Class 12

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Solutions Chapter 10 Computer in Accounting Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Book Keeping & Accountancy Solutions Chapter 10 Computer in Accounting

1. Objective questions:

A. Select the most appropriate alternatives from those given below and rewrite the statements.

Question 1.
The primary document for recording all financial transactions in Tally is the ______________
(a) Journal
(b) Trial sheet
(c) Voucher
(d) File
Answer:
(c) Voucher

Question 2.
______________ displays the balance day-wise for a selected voucher type.
(a) Record book
(b) Ledger book
(c) Journal book
(d) Daybook
Answer:
(d) Daybook

Question 3.
Fixed Deposit A/c comes under ______________ group.
(a) Investments
(b) Current liability
(c) Bank A/c
(d) Current asset
Answer:
(a) Investments

B. Give the word term or phrase which can substitute each of the following statements:

Question 1.
The details of Bills receivable are maintained in this record.
Answer:
Sundry Debtors

Question 2.
Tally software is classified into this category.
Answer:
Mercantile

Question 3.
The short key is used to save or accept the information.
Answer:
Ctrl + A

Question 4.
It is a damaged software, cracked, nearly fully functional.
Answer:
Pirated Software

Question 5.
The process by which all the calculations are automatically done by the accounting software.
Answer:
Automation

C. State whether the following statements are true or false with reason:

Question 1.
Alt + D is the short key for delete voucher entries.
Answer:
This statement is True.
To delete voucher entries, people use the Alt + D key.

Question 2.
In Tally, the F6 Function key is for the payment vouchers.
Answer:
This statement is False.
In Tally, the F6 Function key is useful for receipt vouchers.

Question 3.
Legal software is fully functional software without any restriction.
Answer:
This statement is True.
The base of the legal software is fully functional, safe, and legal, so one can use this kind of software without any hesitation and restriction.

Question 4.
Salary Account comes under Indirect expenses.
Answer:
This statement is True.
When the expenses are made for the purchase of goods, and for the manufacturing process, they are known as a direct expense. Salary does not fall in that category and so it comes under the indirect expense category.

Question 5.
Accounting software may not be customized to meet the special requirement of the user.
Answer:
This statement is False.
Customized Accounting software is prepared to meet the special requirement of the user which is not readily available in the market.

D. Answer in One Sentences:

Question 1.
What is CAS?
Answer:
CAS means Computerized Accounting System which helps business firms to implement accounting processes and makes it user friendly with automation.

Question 2.
Write the steps to create a ledger account in tally.
Answer:
Steps to create a ledger account in the tally are as follows:

• From Gateway of Tally Screen, click on accounts info.
• Path gateway to Tally – Accounts Info – Ledgers – Single ledger – Choses create.

Question 3.
How to view reports in Tally?
Answer:
For viewing accounting reports in accounting software to click on the report option and select the Display option.

Question 4.
State the various types of vouchers.
Answer:
Following are the various voucher types:

1. F4 Contra voucher – For cash deposited in the bank and cash withdrawn from the bank, Transfer from one cash A/c to another Cash A/c and Bank to Bank transfer.
2. F5 Payment voucher – For all types of payments are recorded through this voucher type (Cash and Bank) Cash or Bank.
3. F6 Receipt voucher – For Cash and Bank receipts
4. F7 Journal voucher – For non-cash transactions
5. F8 Sales voucher – For cash as well as credit sales
6. F9 Purchase voucher – For cash as well as a credit purchase

Question 5.
Write the steps to create a company.
Answer:
Following are the steps to create a company:

1. After entering into Accounting software Tally, double click on the option, create a company, under company information. Then follow the navigation path.
   Gateway of Tally > Company Info > Create Company
2. Fill in the detailed information in the company creation form, displayed on the screen – Company creation window.

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Solutions Chapter 9 Analysis of Financial Statements Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Book Keeping & Accountancy Solutions Chapter 9 Analysis of Financial Statements

Objective Questions

A. Select the most appropriate alternative from those given below and rewrite the sentences:

Question 1.
Gross Profit Ratio indicates the relationship of gross profit to the ____________
(a) Net cash
(b) Net sales
(c) Net purchases
(d) Gross sales
Answer:
(b) Net sales

Question 2.
Current ratio = \(\frac{……………….}{Current liabilities}\)
(a) Quick assets
(b) Quick liabilities
(c) Current assets
(d) None of these
Answer:
(c) Current assets

Question 3.
Liquid assets = ____________
(a) Current assets + Stock
(b) Current assets – Stock
(c) Current assets – Stock + Prepaid Expenses
(d) None of these
Answer:
(b) Current assets – Stock

Question 4.
Cost of goods sold = ____________
(a) Sales – Gross profit
(b) Sales – Net profit
(c) Sales proceeds
(d) None of these
Answer:
(a) Sales – Gross profit

Question 5.
Net profit ratio is equal to ____________
(a) Operating ratio
(b) Operating net profit ratio
(c) Gross profit ratio
(d) Current ratio
Answer:
(a) Operating ratio

Question 6.
The common size statement requires ____________
(a) common base
(b) journal entries
(c) cashflow
(d) current ratio
Answer:
(a) common base

Question 7.
Bill payable is ____________
(a) long-term loan
(b) current liabilities
(c) liquid assets
(d) net loss
Answer:
(b) current liabilities

Question 8.
Generally current ratio should be ____________
(a) 2 : 1
(b) 1 : 1
(c) 1 : 2
(d) 3 : 1
Answer:
(a) 2 : 1

Question 9.
From financial statement analysis the creditors are specially interested to know ____________
(a) Liquidity
(b) Profits
(c) Sale
(d) Share capital
Answer:
(a) Liquidity

B. Give one word/term/phrase for each of the following statements.

Question 1.
The statement showing the profitability of two different periods.
Answer:
Comparative Income Statement

Question 2.
The ratio measures the relationship between gross profit and net sales.
Answer:
Gross Profit Ratio

Question 3.
Critical evaluation of financial statement to measure profitability.
Answer:
Analysis of Financial Statement

Question 4.
A particular mathematical number showing the relationship between two accounting figures.
Answer:
Ratio

Question 5.
An asset that can be converted into cash immediately.
Answer:
Liquid Asset

Question 6.
The ratio measuring the relationship between net profit and ownership capital employed.
Answer:
ROCE

Question 7.
The statement showing financial position for different periods of the previous year and the current year.
Answer:
Comparative Balance Sheet

Question 8.
Statement showing changes in cash and cash equivalent during a particular period.
Answer:
Cash Flow Statement

Question 9.
Activity related to the acquisition of long-term assets and investment.
Answer:
Financing Activities

Question 10.
The ratio that establishes a relationship between Quick Assets and Current Liabilities.
Answer:
Liquid Ratio

C. State true or false with reasons.

Question 1.
Financial statements include only the Balance Sheet.
Answer:
This statement is False.
Financial statements include Balance Sheet and Profit and Loss A/c. This is because financial statements are prepared by business organisations to find out the efficiency, solvency, profitability, growth, strength, and status of the business. For this, they need information from the balance sheet as well as from Profit and Loss A/c.

Question 2.
Analysis of financial statements is a tool but not a remedy.
Answer:
This statement is True.
Based on analysis of the financial statement one can get an idea of the financial strength and weakness of the business. However, based on this one cannot take decisions about the business on various issues. Hence analysis of financial statements is a tool but not a remedy.

Question 3.
Purchase of fixed assets is operating cash flow.
Answer:
This statement is False.
Purchase of fixed assets is cash flow from investing activities. It is not a day-to-day operations activity like office/selling/distribution finance expenses/activities.

Question 4.
Dividend paid is not a source of funds.
Answer:
This statement is True.
The dividend is always paid on shares issued by a company as an expense. Shares itself is a source of funds. In payment of dividends, cash goes out from the company. It is an outflow of cash and not a source of funds.

Question 5.
Gross profit depends upon net sales.
Answer:
This statement is True.
The gross profit ratio discloses the relation between gross profit and total net sales. The gross profit ratio is an income-based ratio, where gross profit is an income. There is a direct relation between net sales and gross profit. Higher the net sales higher the gross profit.

Question 6.
Payment of cash against the purchase of stock is the use of funds.
Answer:
This statement is True.
Cash payment for the purchase of stock is made from cash balance or/and from bank balance which is a part of the business fund. When stock or materials we purchase we use cash for payment.

Question 7.
Ratio Analysis is useful for inter-firm comparison.
Answer:
This statement is True.
The comparison of the operating performance of a business entity with the other business entities is known as an inter-firm comparison. This ratio analysis assists to know-how and to what extent a business entity is strong or weak as compared to other business entities.

Question 8.
The short-term deposits are considered as cash equivalent.
Answer:
This statement is True.
The short-term deposits are liquid assets. It means deposits are kept for some period (usually less than one year) and they are kept with an intention to get money quickly as and when required.
They are as good as cash and considered as cash equivalent.

Question 9.
Activity ratios and Turnover ratios are the same.
Answer:
This statement is True.
Turnover ratio is an efficiency ratio to check how efficiently a company is using different assets to extract earnings from them.
Activity ratios are financial analysis tools used to measure a business’s ability to convert its assets into cash. From both these definitions, we can say that Activity ratios and Turnover ratios are the same.

Question 10.
The current ratio measures the liquidity of the business.
Answer:
This statement is True.
The current ratio shows the relationship between current assets and current liabilities. If the proportion of current assets is higher than current liabilities, the liquidity position of the business entity is considered good. More liquidity means more short-term solvency. From the above, it is proved that the current ratio measures the liquidity of the business.

Question 11.
Ratio analysis measures profitability efficiency and financial soundness of the business.
Answer:
This statement is True.
With the help of profitability ratios (Gross profit, Net profit, and Operating profit) one can get the idea of profitability efficiency of the firm, and with the help of liquidity ratios (Current ratio and liquid ratio) one can get the idea of solvency or financial soundness of the business.

Question 12.
Usually, the current ratio should be 3 : 1.
Answer:
This statement is False.
Usually, the current ratio should be 2 : 1. It means current assets are double of current liabilities. It shows the short-term solvency of business enterprises.

D. Answer in one sentence only.

Question 1.
Mention two objectives of the comparative statement.
Answer:
Objectives of comparative statements are:

• Compare financial data at two points of time and
• Helps in deriving the meaning and conclusions regarding the changes in financial positions and operating results.

Question 2.
State three examples of cash inflows.
Answer:
Examples of cash inflows are:

• Interest received
• Dividend received
• Sale of asset/investment
• Rent received.

Question 3.
State three examples of cash-out flows.
Answer:
Examples of cash outflows are:

• Interest paid
• Loss on sale of an asset
• Dividend paid
• Repayment of short-term borrowings.

Question 4.
Give the formula of Gross Profit Ratio.
Answer:
Gross profit ratio = \(\frac{\text { Gross profit }}{\text { Net sales }} \times 100\)
Where Gross profit = Net sales – Cost of goods sold.
Cost of goods sold = Opening stock + Purchase – Purchase return + Direct expense – Closing stock
Net sales = Sales – Sales return.

Question 5.
Give the formula of Gross profit.
Answer:
Gross profit = Net sales – Cost of goods sold.
Cost of goods sold = Opening stock + Purchase – Purchase return + Direct expense – Closing stock
Net sales = Sales – Sales return.

Question 6.
Give any three examples of current assets.
Answer:
Cash or cash equivalent short-term lending and advances, expenses paid in advance, taxes paid in advance, etc. are examples of current assets.

Question 7.
Give the formula of the current ratio.
Answer:
Current ratio = \(\frac{\text { Current assets }}{\text { Current liabilities }}\)

Question 8.
Give the formula of quick assets.
Answer:
Quick assets = Current assets – (Stock + Prepaid expense)

Question 9.
State the formula of Cost of goods sold.
Answer:
Cost of goods sold = Opening stock + Purchase – Purchase return + Direct expense – Closing stock

Question 10.
State the formula of Average stock.
Answer:
Average stock = \(\frac{\text { Opentng stock of goods }+\text { Closing stock of goods }}{2}\)

Practical Problems

Question 1.
From the Balance Sheet of Amar Traders as of 31st March 2018 and 31st March 2019 prepare a Comparative Balance Sheet.

Solution:
Comparative Balance Sheet of Amar Traders as of 31st March 2018 and 31st March 2019


Percentage change = \(\frac{\text { Amount of absolute change }}{\text { Amount of previous year }} \times 100\)

Question 2.
From the Balance Sheet of Alpha Limited prepare a Comparative Balance Sheet as of 31st March 2018 and 31st March 2019:
Balance Sheet as of 31st March 2018 and 31st March 2019

Solution:
Comparative Balance Sheet of Alpha Limited as of 31st March 2018 and 31st March 2019

Question 3.
Prepare Comparative Balance Sheet for the year ended 31-3-18 and 31-3-19. Assets & Liabilities as follows:

Solution:
Comparative Balance Sheet as of 31st March 2018 and 31st March 2019

Question 4.
Prepare Comparative Balance Sheet for the year ended 31-3-17 and 31-3-18.

Solution:
Comparative Balance Sheet as of 31st March 2017 and 31st March 2018

Question 5.
Prepare Comparative Income statement of Noha Limited for the year ended 31-3-17 and 31-3-18.

Solution:
Comparative Income Statement of Noha Limited
For the year ended 31st March 2017 and 31st March 2018

Question 6.
Prepare Comparative Income Statement of Sourabh Limited for years ended 31-3-17 and 31-3-18.

Solution:
Comparative Income Statement of Sourabh Limited
For the year ended on 31st March 2017 and 31st March 2018

Question 7.
Following is the Balance Sheet of Sakshi Traders for the year ended 31-3-17 and 31-3-18

Prepare Common Size Balance Sheet for the years 31-03-17 and 31-03-18.
Solution:
Common Size Statement of Balance Sheet of Sakshi Traders as of 31st March 2017 and 31st March 2018

Note: Taking Total borrowed funds and Tota Funds applied as a base (100), Calculation is done.

Question 8.
Prepare Common Size Income Statement for the year ended 31-3-17 and 31-3-18.

Solution:
Common Size Statement for the year ended on 31st March 2017 and 31st March 2018

Note: Taking the amount of sales as base (100) other percentage figures are calculated.

Question 9.
Following is the Balance Sheet of Sakshi Limited. Prepare Cash Flow Statement:

Solution:
Cash Flow Statement
For the year ended 31st March 2017 and 31st March 2018

Question 10.
From the following Balance Sheet of Konal Traders prepare a Cash Flow Statement.

Solution:
Cash Flow Statement
For the year ended on 31st March 2017 and 31st March 2018

Question 11.
A company had following Current Assets and Current Liabilities:
Debtors = ₹ 1,20,000
Creditors = ₹ 60,000
Bills Payable = ₹ 40,000
Stock = ₹ 60,000
Loose Tools = ₹ 20,000
Bank Overdraft = ₹ 20,000
Calculate Current ratio.
Solution:
1. Current Assets = Debtors + Stock + Loose Tools
= 1,20,000 + 60,000 + 20,000
= ₹ 2,00,000

2. Current liabilities = Creditors + Bills payable + Bank overdraft
= 60,000 + 40,000 + 20,000
= ₹ 1,20,000

3. Current ratio = \(\frac{\text { Current assets }}{\text { Current liabilities }}\)
= \(\frac{2,00,000}{1,20,000}\)
= \(\frac{5}{3}\)
= 5 : 3

Question 12.
Current assets of company ₹ 6,00,000 and its Current ratio is 2 : 1. Find Current liabilities.
Solution:
Current ratio = \(\frac{\text { Current assets }}{\text { Current liabilities }}\)
\(\frac{2}{1}=\frac{6,00,000}{\text { Current liabilities }}\)
2 × Current liabilities = 6,00,000 × 1
Current liabilities = \(\frac{6,00,000}{2}\) = ₹ 3,00,000

Question 13.
Current liabilities = ₹ 3,00,000
Working capital = ₹ 8,00,000
Inventory = ₹ 2,00,000
Calculate Quick ratio.
Solution:
Current assets = Current liabilities + Working capital
= 3,00,000 + 8,00,000
= ₹ 11,00,000
Quick assets = Current assets – Inventory
= 11,00,000 – 2,00,000
= ₹ 9,00,000
Quick liability = Current liabilities – Bank O/D = ₹ 3,00,000
Quick ratio = \(\frac{\text { Quick assets }}{\text { Quick liabilities }}\)
= \(\frac{9,00,000}{3,00,000}\)
= \(\frac{3}{1}\)
= 3 : 1

Question 14.
Calculate Gross Profit ratio
Sales = ₹ 2,70,000
Net purchases = ₹ 1,50,000
Sales Ratio = ₹ 20,000
Closing Stock = ₹ 25,000
Operating Stock = ₹ 45,000
Solution:
Net sales = Sales – Sales return
= 2,70,000 – 20,000
= ₹ 2,50,000
Cost of goods sold = Opening stock + Net purchase – Closing stock
= 45,000 + 1,50,000 – 25,000
= ₹ 1,70,000
Gross profit = Net sales – Cost of goods sold
= 2,50,000 – 1,70,000
= ₹ 80,000
Gross Profit ratio = \(\frac{\text { Gross profit }}{\text { Net sales }} \times 100\)
= \(\frac{80,000}{2,50,000} \times 100\)
Gross profit ratio = 32%

Question 15.
Calculate Net Profit ratio from the following:
Sales = ₹ 3,80,000
Cost of goods sold = ₹ 2,60,000
Indirect expense = ₹ 60,000
Solution:
Sales = ₹ 3,80,000
Less: Cost of goods sold = ₹ 2,60,000
Gross profit = ₹ 1,20,000
Less: Indirect expense = ₹ 60,000
Net profit = ₹ 60,000
Net profit ratio = \(\frac{\text { Net profit }}{\text { Sales }} \times 100\)
= \(\frac{60,000}{3,80,000} \times 100\)
= 15.79%

Question 16.
Calculate Operating ratio:
Cost of goods sold = ₹ 3,50,000
Operating expense = ₹ 30,000
Sales = ₹ 5,00,000
Sales return = ₹ 30,000
Solution:
Net sales = Sales – Sales return
= 5,00,000 – 30,000
= ₹ 4,70,000
Operating ratio = \(\frac{\text { Cost of goods sold }+\text { Operating expense }}{\text { Net sales }} \times 100\)
= \(\frac{3,50,000+30,000}{4,70,000} \times 100\)
= \(\frac{3,80,000}{4,70,000} \times 100\)
= 80.85%

Question 17.
Calculate Current ratio.
1. Current assets = ₹ 3,00,000
2. Current liabilities = ₹ 1,00,000
Solution:
Current ratio = \(\frac{\text { Current assets }}{\text { Current liabilities }}\)
= \(\frac{3,00,000}{1,00,000}\)
= \(\frac{3}{1}\)
= 3 : 1

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 7 Elements of Groups 16, 17 and 18 Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 7 Elements of Groups 16, 17 and 18

1. Select appropriate answers for the following.

Question i.
Which of the following has the highest electron gain enthalpy?
A. Fluorine
B. Chlorine
C. Bromine
D. Iodine
Answer:
B. Chlorine

Question ii.
Hydrides of group 16 are weakly acidic. The correct order of acidity is
A. H₂O > H₂S > H₂Se > H₂Te
B. H₂Te > H₂O > H₂S > H₂Se
C. H₂Te > H₂Se > H₂S > H₂O
D. H₂Te > H₂Se > H₂O > H₂S
Answer:
C. H₂Te > H₂Se > H₂S > H₂O

Question iii.
Which of the following element does not show oxidation state of +4 ?
A. O
B. S
C. Se
D. Te
Answer:
A. O

Question iv.
HI acid when heated with conc. H₂SO₄ forms
A. HIO₃
B. KIO₃
C. I₂
D. KI
Answer:
C. I₂

Question v.
Ozone layer is depleted by
A. NO
B. NO₂
C. NO₃
D. N₂O₅
Answer:
A. NO

Question vi.
Which of the following occurs in liquid state at room temperature?
A. HIO₃
B. HBr
C. HCl
D. HF
Answer:
D. HF

Question vii.
In pyrosulfurous acid oxidation state of sulfur is
A. Only +2
B. Only +4
C. +2 and +6
D. Only +6
Answer:
B. Only + 4

Question viii.
Stability of interhalogen compounds follows the order
A. BrF > IBr > ICl > ClF > BrCl
B. IBr > BeF > ICl > ClF > BrCl
C. ClF > ICl > IBr > BrCl > BrF
D. ICl > ClF > BrCl > IBr > BrF
Answer:
C. ClF > ICl > IBr > BrCl > BrF

Question ix.
BrCl reacts with water to form
A. HBr
B. Br₂ + Cl₂
C. HOBr
D. HOBr + HCl
Answer:
D. HOBr + HCl

Question x.
Chlorine reacts with excess of fluorine to form.
A. ClF
B. ClF₃
C. ClF₂
D. Cl₂F₃
Answer:
B. ClF₃

Question xi.
In interhalogen compounds, which of the following halogens is never the central atom.
A. I
B. Cl
C. Br
D. F
Answer:
D. F

Question xii.
Which of the following has one lone pair of electrons?
A. IF₃
B. ICl
C. IF₅
D. ClF₃
Answer:
C. IF₅

Question xiii.
In which of the following pairs, molecules are paired with their correct shapes?
A. [I₃] : bent
B. BrF₅ : trigonal bipyramid
C. ClF₃ : trigonal planar
D. [BrF₄] : square planar
Answer:
A. [I₃] : bent

Question xiv.
Among the known interhalogen compounds, the maximum number of atoms is
A. 3
B. 6
C. 7
D. 8
Answer:
D. 8

2. Answer the following.

Question i.
Write the order of the thermal stability of the hydrides of group 16 elements.
Answer:
The thermal stability of the hydrides of group 16 elements decreases in the order of H₂O > H₂S > H₂Se > H₂Te.

Question ii.
What is the oxidation state of Te in TeO₂?
Answer:
The oxidation state of Te in TeO₂ is + 4.

Question iii.
Name two gases which deplete ozone layer.
Answer:
Nitrogen oxide (NO) released from exhaust systems of car or supersonic jet aeroplanes and chlorofluorocarbons (Freons) used in aerosol sprays and refrigerators deplete ozone layer.

Question iv.
Give two uses of ClO₂
Answer:
(i) ClO₂ is used as a bleaching agent for paper pulp and textiles.
(ii) It is also used in water treatment.

Question v.
What is the action of bromine on magnesium metal?
Answer:
Bromine reacts instantly with magnesium metal to give magnesium bromide.

Question vi.
Write the names of allotropic forms of selenium.
Answer:
Selenium has two allotropic forms as follows :
(i) Red (non-metallic) form
(ii) Grey (metallic) form

Question vii.
What is the oxidation state of S in H₂SO₄.
Answer:
The oxidation state of S in H₂SO₄ is + 6.

Question viii.
The pKa values of HCl is -7.0 and that of HI is -10.0. Which is the stronger acid?
Answer:
For HCl, pK_a = -7.0, hence its dissoClation constant is, K_a = 1 x 10^-7.
For HI pK_a = – 10.0, hence its dissoClation constant is K_a = 1 x 10^-7. Hence HCl dissoClates more than HI.
Therefore HCl is a stronger acid than HI.

Question ix.
Give one example showing reducing property of ozone.
Answer:
Ozone decomposes to liberate nascent oxygen, hence it is a powerful oxidising agent. O_3(g) → O_2(g) + O

For example :
(i) It oxidises lead sulphide (PbS) to lead sulphate (PbSO₄).
pbS_(s) + 4O_3(g) → PbSO_(s) + 4O_2(g)
(ii) Potassium iodide, KI is oxidised to iodine, I₂ in the solution.
2KI_(aq) + H₂O₍₁₎ + O_3(g) → 2KOH_(aq) + I_2(s) + O_2(g)

Question x.
Write the reaction of conc. H₂SO₄ with sugar.
Answer:
Concentrated sulphuric acid when added to sugar, it is dehydrated giving carbon.

The carbon that is left behind is called sugar charcoal and the process is called char.

Question xi.
Give two uses of chlorine.
Answer:
Chlorine is used for :

• for sterilization of drinking water.
• bleaching wood pulp required for the manufacture of paper and rayon, cotton and textiles are also bleached using chlorine.
• in the manufacture of organic compounds like CHCl₃, CCl₄, DDT, dyes and drugs.
• in the extraction of metals like gold and platinum.
• in the manufacture of refrigerant like Freon (i.e., CCl₂F₂).
• in the manufacture of several poisonous gases like mustard gas (Cl-C₂H₄-S-C₂H₄-Cl), phosgene (COCl₂) used in warfare.
• in the manufacture of tear gas (CCl₃NO₂).

Question xii.
Complete the following.
1. ICl₃ + H₂O …….. + …….. + ICl
2. I₂ + KClO₃ ……. + KIO₂
3. BrCl + H₂O ……. + HCl
4. Cl₂ + ClF₃ ……..
5. H₂C = CH₂ + ICl …….
6. XeF₄ + SiO₂ ……. + SiF₄
7. XeF₆ + 6H₂O …….. + HF
8. XeOF₄ + H₂O ……. + HF
Answer:
1. 2ICI₃ + 3H₂O → 5HCl + HlO₃ + ICl
2. I₂ + KCIO₃ → ICl + KIO₃
3. BrCl + H₂O → HOBr + HCl
4. Cl₂ + C1F₃ → 3ClF
5. CH₂ = CH₂ + ICl → CH₂I – CH₂Cl
6. 2XeF₆ + SiO₂ → 2XeOF₄ + SiF₄
7. XeF₆ + 3H₂O → XeO₃ + 6HF
8. XeOF₄ + H₂O→  XeO₂F₂ + 2HF

Question xiii.
Match the following
A – B
XeOF₂ – Xenon trioxydifluoride
XeO₂F₂ – Xenon monooxydifluoride
XeO₃F₂ – Xenon dioxytetrafluoride
XeO₂F₄ – Xenon dioxydifluoride
Answer:
XeOF₂ – Xenon monooxydifluoride
XeO₂F₂ – Xenon dioxydifluoride
XeO₃F₂ – Xenon trioxydifluoride
XeO₂F₄ – Xenon dioxytetrafluoride

Question xiv.
What is the oxidation state of xenon in the following compounds?
XeOF₄, XeO₃, XeF₅, XeF₄, XeF₂.
Answer:

Compound	Oxidation state of Xe
XeOF4	+ 6
XeO3	+ 6
XeF6	+ 6
XeF4	+ 4
XeF2	+ 2

3. Answer the following.

Question i.
The first ionisation enthalpies of S, Cl and Ar are 1000, 1256 and 1520 kJ/mol^-1, respectively. Explain the observed trend.
Answer:
(i) The atomic number increases as, ₁₆S < ₁₇Cl < ₁₈Ar₁.
(ii) Due to decrease in atomic size and increase in effective nuclear charge, Cl binds valence electrons strongly.
(iii) Hence ionisation enthalpy of Cl (1256 kJ mol^-1) is higher than that of S(1000 kJ mol^-1)
(iv) Ar has electronic configuration 3s²3p⁶. Since all electrons are paired and the octet is complete, it has the highest ionisation enthalpy, (1520 kJ mol^-1)

Question ii.
“Acidic character of hydrides of group 16 elements increases from H₂O to H₂Te” Explain.
Answer:
(i) The thermal stability of the hydrides of group 16 elements decreases from H₂O to H₂Te. This is because the bond dissociation enthalpy of the H-E bond decreases down the group.
(ii) Thus, the acidic character increases from H₂O to H₂Te.

Question iii.
How is dioxygen prepared in laboratory from KClO₃?
Answer:
By heating chlorates, nitrates and permanganates.
Potassium chlorate in the presence of manganese dioxide on heating decomposes to form potassium chloride and oxygen.

Question iv.
What happens when
a. Lead sulfide reacts with ozone (O₃).
b. Nitric oxide reacts with ozone.
Answer:
(i) It oxidises lead sulphide (PbS) to lead sulphate (PbSO₄) changing the oxidation state of S from – 2 to +6.
PbS_(s) + 4O_3(g) → PbSO_(s) + 4O_2(g)

(ii) Ozone oxidises nitrogen oxide to nitrogen dioxide.
NO_(g) + O_3(g) → NO_2(g) + O_2(g)

Question v.
Give two chemical reactions to explain oxidizing property of concentrated H₂SO₄.
Answer:
Hot and concentrated H₂SO₄ acts as an oxidising agent, since it gives nascent oxygen on heating.

Question vi.
Discuss the structure of sulfur dioxide.
Answer:
(i) SO₂ molecule has a bent V shaped structure with S-O-S bond angle 119.5° and bond dissoClation enthalpy is 297 kJ mol^-1.
(ii) Sulphur in SO₂ is sp² hybridised forming three hybrid orbitals. Due to lone pair electrons, bond angle is reduced from 120° to 119.5°.
(iii) In SO₂, each oxygen atom is bonded to sulphur by σ and a π bond.
(iv) a bond between S and O are formed by sp²-p overlapping.
(v) One of π bonds is formed by pπ – pπ overlapping while other n bond is formed by pπ – dπ overlap.
(vi) Due to resonance both the bonds are identical having observed bond length 143 pm due to resonance,

Question vii.
Fluorine shows only -1 oxidation state while other halogens show -1, +1, +3, +5 and +7 oxidation states. Explain.
Answer:

• Halogens have outer electronic configuration ns² np⁵.
• Halogens have tendency to gain or share one electron to attain the stable configuration of nearest inert element with configuration ns²np⁶.
• Hence they are monovalent and show oxidation state – 1.
• Since fluorine does not have vacant d-orbital, it shows only one oxidation state of – 1 while all other halogens show variable oxidation states from – 1 to +7.
• These oxidation states are, – 1, +1, + 3, +5 and + 7. Cl and Br also show oxidation states + 4 and + 6 in their oxides and oxyaClds.

Question viii.
What is the action of chlorine on the following
a. Fe
b. Excess of NH₃
Answer:
(a) Chlorine reacts with Fe to give ferric chloride.
2Fe + 3Cl₂ → 2FeCl₃

(b) Chlorine reacts with the excess of ammonia to form ammonium chloride, NH₄Cl and nitrogen.

Question ix.
How is hydrogen chloride prepared from sodium chloride?
Answer:

1. In the laboratory, hydrogen chloride, HCl is prepared by heating a mixture of NaCl and concentrated H₂SO₄.
   
2. Hydrogen chloride gas, is dried by passing it through a dehydrating agent like concentrated H₂SO₄ and then collected by upward displacement of air.

Question x.
Draw structures of XeF₆, XeO₃, XeOF₄, XeF₂.
Answer:

Question xi.
What are interhalogen compounds? Give two examples.
Answer:
Interhalogen compounds : Compounds formed by the combination of atoms of two different halogens are called interhalogen compounds. In an interhalogen compound, of the two halogen atoms, one atom is more electropositive than the other. The interhalogen compound is regarded as the halide of the more electropositive halogen.
For example ClF, BrF₃, ICl

Question xii.
What is the action of hydrochloric acid on the following?
a. NH₃
b. Na₂CO₃
Answer:
a. Hydrochloric acid reacts with ammonia to give white fumes of ammonium chloride.
NH₃ + HCl → NH₄Cl

b. Hydrochloric acid reacts with sodium carbonate to give sodium chloride, water with the liberation of carbon dioxide gas.
Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

Question xiii.
Give two uses of HCl.
Answer:
Hydrogen chloride (OR hydrochloric acid) is used :

• in the manufacture of chlorine and ammonium chloride,
• to manufacture glucose from com, starch
• to manufacture dye
• in mediClne and galvanising
• as an important reagent in the laboratory
• to extract glue from bones and for the purification of bone black.
• for dissolving metals, Fe + 2HCl_(aq) → FeCl₂ + H_2(g)

Question xiv.
Write the names and structural formulae of oxoacids of chlorine.
Answer:

Question xv.
What happens when
a. Cl₂ reacts with F₂ in equal volume at 437 K.
b. Br₂ reacts with excess of F₂.
Answer:
(a) Cl₂ reacts with F₂ in equal volumes at 437 K to give chlorine monofluoride ClF.

(b) Br₂ reacts with excess of F₂ to give bromine trifluoride BF₃.

Question xvi.
How are xenon fluorides XeF₂, XeF₄ and XeF₆ obtained ? Give suitable reactions.
Answer:
Xenon fluorides are generally prepared by the direct reaction of xenon and fluorine in different ratios and under appropriate experimental conditions, such as temperature, in the presence of an electric discharge and by a photochemical reaction.
(i) Preparation of XeF₂ :

(ii) Preparation of XeF₄ :

(iii) Preparation of XeF₆ :

Question xvii.
How are XeO₃ and XeOF₄ prepared?
Answer:
Preparation of XeO₃ : Xenon trioxide (XeO₃) is prepared by the hydrolysis of XeF₄ or XeF₆.

• By hydrolysis of XeF₄ :
  3XeF₄ + 6H₂0 → 2Xe + XeO₃ + 12 HF + \(1 \frac{1}{2} \mathrm{O}_{2}\)
• By hydrolysis of XeF₆ :
  XeF₆ + 3H₂O → XeO₃ + 6HF
• Preparation of XeOF₄ :
  Xenon oxytetrafluoride (XeOF₄) is prepared by the partial hydrolysis of XeF₆.
  XeF₆ + H₂O → XeOF₄ + 2HF

Question xviii.
Give two uses of neon and argon.
Answer:
Uses of neon (Ne) :

• Neon is used in the production of neon discharge lamps and signs by filling Ne in glass discharge tubes.
• Neon signs are visible from a long distance and also have high penetrating power in mist or fog.
• A mixture of neon and helium is used in voltage stabilizers and current rectifiers.
• Neon is also used in the production of lasers and fluorescent tubes.

Uses of argon (Ar) :

• Argon is used to fill fluorescent tubes and radio valves.
• It is used to provide inert atmosphere for welding and production of steel.
• It is used along with neon in neon sign lamps to obtain different colours.
• A mixture of 85% Ar and 15% N2 is used in electric bulbs to enhance the life of the filament.

Question xix.
Describe the structure of Ozone. Give two uses of ozone.
Answer:
(A)

• Ozone has molecular formula O₃.
• The lewis dot and dash structures for O₃ are :
  
• Infrared and electron diffraction spectra show that O₃ molecule is angular with 0-0-0 bond angle 117°.
• Both 0-0 bonds are identical having bond length 128 pm which is intermediate between single and double bonds.
• This is explained by considering resonating structures and resonance hybrid.
  

(B) Uses of Ozone :

• Ozone sterilises drinking water by oxidising germs and bacteria present in it.
• It is used as a bleaching agent for ivory, oils, starch, wax and delicate fabrics like silk.
• Ozone is used to purify the air in crowded places like Clnema halls, railways, tunnels, etc.
• In industry, ozone is used in the manufacture of synthetic camphor, potassium permanganate, etc.

Question xx.
Explain the trend in following atomic properties of group 16 elements.
i. Atomic radii
ii. Ionisation enthalpy
iii. Electronegativity.
Answer:
(1) Atomic and ionic radii :

1. As compared to group 15 elements, the atomic and ionic radii of group 16 elements are smaller due to higher nuclear charge.
2. The atomic and ionic radii increase down the group from oxygen to polonium. This is due to the addition of a new shell at each successive elements on moving down the group. The atomic radii increases in the order O < S < Se < Te < Po

(2) Ionisation enthalpy :

• The ionisation enthalpy of group 16 elements has quite high values.
• Ionisation enthalpy decreases down the group from oxygen to polonium. This is due to the increase in atomic volume down the group.
• The first ionisation enthalpy of the lighter elements of group 16 (O, S, Se) have lower values than those of group 15 elements in the corresponding periods. This is due to difference in their electronic configurations.

Group 15 : (valence shell) ns² np_x¹ np_y¹ np_z¹
Group 16 : (valence shell) ns² np_x² np_y¹ np_z¹

Group 15 elements have extra stability of half-filled and more symmetrical orbitals, while group 16 elements acquire extra stability by losing one of paired electrons from npx- orbital forming half-filled p-orbitals.

Hence group 16 elements have lower first ionisation enthalpy than group 15 elements.

(3) Electronegativity :

• The electronegativity values of group 16 elements have higher values than corresponding group 15 elements in the same periods.
• Oxygen is the second most electronegative elements after fluorine. (O = 3.5, F = 4)
• On moving down the group electronegativity decreases from oxygen to polonium.
• On moving down the group atomic size increases, hence nuclear attraction decreases, therefore electro-negativity decreases.

Elements	O	S	Se	Te	Po
Electronegativity	3.5	2.44	2.48	2.01	1.76

4. Answer the following.

Question i.
Distinguish between rhombic sulfur and monoclinic sulfur.
Answer:

Rhombic sulphur	Monoclinic sulphur
1. It is pale yellow.	1. It is bright yellow.
2. Orthorhombic crystals	2. Needle-shaped monoclinic crystals
3. Melting point, 385.8 K	3. Melting point, 393 K
4. Density, 2.069 g/cm3	4. Density: 1.989 g/cm3
5. Insoluble in water, but soluble in CS2	5. Soluble in CS2
6. It is stable below 369 K and transforms to α-sulphur above this temperature.	6. It is stable above 369 K and transforms into β-sulphur below this temperature.
7. It exists as S8 molecules with a structure of a puckered ring.	7. It exists as S8 molecules with a structure of a puckered ring.
8. It is obtained by the evaporation of roll sulphur in CS2	8. It is prepared by melting rhombic sulphur and cooling it till a crust is formed. Two holes are pierced in the crust and the remaining liquid is poured to obtain needle-shaped crystals of monoclinic sulphur (β-sulphur).

Question ii.
Give two reactions showing oxidizing property of concentrated H₂SO₄.
Answer:
Hot and concentrated H₂SO₄ acts as an oxidising agent, since it gives nascent oxygen on heating.

Question iii.
How is SO₂ prepared in the laboratory from sodium sulfite? Give two physical properties of SO₂.
Answer:
(A) Laboratory method (From sulphite) :

• Sodium sulphite on treating with dilute H₂SO₄ forms SO₂.
  Na₂SO₃ + H₂SO_4(aq) → Na₂SO₄ + H₂O₍₁₎ + SO_2(g)
• Sodium sulphite, Na₂SO₃ on reaction with dilute hydrochloric acid solution forms SO₂.
  Na₂SO_3(aq) + 2HCl_(aq) → 2NaCl_9aq0 + H₂O₍₁₎ + SO_2(g)

(B) Physical properties of SO₂

• It is a colourless gas with a pungent smell.
• It is highly soluble in water and forms sulphurous acid, H₂SO₃ SO_2(g) + H₂O₍₁₎ → H₂SO_3(aq)
• It is poisonous in nature.
• At room temperature, it liquefies at 2 atmospheres. It has boiling point 263K.

Question iv.
Describe the manufacturing of H₂SO₄ by contact process.
Answer:
Contact process of the manufacture of sulphuric acid involves following steps :

(1) Preparation of SO₂ : Sulphur or pyrite ores like iron pyrites, FeS₂ on burning in excess of air, form SO₂.

(2) Oxidation of SO₂ to SO₃ : SO₂ is oxidised to SO₃ in the presence of a heterogeneous catalyst V₂O₅ and atmospheric oxygen. This oxidation reaction is reversible.

To avoid the poisoning of a costly catalyst, it is necessary to make SO₂ free from the impurities like dust, moisture, As₂O₃ poison, etc.

The forward reaction is exothermic and favoured by increase in pressure. The reaction is carried out at high pressure (2 bar) and 720 K temperature. The reacting gases, SO₂ and O₂ are taken in the ratio 2:3.

(3) Dissolution of SO₃ : SO₃ obtained from catalytic converter is absorbed in 98%. H₂SO₄ to obtain H₂S₂O₇, oleum or fuming sulphuric acid.


Flow diagram for the manufacture of sulphuric acid

Question 7.1 (Textbook Page No 141)

12th Chemistry Digest Chapter 7 Elements of Groups 16, 17 and 18 Intext Questions and Answers

Question 1.
Elements of group 16 generally show lower values of first ionisation enthalpy compared to the elements of corresponding period of group 15. Why?
Answer:
Group 15 elements have extra stable, half filled p-orbitals with electronic configuration (ns²np³). Therefore more amount of energy is required to remove an electron compared to that of the partially filled orbitals (ns²np⁴) of group 16 elements of the corresponding period.

Question 7.2 (Textbook Page No 141)

Question 1.
The values of first ionisation enthalpy of S and Cl are 1000 and 1256 kJ mol^-1, respectively. Explain the observed trend.
Answer :
The elements S and Cl belong to second period of the periodic table.
Across a period effective nuclear charge increases and atomic size decreases with increase in atomic number. Therefore the energy required for the removal of electron from the valence shell (I.E.) increases in the order S < Cl.

Question 7.4 (Textbook Page No 141)

Question 1.
Fluorine has less negative electron gain affinity than chlorine. Why?
Answer :
The size of fluorine atom is smaller than chlorine atom. As a result, there are strong inter electronic repulsions in the relatively small 2p orbitals of fluorine and therefore, the incoming electron does not experience much attraction. Thus fluorine has less negative electron gain affinity than chlorine.

Try this… (Textbook Page No 140)

Question 1.
Explain the trend in the following properties of group 17 elements.

(1) Atomic size,
(2) Ionisation enthalpy,
(3) Electronegativity,
(4) Electron gain enthalpy.
Answer:
(1) Atomic size :

• Atomic and ionic radii increase down the group as atomic number increases due to the addition of new electronic valence shell to each succeeding element.
• The atomic radii increase in the order F < Cl < Br < 1
• Halogens possess the smallest atomic and ionic radii in their respective periods since the effective nuclear charge experienced by valence electrons in halogen atoms is the highest.

(2) Ionisation enthalpy :

• The ionisation enthalpies of halogens are very high due to their small size and large nuclear attraction.
• The ionisation ethalpies decrease down the group since the atomic size increases.
• The ionisation enthalpy decreases in the order F > Cl > Br > I.
• Among halogens fluorine has the highest ionisation enthalpy due to its smallest size.

Element	F	Cl	Br	I
Ionisation enthalpy kJ/mol	1680	1256	1142	1008

(3) Electronegativity :

• Halogens have the highest values for electronegativity due to their small atomic radii and high effective nuclear charge.
• Each halogen is the most electronegative element of its period.
• Fluorine has the highest electronegativity as compared to any element in the periodic table.
• The electronegativity decreases as,
  F > Cl > Br > I
  4.0 3.2 3.0 2.7 (electronegativity)

(4) Electron gain enthalpy (Δ_egH) :

• The halogens have the highest negative values for electron gain enthalpy.
• Electron gain enthalpies of halogens are negative indicating release of energy.
• Halogens liberate maximum heat by gain of electron as compared to other elements.
• Since halogens have outer valence electronic configuration, ns² np⁵, they have strong tendency to accept an electron to complete an octet and acquire electronic configuration of the nearest inert elements.
• In case of fluorine due to small size of 2 p-orbitals and high electron density, F has less negative electron gain enthalpy than Cl.
  F_(g) + e^– → F^–_(g) Δ_egH = – 333 klmol^-1
  Cl_(g) + e^– → Cl^–_(g) Δ_egH = – 349 kJ mol^-1
• The variation in electron gain enthalpy is in the order of, Cl > F > Br > I.

Question 2.
Oxygen has less negative electron gain enthalpy than sulphur. Why?
Answer:

• Oxygen has a smaller atomic size than sulphur.
• It is more electronegative than sulphur.
• It has a larger electron density.
• Due to high electron density, oxygen does not accept the incoming electron easily and therefore has less electron gain enthalpy than sulphur.

Question 7.3 (Textbook Page No 141)

Question 1.
Why is there a large difference between the melting and boiling points of oxygen and sulphur?
Answer :
Oxygen exists as diatomic molecule (O₂) whereas sulphur exists as polyatomic molecule (S₈). The van der Waals forces of attraction between O₂ molecules are relatively weak owing to their much smaller size. The large van der Waals attractive forces in the S₈ molecules are due to large molecular size. Therefore oxygen has low m.p. and b.p. as compared to sulphur.

Question 7.5 (Textbook Page No 141)

Question 1.
Bond dissoClation enthalpy of F₂ (158.8 kj mol^-1) is lower than that of Cl₂ (242.6 kj mol^-1) Why?
Answer :
Fluorine has small atomic size than chlorine. The lone pairs on each F atom in F₂ molecule are so close together that they strongly repel each other, and make the F – F bond weak. Thus, it requires less amount of energy to break the F – F bond. In Cl₂ molecule the lone pairs on each Cl atom are at a larger distance and the repulsion is less.

Thus Cl – Cl bond is comparatively stronger. Therefore bond dissoClation enthalpy of F₂ is lower than that of Cl₂.

Question 7.6 (Textbook Page No 142)

Question 1.
Noble gases have very low melting and boiling points. Why?
Answer :
Noble gases are monoatomic, the only type of interatomic interactions which exist between them are weak van der Waals forces. Therefore, they can be liquefied at very low temperatures and have very low melting or boiling points.

Can you tell? (Textbook Page No 142)

Question 1.
The first member of the a group usually differs in properties from the rest of the members of the group. Why?
Answer:
The first member of a group usually differs in properties from the rest of the members of the group for the following reasons :

• Its small size
• High electronegativity
• Absence of vacant d-orbitals in its valence shell.

Use your brain power! (Textbook Page No 142)

Question 1.
Oxygen forms only OF₂ with fluorine while sulphur forms SF₆. Explain. Why?
Answer:

• Oxygen combines with the most electronegative element fluorine to form OF₂ and exhibits positive oxidation state (+ 2). Since, oxygen does not have vacant J-orbitals it cannot exhibit higher oxidation states.
• Sulphur has vacant d-orbitals and hence can exhibit + 6 oxidation state to form SF₆.

Question 2.
Which of the following possesses hydrogen bonding? H₂S, H₂O, H₂Se, H₂Te
Answer:

• Oxygen being more electronegative, is capable of forming hydrogen bonding in the compound H₂O.
• The other elements S, Se and Te of Group 16, being less electronegative do not form hydrogen bonds.
• Thus, hydrogen bonding is not present in the other hydrides H₂S, H₂Se and H₂Te.

Question 3.
Show hydrogen bonding in the above molecule with the help of a diagram.
Answer:

Try this….. (Textbook Page No 143)

Question 1.
Complete the following tables :

Answer:

Can you tell? (Textbook Page No 146)

Question 1.
What is allotropy?
Answer:
The property of some elements to exist in two or more different forms in the same physical state is called allotropy.

Question 2.
What is the difference between allotropy and polymorphism?
Answer:

• Allotropy is the existence of an element in more than one physical form. It means that under different conditions of temperature and pressure an element can exist in more than one physical forms.
• Coal, graphite and diamond etc., are different allotropic forms of carbon.
• Polymorphism is the existence of a substance in more than one crystalline form.
• It means that under different conditions of temperature and pressure, a substance can form more than one type of crystal. For example, mercuric iodide exists in the orthorhombic and trigonal form.

Question 7.7 (Textbook Page No 146)

Which form of sulphur shows paramagnetic behaviour?
Answer :
In the vapour state, sulphur partly exists as S₂ molecule, which has two unpaired electrons in the antibonding π^* orbitals like O₂. Hence it exhibits paramagnetism.

Try this….. (Textbook Page No 149)

Question 1.
Why water in a fish pot needs to be changed from time to time?
Answer:
A fish pot is an artificial ecosystem and the fish in it are selective and maintained in a restricted environment.

In a fish pot, the unwanted food and waste generated by the fish mix with the water and remain untreated due to lack of decomposers.

Accumulation of waste material will decrease the levels of dissolved oxygen in the water pot.

Hence, it is necessary to change the water from time to time.

Question 7.8 (Textbook Page No 149)

Dioxygen is paramagnetic in spite of having an even number of electrons. Explain.
Answer :
Dioxygen is a covalently bonded molecule.
The paramagnetic behaviour of O₂ can be explained with the help of molecular orbital theory.
Electronic configuration O₂
KK σ(2s)² σ(2s)² σ^*(2p_z)² π(2p_x)² π(2p_x)² π(2p_y)² π^*(2p_x)¹ π^*(2p_y)¹. Presence of two unpaired electrons in antibonding orbitals explains paramagnetic nature of dioxygen.

Question 7.9 (Textbook Page No 150)

High concentration of ozone can be dangerously explosive. Explain.
Answer :
Thermal stability : Ozone is thermodynamically unstable than oxygen and decomposes into O₂. The decomposition is exothermic and results in the liberation of heat (ΔH is – ve) and an increase in entropy (ΔS is positive). This results in large negative Gibbs energy change (ΔG). Therefore high concentration of ozone can be dangerously explosive. Eq O₃ → O₂ + O

Try this…… (Textbook Page No 151)

(a) Ozone is used as a bleaching agent. Explain.
Answer:

• Ozone due to its oxidising property can act as a bleaching agent. O_3(g) → O_2(g) + O
• It bleaches coloured matter. coloured matter + O → colourless matter
• Ozone bleaches in the absence of moisture, so it is also known as dry bleach.
• Ozone can bleach ivory and delicate fabrics like silk.

(b) Why does ozone act as a powerful oxidising agent?
Answer:
Ozone decomposes to liberate nascent oxygen, hence it is a powerful oxidising agent. O_3(g) → O_2(g) + O
For example :

• It oxidises lead sulphide (PbS) to lead sulphate (PbSO₄).
  pbS_(s) + 4O_3(g) → PbSO_(s) + 4O_2(g)
• Potassium iodide, KI is oxidised to iodine, I₂ in the solution.
  2KI_(aq) + H₂O₍₁₎ + O_3(g) → 2KOH_(aq) + I_2(s) + O_2(g)

Question 7.10 : (Textbook Page No 154)

What is the action of concentrated H₂SO₄ on (a) HBr (b) HI
Answer :
Concentrated sulphuric acid oxidises hydrobromic acid to bromine.

2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O
It oxidises hydroiodic acid to iodine.
2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O

Try this….. (Textbook Page No 156)

Question 1.
Give the reasons for the bleaching action of chlorine.
Answer:

• Chlorine acts as a powerful bleaching agent due to its oxidising nature.
• In moist conditions or in the presence of water it forms unstable hypochlorous acid, HOCl which decomposes giving nascent oxygen which oxidises the vegetable colouring matter of green leaves, flowers, litmus, indigo, etc.
  Cl₂ + H₂O → HCl + HOCl
  HOCl → HCl + [O]
  Vegetable coloured matter + [O] → colourless matter.

Question 2.
Name two gases used in war.
Answer:
Phosgene : COCl₂
Mustard gas: Cl – CH₂ – CH₂ – S – CH₂ – CH₂ – Cl

Use your brain power! (Textbook Page No 157)

Question 1.
Chlorine and fluorine combine to form interhalogen compounds. The halide ion will be of chlorine or fluorine?
Answer:
Among the- two halogens, chlorine is more electropositive than fluorine (Electronegativity values: F = 4.0, Cl = 3.2)

The interhalogen compound is regarded as the halide of the more electropositive halogen. Hence, the interhalogen compound is the fluoride of chlorine, i.e. chlorine monofluoride, CiF.

Question 2.
Why does fluorine combine with other halogens to form maximum number of fluorides?
Answer:
Since fluorine is the most electronegative element and has the smallest atomic radius compared to other halogen fluorine forms maximum number of fluorides.

Use your brain power! (Textbook Page No 158)

Question 1.
What will be the names of the following compounds: ICl, BrF?
Answer:
ICl : Iodine monochloride
BrF : Bromine monofluoride

Question 2.
Which halogen (X) will have maximum number of other halogen (X ) attached?
Answer:
The halogen Iodine (I) will have the maximum number of other halogens attached.

Question 3.
Which halogen has tendency to form more interhalogen compounds?
Answer:
The halogen fluorine (F) has the maximum tendency to form more interhalogen compounds as it has a small size and more electronegativity.

Question 4.
Which will be more reactive?
(a) ClF₃ or ClF,
(b) BrF₅ or BrF
Answer:
ClF₃ is more reactive than ClF, while BrF₅ is more reactive than BrF. Both ClF₃ and BrF₅ are unstable compared to ClF and BrF respectively due to steric hindrance hence are more reactive.

Question 5.
Complete the table :

Formula	Name
ClF	Chlorine monofluoride
ClF3	…………………………………
…………………………………	Chlorine pentachloride
BrF	…………………………………
…………………………………	Bromine pentafluoride
ICl	…………………………………
ICl3	…………………………………

Answer:

Formula	Name
ClF	Chlorine monofluoride
ClF3	Chlorine trifluoride
CIF5	Chlorine pentafluoride
BrF	Bromine monofluoride
BrF5	Bromine pentafluoride
ICl	Iodine monochloride
ICl3	Iodine trichloride

Use your brain power! (Textbook Page No 159)

Question 1.
In the special reaction for ICl, identify the oxidant and the reductant? Denote oxidation states of the species.
Answer:

Potassium chlorate, KClO₃ is the oxidising agent or oxidant and iodine is the reducing agent or reductant.

Use your brain power! (Textbook Page No 162)

Question 1.
What are missing entries?

Formula	Name
XeOF2 …………… XeO3F2 XeO2F4	Xenon monooxyfluoride Xenon dioxydifluoride …………………………………….. ……………………………………..

Answer:

Formula	Name
XeOF2 XeO2F2 XeO3F2 XeO2F4	Xenon monooxydifluoride Xenon dioxydifluoride Xenon trioxydifluoride Xenon dioxytetrafluoride

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Solutions Chapter 8 Company Accounts – Issue of Shares Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Book Keeping & Accountancy Solutions Chapter 8 Company Accounts – Issue of Shares

1. Objective Questions:

A. Select the appropriate answer from the alternative given below and rewrite the sentence.

Question 1.
The balance of Share Forfeiture A/c is transferred to ______________ Account after re-issue of these share.
(a) Reserve Capital
(b) Capital Reserve
(c) Profit & Loss
(d) Share Capital
Answer:
(b) Capital Reserve

Question 2.
Premium received on issue of shares is shown to ______________
(a) Liability side of Balance Sheet
(b) Asset side of Balance Sheet
(c) Profit & Loss A/c debit side
(d) Profit & Loss A/c credit side
Answer:
(a) Liability side of Balance Sheet

Question 3.
Shareholders get ______________ on shares.
(a) interest
(b) commission
(c) rent
(d) dividends
Answer:
(d) dividends

Question 4.
The document inviting to subscribe the shares of a company is ______________
(a) Prospectus
(b) Memorandum of Association
(c) Articles of Association
(d) Share certificate
Answer:
(a) Prospectus

Question 5.
As per SEBI guidelines, minimum amount payable on share application should be ______________ Nominal Value of shares.
(a) 10%
(b) 15%
(c) 2%
(d) 5%
Answer:
(d) 5%

Question 6.
When shares are forfeited the Share Capital Account is ______________
(a) credited
(b) debited
(c) neither debited nor credited
(d) None of the given
Answer:
(b) debited

Question 7.
The liability of shareholder in Joint Stock Company is ______________
(a) joint and several
(b) limited
(c) unlimited
(d) huge
Answer:
(b) limited

Question 8.
The Share Capital which a company is authorized to issue by its Memorandum of Association is ______________
(a) Nominal Capital/Authorised Capital
(b) Issued Capital
(c) Paid-up Capital
(d) Reserve Capital
Answer:
(a) Nominal Capital/Authorised Capital

Question 9.
The unpaid amount on allotment and calls may be transferred to ______________ Account.
(a) Calls-in-Advance
(b) Calls
(c) Calls-in-Arrears
(d) Allotment
Answer:
(c) Calls-in-Arrears

Question 10.
There must be provision in ______________ for forfeiture of shares.
(a) Articles of Association
(b) Memorandum of Association
(c) Prospectus
(d) Balance Sheet
Answer:
(a) Articles of Association

B. Give one word/term/phrase for each of the following statements.

Question 1.
Amount called up on shares by the company but not received.
Answer:
Calls-in-Arrears

Question 2.
Issue of share at its face value.
Answer:
Issue at par

Question 3.
The person who purchases the shares of a company.
Answer:
Shareholder

Question 4.
The form of business organisation where a huge amount of capital can be raised.
Answer:
Joint-stock company

Question 5.
The capital is subscribed by the public.
Answer:
Subscribed capital

Question 6.
The shares having preferential rights at the time of winding up of the company.
Answer:
Preference shares

Question 7.
The shares on which dividend is not fixed.
Answer:
Equity shares

Question 8.
The part of subscribed capital is not called up by the company.
Answer:
Uncalled capital

C. State true or false with reasons.

Question 1.
Directors can forfeit the shares for any reason.
Answer:
This statement is False.
After paying money on share application, When share applicant fails to pay the call money or premium on shares in spite of repeated reminders and warnings directors/company can forfeit the shares.

Question 2.
Once the application money is received, directors can immediately proceed with the allotment of shares.
Answer:
This statement is False.
Directors can proceed for allotment of shares only after receiving the minimum subscription amount of the issued amount by cheque or other instrument complying with all legal requirements.

Question 3.
Joint-stock company forms of business organisations came into existence after the industrial revolution.
Answer:
This statement is True.
As the volume and scale of trade and industry expanded, especially after the industrial revolution, a very large unit of the commercial organisation requiring large capital and greater managerial skill, called Joint-stock company came into existence.

Question 4.
Equity shareholders get a guaranteed rate of dividend every year.
Answer:
This statement is False.
One of the features of equity shares is the rate of dividend payable on equity shares keeps on changing from one year to another. So, there is no question of guaranteed dividend every year for equity shareholders.

Question 5.
The face value of shares and market value of shares is always the same.
Answer:
This statement is False.
Face value of shares means the issue price of shares while the market value of shares means the trading price of shares at the stock exchange. The face value of shares remains the same and fixed. However, market price changes as per the performance of the company. Hence face value and market value of shares is not the same.

Question 6.
Sweat shares are issued to the public.
Answer:
This statement is False.
Sweat shares are issued by a company to its directors or employees at a discount or for consideration other than cash. Sweat shares are not issued to the public.

D. State whether you agree or disagree with the following statements.

Question 1.
In the case of Pro-rata allotment the excess application money received must be refunded.
Answer:
Disagree

Question 2.
Calls-in-Advance account is shown on the asset side of the Balance Sheet.
Answer:
Disagree

Question 3.
The Authorised Capital is also known as Nominal Capital.
Answer:
Agree

Question 4.
Paid-up capital can be more than Called-up Capital.
Answer:
Disagree

Question 5.
The joint-stock company can raise a huge amount of capital.
Answer:
Agree

Question 6.
When shares are Forfeited Shares Capital Account is credited.
Answer:
Disagree

Question 7.
Directors can re-issue forfeited shares.
Answer:
Agree

Question 8.
When the issued price of a share is ₹ 12 and face value is ₹ 10, the share is said to be issued at a premium.
Answer:
Agree

Question 9.
A public limited company can issue its share without issuing its prospectus.
Answer:
Disagree

Question 10.
Shares can be issued for consideration other than cash.
Answer:
Agree

E. Answer in one sentence only.

Question 1.
What are Preference Shares?
Answer:
Preference Shares are a type of share which enjoys priority or preference over equity share for the repayment of dividends at a predetermined fixed rate and for the repayment of capital.

Question 2.
What is Registered Capital?
Answer:
Registered Capital or Authorised Capital means the maximum limit up to which a company is authorized to raise share capital.

Question 3.
What is Reserve Capital?
Answer:
Reserve Capital is that part of the subscribed capital which is reserved to be called up only at the time of winding up or liquidation of the company.

Question 4.
What is Over Subscription of Shares?
Answer:
When a company received more applications of shares than those actually offered or issued to the public, known as Over Subscription of Shares.

Question 5.
Which account is debited when share first call money is received?
Answer:
The bank account will be debited when share first call money is received.

Question 6.
When are shares allotted on a pro-rata basis?
Answer:
Shares are said to be allotted on a pro-rata basis when the applications are received for more shares than the number of shares issued and shares are allotted in the proportion to the number of shares applied for.

Question 7.
What is Forfeiture of Shares?
Answer:
When a shareholder fails to pay the call money or premium on the shares in spite of repeated reminders and warnings, the company forfeits the shares of such defaulters known as forfeiture of shares.

Question 8.
What is Calls-in-Arrears?
Answer:
Non-payment of allotment or call money by the applicants in spite of repeated reminders are called Calls-in-Arrears.

Question 9.
What do you mean by Shares Issued at Premium?
Answer:
When shareholders are supposed to pay a price higher than the face value of the shares, their shares are said to be issued at a premium.

Question 10.
What is Paid-up Capital?
Answer:
Part of the called-up capital which is actually paid by the shareholders is called Paid-up Share Capital.

F. Complete the following sentences.

Question 1.
When the face value of the share is ₹ 100 and the issued price is ₹ 120, then it is said that the shares are issued at ______________
Answer:
premium

Question 2.
______________ Capital is the capital which a company is authorized to issue by its Memorandum of Association.
Answer:
Authorized

Question 3.
The difference between Called-up Capital and Paid-up Capital is known as ______________
Answer:
Calls-in-Arrears

Question 4.
______________ shareholders get fixed rate of dividend.
Answer:
Preference

Question 5.
______________ shareholders are the real owners of the company.
Answer:
Equity

Question 6.
______________ form of business organisation in which capital is raised through the issue of shares.
Answer:
Joint-stock company

Question 7.
______________ Capital is the part of Issued capital which is subscribed by the public.
Answer:
Subscribed

Question 8.
The part of Authorised Capital which is not issued to the public is known as ______________ Capital.
Answer:
Unissued

G. Calculate the following.

Question 1.
One shareholder holding 500 equity shares paid share application money @ ₹ 3, Allotment money @ ₹ 4 per share and failed to pay a final call of ₹ 3 per share his share was forfeited calculate the amount of forfeiture.
Solution:
Amount of forfeiture = Amount received by the company (In case of non-payment of ‘calls’)
Here, shareholders paid ₹ 3 per share on application and ₹ 4 per share on the allotment on 500 shares.
So, total amount received by company = 500 × ₹ 3 + 500 × ₹ 4
= 1,500 + 2,000
= ₹ 3,500
∴ Amount of share forfeiture = ₹ 3,500.

Question 2.
10,000 equity shares of ₹ 10 each issued at a 10% premium. Calculate the total amount of share premium.
Solution:
Equity shares = 10,000
Face value = ₹ 10 per share
Premium @ 10% = 10,000 × 10 × \(\frac{10}{100}\) = ₹ 10,000
So, premium 10,000 shares of ₹ 10 each at 10% = ₹ 10,000

Question 3.
The company received excess applications for 5000 shares @ ₹ 4 per share. The application of 1000 shares was rejected and a pro-rata allotment was made. Calculate the amount of application money adjusted with allotment.
Solution:
Excess application money received for 5000 shares @ ₹ 4 per share = ₹ 20,000
Less: Application of 1000 shares rejected and money refunded = ₹ 4,000
Excess money received to be adjusted with allotment = ₹ 16,000

Question 4.
80,000 equity shares of ₹ 10 each issued and fully subscribed and called up at 20% premium. Calculate the amount of Equity Share capital.
Answer:
Equity Share capital = No. of equity shares × face value of each share
= 80,000 × ₹ 10
= ₹ 8,00,000
Note: Equity Share capital has no concern with premium or discount amount.

Question 5.
Directors issued 20,000 equity shares of ₹ 100 each at par. These were fully subscribed and called up. All money was received except one shareholder holding 100 equity shares failed to pay a final call of ₹ 20 per share. Calculate the amount of Paid-up capital of the company.
Solution:
Fully subscribed and called-up amount = 20,000 equity shares × ₹ 100 each share
= ₹ 20,00,000
But one share holder failed to pay final call of ₹ 20 per share of 100 equity shares means
Non-payment of shares = 100 equity shares × ₹ 20 per share = ₹ 2,000
∴ Total Paid-up capital amount = ₹ 20,00,000 – ₹ 2,000 = ₹ 19,98,000

Question 6.
The company sends a regret letter for 100 shares and an Allotment letter to 25,000 shareholders. Application money per ₹ 20 per share. Calculate the amount of application money that the company is refunding.
Solution:
The company sends a Regret letter for 100 shares for ₹ 20 per share application money received i.e. only that much amount the company will refund.
Amount of refund = No. of shares × Value of per share
= 100 × ₹ 20
= ₹ 2,000

Practical Problems

Question 1.
Vijay Ltd. was registered with an authorized capital of ₹ 15,00,000 divided into 1,50,000 equity shares of ₹ 10 each.
The company issued 1,00,000 equity shares of ₹ 10 each at a premium of ₹ 2 per share. The company received applications for 80,000 equity shares and was allotted the shares.
The company received application money ₹ 3 per share, allotment money ₹ 4 per share
(Including premium) and first, call money ₹ 3 per share.
The Directors have not made the final call of ₹ 2 per share. All money was received except one shareholder holding 500 shares did not pay the first call.
Show Authorised Capital, Issued Capital, Subscribed Capital, Called-up Capital,
Paid-up Capital, Calls in Arrears, and Share Premium amount in the company balance sheet.
Solution:
In the books of Vijay Ltd.
Balance Sheet as on ______________

Working Notes:
1. Bank balance at the end = Amount received on application + Amount received on allotment + Amount received on 1st call + Premium amount received
= 80,000 × 3 + 80,000 × 2 × 79,500 × 3 + 80,000 × 2
= 2,40,000 + 1,60,000 + 2,38,500 + 1,60,000
= ₹ 7,98,500

2. Directors have not made the final call of ₹ 2 per share means total called-up amount = ₹ 10 – ₹ 2 = ₹ 8

3. Calls-in-Arrears on 500 shares at ₹ 3 = ₹ 1,500 of the first call

4. Share premium on 80,000 shares @ ₹ 2 received at allotment stage i.e. share premium amount = 80,000 x ₹ 2 = ₹ 1,60,000

Question 2.
Anand Company Limited issued 1,00,000 preference shares of ₹ 10 each payable as-
On Application ₹ 4
On Allotment ₹ 3
On First call ₹ 2
On Second & Final call?
The company received applications for all these shares and received all money.
Pass Journal Entries in the books of Anand Company Ltd.
Solution:
Journal Entries in the books of Anand Company Ltd.

Question 3.
Rohini Company Limited issued 25,000 equity shares of ₹ 100 each payable as follows:
On Application ₹ 20
On Allotment ₹ 30
On First call ₹ 20
On the Second & Final call ₹ 30
The application was received for 22,000 equity shares and allotment of shares was made to them. All money was received by the company.
Pass Journal Entries in the books of Rohini Co. Ltd.
Solution:
Journal Entries in the books of Rohini Company Limited

Question 4.
Deepak Manufacturing Co. Ltd. issued a prospectus inviting applications for 1,00,000 equity shares of ₹ 10 each payable as follows :
₹ 2 on Application
₹ 4 on Allotment
₹ 2 on the First call
₹ 2 on Final call
The application was received for 1,20,000 equity shares. The Directors decided to reject excess applications and refunded application money on that. The company received all money.
Pass Journal Entries in the books of a company.
Solution:
Journal Entries in the books of Deepak Manufacturing Co.Ltd

Question 5.
Sucheta Company Limited issued ₹ 20,00,000 new capital divided into ₹ 100 equity shares at a premium of ₹ 20 per share payable as ₹ 10 on Application, ₹ 40 on Allotment and ₹ 10 premium ₹ 50 on Final call and ₹ 10 premium.
The issue was oversubscribed to the extent of 26,000 equity shares. The applicants on 2,000 shares were sent a letter of regret and their application money was refunded.
The remaining applicants were allotted shares on a Pro-rata basis. All the money due on Allotment and Final call was only received.
Make necessary Journal Entries in the books of Sucheta Company Ltd.
Answer:
Solution:
Journal Entries in the books of Sucheta Company Limited

Working Note:
Calculation of Application money transferred to Share Allotment:
Application money received (26,000 × 10) = 2,60,000
Less: Application money refunded (2,000 × 10) = 20,000
Less: Application money transferred to Share Capital: (20,000 × 10) = 2,00,000
Excess money received on application transferred to Share Allotment = 40,000
Bifurcation of calls amount:

Question 6.
Suhas Limited issued 10,000 equity shares of ₹ 10 each at a premium of ₹ 2 per share payable ₹ 3 on application, ₹ 5 (including premium) on the allotment, and the balance in two calls of an equal amount. Applications were received for 11,000 equity shares and pro-rata allotment was made for all the applicants. The excess application money was adjusted towards allotment.
Mrs. Shobha who was allowed 200 equity shares failed to pay F/F/C and her shares were forfeited after the final call.
Show Journal Entries in the books of Suhas Ltd. and also show its presence in Balance Sheet.
Solution:
Journal Entries in the books of Suhas Limited

Balance Sheet of Suhas Limited

Working Notes:
1. Excess amount received at the time of application ₹ 3,000 adjusted at allotment stage, so allotment amount received in the bank is ₹ 47,000.

2. Amount called-up per share: ₹ 3 on application, ₹ 5 (including premium) on allotment i.e. ₹ 2 premium + ₹ 3 capital and balance amount ₹ 4 in two calls of the equal amount i.e. ₹ 2 on the first call and ₹ 2 on final call.

3. Mrs. Shobha was not able to pay F/F/C i.e. first and final call means 200 × ₹ 2 first call money = ₹ 400 and 200 × ₹ 2 final call money = ₹ 400.
Mrs. Shobha paid ₹ 6 per share towards capital which the company received and the company has the right to forfeit only paid amount means the company forfeited ₹ 1,200 of Mrs. Shobha.

Question 7.
Subhash Company Limited issues 2000 Equity shares of ₹ 100 each payable as ₹ 30 on application, ₹ 30 on the allotment, ₹ 40 on first and final call.
All the shares were subscribed and duly allotted. The company made all the calls. All cash was duly received except the first and final call on 100 equity shares. These shares were forfeited by the company and were re-issued as fully paid for ₹ 75 per share.
Show the Journal Entries in the books of Subhash Company Ltd.
Solution:
Journal Entries in the books of Subhash Company Limited

Working Notes:
1. Amount forfeited by the company on 100 shares forfeited = 100 × (30 + 30)
= 100 × 60
= ₹ 6,000

2. Calls-in-Arrears = 100 × 40 = ₹ 4,000.

3. Amount received on re-issue of 100 forfeited shares = 100 × 75 = ₹ 7,500.
Balance of ₹ 2,500 (i.e. loss 25 × 100) is transferred to Share Forfeiture A/c.

4. Amount transfer from Share Forfeiture A/c to Capital Reserve is ascertained by preparing Share Forfeiture A/c.

Question 8.
Pass Journal Entries for the forfeiture and re-issue of shares in the following cases:
(A) Asha Ltd. forfeited 100 equity shares of ₹ 20 each fully called-up for non-payment of the first call of ₹ 3 per share and final call of ₹ 5 per share. 80 shares of these were re-issued at ₹ 15 per share as fully paid.
(B) Bhakti Ltd. forfeited 100 equity shares of ₹ 10 each, ₹ 6 called-upon which the shareholder paid application and allotment of ₹ 5 per share. Of these 80 shares were re-issued as fully paid-up for ₹ 16 per share.
(C) Konark Ltd. forfeited 50 shares of ₹ 10 each, ₹ 8 called-up. The shareholder failed to pay the first call of ₹ 3 per share. Later on, 30 shares of these were re-issued at ₹ 7 per share.
Solution:
Journal Entries [For Asha Ltd.]

Working Notes for A:
1. Out of 100 forfeited shares, 80 shares were re-issued accordingly Equity Share Capital A/c is debited and credited.
2. To find the proportionate amount for Forfeiture A/c:
For 100 shares-share forfeiture amount = ₹ 1,200
∴ 80 shares – share forfeiture amount = ₹ 960
Now, out of this ₹ 960 we used ₹ 400 from Share Forfeiture A/c at the time of re-issue of shares.
So, balance of Share Forfeiture A/c = ₹ 960 – ₹ 400 = ₹ 560

Journal Entries [For Bhakti Ltd.]

Working Notes for B:
1. Out of 100 forfeited shares, 80 shares were re-issued accordingly Equity Share Capital A/c is debited for ₹ 600 and credited for ₹ 480.

2. The proportionate amount debited to Forfeiture A/c:
For 100 shares-share forfeiture amount debited = ₹ 500 1 Qn
∴ 80 shares – share forfeiture amount = ₹ \(\frac{80}{100} \times \frac{500}{1}\) = ₹ 400
Now, shares were re-issued at ₹ 6 per share which is a called-up amount.
∴ The proportionate amount for Forfeiture A/c ₹ 400 will be transferred to Capital Reserve A/c.

Journal Entries (For Konark Ltd.)

Working Note for C:
The proportionate amount debited to Forfeiture A/c:
For 50 shares – share forfeiture amount debited is ₹ 250
∴ 30 shares-share forfeiture amount = ₹ \(\frac{30}{50} \times 250\) = ₹ 150
Out of this ₹ 30 used for re-issue of forfeited shares.
∴ Balance of Share Forfeiture A/c = ₹ 150 – ₹ 30 = ₹ 120.

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Solutions Chapter 7 Bills of Exchange Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Book Keeping & Accountancy Solutions Chapter 7 Bills of Exchange

Objective Questions

A. Select the correct option and rewrite the sentence:

Question 1.
The person on whom a bill is drawn is called a ______________
(a) Drawee
(b) Payee
(c) Drawer
(d) Acceptor
Answer:
(a) Drawee

Question 2.
Before acceptance the bill is called a ______________
(a) Order
(b) Request
(c) Draft
(d) Instrument
Answer:
(c) Draft

Question 3.
When the due date of the bill drawn falls due on a public holiday, the payment must be made on the ______________ day.
(a) same
(b) preceding
(c) next
(d) any
Answer:
(b) preceding

Question 4.
The due date of the bill drawn for 2 months on 23rd Nov. 2019 will be ______________
(a) 23rd Jan. 2020
(b) 25th Jan. 2019
(c) 26th Jan. 2019
(d) 25th Jan. 2020
Answer:
(d) 25th Jan. 2020

Question 5.
Noting charges are borne by ______________
(a) Notary Public
(b) Drawee
(c) Drawer
(d) Endorsee
Answer:
(b) Drawee

Question 6.
There are ______________ parties to bill of exchange.
(a) five
(b) four
(c) three
(d) two
Answer:
(c) three

Question 7.
When a bill is drawn for 2 months after date on 3rd Jan. 2020, its due date will be ______________
(a) 3rd Jan. 2020
(b) 3rd Mar. 2020
(c) 5th Mar. 2020
(d) 6th Mar. 2020
Answer:
(d) 6th Mar. 2020

Question 8.
Notary Public is ______________
(a) Govt. Officer
(b) Drawer
(c) Payee
(d) Endorsee
Answer:
(a) Govt. Officer

Question 9.
When Acceptor or Drawee does not pay the amount of bill to the holder on the due date it is known as ______________ the bill.
(a) returning
(b) discounting
(c) honouring
(d) dishonouring
Answer:
(d) dishonouring

Question 10.
The person who accepts the bill treats the bill as ______________
(a) Bills Payable
(b) Promissory Note
(c) Draft
(d) Bills Receivable
Answer:
(a) Bills Payable

B. Write the word/phrase/term, which can substitute each of the following statements:

Question 1.
Three extra days are allowed over and above the term of the bill.
Answer:
Grace days

Question 2.
Fees charged by Notary Public for getting the fact of dishonour noted.
Answer:
Noting Charges

Question 3.
A person who is entitled to receive the amount of bill of exchange.
Answer:
Payee

Question 4.
A person in whose favour a bill endorsed.
Answer:
Endorsee

Question 5.
Officer appointed by the government for noting of dishonour of bill.
Answer:
Notary Public

Question 6.
Cancellation of the bill on maturity in return for a new bill for an extended period of credit.
Answer:
Renewal of Bill

Question 7.
Bill of exchange drawn and accepted without any valuable consideration.
Answer:
Accommodation bill

Question 8.
A person who is in possession of the Bill of Exchange.
Answer:
Holder

Question 9.
Conversion of Bill of Exchange into its present value.
Answer:
Discounting of the bill

Question 10.
The amount is not recoverable from Drawee on account of insolvency.
Answer:
Bad debts

C. State whether the following statements are True or False with reasons:

Question 1.
An Inland bill is one that is drawn in one country and payable in another country.
Answer:
This statement is False.
Inland bill means, a bill drawn, accepted, and made payable within the territory of one and same, country. So, a bill is drawn in one country and payable in another country can’t be an inland bill.

Question 2.
Retirement of the bill means payment of the bill before the due date.
Answer:
This statement is True.
Payment of the bill, by the acceptor of the bill to the holder of the bill before the due date, is known as Retirement of the bill. So retirement of the bill means payment of the bill before the due date.

Question 3.
Drawee can transfer the ownership of the bill.
Answer:
This statement is False.
Drawee is a debtor. He has to pay the amount of the bill to its holder on the due date. Hence he cannot transfer its ownership to other people. The drawer can transfer the ownership of the bill as he is the owner of the bill.

Question 4.
Acceptance of the bill without making any changes in the terms of the bill is called qualified acceptance.
Answer:
This statement is False.
Acceptance of the bill with some changes as regards the terms, amount, place, etc. of a bill is known as qualified acceptance. Acceptance of the bill without making changes as regards the term is called general acceptance.

Question 5.
Discounting is a device to convert the bill into its present value.
Answer:
This statement is True.
When the drawer or holder of the bill approaches the bank to discount the bill, the bank pays the bill amount after deducting a certain amount (which is known as discounting charges). It means conversion of the bill into its present value in cash. So, we can say that discounting is a device to convert the bill into its present value.

Question 6.
A bill of exchange must be presented to the acceptor on the due date.
Answer:
This statement is True.
To get the payment of the bill from the acceptor, the holder of the bill is required to present it to the acceptor on its due date. Acceptor either honours the bill or dishonours the bill.

Question 7.
If a bill is discounted by the holder, no entry is passed in his book when the bill is honoured on the due date.
Answer:
This statement is True.
On discounting the bill the holder gives the possession of the bill to the bank. On the maturity date, the bank has to present the bill to the drawee to collect the payment. When the discounted bill is honoured, the transaction takes place between drawee and bank.

Question 8.
Noting charges are to be borne by the drawer.
Answer:
This statement is False.
Noting charges are to be borne by the drawee only as due to his act of non-payment, the bill is dishonoured and the drawer is not able to get money on its due date.

Question 9.
If a bill is drawn payable ‘on demand’ no grace days are allowed.
Answer:
This statement is True.
‘On demand’ means the amount of the bill is to be paid by drawee immediately on presentation of the bill as no time period is mentioned on it. In demand bill, 3 days grace is not allowed by law.

Question 10.
There are three parties to a promissory note.
Answer:
This statement is False.
There are only two parties to a promissory note, i.e. Drawer or maker of the note and drawee or payee of the note.

D. Find the odd one:

Question 1.
(a) Retaining
(b) Noting
(c) Discounting
(d) Endorsing
Answer:
(b) Noting

Question 2.
(a) Trade bill
(b) Accommodation bill
(c) After date bill
(d) Demand bill
Answer:
(d) Demand bill

Question 3.
(a) Notary public
(b) Drawer
(c) Drawee
(d) Payee
Answer:
(a) Notary public

Question 4.
(a) Discounting charges
(b) Rebate
(c) Bank charges
(d) Noting charges
Answer:
(d) Noting charges

Question 5.
(a) Stamp
(b) Acceptance
(c) Draft
(d) Amount
Answer:
(c) Draft

E. Complete the sentences:

Question 1.
Making payment of bill before the due date of maturity is known as ______________
Answer:
Retirement of Bill

Question 2.
A person whose liabilities are more than his assets and is not in a position to pay off his liabilities is ______________
Answer:
Insolvent person

Question 3.
Amount that cannot be paid by acceptor on account of insolvency is known as ______________
Answer:
Deficiency

Question 4.
A bill of exchange payable after certain period is known as ______________
Answer:
After date bill

Question 5.
A bill which is drawn and accepted with valuable consideration is known as ______________
Answer:
Trade Bill

Question 6.
A person who draws the bill of exchange is known as ______________
Answer:
Drawer

Question 7.
A bill whose due date is calculated from the date of acceptance is known as ______________
Answer:
After sight bill

Question 8.
Recording the fact of dishonour of bill is known as ______________
Answer:
Noting

Question 9.
When drawee accepts the bill payable at a particular place only, it is known as ______________
Answer:
qualified acceptance as to place

Question 10.
Fees charged by the bank for collection of bill on behalf of holder is ______________
Answer:
bank charges

F. Answer in a sentence:

Question 1.
What do you mean by Bill of Exchange?
Answer:
A Bill of Exchange is a written order signed by the drawer, directing a certain person to pay a certain sum of money on-demand or on a certain future date to a certain person or as per his order.

Question 2.
What are Days of Grace?
Answer:
The three extra days allowed to the drawee or the acceptor of a bill for making payment on it are called Days of Grace.

Question 3.
What do you mean by Discounting a Bill of Exchange?
Answer:
Encashment of a bill of exchange with the bank for certain cash which is less than the face value of the bill, before its due date by its drawer or holder is called Discounting of a Bill of Exchange.

Question 4.
What is Noting of the Bill?
Answer:
Noting of a Bill of Exchange is the recording of the facts of its dishonour by a Notary public.

Question 5.
What are Noting Charges?
Answer:
Noting Charges are the fees charged by the Notary public for noting the facts of dishonour on the face of the bill and in his official register.

Question 6.
What is the relationship between drawer and drawee?
Answer:
The relationship between the drawer and the drawee is that of the creditor and debtor.

Question 7.
Who is the Payee of the Bill?
Answer:
The Payee of a Bill is the person to whom the bill is made payable or in whose favour the bill is drawn.

Question 8.
What do you mean by Rebate?
Answer:
Any concession or discount in monetary terms given by the holder of the bill of exchange to the drawee or acceptor, when a bill is retired is called a Rebate.

Question 9.
What is the Legal Due Date?
Answer:
The date which is arrived at after adding three days of grace to the nominal due date is known as Legal Due Date.

Question 10.
What are Bills Payable on Demand?
Answer:
When the amount of bill is payable by a drawee on the presentation of a bill, in which time period is not mentioned and grace days are not allowed is known as Bills Payable on Demand.

G. Do you agree or disagree with the following statements:

Question 1.
A bill of exchange is a conditional order.
Answer:
Disagree

Question 2.
The party which is ordered to pay the amount is known as the payee.
Answer:
Disagree

Question 3.
The person in whose favour the bill is endorsed is known as the endorsee.
Answer:
Agree

Question 4.
Rebate or discount given on retiring a bill is an income to the Drawee.
Answer:
Agree

Question 5.
A bill from the point of view of the debtor is called Bills payable.
Answer:
Agree

Question 6.
In case of bill drawn payable ‘on demand,’ no grace days are allowed.
Answer:
Agree

Question 7.
A bill is required to be accepted by Drawer.
Answer:
Disagree

Question 8.
A bill of exchange need not be dated.
Answer:
Disagree

Question 9.
A bill before acceptance is called Promissory Note.
Answer:
Disagree

Question 10.
Renewal is requested by the drawee to extend the credit period of the bill.
Answer:
Agree

H. Calculations:

Question 1.
Ganesh draws a bill for ₹ 40,260 on 15th Jan. 2020 for 50 days. He discounted the bill with the Bank of India @ 15 % p.a. on the same day. Calculate the amount of discount.
Solution:
Discount = Amount of Bill × \(\frac{\text { Rate }}{100} \times \frac{\text { Unexpired days }}{366}\)
= 40,260 × \(\frac{15}{100} \times \frac{50}{366}\)
= ₹ 825
(Note: 2020 is a Leap year, so the total number of days = 366)

Question 2.
Shefali Traders drew a bill on Maya for ₹ 30,000 on 1st Oct. 2019 payable after 3 months.
Calculate the amount of discount in the following cases:
(i) Shefali Traders discounted the bill on the same day @ 12 % p.a.
(ii) Shefali Traders discounted the bill on 1st Nov. 2019 @ 12 % p.a.
(iii) Shefali Traders discounted the bill on 1st Dec. 2019 @ 12 % p.a.
Solution:
Discount = Amount of Bill × \(\frac{\text { Rate }}{100} \times \frac{\text { Unexpired days }}{365}\)
(i) Discount = 30,000 × \(\frac{12}{100} \times \frac{3}{12}\) = ₹ 900
(ii) Discount = 30,000 × \(\frac{12}{100} \times \frac{2}{12}\) = ₹ 600
(iii) Discount = 30,000 × \(\frac{12}{100} \times \frac{1}{12}\) = ₹ 300

Question 3.
Veena who had accepted Sudha’s bill for ₹ 28,000 was declared bankrupt and only 35 paise in a rupee could be recovered from her estate. Calculate the amount of bad debts.
Solution:
From Veena, only 35 paise in a rupee could be recovered i.e. 65 paise in a rupee is bad debt for Sudha. So 65% of ₹ 28,000 = ₹ 18,200 is the amount of bad debts.

Question 4.
Nitin renewed his acceptance for ₹ 72,000 by paying ₹ 22,000 in cash and accepting a new bill for the balance plus interest @ 18%. p.a. for 4 months. Calculate the amount of the new bill.
Selution:
For Nitin,
Total outstanding = ₹ 72,000
Nitin paid in cash= ₹ 22,000
Remaining dues = ₹ 50,000
Now, on this ₹ 50,000 we have to calculate interest @ 18% for 4 months
I = \(\frac{\mathrm{PRN}}{100}\)
= 50,000 × \(\frac{18}{100} \times \frac{4}{12}\)
= ₹ 3,000
So, amount of new bill = Remaining dues + Interest
= 50,000 + 3,000
= ₹ 53,000

Question 5.
Nisha’s acceptance for ₹ 16,850 sent to the bank for the collection was honoured and bank charges debited were ₹ 125. Find out the amount actually received by Drawer.
Solution:
Bill of ₹ 16,850 sent to the bank for collection and it is honoured and bank charges = ₹ 125
So, actual amount received by drawer = 16,850 – 125 = ₹ 16,725.

Question 6.
A bill of ₹ 16,000 was drawn by Keshav on Gopal on 12th June 2019 for 2 months, what will be the due date, if all of sudden, the legal due date is declared as an emergency holiday?
Solution:
Consider immediate or next working day as the due date in case the legal due date is declared as an emergency holiday.

∴ The legal due date is 16th August 2019 (The next day).

I. Prepare the following specimens:

Question 1.
Prepare a bill of exchange from the following information:
Drawer: Shankar, Vadodara, Gujarat
Drawee: Vinayak, Somwar Peth, Pune
Amount: ₹ 16,000
Period: 3 months
Date of Bill: 6th Sept. 2019
Date of acceptance: 11th Sept. 2019
Solution:

Question 2.
Prepare a bill of exchange from the following information:
Drawer: Dinesh, P. R. Road, Andheri (West)
Drawee: Mahesh, L. B. S. Road, Mulund
Payee: Amit, Thane (West)
Amount: ₹ 9,500
Period of Bill: 4 months after sight
Date of Bill: 26th Nov. 2019
Date of acceptance: 29th Nov. 2019
Solution:

Question 3.
Kantilal, 343/D, Palm Heights, Jogeshwari, drew a bill on 10th Oct. 2019 for ₹ 63,490 for 45 days after the date on Shantilal, B2, Himalaya Towers, Baramati, payable to Priyanka, Satara. The bill was accepted on 13th Oct. 2019 for 60 days.
Prepare a format of bill of exchange from the above details.
Solution:

Question 4.
Prepare a format of bill exchange from the following:
Rahul Sane, 86-D, Raviwar Peth, Nagpur accepted the bill drawn on him by Prithviraj, Icon Heights, Wardha for ₹ 87,000 on 30th July 2019.
The bill was drawn on 26th July 2019 for ₹ 1,00,000 for 90 days after the date.
Solution:

Question 5.
Prepare a format of bill of exchange from the following.
Drawer: Kashmira Shah, Partner M/S Shah, and Shah, 2 – C, Matruchhaya Building, Akola
Drawee: Dhanashree Traders, Bangalore Road, Belgaum (Signed by Jayshree, Partner)
Payee: M/S Janki Traders, Akola
Amount: ₹ 64,500
Period of Bill: 3 months
Date of drawing: 12th Sept. 2019
Date of acceptance: 15th Sept. 2019
Solution:

Question 6.
Prepare a format Bill of Exchange with imaginary Drawer, Drawee, Address, Amount, Dates.
Drawer: Dhanesh Shah, 24/c, Amir Mahal, Borivali, Mumbai
Drawee: Kalpana Shah, 33, Sharadashram, Dadar (West), Mumbai
Amount: ₹ 80,500
Period: 60 days
Date of the bill: 2nd December 2020
Accepted on: 5th December 2020
Solution:

J. Complete the following Table.

Question 1.

Answer:

Question 2.

Answer:

S.No.	Date of Drawing	Date of Acceptance	Tenure	Type	Nominal due Date	Legal due Date
(i)	3rd January, 2020	5th January, 2020	45 days	after date	17th Feb. 2020	20th Feb. 2020
(ii)	9th April, 2019	12th April, 2019	4 months	after sight	12th Aug. 2019	14th Aug. 2019
(iii)	23rd November, 2019	23rd November, 2019	2 months	after date	23rd Jan. 2020	25th Jan. 2020
(iv)	16th August, 2019	20th August, 2019	4 months	after sight	20th Dec. 2019	23rd Dec. 2019
(v)	23rd December, 2018	24th December, 2018	60 days	after date	21st Feb. 2019	24th Feb. 2019

Practical Problems

Question 1.
On 1st Jan., 2020 Hemant sold goods of ₹ 18,500 to Nitin. On the same date Hemant drew a bill of exchange for ₹ 18,500 at 2 months. On the due date the bill was duly honoured.
Give Journal Entries in the books of Hemant and Nitin. Prepare Hamant’s Account in the books of Nitin.
Solution:
In the books of Hemant
Journal Entries

In the books of Nitin
Journal Entries

Question 2.
Neha sold goods to Rohan ₹ 42,000 on 6th Sept. 2019. Neha drew a bill of exchange at 3 months for the amount which was accepted by Rohan. Neha discounted the bill with her bankers at ₹ 41,000. On the due date of the bill Rohan dishonoured the bill and bank paid ₹ 300 as Noting Charges.
Show Journal Entries in the books of Neha and Rohan.
Solution:
In the books of Neha
Journal Entries

In the books of Rohan
Journal Entries

Question 3.
Jyoti owes ₹ 31,000 to Swati for which she draws a bill on Jyoti for 2 months. The bill was duly accepted by Jyoti. Swati sends the bill to bank for collection. Jyoti honoured the bill on the due date and bank charges ₹ 475 as bank charges.
Give Journal Entries in the books of Swati.
Solution:
In the books of Swati
Journal Entries

Question 4.
Pankaj purchased goods of ₹ 20,000 from Omprakash on credit on 15th April, 2019. Omprakash draws After Sight bill for the amount due on Pankaj for 3 months which was accepted by Pankaj on 18th April, 2019. On 20th April, 2019 Omprakash endorsed the bill to his creditor Jagdish in full settlement of his amount ₹ 21,000. On the due date the bill was dishonoured by Pankaj.
Give Journal Entries in the books of Omprakash, Pankaj and Jagdish.
Solution:
In the books of Omprakash
Journal Entries

In the books of Pankaj
Journal Entries

In the books of Jagdish
Journal Entries

Question 5.
Siddhant sold goods to Sudhir of ₹ 43,800 on 18th March, 2019. Siddhant draws a bill on Sudhir on the same day for ₹ 43,800 for 3 months which was duly accepted by Sudhir. Siddhant discounted the bill on the same day at 8% p.a. The bill was dishonoured on the due date and Sudhir requested Siddhant to accept ₹ 13,800 and interest in cash on remaining amount at 12% p. a. Siddhant agreed and for the balance amount accepted a new bill at 2 months. Before the due date of new bill Sudhir retired the bill by paying ₹ 29,700.
Pass necessary Journal Entries in the books of Siddhant.
Solution:
In the books of Siddhant
Journal Entries

Working Notes:
1. March 18, Discount = 43,800 × \(\frac{3}{12} \times \frac{8}{100}\) = ₹ 876

2. March 21, calculation of interest balance amount:
I = \(\frac{\mathrm{PRN}}{100}\)
= 30,000 × \(\frac{12}{100} \times \frac{2}{12}\) (for 2 months on remaining amount ₹ 30,000)
= ₹ 600

3. Before due date bill was retired by Sudhir by paying ₹ 300 less which is considered as discount and as date is not given, here it is not recorded.

Question 6.
Sangeeta accepted a bill for ₹ 18,000 drawn by Geeta at 3 months. Geeta discounted the bill for ₹ 17,400. Before the due date Sangeeta approached Geeta for renewal of the bill. Geeta agreed on the condition that Sangeeta should pay ₹ 6,000 immediately and for the balance she should accept a new bill for 4 months along with interest ₹ 550. The arrangements were carried through. But on the due date of new bill Sangeeta became insolvent and 35 paise in a rupee could be recovered from her estate.
Give Journal Entries in the books of Sangeeta and prepare Sangeeta’s Account in the books of Geeta.
Solution:
In the books of Sangeeta
Journal Entries

Working Notes:
1. It is advisable to write journal entries in the books of Geeta also to get entries in ‘Sangeeta’s Account’ property.
In the books of Geeta
Journal Entries

Question 7.
Priyanka owed Meena ₹ 18,000, Priyanka accepted a bill drawn by Meena for the amount at 4 months. Meena endorsed the same bill to Sagar. Before due date Priyanka approached Meena for renewal of bill. Meena agreed on condition that ₹ 6,000 be paid immediately together with interest on the remaining amount of 8% p.a. for 3 months and Priyanka should accept a new bill for the balance amount. These arrangements were carried through. However, before the due date Priyanka became insolvent and only 50% of the amount could be recovered from her estate.
Give Journal Entries in the books of Meena.
Solution:
In the books of Meena
Journal Entries

Working Note:
Calculation of interest on remaining amount ₹ 12,000 @ 8 % p.a. and for 3 months
I = \(\frac{\mathrm{PRN}}{100}\)
= 12,000 × \(\frac{8}{100} \times \frac{3}{12}\)
= ₹ 240

Question 8.
Seema purchased goods from Roma on credit on 1st August, 2019 for ₹ 37,000. Seema accepts bill for 2 months drawn by Roma for the same amount. On the same day, Roma discounts the bill with the bank for ₹ 36,200 on 3rd August, 2019. On the due date the bill is dishonoured and Noting Charges of ₹ 160 is paid by the bank. Seema pays ₹ 19,000 and Noting Charges in cash immediately. A new bill is drawn by Roma for the balance including interest ₹ 650 for 2 months, which is accepted by Seema. The new bill is retired one month before the due date at a rebate of ₹ 300.
Give Journal Entries in the books of Seema and prepare Seema’s Account in the books of Roma.
Solution:
In the books of Seema
Journal Entries

Question 9.
Uday purchased goods from Shankar on credit for ₹ 35,000 at 10 % trade discount. Uday paid ₹ 1,500 immediately and for the balance accepted a bill for 3 months. Before due date Uday approached Shankar with a request to renew the bill. Shankar agreed but with condition that Uday should accept a new bill for 3 months including interest at 12% p.a.
Give Journal Entries in the books of Shankar.
Solution:
In the books of Shankar
Journal Entries

Working Note:
I = \(\frac{\text { PRN }}{100}\)
= 30,000 × \(\frac{3}{12} \times \frac{12}{100}\)
= ₹ 900

Question 10.
Sagar drawn an after sight bill on 21st Nov., 2019 for ₹ 21,000 at 3 months on Prasad. The bill is discounted by Sagar at 8% p.a. with his bank. On maturity. Prasad finds himself unable to make payment of the bill and requests Sagar to renew it. Sagar accepts the request and draws a new bill at one month for ₹ 21,750 including interest which was duly accepted by Prasad. Sagar deposits the bill into bank for the collection. Prasad honours the bill on the due date and bank charges ₹ 250 as bank charges.
Pass necessary Journal Entries in the books of Sagar and prepare Sagar’s Account in the books of Prasad.
Solution:
In the books of Sagar
Journal Entries

Question 11.
Journalise the following transaction in the books of Abhishek:
(a) Siddhant informs Abhishek that Vineet’s acceptance for ₹ 23,000 endorsed to Siddhant has been dishonoured. Noting Charges amounted to ₹ 430.
(b) Kajal renews her acceptance to Abhishek for ₹ 39,000 by paying ₹ 3,000 in cash and accepting a fresh bill for the balance along with interest at 11.5% p.a. for 3 months.
(c) Radhika retired her acceptance to Abhishek for ₹ 23,000 by paying ₹ 22,250 by cheque.
(d) Abhishek sent a bill of Subodh for ₹ 9,000 to bank for collection. Bank informed that the bill has been dishonoured by Subodh.
Solution:
In the books of Abhishek
Journal Entries

Working Note:
Amount of interest = 36,000 × \(\frac{3}{12} \times \frac{11.5}{100}\) = ₹ 1,035.

Question 12.
Journalise the following transaction in the books of Narendra:
(a) Narendra retires his acceptance to Upendra by paying ₹ 4,000 in cash and endorsing a bill accepted by Ramlal for ₹ 5,000.
(b) Vikram’s acceptance to Narendra ₹ 6,000 retired one month before the due date at rebate of 12% p.a.
(c) Dilip renews his acceptance to Narendra for ₹ 12,000 by paying ₹ 4,000 in cash and accepting a fresh bill for the balance plus interest at 12% p.a. for 3 months.
(d) Bank informed Narendra that, Kartik’s acceptance for ₹ 13,000 to Narendra, discounted with the bank was dishonoured and Noting Charges paid by bank ₹ 140.
Solution:
In the books of Narendra
Journal Entries

Question 13.
Journalise the following transaction in the books of Bharti:
(a) Bank informed that Amit’s acceptance for ₹ 15,750 sent to bank for collection was honoured and bank charges debited were ₹ 150.
(b) Nitin renewed his acceptance for ₹ 22,200 by paying ₹ 2,200 in cash along with interest on balance amount at 10% and accepted a fresh bill for the balance for 3 months.
(c) Dhanshri who had accepted Bharti’s bill for ₹ 17,500 was declared insolvent and only 40% of the amount due could be recovered from her estate.
(d) Discharged our acceptance to Savita for ₹ 9,450 by endorsing Pravin’s acceptance to us ₹ 9,000.
Solution:
In the books of Bharti
Journal Entries

Question 14.
Journalise the following transaction in the books of Sudha:
(a) Endorsed Sonali’s acceptance at 2 months for ₹ 6,000 in favour of Urmila and paid cash ₹ 3,500 in full settlement of her account ₹ 10,000.
(b) Discounted 2 months acceptance of Surya for ₹ 7,800 with bank at 10% p.a.
(c) Bank informed that Anuradha’s acceptance of ₹ 4,800 which was discounted was dishonoured and bank paid Noting Charges ₹ 125.
(d) Pooja honoured her acceptance of ₹ 16,400 which was deposited into bank for collection.
Solution:
In the books of Sudha
Journal Entries

Question 15.
Journalise the following transaction in the books of Mrunal:
(a) Bank informed that Aishwarya’s acceptance of ₹ 24,000 which was discounted had been dishonoured and bank paid Noting Charges ₹ 220. Bill was renewed at the request of Aishwarya for 2 months with interest of ₹ 480.
(b) Received ₹ 4,630 from private estate of Ankur who was declared insolvent against bill accepted by him for ₹ 6,000.
(c) Accepted a bill of ₹ 15,000 at 3 months drawn by Anushka for the amount due to her ₹ 20,000 and balance paid by cheque.
(d) Dishonoured our acceptance to Vivek ₹ 27,000 and Noting Charges paid by Vivek ₹ 700.
Solution:
In the books of Mrunal
Journal Entries

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 8 Respiration and Circulation Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 8 Respiration and Circulation

1. Multiple choice questions

Question 1.
The muscular structure that separates the thoracic and abdominal cavity is …………………..
(a) pleura
(b) diaphragm
(c) trachea
(d) epithelium
Answer:
(b) diaphragm

Question 2.
What is the minimum number of plasma membrane that oxygen has to diffuse across to pass from air in the alveolus to haemoglobin inside a R.B.C.?
(a) two
(b) three
(c) four
(d) five
Answer:
(a) two

Question 3.
…………………. is a sound producing organ.
(a) Larynx
(b) Pharynx
(c) Tonsils
(d) Trachea
Answer:
(a) Larynx

Question 4.
The maximum volume of gas that is inhaled during breathing in addition to T.V. is …………………..
(a) residual volume
(b) IRV
(c) GRV.
(d) vital capacity
Answer:
(b) IRV

Question 5.
………………….. muscles contract when the external intercostals muscles contract.
(a) Internal abdominal
(b) Jaw
(c) Muscles in bronchial walls
(d) Diaphragm
Answer:
(d) Diaphragm

Question 6.
Movement of cytoplasm in unicellular organisms is called …………………..
(a) diffusion
(b) cyclosis
(c) circulation
(d) thrombosis
Answer:
(b) cyclosis

Question 7.
Which of the following animals do not have closed circulation?
(a) Earthworm
(b) Rabbit
(c) Butterfly
(d) Shark
Answer:
(c) Butterfly

Question 8.
Diapedesis is performed by …………………..
(a) erythrocytes
(b) thrombocytes
(c) adipocytes
(d) leucocytes
Answer:
(d) leucocytes

Question 9.
Pacemaker of heart is …………………..
(a) SA node
(b) AV node
(c) His bundle
(d) Purkinje fibers
Answer:
(a) SA node

Question 10.
Which of the following is without nucleus?
(a) Red blood corpuscle
(b) Neutrophil
(c) Basophil
(d) Lymphocyte
Answer:
(a) Red blood corpuscle

Question 11.
Cockroach shows which kind of circulatory system?
(a) Open
(b) Closed
(c) Lymphatic
(d) Double
Answer:
(a) Open

Question 12.
Diapedesis can be seen in …………………..
(a) RBC
(b) WBC
(c) Platelet
(d) neuron
Answer:
(b) WBC

Question 13.
Opening of inferior vena cava is guarded by …………………..
(a) bicuspid valve
(b) tricuspid valve
(c) Eustachian valve
(d) Thebesian valve
Answer:
(c) Eustachian valve

Question 14.
…………………. wave in ECG represent atrial depolarization.
(a) P
(b) QRS complex
(c) Q
(d) T
Answer:
(a) P

Question 15.
The fluid seen in the intercellular spaces in Human is …………………..
(a) blood
(b) lymph
(c) interstitial fluid
(d) water
Answer:
(b) lymph

2. Match the columns

Question 1.
Respiratory surface Organism

Respiratory surface	Organism
(1) Plasma membrane	(a) Insect
(2) Lungs	(b) Salamander
(3) External gills	(c) Bird
(4) Internal gills	(d) Amoeba
(5) Trachea	(e) Fish

Answer:

Respiratory surface	Organism
(1) Plasma membrane	(d) Amoeba
(2) Lungs	(c) Bird
(3) External gills	(b) Salamander
(4) Internal gills	(e) Fish
(5) Trachea	(a) Insect

3. Very Short Answer Questions

Question 1.
Why does trachea have ‘C’-shaped rings of cartilage?
Answer:
Trachea is supported by ‘C’-shaped rings of J cartilage which prevent it from collapsing and always keep it open.

Question 2.
Why is respiration in insect called direct respiration?
Answer:
Respiration in insect is called direct because tracheal tubes exchange O₂ and CO₂ directly with the haemocoel which then exchange them with tissues.

Question 3.
Why is gas exchange very rapid at alveolar level?
OR
Why does gas exchange in the alveolar region very rapid?
Answer:
Gas exchange is very rapid at alveolar level because numerous alveoli (about 700 millions) in the lungs provide large surface area for gaseous exchange.

Question 4.
Name the organ which prevents the entry of food into the trachea while eating.
Answer:
Epiglottis prevents the entry of food into trachea while eating.

4. Short Answer Questions

Question 1.
Why is it advantageous to breathe through the nose than through the mouth?
Answer:
Breathing through nose is better than breathing through the mouth because of the following reasons:

1. The nostrils are smaller than the mouth so air exhaled through the nose creates a backflow of air into the lungs.
2. As we exhale more slowly through the nose than we do through the mouth, the lungs have more time to extract oxygen from the air that we have already taken in.
3. The hairs inside nostrils filter any dust particles and microbes in the air and it only lets the clean air pass through.
4. The air gets warm and humidified in nostrils as it passes into our bodies.
5. Moreover breathing through the mouth can dry the oral cavity and lead to bad breath, gum disease and tooth decay.

Question 2.
Identity the incorrect statement and correct it.
(a) A respiratory surface area should have a. large surface area.
(b) A respiratory surface area should be kept dry.
(c) A respiratory surface area should be thin, may be 1 mm or less.
Answer:
Statement (a) and statement (c) are correct whereas statement (b) is incorrect. A respiratory surface area should be kept moist, is the correct statement.

Question 3.
Given below are the characteristics of some modified respiratory movement. Identify them.
a. Spasmodic contraction of muscles of expiration and forceful expulsion of air through nose and mouth.
Answer:
Sneezing

b. An inspiration followed by many short convulsive expiration accompanied by facial expression.
Answer:
Laughing, Crying.

Question 4.
Blood plasma.
Answer:

1. Plasma is a straw coloured, slightly alkaline viscous fluid part of the blood, having 90-92% water and 8-10% soluble proteins.
2. Serum albumin, serum globulin, heparin, fibrinogen and prothrombin are the plasma proteins which form 7% of the plasma.
3. Glucose, amino acids, fatty acids and glycerol are the nutrients dissolved in plasma.
4. Nitrogenous wastes (urea, uric acid, . ammonia and creatinine) and respiratory gases (oxygen and carbon dioxide) is present in plasma.
5. Enzymes and hormones too are transported Ada plasma.
6. Inorganic minerals are also present in plasma such as bicarbonates, chlorides, phosphates and sulphates of sodium, potassium, calcium and magnesium.

Question 5.
Blood clotting/Coagulation of blood.
OR
Explain blood clotting in short.
Answer:

1. The process of converting the liquid blood into a semisolid form is called blood clotting or coagulation.
2. The process of clotting may be initiated by contact of blood with any foreign surface (intrinsic process) or with damaged tissue (extrinsic process).
3. Intrinsic and extrinsic processes involve interaction of various substances called clotting factors by a step wise or cascade mechanism.
4. There are in all twelve clotting factors numbered as I to XII (factor VI is not in active use).
5. Interaction of these factors in a cascade manner leads to formation of enzyme, Thromboplastin which helps in the formation of enzyme prothrombinase.
6. Prothrombinase inactivates heparin and also converts inactive prothrombin into active thrombin.
7. Thrombin converts soluble blood protein- fibrinogen into insoluble fibrin. Fibrin forms a mesh in which platelets and other blood cells are trapped to form the clot.
8. These reactions occur in 2 to 8 minutes. Therefore, clotting time is said to be 2 to 8 minutes.

Question 6.
Describe pericardium.
Answer:

1. Pericardium is the double layered peritoneum that encloses the heart. It consists of two layers, viz. fibrous pericardium and serous pericardium.
2. Fibrous pericardium is the outer layer having tough, inelastic fibrous connective tissue whereas serous pericardium is the v inner double layered membrane. It has in turn an outer parietal layer and inner visceral layer.
3. Parietal layer of serous pericardium lies on the inner side of fibrous pericardium.
4. Visceral layer also known as epicardium adheres to heart and thus forms outer covering over the heart.
5. There is a pericardial fluid in the pericardial space which is present in between the parietal and visceral layers of serous pericardium.

Question 7.
Describe valves in the human heart.
Answer:
Human heart has following main valves:

1. Tricuspid valve : Tricuspid valve is present between the right atrium and right ventricle. It has three cusps or flaps. It prevents the backflow of blood into right atrium.
2. Bicuspid valve : Bicuspid valve, also called mitral valve is present between the left atrium and left ventricle. It has two flaps. It prevents the backflow of blood in left atrium. Both tricuspid and bicuspid valves are attached to papillary muscles with tendinous chords or chordate tendinae to prevent valves from turning back into atria at the time of systole.
3. Semilunar valve : These are present at the opening of pulmonary artery and systemic aorta. They prevent the back flow of blood when ventricles undergo systole.
4. Thebesian valve : Thebesian valve is present at the opening of coronary sinus.
5. Eustachian valve : Eustachian valve is present at the opening of inferior vena cava.

Question 8.
What is the role of papillary muscles and chordae tendinae in human heart?
Answer:

1. Papillary muscles are large and well- developed muscular ridges present along the inner surface of the ventricles.
2. Bicuspid and tricuspid valves are attached to papillary muscles of ventricles by chordae tendinae.
3. Chordae tendinae are inelastic fibres present in the lumen of ventricles.
4. The chordae tendinae prevent the valves from turning back into the atria during the contraction of ventricles and regulate the opening and closing of bicuspid and tricuspid valves.

Question 9.
Explain in brief the factors affecting blood pressure.
Answer:

1. Cardiac output : Normal cardiac output is 5 lit/min. Increase in cardiac output increases systolic pressure.
2. Peripheral resistance : Peripheral resistance depends upon the diameter of blood vessels. Decrease in diameter of arterioles and capillaries under the effect of vasopressin cause increase in peripheral resistance and thereby increase in blood pressure.
3. Blood volume : Loss of blood in accidents decreases blood volume and thus cause decrease in blood pressure.
4. Viscosity of blood : Blood pressure is directly proportional to viscosity of blood.
5. Age : Blood pressure increases with age due to increase in inelasticity of blood vessels.
6. Venous return : Amount of blood brought to the heart via the veins per unit time is called the venous return and it is directly proportional to blood pressure.
7. Length and diameter of blood vessels : Blood pressure is directly proportional to the total length of the blood vessel. Blood pressure can also be affected by vasoconstriction or vasodilation.
8. Gender : Females have slightly lower BP than males of her age before menopause. However, the risk of high B. P increases in the females after menopause sets in.

5. Give Scientific Reason

Question 1.
Closed circulation is more efficient than open circulation.
Answer:

1. Closed circulation considerably enhances the speed, precision and efficiency of circulation.
2. The blood flows more rapidly, it takes less time to circulate through the closed system and return to the heart.
3. This fastens the supply and removed of materials to and from the tissues by the blood as compared to open circulation.
4. In open circulation, there are no blood vessels such as arteries or veins, to pump the blood.
5. Therefore, the blood pressure is very low.
6. Organisms with an open circulatory system typically have a relatively high volume of hemolymph and low blood pressure. Closed circulation is thus more efficient than open circulation.

Question 2.
Human heart is called as myogenic and autorhythmic?
Answer:

1. The heart shows auto rhythmicity because the impulse for its rhythmic movement develops inside the heart. Such heart is called myogenic.
2. Some of the cardiac muscle fibres become auto rhythmic (self-excitable) and start generating impulse during development.
3. These autorhythmic fibres perform two important function, viz. acting as a pacemaker and setting the rhythm for heart.
4. They also form conducting system for conduction of nerve impulses throughout the heart muscles.

Question 3.
In human heart, the blood flows only in one direction.
Answer:

1. In veins there are valves, which prevent the back flow of the blood.
2. In arteries, blood flows with unidirectional pressure.
3. Hence the circulation takes place only in one direction.

Question 4.
Arteries are thicker than veins.
Answer:

1. Arteries have relatively thick walls to enable them to withstand the high pressure of blood ejected from the heart.
2. Arteries expand when the pressure increases as the heart pushes blood out but then recoil (shrink) Wn the pressure decreases when the heart relaxes between heartbeats.
3. This expansion and recoiling occurs to maintain a smooth blood flow.
4. Veins, on the other hand, have thinner walls and larger lumen veins have no need for thick walls as then need not have to withstand high pressure like arteries.
5. Moreover, as veins transport relatively low pressure blood, they are commonly equipped with valves to promote the unidirectional flow of blood towards the heart.

Question 5.
Left ventricle is thick than all other chambers of heart.
OR
Left ventricle has thicker wall than the right ventricle.
Answer:

1. Left ventricle pumps oxygenated blood to all parts of the body. Therefore, there is greater pressure from the blood in left ventricle.
2. Right ventricle sends deoxygenated blood to lungs for oxygenation. This does not put more pressure and lungs are in vicinity of the heart.
3. Due to these functional differences between the two ventricles, left ventricle has thicker wall than that of the right ventricle.

6. Distinguish Between

Question 1.
Open circulation and Closed circulation
Answer:

Open circulation	Closed circulation
1. In open circulation, blood flows through large open spaces and channels called lacunae and haemocoels among the tissues.	1. In closed circulation, blood flows through a network of blood vessels all over the body.
2. Tissues are in direct contact with the blood.	2. Blood does not come in direct contact with tissue.
3. Blood flows with low pressure and usually does not contain any respiratory pigment like haemoglobin.	3. Blood flows with high pressure and contains respiratory pigment like haemoglobin.
4. Exchange of material takes place directly between blood and cells or tissues of the body.	4. Exchange of material takes place between blood and body tissues through an intermediate fluid called lymph.
5. Volume of blood flowing through a tissue cannot be controlled as blood flows out in open space.	5. Volume of blood can be regulated by the contraction and relaxation of the smooth muscles of the blood vessels.
6. Open circulatory system is found in arthropods and some molluscs.	6. Closed circulatory system is found in annelids, echinoderms and all vertebrates.

Question 2.
Arteries and veins.
Answer:

Arteries	Veins
1. The blood vessels that arise from the heart and carry blood away from heart are called arteries.	1. The blood vessels that bring blood to the heart are called veins.
2. Arteries are thick walled blood vessels, situated in deep layers in the body.	2. Veins cure thin walled blood vessels, situated superficially in the body.
3. Arteries do not have valves.	3. Veins have valves.
4. Tunica adventitia, the outermost layer of arteries is thick and elastic.	4. Tunica externa, the outermost layer of veins is thin.
5. Tunica media is very thick and contain elastic fibres.	5. Tunica media is thin layer and contain involuntary muscle fibres.
6. The lumen of arteries is small.	6. The lumen of the veins is very spacious.
7. With the exception of pulmonary arteries, all other arteries carry oxygenated blood.	7. With the exception of pulmonary veins, all other veins carry deoxygenated blood.
8. Blood in the arteries show high blood pressure.	8. Blood in the veins show lesser blood pressure.

Question 3.
Blood and Lymph.
Answer:

Blood	Lymph
1. Contains blood plasma with proteins and all three types of blood cells namely RBCs, WBCs and blood platelets.	1. Contains blood plasma without blood proteins, RBCs and platelets and contains lymphocytes.
2. Red in colour due to presence of RBCs.	2. Light yellow in colour and does not contain RBCs.
3. Carries oxygen in the body.	3. Does not carry oxygen.
4. The flow of blood in blood vessels is fast.	4. The flow of lymph in lymph capillaries is slow.
5. Lymphocytes are present.	5. Lymphocytes are present, more in number than those present in the blood.

Question 4.
Blood capillary and Lymph capillary.
Answer:

Blood capillary	Lymph capillary
1. Reddish, easy to observe.	1. Colourless, difficult to observe.
2. Joined to arterioles at one end and to venules at another end.	2. Blind (closed at the tip).
3. Narrower than lymph capillaries.	3. Wider than blood capillaries.
4. Wall consists of normal endothelium and basement membrane.	4. Wall consists of thin endothelium and poorly developed basement membrane.
5. Contains red blood.	5. Contains colourless lymph.
6. Have relatively high pressure.	6. Have relatively low pressure.

Question 5.
Intrinsic and Extrinsic process of clotting.
Answer:

Intrinsic process	Extrinsic process
1. The intrinsic pathway requires only clotting factors found within the blood itself – in particular, clotting factor XII (Hageman factor) from the platelets.	1. The extrinsic pathway is initiated by factors external to the blood, in the tissues adjacent to damaged blood vessel – in particular, it is initiated by clotting factor III, thromoboplastin from the damaged tissues.
2. It is a longer, multistep process and it takes a little longer for the blood to clot by this mechanism.	2. It involves fewer chemical reaction steps and produce a clot a little more quickly than the intrinsic pathway.

7. Long Answer Questions

Question 1.
Smita was working in a garage with the doors closed and automobiles engine running. After some time she felt breathless and fainted. What would be the reason? How can she be treated
OR
While working with the car engine in a closed garage, John suddenly felt dizzy and fainted what is the possible reason?
Answer:

1. As Smita and John were working with the car engine running in a closed garage, they must be suffering from carbon monoxide poisoning.
2. Carbon monoxide (CO) is a highly toxic gas produced when fuels burn incompletely from automobile engines.
3. Because of strong affinity of haemoglobin with carbon monoxide, it readily combines with carbon monoxide to from a stable compound, carboxyhaemoglobin. Thus, less haemoglobin is available for oxygen transport depriving the cells of oxygen.
4. Exposure to carbon monoxide can usually leads to throbbing headache, drowsiness, breathlessness and often person gets fainted. In extreme cases carbon monoxide poisoning usually leads to unconsciousness, convulsions, cardiovascular failure, coma and eventually death.

The breathless persons can be treated by following method:

1. Oxygen treatment : The best way to treat carbon monoxide poisoning is to breathe in pure oxygen (high-dose oxygen treatment)
2. Oxygen chamber : Doctor may temporarily place her in a pressurized oxygen chamber (also known as a hyperbaric oxygen chamber)

Question 2.
Shreyas went to a garden on a wintry morning. When he came back, he found it difficult to breath and stated wheezing. What could be the possible condition and how can he be treated?
Answer:
(1) It indicates that Shreyas might be suffering from allergic reactions. He may have come in contact with allergens such as pollen, dust, pet dander or other environmental substances on his way in the garden. Or Shreyas may be already a patient of Asthma and his symptoms may have aggrevated due to wintry climate.

(2) If a person is allergic to a substance, such as pollen, his immune system reacts to the substance as if it was foreign and harmful, and tries to destroy it.

(3) The body reacts to these allergens by making and releasing substances known as IgE antibodies. These IgE antibodies attach to most cells in the body which release histamine. Histamine is the main substance responsible for pollen allergy symptoms such as difficulty in breathing, wheezing, sneezing, itchy throat, etc.

(4) Treatment : There are several drugs to treat the allergic reactions:

• Antihistamines such as cetirizine or diphenhydramine.
• Decongestants, such as pseudoephedrine or oxymetazoline.
• Medications that combine an antihistamine and decongestant such as Actifed and Claritin-D.

Question 3.
Why can you feel a pulse when you keep a finger on the wrist or neck but not when you keep them on a vein?
Answer:
(1) When the heart contracts, it creates pressure that pushes blood out of heart. This pressure acts like a wave. This “wave” of pressure is the pulse you feel. But this pressure is not constant.

(2) When the heart pumps the blood out of it at the time of systole, there is maximum pressure in the arteries. This pressure weakens considerably when it reaches capillaries, and so the veins which are away from the heart are under less pressure. Due to low pressure veins have valves to prevent backflow of blood.

(3) The pressure in the arteries can be felt every time the heart beats, especially in arteries which come to surface of the body like that of the wrist and neck but not in veins.

(4) The pressure in veins is always weaker than in arteries, resulting in a weaker pulse to the point that it is undetectable by touch
alone.

(5) Owing to this, when we keep finger on the arteries of wrist or neck, we feel a pulse but not when we keep it on a vein.

Question 4.
A man’s pulse rate is 68 and cardiac output is 5500 cm3. Find the stroke volume.
Answer:
Cardiac output is the volume of blood pumped out per min for a normal adult human being it is calculated as follows:
Cardiac output = Heart rate × Stroke volume
Given : Cardiac output = 5500 cm³
Pulse rate = Heart rate = 68
By using these values stroke volume of is calculated as follows:
∴ Cardiac output = Heart rate × Stroke volume
∴ Stroke volume = Cardiac output/Heart rate
= 5500/68
= Approx. 80. ∴ Stroke volume is 80 ml.

Question 5.
Which blood vessel leaving from the heart will have the maximum content of oxygen and why?
Answer:

1. The Aorta leaving the heart from left ventricle carry the maximum content of oxygen.
2. Deoxygenated blood becomes oxygenated in the pulmonary capillaries surrounding the alveoli of lungs. The oxygenated blood from lungs is collected by the four pulmonary veins.
3. These pulmonary veins carry that oxygenated blood to left atrium of heart. During atrial systole that blood is carried to left ventricle.
4. Left ventricle then pumps that oxygenated blood to Aorta during ventricular systole. Therefore, aorta has the maximum content of oxygen.

Question 6.
If the duration of the atrial ‘systole is 0.1 second and that of complete diastole is 0.4 second, then how does one cardiac cycle complete in 0.8 second?
Answer:

1. The time duration required to complete one cardiac cycle is 0.8 second.
2. Cardiac cycle is divided into three important phases, viz, atrial systole, ventricular systole and joint diastole.
3. Atrial systole in normal condition lasts for 0.1 second, ventricular systole follows atrial systole and lasts for 0.3 second whereas joint diastole or complete diastole lasts for about 0.4 second.
4. In this way one cardiac cycle is completed in 0.8 second.

Question 7.
How is blood kept moving in the large veins of the legs?
Answer:
1. When heart undergoes systole, it pushes the blood with pressure in aorta. This pressure moves the entire circulation of the blood throughout the body. Aorta gives rise to dorsal aorta after supplying to upper parts of body. Then it divides into two arteries which enter two legs. The blood is forced to move in the legs due to blood pressure and also aided by gravity.

2. In addition, the muscles in legs help transport blood back to our heart. As the muscles of our body contract and relax to move our limbs, they squeeze the blood in veins and the blood is then pushed towards the heart.

3. The veins in legs also have valves to keep this process going and prevent blood from flowing back down towards the feet.

4. In this way blood is kept moving in the large veins of the legs.

Question 8.
Describe histological structure of artery, vein and capillary.

Answer:
Histological structure of artery and vein.

1. Artery is a thick walled blood vessel that carries oxygenated blood. (Exception is pulmonary artery which carries deoxygenated blood from heart to lungs for oxygenation.)
2. All the arteries arise from heart and carry blood away from the heart.
3. Each artery is made up of three layers, viz. tunica externa, tunica media and tunica interna.
4. Tunica externa or adventitia is the thickest layer of all. It is the outermost coat made up of connective tissue with elastic and collagen fibres.
5. Tunica media is the middle coat made up of smooth muscle fibres and elastic fibres. It withstands high blood pressure during ventricular systole. It is also thick.
6. Tunica interna or intima is the innermost coat made of endothelium and elastic layer.

Histology of Capillaries:

1. Capillaries are the smallest and thinnest blood vessels. Capillaries are formed by the division and re-division of the arterioles.
2. The wall of the capillary is made up of endothelium or squamous epithelium.
3. The capillary wall is permeable to water and dissolved substances.
4. Exchange of respiratory gases, nutrients, excretory products, etc. takes place through the capillary wall.
5. Capillaries unite to form venules.

Question 9.
What is blood pressure? How is it measured? Explain factors affecting blood pressure.
Answer:
1. Blood pressure:

1. The pressure exerted by blood on the wall of the blood vessels is called blood pressure. Pressure exerted by blood on the wall of arterial wall is arterial blood pressure. Blood pressure is described in two terms viz. systolic blood pressure and diastolic blood pressure.
2. Systolic blood pressure is the pressure exerted on arterial wall during ventricular contraction (systole). For a normal healthy adult the average value is 120 mmHg.
3. Diastolic blood pressure is the pressure on arterial wall during ventricular relaxation (diastole). For a normal healthy adult it is 80 mmHg.
4. B. E = SP/DP = 120/80 mmHg. Blood pressure is normally written as 120/80 mmHg. Difference between systolic and diastolic pressure is called pulse pressure normally, it is 40 mmHg.

2. Measurement of blood pressure:

1. Blood pressure is measured with the help of an instrument called sphygmomanometer.
2. The instrument consists of inflatable rubber bag cuff covered by a cotton cloth. It is connected with the help of tubes to a mercury manometer on one side and a rubber bulb on the other side.
3. During measurement, the person is asked to lie in a sleeping position. The instrument is placed at the level of heart and the cuff is tightly wrapped around upper arm.
4. The cuff is inflated till the brachial artery is blocked due to external pressure. Then pressure in the cuff is slowly lowered till the first pulsatile sound is produced. At this moment, pressure indicated in manometer is systolic pressure. Sounds heard during this measurement of blood pressure are called as Korotkoff sounds.
5. Pressure in the cuff is further lowered till any pulsatile sound cannot be heard due to smooth blood flow. At this moment, pressure indicated in manometer is diastolic pressure an optimal blood pressure (normal) level reads 120/80 mmHg.

3. Factors affecting blood pressure:

1. Cardiac output : Normal cardiac output is 5 lit/min. Increase in cardiac output increases systolic pressure.
2. Peripheral resistance : Peripheral resistance depends upon the diameter of blood vessels. Decrease in diameter of arterioles and capillaries under the effect of vasopressin cause increase in peripheral resistance and thereby increase in blood pressure.
3. Blood volume : Loss of blood in accidents decreases blood volume and thus cause decrease in blood pressure.
4. Viscosity of blood : Blood pressure is directly proportional to viscosity of blood.
5. Age : Blood pressure increases with age due to increase in inelasticity of blood vessels.
6. Venous return : Amount of blood brought to the heart via the veins per unit time is called the venous return and it is directly proportional to blood pressure.
7. Length and diameter of blood vessels : Blood pressure is directly proportional to the total length of the blood vessel. Blood pressure can also be affected by vasoconstriction or vasodilation.
8. Gender : Females have slightly lower BP than males of her age before menopause. However, the risk of high B. P increases in the females after menopause sets in.

Question 10.
Describe human blood and give its functions.
Answer:
Blood Composition:

1. Blood is a red coloured fluid connective tissue derived from embryonic mesoderm.
2. It has two components – the fluid plasma (55%) and the formed elements i.e. blood cells (44%).
3. Plasma is a straw coloured, slightly alkaline and viscous fluid having 90% water and 10% solutes such as proteins, nutrients, nitrogenous wastes, salts, hormones, etc.
4. Blood corpuscles are of three types, viz. erythrocytes (RBCs), white blood corpuscles (WBCs) and thrombocytes (platelets).
   

(5) Red blood corpuscles or Erythrocytes:

1. Erythrocytes or red blood corpuscles. They are circular, biconcave, enucleated cells.
2. The RBC size : 7 pm in diameter and 2.5 pm in thickness.
3. The RBC count : 5.1 to 5.8 million RBCs/ cu mm of blood in an adult male and 4.3 to 5.2 million/cu mm in an adult female.
4. The average life span of RBC : 120 days.
5. RBCs are formed by the process of erythropoiesis. In foetus, RBC formation takes place in liver and spleen whereas in adults it occurs in red bone marrow.
6. The old and worn out RBCs are destroyed in liver and spleen.
7. Polycythemia is an increase in number of RBCs while erythrocytopenia is decrease in their (RBCs) number.

Functions of RBCs:

1. Transport of oxygen from lungs to tissues and carbon dioxide from tissues to lungs with the help of haemoglobin.
2. Maintenance of blood pH as haemoglobin acts as a buffer.
3. Maintenance of the viscosity of blood.

(6) White blood corpuscles / Leucocytes:

1. Leucocytes or White Blood Corpuscles (WBCs) are colourless, nucleated, amoeboid and phagocytic cells.

2. Their size ranges between 8 to 15 pm. Total WBC count is 5000 to 9000 WBCs/cu mm of blood. The average life span of a WBC is about 3 to 4 days.

3. They are formed by leucopoiesis in red bone marrow, spleen, lymph nodes, tonsils, thymus and Payer’s patches, whereas the dead WBCs are destroyed by phagocytosis in blood, liver and lymph nodes.

4. Leucocytes are mainly divided into two types, viz., granulocytes and agranulocytes.

5. Granulocytes : Granulocytes are cells with granular cytoplasm and lobed nucleus. Based on their staining properties and shape of nucleus, they are of three types, viz. neutrophils, eosinophils and basophils.

(I) Neutrophils:

1. In neutrophils, the cytoplasmic granules take up neutral stains.
2. Their nucleus is three to five lobed.
3. It may undergo changes in structure hence they are called polymorphonuclear leucocytes or polymorphs.
4. Neutrophils are about 70% of total WBCs.
5. They are phagocytic in function and engulf microorganisms.

(II) Eosinophils or acidophils:

1. Cytoplasmic granules of eosinophils take up acidic dyes such as eosin. They have bilobed nucleus.
2. Eosinophils are about 3% of total WBCs.
3. They are non-phagocytic in nature.
4. Their number increases (i.e. eosinophilia) during allergic conditions.
5. They have antihistamine property.

(III) Basophils:

1. The cytoplasmic granules of basophils take up basic stains such as methylene blue.
2. They have twisted nucleus.
3. In size, they are smallest and constitute about 0.5% of total WBCs.
4. They too are non-phagocytic.
5. Their function is to release heparin which acts as an anticoagulant and histamine that is involved in inflammatory and allergic reaction.

6. Agranulocytes : There are two types of agranulocytes, viz. monocytes and lymphocytes. Agranulocytes do not show cytoplasmic granules and their nucleus is not lobed. They are of two types, viz. lymphocytes and monocytes.
(I) Lymphocytes:

1. Agranulocytes with a large round nucleus are called lymphocyte.
2. They are about 30% of total WBCs.
3. Agranulocytes are responsible for immune response of the body by producing antibodies.

(II) Monocytes:

1. Largest of all WBCs having large kidney shaped nucleus are monocytes. They are about 5% of total WBCs.
2. They are phagocytic in function.
3. They can differentiate into macrophages for engulfing microorganisms and removing cell debris. Hence they are also called scavengers.
4. At the site of infections they are seen in more enlarged form.

(7) Thrombocytes/Platelets:

1. Thrombocytes or platelets are non- nucleated, round and biconvex blood corpuscles.
2. They are smallest corpuscles measuring about 2.5 to 5 mm in diameter with a count of about 2.5 lakhs/cu mm of blood.
3. Their life span is about 5 to 10 days.
4. Thrombocytes are formed from megakaryocytes of bone marrow. They break from these cells as fragments during the process of thrombopoiesis.
5. Thrombocytosis is the increase in platelet count while thrombocytopenia is decrease in platelet count.
6. Thrombocytes possess thromboplastin which helps in clotting of blood.
7. Therefore, at the site of injury platelets aggregate and form a platelet plug. Here they release thromboplastin due to which further blood clotting reactions take place.

(8) Functions of blood:

1. Transport of oxygen and carbon dioxide
2. Transport of food
3. Transport of waste product
4. Transport of hormones
5. Maintenance of pH
6. Water balance
7. Transport of heat
8. Defence against infection
9. Temperature regulation
10. Blood clotting/coagulation
11. Helps in healing

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Solutions Chapter 6 Dissolution of Partnership Firm Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Book Keeping & Accountancy Solutions Chapter 6 Dissolution of Partnership Firm

1. Objective Questions.

A. Select the most appropriate answer from the alternatives given below and rewrite the sentences.

Question 1.
In case of dissolution assets and liabilities cire transferred to ______________ Account.
(a) Bank Account
(b) Partner’s Capital Account
(c) Realisation Account
(d) Partner’s Current Account
Answer:
(c) Realisation Account

Question 2.
Dissolution expenses are credited to ______________ Account.
(a) Realisation Account
(b) Cash/Bank Account
(c) Partner’s Capital Account
(d) Partner’s Loan Account
Answer:
(b) Cash/Bank Account

Question 3.
Deficiency of insolvent partner will be suffered by solvent partners in their ______________ ratio.
(a) capital ratio
(b) profit sharing ratio
(c) sale ratio
(d) liquidity ratio
Answer:
(b) profit sharing ratio

Question 4.
If any asset is taken over by partner from firm his Capital Account will be ______________
(a) credited
(b) debited
(c) added
(d) divided
Answer:
(b) debited

Question 5.
If any unrecorded liability is paid on dissolution of the firm ______________ account is debited.
(a) Cash/Bank Account
(b) Realisation Account
(c) Partner’s Capital Account
(d) Loan Account
Answer:
(b) Realisation Account

Question 6.
Partnership is completely dissolved when the partners of the firm become ______________
(a) solvent
(b) insolvent
(c) creditor
(d) debtors
Answer:
(b) insolvent

Question 7.
Assets and liabilities are transferred to Realisation Account at their ______________ values.
(a) market
(b) purchase
(c) sale
(d) book
Answer:
(d) book

Question 8.
If the number of partners in a firm falls below two, the firm stands ______________
(a) dissolved
(b) established
(c) realisation
(d) restructured
Answer:
(a) dissolved

Question 9.
Realisation Account is ______________ on realisation of asset.
(a) debited
(b) credited
(c) deducted
(d) closed
Answer:
(b) credited

Question 10.
All activities of partnership firm ceases on ______________ of firm.
(a) dissolution
(b) admission
(c) retirement
(d) death
Answer:
(a) dissolution

B. Write a word/phrase/term which can substitute each of the following statements.

Question 1.
Debit balance of Realisation Account.
Answer:
Realization Loss

Question 2.
Winding up of partnership business.
Answer:
Dissolution of Partnership

Question 3.
An account is opened to find out the profit or loss on sale of assets and settlement of liabilities.
Answer:
Realization A/c

Question 4.
Debit balance of an Insolvent Partner’s Capital Account.
Answer:
Capital Deficiency

Question 5.
The credit balance of the Realisation Account.
Answer:
Realization Profit

Question 6.
Conversion of asset into cash on the dissolution of the firm.
Answer:
Realisation

Question 7.
Liability is likely to arise in the future on the happening of certain events.
Answer:
Contingent Liabilities

Question 8.
Assets that are not recorded in the books of accounts.
Answer:
Unrecorded Assets

Question 9.
The account shows the realization of assets and discharge of liabilities.
Answer:
Realization A/c

Question 10.
Expenses incurred on the dissolution of the firm.
Answer:
Dissolution/Realisation Expenses

C. State whether the following statements are True or False with reasons.

Question 1.
The firm must be dissolved on the retirement of a partner.
Answer:
This statement is False.
On the retirement of a partner, if the partnership agreement allows, then the remaining partner can continue the business activities. It means the firm is not to dissolve.

Question 2.
On dissolution Cash/Bank Account is closed automatically.
Answer:
This statement is True.
As the firm is dissolved, there is no question of any business activities to be carried out further and so Cash/Bank Account is also not necessary. Therefore on dissolution Cash/Bank Account is closed automatically.

Question 3.
On dissolution, Bank overdraft is transferred to Realisation Account.
Answer:
This statement is True.
As a sundry liability of the business, bank overdraft is a liability of a firm and hence, it is transferred to Realisation Account at the time of dissolution and paid a third party Liability.

Question 4.
A solvent partner having a debit balance to his Capital Account does not share the deficiency of insolvent partner Capital Account.
Answer:
This statement is False.
In the partnership, the partner’s liability is unlimited so, a solvent partner having a debit balance to his Capital Account should share the deficiency of the insolvent partner capital account.

Question 5.
At the time of dissolution of the partnership, all assets should be transferred to Realisation Account.
Answer:
This statement is False.
At the time of dissolution of the partnership, the cash account and Bank A/c are not transferred to Realisation A/c. Similarly, if an asset is taken over by a partner or by any creditor then that asset is transferred to the concerned person’s account and not to the Realisation Account.

Question 6.
The debit balance of an insolvent partner’s Capital Account is known as a capital deficiency.
Answer:
This statement is True.
Debit balance of Partners’ Capital Account means the excess of drawings than the capital credit balance. In the case of an insolvent partner, the debit balance of the Capital Account means liabilities which he cannot pay. It means capital deficiency.

Question 7.
At the time of dissolution, a loan from a partner will be transferred to Realisation Account.
Answer:
This statement is False.
At the time of dissolution, a loan from a partner will be paid after the payment of liabilities of third parties to the firm. It is not transferred to Realisation Account. Partner’s Loan A/c is separately opened and paid accordingly.

Question 8.
Dissolution takes place when the relationship among the partners comes to an end.
Answer:
This statement is True.
As per definition, Dissolution means to wind up or to close down, and it is possible only when relations among the partners in a partnership firm come to an end.

Question 9.
The insolvency loss at the time of dissolution of the firm is shared by the solvent partners in their profit sharing ratio.
Answer:
This statement is True.
In the partnership, partners’ liability is unlimited and in case of insolvency loss, legally solvent partners are ultimately liable and are suppose to bear the loss of an insolvent partner in their profit sharing ratio.

Question 10.
Realization loss is not transferred to insolvent partner’s Capital Account.
Answer:
This statement is False.
All partners of the firm are responsible for Loss on realization and hence loss on realization is supposed to be transferred to all Partners’ Capital Account, without any discrimination of solvent or insolvent.

D. Calculate the following:

Question 1.
Vinod, Vijay, and Vishal are partners in a firm sharing profit and losses in the ratio of 3 : 2 : 1. Vishal becomes insolvent and his capital deficiency is ₹ 6000. Distribute the capital deficiency among the solvent partner.
Answer:
Here, capital deficiency of ₹ 6000 is to be distributed among continuing partners in their profit and loss sharing ratio, i.e. 3 : 2
Share of deficiency for Vinod = 6,000 × \(\frac{3}{5}\) = ₹ 3,600
Share of deficiency for Vijay = 6,000 × \(\frac{2}{5}\) = ₹ 2,400
Vinod and Vijay will bear ₹ 3,600 and ₹ 2,400 of Vishal’s capital deficiency.

Question 2.
Creditors ₹ 30,000, Bills Payable ₹ 20,000, and Bank Loan ₹ 10,000. Available Bank balance ₹ 40,000. What will be the amount that creditors will get in case of all partner’s insolvency?
Answer:
Ratio of creditors, Bills payable and Bank Loan = 30,000 : 20,000 : 10,000 i.e., 3 : 2 : 1
Amount received by creditors = \(\frac{3}{3+2+1}\) × 40,000
= \(\frac{3}{6}\) × 40,000
= ₹ 20,000.

Question 3.
Insolvent Partner Capital A/c debit side total is ₹ 10,000 and credit side total is ₹ 6,000. Calculate deficiency.
Answer:
Deficiency of insolvent partner = Debit side total – Credit side total
= 10,000 – 6,000
= ₹ 4,000.

Question 4.
Insolvent Partners Capital A/c debit side is ₹ 15,000 and insolvent partner brought cash ₹ 6,000. Calculate the amount of insolvency loss to be distributed among the solvent partners.
Answer:
₹ 9,000 (15,000 – 6,000) is the amount of insolvency loss to be distributed among the solvent partners.

Question 5.
The realization profit of a firm is ₹ 6,000, partners share profit and loss in the ratio of 3 : 2 : 1. Calculate the amount of realization profit to be credited to Partners’ Capital A/c.
Answer:
Distribution of ₹ 6,000 in 3 : 2 : 1 ratio
6,000 × \(\frac{3}{6}\) = ₹ 3,000, 6,000 × \(\frac{2}{6}\) = ₹ 2,000, 6,000 × \(\frac{1}{6}\) = ₹ 1,000
Amount of realisation profit ₹ 3,000, ₹ 2,000 and ₹ 1,000 is to be credited to Partner’s Capital A/c respectively.

E. Answer in one sentence only.

Question 1.
What is the dissolution of the partnership firm?
Answer:
Dissolution of the partnership firm means complete closure of business activities and stoppage of partnership relations among all the partners.

Question 2.
When is Realisation Account opened?
Answer:
Realisation Account is opened at the time of dissolution of the partnership firm.

Question 3.
Which accounts are not transferred to Realisation Account?
Answer:
Cash/Bank balance, Reserve funds, Profit and Loss A/c balance, Partners’ Loan accounts, etc. are not transferred to Realisation Account.

Question 4.
Who is called an insolvent person?
Answer:
Whose capital A/c shows debit balance and who is not in a position to meet his capital deficiency even from his private property is called an insolvent person.

Question 5.
What is capital deficiency?
Answer:
The debit balance of the insolvent partner’s Capital Account which the insolvent partner cannot pay is called a capital deficiency.

Question 6.
In what proportion is the balance on Realisation Account transferred to Partners Capital/Current Accounts?
Answer:
The balance on the Realisation Account is transferred to Partners Capital/Current Accounts in their profit sharing ratio.

Question 7.
Who should bear the capital deficiency of insolvent partners?
Answer:
The capital deficiency of insolvent partners should be borne by the solvent partners.

Question 8.
Which account is debited on repayment of partner’s loan?
Answer:
Partner’s Loan Account is debited on repayment of partner’s loan.

Question 9.
Which account is debited on payment of dissolution expenses?
Answer:
Realisation Account is debited on payment of dissolution expenses.

F. Complete the table.

Question 1.

Answer:

Practical Problems

(Simple Dissolution)

Question 1.
Ganesh and Kartik are partners sharing profits and losses equally. They decided to dissolve the firm on 31st March 2018. Their Balance Sheet was as under:
Balance Sheet as of 31st March 2018

Assets were realised as under:
Building ₹ 82,000, Debtors ₹ 22,000, Stock ₹ 20,000. Bills Receivable ₹ 3,200 and Ganesh agreed to take over Furniture for ₹ 10,000. Realisation Expenses amounted to ₹ 2,000.
Show Realisation A/c, Partners’ Capital A/c, and Cash A/c.
Solution:
In the books of Ganesh and Kartik


Working Notes:
1. Amount paid to Ganesh and Kartik are ₹ 27,600 and ₹ 77,600 respectively.
2. Loss on Realisation and Reserve fund amounts are equally distributed.
3. Furniture is taken over by Ganesh so his Capital A/c is debited.

Question 2.
Leela, Manda, and Kunda are partners in the firm ‘Janki Stores’ sharing profits and losses in the ratio of 3 : 2 : 1 respectively. On 31st March 2018, they decided to dissolve the firm when their Balance Sheet was as under.
Balance Sheet as of 31st March 2018

Leela agreed to take over the Building at ₹ 1,23,600. Manda took over Goodwill, Stock, and Debtors at book values and agreed to pay Creditors and Bills payable. Motor car and Machinery realized ₹ 1,51,080 and ₹ 31,680 respectively. Investments were taken by Kunda at an agreed value of ₹ 55,440. Realisation expenses amounted to ₹ 6,800.
Pass necessary entries in the books of ‘Janki Stores’.
Solution:
In the books of ‘Janki Stores’
Journal Entries


Working Notes:
In the books of Leela, Manda, and Kunda

Question 3.
Shailesh and Shashank were partners sharing profits and losses in the ratio of 3 : 2. Their Balance Sheet as of 31st March 2019 was as follows:
Balance Sheet as of 31st March 2019

The firm was dissolved on the above date and the assets realised as under:
1. Plant ₹ 8,000, Building ₹ 6,000, Stock ₹ 4,000 and Debtors ₹ 12,000.
2. Shailesh agreed to pay off the Bills Payable.
3. Creditors were paid in full.
4. Dissolution expenses were ₹ 1,400.
Prepare Realisation A/c, Partners’ Current A/c, Partners’ Capital A/c, and Bank A/c.
Solution:
In the books of Shailesh and Shashank

Question 4.
Asha, Usha, and Nisha were partners sharing profits and losses in the ratio of 2 : 2 : 1. The following is the Balance Sheet as of 31st March 2019.
Balance Sheet as of 31st March 2019

On the above date, the partners decided to dissolve the firm.
1. Assets were realised at: Machinery ₹ 90,000, Stock ₹ 36,000, Investment ₹ 42,000 and Debtors ₹ 90,000.
2. Dissolution expenses were ₹ 6,000.
3. Goodwill of the firm realized ₹ 48,000.
Pass Journal Entries to close the books of the firm.
Solution:
In the books of Asha, Usha, and Nisha
Journal Entries


Working Notes:
In the books of Asha, Usha, and Nisha

Question 5.
Seeta and Geeta are partners in the firm sharing profits and losses in the ratio of 4 : 1. They decided to dissolve the partnership on 31st March 2020 on which date their Balance Sheet stood as follows:
Balance Sheet as of 31st March 2020

Additional Information:
1. Plant and Stock took over by Seeta at ₹ 78,000 and ₹ 22,000 respectively.
2. Debtors realised 90% of the book value and Trademark at ₹ 5,000 and Goodwill was realised for ₹ 27,000.
3. Unrecorded assets estimated at ₹ 4,500 were sold for ₹ 1,500.
4. ₹ 1,000 Discounts were allowed by creditors while paying their claim.
5. The Realisation expenses amounted to ₹ 3,500.
You are required to prepare Realisation A/c, Cash A/c, and Partners’ Capital A/c.
Solution:
In the books of Seeta and Geeta


Working Notes:
1. Bank Loan is an external liability of the firm and therefore it is transferred to Realisation A/c.
2. Amount recovered from Debtors = 90% of Gross Debtors = \(\frac {90}{100}\) × 48,000 = ₹ 43,200.
3. Amount paid to creditors = Value of Creditors – Discount given = 35,000 – 1,000 = ₹ 34,000.
4. Sale of unrecorded assets for ₹ 1,500 is recorded on the credit side of Realisation A/c and debit side of Cash A/c.
5. It is presumed that Furniture realised nothing.

Question 6.
Sangeeta, Anita, and Smita were in partnership sharing profits and losses in the ratio 2 : 2 : 1. Their Balance Sheet as of 31st March 2019 was as under:
Balance Sheet as of 31st March 2019

They decided to dissolve the firm as follows:
1. Assets realised as; Land recovered ₹ 1,80,000; Goodwill for ₹ 75,000; Loans and Advance realised ₹ 12,000; 10% of the Debts proved bad.
2. Sangeeta took Plant at book value.
3. Creditors and Bills payable paid at 5% discount.
4. Sandhya’s loan was discharged along with ₹ 6,000 as interest.
5. There was a contingent liability in respect of bills of ₹ 1,00,000 which was under discount. Out of them, a holder of one bill of ₹ 20,000 became insolvent.
Show Realisation Account, Partners’ Capital Account, and Bank Account.
Solution:
In the books of Sangeeta, Anita, and Smita


Working Notes:
1. Amount paid towards Sandhya’s Loan = Loan amount + Interest due on loan
= 1,20,000 + 6,000
= ₹ 1,26,000

2. Amount received from Debtors = Debtors – Bad debts
= 1,25,000 – 10% of 1,25,000
= 1,25,000 – 12,500
= ₹ 1,12,500

3. Amount paid to Creditors = Creditor – 5% discount
= 1,20,000 – 5% on 1,20,000
= 1,20,000 – 6,000
= ₹ 1,14,000

4. Amount paid towards Bills payable = Bills payable – 5% discount
= 20,000 – 5% on 20,000
= 20,000 – 1,000
= ₹ 19,000

5. Bill of ₹ 1,00,000 was discounted with the Bank. On the due date, bank could not recover ₹ 20,000 from one bill holder as he was declared insolvent. Therefore, we are required to settle that contingent liability of ₹ 20,000.

Question 7.
Saiesh, Sumit, and Hemant were in partnership sharing Profits and Losses in the ratio 2 : 2 : 1. They decided to dissolve their partnership firm on 31st March 2019 and their Balance Sheet on that date stood as;
Balance Sheet as of 31st March 2019

It was agreed that;
1. Sailesh to discharge Loan and to take Debtors at book value.
2. Plant realised ₹ 1,35,000.
3. Stock realised ₹ 72,000.
4. Creditors were paid off at a discount of ₹ 45.
Show Realisation Account, Partners’ Capital Account, and Bank Account.
Solution:
In the books of Sailesh, Sumit, and Hemant

(When one partner become Insolvent)

Question 8.
Sitaram, Gangaram, and Rajaram are partners sharing profits and losses in the ratio of 4 : 2 : 3. On 1st April 2019 they agreed to dissolve the partnership, their Balance Sheet was as follows:
Balance Sheet as of 31st March 2019

The assets realised: Building ₹ 46,750; Machinery ₹ 18,550; Furniture ₹ 9,600; Investment ₹ 10,650; Bill Receivable and Debtors ₹ 20,750. All the liabilities were paid off. The cost of realisation was ₹ 800. Rajaram becomes bankrupt and ₹ 1,100 only was recovered from his estate.
Show Realisation Account, Bank Account, and Capital Account of the partners.
Solution:
In the books of Sitaram, Gangaram and Rajaram


Working Notes:
1. ₹ 1,100 is recovered from Rajaram’s estate which is recorded on the credit side of Rajaram’s Capital Account and on the debit side of Bank A/c.

2. Capital deficiency of Rajaram = Debit total of Capital A/c – Credit total of Capital A/c
= 18,000 – 15,900
= ₹ 2,100
The deficit amount of Rajaram A/c ₹ 2,100 is distributed among continuing partners’ in 2 : 1 ratio.

Question 9.
Following is the Balance Sheet of Vaibhav, Sanjay, and Santosh
Balance Sheet as of 31st March 2019

Santosh is declared insolvent so the firm is dissolved and assets realised as follows:
1. Stock and Debtors ₹ 54,000, Goodwill – NIL, Machinery at book value.
2. Creditors allowed a discount of 10%.
3. Santosh could pay only 25 paise in the rupee of the balance due.
4. Profit sharing ratio was 8 : 4 : 3.
5. A contingent liability against the firm ₹ 9,000 is cleared.
Give Ledger Account to close to books of the firm.
Solution:
In the books of Vaibhav, Sanjay, and Santosh


Working Notes:
1. Contingent liability paid, so Realisation A/c is debited and Bank A/c is credited.
2. Santosh could pay only 25 paise in a rupee of the balance due i.e.
Balance due from Santosh (Debit side of Partners Capital A/c) = ₹ 10,560
25% of ₹ 10,560 = ₹ 2,640 (Amount recorded on debit side of Bank A/c)
Capital deficiency of Santosh = 10,560 – 2,640 = ₹ 7,920
₹ 7,920 to be distributed among continuing partner in their profit-loss ratio = 8 : 4 i.e. 2 : 1.
7,920 × \(\frac{2}{3}\) = ₹ 5,280
7,920 × \(\frac{1}{3}\) = ₹ 2,640

(When Two Partners become Insolvent)

Question 10.
Shweta, Nupur, and Sanika are partners sharing profits and losses in the ratio of 3 : 2 : 1. Their Balance Sheet as of 31st March 2019 was as follows:
Balance Sheet as of 31st March 2019

The firm is dissolved on 31st March 2019. Sundry assets realised @ 60% of its book value. Realisation expenses ₹ 2,000 paid by Shweta. Nupur and Sanika both are insolvent.
Nupur’s private estate has got a surplus of ₹ 3,000 and that of Sanika ₹ 8,000.
Show necessary Ledger Accounts to close the books of the firm.
Solution:
In the books of Shweta, Nupur and Sanika

(When All Partners become Insolvent)

Question 11.
Following is the Balance Sheet as of 31st March 2019 of a firm having three partners Priti, Priya, and Prachi.
Balance Sheet as of 31st March 2019

The firm was dissolved due to the insolvency of all the partners. Machinery was sold for ₹ 18,000, while Furniture fetched ₹ 14,000, Stock realized ₹ 35,000. Realisation expenses amounted to ₹ 2,000. Nothing could be recovered from Priya and Prachi, but ₹ 3,400 could be collected from Priti’s private estate.
Close the books of accounts of the firm.
Solution:
In the books of Priti, Priya, and Prachi



Working Notes:
1. Amount paid to loan from sale of machinery = ₹ 18,000
Balance of Loan 30,000 – 18,000 = ₹ 12,000

2. Ratio of Trade creditors and Loan = 50,000 : 12,000
= 50 : 12
= 25 : 6

3. Balance of cash available = 10,000 + 67,000 + 3,400 – 18,000 – 2,000
= 80,400 – 20,000
= ₹ 60,400
Amount paid towards loan = \(\frac{6}{31} \times \frac{60,400}{1}\) = ₹ 11,690
Amount paid to Trade creditors = \(\frac {25}{31}\) × 60,400 = ₹ 48,710
Amount paid towards loan = 18,000 + 11,690 = ₹ 29,690.

Question 12.
Shashwat and Shiv are equal partners. Their Balance Sheet stood as under:
Balance Sheet as of 31st March 2019

Due to weak financial position, all partners were declared bankrupt.
The Assets were realised as follows:
Stock ₹ 3,500, Furniture ₹ 2,000, Debtors ₹ 5,000 and Machinery ₹ 7,000.
The cost of collection and distributing the estate amounted to ₹ 1,500. Shashwat’s private estate is not sufficient even to pay his private debts, whereas in Shiv’s private estate there is a surplus of ₹ 500.
Prepare necessary Ledger Accounts to close the books of the firm.
Solution:
In the books of Shashwat and Shiv


Working Note:
As partners we’re not able to pay their loss amount, a difference of amount is considered as deficiency of partners.

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Solutions Chapter 5 Reconstitution of Partnership (Death of Partner) Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Book Keeping & Accountancy Solutions Chapter 5 Reconstitution of Partnership (Death of Partner)

1. Objective questions:

A. Select the most appropriate answer from the alternative given below and rewrite the sentences.

Question 1.
Benefit Ratio is the ratio in which _______________
(a) The old partner gain on the admission of a new partner
(b) The Goodwill of a new partner on admission is credited to old partners
(c) The continuing partners’ benefits on retirement or death of a partner
(d) All partners are benefitted.
Answer:
(c) The continuing partner’s benefits on retirement or death of a partner

Question 2.
The ratio by which existing partners are benefitted _______________
(a) gain ratio
(b) sacrifice ratio
(c) profit ratio
(d) capital ratio
Answer:
(a) gain ratio

Question 3.
Profit and Loss Suspense Account is shown in the new Balance Sheet on _______________ side.
(a) debit
(b) credit
(c) asset
(d) liabilities
Answer:
(c) asset

Question 4.
Death is a compulsory _______________
(a) Dissolution
(b) Admission
(c) Retirement
(d) Winding up
Answer:
(c) Retirement

Question 5.
The balance on the Capital Account of a partners, on his death is transferred to _______________ Account.
(a) Relatives
(b) Legal Heir’s Loan/Executors Loan
(c) Partners’ Capital
(d) Partners’ Loan
Answer:
(b) Legal Heir’s Loan/Executors Loan.

B. Write a word, term, phrase, which can substitute each of the following statements.

Question 1.
Excess of credit side over the debit side of Profit and Loss Adjustment Account.
Answer:
Profit

Question 2.
A person who represents the deceased partner on the death of the partner.
Answer:
Legal Heir’s or Executor

Question 3.
Accumulated past profit kept in the form of reserve.
Answer:
Reserve fund or General reserve

Question 4.
The partner who died.
Answer:
Deceased partner

Question 5.
The proportion in which the continuing partners benefit due to the death of a partner.
Answer:
Gain/Benefit ratio

C. State whether the following statements are True or False with reasons.

Question 1.
A deceased partner is not entitled to the Goodwill of the firm.
Answer:
This statement is False.
A deceased partner’s contribution was there in the development of business and goodwill is the value of the business in terms of money. Hence, a deceased partner is entitled to receive goodwill from the firm.

Question 2.
A deceased partner is entitled to his share of General Reserve.
Answer:
This statement is True.
General reserve is created out of past undistributed profit. Past profit is earned due to the efforts and hard work of all the partners including the partner who is now dead. Hence a deceased partner has right on it and therefore a deceased partner is entitled to receive his share of General reserve.

Question 3.
If goodwill is written off, a Deceased Partner’s Capital Account is debited.
Answer:
This statement is False.
When the benefits of goodwill are given to the deceased partner, his capital account is credited and when such goodwill is written off, capital accounts of remaining partners are debited.

Question 4.
After the death of a partner, the entire amount due to the deceased partner is paid to the legal representative of the deceased partner.
Answer:
This statement is True.
After the death of a partner, the entire amount due to the deceased partner is paid to the legal representative of the deceased partner as he is the only person who has the legal right to that amount.

Question 5.
For recording the profit or loss up to the death, the Profit and Loss Appropriation Account is operated.
Answer:
This statement is False.
For recording the profit or loss up to the death, the Profit and Loss suspense Account is created and operated. This is because final accounts cannot be prepared on the date of death of a partner. Till that period a separate account called Profit and Loss Suspense A/c is prepared.

D. Fill in the blanks and rewrite the following sentence.

Question 1.
Deceased Partners’ Executors Account is shown on the _______________ side of the Balance Sheet.
Answer:
Liabilities

Question 2.
On the death of a partner, a ratio in which the continuing partners get more share of profits in future is called as _______________ Ratio.
Answer:
Gain

Question 3.
Deceased partners share of profit up to the death is shown on _______________ side of Balance Sheet.
Answer:
Assets

Question 4.
Benefit Ratio = New Ratio – _______________
Answer:
Old Ratio

Question 5.
When Goodwill is raised at its full value and it is written off _______________ Account is to be credited.
Answer:
Goodwill

E. Answer in one sentence only.

Question 1.
What is Gain Ratio?
Answer:
The profit-sharing ratio which is acquired by the surviving or continuing partners on account of the death of any partner is called gain ratio or benefit ratio.

Question 2.
In which ratio general reserve is distributed on the death of a partner?
Answer:
General reserve is distributed on the death of a partner in their old profit sharing ratio.

Question 3.
To whom do you distribute general reserve on the death of a partner?
Answer:
On the death of a partner general reserve is distributed among all partners in their old profit and loss ratio.

Question 4.
How the death of a partner is a compulsory retirement?
Answer:
After the death of a partner, the business is not able to get any kind of services from the deceased partner and so we can say that the death of a partner is like a compulsory retirement.

Question 5.
To which account profit is to be transferred up to the date of his death?
Answer:
Profit of the deceased partner, up to the date of his death, is transferred to his Legal Heir’s/Executor’s Account.

Practical Problems

Question 1.
Rajesh, Rakesh, and Mahesh were equal Partners on 31st March 2019. Their Balance Sheet was as follows 31st March 2019.
Balance Sheet as of 31st March 2019

Mr. Rajesh died on 30th June 2019 and the following adjustment was agreed as:
1. Furniture was to be adjusted to its market price of ₹ 3,40,000.
2. Land and Building were to be depreciated by 10%.
3. Provide R.D.D. @ 5% on debtors.
4. The profit up to the date of death of Mr. Rajesh is to be calculated on the basis of last year’s profit which was ₹ 1,80,000.
Prepare:
1. Profit and Loss Adjustment A/c
2. Partners’ Capital Account
3. Balance Sheet of the continuing firm.
Solution:
In the books of the Partnership Firm


Balance Sheet as of 1st July 2019

Working Note:
The profit of the firm of last year was ₹ 1,80,000.
Proportionate profit up to the date of death for Rajesh is as follows
= 1,80,000 × \(\frac{3}{12} \times \frac{1}{3}\) (Period) (P & L ratio)
= ₹ 15,000 (Profit and Loss Suspense A/c)

Question 2.
Rahul, Rohit, and Ramesh are in a business sharing profits and losses in the ratio of 3 : 2 : 1 respectively. Their Balance Sheet as of 31st March, 2017 was as follows:
Balance Sheet as of 31st March 2017

On 1st October 2017, Ramesh died and the Partnership deed provided that
1. R.D.D. was maintained at 5% on Debtors.
2. Plant and Machinery and Investment were valued at ₹ 80,000 and ₹ 4,10,000 respectively.
3. Of the creditors an item of ₹ 6,000 was no longer a liability and hence was properly adjusted.
4. Profit for 2017-18 was estimated at ₹ 1,20,000 and Ramesh’s share in it up to the date of his death was given to him.
5. Goodwill of the firm was valued at two times the average profit of the last five years, which were
2012-13 – ₹ 1,80,000
2013-14 – ₹ 2,00,000
2014-15 – ₹ 2,50,000
2015-16 – ₹ 1,50,000
2016-17 – ₹ 1,20,000
Ramesh’s share in it was to be given to him.
6. Salary ₹ 5,000 p.m. was payable to him.
7. Interest on capital at 5% i.e. was payable and on Drawings ₹ 2,000 were charged.
8. Drawings made by Ramesh up to September 2017 were ₹ 5,000 p.m.
Prepare Ramesh’s Capital A/c showing the amount payable to his executors.
Give working of Profit and Goodwill.
Ramesh Capital Balance ₹ 3,41,000
Solution:
In the books of the Partnership Firm

Working Notes:
1. Calculation of share of Goodwill:
(a) Average profit = \(\frac{Total Profit}{No. of years}\)
= \(\frac{1,80,000+2,00,000+2,50,000+1,50,000+1,20,000}{5}\)
= \(\frac{9,00,000}{5}\)
= ₹ 1,80,000

(b) Goodwill = Average profit × No. of years
= 1,80.000 × 2
= ₹ 3,60,000

(c) Share of Goodwill to Ramesh = Goodwill of the firm × Ramesh’s share
= 3,60,000 × \(\frac{1}{6}\)
= ₹ 60,000

2. Calculation of share of profit due to Ramesh:
Share of profit = Last year profit × Share of profit × Period
= 1,20,000 × \(\frac{1}{6} \times \frac{6}{12}\)
= ₹ 10,000 (Profit and Loss Suspense A/c)

3. Interest on Capital is calculated for six months.
∴ Interest = 2,40,000 × \(\frac{6}{12} \times \frac{5}{100}\) = ₹ 6,000

4.

Question 3.
Ram, Madhav, and Keshav are partners sharing profit and losses in the ratio 5 : 3 : 2 respectively. Their Balance Sheet as of 31st March, 2018 was as follows:
Balance Sheet as of 31st March 2018

Keshav died on 31st July 2018 and the following adjustments were agreed by as per the partnership deed.
1. Creditors have increased by ₹ 10,000.
2. Goodwill is to be calculated at 2 years purchase of average profits of 5 years.
3. The profits of the preceding 5 years was
2013-14 – ₹ 90,000
2014-15 – ₹ 1,00,000
2015-16 – ₹ 60,000
2016-17 – ₹ 50,000
2017-18 – ₹ 50,000 (Loss)
Keshav’s share in it was to be given to him.
4. Loose Tools and livestock were valued at ₹ 80,000 and ₹ 1,20,000 respectively.
5. R.D.D. was maintained at ₹ 10,000.
6. Commission ₹ 2,000 p.m. was payable to Keshav. Profit for 2018-19 was estimated at ₹ 45,000 and Keshav’s share in it up to the date of his death was given to him.
Prepare Revaluation A/c, Keshav’s Capital A/c showing the amount payable to his executors.
Solution:
In the books of the Partnership Firm

Working Notes:
1. Calculation of share of Goodwill:
(a) Average profit = \(\frac{\text { Total profit }}{\text { No. of years }}\)
= \(\frac{90,000+1,00,000+60,000+50,000-50,000}{5}\)
= \(\frac{2,50,000}{5}\)
= ₹ 50,000

(b) Goodwill = Average profit × No. of years
= 50,000 × 2
= ₹ 1,00,000

(c) Share of Goodwill to Keshav = Goodwill of the firm × Keshav’s share
= 1,00,000 × \(\frac{2}{10}\)
= ₹ 20,000

2. Calculation of share of profit due to Keshav
Share of profit = Last year profit × Share of Keshav × Period
= 45,000 × \(\frac{2}{10} \times \frac{4}{12}\)
= ₹ 3,000 (Profit and Loss Suspense Account)

Question 4.
Virendra, Devendra, and Narendra were partners sharing profit and losses in the ratio of 3 : 2 : 1. Their Balance Sheet as of 31st March 2019 was as follows.
Balance Sheet as of 31st March 2019

Mr. Virendra died on 31st August 2019 and the partnership deed provided that the event of the death of Mr. Virendra his executors be entitled to be paid out.
1. The capital to his credit at the date of death.
2. His proportion of Reserve at the date of last Balance Sheet.
3. His proportion of Profits to date of death is based on the average profits of the last four years.
4. His share of Goodwill should be calculated at two years purchase of the profits of the last four years for the year ended 31st March were as follows:
2016 – ₹ 40,000
2017 – ₹ 60,000
2018 – ₹ 70,000
2019 – ₹ 30,000
5. Mr. Virendra has drawn ₹ 3,000 p.m. to date of death, There is no increase and decrease in the value of assets and liabilities.
Prepare Mr. Virendra Executors A/c.
Solution:
In the books of the Partnership Firm

Working Notes:
1. Calculation of share of profit:
(a) Average Profit = \(\frac{\text { Total profit }}{\text { No. of years }}\)
= \(\frac{40,000+60,000+70,000+30,000}{4}\)
= \(\frac{2,00,000}{4}\)
= ₹ 50,000

(b) Goodwill = Average profit × No. of years
= 50,000 × 2
= ₹ 1,00,000

(c) Share of Goodwill to Virendra = Goodwill of the firm × Virendra’s share
= 1,00,000 × \(\frac{3}{6}\)
= ₹ 50,000

2. Share of profit due to Virendra
Share of profit = Last year profit × Share of Virendra × Period
= 50,000 × \(\frac{3}{6} \times \frac{5}{12}\)
= ₹ 10,417 (Profit and Loss Suspense A/c)

Question 5.
The Balance Sheet of Sohan, Rohan, and Mohan who were sharing profits and losses in the ratio of 3 : 2 : 1 is as follows:
Balance Sheet as of 31st March 2019

Mr. Rohan died on 1st October 2019 and the following adjustments were made:
1. Goodwill of the firm is valued at ₹ 30,000.
2. Land and Building and Machinery were found to be undervalued by 20%.
3. Investments are valued at ₹ 60,000.
4. Stock to be undervalued by ₹ 5,000 and a provision of 10% as debtors were required.
5. Patents were valueless.
6. Mr. Rohan was entitled to share in profits up to the date of death and it was decided that he may be allowed to retain his drawings as his share of profit. Rohan’s drawings till the date of death were ₹ 25,000.
Prepare Partners’ Capital Accounts.
Solution:
In the books of the Partnership firm

Working Notes:
1.

2. Firm’s goodwill = ₹ 30,000.
DistrIbute among partners in their profit and loss ratio 3 : 2 : 1.

3. Revised value of Land & Building = \(\frac{\text { Book value }}{(100-20)} \times 100\)
= \(\frac{40,000}{80} \times 100\)
= ₹ 50,000.
∴ Increase In the value of Land & Building = Revised value – Book value
= 50,000 – 40,000
= ₹ 10,000.

4. Revised value of Machinery = \(\frac{\text { Book value }}{(100-20)} \times 100\)
= \(\frac{80,000}{80} \times 100\)
= ₹ 1 ,00,000.
∴ Increase in the value of Machinery = 1,00,000 – 80,000 = ₹ 20,000.

5. Patents were valueless means it is a loss for the business.

6. Rohan’s share In profit is ₹ 25,000 and his drawings are ₹ 25,000. Rohan is allowed to retain his drawings as his share of profit. Means write ₹ 25,000 as drawings on the debit side and write ₹ 25,000 as Profit and Loss Suspense A/c on the Credit side of Partners’ Capital A/c.

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Solutions Chapter 4 Reconstitution of Partnership (Retirement of Partner) Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Book Keeping & Accountancy Solutions Chapter 4 Reconstitution of Partnership (Retirement of Partner)

A. Select the most appropriate alternatives from those given below and rewrite the sentence.

Question 1.
The profit or loss from revaluation on retirement of partner is shared by ______________
(a) the remaining partners
(b) all the partners
(c) only retiring partner
(d) bank
Answer:
(b) all the partners

Question 2.
Descrease in the value of assets should be ______________ to Profit and Loss Adjustment Account.
(a) debited
(b) credited
(c) added
(d) equal
Answer:
(a) debited

Question 3.
The balance of the capital account of retired partner is transferred to his ______________ account if it is not paid.
(a) loan
(b) personal
(c) current
(d) son’s
Answer:
(a) loan

Question 4.
Gain ratio = ______________ Ratio less Old Ratio.
(a) New
(b) Equal
(c) Capital
(d) Sacrifice
Answer:
(a) New

Question 5.
New Ratio = Old Ratio + ______________ Ratio.
(a) Gain
(b) Capital
(c) Sacrifice
(d) Current
Answer:
(a) Gain

Question 6.
Apte, Bhate and Chitale are sharing 1/2, 3/10, and 1/5 if Apte retire their new ratio will be ______________
(a) 5 : 2
(b) 3 : 2
(c) 5 : 3
(d) 2 : 5
Answer:
(b) 3 : 2

B. Write the word, term, phrase, which can substitute each of the following statement.

Question 1.
Credit balance of Profit and Loss Adjustment Account.
Answer:
Profit on Revaluation Accounts

Question 2.
The ratio in which the continuing partners are benefited due to retirement of partner.
Answer:
Gain Ratio

Question 3.
Debit balance of Revaluation Account.
Answer:
Loss on Revaluation

Question 4.
The ratio which is obtained by deducting Old Ratio from New Ratio.
Answer:
Gain Ratio

Question 5.
Money value of business reputation earned by the firm over a number of years.
Answer:
Goodwill

Question 6.
Partner’s Account where Loss or Profit on revaluation is transferred.
Answer:
Capital/Current Account

C. State whether the following statement are true or false with reasons.

Question 1.
Gain ratio means New ratio minus Old ratio.
Answer:
This statement is True.
As per definition, profit sharing ratio which is acquired by the continuing partners from the retiring partner is called gain ratio. If gain ratio added to old ratio we will get New ratio. It means New ratio = Old ratio + Gain ratio by interchanging the terms, we will get Gain ratio = New ratio – Old ratio.

Question 2.
Retiring partner’s share in profit up to the date of his retirement will be debited to Profit and Loss Suspense Account.
Answer:
This statement is True.
If a partner retires from the firm during the accounting year, the profit or loss for the period from the date of last balance sheet to the date of retirement is calculated on the basis of last year’s profit or average profit and it is credited to retiring partner’s capital A/c and for time being it debited to new account called Profit and Loss Expense A/c. This is because final accounts cannot be prepared on any date during the accounting year.

Question 3.
On retirement of a partner, sacrifice ratio is considered.
Answer:
This statement is False.
On retirement of a partner, his share is acquired by continuing partners in certain proportion and it is nothing but gain for them. Therefore, on retirement of a partner instead of sacrifice ratio gain ratio is considered.

Question 4.
Retiring partner is called an outgoing partner.
Answer:
This statement is True.
When a person retires from the firm due to health issues, financial issues or personal reasons then it is known as person retires from the business and for the business, he is an outgoing partner.

Question 5.
On retirement of a partner, remaining partner will share the goodwill in their profit sharing ratio.
Answer:
This statement is False.
On retirement of a partner, after giving retiring, partner’s share in goodwill and if goodwill is written off, then remaining partners will adjust the goodwill in their new profit sharing ratio. (If raised to full extent and written off)

Question 6.
Retiring partner is not entitled to share in general reserve and accumulated profit.
Answer:
This statement is False.
General reserve and accumulated profit are created out of past undistributed profit, such profits are the outcome of hard work of all the partners including retiring partner. Hence, retiring partner’s has right to share general reserve and accumulated profit. He is therefore, entitled to get share in general reserve and accumlated profit.

D. Fill in the blanks and rewrite the following sentence:

Question 1.
New Ratio (less) ______________ = Gain ratio.
Answer:
Old ratio

Question 2.
Retiring partner’s share of goodwill is ______________ to remaining Partner’s Capital Account.
Answer:
debited

Question 3.
Revaluation A/c is also known as ______________ Account.
Answer:
Profit and Loss Adjustment

Question 4.
On retirement, the balance at a Current Account of a partner is transferred to his ______________ Account.
Answer:
Capital

Question 5.
A proportion in which the continuing partners get the share of retiring partner is known as ______________ Ratio.
Answer:
Gain

E. Answer in one sentence.

Question 1.
What is meant by Retirement of a Partner?
Answer:
Retirement of a partner refers to a process in which a partner leaves the firm or severes his relations with other partners on account of his old age, continued ill health, loss of interest in the firm, misunderstanding amongst the partners, etc.

Question 2.
What is Benefit Ratio?
Answer:
Profit sharing ratio which is acquired by the continuing partners on account of retirement or death of a partner is called Benefit Ratio or Gain Ratio.

Question 3.
What is New Ratio?
Answer:
The ratio in which profits or losses are shared by the continuing partners after retirement of a partner is called New Profit Sharing Ratio.

Question 4.
How is the amount due to the retiring partner settled?
Answer:
The amount due to a retiring partner is settled as per the terms of partnership agreement or otherwise mutually agreed upon either in lumpsum or in instalments.

Question 5.
How is Gain Ratio calculated?
Answer:
Gain ratio is calculated at the time of retirement of a partner by deducting old ratio from new ratio.

Question 6.
Why is retiring partner’s capital account credited with goodwill?
Answer:
Goodwill is an intangible assets or benefits accrued to the firm and its benefits are transferred to retiring partner’s Capital A/c by giving credit.

Practical Problems

Question 1.
The Balance Sheet of Mr Mama, Kaka and Mr Baba who shared profits and losses as 4 : 3 : 3 respectively.
Balance Sheet as on 31st March, 2018

Kaka retires on 1st April, 2018 on the following terms.
1. The share of Kaka in Goodwill of the firm is valued at ₹ 2,700.
2. Furniture to be depreciated by 10% and Motor car by 12.5%.
3. Live Stock to be appreciated by 10% and Plant by 20%.
4. A provision of ₹ 2,000 to be made for a claim of compensation.
5. R.D.D. is no longer necessary.
6. The amount payable to Kaka should be transferred to his Loan A/c.
Prepare Profit and Loss Adjustment A/c, Partners’ Capital A/cs and Balance Sheet of the new firm.
Solution:
In the books of Partnership Firm

Balance Sheet as on 1st April, 2018


Working Notes:
1. R.D.D. is no longer require means it is a gain for firm.
2. A provision of ₹ 2,000 to be made for a claim of compensation, ₹ 2,000 is recorded on debit side of Profit and Loss Adjustments A/c and then on liability side of Balance Sheet.
3. Total payable amount to Kaka ₹ 20,175 is recorded as Kaka’s Loan A/c.

Question 2.
The Balance Sheet of Ram, Shyam and Ghanshyam sharing profits and losses in 3 : 2 : 1 respectively and their position on 31-3-19 were as follows:
Balance Sheet as on 31st March, 2019

Ghanshyam retired on 1st April, 2019 on the following terms:
1. Building and Investment to be appreciated by 5% and 10% respectively.
2. Provision for Doubtful Debts to be created at 5% on Debtors.
3. The provision of ₹ 3,000 be made in respect of Outstanding Salary.
4. Goodwill of the firm is valued at ₹ 90,000 and partners decide that goodwill should be written back.
5. The amount payable to the retiring partner be transferred to his Loan A/c.
Prepare: Profit and Loss Adjustment A/c, Partners’ Capital A/c, Balance Sheet of new firm.
Solution:
In the books of Partnership Firm


Balance Sheet as on 1st April, 2019

Working Notes:
1. Provision of ₹ 3,000 for outstanding salary is recorded on debit side of Profit and Loss Adjustment A/c and then on the Liability side of Balance Sheet.
2. Goodwill of the firm is valued at ₹ 90,000 and share of retiring partner in it is ₹ 15,000 (\(\frac{1}{6}\)th part) and it is to be written back means it is to be shared by remaining partners in their profit-loss ratio.

Question 3.
The Balance Sheet of the Anu, Renu and Dinu is as follows, and the partners are sharing profits and losses in the proportion of 2 : 2 : 1 respectively.
Balance Sheet as on 31st March, 2019

Dinu retires from the firms on 1st April, 2019 on the following terms:
1. The assets are to be revalued as : freehold property ₹ 30,000, Machinery ₹ 5,000, Furniture ₹ 12,000, All debtors are good.
2. Goodwill of the firm be valued at thrice the average profit for preceding five years. Profits of the firm for the year.
2014-15 – ₹ 14,500
2015-16 – ₹ 10,500
2016-17 – ₹ 10,000
2017-18 – ₹ 16,000
2018-19 – ₹ 10,000
3. Dinu should be paid ₹ 3,000 by cheque.
4. The Balance of Dinu’s Capital A/c should be kept in the business as loan.
Prepare: Profit and Loss Adjustment A/c, Capital Accounts of Partners, Balance Sheet of the new firm.
Solution:
In the books of Partnership Firm

Balance Sheet as on 1st April 2019

Working Notes:
1. Average profit = \(\frac{\text { Total Profit }}{\text { No. of years }}\)
= \(\frac{1000+10,500+10,000+16,000+10,000}{5}\)
= \(\frac{47,500}{5}\)
= ₹ 9,500
Goodwill = Avg. profit × No. of years
= 9,500 × 3 years
= ₹ 28,500
Goodwill value given in balance sheet = ₹ 30,000
New value arrived at = ₹ 28,500
Loss due to revaluation = ₹ 1,500
To be recorded in P & L Adj. A/c – Dr. Side.
In asset side of Balance sheet, write ₹ 28,500 for Goodwill.

2. Balance of Bank A/c = Opening Balance – Cheque given to Dinu
= 5,000 – 3,000
= ₹ 2,000

Question 4.
Rohan, Rohit and Sachin are partners in a firm sharing profits and losses in the proportion 3 : 1 : 1 respectively. Their balance sheet as on 31st March, 2018 is as shown below:
Balance Sheet as on 31st March, 2018

On 1st April, 2018 Sachin retired and the following adjustments have been agreed upon:
1. Goodwill was revalued on ₹ 50,000.
2. Assets and Liabilities were revalued as follows:
Debtors ₹ 50,000, Live stock ₹ 45,000, Building ₹ 1,25,000, Plant and Machinery ₹ 30,000, Motor truck ₹ 95,000 and Creditors ₹ 30,000.
3. Rohan and Rohit contributed additional capital through Net Banking of ₹ 50,000 and ₹ 25,000 respectively.
4. Balance of Sachin’s Capital Account is transferred to his Loan Account.
Give Journal entries in the books of new firm.
Solution:
Journal entries in the books of Partnership Firm

Working Notes:
1. Calculation of Profit on Revaluation of Assets and Liabilities.

Question 5.
Shah, Lodha and Dhole were partners sharing profits and losses in the ratio of 4 : 3 : 3. Their Balance Sheet as on 31st March, 2019 is given below:
Balance Sheet as on 31st March, 2019

On 1st April, 2019 Mr. Lodha retired from the firm on the following terms:
1. Goodwill is to be valued at an average profits and losses of the last five years which were as follows:
Year – Profit/Loss
2015 – ₹ 35,000
2016 – ₹ 20,000
2017 – ₹ 30,000
2018 – ₹ 20,000
2019 – ₹ 25,000
2. Computers to be depreciated by 10%.
3. Furniture to be revalued at ₹ 27,500.
4. Vehicles appreciated by 20%.
5. R.D.D. was no longer necessary.
6. Shah and Dhole will share the future profits and losses in the ratio of 2 : 1.
7. It was decided that goodwill should not appear in the books of a new firm and amount payable to Lodha is to be transferred to his Loan A/c.
Prepare: Profit and Loss Adjustment A/c, Partners’ Capital Accounts, Balance Sheet of new firm.
Solution:
In the books of Partnership Firm

Balance Sheet as on 1st April 2019

Working Note:
Average profit = \(\frac{\text { Total Profit }}{\text { No. of Years }}\)
= \(\frac{35,000+20,000+30,000+20,000+25,000}{5}\)
= \(\frac{1,30,000}{5}\)
= ₹ 26,000
∴ Goodwill = ₹ 26,000
Goodwill should not appear in the books of accounts.
Therefore, ₹ 26,000 credited in Partners’ Capital Account in partners’ old profit and loss ratio. ₹ 26,000 will be debited in Partners’ Capital Account in partners’ new profit-loss ratio.

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 8 Transition and Inner Transition Elements Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 8 Transition and Inner Transition Elements

Question 1.
What are d-block elements? Give their general electronic configuration.
Answer:
Definition : d-block elements are defined as the elements in which the differentiating electron enters d-orbital of the penultimate shell i.e. (n – 1) d-orbital where ‘n is the last shell.

The general electronic configuration can be represented as, (n – n) d^{n – 10}, ns^{n – 2}

Question 2.
What is the position of the transition elements in the periodic table?
Answer:
The transition elements are placed in periods 4 to 7 and groups 3 to 12 of the periodic table.

Question 3.
In which block of the modern periodic table are the transition elements placed?
Answer:
Transition elements are placed in d-block of the modern periodic table.

Question 4.
Why are most of the d-block elements called transition elements?
Answer:

• d-block elements have electronic configuration,(n – n) d^{n – 10}, ns^{l – 2}. They are all metals.
• In the periodic table, they are placed between the ,s-block and p-block elements, i.e., in the groups between 2 and 13.
• They show characteristic properties which are intermediate between those of the elements of s-block and p-block.
• Hence, they show a transition in the properties from those of the most electropositive .v-block elements and less
• electropositive (or electronegative) p-block elements.
• Therefore, most of the d-block elements are called transition elements.

Question 5.
How many series of d-block elements are present in the long-form periodic table? Give their general electronic configuration.
Answer:
There are four series of d-block elements which are placed between 5 and p-block elements in the long-form periodic table as follows :

d-series	Period	Electronic configuration
(1) 3d-series	fourth	[Ar] 3d1 – 10, 4s1 – 2
(2) 4d-series	fifth	[Kr] 4d1 – 10, 5s1 – 2
(3) 5d-series	sixth	[Xe] 4f14 5d1 – 10 6s1 – 2
(4) 6d- series	seventh	[Rn] 5f14 6d1 – 10 7s2

Modern periodic table :

Question 6.
Represent the elements in the four series of transition elements.
Answer:

Question 7.
In which period of the periodic table, will an element, be found whose differentiating electron is a 4d-electron?
Answer:
An element whose differentiating electron is a 4d-electron will be present in fifth period of the periodic table.

Question 8.
Write the condensed electronic configuration of each series of transition elements.
Answer:
Condensed Electronic Configuration of Transition Elements

Question 9.
Write expected and observed electronic configuration of 3d-series block elements.
Answer:
Electronic configuration of 3d-series of d-block elements

Question 10.
Explain why transition elements with electronic configuration 3d44s2 and 3d94s2 do not exist.
Answer:
(1) d-orbitals are degenerate orbitals and they acquire extra stability when half-filled (3d⁵) or completely filled (3d¹⁰). Hence 3d⁴ and 3d⁹ electronic configurations are less stable.
(2) The energy difference between 3d and 4.s’ subshells is very low, hence there arises a transfer of one electron from 45 orbital to 3d orbital.
The electronic configuration changes as,
3d⁴, 4s² → 3d⁵ 4s¹
3d⁹, 4s² → 3d¹⁰ 45¹
Therefore transition elements, with electronic configurations 3d⁴, 4s² and 3d⁹, 4s² do not exist.

Question 11.
Write observed electronic configuration of elements from first transition series having half-filled d-orbitals.
Answer:
There are two elements namely Cr and Mn which have half-filled d-orbitals.
₂₄Crls²2s²2p⁶3s²3p⁶3d⁵4s¹
₂₅Mnls²2s²2p⁶3s²3p⁶3d⁵4s²

Question 12.
Explain the variable oxidation states of metals of first transition series.
Answer:

• The transition metals (or, elements) exhibit variable oxidation states due to their electronic configuration, (n – 1) d^{1 – 10} ns^{1 – 2} for the first row.
• They show only positive oxidation states due to loss of electrons from outer 45-orbital and the penultimate 3rf-orbital.
• Loss of one 45 electron forms M+ ion. Loss of two 45 electrons form M²⁺ ion.
• +2 is the common oxidation state of these elements.
• Higher oxidation states are due to loss of 3 d-electrons along with 45 electrons.
• As the number of unpaired electrons increases, the number of oxidation states shown by the element also increases.
• Sc has only one unpaired electron and it shows two oxidation states ( + 2 and + 3)
• Mn with 5 unpaired d electrons show six different oxidation states. They are +2, +3, +4, +5, +6 and + 7. Thus Mn has the highest oxidation state.
• From Fe onwards variable oxidation states decreases as the number of unpaired electron decreases.
• The last element in the series, Zn shows only one oxidation state ( + 2).

Question 13.
Show different oxidation states of 3d-series of transition elements.
Answer:
The following table shows, different oxidation states of 3d-series of transition elements.

Question 14.
Write oxidation states and outer electronic configuration of first transition series elements.
Answer:
Oxidation states of first transition series elements

Question 15.
Zinc shows only one oxidation slate. Explain.
Answer:

• The electronic conliguration of zinc is, ₃₀Zn Is² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² or [Ar] 3d¹⁰ 4s².
• Due lo loss of two electrons from 4s suhshell Zn shows oxidation state +2. with elcctronic configuration. [Ar] ¹⁸3d¹⁰.
• Since Zn⁺² acquires an extra stability of completely fIlled 3d¹⁰ orbital. it shows only one oxidation state + 2.

Question 16.
Why is manganese more stable in the + 2 state than the + 3 state and the reverse is true for iron?
Answer:

• The electronic configuration of Mn is ₂₅Mn [Ar] 3d⁵ 4s²
• In + 2 and + 3 oxidation states, the electronic configuration of Mn is, Mn²⁺ [Ar] 3d⁵ and Mn³⁺ [Ar] 3d⁴
• Since half-filled d-orbital (3d⁵) has more stability and lower energy than 3d⁴, Mn²⁺ is more stable than Mn³⁺.
• The electronic configuration of Fe is ₂₆Fe [Ar] 3d⁶ 4s² In + 2 and + 3 oxidation states of Fe, the electronic configuration is, Fe²⁺ [Ar] 3d⁶ and Fe³⁺ [Ar] 3d⁵ Since half-filled orbital is more stable, + 3 state of Fe is more stable than + 2 state.

Question 17.
What are the electronic configurations of various ions of 3d-elements?
Answer:
Electronic configuration of various ions of 3d elements

Question 18.
Scandium shows only two oxidation states. Explain.
Answer:
Scandium has electronic configuration, ₂₁Sc : Is², 2s², 2p⁶, 3s², 3p⁶, 3d¹, 4s² Sc shows only two oxidation states namely + 2 and + 3.

• Due to the loss of two electrons from the 4s-orbital, Sc acquires + 2 oxidation state Sc^{2 +} : Is² 2s² 2p⁶ 3s² 3p⁶ 3d¹.
• Due to the loss of one more electron from the 3d-orbital, it acquires + 3 oxidation state with the extra stability of an inert element ₁₈Ar. Sc⁺³ : Is² 2s² 2p⁶ 3s² 3p⁶.
• Since Sc³⁺ acquires extra stability of inert element [Ar]¹⁸, it does not form higher oxidation state.

Question 19.
Write different oxidation states of iron.
OR
Write the electronic configurations of
(i) Fe
(ii) Fe²⁺ and
(iii) Fe³⁺.
Answer:
Oxidation states of iron are +2, +3, +4, +5, +6.
(i) ₂₆Fe : ls²2s²2p⁶3s²3p⁶3d⁶4s²
(ii) Fe²⁺ : Is² 2s² 2p⁶ 3s² 3p⁶ 3d⁶
(iii) Fe³⁺ : Is² 2s² 2p⁶ 3s² 3p⁶ 3d⁵.

Question 20.
Explain different oxidation states of chromium.
Answer:

• The observed electronic configuration of chromium is, ₂₄Cr [Ar] 3d⁵ 4s¹.
• Different possible oxidation states of Cr are 4-1 (3d⁵), + 2 (3d⁴), + 3 (3d³), + 4 (3d²), + 5 (3d¹) and + 6 (3d°).
• Although in + 1 state, Cr gets extra stability of half-filled 3d⁵-orbital, it does not exhibit + 1 state in common except with pyridine.
• Cr⁺² has few stable salts like CrCl₂, CrSO₄ while Cr⁺³ forms very stable salts like CrCl₃.
• Cr⁺⁴ and Cr ^{+ 5} are unstable oxidation states.
• Cr⁺⁶ is the most stable state due to inert gas [Ar] electronic configuration and form the salts like K₂Cr₂O₇.

Question 21.
Manganese shows variable oxidation states. Give reasons.
Answer:

• Manganese (₂₅Mn) has electronic configuration. ₂₅Mn [Ar]¹⁸ 3d⁵ 4s².
• Mn has stable half-filled d-subshell.
• Due to a small difference in energy between 3d and 4s-orbitals, Mn can lose or share electrons from both the orbitals, hence shows variable oxidation states.
• Mn shows oxidation states ranging from + 2 to + 7.

Question 22.
Write the different oxidation states of manganese. Why is + 2 oxidation state of manganese more stable than Mn³⁺?
Answer:

• The different oxidation states of Mn are +2, +3, +4, +5, + 6 and +7.
• The electronic configuration of Mn is Is² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²
• + 2 oxidation state is very stable due to higher stability of half-filled 3d orbital.
• Mn³⁺ has electronic configuration, ls²2s² 2p⁶3s²3p⁶3d^A which is less stable.

Question 23.
Write the physical properties of first transition series.
Answer:
Physical properties of first transition series :

• All transition elements of the first series are metals.
• Except Zn, they are very hard and have low volatility.
• They show characteristic properties of metals. They are lustrous, malleable and ductile.
• They are good conductors of heat and electricity.
• They have high melting points and boiling points.
• Except Zn and Mn, they have one or more typical metallic structures at normal temperatures.

Question 24.
Which elements in the transition elements, 3d-series has
(i) the lowest density
(ii) the highest density?
Answer:
In 3d transition elements,
(i) Scandium (Sc) has lowest density and
(ii) Zinc (Zn) has the highest density.

Question 25.
Explain the variation in density of d-block elements.
Answer:
The densities of d-block elements are higher than 5-block elements due to higher nuclear charge which results in reduction in atomic size.

Question 26.
Explain the variation in melting points of the transition elements.
Answer:

1. All transition elements are metals and the strength of metallic bonding increases as the number of unpaired electrons increases.
2. In transition elements as atomic number increases, the number of unpaired electrons increases from (n – 1)d¹ to (n – 1 )d⁵.
   For example in 3d-series, melting points increase from ₂₁Sc to ₂₄Cr in 4d-series from ₃₉Y to ₄₂Mo, and in 5d-series from ₇₂Hf to ₇₄W.
3. After (n – l)d⁵ electronic configuration, the electrons start pairing, hence the number of unpaired electrons decrease, hence metallic character, melting points decrease from (n-1 )d6 to (n – 1)d¹⁰.
4. In all transition series the melting point increases steadily up to d⁵ configuration and after this melting point decreases regularly.
   

Question 27.
The first ionisation enthalpies of third transition series elements are much higher than those of the elements of first and second transition series. Explain.
Answer:

1. Third transition series elements have electronic configuration, 4f¹⁴ 5d^{1 – 10} 6s².
2. Thus, atoms of third series elements possess filled 4f-orbitals.
3. 4f-orbitals due to their diffused shape, exhibit poor shielding effect and give rise to lanthanide contraction. Hence the valence electrons experience greater nuclear attraction and greater amount of energy is required to ionise the elements of third transition series namely (H_f to Au).
4. Therefore the ionisation enthalpies of third transition series elements are much higher than those of the first and second transition series.

Question 28.
Explain the metalic character of transition metals.
Answer:

• All the transition elements are metals.
• They are hard, lustrous, malleable, ductile and they have high tensile strength.
• They have high melting points and boiling points.
• Their metallic character is due to vacant or partially filled (n – 1) d-orbitals, and they involve both metallic and covalent bonding.
• Since the strength of metallic bonds depends upon the number of unpaired electrons, it increases up to middle i.e., up to (n – 1 )d⁵, hence accordingly melting points and boiling points also increase.
• After (n – l)d⁵ configuration, the electrons start pairing, hence the metallic strength, melting points and boiling points decrease with the increase in atomic number.

Question 29.
How does metallic character vary in 3d transition elements?
Answer:

1. In 3d-series elements as atomic number increases from scandium (Sc [Ar]¹⁸ 3d¹ 4s²) the number of unpaired electrons increases up to 3d⁵ in chromium.
2. As the number of unpaired electrons increases, the metallic character increases, hence the melting points and boiling points increase from ₂₁Sc(3d¹) to ₂₄Cr (3d⁵).
3. After chromium the number of unpaired electrons goes on decreasing due to the pairing of electrons, hence metallic character, melting points and boiling points decrease from ₂₅Mn to ₂₉Cu.
4. Zinc has all electrons paired, hence it is soft, has a low melting and boiling points.

Question 30.
Which are the common arrangement of the atoms in the structure of transition metals?
Answer:
Most of the transition metals have simple hexagonal closed packed (hep), cubic closed packed (ccp) or body centred cubic (bcc) lattices.

Question 31.
Why do the compounds of transition metals exhibit magnetic properties?
Answer:
The compounds of transition metals exhibit magnetic properties due to the presence of unpaired electrons in their atoms or ions.

Question 32.
What is the cause of paramagnetism and ferromagnetism?
Answer:
Paramagnetism and ferromagnetism is due to the presence of unpaired electrons in species.

Question 33.
When does species become diamagnetic?
Answer:
When there is no unpaired electron, i.e. all electron spins are paired, the species become diamagnetic.

Question 34.
How do metals Fe, Co, Ni acquire permanent magnetic moment?
Answer:
The transition metals Fe, Co and Ni are ferromagnetic. When the magnetic field is applied, all the unpaired electrons in these metals (and their compounds) align in the direction of the applied magnetic field. Due to this the magnetic susceptibility is enhanced and these metals can be magnetised, that is, they acquire permanent magnetic properties.

Question 35.
In which oxidation state, is vanadium diamagnetic?
Answer:

• The electronic configuration of vanadium is, ₂₃V [Ar] 3d³ 4s².
• In +5 oxidation state, the electronic configuration is, V⁵⁺ [Ar].
• Since in V⁵⁺ state, vanadium has all electrons paired, it is diamagnetic.

Question 36.
How is a magnetic moment expressed?
Answer:
The magnetic moment is expressed in Bohr magneton (B.M.). It is denoted by μ.

Question 37.
What is Bohr magneton (B.M.)?
Answer:
Bohr magneton (B.M.) is a unit of magnetic moment :
\(1 \mathrm{~B} . \mathrm{M} .=\frac{e h}{4 \pi m_{\mathrm{e}} c}\)
where, h : Planck’s constant (h = 6.626 x 10^-34 Js)
e : electronic charge (1.60218 x 10^-19 C)
me : mass of an electron (9.109 x 10^-31 kg)
c : velocity of light. (2.998 x 10⁸ ms^-1)

Question 38.
Explain the magnetic properties of transition (or d-block) elements.
Answer:

• Most of the transition metal ions and their compounds are paramagnetic in nature due to the presence of one or more unpaired electrons in their (n – 1)d-orbitals. Hence they are attracted in the magnetic field.
• As the number of unpaired electrons increases from 1 to 5 in J-orbitals, the paramagnetic character and magnetic moment increase.
• The transition elements or their ions having all electrons paired are diamagnetic and they are repelled in the magnetic field.
• Metals like Fe, Co and Ni possess very high paramagnetism and acquire permanent magnetic moment hence they are ferromagnetic.

Question 39.
Explain the effective magnetic moment of the species.
Answer:

• The magnetic moment in the species arises due to the presence of unpaired electrons.
• The magnetic moment depends upon the sum of orbitals and spin contribution for each unpaired electron present in the species.
• In transition metal ions, the contribution of orbital magnetic moment is suppressed by the electrostatic field of other atoms, molecules or ions surrounding the metal ion in the compound.
• Hence the net or effective magnetic moment arises mainly due to spin of electrons. The effective magnetic moment μ_eff, of a paramagnetic substance is given by 1 spin only’ formula represented as, \(\mu=\sqrt{n(n+2)}\) B.M. where n is the number of unpaired electrons.

Question 40.
What is the importance of magnetic moment (μ)?
Answer:

• From the measurements of the magnetic moment (μ) of the species or metal complexes of the first row of transition elements, the number of unpaired electrons can be calculated with the spin-only formula.
• As magnetic moment is directly related to the number of unpaired electrons, value of μ will vary directly with the number of unpaired electrons.
• In 2nd and 3rd transition series, orbital angular moment is significant. Hence spin-only formula for the complexes of 2nd and 3rd transition series is not useful.

Question 41.
Calculate the magnetic moment of the following species :
(1) Cr³⁺
(2) Co
(3) Co³⁺
(4) Cu^{2 +}.
Answer:

Question 42.
Explain : A slight difference in the calculated and observed values of magnetic moments.
Answer:
Magnetic moments are determined experimentally in solution or in solid state where the central atom or ion is hydrated or bound to ligands. Hence a slight difference is observed in calculated and experimentally obtained values of magnetic moment (μ).

Question 43.
Calculate the magnetic moment of a divalent ion in an aqueous solution, if its atomic number is 24.
Answer:
(1) The electronic configuration of divalent inri M²⁺ having atomic number 24 is.

The ion has number of unpaired electrons. n = 4.
By spin only’ formula, the magnetic μ is given by, \(\mu=\sqrt{n(n+2)}=\sqrt{4(4+2)}=4.90 \mathrm{~B} . \mathrm{M}\)
(This M²⁺ ion is Cr²⁺ ion)

Question 44.
When does a substance appear coloured?
Answer:
A substance appears coloured when it absorbs a portion of visible light. The colour depends upon the wavelength of absorption in the visible region of electromagnetic radiation.

Question 45.
Why do the d-block elements form coloured compounds?
Answer:

• Compounds (or ions) of many d-block elements or transition metals are coloured.
• This is due to the presence of one or more unpaired electrons in (n – 1) d-orbital. The transition metals have incompletely filled (n – 1) cf-orbitals.
• The energy required to promote one or more electrons within the d-orbitals involving d-d transitions is very low.
• The energy changes for d-d transitions lie in visible region of electromagnetic radiation.
• Therefore transition metal ions absorb the radiation in the visible region and appear coloured.
• Colour of ions of d-block elements depends on the number of unpaired electrons in (n – 1) d-orbital. The ions having equal number of unpaired electrons have similar colour.
• The colour of metal ions is complementary to the colour of the radiation absorbed.

Question 46.
How is complementary colour of a compound identified?
Answer:

1. The transition metal ions absorb the radiation in the visible region and appeared coloured.
2. Metal ion absorbs radiation of certain wavelength from the visible region. Remaining light is transmitted and the observed colour corresponds to the complementary colour of the light observed.
3. The complementary colour can be identified (with the diagram given).

For example if red colour is absorbed then transmitted complementary colour is green.

Question 47.
Write outer electronic configuration (d-orbital) and colour of 3d-series of transition metal ions.
Answer:
Colour of 3d-transition metal ions

Question 48.
Mention the factors on which the colour of a transition metal ion depends.
Answer:
The factors on which the colour of transition metal ion depends are as follows :

• The presence of incompletely filled d-orbitals in metal ions. (The compounds with the configuration d° and d’0 are colourless.)
• The presence of unpaired electrons in d-orbitals.
• d → d transitions of electrons due to absorption of radiation in the visible region.
• Nature of groups (anions) (or ligands) linked to the metal ion in the compound or a complex.
• Type of hybridisation in metal ion in the complex.
• Geometry of the complex of the metal ion.

Question 49.
Give reasons : Zinc salts are colourless.
Answer:

• Colour of the ions of d-block elements depends on the number of unpaired electrons in (n – 1) d-orbitals.
• Zinc forms salts of Zn²⁺ ions.
• The electronic configuration of Zn⁺² is [Ar] 3d¹⁰.
• Since Zn⁺² does not have unpaired electrons in 3d-orbital, d→d transition cannot take place, hence, Zn⁺² ions form colourless salts.

Question 50.
Explain : The compounds of Cu(II) are coloured.
Answer:

• The electronic configuration of ₂₉Cu [Ar] 3d¹⁰ 4s¹ and Cu²⁺ [Ar] 3d⁹.
• In copper compounds Cu²⁺ ions have incompletely filled 3d-orbital (3d⁹).
• Due to the presence of one unpaired electron in 3 d-orbital, Cu²⁺ ions absorb red light from visible spectrum and emit blue radiation due to d → d transition. Therefore, copper compounds are coloured.

Question 51.
Explain why the solution of Ti³⁺ salt is purple in colour.
OR
Why is Ti³⁺ coloured? (atomic number Ti = 22)
Answer:

• Ti²⁺ ions in the aqueous solution exist in the hydrated complex form as [Ti(H₂O)₆]²⁺.
• The electronic configuration of Ti is, ₂₂Ti [Ar]¹⁸ 3d² 4s² and Ti³⁺ [Ar]¹⁸ 3d¹. Hence in complex, Ti³⁺ has one unpaired electron in 3d subshell.
• Initially, the 3d electron occupies lower energy d-orbital (in t_2g-orbitals).
• On the absorption of radiations of about 500 nm in yellow green region by a complex, 3d¹ electron is excited to the higher energy d-orbital (e_g-orbitals).
• When the electron returns back to the lower energy d-orbital (t_2g), it transmits radiation of complementary colour i.e. red blue or purple colour. Hence, the solution of hydrated Ti³⁺ is purple.

Question 52.
What will be the colour of Cd²⁺ salts? Explain.
Answer:

• The electronic configuration of, ₄₈Cd [Kr]³⁶ 3d¹⁰ 5s² and Cd²⁺ [Kr]³⁶ 3d¹⁰.
• Cd²⁺ ions have completely filled 3d subshell and there are no unpaired electrons in 3d-orbital.
• Hence d → d transition is not possible.
• Therefore, Cd²⁺ ions do not absorb radiations in the visible region and the salts of Cd²⁺ ions are colourless (or white).

Question 53.
Indicate which of the ions may be coloured- V³⁺, Sc³⁺, Cr³¹, Cu²⁺, Ti³⁺, Cu⁺
Answer:

• V³⁺ [Ar]¹⁸ 3d²-((green)
  Since there are two unpaired electrons available, for d → d transition, it will show a Green colour.
• Sc³⁺ [Ar]¹⁸ 3d° (colourless/white).
  Since there are no unpaired electrons in the 3d subshell, it will not show colour.
• Cr³⁺ [Ar]¹⁸ 3d³ – (violet)
  There are three unpaired electrons in the 3d subshell, hence due to d → d transition, it will show violet colour.
• Cu²⁺ [Ar]¹⁸ 3d⁹ (blue)
  It has one unpaired electron that can undergo a d → d transition, hence it will show the colour blue.
• Ti³⁺ [Ar]¹⁸ 3d¹ (purple)
  It has one unpaired electron that can undergo a d → d transition, hence it will show the colour purple.
• Cu¹⁺ [Ar]¹⁸ 3d¹⁰ (colourless)
  There are no unpaired electrons in the 3d subshell, hence it will not show colour.

Question 54.
Explain why is cobalt chloride pink in colour when dissolved in water but turns deep blue when treated with concentrated hydrochloric acid.
Answer:

• The electronic configuration of ₂₇Co : [Ar] 3d¹4s² and Co²⁺ [Ar] 3d¹.
• When dissolved in water cobalt chloride, Co²⁺ forms pink complex, [Co(H₂O)₆]²⁺.
• The complex has octahedral geometry.
• Due to absorption of radiation in the visible region and d – d transition, it forms pink coloured solution.
• When CoCl₂ solution is treated with concentrated HCl solution it turns deep blue.
• This change is due to the formation of another complex, [CoC1₄]²⁺ which has a tetrahedral geometry.
• Thus due to a change in geometry of the complex formed the colour of the solution changes from pink to deep blue.

Question 55.
Explain the catalytic properties of the rf-block or transition metals.
Answer:

• d-block elements or transition metals and their compounds or complexes influence the rate of a chemical reaction and hence act as catalysts.
• In homogeneous catalysis a catalyst forms an unstable intermediate compound which decomposes into products and regenerates the catalyst. But transition metals involve heterogeneous catalysis.
• The transition metals have incompletely filled d-subshells which adsorb reactants on the surface and provide a large surface area for the reactants to react.
• Since transition metals have variable oxidation states they are very good catalysts.
• Hence, compounds of Fe, Co, Ni, Pt, Pd, Cr etc are used as catalysts in many reactions.

Question 56.
Explain the use of different transition metals as catalysts.
Answer:
The transition metals are very good catalysts.

• MnO₂ is used as a catalyst in the decomposition of KClO₃.
  
• In the manufacture of ammonia by Haber’s process, Mo/Fe is used as a catalyst.
  
• In the synthesis of gasoline by Fischer Tropsch process, Co-Th alloy is used as a catalyst.
• Finely divided Ni (formed by reduction of heated oxide in hydrogen) is very efficient catalyst in hydrogenation of ethene to ethane at 140 °C.
  
• Commercially, hydrogenation with Ninkel as catalyst is used to convert inedible oils into solid fat for the production of margarine.
• In the contact process of industrial production of sulphuric acid, sulphur dioxide and oxygen (from air) react reversibly over a solid catalyst of platinised asbestos.
  
• Carbon dioxide and hydrogen are formed by the reaction of carbon monoxide and steam at 500 °C with Fe-Cr catalyst.
  

Question 57.
What are interstitial compounds of transition metals?
Answer:

• The interstitial compounds of the transition metals are those which are formed when small atoms like H, C or N are trapped inside the interstitial vacant spaces in the crystal lattices of the metals.
• Sometimes, sulphides and oxides are also trapped in the crystal lattices of transition elements.
• Presence of these elements in the crystal lattices of metals provide new properties to the metals.

Question 53.
Give one example of an interstitial compound.
Answer:
Steel and cast iron are examples of interstitial compounds of carbon and iron.

Question 54.
Give examples of interstitial compounds where the property of the transition metal is changed.
Answer:
Steel and cast iron are interstitial compounds of carbon and iron (carbides of iron). Due to the presence of carbon, the malleability and ductility of iron is reduced while its tenacity increases.

Question 55.
What are the properties of the interstitial compounds of transition metals?
Answer:

• The chemical properties of the interstitial compounds are the same as that of parent transition metals.
• They are hard and show the metallic properties like electrical and thermal conductivity, lustre, etc.
• Since metal-non-metal bonds in the interstitial compounds are stronger than metal-metal bonds in pure metals, the compounds have very high melting points, higher than the pure metals.
• They have lower densities than the parent metal.
• The interstitial compounds containing hydrogen (i.e., hydrides of metals) are powerful reducing agents.
• The compounds containing carbon, hence behaving as carbides, are chemically inert and extremely hard like diamond.
• In these compounds, malleability and ductility are changed. For example steel and cast iron.

Question 56.
What are interstitial compounds? Why do these compounds have higher melting points than corresponding pure metals?
Answer:

1. The interstitial compounds of the transition metals are those which are formed when small atoms like H, C or N are trapped inside the interstitial vacant spaces in the crystal lattice of the metals.
2. Since metal-nonmetal bonds in the interstitial compounds are stronger than metal-metal bonds in pure metals, the compounds have very high melting points, higher than the pure metals.

Question 57.
Explain the formation of alloys of transition metals.
Answer:

• The transition metals form a large number of alloys among themselves, which are hard with high melting points.
• During alloy formation atoms of one metal are distributed randomly in the lattice of another metal.
• The metals with similar atomic radii and similar properties readily form alloys.
• These alloys have industrial importance.
• The alloys can be ferrous alloys or nonferrous alloys.

Question 58.
How are the transition metal alloys classIfied?
Answer:
The transition metal alloys are classified into

• Ferrous alloys
• Nonferrous alloys.

Question 59.
Explain what are
(1) ferrous alloys and
(2) nonferrous alloys.
Answer:

1. Ferrous alloys: In ferrous alloys, atoms of other elemems are distributed randomly in atoms of iron in the mixture. As the percentage of iron is more in these alloys, they are termed as ferrous alloys. For expamle : nickel steel, chromium steel, stainless steel, (All steels have abot 2% carbon)
2. onferrous alloys : These are formed by mixing atoms of transition metal other than iron with a non transition elemeni. For example, brass is an alloy of Cu and Zn. Bronze is an alloy of Cu and Sn.

Question 60.
What are the uses of alloys?
Answer:

Name of alloy	Important use in industry
(1) Bronze (Cu + Sn)	In making statues, medals and trophies (as it is tough, strong and corrosion-resistant)
(2) Cupra-nickel (Cu + Ni)	In making machinery parts of marine ships, boats, marine conden­ser tubes.
(3) Stainless steel	In the construction of the outer fuselage of ultra-high-speed aircraft.
(4) Nichrome : (Ni+ Cr in the ration 80 : 20)	For gas turbine engines.
(5) Titanium alloys	For ultra-high-speed flight, fireproof bulkheads and exhaust shrouds (as they withstand high temperatures).

Question 61.
Write the preparation of potassium permanganate.
Answer:
Potassium permanganate (KMnO₄) is prepared in the following steps,

(1) Chemical Oxidation : When finely divided manganese dioxide (Mn0₂) is heated strongly with fused caustic potash (KOH) and an oxidising agent potassium chlorate (KCIO₃), dark green potassium manganate (K₂MnO₄) is obtained. (In neutral or acidic medium K₂MnO₄ disproportionates.)

The liquid is filtered through glass wool or sintered glass and evaporated. Potassium manganate crystallises as small, blackish crystals.

(2) Oxidation of K₂MnO₄ by
(i) Electrolytic oxidation : An alkaline solution of manganate ion is electrolysed between iron electrodes separated by a diaphragm. Manganate ion \(\left(\mathrm{MnO}_{4}^{2-}\right)\) undergoes oxidation at anode forming permanganate ion \(\left(\mathrm{MnO}_{4}^{-}\right)\). Oxygen evolved at anode converts \(\left(\mathrm{MnO}_{4}^{2-}\right)\) to \(\left(\mathrm{MnO}_{4}^{-}\right)\).

The overall reaction is as follows :
2K₂MnO₄ + H₂O + [O] → 2KMnO₄ + 2KOH

The electrolytic solution is filtered and evaporated to obtain deep purple black crystals of KMn0₄.

(ii) By passing CO₂ through the solution of K₂MnO₄ :
3K₂MnO₄ + 4CO₂ + 2H₂O → 2KMnO₄ + MnO₂ + 4 KHCO₃

Question 62.
What is meant by the disproportionation of an oxidation state? Explain giving example of manganese.
Answer:

1. Disproportionation reaction is a chemical reaction in which atom or an ion of an element forms two or more species having different oxidation states, one lower and one higher.
2. Manganese (Mn) shows different oxidation states + 2 to +7.
3. When one oxidation state, lower or higher oxidation state becomes unstable as compared to another oxidation state, it undergoes disproportionation reaction.
4. For example, + 6 oxidation state of Mn is less stable than + 7 and + 4.
   • Hence, in acidic medium \(\mathrm{Mn}^{6+} \text { in } \mathrm{MnO}_{4}^{2-}\) undergoes disproportionation reaction.
     
   • In neutral medium green K₂MnO₄ disproportionates to KMn04 and MnO₂.
     

Question 63.
Give examples of oxidising reactions of KMnO₄.
Answer:
(1) KMnO₄ in acidic medium :

(2) KMnO₄ in neutral or alkaline medium in neutral or weakly alkaline medium :
(i) Iodide is oxidised to iodate ion.

(ii) Thiosulphate ion is oxidised to sulphate ion.

(iii) Manganous salt is oxidised to MnO₂.

Question 64.
Balance the following equations :
KI + KMnO₄ + H₂SO₄ → K₂SO₄ + MnSO₄ + 8H₂O + I₂
H₂S + KMnO₄ + H₂SO₄ → K₂SO₄ + MnSO₄ + H₂O + S.
Answer:
10 KI + 2KMnO₄ + 8H₂SO_{4 →} 6K₂SO₄ + 2MnSO₄ + 8H₂O + 5I₂
5H₂S + 2KMnO₄ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 8H₂O + 5S.

Question 65.
Give the uses of potassium permanganate.
Answer:
Uses of potassium permanganate :

• as an antiseptic.
• as a powerful oxidising agent in laboratory and industry.
• in the detection of unsaturation in organic compounds in the laboratory. (Baeyer’s reagent, alkaline KMnO₄).
• for detecting halides in qualitative analysis.
• in volumetric analysis for the estimation of H₂O₂, FeSO₄ etc.)

Question 66.
Write the formula of chromite ore.
Answer:
FeOCr₂O₃.

Question 67.
How is potassium dichromate manufactured from chromite ore (FeOCr₂O₃)?
Answer:
Manufacture of potassium dichromate (K₂Cr₂O₂) from chrome iron ore (FeOCr₂O₃) involves following steps :
(1) Concentration of ore : The chromite ore (FeOCr₂O₃) is powdered and washed with current of water.
(2) Conversion of chromite ore into sodium chromate : The concentrated ore is mixed with anhydrous sodium carbonate (Na₂CO₃) and a flux of lime in excess air and heated in a reverberatory furnace.

Sodium chromate (Na₂CrO₄) formed in the reaction is then extracted with water so that Na₂CrO₄ dissolves into solution and insoluble substances separate out.
(3) Conversion of Na2CrO₄ into sodium dichromate (Na₂Cr₄O₇) : Na₂CrO₄ solution is acidified with concentrated H₂SO₂, so that sodium chromate is converted into sodium dichromate.

Less soluble sodium sulphate crystallises out as Na₂SO₄.10H₂O. which is filtered off.
(4) Conversion of Na₂Cr₂O₇ into K₂Cr₂O₇ : Concentrated solution of Na₂Cr₂O₇ is treated with KCl on by double decomposition, K₂Cr₂O₇ is obtained.
Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl
On concentrating and cooling the solution, less soluble orange coloured K₂Cr₂O₇ crystallises out which is filtered and purified by recrystallisation.

Question 68.
What happens when hydrogen sulphide gas (H₂S) is passed through acidified K₂Cr₂O₇ solution?
Answer:
When hydrogen sulphide (H₂S) gas is passed into solution of K₂Cr₂O₇, H₂S is oxidised to a pale yellow solid (precipitate) of sulphur. Orange coloured solution becomes green due to formation of chromic sulphate (green coloured).

In the reaction, H₂S is oxidised to S and K₂Cr₂O₇ is reduced to Cr₂(SO₄)₃.

Question 69.
What are the common physical properties of d-block elements?
Answer:
The common physical properties of d-block elements are :

• All d-block elements are lustrous and shining.
• They are hard and have high density
• They have high melting and boiling points.
• They are good electrical and thermal conductors.
• They have high tensile strength and malleability.
• They can form alloys with transition and nontransition elements
• Many metals and their compounds are paramagnetic.
• Most of the metals are efficient catalysts.

Question 70.
What are the common chemical properties of d-block elements?
Answer:
The common chemical properties of the d-block elements are :

• All d-block elements are electropositive metals.
• They exhibit variable oxidation states and form coloured salts and complexes.
• They are good reducing agents.
• They form insoluble oxides and hydroxides.
• Iron, cobalt, copper, molybdenum and zinc are biologically important metals.
• They catalyse biological reactions.

Question 71.
Give examples to show that elements of first row of d-block elements differ from second and third row with respect to the stabilisation of higher oxidation states.
Answer:

• Highest oxidation state for the first row element is + 7 as in Mn.
  For the second row, the highest oxidation state is + 8 as in Ru (RuO₄).
  For the third row, the highest oxidation state is + 8 as in Os (OsO₄).
• Compounds of Mo(V) of 2nd row and W(VI) of 3rd row of transitional elements are more stable than Cr(VI) and Mn (VIII) of first row elements.

Question 72.
How do metals occur in nature?
Answer:
In nature, few metals occur in earth’s crust in free state or native state while other metals occur in the combined form.
(1) Elements in free state or native state : The metals which are non-reactive with air, water, CO₂ and non-metals occur in free state or native state. For example, gold, platinum, palladium occur in free state. Metals like Cu, Ag and Hg occur partly in the free state.

(2) Combined form : The metals which are reactive occur in the combined state with other elements forming compounds like oxides, sulphides, sulphates, carbonates, silicates, etc.

Question 73.
What are minerals?
Answer:
Minerals : They are naturally occurring chemical substances in the earth’s crust containing metal in free state or in combined form and obtainable from mining are called minerals. For example, haematite Fe203, galena PbS, etc.

Question 74.
What are ores?
Answer:
Ores : The minerals containing a high percentage of metals from which metals can be profitably extracted are called ores.
[Note : Every ore is a mineral but every mineral is not an ore.]

Question 75.
Write names of minerals and ores of Iron, Copper and Zinc.
Answer:

Metals	Mineral	Ore
Iron	Haematite Fe2O3 Magnetite Fe3O4 Limonite 2Fe2O3, 3H2O Iron pyrites FeS2 Siderite FeCO3	Haematite
Copper	Chalcopyrite CuFeS2 Chalcocite Cuprite Cu2O	Chalcopyrite Chalcocite
Zinc	Zinc blende ZnS Zincite ZnO Calamine ZnCO3	Zinc blende

Question 76.
What is metallurgy?
Answer:
Metallurgy : The process of extraction of metal in a pure state from its ore is called metallurgy.

Question 77.
Define the following:
(1) Pyrometallurgy
(2) Hydrometallurgy
(3) Electrometallurgy.
Answer:

1. Pyrometallurgy : It is a process of extraction of metal from metal oxide from concentrated ore by reduction with a suitable reducing agent like carbon, hydrogen, aluminium, etc. at high temperature.
2. Hydrometallurgy : It is a process of extraction of metals by converting their ores into aqueous solutions of metal compounds and reducing them by suitable reducing agents.
3. Electrometallurgy : It is a process of extraction of highly electropositive metals like Na, K, Al, etc. by electrolysis of fused compounds of the metals where metal ions are reduced at cathode forming metals.

Question 78.
What is gangue?
Answer:
Gangue : The earthly and undesired impurities of various substances like sand (SiO₂), metal oxides, etc. present in the ore are called gangue or matrix.

Question 79.
Define concentration of an ore.
Answer:
Concentration : A process of removal of gangue or unwanted impurities from the ore is called concentration of an ore. It is also called benefaction or dressing of an ore.

Question 80.
What are common methods of concentration of an ore?
Answer:
The concentration of an ore involves different methods depending upon the differences in physical properties of compounds or the metal present and the nature of the gangue.

The common methods of concentration of ore are as follows :

1. Gravity separation or hydraulic washing :
   This can be carried out by two processes as follows :
   • Hydraulic washing by using Wilfley’s table method
   • Hydraulic classifier methods.
2. Magnetic separation
3. Froth floatation process.
4. Leaching.

The method depends upon the nature of ore.

Question 81.
What is leaching?
Answer:
Leaching : ft is a (chemical) process used in the concentration of an ore by extracting soluble material from an insoluble solid by dissolving in a suitable solvent. This method is used in the concentration process of ores of Al, Ag, Au, etc.

Question 82.
What is roasting of an ore?
Answer:
Roasting : It is a process of strongly heating a concentrated ore in the excess of air below melting point of metal, to convert it into oxide form. It is used for a sulphide ore. For example, ZnS ore on roasting forms ZnO.

Question 83.
Write an equation to show how zinc blende (ZnS) is converted to ZnO.
Answer:
When zinc blende is roasted, it is converted to ZnO.
\(\mathrm{ZnS}+\mathrm{O}_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{ZnO}+\mathrm{SO}_{2}\)

Question 84.
Explain the term : Smelting
Answer:
Smelting : The process of extraction of a metal from its ore by heating and melting at high temperature is called smelting. Reduction of ore is carried out during smelting.

Question 85.
What is calcination?
Answer:
Calcination is a process in which the ore is heated to a high temperature below the melting point of the metal in the absence of air or limited supply of air in a reverberatory furnace.

It is generally used for carbonate and hydrated oxides to convert them into anhydrous oxides.

Question 86.
Define the terms :
(1) Flux
(2) Slag
Answer:
(1) Flux : A flux is a chemical substance which is added to the concentrated ore during smelting in order to remove the gangue or impurities by chemical reaction forming a fusible mass called slag.
(2) Slag : It is a waste product formed by combination of a flux and gangue (or impurities) during the extraction of metals by smelting process.

Iron is the fourth most abundant element in the earth’s crust.

Question 87.
What is the composition of haematite ore?
Answer:
Composition of Haematite ore is Fe2O₃ + SiO₂ + Al₂O₃ + phosphates

Question 88.
Which impurities (gangue) are present in haematite ore?
Answer:
SiO₂ and Al₂O₃ are the impurities present in the haematite ore.

Question 89.
Which reducing agents are used to reduce haematite ore into metallic iron?
Answer:
Haematite ore is reduced using coke and CO. Carbon in the coke is converted to carbon monoxide. Carbon and carbon monoxide together reduce Fe203 to metallic iron.

Fe₂O₃ + 3C → 2Fe + 3CO.
Fe₂O₃ + 3CO → 2Fe + 3CO₂.

Question 90.
Why is limestone used in the extraction of iron?
Answer:

• The ore of iron contains acidic gangue or impurity of silica, SiO₂.
• To remove silica gangue, basic flux like calcium oxide CaO, is required, which is obtained from the decomposition of limestone, CaCO₃. \(\mathrm{CaCO}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_{2}\)
• Silica reacts with CaO and forms a fusible slag of CaSiO₃.
  \(\mathrm{SiO}_{2}+\mathrm{CaO} \stackrel{\Delta}{\longrightarrow} \mathrm{CaSiO}_{3}\)

Therefore in the extraction of iron, lime is used.

Question 91.
Name the furnace in which iron is extracted from Haematite ore.
Answer:
Extraction of iron is carried out in Blast furnace.

Question 92.
Explain the extraction of iron from haematite.
Answer:
Iron is mainly extracted from haematite, Fe₂O₃ by reduction process.
Haematite ore contains silica (SiO₂), alumina (Al₂O₃) and phosphates as impurity or gangue.

Coke is used for the reduction of ore.

To remove acidic gangue SiO₂, a basic flux CaO is used which is obtained from lime stone CaCO₃.

The extraction process involves following steps :
(1) Concentration of an ore : The powdered ore is concentrated by gravity separation process by washing it in a current of water. The lighter impurities (gangue) are carried away leaving behind the ore.
(2) Roasting : The concentrated ore is heated strongly in a limited current of air. During this, moisture is removed and the impurities like S, As and phosphorus are oxidised to gaseous oxides which escape.

After roasting, the ore is sintered to form small lumps.
(3) Reduction (or smelting) : The roasted or calcined ore is then reduced by heating in a blast furnace.

The blast furnace is a tall cylindrical steel tower about 25 m in height and has a diameter about 5-10m lined with fire bricks inside.

Blast furnace has three parts :

• the hearth,
• the bosh and
• the stack.

At the top, there is a cup and cone arrangement to introduce the ore and at the bottom, tapping hole for withdrawing molten iron and an outlet to remove a slag.

The roasted ore is mixed with coke and limestone in the approximate ratio of 12 : 5 : 3.

A blast of hot air at about 1000 K is blown from downwards to upwards by layers arrangement. The temperature range is from bottom 2000 K to 500 K at the top. The charge of ore from top and the air blast from bottom are sent simultaneously. There are three zones of temperature in which three main chemical reactions take place.

(i) Zone of combustion : The hot air oxidises coke to CO which is an exothermic reaction, due to which the temperature of furnace rises.
C + ^1/2 O₂ → CO ΔH= – 220kJ
Some part of CO dissociates to give finely divided carbon and O₂.
2CO → 2C + O₂
The hot gases with CO rise up in the furnace and heats the charge coming down. CO acts as a fuel as well as a reducing agent.

(ii) Zone of reduction : At about 900 °C, CO reduces Fe₂O₃ to spongy (or porous) iron.
Fe₂O₃ + 3CO → 2Fe + 3CO₂
Carbon also reduces partially Fe₂0₃ to Fe.
Fe₂O₃ + 3C → 2Fe + 3CO

(iii) Zone of slag formation : At 1200 K limestone, CaCO₃ in the charge, decomposes and forms a basic flux CaO which further reacts at 1500 K with gangue (SiO₂, Al₂O₃) and forms a slag of CaSiO₃ and Ca₃AlO₃.
CaCO₃ + CaO + CO₂.
CaO + SiO₂ → CaSiO₃
12CaO + 2Al₂O₃ → 4Ca₃AlO₃ + 3O₂

The slag is removed from the bottom of the furnace through an outlet.

(iv) Zone of fusion : The impurities in ore like MnO₂ and Ca₃(PO₄)₂ are reduced to Mn and P while SiO₂ is reduced in Si. The spongy iron moving down in the furnace melts in the fusion zone and dissolves the impurities like C, Si, Mn, phosphorus and sulphure. The molten iron collects at the bottom of furnace. The lighter slag floats on the molten iron and prevents its oxidation.

The molten iron is removed and cooled in moulds. It is called pig iron or cast iron. It contains about 4% carbon.

Question 93.
Write the reaction involved in the zone of reduction in blast furnace during extraction of iron.
Answer:
Zone of reduction : At about 900 °C, CO reduces Fe₂O₃ to spongy (or porous) iron.
Fe₂O₃ + 3CO → 2Fe + 3CO₂
Carbon also reduces Fe₂O₃ to Fe.
Fe₂O₃ + 3C → 2Fe + 3CO

Question 94
Write reactions involved at different temperatures in the blast furnace.
Answer:

Temperature K	Change taking place in the blast furnace	Reactions
1. 500 K	Haematite ore loses moisture	ore xH2O → ore
2. 900 K	Reduction of ore by CO	Fe2O3 + 3CO → 2Fe + 3CO
3. 1200K	Limestone decomposes	CaCO3 → CaO + CO2
4. 1500K	Reduction of ore by C	Fe2O3 + 3C → 2Fe + 3CO
5. 1600 K	(i) Reduction of FeO by C (ii) Fusion of iron and slag formation	FeO + C → Fe + CO CaO + SiO2 → CaSiO3
6. 2000 K	Combustion of coke	2C + O2 → CO

Question 95.
What is the action of carbon on Fe203 in blast furnace?
Answer:
Fe₂O₃ + 3C → 2Fe + 3CO

Question 96.
What is refining of metals?
Answer:
Refining of metals : The purification of impure or crude metals by removing metallic and nonmetallic impurities is known as refining of metals. H

Question 97.
How is pure iron obtained from crude iron?
Answer:
Pure iron can be obtained by electrolytic refining.

Question 98.
Name the methods of refining of metals.
Answer:
Methods of refining of metals :

• Electrorefining
• Liquefaction
• Distillation
• Oxidation m

Question 99.
What are the factors that govern the choice of extraction technique of metals?
Answer:
The choice of extraction technique is governed by the following factors.

• Nature of ore
• Availability and cost of reducing agent. (Generally, cheap coke is used).
• Availability of hydraulic power.
• Purity of metal required.
• Value of by-products. For example. SO₂ obtained during the roasting of sulphide ores is important for the manufacture of H₂SO₄.

Question 100.
Which are the commercial forms of iron?
Answer:
Commercial forms of iron are :

• Cast iron
• wrought iron
• steel. H

Question 101.
(A) What are f-block elements?
(B) What are inner transition elements?
Answer:
(A)

• Elements in which differentiating electron enters into the pre-penultimate shell the (n – 2) f-orbital are known as f.block elements.
• They include 28 elements with atomic numbers ranging from 58-71 and atomic numbers 90 to 103 collectively.
• There are two f-series or two f-block elements, namely 4f and 5f series.
• The f-block includes two inner transition series namely the lanthanoid series. Cerium (58) to LuteUum (71) or the 4 f-block elements and the actinoid series. Thorium (90) to I.awrencium (103) or the 5f block elements.

(B) f-block elements are called inner transition elements since f-orbital lies much inside the f-orbital in relation to the transition metals, These elements have 1 to 14 electrons in their f-orbital.

Question 102.
What are fIrst inner transition elements?
Answer:

1. 4f-hlock elements are called (first) inner transition elements and have partly filled inner orbitaIs or (4f) orbitais.
2. They have general outer electronic configuration \((n-2) f^{1-14},(n-1) d^{0-1}, n s^{2}\).
3. There are two f-series, namely 4f and 5f series, called lanthanoids and acùnoids respectively.
4. They shos intermediate properties as compared to electropositive s-block elements and electronegative p-block elements. Hence they are called (first) inner transition elements.

Question 103.
What are lanthanoids (or lanthanides)?
OR
What is the lanthanoid series?
Answer:

• Lanthanoids or Lanthanoid series or Lanthanones : The series of fourteen elements from ₅₈Ce to ₇₁Lu in which a differentiating electron enters 4f sub-shell and follows lanthanum is called lanthanoid series and the elements are called lanthanoids.
• They have general electronic configuration, [Xe] 4f^1-14 ,5d^0-1, 6s².
• They follow Lanthanum (Z = 57) in 3d-series.

Question 104.
What are rare earths?
Answer:

• Lanthanoids or 4f-block elements are called rare earths.
• Lanthanoids are never found in free state, and their minerals are not pure.
• They exhibit similar chemical properties hence cannot be extracted and separated by normal metallurgical processes.
• Lanthanoid metals are available on small scale. Therefore they are called rare earths.

Question 105.
Explain the position of lanthanoids in the periodic table.
OR
How is the position of lanthanoids justified?
Answer:

1. Position of Lanthanoids in the periodic table : Group – 3; Period – 6.
2. They interrupt the third transition series of t/-block elements (i.e. 5 d series) in the sixth period.
3. They are 14 elements from ₅₈Ce to ₇₁Lu and their position is in between La and Hf. Since they follow lanthanum, they are called lanthanoids.
4. They are called 4f-series elements and for the convenience, they are placed separately below the main periodic table.
5. The actual position of lanthanoids is in between Lanthanum (Z = 57) and Hafnium (Z = 72).
6. Their position is justified due to following reasons :
   • All these elements have the same electronic configuration in ultimate and penultimate shells, one electron in 5d-orbital and two electrons in 6s-orbital.
   • Group valence of all lanthanoids is 3.
   • All lanthanoids from ₅₈Ce to ₇₁Lu have similar physical and chemical properties.

Question 106.
Explain the meaning of inner-transition series.
Answer:
A series of f-block elements having electronic configuration (n – 2)f^1-14 (n – I) d^0-1 ns² placed separately in the periodic table represents inner transition series. The f-orbitals lie much inside the e/ orbitals.

Since the last electron enters pre-penultimate shell, these elements are inner transition elements.

There are two inner transition series as follows :
4f-series ₅₈Ce → ₇₁Lu
5f-series ₉₀Th → ₁₀₃Lr

Question 107.
Draw a skeletal diagram of the periodic table to show the position of d and/- block elements.
Answer:

Question 108.
What are the properties of lanthanoids?
Answer:

• Lanthanoids are soft metak with silvery white colour, Colour and brightness reduces on exposure to air.
• They are good conductors of heat and electricity.
• Except promethium (P_m), all are non-radioactive in nature.
• The atomic and ionic radii decrease from La to Lu. (Lanthanoid contraction).
• Coordination numbers arc greater than 6.
• They are paramagnetic.
• They become ferromagnetic at lower temperature.
• Their magnetic and optical properties are independent of environment.
• They are called rare earths as their exiractioli was difficult.
• They are abundant in earth’s crust
• All lanthanoids fonn hydroxides which are ionic and basic. l3asicity decreases with atomic number,
• They react with nitrogen to give nitrides and with halogen to give halides.
  
• When heated with carbon at very high temperature give carbides
  

Question 109.
Explain the variations in ionisation enthalpy of lanthanoids.
Answer:

• The first ionisation enthalpy of lanthanoids is nearly same. It is very high for Gd and Yb.
• The ionisation enthalpy increases from first (IE₁] to third (IE₃).

First, second and third ionization enthalpies of lanthanoids in kj/mol

Lanthanoid	IE1	IE2	IE3
La	538.1	1067	1850.3
Ce	528.0	1047	1949
Pr	523.0	1018	2086
Nd	530.0	1034	2130
Pm	536.0	1052	2150
Sm	543.0	1068	2260
Eu	547.0	1085	2400
Gd	592.0	1170	1990
Tb	564.0	1112	2110
Dy	572.0	1126	2200
Ho	581.0	1139	2200
Er	589.0	1151	2190
Tm	596.7	1163	2284
Yb	603.4	1175	2415
Lu	523.5	1340	2022

Question 110.
Give the general electronic configuration of 4f-series elements (OR lanthanoids).
Answer:

• The general electronic configuration of 4f-series elements is, Ln[Xe]⁵⁴ 4f^1-14 5d^0-1 6s² where Ln is a lanthanoid.
• Xenon has electronic configuration, [Xe] : Is² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶.
• In lanthanoids, the differentiating electron enters prepenultimate shell, 4f m

Question 111.
What are the important features of the electronic configuration of lanthanoids?
Answer:

1. Lanthanoids show two types of electronic configurations
   (a) an expected or idealized
   (b) an observed electronic configuration.
   In the idealized electronic configuration, the filling of the 4/-orbitals is regular but in the observed configuration, there is the shift of a single electron from 5d to 4/ sub-shell.
2. Lanthanum (57) has an electronic configuration [Xe] 4f° 5d¹6s². It does not have any f-electron.
3. The next incoming electron does not enter the 5d sub-shell but goes to the 4f sub-shell.
4. 14 electrons are progressively filled in the 4f sub-shell as the atomic number increases by one unit from La to Lu.
5. La, Gd and Lu are the only elements which possess one electron in a 5d orbital, while in all other lanthanoids the 5d sub-shell is empty.
6. La-(4f°), Gd-(4f⁷) and Lu-(4f¹⁴) posses extra stability due to their empty, half-filled and completely filled 4f-orbitals respectively.
7. The 4f-electrons in the prepenultimate shell are shielded by the outermost higher orbitals, 5s², 5p⁶, 5d¹, 6s², i.e. by eleven electrons, hence they are less effective in chemical bonding.

Electronic configuration (Idealised and observed)

[Xe]⁵⁴ ls²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶

Question 112.
Write the expected electronic configuration of (a) Nd (Z = 60) (b) Tm (Z = 69).
Answer:
Expected electronic configuration :
(a) Nd = [Xe] 4f³ 5d¹ 6s²
(b) Tm= [Xe] 4f¹⁴5d¹6s²

Question 113.
Write electronic configurations of
(i) Gd
(ii) Yb.
Answer:
(i) ₆₄Gd [Xe] 4f⁷5d¹6s² (Observed)
(ii) ₇₀Yb [Xe] 4f¹⁴5d°6s² (Observed)

Question 114.
Write expected and observed electronic configurations of
(i) Ce
(ii) Tb.
Answer:

Element	Expected (Idealised)	Observed
(i) 58Ce	[Xe] 4f15d16s2	[Xe] 4f25d°6s2
(ii) 65Tb	[Xe] 4f85d16s2	[Xe] 4f95d°6s2

Question 115.
Why are the expected and observed ground state electronic configurations of gadolinium and lawrencium same?
Answer:

• The degenerate orbitals like 4f and 5f acquire extra stability when they are half filled (4f⁷) or completely filled (5f¹⁴).
• The expected and observed electronic configuration of gadolinium is, ₆₄Gd [Xe] 4f⁷ 5d¹ 6s².
• The expected and observed electronic configuration of lawrencium is ₁₀₃Lr [Rn] 5f¹⁴ 6d¹ 7s².

Question 116.
Explain oxidation states of lanthanoids.
Answer:

• The common oxidation state of the Lanthanoids is 3 + due to the loss of 2 electrons from outermost 6s orbital and one electron from the penultimate 5d sub-shell.
• Gd³⁺ and Lu³⁺ show extra stability due to their half-filled and completely filled f-orbitals, Gd³⁺ = [Xe]4f⁷, Lu³⁺ = [Xe]4f¹⁴
• Ce and Tb attain the 4f° and 4f⁷configurations in the 4 + oxidation states. Eu and Yb attain the 4f⁷ and 4f¹⁴ configurations in the 2 + oxidation states. Sm and Tm also show the 2+ oxidation state although their stability can be explained based on thermodynamic factors.
• Some lanthanoids show 2 + and 4 + oxidation states even though they do not have stable electronic configuration of 4f°, 4f⁷ or 4f¹⁴. E.g. Pr⁴⁺ (4f¹), Nd²⁺ (4f⁴), Sm²⁺ (4f⁶), Dy⁴⁺ (4f⁸) etc

Question 117.
Write the. electronic configuration of the following ions :
(1) La^{3 +} ;
(2) Gd³⁺;
(3) Eu³⁺;
(4) Ce³⁺.
Answer:
(1) La^{3 +} = [Xe]
(2) Gd³⁺ = [Xe] 4f⁷
(3) Eu³⁺ = [Xe] 4f⁶
(4) Ce³⁺ = |Xe] 4f¹

Question 118.
Write the electronic configuration of
(1) Nd²⁺
(2) Nd³⁺
(3) Nd⁴⁺.
Answer:
(1) Nd²⁺ [Xe] 4f⁴
(2) Nd³⁺ [Xe] 4f³
(3) Nd⁴⁺ [Xe] 4f²

Question 119.
Among the following lathanoids, which elements show only one oxidation state 3 +? Why? Dy, Gd, Yb, Lu.
Answer:
Gd and Lu show only one oxidation state 3 +, since they acquire electronic configurations with extra stability namely 4f⁷ and 4f¹⁴ respectively.

Question 120.
Write the expected electronic configurations of :
(1) europium (Z = 63),
(2) erbium (Z = 68).
Answer:
(1) Europium (₆₃Eu) [Xe]⁵⁴4f⁶ 5d¹ 6s²
(2) Erbium (₆₈Er) [Xe]⁵⁴4f¹¹ 5d¹ 6s²

Question 121.
Why does lanthanum form La³⁺ ion, while cerium forms Ce⁴⁺ ion? (Atomic number La = 57 and Ce = 58).
Answer:

1. Electronic configuration Lanthanum is La [Xe] 4f° 5d¹ 6s². By losing three electrons, La acquires stable electronic configuration of Xe and forms La³⁺.
2. Electronic configuration of Cerium is Ce [Xe] 4f¹ 5d¹ 6s². By losing four electrons, Ce acquires stable electronic configuration of Xe and forms Ce⁴⁺.

Question 122.
₆₃EU and ₇₀Yb show 2 + oxidation state. Explain.
Answer:
₆₃EU has electronic configuration, [Xe] 4f⁷ 5d°6s². By losing 2 electrons from 6s orbital, it acquires stable configuration and 4f-orbital is half-filled.
₇₀Yb has electronic configuration, [Xe] 4f¹⁴ 5d° 6s². By losing 2 electrons from 6 s orbital, it acquires stable configuration and 4/-orbital is completely filled.
Hence Eu and Yb show 2 + oxidation states.

Question 123.
Display electronic configuration, atomic and ionic radii of lanthanoids.
Answer:
Answers are given in bold.

Electronic configuration and atomic ionic radii of lanthanoids

Question 124.
Explain the trend in atomic and ionic sizes of lanthanoids.
Answer:

• From ₅₇La (187 pm) to first element of 4f-series ₅₈Ce (183 pm), the contraction in atomic radius is very large, 4 pm.
• But from Ce onwards as atomic number increases atomic radius decreases very steadily so that total decrease in atomic radius from Ce to Lu is only 10 pm.
• In case of tripositive ions due to large pull by nucleus, the decrease in ionic radii is slightly more, i.e. 18 pm. For example, Ce³⁺ (103 pm) to Lu³⁺ (85 pm ).
• Hence all lanthanoids have similar properties. Therefore they cannot be separated from each other easily by normal metallurgical methods but require special methods.
  

Question 125.
What is meant by lanthanoid contraction?
Answer:
Lanthanoid contraction : The gradual decrease in atomic and ionic radii of lanthanoids with the increase in atomic number is called lanthanoid contraction.

Question 153.
Explain the causes of the lanthanoid contraction.
Answer:
The causes of the lanthanoid contraction are as follows :

• As the atomic number of lanthanoids or 4f-block elements increases the positive nuclear charge increases and correspondingly electrons are added to the prepenultimate 4f sub-shell.
• The attraction of nucleus on 4 f-electrons increases with the increase in atomic number.
• The outer eleven electrons namely, 5s², 5p⁶, 5d³ and 6s² do not shield inner 4 f-electrons from the nucleus.
• There is imperfect shielding of each 4f-electron from other 4 f-electrons.
• As compared to d sub-shell, the extent of shielding for 4 f-electrons is less.
• Due to these cumulative effects, 4 f-electrons experience greater nuclear attraction and hence valence shell is pulled towards the nucleus to the greater extent decreasing atomic and ionic radii appreciably.
• From 57La to 58Ce, there is a sudden contraction in atomic radius from 187 pm to 183 pm but the further decrease up to the last 4f-element, ₇₁Lu is comparatively low (about 10 pm).

Question 126.
Explain lanthanoid contraction effect with respect to (1) decrease in basicity, (2) ionic radii of post-lanthanoids.
Answer:
The lanthanoid contraction has a definite effect on the properties of lanthanoids as well as on the properties of post-lanthanoid elements.
(1) Decrease in basicity :

• In lanthanoids due to lanthanoid contraction, as the atomic number increases, the size of the lanthanoid atoms and their try positive ions decreases, i.e. from La³⁺ to Lu³⁺.
• As size of the cation decreases, according to Fajan’s rule, the polarizability increases and thus the covalent character of the M-OH bond increases, and ionic character decreases.
• Therefore the basic nature of the hydroxides decreases.
• Basicity and ionic character decrease in the order La(OH)₃ > Ce(OH)₃ > … Lu(OH)₃.

(2) Ionic radii of post-lanthanoids :

• Elements following the lanthanoids in the 6th period (third transition series, i.e. 5d-series) are known as post-lanthanoids.
• Due to lanthanoid contraction the atomic radii (size) of elements which follow lanthanum in the 6th period (3rd transition series – Hf, Ta, W, Re)-are similar to the elements of the 5th period (4d-series Zr, Nb Mo, Tc).
• Due to similarity in their size, post-lanthanoid elements (5d-series) have closely similar properties to the elements of the 2nd transition series (4d-series) which lie immediately above them.
• Pairs of elements namely Zr-Hf(Gr-4), Nb-Ta (Gr-5), Mo-W(Gr-6), Tc-Re (Gr-7) are called chemical twins since they possess almost identical sizes and similar properties.

Question 127.
Why do lanthanoids form coloured compounds?
Answer:

• The colour in lanthanoid ions is due to the presence of unpaired electrons in partially filled 4f sub-shells.
• Due to the absorption of radiations in the visible region there arises the excitations of the unpaired electrons from f-orbital of lower energy to the f-orbital of higher energy-giving f → f transitions.
• The observed colour is complementary to the colour of the light absorbed.
• The colour of try positive ions (M³⁺) depends upon the number of unpaired electrons in f-orbitals. Hence the lanthanoid ions having equal number of unpaired electrons have similar colour.
• The colours of M³⁺ ions of the first seven lanthanoids, La³⁺ to Eu³⁺ are similar to those of seven elements Lu³⁺ to Tb³⁺ in the reverse order.

Question 128.
Explain, why Ce³⁺ ion is colourless.
Answer:

• The electronic configuration of Ce³⁺ is, [Xe] 4f⁷
• Even though there is one unpaired electron in 4f sub-shell, the f → f transition involves very low energy. Hence, Ce³⁺ ion does not absorb radiation in the visible region.

Therefore Ce³⁺ ion is colourless.

Question 129.
Explain why Gd³⁺ is colourless.
Answer:

• Gd³⁺ has electronic configuration, [Xe] 4f⁷
• Due to extra stability of half filled orbital, it does not allow f → f transition, and hence does not absorb radiations in the visible region.

Hence Gd³⁺ is colourless.

Question 130.
The salts of (1) La³⁺ and (2) Lu³⁺ are colourless. Explain.
Answer:
(1) (i) La³⁺ has electronic configuration, [Xe] 4f°
(ii) Since there are no unpaired electrons in 4 f-orbital, f → f transition is not possible. Hence La³⁺ ions do not absorb radiations in visible region, and they are colourless.

(2) (i) LU³⁺ has electronic configuration [Xe] 4 f¹⁴
(ii) Since there are no unpaired electrons in 4f-orbital, f → f transition is not possible. Hence Lu³⁺ ions do not absorb radiations in visible region and they are colourless.

Question 131.
Explain giving examples, the colour of nf electrons is about the same as those having (14-n) electrons.
Answer:
(1) Consider Pr³⁺ and Tm³⁺ ions.
Tm³⁺ (4f¹²) has nf electron 12 electrons.
Pr²⁺ (4f²) has (14 – n) = (14 – 2) = 12 electrons. Both, Tm³⁺ and Pr³⁺ are green.

(2) Consider Nd³⁺ and Er³⁺ ions. Er³⁺ (4f¹¹) has nf electrons 11.
Nd³⁺ (4f³) has (14 – n) is (14 – 3) = 11 electrons. These both ions Er³⁺, Na³⁺ are pink in colour.

Question 132.
Lu³⁺ has observed magnetic moment zero. How many unpaired electrons are present?
Answer:
Since magnetic moment is zero, it has no unpaired electrons.

Question 133.
What are the application of lanthanoids?
Answer:

1. Lanthanoid compounds are used inside the colour television tubes and computer monitor. For example mixed oxide (Eu, Y)₂ O₃ releases an intense red colour when bombarded with high energy electrons.
2. Lanthanoid ions are used as active ions in luminescent materials. (Optoelectronic application)
3. Nd : YAG laser is the most notable application. (Nd : YAG = neodymium doped ytterium aluminium garnet)
4. Erbium doped fibre amplifiers are used in optical fibre communication systems.
5. Lanthanoids are used in cars, superconductors and permanent magnets.

Question 134.
What are actinoids? Give their general electronic configuration.
Answer:

• Actinoids : The series of fourteen elements from ₉₀Th to ₁₀₃Lr which follow actinium (₈₉Ac) and in which differentiating electrons are progressively filled in 5f-orbitals in prepenultimate shell are called actinoids.
• Their general electronic configuration is, [Rn]⁸⁶ 5f^1-14 6d^0-1 7s².

Question 135.
Why are actinoids called inner transition elements?
Answer:

• Actinoids are 5f-series elements in which electrons progressively enter into 5f-orbitals, which are inner orbitals.
• They have electronic configuration [Rn]⁸⁶ 5f^1-14 6d^0-1 7s².
• They show intermediate properties as compared to electropositive 5-block elements and electronegative p-block elements. Hence they are called second inner transition elements.

Question 136.
Explain the position of actinoids in the periodic table.
OR
What is the position of actinoids in the periodic table?
Answer:

• Position of actinoids in the periodic table : Group-3; Period-7.
• They interrupt the fourth transition series (6d series) in the seventh period in the periodic table.
• After Actinium, 89Ac which has electronic configuration [Rn] 6d¹7s², the electrons enter progressively 5f orbital and they have general electronic configuration, [Rn] 5f^{1 – 14} 6d^{0 – 1} 7s².
• They are fourteen elements from ₉₀Th to ₁₀₃Lr and since they follow actinium, they are called actinoids.
• They are called 5f series or second inner transition series elements and for the convenience they are placed separately below the periodic table.

Question 137.
Write idealised and observed electronic configuration of actinoids.
Answer:

Question 138.
Explain the oxidation states of actinoids.
Answer:

• Due to availability of electrons in 5f, 6d and 7s sublevels, lanthanoids show varied oxidation states.
• The most common oxidation state is + 3 due to loss of one electron from 6d and two electrons from 6s-orbitals.
• Ac, Th and Am show + 2 oxidation state.
• Th, Pa, U, Np, Pu, Am and Cm show + 4 oxidation state.
• Np and Pu show the highest oxidation state + 7.
• U, Np, Bk, Cm and Am show stable oxidation state + 4.
• In + 6 oxidation state, due to high charge density the actinoid ions form oxygenated ions, e.g. \(\mathrm{UO}_{2}^{+}, \mathrm{NpO}_{2}^{+},\) etc.

Question 139.
Why do actinoids show variable oxidation states?
Answer:

• The large number of variable oxidation states of actinoids is due to very small energy difference between 5f, 6d and 7s subshells.
• The electronic configuration of actinoids is, [Rn] 5f^1-14 6d^0-1, 7s²
• Due to the loss of three electrons from 6d¹ and 7s², the common oxidation state is + 3, but due to further loss of electrons from 5f subshell, actinoids show higher oxidation states.
• The variable oxidation states are + 2 to + 7.

Electronic configuration of actinoids and their ionic radii in + 3 oxidation state

Question 140.
What is meant by actinoid contraction?
Answer:
Actinoid contraction: The gradual decrease in atomic and ionic radii of actinoids with the increase in atomic number is called actinoid contraction.

Question 141.
The extent of actinoid contraction is greater than lanthanoid contraction. Explain Why?
Answer:

• The electronic configurations of :
  Lanthanoids [Xe] 4f^{1 – 14} 5d^{0 – 1} 6s²
  Actinoids [Rn] 5f^{1 – 14}, 6d^{0 – 1} 7s²
• The mutual screening offered in case of 5f-electrons is less than that in the 4f-electrons.
• Hence, the outer orbitals are pulled to the greater extent by nuclei in actinoids (5f-series) than in lanthanoids (4f-series).
• Therefore, actinoid contraction is greater than lanthanoid contraction.

Question 142.
Describe the important properties of actinoids.
Answer:
Properties of actinoids :

• Actinoids are silvery white ( similar to lanthanoids).
• They are highly reactive radioactive elements.
• Most of these elements are not found in nature. They are radioactive and man made.
• They experience decrease in the atomic and ionic radii from Ac to Lw, known as actinoid contraction.
• The common oxidation state is +3. Elements of the first half of the series exhibit higher oxidation states.

Question 143.
What are the applications of actinoids?
Answer:

• Thorium oxide (ThO₂) with 1% CeO₂ is used as a major source of indoor lighting, as well as for outdoor camping.
• Uranium is used in the nuclear reactors.
• The isotopes of Thorium and Uranium have very long half-life, so that we get very negligible radiation from them: Hence they can be used safely.

Question 144.
What are transuranic elements?
Answer:

• The man-made elements heavier titan Uranium (Z = 92) in the Actinoid señes are called transuranic elements.
• These are synthetically or artificially prepared (man-made) elements starting from Neptunium (Z= 93).
• Transuranic elements arc generally considered to be from Neptunium (Z = 93) to Lawrencium (Z = 103).
• Recently elements from atomic number 104 (Rf) to atomic number 118 (Og) or (Uuo) in 6 d series have also been identified as transuranic elements.
• All transuranic elements are radioactive.

Question 145.
What are post actinoid elements?
Answer:

• Elements from atomic number 104 to 118 are called postactinoid elements.
• The post actinoid elements known so far are transition metals.
• They can be synthesised in the nuclear reactions.
• As they have very short half life period, it is difficult to study their chemistry.
• Ruiherfordium forms a chloride (RfCl₄) similar to zirconium and hafnium in + 4 oxidation state.
• Dubniurn resembles niobium and protactinium.

Question 146.
Name the transuranic elements.
Answer:
Names of transuranic elements

Name	Symbol	Atomic number
Neptunium	Np	93
Plutonium	Pu	94
Americium	Am	95
Curium	Cm	96
Berkelium	Bk	97
Californium	Cf	98
Einsteinium	Es	99
Ferminum	Fm	100
Mendelevium	Md	101
Nobelium	No	102
Lawrencium	Lr	103
Rutherfordium	Rf	104
Dubnium	Db	105
Seaborgium	Sg	106
Bohrium	Bh	107
Hassium	Hs	108
Meitnerium	Mt	109
Darmstadtium	Uun/Ds	110
Roentgenium	Uuu/Rg	111
Copernicium	Uub/Cn	112
Ununtrium	Uut	113
Ununquadium	Uuq	114
Ununpentium	Uup	115
Ununhexium	Uuh	116
Ununseptium	Uus	117
Ununoctium	Uuo	118

In the transuranic elements, elements from atomic number 93 to 103 are actinoids and from atomic number 104 to 118 are called postactinoid elements.

Question 147.
What are the similarities between lanthanides and actinides.
Answer:
Lanthanides and actinides show similarities as follows :

• Both, lanthanides and actinides show+ 3 oxidation state.
• In both the series, the f-orbitals are filled gradually.
• Ionic radius of the elements in both the series decreases with increase in atomic number.
• Electronegativity in both the series is low for all the elements.
• They all are highly reactive.
• The nitrates, perchlorates and sulphates of all elements are soluble while their hydroxides, theorides and carbonates
  are insoluble.

Question 148.
Differentiate between lanthanoids and actinoids.
Answer:

Lanthanoids	Actinoids
Electronic configuration [Xe] 4f1-14 5d0-1, 6s2	Electronic configuration [Rn] 5f1-14 6d0-1, 7s2
The differentiating electron enters the 4f subshell.	The differentiating electron enters the 5f subshell.
Except for Promethium all other elements occur in nature.	Except for Uranium and Thorium, all others are synthesized in the laboratory.
The binding energy of 4f electrons is higher.	5f-orbitals have lower binding energy.
Only Promethium is radioactive.	All elements are radioactive.
Besides 3 + oxidation state they show 2 + and 4 + oxidation states.	Besides 3 + oxidation state they show 2 + , 4 + , 5 + , 6 + , 7 + oxidation states.
They have a less tendency to form complexes.	They have greater tendency to form complexes.
Many lanthanoid ions are colourless. Their colour is not as deep and sharp as actinoids.	Actinoids are coloured ions. Their colour is deep, e.g. U3+ is red and U4+ is green.
Lanthanoids cannot form oxo-cations.	Actinoids form oxo-cations such as – UO2+, PuO2+, UO22+, PuO22+.
Lanthanoid hydroxides are less basic.	Actinoid hydroxides are more basic.
Lanthanoid contraction is relatively less.	Actinoid contraction from element to element is comparatively more.
Mutual shielding of 4f electrons is more.	Mutual shielding effect of 5f electrons is less.

Question 149.
Compare Pre-transition metals, Lanthanoid and transition metals.
Answer:

Multiple Choice Questions

Question 150.
Select and write the most appropriate answer from the given alternatives for each sub-question :

1. In transition elements, the different electron enters into
(a) ns subshell
(b) np subshell
(c) (n – 1) d subshell
(d) (n – 2)f subshell
Answer:
(c) (n – 1) d subshell

2. Chromium (Z = 24) has electronic configuration
(a) [Ar]4d^A 4s²
(b) [Ar] 4d⁵ 45¹
(c) [Ar] 3d⁵ 3s¹
(d) [Ar] 3d⁵ 4s¹
Answer:
(d) [Ar] 3d⁵ 4s¹

3. Manganese achieves the highest oxidation state in its compounds
(a) Mn₃O₄
(b) KMnO₄
(c) K₂MnO₄
(d) MnO₂
Answer:
(b) KMnO₄

4. The group which belongs to transition series is
(a) 2
(b) 7
(c) 13
(d) 15
Answer:
(b) 7

5. The last electron of transition element is called
(a) s-electron
(b) p-electron
(c) d-electron
(d) f-electron
Answer:
(c) d-electron

6. Which one of the following elements does NOT belong to first transition series?
(a) Fe
(b) V
(c) Ag
(d) Cu
Answer:
(c) Ag

7. The incomplete d-series is
(a) 3d
(b) 4d
(c) 5d
(d) 6d
Answer:
(d) 6d

8. The electronic configuration of Sc is
(a) [Ar] 3d² 4s²
(b) [Ar] 3d¹ 4s²
(c) [Kr] 3d¹ 4s²
(d) [Kr] 3d² 4s¹
Answer:
(b) [Ar] 3d¹ 4s²

9. The observed electronic configuration of copper is
(a) [Ar]¹⁸ 3d⁹ 4s²
(b) [Kr] 3d¹⁰ 45¹
(c) [Kr] 3d⁹ 4s²
(d) [Ar] 3d¹⁰ 45¹
Answer:
(d) [Ar] 3d¹⁰ 45¹

10. Fe belongs to the
(a) 3d-transition series elements
(b) 4d-transition series elements
(c) 5d-transition series elements
(d) 6d-transition series elements
Answer:
(a) 3d-transition series elements

11. Which one of the following elements does not exhibit variable oxidation states?
(a) Iron
(b) Copper
(c) Zinc
(d) Manganese
Answer:
(c) Zinc

12. In KMnO₄, oxidation number of Mn is
(a) 2+
(b) 4 +
(c) 6 +
(d) 7+
Answer:
(d) 7+

13. Which one of the following transition elements shows the highest oxidation state?
(a) Sc
(b) Ti
(c) Mn
(d) Zn
Answer:
(c) Mn

14. The colour of transition metal ions is due to
(a) s → s transition
(b) d → d transition
(c) p → p transition
(d) f → f transition
Answer:
(b) d → d transition

15. Which one of the following compounds is expected to be coloured?
(a) AgNO₃
(b) CuSO₄
(c) ZnCl₂
(d) CuCl
Answer:
(b) CuSO₄

16. The metal ion which is NOT coloured, is
(a) Fe³⁺
(b) V²⁺
(c) Zn²⁺
(d) Ti³⁺
Answer:
(c) Zn²⁺

17. A pair of coloured ion is
(a)Cu²⁺, Zn²⁺
(b)Cr³⁺ , Cu⁺
(c) Cd²⁺, Mn⁵⁺
(d) Fe²⁺, Fe³⁺
Answer:
(d) Fe²⁺, Fe³⁺

18. The highest oxidation state is shown by
(a) Fe
(b) Mn
(c) Os
(d) Cr
Answer:
(c) Os

19. Transition elements are good catalysts since
(a) they show variable oxidation states
(b) they have partially filled d-orbitals
(c) they have low I.P
(d) they have small atomic radii
Answer:
(a) they show variable oxidation states

20. Highest magnetic moment is shown by the ion
(a) V³⁺
(b) Co³⁺
(c) Fe³⁺
(d) Cr³⁺
Answer:
(c) Fe³⁺

21. The most common oxidation state of lanthanoids is
(a) +4
(b) +3
(c) +6
(d) +2
Answer:
(b) +3

22. Which one of the following elements belong to the actinoid series?
(a) Cerium
(b) Lutetium
(c) Thorium
(d) Lanthanum
Answer:
(c) Thorium

23. The total number of elements in each of f-series is
(a) 10
(b) 12
(c) 14
(d) 15
Answer:
(c) 14

24. The general electronic configuration of Lanthanoids is
(a) [Xe] 4f^{1 – 14} 5d^{0 – 1} 6s²
(b) [Xe] 4f^{2 – 14} 5d^{0 – 1} 6s²
(c) [Xe] 4f^{1 – 13} 5d^{0 – 1} 6s²
(d) [Xe] 4f^{0 – 14} 5d^{0 – 1} 6s¹
Answer:
(a) [Xe] 4f^{1 – 14} 5d^{0 – 1} 6s²

25. f-block elements are called ………………….
(a) transition elements
(b) representative elements
(c) inner transition elements
(d) alkalin earth metals
Answer:
(c) inner transition elements

26. Actinoids form coloured salts due to the transition of electrons in
(a) d – d
(b) f – f
(c) f – d
(d) s – f
Answer:
(b) f – f

27. In the periodic table, Gadolinium belongs to
(a) 4th Group 6th period
(b) 4th group 4th period
(c) 3rd group 5th period
(d) 3rd group 7th period.
Answer:
(d) 3rd group 7th period.

28. The transuranic elements are prepared by
(a) addition reaction
(b) substitution reactions
(c) decomposition reaction
(d) nuclear reactions
Answer:
(d) nuclear reactions