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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 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 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

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Solutions Chapter 3 Reconstitution of Partnership (Admission of Partner) Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Book Keeping & Accountancy Solutions Chapter 3 Reconstitution of Partnership (Admission of Partner)

1. Objetive type questions.

A. Select the most appropriate alternatives from the following and rewrite the sentences.

Question 1.
Anuj and Eeshan are two partners sharing profits and losses in the ratio of 3 : 2. They decided to admit Aaroh for 1/5th share, the new profit sharing ratio will be __________
(a) 12 : 8 : 5
(b) 4 : 3 : 1
(c) 12 : 8 : 1
(d) 12 : 3 : 1
Answer:
(a) 12 : 8 : 5

Question 2.
Excess of proportionate capital over actual capital represents __________
(a) equal capital
(b) surplus capital
(c) deficit capital
(d) gain
Answer:
(c) deficit capital

Question 3.
__________ is credited when unrecorded asset is brought into business.
(a) Revaluation Account
(b) Balance Sheet
(c) Trading Account
(d) Partners Capital Account
Answer:
(a) Revaluation Account

Question 4.
When goodwill is withdrawn by the partner __________ account is credited.
(a) Revaluation
(b) Cash/Bank
(c) Current
(d) Profit and Loss Adjustment
Answer:
(b) Cash/Bank

Question 5.
If asset is taken over by the partner __________ Account is debited.
(a) Revaluation
(b) Capital
(c) Asset
(d) Balance Sheet
Answer:
(b) Capital

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

Question 1.
The method under which calculation of goodwill is done on the basis of extra profit earned above the normal profit.
Answer:
Super Profit Method

Question 2.
An account is opened to adjust the value of assets and liabilities at the time of admission of a partner.
Answer:
Revaluation A/c or Profit and Loss A/c

Question 3.
The reputation of a business is measured in terms of money.
Answer:
Goodwill

Question 4.
The ratio in which general reserve is distributed to the old partners.
Answer:
Old Ratio

Question 5.
Name the method of the treatment of goodwill where a new partner will bring his share of goodwill in cash.
Answer:
Premium Method

Question 6.
The proportion in which old partners make a sacrifice.
Answer:
Sacrifice Ratio

Question 7.
Capital employed × NRR/100 = __________
Answer:
Normal Profit

Question 8.
An Account is debited when the partner takes over the asset.
Answer:
Partner’s Capital A/c or Partner’s Current A/c

Question 9.
Profit and Loss Account balance appearing on the liability side of the Balance Sheet.
Answer:
Undistributed Profit or Accumulated Profit

Question 10.
Old ratio – New ratio = __________
Answer:
Sacrifice Ratio

C. State True or False with reasons:

Question 1.
A new partner can bring capital in cash or kind.
Answer:
This statement is True.
As per the provision of partnership deed, when any person is admitted in the firm, he has to bring some amount as capital which can be in cash or in-kind of assets to get rights in the assets and definite share in the future profit of the firm.

Question 2.
When goodwill is paid privately to the partners, it is not recorded in the books.
Answer:
This statement is True.
When goodwill is paid privately to the partners, by a newly admitted person, then in such case no transaction takes place in the business, and the firm as such is not all benefited. Hence it is not recorded in the books of accounts.

Question 3.
The gain ratio is calculated at the time of admission of a partner.
Answer:
This statement is False.
At the time of admission of a person, in the business, sacrifices are made by the old partners in favour of the new partner. It means there is no question of any gain to the partners, so we can say that the Gain ratio is not calculated at the time of admission of a partner.

Question 4.
Revaluation profit is distributed among all partners including new partners.
Answer:
This statement is False.
Revaluation profit arises due to efforts and hardworking of the old partners in the past and hence profit earned on revaluation of assets and liabilities at the time of admission of a person as a partner in the business belongs to old partners. So, such profit is not distributed among all partners including the new partners. It is distributed only among old partners.

Question 5.
Change in the relationship between the partners is called as Reconstitution of Partnership.
Answer:
This statement is True.
When any person joins the business as a partner, a change in the relationship takes place. The old agreement is terminated and a new agreement is prepared. There is the change in profit or loss sharing ratio and relationship of the partners which is known as Reconstitution of Partnership.

Question 6.
New partners always bring their share of goodwill in cash.
Answer:
This statement is False.
When a new person is admitted to the partnership firm, the old partners surrender a certain share in profit and give it to a new partner. In exchange for that new partner is required to bring goodwill in cash or in kind. If he is unable to bring cash for goodwill, then Goodwill is raised and adjusted to the new partner’s capital A/c.

Question 7.
When the goodwill is written off, the goodwill account is debited.
Answer:
This statement is False.
To write off goodwill means to decrease or wipe out the value of goodwill. When goodwill as an asset of the business is raised, Goodwill A/c is debited in the books of Account. Conversely, when Goodwill is written off from the business, the Goodwill A/c is credited in the books of business.

Question 8.
The new ratio minus the old ratio is equal to the sacrifice ratio.
Answer:
This statement is False.
When a new partner is admitted, old partners have to sacrifice their profit share in favour of the new partner and their old ratio gets reduced and whatever ratio is left becomes a new ratio. Hence, as per equation:
New Ratio = Old Ratio – Sacrifice Ratio.
By interchanging the terms,
Sacrifice Ratio = Old Ratio – New Ratio.

Question 9.
Usually, when a new partner is admitted to the firm, there will be an increase in the capital of the firm.
Answer:
This statement is True.
When a new partner is admitted to the firm, he brings his share of capital and goodwill, in cash or in-kind, to enjoy the right of sharing the future profit, and hence there will be an increase in the capital of the firm.

Question 10.
Cash/Bank Account is credited when goodwill is withdrawn by the old partners.
Answer:
This statement is True.
When a new partner brings his share of goodwill, old partners have the right to withdraw it in cash. Therefore, when old partners withdraw the amount of goodwill, cash goes out from the firm and not goodwill. Hence Cash/Bank A/c is credited.

D. Find the odd one.

Question 1.
General reserve, Creditors, Machinery, Capital
Answer:
Machinery

Question 2.
Decrease in Furniture, Patents wrote off, Increase in Bills payable, R.D.D. written off
Answer:
R.D.D. written off

Question 3.
Super profit method, Valuation method, Average profit method, Fluctuating capital method
Answer:
Fluctuating capital method

E. Calculate the following:

Question 1.
A and B are partners in a firm sharing profit and losses in the ratio of 1 : 1. C is admitted. A surrenders 1/4th share and B surrenders 1/5th of his share in favour of C. Calculate new profit sharing ratio.
Solution:
Old ratio of A and B = 1 : 1 or \(\frac{1}{2}\) : \(\frac{1}{2}\)
A’s sacrifice = \(\frac{1}{4} \times \frac{1}{2}=\frac{1}{8}\)
B’s sacrifice = \(\frac{1}{5} \times \frac{1}{2}=\frac{1}{10}\)
Sacrificing ratio of A and B = \(\frac{1}{8}\) : \(\frac{1}{10}\) = 5 : 4
C’s share = A’s share + B’s share = \(\frac{1}{8}+\frac{1}{10}=\frac{5+4}{40}=\frac{9}{40}\)
A’s new share = Old ratio – Sacrifice ratio = \(\frac{1}{2}-\frac{1}{8}=\frac{4-1}{8}=\frac{3}{8}\)
B’s new share = Old ratio – Sacrifice ratio = \(\frac{1}{2}-\frac{1}{10}=\frac{5-1}{10}=\frac{4}{10}\)
Therefore, New ratio of A, B and C = \(\frac{3}{8}: \frac{4}{10}: \frac{9}{40}\) = 15 : 16 : 9
(Making denominator equal)

Question 2.
Anika and Radhika are partners sharing profit in the ratio of 5 : 1. They decide to admit Sanika to the firm for 1/5th share. Calculate the Sacrifice ratio of Anika and Radhika.
Solution:
Balance = 1 – share of new partner
= 1 – \(\frac{1}{5}\)
= \(\frac{4}{5}\) (Remaining share)
New ratio = Old ratio x Balance of 1
Anika’s New ratio = \(\frac{5}{6} \times \frac{4}{5}=\frac{20}{30}\)
Radhika’s New ratio = \(\frac{1}{6} \times \frac{4}{5}=\frac{4}{30}\)
Sanika’s New ratio = \(\frac{1}{5} \times \frac{6}{6}=\frac{6}{30}\) (Making denominator equal)
∴ New Profit and Loss ratio = \(\frac{20}{30}: \frac{4}{30}: \frac{6}{30}\) = 20 : 4 : 6 i.e. 10 : 2 : 3
Sacrifice ratio = old ratio – New ratio
Anika’s Sacrifice ratio = \(\frac{5}{6}-\frac{20}{30}=\frac{25-20}{30}=\frac{5}{30}\)
Radhika’s Sacrifice ratio = \(\frac{1}{6}-\frac{4}{30}=\frac{5-4}{30}=\frac{1}{30}\)
∴ Sacrifice ratio = \(\frac{5}{30}: \frac{1}{30}\) = 5 : 1

Question 3.
Pramod and Vinod are partners sharing profits and losses in the ratio of 3 : 2. After the admission of Ramesh the New ratio of Pramod, Vinod and Ramesh are 4 : 3 : 2. Find out the Sacrifice ratio.
Solution:
Sacrifice Ratio = Old ratio – New ratio
Pramod’s Sacrifice ratio = \(\frac{3}{5}-\frac{4}{9}=\frac{27-20}{45}=\frac{7}{45}\)
Vinod’s Sacrifice ratio = \(\frac{2}{5}-\frac{3}{9}=\frac{18-15}{45}=\frac{3}{45}\)
∴ Sacrifice ratio = \(\frac{7}{45}: \frac{3}{45}\) = 7 : 3.

F. Answer in one sentence.

Question 1.
What is Revaluation Account?
Answer:
An account opened and operated by any partnership firm for recording changes in the value of assets and liabilities and to ascertain profit or loss made on revaluation of assets and liabilities is called Revaluation Account.

Question 2.
What is meant by Reconstitution of Partnership?
Answer:
Reconstitution of partnership means a change in the relationship between/among partners and in the form of partnership.

Question 3.
Why is the new partner admitted?
Answer:
A new partner is admitted to the existing partnership firm to increase the capital resources of the firm and to secure advantages of a new entrant’s skill and business connections, i.e. goodwill.

Question 4.
What is the sacrifice ratio?
Answer:
A ratio that is surrendered or given up by the old partners in favour of a newly admitted partner is called the sacrifice ratio.

Question 5.
What do you mean by raising the goodwill at the time of admission of a new partner?
Answer:
Raising the Goodwill at the time of admission of a new partner means debiting Goodwill Account up to the value it is raised and crediting. Old partners’ Capital Accounts in their old ratio in the books of the firm.

Question 6.
What is the super profit method of calculation of goodwill?
Answer:
Super profit method of calculation of Goodwill is a method in which Goodwill is valued at a certain number of years purchases of the super profit of the partnership firm.

Question 7.
When is the ratio of sacrifice calculated for the distribution of goodwill?
Answer:
The ratio of sacrifice is calculated when the benefits of goodwill contributed by a new partner in cash is to be transferred to existing partners’ Capital/Current Account.

Question 8.
What is the treatment of accumulated profits at the time of admission of a partner?
Answer:
Accumulated profits at the time of admission of a partner are transferred to old partners’ Capital/Current Accounts in their old profit sharing ratio.

Question 9.
State the ratio in which the old partner’s Capital A/c will be credited for goodwill when the new partner does not bring his share of goodwill in cash.
Answer:
When the new partner does not bring his share of goodwill in cash, Goodwill is raised up to a certain value and credited to old partners’ Capital/Current A/cs in their old profit sharing ratio.

Question 10.
What does the excess of debit over credits in the Profit and Loss Adjustment Account indicate?
Answer:
The excess of debit over credits in the Profit and Loss Adjustment Account indicates loss on revaluation of assets and liabilities.

G. Complete the table.

Question 1.

Answer:
Average Profit = \(\frac{Total Profit}{Number of years}\)

Question 2.

Answer:
Normal Profit = Capital Employed × \(\frac{NPR}{100}\)

Question 3.

Answer:
The stock shown in Balance Sheet → Stock undervalued by 20% → Cost of Stock
₹ 1,60,000 → ₹ 40,000 → ₹ 2,00,000

Practical Problems

Question 1.
Vikram and Pradnya share profits and losses in the ratio 2 : 3 respectively. Their Balance Sheet as of 31st March 2018 was as under.
Balance Sheet as of 31st March 2018

They agreed to admit Avani as a partner on 1st April 2018 on the following terms:
1. Avani shall have 1/4th share in future profits.
2. He shall bring ₹ 37,500 as his capital and ₹ 30,000 as his share of goodwill.
3. Land and building to be valued at ₹ 45,000 and furniture to be depreciated by 10%.
4. Provision for bad and doubtful debts is to be maintained at 5% on the Sundry Debtors.
5. Stock to be valued ₹ 82,500.
The Capital A/c of all partners to be adjusted in their new profit and loss ratio and excess amount be transferred to their loan accounts.
Prepare Profit and Loss Adjustment Account, Capital Accounts, and New Balance Sheet.
Solution:
In the books of Partnership Firm

Balance Sheet as of 1st April 2018

Working Notes:
1. Calculation of new profit ratio = 1 – share of new partner
= 1 – \(\frac{1}{4}\)
= \(\frac{3}{4}\) (Remaining share)
New ratio = old ratio × balance 1 (Remaining share)
Vikram’s new ratio = \(\frac{2}{5} \times \frac{3}{4}=\frac{6}{20}\)
Pradnya’s new ratio = \(\frac{3}{5} \times \frac{3}{4}=\frac{9}{20}\)
Avani’s ratio = \(\frac{1}{4}=\frac{1}{4} \times \frac{5}{5}=\frac{5}{20}\)
∴ New profit sharing ratio = 6 : 9 : 5.
Capital amount adjusted in their new profit and loss ratio:
Total Capital of the Partnership Firm = (Reciprocal of New Partner’s Share) × (Capital of New Partner)
= (Reciprocal of \(\frac{1}{4}\)) × 37,500
= 4 × 37,500
= ₹ 1,50,000
Vikram’s Capital balance = (Vikram’s New Ratio) × (Total Capital of the firm)
= \(\frac{6}{20}\) × 1,50,000 = ₹ 45,000
Pradnya’s Capital balance = \(\frac{9}{20}\) × 1,50,000 = ₹ 67,500

Question 2.
Amalendu and Sameer share profits and losses in the ratio 3 : 2 respectively. Their Balance Sheet as of 31st March 2017 was as under:
Balance Sheet as of 31st March 2017

On 1st April 2017 they admit Paresh into partnership. The term being that:
1. He shall pay ₹ 16,000 as his share of Goodwill 50% amount of Goodwill shall be withdrawn by the old partners.
2. He shall have to bring in ₹ 20,000 as his Capital for 1/4 share in future profits.
3. For the purpose of Paresh’s admission it was agreed that the assets would be revalued as follows:
A. Land and Building is to be valued at ₹ 60,000.
B. Plant and Machinery to be valued at ₹ 16,000.
C. Stock valued at ₹ 20,000 and Furniture and Fixtures at ₹ 4,000.
D. A Provision of 5% on Debtors would be made for Doubtful Debts.
Pass the necessary Journal Entries in the books of a new firm.
Solution:
Journal entries in the books of Partnership Firm


Working Notes:

Question 3.
Vasu and Viraj share profits and losses in the ratio of 3 : 2 respectively. Their Balance Sheet as on 31st March, 2019 was as under:
Balance Sheet as on 31st March, 2019

They admit Hari into partnership on 1-4-2019. The terms being that:
1. He shall have to bring in ₹ 60,000 as his Capital for 1/4 share in future profits.
2. Value of Goodwill of the firm is to be fixed at the average profits for the last three years.
The Profit were:
2015-16 – ₹ 48,000
2016-17 – ₹ 81,000
2017-18 – ₹ 73,500
Hari is unable to bring the value of Goodwill in cash. It is decided to raise Goodwill in the books of accounts.
3. Reserve for Doubtful debts is to be created at ₹ 1,500.
4. Closing stock is valued at ₹ 22,500.
5. Plant and Building are to be depreciated by 5%.
Prepare Profit and Loss-Adjustment A/c, Capital Accounts of Partners and Balance Sheet of the new firm.
Solution:
In the books of the firm __________

Balance Sheet as on 1st April 2019

Working Notes:
1. Average Profit = \(\frac{Total Profit}{No. of years}\)
= \(\frac{48,000+81,000+73,500}{3}\)
= ₹ 67,500
∴ Goodwill value = ₹ 67,500
Vasu’s share in Goodwill = ₹ 40,500 (67,500 × \(\frac{3}{5}\))
Viraj’s share in Goodwill = ₹ 27,000 (67,500 × \(\frac{2}{5}\))

2. Hari is not able to bring a share in goodwill and it is decided to raise the goodwill in the book.
Therefore, Goodwill is recorded in the Asset side ₹ 67,500.

Question 4.
Mr. Deep & Mr. Karan were in partnership sharing profits & losses in the proportion of 3 : 1 respectively. Their Balance Sheet on 31st March 2018 stood as follows:
Balance Sheet as of 31st March 2018

They admit Shubham into Partnership on 1 April 2018. The terms being that:
1. He shall have to bring in ₹ 20,000 as his capital for 1/5 share in future profits & ₹ 10,000 as his share of Goodwill.
2. A provision for 5% doubtful debts to be created on sundry debtors.
3. Furniture to be depreciated by 20%.
4. Stock should be appreciated by 5% and Building be appreciated by 20%.
5. Capital A/c of all partners be adjusted in their new profit sharing ratio through cash account.
Prepare Profit and Loss-Adjustment A/c, Partners’ Capital A/c, Balance Sheet of the new firm.
Solution:
In the books of the firm __________


Balance Sheet as of 1st April 2018

Working Note:
Calculation of new ratio : Balance of 1 = 1 – share of new partner
= 1 – \(\frac{1}{5}\)
= \(\frac{4}{5}\) (Remaining share)
New ratio = Old ratio × balance 1 (Remaining share)
Deep’s new ratio = \(\frac{3}{4} \times \frac{4}{5}=\frac{3}{5}\)
Karan’s new ratio = \(\frac{1}{4} \times \frac{4}{5}=\frac{1}{5}\)
Shubham’s new ratio = \(\frac{1}{5}=\frac{1}{5}\)
∴ New profit and loss sharing ratio = 3 : 1 : 1
Capital amount to be adjusted in Partner’s new profit and loss ratio:
Total Capital of the firm = (Reciprocal of New partner’s share) × (New partner’s capital)
= 5 × 20,000
= ₹ 1,00,000
Deep’s capital balance = \(\frac {3}{5}\) × 1,00,000 = ₹ 60,000
Karan’s capital balance = \(\frac {1}{5}\) × 1,00,000 = ₹ 20,000

Question 5.
Mr. Kishor & Mr. Lai were in partnership sharing profits & losses in the proportion of 3/4 and 1/4 respectively.

They decided to admit Ram on 1 April 2018 on the following terms:
1. He should be given 1/5th share in profit and for that, he brought in ₹ 60,000 as capital through RTGS.
2. Goodwill should be raised at ₹ 60,000.
3. Appreciate Land and Building by 20%.
4. Furniture and Stock are to be depreciated by 10%.
5. The capitals of all partners should be adjusted in their new profit sharing ratio through Bank A/c.
Pass necessary Journal Entries in the books of the partnership firm and a Balance Sheet of the new firm.
Solution:
Journal entries in the books of the firm


Balance Sheet as of 1st April 2018

Working Notes:
1.

2. Calculation of new profit sharing ratio:
New Ratio = (Balance of 1) × (old ratio)
Kishor’s New ratio = \(\left(1-\frac{1}{5}\right) \times \frac{3}{4}=\frac{4}{5} \times \frac{3}{4}=\frac{3}{5}\)
Lal’s New ratio = \(\left(1-\frac{1}{5}\right) \times \frac{1}{4}=\frac{4}{5} \times \frac{1}{4}=\frac{1}{5}\)
Ram’s ratio = \(\frac{1}{5}\)

3. Total capital of the firm = (Reciprocal of Ram’s ratio) × (His capital contribution)
= \(\frac{5}{1}\) × 60,000 = ₹ 3,00,000
Kishor’s new closing capital balance = 3,00,000 × \(\frac{3}{5}\) = ₹ 1,80,000
Lai’s new closing capital balance = 3,00,000 × \(\frac{1}{5}\) = ₹ 60,000
Ram’s new closing capital balance = ₹ 60,000

Question 6.
Vrushali and Leena are equal partners in the business. Their Balance Sheet as of 31st March 2013 stood as under.
Balance Sheet as of 31st March 2018

They decided to admit Aparna on 1st April 2018 on the following terms:
1. The Machinery and Building be depreciated by 10%. Reserve for Doubtful Debts to be increased by ₹ 5,000.
2. Bills receivable are taken over by Vrushali at a discount of 10%.
3. Aparna should bring ₹ 60,000 as capital for her 1/4th share in future profits.
4. The Capital accounts of all the partners be adjusted in proportion to the new profit sharing ratio by opening the Current accounts of the partners.
Prepare Profit and Loss-Adjustment A/c, Partners’ Capital A/c, Balance Sheet of the new firm.
Solution:

Balance Sheet as on 1st April 2018

Working Notes:
1. R.D.D. to be increased by ₹ 5,000 means subtract ₹ 5,000 from Debtors.

2. Bills receivable taken by Vrushali at 10 % discount i.e. 12,000 – 1,200 = ₹ 10,800.
Write this amount on the debit side of the partners’ Capital Account in Vrushali’s column.

3. Calculation of new ratio = 1 – share of new partner
= 1 – \(\frac{1}{4}\)
= \(\frac{3}{4}\) (Remaining share)
New ratio = Old ratio × Balance 1 (Remaining Share)
Vrushali’s new ratio = \(\frac{1}{2} \times \frac{3}{4}=\frac{3}{8}\)
Leena’s new ratio = \(\frac{1}{2} \times \frac{3}{4}=\frac{3}{8}\)
Aparna’s ratio = \(\frac{1}{4}\)
∴ Partner’s new profit and loss ratio = \(\frac{3}{8}: \frac{3}{8}: \frac{1}{4}\) = 3 : 3 : 2
Now, capital amount to be adjusted in partners new profit and loss ratio.
Total capital of the firm = (Reciprocal of New Partner’s Share) × (New Partner’s Capital)
= (Reciprocal of \(\frac{1}{4}\)) × 60,000
= 4 × 60,000
= ₹ 2,40,000
Vrushali’s capital balance = \(\frac {3}{8}\) × 2,40,000 = ₹ 90,000
Leela’s capital balance = \(\frac {3}{8}\) × 2,40,000 = ₹ 90,000
The deficit of these capital balances is to be adjusted through the Current account.
To keep the balance of Vrushali’s and Leena’s capital ₹ 90,000 each, deficit of ₹ 53,850 and ₹ 58,050 are incurred which is transferred to the respective Partner’s Current A/cs and recorded on the Asset side of Balance Sheet [As it is to be recovered from Partners].

Question 7.
The balance sheet of Medha and Radha who share profit and loss in the ratio 3 : 1 is as follows:
Balance Sheet as of 31st March 2018

They decided to admit Krutika on 1st April 2018 on the following terms:
1. Krutika is taken as a partner on 1st April 2017. She will pay 40,000 as her capital for 1/5th share in future profits and ₹ 2,500 as goodwill.
2. 5% provision for bad and doubtful debt be created on debtors.
3. Furniture be depreciated by 20%.
4. Stocks be appreciated by 5% and plant & machinery by 20 %.
5. The Capital accounts of all partners be adjusted in their new profit sharing ratio by adjusting the amount through a loan.
6. The new profit sharing ratio will be 3/5 : 1/5 : 1/5 respectively.
You are required to prepare Profit and Loss-Adjustment A/c, Partners’ Capital A/c, Balance Sheet of the new firm.
Solution:


Balance Sheet as of 1st April 2018

Working Notes:
1. Total capital of the firm = (Reciprocal of New Partner’s Profit Sharing ratio) × (Capital contributed by new partner)
= (Reciprocal of \(\frac{1}{5}\)) × 40,000
= 5 × 40,000
= ₹ 2,00,000
Medha’s closing capital, balance = \(\frac{3}{5}\) × 2,00,000 = ₹ 1,20,000
Radha’s closing capital balance = \(\frac{1}{5}\) × 2,00,000 = ₹ 40,000

2. General reserve is distributed among old partners in their old profit and loss ratio.

3. Cash Balance = 78,000 + 40,000 + 2,500 = ₹ 1,20,500 [Amount brought in by new partner.]

Question 8.
The Balance Sheet of Sahil and Nikhil who share profits in the ratio of 3 : 2 as of 31st March 2017 is as follows:
Balance Sheet as of 31st March 2017

Varad admitted on 1st April 2017 on the following terms:
1. Varad was to pay ₹ 1,00,000 for his share of capital.
2. He was also to pay ₹ 40,000 as his share of goodwill.
3. The new profit sharing ratio was 3 : 2 : 3.
4. Old partners decided to revalue the assets as follows:
Building ₹ 1,00,000. Furniture ₹ 48,000, Debtors ₹ 38,000 (in view of likely bad debts)
5. It was found that there was a liability for ₹ 3,000 for goods in March 2017 but recorded on 2nd April 2017.
You are required to prepare:
(a) Profit and Loss-Adjustment account
(b) Capital accounts of the partners
(c) Balance Sheet after the admission of Varad.
Solution:


Balance Sheet as of 1st April 2017

Working Notes:
1. Cash in hand = Opening balance + Varad’s capital + Varad’s goodwill (amount brought in)
= 20,000 + 1,00,000 + 40,000
= ₹ 1,60,000

2. Sacrifice ratio = Old ratio – New ratio
Sahil’s sacrifice = \(\frac{3}{5}-\frac{3}{8}=\frac{24-15}{40}=\frac{9}{40}\)
Nikhil’s sacrifice = \(\frac{2}{5}-\frac{2}{8}=\frac{16-10}{40}=\frac{6}{40}\)
i.e. sacrifice ratio = \(\frac{9}{40}: \frac{6}{40}\) = 9 : 6 = 3 : 2.
Goodwill is distributed among old partners in the sacrifice ratio.

Question 9.
Mr. Amit and Baban share profits and losses in the ratio 2 : 3 respectively. Their Balance Sheet as of 31st March 2018 was as under:
Balance Sheet as of 31st March 2018

They agreed decided to admit Kamal on 1st April 2018 on the following terms:
1. Kamal shall have 1/4th share in future profits.
2. She shall bring 50,000 as her capital and 40,000 as her share of goodwill.
3. Land and building to be valued at 60,000 and furniture to be depreciated by 10%.
4. Provision for bad and doubtful debts is to be maintained at 5% on the sundry debtors.
5. Stocks to be valued at 1,10,000.
The Capital A/c of all partners to be adjusted in their new profit and loss ratio and excess amount be transferred to their loan accounts.
Prepare Profit and Loss-Adjustment A/c, Capital A/cs, and New Balance Sheet.
Solution:

Balance Sheet as of 1st April 2018

Working Notes:
1. Cash balance = Opening balance + Amount brought in by Kamal
= 1,10,000 + 50,000 + 40,000
= ₹ 2,00,000

2. For calculation of new profit and loss ratio:
Calculation of new profit ratio = 1 – share of new partner
= 1 – \(\frac{1}{4}\)
= \(\frac{3}{4}\) (Remaining share)
New ratio = old ratio × balance 1 (Remaining share)
Amit’s new ratio = \(\frac{2}{5} \times \frac{3}{4}=\frac{6}{20}\)
Baban’s new ratio = \(\frac{3}{5} \times \frac{3}{4}=\frac{9}{20}\)
Kamal’s ratio = \(\frac{1}{4}=\frac{1}{4} \times \frac{5}{5}=\frac{5}{20}\)

3. New profit and loss ratio = 6 : 9 : 5
Capital amount adjusted in their new profit and loss ratio by taking new partner Kamal’s capital (₹ 50,000) as a base.
For part 5 capital = ₹ 50,000 (Kamal’s capital)
For part 6 capital = ₹ 60,000 (Amit’s capital)
For part 9 capital = ₹ 90,000 (Baban’s capital)

4. After keeping these capital balances difference of the amount of Amit’s capital ₹ 63,520 and of Baban’s capital ₹ 45,280 are taken as partner’s loan to the firm and as a liability of the firm it is recorded in the Liabilities side of the Balance Sheet.

Question 10.
The following is the Balance Sheet of Om and Jay on 31st March 2018, they share profits and losses in the ratio 3 : 2.
Balance Sheet as of 31st March 2018

They take Jagdish into partnership on 1st April 2018. The terms being:
1. Jagdish should pay ₹ 3,000 as his share of Goodwill. 50% of goodwill withdrawn by partners in cash.
2. He should bring ₹ 9,000 as capital for 1/4th share in future profits.
3. Building to be valued at 18,000, Machinery and Furniture to be reduced by 10 %.
4. A provision of 5% on debtors to be made for doubtful debts.
5. Stock to be taken at the value of ₹ 15,000.
Prepare Profit and Loss A/c, Partners’ Current A/c, Balance Sheet of the new firm.
Solution:

Balance Sheet as of 1st April 2018

Working Notes:
1.

2. Write partner’s capital accounts balance as fixed capital balance in the Balance Sheet and transferred current account balance in the Balance Sheet as Partners Current A/c.

3. As shown in the cash account partners’ withdrew half amount of goodwill amount share.