a reference, the rest can be arranged in three different ways also show Cis – trans isomerism.

Question 2.
Explain geometrical isomerism in octahedral complexes.
Answer:

• Octahedral complexes of the type [MA₂B₄]^n±, [M(XX)₂B₂]^n± show Cis- trans isomerism Here A and B are monodentate ligands and XX is a bidentate ligand with two same kind of donor atoms.
• In the above diagram the positions (1,2), (1.3) (1,4) (1,5), (2,3) (2,5), (2,6) (3,4) (3,6), (4,5), (4,6) and (5,6) are identical and if two similar groups are present in any one of these positions, the isomer is called as Cis isomer.
• Similarly, positions (1, 6), (2,4) and (3, 5) are identical and if two similar groups are present in these positions, the isomer is called as trans-isomer.
  (ex) [Co (NH₃)₄Cl₂]⁺

Question 3.
Write about facial and meridional isomers.
Answer:

• Octahedral complex of the type [MA₃B₃]^n± shows geometrical isomerism
• If the three similar ligands (A) are present in the corners of one triangular face of the octahedron and the other three ligands (B) are present in the opposing triangular face, then the isomer is called as a facial isomer (fac isomer)
• If the three similar ligands are present around the meridian which is an imaginary semicircle from one apex of the octahedron to the opposite apex, the isomer is called as a meridional isomer (mer isomer)
• This is called meridional because each set of ligands can be regarded as lying on a meridian of an octahedron

Question 4.
In a tetrahedral crystal field, draw the figure to show splitting of d-orbitals (PTA – 6)
Answer:

Question 5.
Write the use of complex formation in photography.
Answer:
In photography, when the developed film is washed with sodium thiosulphate solution (hypo), the negative film gets fixed. Undecomposed AgBr forms a soluble complex called
sodiumdithiosulphatoargentate(I) which can be easily removed by washing the film with water.
AgBr + 2Na₂S₂O₃ → Na₃[Ag (S₂O₃)₂] + 2NaBr

Question 6.
Write the use of metal complexes in biological systems.
Answer:

• RBC is composed of heme group which is a complex of Fe²⁺ and Porphyrin ligand. It carries oxygen from lungs to tissues.
• Chlorophyll, a green pigment present in green plants, algae is a complex of Mg²⁺ and corrin ring, a modified porphyrin ligand. It is responsible for the photosynthesis and the conversion of CO₂ and water into carbohydrates and oxygen.
• The only vitamin with a metal ion Vitamin B₁₂ (Cyanocobalamine) is a complex of Co⁺ and porphyrin ligand.
• Many enzymes which regulate biological processes are metal complexes.
  (ex) Carboxypeptidase is a protease enzyme that hydrolytic enzyme important in digestion is a zinc complex with protein-ligand.

Question 7.
How is Cisplatin used as an antitumour drug in cancer treatment?
Answer:

• C is platin is a square planar coordination complex Cis – [Pt (NH₃)₂ Cl₂]
• It is a platinum-based anti-cancer drug
• This drug undergoes hydrolysis and reacts with DNA to produce various cross-links.
• These cross-links hinder the DNA replication and transcription, which results in cell growth inhibition and ultimately cell death.
• It also cross-links with cellular proteins and inhibits mitosis.

IX. Five Mark Questions

Question 1.
How are coordination complexes classified?
Answer:
Coordination complexes are classified into two types.
I. Classification based on the net charge on the complex
A coordination compound in which the complexion
(i) Carries a net positive charge is called a cationic complex. (Ex). [Ag(NH₃)2]⁺
(ii) Carries a net negative charge is called an anionic complex. (Ex). [Ag(CN)₂]^–
(iii) Bears with no net charge is called a neutral complex. (Ex). [Ni(CO)₄]

II. Classification based on kind of ligands.
A coordination compound in which
(i) The central metal ion/ atom is coordinated to only one kind of ligand is called a homoleptic complex. (Ex). [Ag(NH₃)₂]⁺
(ii) The central metal ion/ atom is coordinated to more than one kind of ligands is called a heteroleptic complex. (Ex). [Co (NH₃)₅ Cl]²⁺

Question 2.
Write the steps involved in the IUPAC nomenclature of coordination compounds.
Answer:
i) The cation is named first, followed by the anion regardless of whether the ion is simple or complex.
ii) The simple ions are named as in other ionic compounds.
iii) To name a complexion, the ligands are named first followed by the central metal atom/ ion. If more than one ligand is present they are named in alphabetical order.

(a) Naming the ligands:

• Name of anionic ligands end with the letter ‘O’.
  Name of cationic ligands end with ‘ium’.
• Neutral ligands are usually called with their molecular names with fewer exception namely H₂O (aqua), NH₃ (ammine) etc.,
• K-term is used to denote an ambidentate ligand in which more than one coordination mode is possible.
• If the coordination entity contains more than one ligand of a particular type, the multiples of ligand (2,3,4, …..) is indicated by adding Greek prefixes (di,tri,tetra ….) to the name of the ligand.
• If the name of the ligand contains a Greek prefix (eg. ethylene diamine) alternate prefixes (bis, tris, tetrakis ) can be used.
• These numerical prefixes are not taken into account for alphabetising the name of ligands.

(b) Naming the central metal atom/ion.

• In cationic/neutral complexes, the element name is used as such for naming the central metal atom/ion.
• In anionic complexes a suffix ‘ate’ is used along with the element name or its Greek or Latin name if any.

Question 3.
Explain structural isomerism exhibited by coordination compounds.
Answer:

• The coordination compounds with same formula, but have different connections among their constituent atoms are called structural isomers or constitutional isomers.
• It is of four types.

i) Linkage isomers:
This type arises when an ambidentate ligand is bonded to the central metal atom/ion through either of its two different donor atoms.
(ex). [Co (NH₃)₅ NO₂]²⁺ – nitritoK-N ligand N – attached.
[Co (NH₃)₅ ONO]²⁺ – nitrito K-O ligand O – attached.

ii) Coordination isomers:
This type arises in the coordination compounds having both the cation and anion as complex ion.

The interchange of one or more ligands between the cationic and the anionic coordination entities result in different isomers.
(ex). [Co (NH₃)₆] [Cr (CN)₆] &
[Cr (NH₃)₆] [CO (CN)₆]

iii) Ionisation isomers:

• This type arises when an ionisable counter ion itself can act as a ligand.
• The exchange of such counterions with one or more ligands in the coordination entity will result in ionisation isomers.
• These isomers will give different ions in solution.
  (ex). [Co (NH₃)₄ Br₂] Cl &
  [Co (NH₃)₄ Cl Br] Br.

iv) Solvate isomers:
This arises by the exchange of free solvent molecules such as water, ammonia, alcohol etc., in the crystal lattice with a ligand in the coordination entity.

If the solvent molecule is water, these isomers are called hydrate isomers. (ex).[Cr(H20)6]Cl3;
[Cr (H₂O)₅ Cl] Cl₂ H₂O;
[Cr (H₂O)₄ Cl₂] Cl 2H₂O.

Question 4.
Explain the main assumption of VBT of coordination compounds (PTA – 1)
Answer:

• The ligand → Metal bond in coordination complex is an aicnl in nature, it is formed by sharing ol electrons between theienlrul metal atom and the ligand.
• Each ligand should have at least one filled orbital containing a lone pair of electrons.
• To accept the electron pairs donated by the ligands, the central metal ion must contain vacant orbitals equal to its coordination number.
• This vacant orbital of central metal atom undergo hybridisation.
• The vacant hybridized orbitals of central metal ion linearly overlap with filled orbital of the ligands forming coordinate covalent sigma bonds.
• The hybridized orbitals are directional and give definite geometry to the complexion.
• In octahedral complexes, if (n-1) d orbitals are involved in hybridisation, they are called inner orbital complexes or low spin complexes or spin paired complexes. If nd orbitals are involved in hybridisation, they are called as outer orbital complexes, or high spin complexes or spin-free complexes.
• If the metal ion contains unpaired electrons the complex is paramagnetic, if the electrons are paired, the complex is diamagnetic.
• Ligands such as CO, CN^–, en and NH₃ which cause pairing of electrons are called strong field ligands.
• Ligands such as, F^–, Cl^–, Br^– which do not cause pairing of electrons are called weak field ligands.
• Greater the overlapping between the ligand orbitals and the hybridised metal orbital, greater is the bond strength.
• The relation between crystal field splitting energies of octahedral and tetrahedral field is \(\Delta t=\frac{4}{9} \Delta 0\)

Question 5.
Write the salient features of Crystal Field Theory.
Answer:

• Crystal field theory assumes that the bond between the ligand and the central metal atom is purely ionic, ie. the bond is formed due to the electrostatic attraction between the electron-rich ligand and the electron-deficient metal.
• In case of charged metal ions or ligands, they are considered as point charges and in case of neutral metal atoms or ligands they are considered as electric dipoles.
• The complex formation is considered as the following series of hypothetical steps.

Step-1:
In an isolated gaseous state, all the five ‘d’ orbitals of the central metal ion are degenerate. Initially, the ligands form a spherical field of negative charge around the metal. In this field the energies of all the five d orbitals will increase due to the repulsion between the electrons of the metal and the ligand.

Step-2:
The ligands are approaching the metal atom in actual bond directions. In an octahedral field the central metal ion is located at the origin and the six ligands are coming from +x, -x, +y, -y, +z and -z directions.
Orbitals lying along the axes d_x²-y² and d_z² (e_g) will experience strong repulsion and raise in energy to a greater extent than the orbitals with lobes directed between the axes (d_xy/ d_yz, d_zx) (t_2g). This splitting of degenerated orbitals into two sets is called crystal field splitting.

Step-3:
Upto this point the complex formation would not be favoured. When the ligands approach further, there will be an attraction between the negatively charged electron and the positively charged metal ion resulting in a net decrease in energy. This decrease in energy is the driving force for the complex formation.

In the case of the tetrahedral complex, none of these orbital is directly pointing to the ligands. The d_xy, d_yz, d_zx (t_2g) orbitals are nearer to the ligands and hence they interact to a greater extent with the ligand orbitals, therefore d_xy, d_yz, d_zx (t_2g) orbitals become high energy orbitals than d_x²-y² and d_z² (e_g) orbitals in a tetrahedral complex.
Thus the mode of splitting of d orbitals in an octahedral complex is just the reverse of that observed in a tetrahedral complex.

Question 6.
Explain the classification of metallic carbonyls.
Answer:
Metallic carbonyls are classified in two different ways.

I. Classification based on the number of metal atoms present
a) Mononuclear carbonyls:
These contain only one metal atom.
(ex). [Ni (CO)₄]

b) Polynuclear carbonyls
These contain two or more metal atoms. They may be homonuclear or heteronuclear.
(ex). [Co₂ (CO)₈]; [MnCo (CO)₉]

II. Classification based on the structure:
a) Non bridged metal carbonyls.
These do not contain any bridging carbonyl ligands.
Those contain only terminal carbonyls.
(ex)
Those contain terminal carbonyls and metal-metal bonds

b) Bridged carbonyls:
These contain one or more bridging carbonyls along with terminal carbonyl ligands and one or more metal-metal bonds, (ex).

Question 7.
Write the IUPAC names for the following complexes. (MARCH 2020)

1. [Co (NH₃)₅Cl]²⁺
2. K₃ [Fe (C₂O₄)₃]
3. [Co <NH₃)₆] [Cr(CN)₆]
4. [Cr (H₂O)₄ Cl₂] Cl 2H₂O
5. [Ft (Py)₄] [Pt Cl₄]
6. [Zn(NCS)₄]^2-
7. [Ag(NH₃)₂]²⁺

Answer:

1. Pentaamminechloridocobalt (III)ion
2. Potassium trioxalato ferrate (III)
3. Hexaammine Cobalt (III) hexa cyanido K-C chromate (III)
4. Tetra aqua dichlorido chromium (III) chloride dihydrate
5. Tetrapyridine platinum (II) tetra chloridoplatinate (II)
6. Tetrathiocyanato – kN Zincate (II) ion
7. Diammine Silver (I) ion

Question 8.
Write the formula for the following coordination compounds.

1. Tris (ethylenediamine) Chromium (III) Chloride
2. Potassium tetracyanido K-C nickelate (II)
3. Ammine bromido chlorido nitrito – kN platinate (II) ion
4. Dichlorido bis (ethane -1, 2 – diamine) Platinum (IV) nitrate
5. Hexa aqua manganese (II) Phosphate

Answer:

1. [Cr (en)₃] Cl₃
2. K₂[Ni(CN)₄]
3. [Pt (NH₃) BrCl No₂]^–
4. [Pt (en)₂ Cl₂] (N0₃)₂
5. [Mn (H₂O)₆]₃ (PO₄)₂

Question 9.
What will be the correct order for the wavelengths of absorption in the visible region and explain for the following (PTA – 3)
[Ni(NO₂)₆]^4-; [Ni(NH₃)₆]²⁺; [Ni(H₂O)₆]²⁺
Answer:
The central metal ion in all three complexes is the same. Therefore, absorption in the visible region depends on the ligands. The order in which the CFSE values of the ligands increases in the spectrochemical series is as follows:
H₂O < NH₃ < NO₂^–
Thus, the amount of crystal-field splitting observed will be in the following order.
∆_0(H2O) < ∆_0(NH3) < ∆_0(NO2^–)
Hence, the wavelengths of absorption in the visible region will be in the order.
[Ni(H₂O)₆]²⁺ >[Ni(NH₃)₆]²⁺ > [Ni(NO₂)₆]^4-

Question 10.
Answer all the questions for the complex [Fe(en)₂Cl₂]Cl₂ (PTA -6)

1. Oxidation number of Fe
2. Hybridisation and shape
3. Magnetic behaviour
4. Number of geometric isomers
5. Whether there may be optical isomer also?
6. IUPAC name

Answer:

1. Oxidation number of Fe is +3
2. Hybridisation and shape is -d²sp³
3. Paramagnetic due to presence of three unpaired electrons
4. Two cis and trans isomers
5. Yes, cis isomer will also show optical isomerism
6. Dichlorido bis (ethane 1, 2 diamine) iron (III) chloride (or)
7. Dichloro bis (ethylenediamine) iron (III) chloride.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 10 Ordinary Differential Equations Ex 10.2 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.2

Question 1.
Express each of the following physical statements in the form of differential equation.
(i) Radium decays at a rate proportional to the amount Q present.
(ii) The population P of a city increases at a rate propotional to the product of population and to the difference between 5,00,000 and the population.
(iii) For a certain substance, the rate of change of vapor pressure P with respect to temperature T is proportional to the vapor pressure and inversely proportional to the square of the temperature.
(iv) A saving amount pays 8% interest per year compound continuously. In addition, the income from another investment is credited to the amount continuously at the rate of Rs 400 per year.
Solution:
(i) If at anytime t, The amount of Radium present is Q. The rate at which Q is decreasing is \(\frac { dQ }{ dt }\) and this is negative. This rate of decrease or decay is found to be proportional to Q itself. Hence we have the law, \(\frac { dQ }{ dt }\) = -kQ where k is a positive constant. Therefore \(\frac { dQ }{ dt }\) + kQ = 0 which is a differential equation.

(ii) The rate of change of population increases with respect to time t, is \(\frac { dP }{ dt }\) & the rate of population is proportional to the product of population is \(\frac { dP }{ dt }\) = kP & the also the difference between 5,00,000 & the population is \(\frac { dP }{ dt }\) = kP(5,00,000 – P) is required differential equation.

(iii) The rate of change of vapor pressure P with respect to time t is \(\frac { dP }{ dt }\) & the rate of increase vapor pressure is P at time T is proportibnal to the vapor pressure and also is inversely proportional to the square of the temperature is \(\frac { dP }{ dt }\) = \(\frac { kP }{ T^2 }\)
since \(\frac { k }{ T^2 }\) = α
\(\frac { dP }{ dt }\) = α P ⇒ \(\frac { dP }{ dt }\) – α P = 0 is a required differential equation.

(iv) Let x be the amount. Amount varies from every year, (ie) Amount varies with respect to time t is \(\frac { dx }{ dt }\) & in addition the income from other source credited Rs.400 continuously for every year.
∴ \(\frac { dx }{ dt }\) = \(\frac { 8 }{ 100 }\) × x + 400
⇒ \(\frac { dx }{ dt }\) = \(\frac { 2x }{ 25 }\) + 400
⇒ ⇒ \(\frac { dx }{ dt }\) – \(\frac { 2 }{ 25 }\)x – 400 = 0 is a required differential equation.

Question 2.
Assume that a spherical rain drop evaporates at a rate proportional to its surface area. Form a differential equation involving the rate of change of the radius of the rain drop.
Solution:
The rate of change of the radius of the rain drop evaporates with respect to time t is \(\frac { dx }{ dt }\) = -k i.e The drop is decreasing so put negative sign. Therefore \(\frac { dx }{ dt }\) = -k is a required differential equation.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 10 Ordinary Differential Equations Ex 10.1 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.1

Question 1.
For each of the following equations, determine its order, degree (if exists)
(i) \(\frac { dy }{ dx }\) + xy = cot x
(ii) (\(\frac { d^3y }{ dx^3 }\))^2/3 – 3 \(\frac { d^2y }{ dx^2 }\) + 5\(\frac { dy }{ dx }\) + 4 = 0
(iii) (\(\frac { d^2y }{ dx^2 }\))² + (\(\frac { dy }{ dx }\))² = x sin (\(\frac { d^2y }{ dx^2 }\))
(iv) \(\sqrt{\frac { dy }{ dx }}\) – 4 \(\frac { dy }{ dx }\) – 7x = 0
(v) y(\(\frac { dy }{ dx }\)) = \(\frac { x }{ (\frac { dy }{ dx })+(\frac { dy }{ dx })^3 }\)
(vi) x²\(\frac { d^2y }{ dx^2 }\) + [1 + (\(\frac { dy }{ dx }\))²]^1/2 = 0
(vii) (\(\frac { d^2y }{ dx^2 }\))³ = \(\sqrt{1+(\frac { dy }{ dx })}\)
(viii) \(\frac { d^2y }{ dx^2 }\) = xy + cos (\(\frac { dy }{ dx }\))
(ix) \(\frac { d^2y }{ dx^2 }\) + 5 \(\frac { dy }{ dx }\) + ∫ ydx = x³
(x) x = e^{xy(\(\frac { dy }{ dx }\))}
Solution:
(i) \(\frac { dy }{ dx }\) + xy = cot x
In the given equation, the highest order derivative is \(\frac { dy }{ dx }\) only its power is 1
∴ Its order = 1 & degree = 1

(ii) (\(\frac { d^3y }{ dx^3 }\))^2/3 – 3 \(\frac { d^2y }{ dx^2 }\) + 5\(\frac { dy }{ dx }\) + 4 = 0
Taking power 3 on both sides, we get
(\(\frac { d^3y }{ dx^3 }\))² = (3 \(\frac { d^2y }{ dx^2 }\) – 5\(\frac { dy }{ dx }\) – 4)³
In the equation (1), the highest order derivative is \(\frac { d^3y }{ dx^3 }\) and its power is 2.
∴ Its order = 3 & degree = 2

(iii) (\(\frac { d^2y }{ dx^2 }\))² + (\(\frac { dy }{ dx }\))² = x sin (\(\frac { d^2y }{ dx^2 }\))
In the equation, the highest order derivative is \(\frac { d^2y }{ dx^2 }\) and its order is 2.
It has a term sin (\(\frac { d^2y }{ dx^2 }\)), so its degree is not defined or degree does not exist.

on squaring both sides,
\(\frac { dy }{ dx }\) = 16 (\(\frac { dy }{ dx }\))² + 49 x² + 56x \(\frac { dy }{ dx }\)
clearly, it is a differential equation of order = 1 & degree = 2.

In this equation, the highest order derivative is \(\frac { dy }{ dx }\) & its power is 4.
∴ Its order = 1 & degree = 4

In this equation, the highest order derivative is \(\frac { d^2y }{ dx^2 }\) & its power is 2.
∴ Its order = 2 & degree = 2

In this equation, the highest order derivative is \(\frac { d^2y }{ dx^2 }\) & its power is 6.
∴ Its order = 2 & degree = 6

(viii) \(\frac { d^2y }{ dx^2 }\) = xy + cos (\(\frac { dy }{ dx }\))
In this equation, the highest order derivative is \(\frac { d^2y }{ dx^2 }\) & its power is 2.
It has a term cos(\(\frac { dy }{ dx }\)), so its degree is not defined or degree does not exist.

(ix) \(\frac { d^2y }{ dx^2 }\) + 5 \(\frac { dy }{ dx }\) + ∫ ydx = x³
differentiating with respect to x, we get
\(\frac { d^3y }{ dx^3 }\) + 5 \(\frac { d^2y }{ dx^2 }\) + y = 3x²
In this equation the highest order derivative is \(\frac { d^3y }{ dx^3 }\) & its power is 1
∴ Its order = 3 & degree = 1

(x) x = e^{xy(\(\frac { dy }{ dx }\))}
In this equation the highest order derivative is \(\frac { dy }{ dx }\) & its order is 1
It has the term e^{xy(\(\frac { dy }{ dx }\))}
So its degree is not defined or degree does not exist.

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• Chapter 1 Principles of Management
• Chapter 2 Functions of Management
• Chapter 3 Management By Objectives (MBO) and Management By Exception (MBE)
• Chapter 4 Introduction to Financial Markets
• Chapter 5 Capital Market
• Chapter 6 Money Market
• Chapter 7 Stock Exchange
• Chapter 8 Securities Exchange Board of India (SEBI)
• Chapter 9 Human Resource Management
• Chapter 10 Recruitment Methods
• Chapter 11 Employee Selection Process
• Chapter 12 Employee Training Method
• Chapter 13 Concept of Market and Marketer
• Chapter 14 Marketing and Marketing Mix
• Chapter 15 Recent Trends in Marketing
• Chapter 16 Consumerism
• Chapter 17 Rights, Duties & Responsibilities of Consumers
• Chapter 18 Grievance Redressal Mechanism
• Chapter 19 Environmental Factors
• Chapter 20 Liberalization, Privatization and Globalization
• Chapter 21 The Sale of Goods Act, 1930
• Chapter 22 The Negotiable Instruments Act, 1881
• Chapter 23 Elements of Entrepreneurship
• Chapter 24 Types of Entrepreneurs
• Chapter 25 Government Schemes for Entrepreneurial Development
• Chapter 26 Companies Act, 2013
• Chapter 27 Company Management
• Chapter 28 Company Secretary

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Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Accountancy Guide Pdf Chapter 7 Company Accounts Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Accountancy Solutions Chapter 7 Company Accounts

12th Accountancy Guide Company Accounts Text Book Back Questions and Answers

I Multiple Choice Questions

Choose the correct answer

Question 1.
A preference share is one
(i) which carries preferential right with respect to payment of dividend at fixed rate
(ii) which carries preferential right with respect to payment of capital on winding up
(a) Only (i) is correct
(b) Only (ii) is correct
(c) Both (i) and (ii) are correct
(d) Both (i) and (ii) are incorrect
Answer:
(c) Both (i) and (ii) are correct

Question 2.
That part of share capital which can be called up only on the winding up of a company is called:
(a) Authorised capital
(b) Called up capital
(c) Capital reserve
(d) Reserve capital
Answer:
(d) Reserve capital

Question 3.
At the time of forfeiture, share capital account is debited with
(a) Face value
(b) Nominal value
(c) Paid up amount
(d) Called up amount
Answer:
(d) Called up amount

Question 4.
After the forfeited shares are reissued, the balance in the forfeited shares account should be transferred to
(a) General reserve account
(b) Capital reserve account
(c) Securities premium account
(d) Surplus account
Answer:
(b) Capital reserve account

Question 5.
The amount received over and above the par value is credited to
(a) Securities premium account
(b) Calls in advance account
(c) Share capital account
(d) Forfeited shares account
Answer:
(a) Securities premium account

Question 6.
Which of the following statement is false?
(a) Issued capital can never be more than the authorised capital
(b) In case of under subscription, issued capital will be less than the subscription capital
(c) Reserve capital can be called at the time of.winding up
(d) Paid up capital is part of called up capital
Answer:
(b) In case of under subscription, issued capital will be less than the subscription capital

Question 7.
When shares are issued for purchase of assets, the amount should be credited to
(a) Vendors A/c
(b) Sundry assets A/c
(c) Share capital A/c
(d) Bank A/c
Answer:
(c) Share capital A/c

Question Match the pair and identify the correct option
(1) Under subscription – (i) Amount prepaid for calls
(2) Over subscription – (ii) Subscription above the offered shares
(3) Calls in arrear – (iii) Subscription below the offered shares
(4) Calls in advance -(iv) Amount unpaid on calls

Answer:
C

Question 9.
If a share ₹ 10 on the which ₹ 8 has been paid up is forfeited. Minimum reissue price is
(a) ₹ 10 per share
(b) ₹ 8 per share
(c) ₹ 5 per share
(d) ₹ 2 per share
Answer:
(d) ₹ 2 per share
Hint:
One share Rs. 10
(-) Paid up value Rs. 8 Each
Minimum Reissue Price 2 Each

Question 10.
Supreme Ltd. forfeited 100 shares of ₹ 10 each for non-payment of final call of ₹ 2 per share. All these shares were re-issued at ₹ 9 per share. What amount will be transferred to capital reserve account?
(a) ₹ 700
(b) ₹ 800
(c) ₹ 900
(d) ₹ 1,000
Answer:
(a) ₹ 700

II Very Short Answer Questions

Question 1.
What is a share?
Answer:
The capital of a company is divided into small units of a fixed amount. These units are called shares. There are two types

1. Preference shares and
2. Equity shares.

Question 2.
What is Over-Subscription?
Answer:
When the number of shares applied for is more than the number of shares offered for subscription, it is said to be oversubscription.

Question 3.
What is meant by calls in arrears?
Answer:
When a shareholder fails to pay the amount due on the allotment or on calls, the amount remaining unpaid is known as calls in arrears.

Question 4.
Write a short note on the securities premium account.
Answer:
When a company issues shares at a price more than the face value (nominal value), the shares are said to be issued at a premium. The excess is called a premium amount and is transferred to a securities premium account.
Securities premium account is shown under reserves and surplus as a separate head in the Note to Account to the balance sheet.

Question 5.
Why are the shares forfeited?
Answer:
When a shareholder defaults in making payment of allotment/call money, the shares may be forfeited. On forfeiture, the share allotment is cancelled and to that extent, paid-up share capital reduced.

III Short Answer Questions

Question 1.
State the differences between preference shares and equity shares.
Answer:
(i) Preference shares
Preference shares are the shares which have the following two preferential right over, the equity shares:

• Preference towards the payment of dividend at a fixed rate during the lifetime of the company and.
• Preference towards the repayment of capital on winding up of the company.

(ii) Equity shares
Equity shares are those shares which do not preference shares. These shares do not enjoy any preferential rights. The rate of dividend is not fixed on equity shares and it depends upon the profits earned by the company. In case of winding up of a company equity shareholders are paid after the payment are made to preference shareholders. Equity shares are also known as ordinary shares.

Question 2.
Write a brief note on calls in advance.
Answer:
The excess amount paid over the called up value of a share is known as calls in advance. It is the excess money paid on application or allotment or calls. Such excess amounts can be returned or adjusted towards future payments. If the company decides to adjust such amount towards future payment, the excess amount is transferred to a separate account called a calls-in advance account.

Question 3.
What is a reissue of forfeited shares?
Answer:
Shares forfeited can be reissued by the company. The shares can be reissued at any price.
But, the reissue price cannot be less than the amount unpaid on forfeited shares.
When forfeited shares are reissued at a loss, such loss is to be debited to the forfeited shares account.
when forfeited shares are reissued at a premium, the amount of such premium will be credited to the securities premium account.

Question 4.
Write a short note on

1. Authorized capital
2. Reserve capital

Answer:

1. Authorised Capital: According to Sec 2(8) of the Companies Act, “It means such capital as is authorized by the memorandum of a company to be the maximum amount of share capital of the company”.

2. Reserve Capital: Sec 65 of Companies Act 2013, only an unlimited company having share capital while converting into a limited company, may have a reserve capital. The company reserve a part of its subscribed capital to be called up only at the time of winding up. It is called reserve capital.

Question 5.
What is meant by the issue of shares for consideration other than cash?
Answer:
A company may issue shares for consideration other than cash when the company acquires fixed assets such as land and buildings, machinery, etc.

IV Exercises

Question 1.
Progress Ltd. issued 50,000 ordinary shares of ₹ 10 each, payable ₹ 2 on the application, ₹ 4 on allotment ₹ 2 on first call and ₹ 2 on final call. All the shares are subscribed and the amount was duly received. Pass journal entries.
Solution :

Under Subscription

Question 2.
Sampath company issued 25,000 shares at ₹ 10 per share payable ₹ 3 on the application, ₹ 4 on allotment; ₹ 3 on first and final call. The public subscribed for 24,000 shares. The directors allotted all the 24,000 shares and received the money duly. Pass necessary journal entries.
Solution:

Over Subscription

Question 3.
Saranya Ltd. issued 20,000 equity shares of ₹ 10 each to the public at par. The details of the amount payable on the shares are as follows:
On application – ₹ 3 per share
On allotment – ₹ 4 per share
On first and final call – ₹ 3 per share
Application money was received on 30,000 shares. Excess application money was refunded immediately. Pass journal entries are as follows:
Solution:

Question 4.
Gaja Lid issued 40,000 shares of ₹ 10 each of the public payable ₹ 2 on the application, ₹ 5 on the allotment, and ₹ 3 on the first and final call. The application was received for 50,000 shares. The Directors decided to allot 40,000 shares on a pro-rata basis and a surplus of application money was utilized for allotment. Pass journal entries assuming that the amount due was received.
Solution :

Question 5.
Lalitha Ltd. offered 30,000 equity shares ₹10 each to the public payable ₹ 2 per share on the application, ₹ 3 on share allotment, and the balance when required. Applications for 50,0 shares were received on which directors allotted as:
Applicants for 10,000 shares Full
Applicants for 35,000 shares 20,000 shares (excess money will be utilized for allotment
Applicants for 5,000 shares Nil
All the money due was received. Pass journal entries upto the receipt of allotment.
Solution :

Calls In advance

Question 6.
Das Ltd, offered 50,000 equity shares of ₹ 1 0 each to the public payable as follows: On applications were ₹ 4; on allotment ₹ 3; on first call 1 and on second and final call ₹ 2. Applications were received for 1,00,000 shares. All the applicants were allotted 1 share for every two shares applied. Excess application money was used for the amount due on allotment and call. Pass necessary journal entries.
Solution :

Question 7.
Anjali Flour Ltd. with a registered capital of ₹ 4,00,000 in equity shares of ₹ 10 each, issued 30,0 of such shares; payable ₹ 2 per share on the application, ₹ 5 per share on the allotment, and ₹ 3 shares on the first call. The issue was duly subscribed.
All the money payable was duly received but on the allotment, one shareholder paid the entire balance on his holding of 500 shares. Give journal entries to record the I transactions.
Solution :

Calls in Advance

Question 8.
Muthu Ltd. issued 50,000 shares of ₹ 10 each payable as follows; ₹ 2 on the application; ₹ 4 on allotment; ₹ 4 on first and final, call.
All money payable was duly received except one shareholder holding 1,000 shares failed to pay the call money. Pass the necessary journal entries for calls by using calls in the arear account.
Solution :

Answer:
Calls in arrear: ₹ 4,000

Forfeiture os Shares

Question 9.
Arjun was holding 1,000 shares ₹ 10 each of Vanavill Electronics Ltd, issued at par. He paid ₹ 3 on the application, ₹ 4 on the allotment but could not pay the first and final call of ₹ 3. The directors forfeited the shares for nonpayment of call money. Give Journal entry for forfeiture of shares.
Solution :

Answer:
Forfeited shares account: ₹ 7,000

Question 10.
Lakshmi was holding 50 hares of ₹ 10 each on which he paid ₹ 2 on application but could not pay ₹ 4 on the allotment and ₹ 2 on first call. Directors forfeited the shares after the first call. Give journal entry for recording the forfeiture of shares.
Solution :

Answer:
Forfeited shares account: ₹ 100

Reissue of shares

Question 11.
Goutham Ltd. forfeited 500 equity shares of ₹ 10 each issued at par held by Ragav for nonpayment of the final call of ₹ 2 per share. The shares were forfeited and reissued to Madhan at ₹ 8 per share. Show the journal entries for forfeiture and reissue.
Solution :

Answer:
Capital reserve account: ₹ 3,000

Question 12.
Nivetha Ltd. forfeited 1,000 equity shares of ₹ 10 each for non-payament of call of ₹ 4 per share. Of these 800 shares were reissued @ ₹ 7 per share. Pass journal entries for forfeiture and reissue?
Solution :

Answer:
Capital reserve account: ₹ 2,400

Question 13.
Nathiya Textiles Ltd. forfeited 100 shares of ₹ 10 each, ₹ 8 called up, on which Mayuri had paid application and allotment money of 6 per share. Of these 75 shares were reissued to soundayya by receiving ₹ 7 per share. Pass journal entries for forfeiture and reissue?
Solution :

Answer:
Capital reserve account ₹ 375.

Question 14.
Simon Ltd issued 50,000 equity shares of ₹ 10 each at par payable on-application ₹ 1 per share, on allotment ₹ 5 per share, on first call ₹ 2 per share, and on second and final call ₹ 2 per share. The issue was fully subscribed and all the amounts were duly received with exception of 2,000 shares held by chezhian, who failed to pay the second and final call. His shares were forfeited and reissued to Elango at ₹ 8 per share. Journalise the above transactions?
Solution

Answer:
Capital reserve account: ₹ 12,000

Question 15.
Kanchana Ltd. issued 50,000 shares ₹ 10 each payable as under?
On application ₹ 1
On allotment ₹ 5
On first call ₹ 2
On final call ₹ 2
Applications were received for 70,000 shares. Applications for 8,000 shares were rejected and allotment was made proportionately towards the remaining applications. The directories made both the calls and all the amounts were received except the final call on 1,500 shares which were subsequently forfeited. Later 1.200 forfeited shares were reissued by receiving ₹ 8 per share. Give journal entries.
Solution :

Answer:
Capital reserve account: ₹ 7,200

Shares issued premium

Question 16.
Viswanath Furniture Ltd. invited applications for 20,000 shares of ₹ 10 each at a premium of 2 per share payable?
₹ 2 On application
₹ 5 (including premium) on the allotment
₹ 5 On the first and final call
There were oversubscription and applications were received for 30,000 shares and the excess applications were rejected by the directors. Pass the journal entries.
Solution :

Question 17.
United Industries Ltd. issued shares of ₹ 10 each at 10% premium payable ₹ 3 on the application, ₹ 4 on the allotment (including premium), ₹ 2 on the first call, and ₹ 2 on the final call.
Journalise the transections relating to forfeiture of shares for the following situations: Manoj who holds 250 shares failed to pay the second and final call and his shares were forfeited.
Manoj who holds 250 shares field to pay the allotment money and first call and second and final call and his shares were forfeited.
Manoj who holds 250 shares failed to pay the allotment money and first call money and his shares were forfeited after the first call.
Solution:

Answer:
Forfeited shares account: (i) ₹ 2,000; (ii) ₹ 750; (iii) ₹ 750

Question 18.
Kasthuri Ltd. had allotted 20,000 shares to applicants of 30,000 shares on a pro-rata basis. The amount payable was ₹ 1 on the application, ₹ 5 on the allotment (including a premium of ₹ 2 each), and ₹ 2 on the first call and ₹ 2 on final calls, Subin, a shareholder failed to pay the first call and final call on his 500 shares. All the shares were forfeited and out of the 400 shares were re-issued @ ₹ 8 per share. Pass necessary journal entries.
Solution :

Answer:
Capital reserve account: ₹ 1,600

Question 19.
Vairam Ltd. issued 60,000 shares of ₹ 10 each at a premium of ₹ 2 per share payable as follows:
On application ₹ 6
On allotment ₹ 4 (including premium)
On the first and final call ₹ 2
The issue was fully subscribed and the amount due was received except Saritha to whom 1,000 shares were allotted who failed to pay the allotment money and first and final call money. Her shares were forfeited. All the forfeited shares were reissued to Parimala at ₹ 7 per share.
Solution :

Answer:
Capital reserve account: ₹ 3,000

Issue of shares for cash in lumpsum

Question 20.
Abdul Ltd. issue 50,000 shares of ₹ 10 each at a premium of ₹ 2 per share. Pass journal entry if the amount is fully received along with a premium amount of ₹ 2 per share.
Solution :

Answer:
Share capital account ₹ 5,00,000; Securities premium account ₹ 1,00,000

Issue of shares for consideration other than cash

Question 21.
Paradise Ltd. purchased assets of ₹ 4,40,000 from Suguna Furniture Ltd. It issued equity shares of ₹ 10 each fully paid in satisfaction of their claim. What entries will be made if such issue is: (a) at par and (b) a premium of 10%.’
Solution :
Calculation of the number of shares to be issued

Answer:
(a) Share capital account: ₹ 4,40,000;
(b) Share capital account ₹ 4,00,000; Securities premium account ₹ 40,000

12th Accountancy Guide Company Accounts Additional Important Questions and Answers

I Multiple Choice Questions

Question 1.
A company issued 3,00,000 shares of Rs.10 each to the public but only 1,50,000 shares were subscribed. It’s subscribed capital is
(a) Rs. 30,00,000
(b) Rs. 15,00,000
(c) Rs. 10,00,000
Hint:
issued No. of share = 3,00,000
Subscribed No. of share = 1,50,000
Each share Rs. 10
Now ask the Question
No. of Subscribed Capital × Each Share
= 1,50,000 × 10 = ₹ 15,00,000
Answer:
(b) Rs. 15,00,000

Question 2.
The directors of a company forfeited 100 shares of Rs.10 each on which the final call, money of Rs.3 was not paid Later these shares were reissued of Rs.800 capital reserve will be
(a) Rs. 700
(b) Rs.500
(c) Rs. 500

Answer:
(b) Rs.500

Question 3.
When more applications are received than that are offered to the public it is called.
(a) Under subscription
(b) Full subscription
(c) Oversubscription
Answer:
(c) Oversubscription

Question 4.
Captial reserve is shown on the ………….. ride of the B/s
(a) Liability
(b) Assets
(c) Both
Answer:
(a) Liability

Question 5.
The balance of forfeited shares A/c is ……………. is in the B/s
(a) Added to paid-up capital
(b) Added to the authorized capital
(c) deducted from paid to capital
Answer:
(a) Added to paid-up capital

Question 6.
Calls-in-arrears are shown in the B/s as
(a) Deduction from called up capital
(b) Addition to paid-up capital
(c) Addition to issued capital
Answer:
(a) Deduction from called up capital

Question 7.
A company issued 20,000 Es of Rs.100 each to the public, but only 18,000 shares were subscribed, its subscribed capital is Rs.
(a) Rs. 20,00,000
(b) Rs. 18,00,000
(c) Rs. 21,00,000
Hint:
issued No. of share = 20,000
Paid up No. of share = 18,000
Each share = ₹ 10
∴ Now
Paid up Capital is
= 18,000 x 100
= ₹ 18,00,000 .
Answer:
(b) Rs. 18,00,000

Question 8.
A company issued 50,000 shares of Rs. 10 each at a premium of Rs. 2 each …………… is the securities premium amount
(a) Rs. 50,000
(b) Rs. 1,00,000
(c) Rs. 1,50,000
Hint:
Premium – ₹ 2 Each
Securities premium = 50,000 × 2 = ₹ 1,00,000
Answer:
(b) Rs. 1,00,000

Question 9.
Securities premium will appear in the _
(a) Assets
(b) Liability
(c) Assets & Liability
Answer:
(b) Liability

Question 10.
A Company is an association of
(a) workers
(b) merchants
(c) persons
Answer:
(c) persons

III Short Answer Questions

Question 1.
Define Company.
Answer:
According to Lord Justice Lindsey, “A Company is an association of many persons who contribute money or money’s worth to common stock and employ it in some trade or business and who share the profit and loss arising therefrom. The common stock so contributed is denoted in money and is the capital of the company. The persons who contributed in it or from it, or to whom it belongs, are members. The proportion of capital to which each member is entitled is his share”.

Question 2.
What are the characteristic features of the company?
Answer:
Following are the characteristics of a company:

• Voluntary association: A Company is a voluntary association of persons. No law can compel persons to form a company.
• Separate legal entity: Company is an artificial person. It has a separate legal entity which is separate and distinct from its members.
• Common seal: A company may have a common seal which can be affixed on the documents.
• Perpetual succession: A company continues forever. Its continuity is not affected by the changes in its members. It can be wound up only by law.
• Limited liability: The liability of the shareholders of the company is limited to the extent of the face value of the shares held by the shareholders.
• Transferability of shares: The shares of a company are freely transferable except in case of a private company.

Question 3.
Write short notes on

1. Issued capital
2. Subscribed capital
3. Called up capital
4. Paid-up capital

Answer:

1. Issued capital: This represents that part of authorized capital which is offered for subscription.
2. Subscribed capital: It refers to that part of issued capital which has been applied for and also allotted by the company.
3. Called up capital: It refers to that part of subscription capital which has been called up by the company for payment.
4. Paid-up capital: It is that part of called up capital which has been actually paid by the shareholders.

Question 4.
What do you mean by under subscription?
Answer:
All the shares offered to the public may not be subscribed in full. When the number of shares subscribed is less than the number of shares offered, it is known as under subscription.

Question 5.
What are the journal entries passed when shares are issued for cash in lumpsum?
Answer:

IV Exercises

Question 1.
A company issued 20,000 shares of ₹ 10 each payable as follows:
₹ 3 on Application,
₹ 3 on Allotment,
₹ 4 on First and Final call.
All the shares were subscribed and duly paid for. Pass journal entries.
Solution :

Question 2.
Preeti Ltd. invited applications for 5,000 shares of ₹ 10 each payable as follows:
₹ 3 on Application,
₹ 2 on Allotment,
₹ 2 on First call and
₹ 3 Final call
All these shares were subscribed and paid. Pass journal entries.
Solution:
 

Question 3.
A Joint Stock Company had a Nominal Capital of 5,00,000 divided into 5,000 shares ₹ 100 each payable:
₹ 30 per share on Application,
₹ 20 per share on Allotment,
₹ 30 on First call and
₹ 20 Final call
All the shares were subscribed and fully paid for by the public. Pass journal entries.
Solution :

Question 4.
Mary Ltd. Issued 1,000 shares of ₹ 10 each at a premium of ₹ 2 per share payable as follows:
₹ 4 on Application,
₹ 4 on Allotment, (including premium
and the balance when required. All the shares were subscribed for and duly paid. Pass necessary Journal entries.
Solution :

Question 5.
Global Ltd. issued 6,000 shares on ₹ 100 each at a premium of ₹ 20 per share payable as follows:
₹ 30 on Application,
₹ 50 Oon Allotment, (including premium)
₹ 30 on First call and
₹ 10 Final call
All shares were duly subscribed and money due was fully received. Pass journal entries.
Solution :
Journal Entries in the Books of Global Ltd

Question 6.
On 1.1.98 Alpha Ltd issued 1,00,000 shares of ₹ 10 each payable ₹ 2 on application. The company received applications for 1,20,000 shares. The excess applications were rejected. and money refunded Pass necessary entries.
Solution :
Journal Entries in the Books of Alpha Ltd

Question 7.
Good Luck Co. Ltd issued 1,00,000 shares @ 10 each payable:
₹ on Application,
₹ on Allotment,
and the balance when required. 1,50,000 shares were, applied for. The directors rejected the excess applications and refunded the application money. All money was received. Pass entries to record the transactions.
Solution :

Question 8.
Bhagavathi Ltd. issued 10,000 shares of ₹ 10 each at a discount of 10% payable as follows:
on Application, ₹2.50
on Allotment, ₹ 3.00
On the First and Final call ₹ 3.50
All money due was received except the final call on 100 shares which were forfeited by the company after giving due notice. Pass the forfeited entry.
Solution :

Question 9.
Ganthimathi Ltd. was registered with a nominal capital of ₹ 1,00,000 in equity shares of ₹ 10 each. It offered to the public 6,000 shares for subscripted The application was, however, received for 8,000 shares. The directors had to reject the application for 1,000 shares and to return the money received by Theron. The application money received on the other 1,000 shares was adjusted to the allotment account. The amount payable on shares was
₹ 3 per share on Application,
₹ 3 per share on Allotment and the balance on the first and final call
On shareholder holding, 100 shares failed to pay the call money and as a result, his shares were forfeited. Pass the necessary journal entries.
Solution :
Journal Entries in the Books of Ghandhimathi Ltd

Question 10.
Gani Ltd. forfeited 20 shares of ₹ 10 each fully called up, held by Santha for nonpayment of final call of 4 per sháre. These shares were re-issued to Josephin for ₹ 8 per share as fully paid up. Give journal entries for the forfeited and re-issue of shares.
Solution :

Question 11.
A Company forfeited 100 equity shares of ₹ 100 each issued at a premium of 10% (to be paid at the time of allotment) on which first call money of ₹ 30 per share and final call of ₹ 20 were not received. These shares were forfeited and subsequently re-issued at 90 per share. Give necessary journal entries regarding forfeited and re-issue of shares.
Solution :

Question 12.
The Directors of a company forfeited 200 Equity shares of ₹ 10 each fully called upon which the final call of ₹ 2 has not been paid. The shares were re-issued upon payment of ₹ 1,500. Journalise the above transactions.
Solution :

Question 13.
The Director of a company forfeited 100 shares of ₹ 10.each fully called up for non-payment of the First call of ₹ 2 per share and Final call of ₹ 3 per share. 60 of these shares were subsequently re-issued at 6 per share fully paid up. Pass necessary Journal entries to record the above.
Solution :

Question 14.
The directors of Sheeia Ltd. forfeited 2,000 shares ₹ 10 each for non-payment of final call of ₹ 2.50. 1,800 of these shares were re-issued for ₹ 6 per share fully paid up. Give the necessary Journal entries.
Solution :

Amount to be transferred to Capital reserve A/c = Profile on reissue × No.of share reissued
= 3.50 × 1,800
= Rs.6,300

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Accountancy Guide Pdf Chapter 8 Financial Statement Analysis Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis

12th Accountancy Guide Financial Statement Analysis Text Book Back Questions and Answers

I Multiple Choice Questions

Choose the correct answer

Question 1.
Which of the following statements is not true?
(a) Notes and schedules also form part of financial statements.
(b) The tools of financial statement analysis include common -size statement
(c) Trend analysis refers to the study of movement of figures for one years
(d) The common-size statements show the relationship of various items with some common base, expressed as percentage of the common base.
Answer:
(c) Trend analysis refers to the study of movement of figures for one years

Question 2.
Balance sheet provides information about the financial position of a business concern
(a) Over a period of time
(b) As on a particular date
(c) For a period of time
(d) For the accounting period
Answer:
(b) As on a particular date

Question 3.
Which of the following tools of financial statement analysis is suitable when data relating to several years are to be analysed?
(a) Cash flow statement
(b) Common size statement
(c) Comparative statement
Answer:
(d) Trend analysis

Question 4.
The financial statements do not exhibit
(a) Non-monetary data
(b) Past data
(c) Comparative statement
(d) Standard costing
Answer:
(a) Non-monetary data

Question 5.
Which of the following is not a tool of financial statement analysis?
(a) Trend analysis
(b) Common size statement
(c) Comparative statement
Answer:
(d) Standard costing

Question 6.
The term fund’ refers to
(a) Current liabilities
(b) Working capital
(c) Fixed assets
(d) Non-current assets.
Answer:
(b) Working capital

Question 7.
Which of the following statement is not true?
(a) All the limitations of financial statements are applicable to financial statement analysis also.
(b) Financial statement analysis is only the means and not an end.
(c) Expert knowledge is not required in analyzing the financial statements
(d) Interpretation of the analysed data involves personal judgment.
Answer:
(c) Expert knowledge is not required in analyzing the financial statements

Question 8.
A limited company’s sales have increased from ‘ 1,25,000 to 1,50,000. How does this appear in the comparative income statement?
(a) +20%
(b) +120%
(c) -120%
(d) -20%
Hint:
Percentage increase or Decrease = \(\frac{\text { Absolute amount of increase or decrease }}{\text { Year } 1 \text { amount }}\) × 100
Sales increase = ₹ 1,25,000 – ₹ 1,50,000 = ₹ 25,000
\(\frac{25,000}{1,25,000}\) × 100 = +20% = 1,25,000
Answer:
(a) +20%

Question 9.
In a common-size balance sheet, if the percentage of non-current assets is 75, what would be the percentage of current assets?
(a) 175
(b) 125
(c) 25
(d) 100
Hint:
Let Assets = ₹ 100
Non current assets = ₹ 75
∴ Current assets = 100 – 75 = 25
Answer:
(c) 25

Question 10.
Expenses for a business for the first year were ₹ 80,000. In the second year, it was increased to ₹ 88,000. What is the trend percentage in the second year?
(a) 10%
(b) 110%
(c) 90%
(d) 11%
Hint:
Computation of trend percentage = \(\frac{\text { Other year }}{\text { Earliest base year }}\) × 100
Earliest base year x 100
For second year = \(\frac{88,000}{80,000}\) × 100 = 110%
Answer:
(b) 110%

II Very Short Answer Questions

Question 1.
What are the financial statements?
Answer:
Financial statements are the statements prepared by the business concerns at the end of the accounting period to ascertain the operating results and the financial position.

Question 2.
List the tools of financial statement analysis.
Answer:

• Comparative statement
• Common size statements
• Trend analysis
• Funds flow statement
• Cash flow analysis

Question 3.
What is working capital?
Answer:
The term ’fund’ refers to working capital. Working capital refers to the excess of current assets over current liabilities.

Question 4.
When is trend analysis preferred to other tools?
Answer:
Trend analysis discloses the changes in financial and operating data between specific periods when data for more than two years are to be analyzed. It may be difficult to use a comparative statement.

III Short Answer Questions

Question 1.
‘Financial statements are prepared based on past data’. Explain how this is a limitation.
Answer:
The nature of the financial statement is historical. Past cannot be the index of the future and cannot be cent percent basis for future estimation, forecasting, budgeting, and planning.

Question 2.
Write a short note on cash flow analysis?
Answer:
Cash flow analysis is concerned with the preparation of a cash flow statement which shows * the inflow and outflow of cash and cash equivalents in a given period of time. Cash includes cash in hand and demand deposits with banks. Cash equivalents denote short term investments which can be realized easily within a short period of time, without much loss in value. Cash flow analysis helps in assessing the liquidity and solvency of a business concern.

Question 3.
Briefly explain any three limitations of financial statements.
Answer:

1. Lack of qualitative information: Qualitative information, that is non – monetary information is also important for business decisions. For example Efficiency of the employees and efficiency of the management. But this is ignored in financial statements.
2. Record of historical data: Financial statement are prepared based on historical data. They may not reflect the current position.
3. Ignores price level changes: Adjustments for price level changes are not made in the financial statements. Hence financial statements may not reveal the current position.

Question 4.
Explain the steps involved in preparing comparative statements?
Answer:
Following are the steps to be followed in preparation of the comparative statement:

1. Column 1: In this column, particulars of items of income statement or balance sheet are written.
2. Column 2: Enter absolute amount of year 1.
3. Column 3: Enter absolute amount of year 2.
4. Column 4: Show the difference in amounts between year 1 and year 2. If there is an increase in year 2, put plus sign and if there is decrease put minus sign.
5. Column 5: Show percentage increase or decrease of the difference amount shown in column 4 by dividing the amount shown in column 4 (absolute amount of increase or decrease) by column 2 (year 1 amount).

That is,
Percentage increase or decrease = \(\frac{\text { Absolute amount of increase or decrease }}{\text { Year } 1 \text { amount }}\) × 100

Question 5.
Explain the procedure for preparing a common – Size statement.
Answer:
Common-size statements can be prepared with three columns. Following are the steps to be followed in the preparation of the common – size statement.

1. Column 1. In this column, the particulars of items of the income statement or balance sheet are written.
2. Column 2. Enter the absolute amount.
3. Column 3. Choose a common base as 100.

For example Revenue from operations can be taken as the base for income statement and total of the balance sheet can be taken as the base for the balance sheet. Work out the percentage for all the items of column 2 in terms of the common base and enter them in column 3.

IV Exercises

Comparative statement analysis

Question 1.
From the following particulars, prepare a comparative income statement of Arul Ltd.

Solution:
The comparative income statement of Arul Ltd for the year ended 31.3.16 & 31.3.17

Answer :
Revenue from operation: 20 %;
Other income: 200%;
Total revenue: 50%;
Expenses: 25%;
Profit before tax: 100%

Question 2.
From the following particulars, prepare a comparative income statement of Barani Ltd.

Solution:
Computation of % increase for revenue from operation
Percentage increase = \(\frac{15,000}{30,000}\) × 100 = 50%
Comparative Income statement of Barani Ltd. for the year ended 31.3.17 & 31.3.18


Answer :
Revenue from operations: 50%
Other income: 50%
Total revenue: 50%
Expenses: 50%;
Profit before tax: 50%;
Profit after tax: 50%

Question 3.
From the following Particulars, prepare comparative income ‘statement of Daniel Ltd.

Solution:
The comparative Income statement of Daniel Ltd. for the year ended 31.3.16 & 31.3.17

Answer :
Revenue from operations: 25%
Expenses: 10%
Tax: 50%;
Profit before tax: 50%;
Profit after tax: 50%

Question 4.
From the following particulars, prepare a comparative .statement of the financial position of Muthu Ltd.

Solution:
Comparative B/s of Muthu Ltd. As on 31.3.17 & 31.3.18

Answer :
Shareholder’s fund: 10%;
Non-current liabilities: 10%
current liabilities: 10%
Total equity and liabilities: 10%;
Non -current assets: 20%;
Current assets: 30%
Total Assets: 10%

Question 5.
From the following particulars, prepare a comparative statement of the financial position of Kala Ltd.


Solution:



Answer :
Share capital: 20%;
Reserves and surplus: Nil.
Non-Current liabilities: -20%;
Current liabilities: -40%;
Total equity and liabilities: 10%;
Fixed assets: 16%;
Non-current investments: -20%;
Inventories: 25%;
Cash and cash equivalents: 20%;
total assets: 10%

Question 6.
Prepare a common-size income statement for the following particulars Raja Ltd. for the year ended 31st March 2017.

Solution:

Answer :
2016-2017:
Other income: 15%
total revenue: 115%
Expenses: 30%
Profit before tax: 85%

Question 7.
From the following particulars of Maria ltd and Kala Ltd. Prepare a common-Size income statement for the year ended 31st March 2019.

Solution:


Answer :
Maria Ltd: Other Income: 10%; total revenue: 110%; Expenses: 70%; Profit before tax:40%
Kala Ltd: Other income: 15%; Total revenue: 115% Expenses: 60%; Profit before tax:55%

Question 8.
Prepare a common-size income statement for the following particulars of Sam Ltd. table

Solution:

Answer :
2015-16: Other income:20%; Total revenue: 120% Expenses:60%; Profit before tax:60%: Tax: 18%; Profit after tax:42%;
2016-17: Other income: 10%; total revenue: 110%; Expenses:50%; Profit before tax: 60%: Tax:18%;
Profit after tax:42%

Question 9.
Prepare Common-size balance sheet of Meena Ltd. as of 31st March 2018.

Solution:
Common size B/s Meena Ltd as on 31.3.18


Answer :
Shareholder’s fund:50%; Non-current liabilities: 40%; Current liabilities: 10%; Non-current assets:75%; current assets:25%

Question 10.
Prepare a common-size statement of financial position for the following particulars of Rani Ltd.

Solution:


Answer :
2015-16 Shareholder’s fund:60%; Non-current liabilities: 30%; Current liabilities: 10%; Non-current assets:80%; current assets:20%
2016-17 Shareholder’s fund:60%;Non-current liabilities: 25%; Current liabilities: 15%; Non-current assets:80%; Current assets:20%

Question 11.
Prepare a common-size statement of financial position for the following particulars of Yasmin Ltd. and Sakthi Ltd.


Solution:


Answer:
Yasmin Ltd: Share capital 40%; Reserves and surplus: 10%; Noncurrent liabilities:30%; current liabilities:20%. Fixed assets:40%; Non -Current investments: 10% Inventories:40%; Cash & cash equivalents: 10% Sakthi Ltd: Share „ capital: 50%; Reserves and surplus: 10%; Non-current liabilities:30%; Current liabilities: 10%; Fixed assets:50% Non-current investmesits:20%; Inventories: 15% cash & cash equivalents: 15%
Trend Analysis:

Question 12.
From the following particulars, calculate the trend percentages of Kala Ltd.

Solution:

Answer:
2016-17: Revenue from operations: 125%; Other income: 150%; Total revenue: 130% Expenses:145% Profit before tax: 120%
2017-18: Revenue from operations: 150%; Other income 200%; Total revenue: 160% Expenses:175%; Profit before tax: 150%

Question13.
From the following particulars, calculate the Trend percentages of Kavitha Ltd.

Solution:

Answer:
2016-17: Revenue from operations: 125%; Other income: 125%; total revenue:125%; Expenses: 120%; Profit before tax:150%; Tax :150%; Profit after tax: 150%;
2017-2018; Revenue from operations: 150%; Other income:150%; total revenue:150% Expenses:80%; Profit before tax: 500% Tax:500%; Profit after tax:500%

Question 14.
From the following particulars, calculate the trend percentage of Kumar Ltd.

Solution:

Answer :
2016-17: Revenue from operations:90%; Other income: 160%; total revenue: 100%; Expenses:80%; Profit before tax: 150%; Tax: 150%; Profit after tax:150%;
2017-18: Revenue from operations:50%; Other income: 120%; total revenue:60%; Expenses:50%; Profit before tax: 85%; Tax:85%; Profit after tax:85%

Question 15.
From the following particulars, calculate the trend percentages of Anu Ltd.

Solution:


Answer :
year 2: Shareholder’s Fund: 110%, Non-current liabilities: 125%; Current liabilities:80%; Total equity and liabilities: 110%; Non-Current assets: 120%, Current assets:80%; total assets: 10%;
year 3: Shareholder’s fund: 120%; Non- current liabilities: 120%; Current liabilities: 120% Total equity and liabilities: 120%; Non-current assets: 130% Current assets:90%; total assets: 120%

Question 16.
From the following particulars, calculate the trend percentages of Babu Ltd.

Solution:



Answer:
Year 2: Share capital: 127%; Reserves and surplus: 100%; Non-current liabilities: 110%; Current liabilities: 150% Total equity and liabilities: 120%; Fixed assets:118%; Non-current investments: 125%; Inventories:! 10%; Cash & cash equivalents: 150%; total assets:120%;
Year 3: Share capital: 106%; Reserves and surplus :150%; Non-current liabilities: 120%; Current liabilities: 200%; total equity and liabilities: 125%; Fixed assets:103%; Non – Current investments: 150%; Inventories: 120%; Cash & cash equivalents:200%; total assets: 125%

12th Accountancy Guide Financial Statement Analysis Additional Important Questions and Answers

Other Important questions & Answers

Question 1.
…………………………. are the tools of financial analaysis
(a) Comparative statements
(b) Trend analysis
(c) Common size statement
(d) all the above
Answer:
(d) all the above

Question 2.
Analysis of financial statements involves
(a) B/S
(b) Trading A/c
(c) All the above
Answer:

III Short Answer Questions

Question 1.
What are the features of a financial statement?
Answer:
Following are the features of financial statements:

• Financial statements are generally prepared at the end of an accounting period based on transactions recorded in the books of accounts.
• These statements are prepared for the organization as a whole.
• Information is presented in a meaningful way by grouping items of similar nature such as fixed assets.
• Financial statements are prepared based on historical cost.
• Financial statements are prepared based on accounting principle and Accounting Standards, which make financial statements comparable and realistic.
• Financial statements involve personal judgment in a certain case. For example, the selection of the method of depreciation, percentage of reserve, etc.

Question 2.
Explain the significance of financial statements.
Answer:
Financial statements reveal the operating results and financial position of the business concern. The significance of financial statements to various stakeholders is as follows:

• To Management: Financial statements provide information to the management to take decision and to have control over business activities, in various areas.
• To shareholders: Financial statements help the shareholders to know whether the business has potential for growth and to decide to continue their shareholding.
• To potential investors: Financial statements help to value the securities and compare it with those of other business concerns before making their investment decisions.
• To creditors: Creditors can get information about the ability of the business to repay the debts from financial statements.
• To bankers: Information given in the financial statements is significant to the bankers to assess whether there is adequate security to cover the amount of the loan or overdraft.
• To the government: Financial statements involve personal judgment in a certain case. For example, the selection of the method of depreciation, percentage of reserve, etc.
• To employees: Through the financial statements, the employees can assess the ability of the business to pay salaries and whether they have future growth in the concern.

Question 3.
Explain the provisions of the Indian Companies Act 2013 to be followed while preparing the financial statement of or company.
Answer:
Following provisions of the Indian Companies Act, 2013 have to be followed while preparing the financial statements of a company:

• As per Section 2(40), financial statements include balance sheet, profit and loss account/income and expenditure account, cash flow statement, statement of changes in equity and any explanatory note annexed to the above.
• Section 129 (1) of the Indian Companies Act, 2013 states that the financial statements shall give a true and fair view of the state of affairs of the company and shall comply with the Accounting Standards notified under section 133.
• Section 129 (1) also states that the financial statements shall be prepared in the form provided in schedule III of the Indian Companies Act, 2013.

Question 4.
What do you mean by financial statement analysis?
Answer:
Financial statement analysis is a comparison of the various items in the financial statement by establishing and evaluating relationships among them so that, it gives a better understanding of the performance and financial status of the business concern.

Question 5.
What are the objectives of financial statement analysis?
Answer:
Financial statement analysis may be done with any of the following objectives:

• To analyze profitability and earning capacity.
• To study the long term and short term solvency of the business.
• To determine the efficiency in operations and use of assets.
• To determine the efficiency of management and employees.
• To determine the trend in sales, production, etc.
• To forecast for the future and prepare budgets.
• To make inter-firm and intra-firm comparisons.

Question 6.
What are the limitations of fin. State analysis.
Answer:
Following are the limitations of financial statement analysis:
1. All the limitations of financial statements such as ignoring non-monetary information, ignoring price level changes, etc., are applicable to financial statement analysis also.

2. Financial statement analysis is only the means and not an end, that is, it is only a tool in the hands of management and other shareholders. Interpretation of the results has to be done only by the financial analysts with due regard to the internal and external environmental factors.

3. Expert knowledge is required in analyzing the financial statements.
4. Interpretation of the analyzed data involves personal judgments as different experts may give different views.

Question 7.
What do you mean by Horizontal analysis?
Answer:
Horizontal analysis
When figures relating to several years are considered for the purpose of analysis, the analysis is called horizontal analysis. Generally, one year is taken as the base year and the figures relating to the other years are compared with that of the base year. Comparative statements and trend percentages are examples of horizontal analysis.

Question 8.
What do you mean by vertical analysis?
Answer:
Vertical analysis
When figures relating to one accounting year alone are considered for the purpose of analysis, the analysis is called vertical analysis. Here, the relationship is established among items from various financial statements relating to the same accounting period. Preparation of common size statements and computation of ratios are examples of vertical analysis.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 9 Applications of Integration Ex 9.10 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10

Choose the most suitable answer from the given four alternatives:

Question 1.
The value of \(\int_{0}^{2/3}\) \(\frac { dx }{ \sqrt{4-9x^2} }\)
(a) \(\frac { π }{ 6 }\)
(b) \(\frac { π }{ 2 }\)
(c) \(\frac { π }{ 4 }\)
(d) π
Solution:
(a) \(\frac { π }{ 6 }\)
Hint:

Question 2.
The value of \(\int_{-1}^{2}\) |x| dx
(a) \(\frac { 1 }{ 2 }\)
(b) \(\frac { 3 }{ 2 }\)
(c) \(\frac { 5 }{ 2 }\)
(d) \(\frac { 7 }{ 2 }\)
Solution:
(c) \(\frac { 5 }{ 2 }\)
Hint:
|x| = x if x > 0
= -x if x < 0

Question 3.
For any value of n ∈ Z \(\int_{0}^{π}\) e^cos²x cos3 [(2n + 1) x] dx
(a) \(\frac { π }{ 2 }\)
(b) π
(c) 0
(d) 2
Solution:
(c) 0
Hint:

Question 4.
The value of \(\int_{-π/2}^{π/2}\) sin² x cos x dx
(a) \(\frac { 3 }{ 2 }\)
(b) \(\frac { 1 }{ 2 }\)
(c) 0
(d) \(\frac { 2 }{ 3 }\)
Solution:
(d) \(\frac { 2 }{ 3 }\)
Hint:
f(x) = sin²x cos x
f(-x) = sin²(-x) cos(-x) = sin²x cos x
f(x) = f(-x)
f is an even function

Question 5.
The value of \(\int_{-4}^{4}\)[tan^-1(\(\frac { x^2 }{ x^4+1 }\)) + tan^-1(\(\frac { x^4+1 }{ x^2 }\))] dx
(a) π
(b) 2π
(c) 3π
(d) 4π
Solution:
(d) 4π
Hint:

Question 6.
The value of \(\int_{-π/4}^{π/4}\) (\(\frac { 2x^7-3x^5+7x^3-x+1 }{ cos^2x }\)) dx is
(a) 4
(b) 3
(c) 2
(d) 0
Solution:
(c) 2
Hint:

Odd function – 3 Odd function + 7 Odd function – Odd function + even function
= 0 + 2 \(\int_{0}^{π/4}\) sec² x dx
= 2 [tan x]\(_{0}^{π/4}\)
= 2(1 – 0)
= 2

Question 7.
If f(x) = \(\int_{0}^{x}\) cos t dt, then \(\frac { df }{ dx }\) =
(a) cos x – x sin x
(b) sin x + x cos x
(c) x cos x
(d) x sin x
Solution:
(c) x cos x
Hint:
f(x) = \(\int_{0}^{x}\) cos t dt
u = t, dv = cos t dt
u’ = 1, v = sin t
v₁ = -cos t
f(x) = [t sin t + cos t]\(_{0}^{x}\)
f(x) = x sin x + cos x – 1
\(\frac { df }{ dx }\) = x cos + sin x – sin x
\(\frac { df }{ dx }\) = x cos x

Question 8.
The area between y² = 4x and its latus rectum is
(a) \(\frac { 2 }{ 3 }\)
(b) \(\frac { 4 }{ 3 }\)
(c) \(\frac { 8 }{ 3 }\)
(d) \(\frac { 5 }{ 3 }\)
Solution:
(c) \(\frac { 8 }{ 3 }\)
Hint:
y² = 4x
y = 2√x

Question 9.
The value of \(\int_{0}^{1}\) x (1 – x)⁹⁹ dx is
(a) \(\frac { 1 }{ 11000 }\)
(b) \(\frac { 1 }{ 10100}\)
(c) \(\frac { 1 }{ 10010 }\)
(d) \(\frac { 1 }{ 10001 }\)
Solution:
(b) \(\frac { 1 }{ 10100}\)
Hint:

Question 10.
The value of \(\int_{0}^{π}\) \(\frac { dx }{ 1+5^{cosx} }\) is
(a) \(\frac { π }{ 2 }\)
(b) π
(c) \(\frac { 3π }{ 2 }\)
(d) 2π
Solution:
(a) \(\frac { π }{ 2 }\)
Hint:

Adding (1) and (2)

Question 11.
If \(\frac { Γ(n+2) }{ Γn }\) = 90 then n is
(a) 10
(b) 5
(c) 8
(d) 9
Solution:
(d) 9
Hint:

n² + n = 90
n² + n – 90 = 0
(n + 10) (n – 9) = 0
n = 9

Question 12.
The value of \(\int_{0}^{π/6}\) cos³ 3x dx is
(a) \(\frac { 2 }{ 3 }\)
(b) \(\frac { 2 }{ 9}\)
(c) \(\frac { 1 }{ 9 }\)
(d) \(\frac { 1 }{ 3 }\)
Solution:
(b) \(\frac { 1 }{ 10100}\)
Hint:
\(\int_{0}^{π/6}\) cos³ 3xdx = \(\frac { 1 }{ 3 }\) × \(\frac { 2 }{ 3 }\) × 1
= \(\frac { 2 }{ 9 }\)

Question 13.
The value of \(\int_{0}^{π}\) sin⁴ x dx
(a) \(\frac { 3π }{ 10 }\)
(b) \(\frac { 3π }{ 8}\)
(c) \(\frac { 3π }{ 4 }\)
(d) \(\frac { 3π }{ 2 }\)
Solution:
(b) \(\frac { 3π }{ 8}\)
Hint:
(x) = sin⁴x dx
f(π – x) = sin⁴ (π – x) = sin⁴x

Question 14.
The value of \(\int_{0}^{∞}\) e^-3x x² dx
(a) \(\frac { 7 }{ 27 }\)
(b) \(\frac { 5 }{ 27 }\)
(c) \(\frac { 4 }{ 27 }\)
(d) \(\frac { 2 }{ 27 }\)
Solution:
(d) \(\frac { 2 }{ 27 }\)
Hint:

Question 15.
If \(\int_{0}^{a}\) \(\frac { 1 }{ 4+x^2 }\) dx = \(\frac { π }{ 8 }\) then a is
(a) 4
(b) 1
(c) 3
(d) 2
Solution:
(d) 2
Hint:

Question 16.
The volume of solid of revolution of the region bounded by y² = x(a – x) about the x-axis is
(a) πa³
(b) \(\frac { πa^3 }{ 4 }\)
(c) \(\frac { πa^3 }{ 5 }\)
(d) \(\frac { πa^3 }{ 6 }\)
Solution:
(d) \(\frac { πa^3 }{ 6 }\)
Hint:
y² = x(a – x)
To find limit put y = 0
x (a – x) = 0
x = 0, x = a

Question 17.
If f(x) = \(\int_{1}^{x}\) \(\frac { e^{sinx} }{ u }\) du, x > 1 and \(\int_{1}^{3}\) \(\frac { e^{sinx^2} }{ x }\) dx = \(\frac { πa^3 }{ 6 }\) [f(a) – f(1)] then one of the possible value of a is
(a) 3
(b) 6
(c) 9
(d) 5
Solution:
(c) 9
Hint:

Question 18.
The value of \(\int_{0}^{1}\) (sin^-1 x)² dx is
(a) \(\frac { π^2 }{ 4 }\) – 1
(b) \(\frac { π^2 }{ 4 }\) + 2
(c) \(\frac { π^2 }{ 4 }\) + 1
(d) \(\frac { π^2 }{ 4 }\) – 2
Solution:
(d) \(\frac { π^2 }{ 4 }\) – 2
Hint:

Question 19.
The value of \(\int_{0}^{a}\) (\(\sqrt{a^2-x^2}\))³ dx is
(a) \(\frac { πa^3 }{ 16 }\)
(b) \(\frac { 3πa^4 }{ 16 }\)
(c) \(\frac { 3πa^2 }{ 8 }\)
(d) \(\frac { 3πa^4 }{ 8 }\)
Solution:
(b) \(\frac { 3πa^4 }{ 16 }\)
Hint:

Question 20.
If \(\int_{0}^{x}\) f(t) dt = x + \(\int_{x}^{1}\) f(t) dt, then the value of f(1) is
(a) \(\frac { 1 }{ 2 }\)
(b) 2
(c) 1
(d) \(\frac { 3 }{ 4 }\)
Solution:
(a) \(\frac { 1 }{ 2 }\)
Hint:

= 1 + 0 – x f(x)
f(x) + x f(x) = 1
f(x)[1 + x] = 1
when x = 1,
f(1)(2) = 1
f(1) = \(\frac { 1 }{ 2 }\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Accountancy Guide Pdf Chapter 6 Retirement and Death of a Partner Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner

12th Accountancy Guide Retirement and Death of a Partner Text Book Back Questions and Answers

I Multiple Choice Questions

Choose the correct answer

Question 1.
A partner retires from the partnership firm on 30th June. He is liable for all the acts of the firm up to the
(a) End of the current accounting period
(b) End of the previous accounting period
(c) Date of his retirement
(d) Date of his final settlement
Answer:
(c) Date of his retirement

Question 2.
On the retirement of a partner from a partnership firm, accumulated profits and losses are distributed to the partners on the basis of
(a) New profit sharing ratio
(b) Old profit sharing ratio
(c) Gaming ratio
(d) Sacrificing ratio
Answer:
(b) Old profit sharing ratio

Question 3.
On the retirement of a partner, general reserve will be transferred to the
(a) Capital account of all the partners
(b) Revaluation account
(c) Capital account of the continuing partners
(d) Memorandum revaluation account
Answer:
(a) Capital account of all the partners

Question 4.
On revaluation, the increase in liabilities leads to ……………….
(a) Gain
(b) Loss
(c) Profit
(d) None of these
Answer:
(b) Loss

Question 5.
At the time of retirement of a partner, determination of gaining ratio is required
(a) To transfer revaluation profit or loss
(b) To distribute accumulated profits and losses .
(c) To adjust goodwill
(d) None of these
Answer:
(c) To adjust goodwill

Question 6.
The final amount due to a retiring partner is not paid immediately,it is transferred to
(a) BankA/c
(b) Retiring partners capital A/c
(c) Retiring partner s loan A/c
(d) Other partner s capital A/c
Answer:
(c) Retiring partner s loan A/c

Question 7.
‘A’was a partner in a partnership firm. He died on 31st March 2019. The final amount due to him is ? 24,000 which is not paid immediately. It will be.transferred to
(a) As capital account
(b) A’s loan account
(c) As Executors account
(d) As Executors loan account
Answer:
(d) As Executors loan account

Question 8.
A, B and C are partners sharingprofits in the ratio of 2:2:1. On retirement of B, goodwill of the firm was valued as ₹ 30,000. Find the contribution of A and C to compensate B:
(a) ₹ 20,000 and ₹ 10,000
(b) ₹ 8,000 and ₹ 4,000
(c) ₹ 10,000 and ₹ 20,000
(d) ₹ 15,000 and ₹ 15,000
Hint:

∴ Retire partner share will be compensate of‘A’ and ‘Copartners
’A’ Goodwill value = 30,000 x \(\frac{2}{5}\) = 5 12,000
‘B’ Goodwill Value = 30,000 x \(\frac{2}{5}\) = 5 12,000
‘C’ Goodwill Value = 30,000 x \(\frac{1}{5}\) = 5 6,000
Now Compensate A& C
‘B’ Share good will – 12,000 x \(\frac{2}{3}\) = 8,000
12,000 x \(\frac{1}{3}\) = 4,000
Answer:
(b) ₹ 8,000 and ₹ 4,000

Question 9.
A, B and C are partners sharing profits in the ratio of 4:2:3. C retires. The new profit sharing ratio between A and B will be
(a) 4:3
(b) 3:4
(c) 2:1
(d) 1:2
Hint:
OldPartnersABC

Answer:
(c) 2:1

Question 10.
X, Y and Z were partners sharing profits and losses equally. X died on 1st April 2019. Find out the share of X in the profit of 2019 based on the profit of 2018 which showed ₹ 36,000.
(a) ₹ 1,000
(b) ₹ 3,000
(c) ₹ 12,000
(d) ₹ 36,000
Hint:
Partner’s X: Y: Z
Sharing ratio = equally =\(\frac{1}{3}: \frac{1}{3}: \frac{1}{3}\)
Profit = ₹ 36,000
Share of ‘X’
36,000 x \(\frac{1}{3}\) = ₹ 12,000
Answer:
(c) ₹ 12,000

II Very Short Answer Questions

Question 1.
What is meant by the retirement of a partner?
Answer:
When a partner leaves from partnership firm it is known as retirement. The reasons for the retirement of a partner may be illness, old age, and disagreement with other partners, etc.

Question 2.
What is the gaining ratio?
Answer:
The continuing partners may gain a portion of the share of profit of the retiring partner. The gain may be shared by all the partners or some of the partners. The gaining ratio is the proportion of the profit which is gained by the continuing partners.

Question 3.
What is the purpose of calculating the gaining ratio?
Answer:
The purpose of finding the gaining ratio is to bear the goodwill to be paid to the retiring partner.

Question 4.
What Is the journal entry to be passed to transfer the amount due to the deceased partner to the executor of the deceased partner?
Answer:
To transfer the amount due to the deceased partner to the executor or legal representative of the deceased partner.

III. Short Answer Questions

Question 1.
List out the adjustments made at the time of retirement.
Answer:
The Following adjustments are necessary at the time of retirement of a partner:

• Distribution of accumulated profits, reserves, and losses
• Revaluation of assets and liabilities
• Determination of new profit sharing ratio and gaining ratio
• Adjustment for goodwill
• Adjustment for current year’s profit or loss up to the date of retirement
• Settlement of the amount due to the retiring partner

Question 2.
Distinguish between sacrificing ratio and gaining ratio.
Answer:

Question 3.
What are the ways in which the final amount to an outgoing partner can be settled?
Answer:
The amount due to the retiring partner may be settled one of the following ways:

1. Paying the entire amount due immediately in cash
2. Transfer the entire amount due to the loan account of the partner
3. Paying part of the amount immediately in cash and transferring the balance to the loan account of the partner.

IV Exercise

Distribution of accumulated profits, reserves, and losses

Question 1.
Dheena, Surya, and Jankai are partners sharing profits and losses in the ratio of 5:3:2. on 31.3.2018, Dheena retired. On the date of retirement, the books of the firm showed a reserve fund of ₹ 50,000. The pass journal entry to transfer the reserve fund.
Solution:

Answer:
Reserves fund: dheena: ? 25,000(Cr); Surya: ₹ 15,000(Cr.); Janaki: ₹ 10,000(Cr.);

Question 2.
Rosi, Rathi, and Rani are partners of firms sharing profits and losses equally. Rathi retired from the partnership on 1.1.2018. On the date, their balance sheet showed an accumulated loss of ₹ 45,000 on the asset side of the balance sheet. Give the journal [ entry to distribute the accumulated loss.
Solution:

Answer:
Rosi: ₹ 15,000(Dr.); Rathi: ₹ 15,000(Dr.);Rani: ₹ 15,000(Dr.);

Question 3.
Akash, Mukesh, and Sanjay are partners in firm sharing profits and losses in the ratio of 3:2:1. Their balance sheet as on 31st March 2017 is as follows:

Pass journal entry to transfer the accumulated profit and prepare the capital account of the partners
Solution:


Answer:
Akash capital: ₹ 67,000(Cr.); Mugesh Capital: ₹ 78,000(Cr.); Sanjay’s Capital:₹ 39,000(Cr.);

Question 4.
Roja, Neeia ‘and Kanaga are partners, sharing profit and losses in the ratio of 4:3:3. On 1st April 2017, Roja retires and on retirement, the following adjustments are agreed upon.
(i) Increase the value of the Building by ₹ 30,000.
(ii) Depreciate stock by ₹ 5,000, and furniture by ₹ 12,000.
(iii) Provide for an outstanding liability of ₹ 1,000.
Pass journal entries and prepare revaluation account.
Solution:

Answer:
Revaluation profit : ₹ 12,000

Question 5.
Vinoth, Karthi, and Pranav are partners sharing profits and. losses’ in the ratio of 2:2:1 Pranav retires from the partnership on 1st April 2018. The following adjustments are to be made.
(i) increase the value of the land building by ₹ 18,000.
(ii) Reduce the value of machinery by ₹ 15,000.
(iii) A provision would also be made for outstanding expenses for ₹ 8,000.
Give journal entries and prepare a revaluation account.
Solution:

Question 6
Chandru, Vishal, and Ramanan are partners in firms sharing profits and losses equally. Their balance sheet as of 31st March 2018 is as follows.

Ramanan retired on 31st March 2019 subject to the following conditions:
(i) Machinery is valued at ₹ 1,50,000
(ii) Value of furniture brought down by ₹ 10,000
(iii) Provision for doubtful debts should be increased to ₹ 5,000. ,
(iv) investment of ₹ 30,000 not recorded in the books is to be recorded now.
Pass necessary journal entries and prepare revaluation account and capital account of partners.
Solution:

Answer:
Revaluation profit: ₹ 48,000;
Capital account:)
Chandru :₹ 76,000(Cr);
Vishal: ₹ 86,000(Cr);
Ramanan’s loan account: ₹ 86,000 (a))

Question 7.
Kayal, Mala and Neela are partners sharing profits in the ratio of 2:2:1. kayal retires and the new profit sharing ratio between Mala and Neela is 3:2. Calculate the gaining ratio.
Solution:
New Profit Sharing Ratio and Gaining Ratio
Gain Ratio = New Ratio – Old Ratio
Kayal –
Mala = \(\frac { 3 }{ 2 }\) – \(\frac { 2 }{ 5 }\) – \(\frac { 1 }{ 5 }\)
Neela = \(\frac { 2 }{ 5 }\) – \(\frac { 1 }{ 5}\) = \(\frac { 1 }{ 5 }\)
Gaining Ratio = 1:1

Question 8.
Sunil, Sumathi and Sundari are partners sharing profits in the ratio of 3:3:4. Sundari retires and her share is taken up entirely by Sunil. Calculate the new profit sharing ratio and gaining ratio.
Solution:
old ratio = 3:3:4 (Sunil:sumathi:sundar)
New Ratio = old ratio + gaining ratio

Answer:
Gaining ratio:4:0;
New ratio:7:3

Question 9.
Ramu, Somu and Gopu are partners sharing profits in the ratio of 3:5:7. Gopu retires and the share is purchased by Ramu and Somu in the ratio of 3:1. Find the new profit sharing ratio and gaining ratio.
Solution:

Question 10.
Navin, Ravi and Kumar are partners sharing profits in the ratio of 1/2, 1/4 and 1/4 respectively. Kumar retires and his share is taken up by Navin and Ravi equally. Calculate the new profit sharing ratio and gaining ratio.
Solution :

Answer:
New ratio:5:3;
Gaining ratio: 1:1

Question 11.
Mani, Gani and Soni are partners sharing the profits and losses in the ratio of 4:5:6. Mani retires from the firm. Calculate the new profit sharing ratio and gaining ratio.
Mani: Gani: Soni → 4:5:6
[If nothing is mentioned about the new ratio, old ratio of the continuing partners is equal to their new ratio.]
Solution:
Old ratio = 4:5:6
New ratio – 5:6

Answer:
New profit shaving ratio and gaining ratio is 5:6

Question 12.
Rajan, Suman and jegan were partners in firm sharing profits and losses in the ratio of 4:3:2 Suman retired from partnership. The goodwill of the firm on the date of retirement was valued at ₹ 45,000. Pass necessary journal entries for goodwill on the assumption that the fluctuating capital method is followed.
Solution :

Answer:
Suman’s share of goodwill: ₹ 15,000;
Rajan’s capital: ₹ 10,000(Dr);Jegan’s capital: ₹ 5,000(Dr);

Question 13.
Balu, Chandru and Nirmal are partners in a firm sharing profits and losses in the ratio of 5:3:2 on 31st March 2018, Nirmal retires from the firm. On the date of Nirmal’s retirement, goodwill appeared in the books of the firm at ₹ 60,000 By assuming thuctuating capital account, pass the necessary journal entry if the partners decide to
(a) Write off the entire amount of existing goodwill
(b) Write off half of the existing goodwill.
Solution :

Answer:
(a) Balu’s capital: ₹ 30,000(Dr.); Chandru’s capital: ₹ 18,000(Dr.)
Nirmala’s capital: ₹ 12,000(Dr);(b)balu’s capital: ₹ 15,000(Dr);
Chandru’s capital: ₹ 9,000(Dr);Nirmala’s capital: ₹ 6,000(Dr);

Question 14.
Rani, Jaya and Rathi are partners sharing profits and losses in the ratio of 2:2:1. On 31.3.2018, Rathi retired from the partnership. Profit of the preceding years is as follows: 2014: ₹ 10,000; 2015; ₹ 20,000; 2016; ₹ 18,000 and 2017; ₹ 32,000
Find out the share of profit of Rathi for the year 2018 till the date of retirement if
(a) profit is to be distributed on the basis of the previous year’s profit
(b) Profit is to be distributed on the basis of the average profit of the past 4 years.
Also pass necessary journal entries by assuming partners’ capitals are fluctuating.
Solution:
(a) on the basis of previous years profit
Profit of 2017 = Rs.32,000
Date of retirement = 31.3.2018
share of profit of Rathi for 3 months = \(32,000 \times \frac{3}{12} \times \frac{1}{5}\)
= Rs. 1600

Answer:
(a) Rathi’s share of profit: ₹ 1,600; (b) Rathi’s share of profit: ₹ 1,000;
Settlement of amount due to the retiring partner.

Question 15.
Kavin, Madhan, and Ranjith are partners sharing profits and losses in the ratio of 4:3:3 respectively. Kavin retires from the firm on 31st December 2018. On the date of retirement, his capital account shows a credit balance of ₹ 1,50,000. Pass journal entries if:
(a) The amount due is paid off immediately.
(b) The amount due is not paid immediately.
(c) ₹ 1,00,000 is paid and the balance in the future.
Solution:

Answer:
(a) Kavin’s loan:Nil;
(b) Kavin’s loan: ₹ 1,50,000;
(c) Kavin’s loan: ₹ 50,000;

Question 16.
Manju, Chara’and Lavanya are partners in firm sharing profits and losses in the ratio of 5:3:2. The balance sheet as of 31st March, 2018 was as follows:

Manju retired from the partnership firm on 31.3.2018 subject to the following adjustments:
(i) Stock to be depreciated by ₹ 10,000
(ii) Provision for doubtful debts to be created for ₹ 3,000
(iii) Buildings to be appreciated by ₹ 28,000.
Prepare revaluation account and capital accounts of partners after retirement.
Solution:

Answer:
Revaluation profit: ₹ 15,000; Manju’s loan account:
₹ 1,02,000(Cr).Capital account: Charu ₹ 89,OO0(Cr); Lavanya: ₹ 83,000(Cr))

Question 17.
Kannan, Rahim, and John are partners in a firm sharing profits and losses in the ratio of 5:3:2. The balance sheet as of 31st December 2017 was as follows:

John retires on 1st January 2018, subject to the following conditions :
(i) To appreciate building by 10%
(ii) Stock to be depreciated by 5%
(iii) To provide ₹ 1,000 for bad debts
(iv) An unrecorded liability of ₹ 8,000 has been noticed.
(v) The retiring partner shall be paid immediately.
prepare revaluation account, partner’s capital account, and the balance sheet of the firm after retirement.
Solution :

Answer:
Revaluation loss: ₹ 1,000; capital Account; Kannan: ₹ 1,04,500
Rahim: ₹ 82,700 Amount paid to John ₹ 41,800; Balance sheet total: ₹ 2,15,000

Question 18.
Saran, Arun, and Karan are partners in firms sharing profits and losses in the ratio of 4:3:3. The balance sheet as of 31.12.2016 was as follows:

Karan retires on 1.1.2017, subject to the following conditions:
(i) Goodwill of the firm is valued at ₹ 21,000 %
(ii) Machinery to be appreciated by 10%
(iii) Building to be valued at ₹ 80,000
(iv) provision for bad debts to be raised to ₹ 2,000
(v) Stock to be depreciated by ₹ 2,000
(vi) The final amount due to Karan is not paid immediately.
prepare the necessary ledger accounts and show the balance sheet of the firm after retirement. *
Solution:

Gaining ratio
Old ratio = 4:3:3
New ratio = 4:3
Gaining ratio = Saran = New ratio – old ratio

GR = 12:9 or 4:3
Karan’s share of G/w = 21,000 x \(\frac{3}{10}\) = Rs.6,300
K’s g/w In GR = Rs. 6,300
Saran’s share = 6300 x \(\frac{3}{10}\) = Rs.3,600
Aruns share = 6,300 x \(\frac{3}{7}\) = Rs.2,700
Answer:
Revaluation profit: ₹ 21,000; capital A/c: Saran: ₹ 70,800, Arun: ₹ 58,100
Karan’s loan Account ₹ 57,100; Balance sheet total: ₹ 2,21,000

Question 19.
Rajesh, Sathish and Mathan are partners sharing profits and losses in the ratio of 3:2:1. respectively. Their balance sheet as on 31.3.2017 is given below:

Mathan retires on 31st March, 2017 subject to the following conditions :
(i) Rajesh and Sathis will share profits and losses in the ratio of 3:2
(ii) Assets are to be revalued as follows;
Machinery ₹ 4,50,000, Stock ₹ 2,90,000 , Debtors ₹ 1,52,000
(iii) Goodwill of the firm is valued at ₹ 1,20,000
prepare the necessary ledger accounts and the balance sheet immediately after the retirement of Mathan.
Solution:

Gaining ratio = New ratio – old ratio
Rajesh = \(\frac{3}{5}-\frac{3}{6}=\frac{18-15}{30}=\frac{3}{30}\)
Sathish = \(\frac{2}{5}-\frac{2}{6}=\frac{12-10}{30}=\frac{2}{30}\)
GR = 3:2
Mathans share of G/w = 12,000 x \(\frac{1}{6}\) = Rs.20,000
In gaining ratio
Rajeshs share = 20,000 x \(\frac{3}{5}\) = Rs. 12,000
Sathishs share = 12,000 x \(\frac{2}{5}\) = Rs. 8,000

Answer:
Revaluation Profit: ₹ 12,000; Mathan loan: A/c: ₹ 2,92,000 Rajesh’s capital A/c ₹ 4,54,000; Sathish capital A/c: ₹ 3,36,000
Balance sheet total ₹ 13,10,000

Question 20.
Janani and Jamuna are partners sharing profits and losses in the ratio of 3:3:1. respectively. Janaki died on 31st December 2017. The final amount due to her showed a credit balance of ₹ 1,40,000. Pass Journal Entry if
(i) The amount due is paid off immediately.
(ii) The amount due is not paid immediately.
(iii) ₹ 75,000 is paid and the balance in the future.
Solution:

Question 21.
Varsha, Shanthi, and Madhuri are partners in a firm sharing profits in the ratio of 5:4:3. Their balance sheet as of 31st December 2017 is as under:

on 1st January 2018, Madhuri died and on her death, the following arrangements are made:
(i) Stock to be depreciated by ₹ 5,000
(ii) Premises is to be appreciated by 20%
(iii) To provide ₹ 4,000 for bad debts
(iv) The final amount due to Madhuri was not paid.
prepare revaluation account, partners capital account, and the balance sheet of the firm after death.
Solution:

Answer: Revaluation Profit ₹ 15,000;
Varsha’s capital A/c ₹ 1,01,250; Sathish capital A/c: ₹ 77,000
Madhuri’s executors account ₹ 32,750; Balance sheet total: ₹ 2,43,000

Question 22.
Vijayan, Sudhan, and Suman are partners who share profits and losses; in the capital ratio. Their balance sheet as of 31st December 2018 is as follows:

Suman died on 31.3.2019. On the death of Suman, the following adjustments are made:
(i) Building is to be valued at ₹ 1,00,000
(ii) Stock to be depreciated by ₹ 5,000
(iii) Goodwill of the firm is valued at ₹ 36,000
(iv) Share of profit from the closing of the last financial year to the date of death on the basis of the average of the three completed year’s profit before death.
profit for 2016, 2017 and 2018 were ₹ 40,000,₹ 50,000 and ₹ 30,000 respectively.
Prepare the necessary ledger accounts and the balance sheet immediately after the death of Suman.
Solution:

Suman share of G/w
Vijayan : Sudhan : suman
Old ratio = 7:5:3
New ratio = 7:5
Gaining ratio New ratio – old ratio

Answer:
Revaluation Profit: ₹ 15,000; Vijayan’s capital A/c: ₹ 81,200;
Sudhan’s capital A/c ₹ 58,000; Suman’s Executor’s A/c: ₹ 45,800
Balance sheet total: ₹ 2,02,000

12th Accountancy Guide Retirement and Death of a Partner Additional Important Questions and Answers

I. Choose the best answer

Question 1.
In the absence of any specific agreement, between the partners, partners loan to the firms will carry an inters at the rate of …………… %
(a) 5%
(b) 6%
(c) 4%
Answer:
(b) 6%

Question 2.
A,B & C shares profit as 1/2 to A, 1/3 to B, and 1/6 to C. If B retries, then the new profit sharing ratio ………..
(a) 3:1
(b) 3:2
(c) 1:3
Hint:
Old Partner’s A, B, C
Old ratio = \(\frac{1}{2}: \frac{1}{3}: \frac{1}{6}\)
‘B’ Retire Partner
In order to equalise the denomination
\(\frac{3}{6}: \frac{2}{3}: \frac{1}{6}\)
Now profit sharing ration = \(\frac{3}{4}: \frac{1}{4}\). or 3 :1
Answer:
(a) 3:1

Question 3.
At the time of retirement, the revelation profits of the business will be shared by ……………… partners.
(a) all the partners
(b) Continuing partners
(c) Old partners
Answer:
(a) all the partners

Question 4.
At the time of retirement, of partners, the existing partners stand to
(a) gain
(b) loss
(c) no change
Answer:
(a) gain

Question 5.
A,B & C are sharing profits in the ratio of \(\frac{2}{5}\) , \(\frac{2}{5}\) \(\frac{1}{5}\) C retired from business and his share was purchased equally by A and B, Then new profit sharing ratio shall be

Hint:

Answer:
a

Question 6.
If the amount due to the outgoing partner is transferred to loan A/c then he is entitled to interest at ……………….. until it is paid out.
(a) 6%
(b) 5%
(c) 8%
Answer:
(a) 6%

Question 7.
At the time of retirement of a partner calculation of new profit, ratio is
(a) not necessary
(b) necessary
(c) optional
Answer:
(b) necessary

III Short Answer Questions

Question 1.
What are the journal entries to be passed to transfer the accumulated profits, losses & reserves?
Answer:

Question 2.
Write the format of Revaluation A/c.
Answer:
New profit sharing ratio is the agreed proportion in which future profit will be distributed to the continuing partners. If the new profit sharing ratio is not agreed, the continuing partners will share the profits and losses equally.

Question 3.
How will you adjust the share of profits loss of the retaining partner if he retires in between in an accounting year?
Answer:
When a partner retires in between in an accounting year, his share of the current year’s profit or loss upto the date of retirement has to be distributed to the retiring partner. It may be estimated based on the current year’s turnover. The previous year’s profit or the average of the past year’s profit may also be taken as the base to estimate the current year’s profit. The following journal entry is passed

Question 4.
What are the adjustments to be made on the death of a partner?
Answer:
The following adjustments are made on the death of a partner.

• Distribution of accumulated profits, reserves, and losses
• Revaluation of assets and liabilities
• Determination of new profit sharing ratio and gaining ratio
• Adjustment for goodwill
• Adjustment for current year’s profit or loss upto the date of death
• Settlement of the amount due to the deceased partner.

Question 5.
What are the journal entries to be passed for settlement of the amount due to the deceased partner
Answer:
To transfer the amount due to the deceased partner to the executor or legal representative of the deceased partner.

IV Additional Problems:

Question 1.
A, B, and C sharing profits in the ratio of 5:3:2. C retires. Find out the new profit sharing ratio and gaining ratio.
Solution:

Question 2.
G, H, and I are partners sharing profits in the ratio of 5:3:2.1 taken up by G and H equally. Find out the new profit sharing ratio of G and H.
Solution:

Question 3.
J, K and L are partners sharing profits in the ratio of 5:3:2. retires and his share was taken up entirely by K. Find out the new profit sharing ratio and gaining ratio of continuing partners
Solution:

Question 4.
X,Y and Z are partners sharing profits in- the ratio of 5:3:2 Z retires and the ratio between X and Y is 3:2. Find out the gaining ratio
Solution :

Question 5.
Prabha,Kavitha and Meena were partners of a firm sharing profit and loss in the ratio of 3:2:1 Meena Wanted to retire. They decided to revalue the assets and liabilities of the firm as indicated below:
To write down Machinery by Rs 10,000 and Stock by Rs. 4,000
To bring into books as unrecorded investments Rs. 5,000
To Write off Rs.3,000 from sundry creditors as it was no longer liable.
Pass entries to give effect to the above adjustments. Show also Revaluation account.
Solution:

Question 6.
Banumathi, Bharafhi and Shanthi are partners sharing, profits in the ratio of 5:3:2. On April 1, 2005 Shanthi decided to retire. On that date, there was a credit balance of Rs. 60,000 in their profit and loss account. Pass entry.
Solution:

Question 7.
Thangamuthu, Anaimuthu and Vairamuthu are partners sharing profit and loss in the ratio of 3:3:2. Thangamuthu wanted to retire on 1st June 2005, the firms books showed a general reserve of Rs. 40,000. Pass entry.
Solution:

Question 8.
Lalitha, Jothi and Kanaga were partners of a sharing profit and losses in the ratio of 3:2:3, Set out below was their balance sheet as on 31st March 2003.

Lalitha retired from the partnership on 1st April 2004 on the following terms:
Goodwill of the firm was to be valued at Rs.30,000
The assets are to be valued as under Stock Rs. 1,00,000; Furniture Rs.15,000; Plant and Machinery Rs.45,000; Building Rs.1,00,000.
A provision for doubtful debts is created at Rs.4,250.
Lalitha was to be paid off immediately.
Show the journal entries, prepare a revaluation account, capital account, Bank account, and balance sheet of the reconstituted partnership.
Solution:

Question 9.
‘9. A, B and C are partners sharing profits and losses in the ration of 5:3:2 respectively. Retries from the firm on 1st April 2005. After his retirement, his capital account shows a credit balance of Rs.1,35,000 after the necessary adjustment made. Give journal entries, if.
the amount due is paid off immediately.
When the amount due is not paid immediately.
Rs. 45,000 is paid and the balance In the future.
Solution :

Question 10.
Pallavan, Pandian and Chozhan were carrying on partnership business sharing profits in the ratio of 3:2:1. On March 31, 2005, the balance sheet of the firm stood as follows:

Chozhan retried on April 1, 2005 on the following terms:
Building to be appreciated by Rs. 15,000
Provision for doubtful debts to be made at 6% on debtors
Goodwill of the firm is valued at Rs. 18,000
Rs.50,000 to be paid to chozhan immediately and the balance transferred to his loan account.
Prepare Revaluation Account, Capital Accounts, Bank Account and the Balance Sheet after Chozhan’s retirement.
Solution :

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 9 Applications of Integration Ex 9.9 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.9

Question 1.
Find by integration, the volume of the solid generated by revolving about the x axis, the region enclosed by y = 2x², y = 0 and x = 1
Solution:
The region to be revolved is sketched

Since revolution is made about the x axis, the volume of the solid generated is given by

Required volume = \(\frac { 4π }{ 5 }\) cubic units

Question 2.
Find, by integration, the volume of the solid generated by revolving about the x axis, the region enclosed by y = e^-2x, y = 0, x = 0 and x = 1.
Solution:
Since revolution is made about the x axis, the volume of the solid generated is given

Required volume = \(\frac { π }{ 4 }\) [1 – e^-4] cubic units

Question 3.
Find, by integration, the volume of the solid generated by revolving about the y axis, the region enclosed by x² = 1 + y and y = 3.
Solution:
The region to be revolved is sketched.

Since revolution is made about the y axis, the volume of the solid generated is given by

Required volume = 8π cubic units

Question 4.
The region enclosed between the graphs of y = x and y = x² is denoted by R. Find the volume generated when R is rotated through 360° about x axis.
Solution:
The region to be revolved is sketched.

Find the intersecting point of y = x and y = x²
x² = x
x² – x = 0
x (x – 1) = 0 x = 0, x = 1
If x = 0, y = 0, x = 1, y = 1
∴ Points of intersection are (0, 0), (1, 1)

Required volume = \(\frac { 2π }{ 15 }\) cubic units

Question 5.
Find, by integration, the volume of the container which is in the shape of a right circular conical frustum as shown to figure.

Solution:
By using integration we have to find the volume of the frustum. So first find the equation of the curve.
Let A(0, 1) and B(2, 2) be two points. Line joining these two points form a straight line. That straight line revolves around x axis.

Volume of the solid revolves around x axis

Volume of the frustum = \(\frac { 14 }{ 3 }\) π m³

Question 6.
A watermelon has an ellipsoid shape which can be obtained by revolving an ellipse with major axis 20 cm and minor axis 10 cm about its major axis. Find its volume using integration.
Solution:
A watermelon has an ellipsoid shape.
2a = 20
a = 10
2b = 10
b = 5

∴ Volume of the frustum = \(\frac { 1000π }{ 3 }\) cubic units.