Category: Class 12

Students can Download Accountancy Chapter 8 Financial Statement Analysis Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Accountancy Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

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

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

I. Choose the Correct Answer

12th Accountancy 8th Chapter Question 1.
Which of the following statements is not true?
(a) Notes and schedules also form part of financial statements
(b) The tools of financial statement analysis include common – size statement
(c) Trend analysis refers to the study of movement of figures for one year
(d) The common – size statements show the relationship of various items with some common base, expressed as percentage of the common base
Answer:
(c) Trend analysis refers to the study of movement of figures for one year

12th Accountancy Chapter 8 Question 2.
Balance sheet provides information about the financial position of a business concern ………………
(a) Over a period of time
(b) As on a particular date
(c) For a period of time
(d) For the accounting period
Answer:
(b) As on a particular date

12th Accountancy 8th Chapter Solutions Question 3.
Which of the following tools of financial statement analysis is suitable when data relating to several years are to be analysed?
(a) Cash flow statement
(b) Common size statement
(c) Comparative statement
(d) Trend analysis
Answer:
(d) Trend analysis

Class 12 Accountancy Chapter 8 Solutions Question 4.
The financial statements do not exhibit ………………
(a) Non – monetary data
(b) Past data
(c) Short term data
(d) Long term data
Answer:
(a) Non – monetary data

Accountancy Class 12 Chapter 8 Solutions Question 5.
Which of the following is not a tool of financial statement analysis?
(a) Trend analysis
(b) Common size statement
(c) Comparative statement
(d) Standard costing
Answer:
(d) Standard costing

Accounting Chapter 8 Answer Key Question 6.
The term ‘fund’ refers to ………………
(a) Current liabilities
(b) Working capital
(c) Fixed assets
(d) Non – current assets
Answer:
(b) Working capital

Class 12 Accountancy Chapter 8 Question 7.
Which of the following statements is not true?
(a) All the limitations of financial statements are applicable to financial statement analysis also.
(b) Financial statement analysis is only the means and not an end.
(c) Expert knowledge is not required in analysing the financial statements.
(d) Interpretation of the analysed data involves personal judgement.
Answer:
(c) Expert knowledge is not required in analysing the financial statements

Financial Statements Of A Company Class 12 Solutions Question 8.
A limited company’s sales has increased from ? 1,25,000 to? 1,50,000. How does this appear in comparative income statement?
(a) + 20%
(b) + 120%
(c) – 120 %
(d) – 20 %
Answer:
(a) + 20 %

Financial Statements Questions And Answers Pdf Question 9.
In a common-size balance sheet, if the percentage of non – current assets is 75, what would be the percentage of current assets?
(a) 175
(b) 125
(c) 25
(d) 100
Answer:
(c) 25

Financial Statement Analysis Solutions Pdf Question 10.
Expenses of a business for the first year were ₹ 80,000. In the second year, it was increased to ₹ 88,000. What is the trend percentage in the second year?
(a) 10%
(b) 110%
(c) 90%
(d) 11%
Answer:
(b) 110 %

II. Very Short Answer Questions

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

Financial Statement Analysis Chapter 8 Solutions Question 2.
List the tools of financial statement analysis.
Answer:

1. Comparative Statement.
2. Common Size Statement.
3. Trend Analysis.
4. Funds Flow Analysis.
5. Cash Flow Analysis.

Financial Statement Analysis Solutions Question 3.
What is working capital?
Answer:
Working capital statement or schedule of changes in working is prepared to disclose net changes in working capitals on two specific dates (generally two balance sheet dates). It is prepared from current assets and current liabilities.
Working Capital = Current Assets – Current Liabilities

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

III. Short Answer Questions

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

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

Question 3.
Briefly explain any three limitations of financial statements.
Answer:
1. Lack of qualitative information:
Qualitative information, that is non – monetary information is also important for business decisions. For examole Efficiency of the employees and efficiency of the management. But this is ignored in financial statements.

2. Record of historical data:
Financial statement are prepared based on historical data. They may not reflect the current position.

3. Ignores price level changes:
Adjustments for price level changes are not made in the financial statements. Hence financial statements may not reveal the current position.

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

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

Percentage increase (or) decrease =

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

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

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

IV Exercises

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

Answer:
Comparative Income Statement Analysis of Arul Ltd.
for the year ended 31.3.2016 to 31.3.2017

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

Answer:
Comparative Income Statement of Barani Ltd. for the year ended 31^st March 2017 to 31^st March 2018

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

Answer:
Comparative Income Statement of Daniel Ltd. for the year ended 31^st March 2016 to 31^st March 2017

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

Answer:
Comparative Balance Sheet of Muthu Ltd. as on 31^st March 2017 to 31^st March 2018

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

Comparative Balance Sheet of Kala Ltd. as on 31^st March 2017 and 31^st March 2018

Question 6.
Prepare common – size income statement for the following particulars of Raja Ltd. for the year ended 31^st March, 2017.

Answer:

Question 7.
From the following particulars of Maria Ltd. and Kala Ltd. prepare a common – size income statement for the year ended 31^st March, 2019.

Answer:
Common Size Income statement for the year ended 31^st March 2019

Question 8.
Prepare common – size income statement for the following particulars of Sam Ltd.

Answer:
Common Size Income statement of Sam Ltd.

Question 9.
Prepare Common – size balance sheet of Meena Ltd. as on 31^st March, 2018.

Answer:
Common Size Balance sheet of Meena Ltd. on 31^st March 18

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

Answer:
Common Size Balance sheet of Rani Ltd. as on 31^st March 2016 to 31^st March 2017

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

Common Size statement of Balance sheet

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

Answer:
Trend Analysis of Kala Ltd Trend Percentage

Question 13.
From the following particulars, calculate the Trend percentages of Kavitha Ltd.

Answer:
Trend Percentage of Kavitha Ltd (in thousands)

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

Answer:
Trend Percentage of Kumar Ltd (in thousands ₹)

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

Answer:
Trend Percentage of Anu Ltd (In Thousands)

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

Answer:
Trend Percentage of Babu Ltd (In Thousands)

Samacheer Kalvi 12th Accountancy Financial Statement Analysis Additional Questions and Answers

I. Choose the correct answer

Question 1.
Financial statements are meaningful and useful only when they are ……………..
(a) verified
(b) presented to owners
(c) Analysed and Interpreted
(d) Published
Answer:
(c) Analysed and Interpreted

Question 2.
Interpretation of Financial statements includes processes like ……………..
(a) Journalising
(b) Ledger writing
(c) Establishing relationship between the account data
(d) None of these
Answer:
(c) Establishing relationship between the account data

Question 3.
Trend analysis is significant for ……………..
(a) Profit planning
(b) Working capital management
(c) Capital rationing
(d) Forecasting and Budgeting
Answer:
(d) Forecasting and Budgeting

Question 4.
The three most useful general purpose financial statements for management are ……………..
(a) Income statement, Statement of Retained Earning and Balance Sheet
(b) Income statement, Balance sheet, and statement of changes in financial position
(c) Income statemen, Statement of Retained Earnings, and Funds flow statement
(d) Statement of Retained Earnings, Balance sheet and Funds flow statement
Answer:
(b) Income statement, Balance sheet, and statement of changes in financial position

Question 5.
In the case of limited company, the term financial statement includes ……………..
(a) Profit and Loss Account and Balance Sheet
(b) Profit and Loss Account, Profit and loss Appropriation Account and Balance Sheet
(c) Balance Sheet
Answer:
(b) Profit and Loss Account, Profit and loss Appropriation Account and Balance Sheet

Question 6.
The term Current Assets does not include ……………..
(a) Payments in Advance
(b) Bills Receivable
(c) Long – Term Deferred Changes
Answer:
(c) Long – Term Deferred Changes

Question 7.
The following is a recorded fact ……………..
(a) Market value of Investment
(b) Debtors
(c) Replacement cost of machinery
Answer:
(b) Debtors

Question 8.
The term Fixed Assets includes ……………..
(a) Stock – in – Trade
(b) Furniture
(c) Payments in Advance
Answer:
(b) Furniture

II. Fill in the blanks

Question 9.
The two statements which are generally included in the definition of financial statements are ……………..
Answer:
Income statement and Balance sheet.

Question 10.
Income statement …………….. the revenues and costs incurred in the process of earning revenues.
Answer:
Matches.

Question 11.
Balance Sheet is a statement of …………….. of a business at a specific moment of time.
Answer:
Financial position.

Question 12.
Assets and liabilities in a Balance sheet may be arranged either according …………….. order or …………….. order.
Answer:
Liquidity, permanency.

Question 13.
Financial statements disclose only …………….. facts.
Answer:
Monetary.

Question 14.
Profit and loss account also called as …………….. statement.
Answer:
Income.

Question 15.
Rearrangement of figures is necessary for …………….. and ………………
Answer:
Analysis, Interpretation.

Question 16.
Analysis of Financial Statements is meant for deriving additional information for various …………….. parties.
Answer:
interested.

III. True or False

Question 17.
Financial statement analysis is carried out as per Government regulation.
Answer:
False.

Question 18.
Financial statement analysis is performed with the help of various techniques and tools.
Answer:
True.

Question 19.
Common – size statements is a technique used in Financial statement analysis.
Answer:
True.

IV. Short answer questions

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

1. Financial statements are generally prepared at the end of an accouning period based on transactions recorded in the books of accounts.
2. These statements are prepared for the organisation as a whole.
3. Information is presented in a meaningful way by grouping items of similar nature such as fixed assets and current assets.
4. Financial statements are prepared based on historical cost.
5. Financial statements are prepared based on accounting principles and Accounting standards, which make financial statements comparable and realistic.
6. Financial statements involve personal judgement in certain cases. For example, selection of method of depreciation and percentage of reserves, etc.

Question 2.
Explain the significance of financial statements.
Answer:

1. To Management: Financial statements provide information to the management to take decision and to have control over business activities, in various areas.
2. To Shareholders: Financial statements help the shareholders to know whether the business has potential for growth and to decide to continue their share holding.
3. To potential investors: Financial statements help to value the securities and compare it with those of other business concerns before making their investment decisions.
4. To Creditors: Creditors can get information about the ability of the business to repay the debts from financial statements.
5. To Bankers: Information given in the financial statements is significant to the bankers to assess whether there is a adequate security to cover the amount of the loan or overdraft.
6. To Government: Financial statements are significant to government to assess the tax liability of business concerns and to frame and amend industrial policies.
7. To Employees: Through the financial statements, the employees can assess the ability of the business to pay salaries and whether they have future growth in the concern.

Question 3.
Explain the objectives of financial statement analysis.
Answer:

1. To analyse the profitability and earing capacity.
2. To study the long term and short term solvency of the business.
3. To determine the efficiency in operations and use of assets.
4. To determine the efficiency of the management and employees.
5. To determine the trend in sales and production, etc.
6. To forecast for future and prepare budgets.
7. To make inter – firm and intra – firm comparisons.

V. Exercise

Question 1.
From the following prepare “Comparative Statement of Proft and Loss” of Good Service Ltd.

Answer:
Comparative Statement of Profit and Loss for the years ended 31^st March 2012 and 2013

Question 2.
Prepare Common – size statement of Genius Paper work Ltd.

Answer:
Common – size Income Statement for the year ended 31^st March 2016 and 2017

Question 3.
From the following figures compute trend percentage using 2005 as the base year.

Answer:

Question 4.
From the following balance sheet extracts, compute trend percentage and comment on the liquidity positon of X ltd. You may take 1990 as base year.

Answer:
Statement showing Trend Percentages

Are you searching for the Samacheer Kalvi 12th Chemistry Chapter Wise Solutions PDF? Then, get your Samacheer Kalvi 12th Chapter Wise Solutions PDF for free on our website. Students can Download Chemistry Chapter 2 p-Block Elements – I Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Chemistry Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements – I

All concepts are explained in an easy way on our website. So, students can easily learn and practice Tamilnadu State Board 12th Chemistry Chapter 2 p-Block Elements – I Question and Answers. You can enjoy the digitized learning with the help of the Tamilnadu State Board Chemistry Online Material.

Samacheer Kalvi 12th Chemistry p-Block Elements – I TextBook Evalution

I. Choose the correct answer:

12th Chemistry Chapter 2 Book Back Answers Question 1.
An aqueous solution of borax is …………
(a) neutral
(b) acidic
(c) basic
(d) amphoteric
Answer:
(c) basic.

12th Chemistry Lesson 2 Book Back Answers Question 2.
Boric acid is an acid because its molecule …………
(a) contains replaceable H⁺ ion
(b) gives up a proton
(c) combines with proton to form water molecule
(d) accepts OH^– from water, releasing proton.
Answer:
(d) accepts OH^– from water, releasing proton

Hint: B(OH)₃ + H₂O \(\rightleftharpoons\) [B(OH)₄]^–+ H⁺

12th Chemistry Samacheer Kalvi Question 3.
Which among the following is not a borane?
(a) B₂H₆
(b) B₃H₆
(C) B₄H₁₀
(d) none of these
Answer:
(a) B₂H₆
Hint:

• Nido borane – B_nH_4+n
• aracno borane – B_nH_6+n , B₃H₆ is not a borane

Samacheer Kalvi 12th Chemistry Question 4.
Which of the following metals has the largest abundance in the earth’s crust?
(a) Aluminium
(b) calcium
(c) Magnesium
(d) Sodium
Answer:
(a) Aluminium

Samacheer Kalvi Guru 12th Chemistry Question 5.
In diborane, the number of electrons that accounts for banana bonds is …………
(a) six
(b) two
(c) four
(d) three
Answer:
(c) four

Hint: There are two 3c – 2e^– bonds i.e., the bonding in the bridges account for 4 electrons.

Samacheer Kalvi 12th Chemistry Solutions Question 6.
The element that does not show catenation among the following p-block elements is …………
(a) Carbon
(b) silicon
(c) Lead
(d) germanium
Answer:
(c) Lead

12th Chemistry Solutions Samacheer Kalvi Question 7.
Carbon atoms in fullerene with formula C₆₀ have …………
(a) sp³ hybridised
(b) sp hybridised
(c) sp² hybridised
(d) partially sp² and partially sp³ hybridised
Answer:
(c) sp² hybridised

12 Chemistry Samacheer Kalvi Question 8.
Oxidation state of carbon in its hydrides …………
(a) +4
(b) -4
(c) +3
(d) +2
Answer:
(a) +4
Hint: CH₄⁺in which the oxidation state of carbon is 4.

Samacheer 12 Chemistry Solutions Question 9.
The basic structural unit of silicates is …………
(a) (SiO₃)^2-
(b) (SiO₄)^2-
(c) (SiO)^–
(d) (SiO₄)^4-
Answer:
(d) (SiO₄)^4-

Chemistry Class 12 Samacheer Kalvi Question 10.
The repeating unit in silicone is …………

Answer:

Class 12 Chemistry Samacheer Kalvi Question 11.
Which of these is not a monomer for a high molecular mass silicone polymer?
(a) Me₃SiCl
(b) PhSiCl₃
(c) MeSiCl₃
(d) Me₃SiCl₃
Answer:
(a) Me₃SiCl

Samacheerkalvi.Guru 12th Chemistry Question 12.
Which of the following is not sp² hybridised?
(a) Graphite
(b) graphene
(c) Fullerene
(d) dry ice
Answer:
(a) dry ice
Hint: dry ice – solid CO₂ in which carbon is in sp hybridized state

Question 13.
The geometry at which carbon atom in diamond are bonded to each other is …………
(a) Tetrahedral
(b) hexagonal
(c) Octahedral
(d) none of these
Answer:
(a) Tetrahedral

Question 14.
Which of the following statements is not correct?
(a) Beryl is a cyclic silicate
(b) Mg₂SiO₄ is an orthosilicate
(c) SiO₄^4- is the basic structural unit of silicates
(d) Feldspar is not aluminosilicate
Answer:
(d) Feldspar is a three dimensional silicate

Question 15.
AlF₃ is soluble in HF only in the presence of KF. It is due to the formation of ………… [NEET]
(a) K₃[AlF₃H₃]
(b) K₃[AlF₆]
(C) AlH₃
(d) K[AlF₃H]
Answer:
(6)K₃[AlF₆]
Hint: AlF₃ + 3KF → K₃[AlF₆]

Question 16.
Match items in column – I with the items of column – II ans assign the correct code


Answer:
(a) A – 2, B – 1, C – 4, D – 3

Question 17.
Duralumin is an alloy of …………
(a) Cu, Mn
(b) Cu, Al, Mg
(c) Al, Mn
(d) Al, Cu, Mn, Mg
Answer:
(d) Al, Cu, Mn, Mg
Hint: Al – 95% , Cu – 4% , Mn – 0.5% , Mg – 1.1 %

Question 18.
Thermodynamically the most stable form of carbon is …………
(a) Diamond
(b) graphite
(c) Fullerene
(d) none of these
Answer:
(b) graphite

Question 19.
The compound that is used in nuclear reactors as protective shields and control rods is …………
(a) Metal borides
(b) metal oxides
(c) Metal carbonates
(d) metal carbide
Answer:
(a) Metal borides

Question 20.
The stability of +1 oxidation state increases in the sequence …………
(a) Al < Ga < In < Tl
(b) Tl < In < Ga < Al
(c) In < Tl < Ga < Al
(d) Ga< In < Al < Tl
Answer:
(a) Al < Ga < In < Tl

II. Answer the following questions:

Question 1.
Write a short note on anamolous properties of the first element of p-block.
Answer:
In p-block elements the first member of each group differs from the other elements of the corresponding group. The following factors are responsible for this anomalous behaviour.

1. Small size of the first member.
2. High ionisation enthalpy and high electronegativity.
3. Absence of d-orbitals in their valance shell.

The first member of the group-13, boron is a metalloid while others are reactive metals. Moreover, boron shows diagonal relationship with silicon of group -14. The oxides of boron and silicon are similar in their acidic nature.

Question 2.
Describe briefly allotropism in p- block elements with specific reference to carbon.
Answer:
Some elements exist in more than one crystalline or molecular forms in the same physical state. This phenomenon is called allotropism. Most common allotropes of carbon are,

1. Graphite
2. Diamond
3. Fullerenes
4. Carbon nanotubes
5. Graphene.

1. Graphite:

• It is the most stable allotropic form of carbon at normal temperature and pressure.
• It is soft and conducts electricity.
• It is composed of flat two dimensional sheets of carbon atoms.
• Each sheet is a hexagonal net of sp² hybridised carbon atoms with a C – C bond length of 1.41 A.
• Structure of graphite,

2. Diamond:

• It is very hard.
• The carbon atoms in diamond are sp¹ hybridised, with a C – C bond length of 1.54 A.
• In the diamond, carbon atoms are arranged in tetrahedral manner.
• Structure of Diamond,

3. Fullerenes:

• It is newly synthesised allotropes of carbon.
• The C₆₀ molecules have a soccer ball like structure and is called buckminster fullerene or buckyballs.
• It has a fused ring structure consists of 20 six membered rings and 12 five membered rings.
• Each carbon atom is sp² hybridised.
• The C – C bond distance is 1.44 A and C = C distance is 1.38 A.
• Structure of fullerene,

4. Carbon nanotubes:

1. It is recently discovered allotropes, have graphite like tubes with fullerene ends.
2. These nanotubes are stronger than steel and conduct electricity.
3. Structure of Carbon nanotubes.

5. Graphene:

• It has a single planar sheet of sp² hybridised carbon atoms that are densely packed in a honeycomb crystals lattice.
• Structure of Graphene,

Question 3.
Boron does not react directly with hydrogen. Suggest one method to prepare diborane from BF₃.
Answer:
Boron does not react directly with hydrogen. However it forms a variety of hydrides called boranes. Treatment of gaseous boron trifluoride with sodium hydride around 450 K gives diborane.

Question 4.
Give the uses of Borax.
Answer:
Uses of borax:

• Borax is used for the identification of coloured metal ions.
• In the manufacture optical and borosilicate glass, enamels and glazes for pottery.
• It is also used as a flux in metallurgy and also acts as a good preservative.

Question 5.
What is catenation ? describe briefly the catenation property of carbon.
Answer:
Catenation:
It is the phenomenon of an atom to form a strong covalent bond with the atoms of itself. Carbon shares the property of catenation to maximum extent because it is small in size and can form pn-pn multiple bonds to itself. The following conditions are necessary for catenation.

1. The valency of element is greater than or equal to two.
2. Element should have an ability to bond with itself.
3. The self bond must be as strong as its bond with other elements.
4. Kinetic inertness of catenated compound towards other molecules. Carbon possesses all the above properties and forms a wide range of compounds with itself.

Question 6.
Write a note on Fisher tropsch synthesis.
Answer:
The reaction of carbon monoxide with hydrogen at a pressure of less than 50 atm using metal catalysts at 500-700 K yields saturated and unsaturated hydrocarbons.

Question 7.
Give the structure of CO and CO₂.
Answer:
Structure of CO:

Structure of CO₂:

Question 8.
Give the uses of silicones.
Answer:
Uses of silicones:

1. Silicones are used for low temperature lubrication and in vacuum pumps, high temperature oil baths etc.
2. They are used for making water proofing clothes.
3. They are used as insulting material in electrical motor and other appliances
4. They are mixed with paints and enamels to make them resistant towards high temperature, sunlight, dampness and chemicals.

Question 9.
AlCl₃ behaves like a lewis acid. Substantiate this statement.
Answer:
In AlCl₃, Al in electron deficient it needs two electrons to complete octet so it act as lewis acid. AlCl₃ usually exist as a dimer to achieve octet by bridged Cl atom electron deficient compounds are lewis acids.

Question 10.
Describe the structure of diborane.
Answer:
In diborane two BH₂ units are linked by two bridged hydrogens. Therefore, it has eight B-H bonds. However, diborane has only 12 valance electrons and are not sufficient to form normal covalent bonds. The four terminal B-H bonds are normal covalent bonds (two centre – two electron bond or 2c-2e bond). The remaining four electrons have to used for the bridged bonds, i.e. two three centred B-H-B bonds utilise two electrons each.

Hence, these bonds are three centre – two electron bonds. The bridging hydrogen atoms are in a plane as shown in the figure. In dibome, the boron is sp³ hybridised. Three of the four sp³ hybridised orbitals contains single electron and the fourth orbital is empty.

Two of the half filled hybridised orbitals of each boron overlap with the two hydrogens to form four terminal 2c-2e bonds, leaving one empty and one half filled hybridised orbitals on each boron. The Three centre – two electron bonds, B-H-B bond formation involves overlapping the half filled hybridised orbital of one boron, the empty hybridised orbital of the other boron and the half filled 1s orbital of hydrogen.

Question 11.
Write a short note on hydroboration.
Answer:
Diborane adds on to alkenes and alkynes in ether solvent at room temperature. This reaction is called as hydroboration and is highly used in synthetic organic chemistry especially for anti-Markovnikov addition.
B₂H₆ + 3RCH = CHR → B( CH₂ – CH₂R )₃+ 6H₂

Question 12.
Give one example for each of the following:

1. icosogens
2. tetragen
3. prictogen
4. chalcogen

Answer:
1. Icosogens:

• Boron
• Aluminium
• Gallium

2. Tetragen:

• Carbon
• Silicon
• Germanium

3. Prictogen:

• Oxygen
• Sulfur
• Selenium

4. Chalcogen:

• Fluorine
• Chlorine
• Bromine

Question 13.
Write a note on metallic nature of p-block elements.
Answer:

1. The tendency of an element to form a cation by loosing electrons is known as electro¬positive or metallic character.
2. This character depends on the ionisation energy.
3. Generally on descending a group the ionisation energy decreases and hence the metallic character increases.

In p-block, the elements present in lower left part are metals while the elements in the upper right part are non metals. Elements of group 13 have metallic character except the first element boron which is a metalloid, having properties intermediate between the metal and nonmetals. The atomic radius of boron is very small and it has relatively high nuclear charge and these properties are responsible for its nonmetallic character.

In the subsequent groups the non-metallic character increases. In group 14 elements, carbon is a nonmetal while silicon and germanium are metalloids. In group 15, nitrogen and phosphorus are non metals and arsenic & antimony are metalloids. In group 16, oxygen, sulphur and selenium are non metals and tellurium is a metalloid. All the elements of group 17 and 18 are non metals.

Question 14.
Complete the following reactions:
(a) B(OH)₃ + NH₃ →
(b) Na₂B₄O₇ + H₂SO₄+ H₂O →
(c) B₂H₆ + 2NaOH + 2H₂O →
(d) B₂H₆ + CH₃OH →
(e) BF₃ + 9H₂O →
(f) HCOOH + H₂SO₄→
(g) SiCl₄ + NH₃ →
(h) SiCl₄ + C₂H₅OH →
(i) B + NaOH →
(j) H₂B₄O₇ \(\underrightarrow { Red\quad hot }\)

Answer:

Question 15.
How will you identify borate radical?
Answer:
When boric acid or borate salt is heated with ethyl alcohol in presence of concentrated H₂SO₄, an ester triethyl borate is formed. The Vapour of this ester burns with a green edged flame and this reaction is used to identify the presence of borate.

Question 16.
Write a note on zeolites.
Answer:
Zeolites:

1. Zeolites are three dimensional crystalline solids containing aluminium, silicon and oxygen in their regular three dimensional framework.
2. They are hydrated sodium alumino silicates with general formula, Na₂O. (Al₂O₃). x(SiO₂)y(H₂O) (x = 2 to 10; y = 2 to 6)
3. Zeolites have porous structure in which the monovalent sodium ions and water molecules are loosely held.
4. The Si and Al atoms are tetrahederally coordinated with each other through shared oxygen atoms.
5. Zeolites structure looks like a honeycomb consisting of a network of interconnected tunnels and cages.
6. Zeolite crystal to act as a molecular sieve. They helps to remove permanent hardness of water.

Question 17.
How will you convert boric acid to boron nitride?
Answer:
Fusion of urea with boric acid B(OH)₃, in an atmosphere of ammonia at 800 – 1200 K gives

Question 18.
A hydride of 2nd period alkali metal
(A) on reaction with compound of Boron
(B) to give a reducing agent
(C) identify A, B and C.
Answer:

1. A hydride of 2^nd period alkali metal (A) is lithium hydride (LiH).
2. Lithium hydride (A) reacts with diborane (B) to give lithiumborohydride (C) which is act as reducing agent.
   

(A) Lithium hydride – LiH
(B) Diborane – B₂H₆
(C) Lithium borohydride – LiBH₄

Question 19.
A double salt which contains fourth period alkali metal
(A) on heating at 500K gives
(B). Aqueous solution of (B) gives white precipitate with BaCl₂ and gives a red colour compound with alizarin. Identify A and B.
Answer:
1. A double salt which contains fourth period alkali metal (A) is potash alum
K₂SO₄. Al₂(SO₄)₃. 24H₂O

2. On heating potash alum (A) 500 k give anhydrous potash alum (or) burnt alum (B).

3. Aqueous solution of burnt alum, has sulphates ion, potassium ion and aluminium ion. Sulphate ion reacts with BaCl₂ to form white precipitate of Barium Sulphate
(SO₄)₂ + BaCl₂ → BaSO₄ + 2Cl^–
Aluminium ion reacts with alizarin solution to give a red colour compound.

Question 20.
CO is a reducing agent. Justify with an example.
Answer:
Both thermodynamic and kinetic factors make carbon monoxide (CO) a better reducing agent. When CO is used to reduce a metal oxide, it gets oxidized to CO₂ Thermodynamically, CO₂ is much more stable than CO. For example,
CO + Fe₂O₃ → 2Fe + 3CO₂

Samacheer Kalvi 12th Chemistry p-Block Elements – I Evaluate yourself

Question 1.
Why group 18 elements are called inert gases? Write the general electronic configuration of group 18 elements.
Answer:
The elements of group-18 have completely filled s and p orbitals, hence they are more stable and have least reactivity. Therefore group-18 elements are called inert gases. ns₂np₆ is the general electronic configuration of group elements.

Samacheer Kalvi 12th Chemistry p-Block Elements – I Additional Questions

Samacheer Kalvi 12th Chemistry p-Block Elements – I 1 Mark Questions and Answers

I. Choose the correct answer:

Question 1.
More common oxidation state for halogens is …………
(a) +1
(b) +2
(c) -1
(d) -2
Answer:
(c) -1

Question 2.
Electronic configuration of noble gases is …………
(a) ns²
(b) ns²np⁵
(c) ns¹np⁶
(d) ns²np⁶
Answer:
(d) ns²np⁶

Question 3.
Noble gases are chemically inert. This is due to …………
(a) unstable electronic configuration
(b) stable electronic configuration
(c) only filled p-orbital
(d) only filled 5-orbital
Answer:
(b) stable electronic configuration

Question 4.
Consider the following statements.
(i) The first member of the group-13, boron is a metalloid while others are reactive metals.
(ii) The oxides of boron and silicon are similar in their acidic nature.
(iii) Both boron and silicon form metallic hydrides.

Which of the above statement(s) is/are not correct?
(a) (i) only
(b) (ii) only
(c) (ii) and (iii)
(d) (iii) only
Answer:
(d) (iii) only

Question 5.
Which element has a greater tendency to form a chain of bonds with itself?
(a) Boron
(b) Silicon
(c) Tin
(d) Carbon
Answer:
(d) Carbon

Question 6.
Which one of the following is the strongest oxidising agent?
(a) Fluorine
(b) Chlorine
(c) Bromine
(d) Iodine
Answer:
(a) Fluorine

Question 7.
Some elements exist in more than one crystalline or molecular forms in the same physical state is called …………
(a) isomerism
(b) allotropism
(c) isomorphism
(d) isoelectronics
Answer:
(b) allotropism

Question 8.
How many allotropes possible for boron?
(a) 1
(b) 4
(c) 6
(d) 7
Answer:
(c) 6

Question 9.
Important ore of boron is …………
(a) bauxite
(b) borosilicate
(c) borax
(d) P-tetragonal boron
Answer:
(c) borax

Question 10.
Less reactive elements in boron family is …………
(a) Boron
(b) Aluminium
(c) Gallium
(d) Thallium
Answer:
(b) Aluminium

Question 11.
More toxic element in boron family is …………
(a) Boron
(b) Aluminium
(c) Gallium
(d) Thallium
Answer:
(d) Thallium

Question 12.
Boron does not …………
(a) Oxygen
(b) Hydrogen
(c) Acids
(d) Alkali
Answer:
(b) Hydrogen

Question 13.
Borontrifluoride reacts with sodium hydride at 450 K gives …………
(a) diborane
(b) tetraborane
(c) pentaborane
(d) decaborane
Answer:
(a) diborane

Question 14.
2B + N₂ \(\underrightarrow { \triangle }\) A . Identify A
(a) BN₃
(b) B₃N
(c) (BN)₃
(d) BN
Answer:
(d) BN

Question 15.
Boron reacts with fused sodium hydroxide to forms …………
(a) Borax
(b) Boric acid
(c) Sodium borate
(d) Sodium tetraborate
Answer:
(c) Sodium borate

Question 16.
Which isotope is used as moderator in nuclear reactors?
(a) ¹⁰B₅
(b) ⁿC₆
(c) ⁴He₂
(d) ⁴⁰Ca₂
Answer:
(a) ¹⁰B₅

Question 17.
Borax is ………… in nature.
(a) basic
(b) acidic
(c) amphoteric
(d) chemically inert
Answer:
(a) basic

Question 18.
The compound used as a flux in metallurgy?
(a) Boron nitride
(b) Boron oxide
(c) Boron fluoride
(d) Borax
Answer:
(d) Borax

Question 19.
The trialkylborate on reaction with sodiumhydride in tetrahydrofuran to form …………
(a) NaBH₄
(b) Na[BH(OR)₃]
(c) Na[B(OR)₃]
(d) Na[BH(OR)₃]
Answer:
(b) Na[BH(OR)₃]

Question 20.
Compounds used as an eye lotion …………
(a) H₃BO₃
(b) HBO₂
(c) H₂B₄O₇
(d) B₂O₃
Answer:
(a) H₃BO₃

Question 21.
Which one of the following is highly reactive compound?
(a) B₂OH₃
(b) H₂BO₃
(c) HBO₂
(d) B₂H₆
Answer:
(d) B₂H₆

Question 22.
B₂H₆ + 6CH₃OH → A Identify A
(a) B₂O₃
(b) CH₃OB
(c) CH₃OH
(d) B(OCH₃)₃
Answer:
(d) B(OCH₃)₃

Question 23.
Which one of the following is called as inorganic benzene?
(a) B₂H₆
(b) BN
(c) H₂B₄O₇
(d) B₃N₃H₆
Answer:
(d) B₃N₃H₆

Question 24.
Compound contains two centred – two electron bond (2c-2e) is …………
(a) B₆H₁₁
(b) B₅H₉
(c) B₂H₆
(d) B₁₀H₁₁
Answer:
(c) B₂H₆

Question 25.
Diborane reacts with excess ammonia at high temperature to give …………
(a) Boron nitride
(b) Boron oxide
(c) Borazole
(d) Diborane diammonate
Answer:
(c) Borazole

Question 26.
Consider the following statements.
(i) Diborane contains two centre-two electron bond.
(ii) In diborane, the boron has sp³ hybridis ed.
(iii) Diborane has two terminal B – H bonds and four B – H – B bonds.

Which of the above statement(s) is/are correct.
(a) (i) and (iii)
(b) (ii) and (iii)
(c) (i) only
(d) (i) and (ii)
Answer:
(d) (i) and (ii)

Question 27.
Compound used for propellant is …………
(a) BN
(b) H₂B₄O₇
(c) B₂H₂
(d) Borax
Answer:
(c) B₂H₂

Question 28.
Which one of the following is double salt?
(a) Potash alum
(b) Potassium sulphate
(c) Aluminium Sulphate
(d) Ammonium sulphate
Answer:
(a) Potash alum

Question 29.
Consider the following statements.
(i) Alums are more soluble in hot water than in cold water.
(ii) Alums are more soluble in cold water than in hot water.
(iii) Potash alum is employed a styptic agent to arrest bleeding.

Which of the above statement(s) is/are not correct?
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (i), (ii) and (iii)
Answer:
(b) (ii) only

Question 30.
The structure of graphite is …………
(a) planner
(b) hexagonal
(c) octahedral
(d) bucky balls
Answer:
(b) hexagonal

Question 31.
Which one of the following is used as a lubricant?
(a) Graphite
(b) Diamond
(c) Fullerene
(d) Graphene
Answer:
(a) Graphite

Question 32.
Which one of the following carbon allotrope is very hard?
(a) Graphite
(b) Diamond
(c) Fullerene
(d) Graphene
Answer:
(b) Diamond

Question 33.
Recently discovered allotropes of carbon is …………
(a) Graphite
(b) Diamond
(c) Carbon nanotubes
(d) Fullerenes
Answer:
(c) Carbon nanotubes

Question 34.
CO and N₂ mixture is …………
(a) natural gas
(b) producer gas
(c) water gas
(d) LPG
Answer:
(b) producer gas

Question 35.
Syn gas is …………
(a) CO + N₂
(b) CO + H₂
(c) CO₂ + H₂
(d) CO₂ + N₂
Answer:
(b) CO + H₂

Question 36.
Ethene is mixed with carbon monoxide and hydrogen gas to produce propanal is known as …………
(a) Oxo process
(b) McAfee process
(c) Wacker process
(d) Haber process
Answer:
(a) Oxo process

Question 37.
………… is a good reducing agent.
(a) CO
(b) CO₂
(c) [Cr(CO)₆]
(d) [Ni(CO)₄]
Answer:
(a) CO

Question 38.
Critical temperature of CO₂ is ………..
(a) -31°C
(b) -13°C
(c) 31°C
(d) 13°C
Answer:
(c)31°C

Question 39.
Which one of the following compound is important for photosynthesis?
(a) CO
(b) CO₂
(c) COCl₂
(d) C
Anwer:
(b) CO

Question 40.
General empirical formula of silicone is ………..
(a) (R₂SiO)
(b) (RSiO)
(c) (R₂CO)
(d) (RSiH)
Answer:
(a) (R₂SiO)

Question 41.
Consider the following statements.
(i) All Silicones are hydrophilic in nature.
(ii) Silicones are thermal and electrical insulators.
(iii) Chemically silicones are highly reactive.

Which of the above statement(s) is/are not correct?
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (ii) only
Answer:
(b) (i) and (iii)

Question 42.
Silicate contains ……….. silicon and oxygen in units.
(a) [SiO₂]^4-
(b) [SiO₂]
(c) [SiO₂]^2-
(d) [SiO₂]^–
Answer:
(a) [SiO₂]^4-

Question 43.
Ortho silicates are also called as ………..
(a) Ino silicates
(b) Soro silicates
(c) Neso silicates
(d) Cyclic silicates
Answer:
(c) Neso silicates

Question 44.
Example of Ring silicate is ………..
(a) Olivine
(b) Beryl
(c) Spodumene
(d) Asbestos
Answer:
(b) Beryl

Question 45.
Pick out the three dimensional silicates?
(a) Talc
(b) Mica
(c) Quartz
(d) Asbestos
Answer:
(c) Quartz

Question 46.
Compound used to remove the permanent hardness of water is ………..
(a) Zeolite
(b) Feldspar
(c) Talc
(d) Mica
Answer:
(a) Zeolite

II. Fill in the blanks:

1. …………… is the general electronic configuration of tetragen elements.
2. Boron and silicon form …………… hydrides.
3. …………… is most reactive element among the halogens.
4. Most stable oxidation state of aluminium is ……………
5. General formula of metal boride is ……………
6. Boron has the capacity to absorb ……………
7.  …………… is used as a rocket fuel igniter.
8. …………… is essential for the cell walls of plants.
9. …………… is a chemical formula of boron.
10. …………… is used for the identification of coloured metal ions.
11. …………… is a colourless transparent crystal.
12. Boric acid has a …………… structure.
13. Boric acid consists of …………… unit.
14. On heating magnesium boride with HCl a mixture of volatile …………… are obtained.
15. Diborane reacts with methylalcohol to give ……………
16. Diborane has …………… B – H bonds.
17. In diborane, the boron is …………… hybridised.
18.  …………… is inorganic benzene.
19. Boron trifluoride has a …………… geometry.
20. Anhydrous aluminium chloride is a …………… substance.
21. With excess of NaOH, aluminium chloride produces ……………
22. …………… is used for the manufacture of petrol by cracking the mineral oils.
23. …………… is used for water proofing and textiles.
24. ……………is the most stable allotropic form of carbon at normal temperature and pressure.
25. …………… is allotropic form of carbon has aromatic character.
26. Graphene has …………… lattice.
27. Carbonyl chloride is a …………… gas.
28. Equimolar mixture of hydrogen and carbonmonoxide is called as ……………
29. Producer gas is a mixture of …………… and ……………
30. Aqueous solution of carbon dioxide forms ……………
31. …………… are high temperature polymers.
32. The viscosity of silicon oil remains ……………
33.  …………… are used as insulating material.
34. In beryl, each aluminium is surrounded by ……………
35. …………… is carcinogenic silicates.
36. Three dimensional by replacing units by ……………
37. …………… crystal to act as a molecular sieve.
38. Borax is a sodium salt of ……………

Answers:

1. ns²np²
2. covalent
3. Fluorine
4. +3
5. MxBy
6. neutrons
7. Amorphous boron
8. Boron
9. Na₂B₄O₇. 10H₂O
10. Borax
11. Boric acid
12. two dimensional layered
13. [BO₃]^3-
14. borones
15. trimethylborate
16. eight
17. sp³
18. Borazine
19. planar
20. hygroscopic
21. Sodium meta aluminate
22. AlCF
23. Potash alum
24. Graphite
25. Fullerene
26. honey comb nitrogen
27. poisonous
28. water gas
29. carbonmonoxide, nitrogen
30. carbonic acid
31. silicones
32. constant
33. Silicones
34. six oxygen
35. Asbestos
36. [SiO₄]^4-; [AlO₄]^5-
37. Zeolite
38. tetraboric acid

III. Match the following:

Question 1.
(i) Tetragens – (a) Oxygen
(ii) Icosagens – (b) Carbon
(iii) Chalcogens – (c) Nitrogen
(iv) Pnictogens – (d) Boron
Answer:
(i) – (b)
(ii) – (d)
(iii) – (a)
(iv) – (c)

Question 2.
(i) Borax – (a) Na₂B₄O₇
(ii) Prismatic form – (b) Na₂B₄O_7. 5H₂O
(iii) Jeweller borax – (c) Na₂B₄O_7. 10H₂O
(iv) Borax glass – (d) [B₄O₅(OH)₄]₂
Answer:
(i) – (c)
(ii) – (d)
(iii) – (b)
(iv) – (a)

Question 3.
(i) Boron – (a) Optical
(ii) Borax – (b) Neutron absorber
(iii) Boric acid – (c) Welding torches
(iv) Diborane – (d) Eye lotion
Answer:
(i) – (b)
(ii) – (a)
(iii) – (d)
(iv) – (c)

Question 4.
(i) Graphene – (a) Honeycomb crystal
(ii) Diamond – (b) Aromatic character
(iii) Fullerene – (c) Lubricant
(iv) Graphite – (d) Very hard
Answer:
(i) – (c)
(ii) – (d)
(iii) – (b)
(iv) – (a)

Question 5.
(i) Ortho silicate
(ii) Pyro silicate
(iii) Cyclic silicate
(iv) Tecto silicate
Answer:
(i) – (d)
(ii) – (a)
(iii) – (b)
(iv) – (c)

IV. Assertion and reason:

Note:
In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
(a) A and R are correct and R explains A.
(b) A and R are correct and R not explains A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.

Question 1.
Assertion (A) – Noble gases are least reactivity.
Reason (R) – Noble gases have completely filled s and p-orbital and attain stable electronic configuration.
Answer:
(a) A and R are correct and R explains A.

Question 2.
Assertion (A) – Both boron and silicon form covalent hydrides.
Reason (R) – Boron does not shows diagonal relationship with silicon of group 14.
Answer:
(c) A is correct but R is wrong.

Question 3.
Assertion (A) – Fluorine is most reactive element among the halogens.
Reason (R) – Fluorine has minimum bond dissociation energy.
Answer:
(a) A and R are correct and R explains A.

Question 4.
Assertion (A) – Boron combines with halogen to form trihalides at high temperatures.
Reason (R) – Boron does not reacts directly with hydrogen.
(b) A and R are correct and R not explains A.

Question 5.
Assertion (A) – Boron-10 isotope is used as moderator in nuclear reactors.
Reason (R) – Boron has the capacity to absorb neutrons.
Answer:
(a) A and R are correct and R explains A.

Question 6.
Assertion (A) – Diborane is highly reactive.
Reason (R) – At high temperatures, diborane forms higher boranes.
Answer:
(b) A and R are correct and R not explains A.

Question 7.
Assertion (A) – BF₃ reacts with ammonia to form complex.
Reason (R) – BF₃ is a electron deficient compound and accepts electron pairs to form coordinate covalent bonds.
Answer:
(a) A and R are correct and R explains A.

Question 8.
Assertion (A) – Fullerene has aromatic character.
Reason (R) – Some of the fullerenes have e-bonds and delocalised n-bonds.
Answer:
(a) A and R are correct and R explains A.

Question 9.
Assertion (A) – Carbon dioxide is non flammable gas.
Reason (R) – CO₂ critical temperature is 31°C and can be readily liquefied.
Answer:
(b) A and R are correct and R not explains A.

Question 10.
Assertion (A) – Zeolites act as a molecular sieve.
Reason (R) – Zeolite structure is pore/ channel sizes are nearly uniform.
Answer:
(a) A and R are correct and R explains A.

V. Find the odd one out and given the reason:

Question 1.
(a) Icosagens
(b) Tetragens
(c) Alkali metals
(d) Chalcogens
Answer:
(c) Alkali metals
Reason: Alkali metals are s-block elements but other are p-block elements.

Question 2.
(a) Al
(b)B
(c) O
(d) Na
Answer:
(d) Na
Reason: Na is s-block element but others are p-block elements.

Question 3.
(a) Diborane
(b) Borax
(c) Carbonmonoxide
(d) Boric acid
Answer:
(c) Carbonmonoxide
Reason: Carbonmonoxide is not a compound of boron.

Question 4.
(a) B₃N₃H₆
(b) B₄H₁₀
(c) B₅H₉
(d) B₅H₁₁
Answer:
(a) B₃N₃H₆
Reason: B₃N₃H₆ not a higher borane

Question 5.
(a) Potash alum
(b) Sodium alum
(c) Burnt alum
(d) Ammonium alum
Answer:
(c) Burnt alum
Reason: Burnt alum does not having water molecule.

Question 6.
(a) Graphite
(b) Borax
(c) Fullerene
(d) Diamond
Answer:
(b) Borax
Reason: Borax is not a allotropic form of carbon.

Question 7.
(a)B
(b) Ga
(c) In
(d) N
Answer:
(d) N
Reason: N is not a icosagens.

VI. Find out the correct pair:

Question 1.
(a) Ortho sililcate – Soro silicate
(b) Pyro silicate – Neso silicate
(c) Chain silicate – Pyroxenes
(d) Double chain silicate – Ring silicate
Answer:
(c) Chain silicate – Pyroxenes

Question 2.
(a) Graphene – 1.54 A
(b) Diamond – 1.40 A
(c) Fullerene – 1.38 A
(d) Graphite – 1.40 A
Answer:
(d) Graphite – 1.40 A

Question 3.
(a) Diborane – welding torches
(b) AlCl₃ – Eye lotion
(c) Borax – pigments
(d) Boric acid – optical
Answer:
(a) Diborane – welding torches

Question 4.
(a) Inert gas – ns²np²
(b) Chalcogens – ns²np¹
(c) Pnictogens – ns²np⁴
(d) Tetragens – ns²np⁴
Answer:
(c) Pnictogens – ns²np³

Question 5.
(a) Icosagen – Oxygen
(b) Tetragen – Carbon
(c) Chalcogen – Fluorine
(d) Halogen – Boron
Answer:
(b) Tetragen – Carbon

VII. Find out the incorrect pair:

Question 1.
(a) Tetragens – ns²np²
(b) Icosagens – ns²np¹
(c) Chalcogens – ns²np⁴
(d) Halogens – ns²np⁶
Answer:
(d) Halogens – ns²np⁶

Question 2.
(a) Boron is BF₃ – +3
(b) Carbon in CO₂ – +4
(c) Nitrogen in N₂O₅ – +5
(d) Fluorine in OF₂ – +4
Answer:
(d) Fluorine in OF₂ – +4

Question 3.
(a) Nitrogen – Tetragens
(b) Oxygen – Chalcogens
(c) Tin – Tetragens
(d) Gallium – Icosagens
Answer:
(a) Nitrogen – Tetragens

Question 4.
(a) Boron – Moderator
(b) Borax – Eye lotion
(c) Boric acid – Antiseptic
(d) Diborane – Propellant
Answer:
(b) Borax – Eye lotion

Question 5.
(a) McAfee process – AlCl₃
(b) Burnt alum – K₂SO₄-Al₂(SO₄)₃
(c) Oxo process – Propanal
(d) Fischer tropsch Synthesis – HCOOH
Answer:
(d) Fischer tropsch Synthesis – HCOOH

Question 6.
(a) Soro silicate – Thortveitite
(b) Ring silicate – Beryl
(c) Sheet silicate – Asbestos
(d) Neso silicate – Olivine
Answer:
(c) Sheet silicate – Asbestos

Question 7.
(a) Pyroxenes – Phenacite
(b) Amphiboles – Asbestos
(c) Phyllo silicate – Mica
(d) Tecto silicate – Quartz
Answer:
(a) Pyroxenes – Phenacite

Samacheer Kalvi 12th Chemistry p-Block Elements – I 2 Mark Questions and Answers

Question 1.
What are Wade’s Rule?
Answer:
Wade’s rules are used to rationalize the shape of borane clusters by calculating the total number of skeletal electron pairs (SEP) available for cluster bonding.

Question 2.
Why-1 oxidation state is more common in halogens. Explain.
Answer:
The halogens have a strong tendency to gain an electron to give a stable halide ion with completely filled electronic configuration (ns₂sp₆) and hence-1 oxidation state is more common in halogens.

Question 3.
Why boron has non-metallic character?
Answer:
The atomic radius of boron is very small and it has relatively high nuclear charge and these properties are responsible for its non-metallic character.

Question 4.
Define inert pair effect.
Answer:
In heavier post-transition metals, the outer s-electron(m) have a tendency to remain inert and show reluctance to take part in the bonding (only p-orbital involved in chemical bonding), which is known as inert pair effect.

Question 5.
Mention the allotropes of boron.
Answer:

1. Amorphous boron
2. a-rhombohedral boron
3. p-rhombohedral boron
4. γ – rhombohedral boron
5. α – tetragonal boron
6. β – tetragonal boron

Question 6.
List out the allotropes of tin.
Answer:

1. Grey tin
2. White tin
3. Rhombic tin
4. Sigma tin

Question 7.
Mention the allotropes of phosphorous?
Answer:

1. White phosphorous
2. Red phosphorous (v)
3. Scarlet phosphorous
4. Violet phosphorous
5. Black phosphorous

Question 8.
Give the two allotropes of sulphur.
Answer:

1. Rhombus
2. Monoclinic

Question 9.
Why boron compounds are covalent in nature?
Answer:
Many of boron compounds are electron deficient and has unusual type of covalent bonding, which is due to its small size, high ionisation energy and similarity in electronegativity with carbon and hydrogen.

Question 10.
Define Borax is basic in nature.
Answer:
Borax is basic in nature and its solution in hot-water is alkaline as it dissociates into boric acid and sodium hydroxide,
Na₂B₄O₇ + 7H₂O → 4H₃BO₃ + 2NaOH

Question 11.
Explain action of heat on borax.
Answer:
On heating borax, it forms a transparent borax beads.

Question 12.
What happen when borax treated with ammonium chloride?
Answer:
When borax treated with ammonium chloride, it forms boron nitride.

Question 13.
What happen when boric acids reacts with sodium hydroxide?
Answer:
Boric acid reacts with sodium hydroxide to form sodium metaborate and sodium tetraborate.

Question 14.
Identify A and B from the following reaction,

Answer:

Question 15.
Explain the action of air on diborane.
Answer:
At room temperature pure diborane does not react with air or oxygen but in impure form it gives [B2O₃]^3- along with large amount of heat.

Question 16
Explain the action of air on diborane.
Answer:
At room temperature pure diborane does not react with air or oxygen but in impure form it gives B₂O₃ along with large amount of heat.
B₂H₆+ 3O₂ → B₂O₃ + 3H₂O

Question 17.
How will you convert diborane into sodium borohydride?
Answer:
Diborane reacts with sodium hydride in the presence of diglyme to give sodim borohydride.

Question 18.
Mention the uses o boron trifluoride.
Answer:

1. Boron trifluoride is used for preparing HBF₄, a catalyst in organic chemistry
2. It is also used as a fluorinating reagent.

Question 19.
What is MCA fee process?
Answer:
Aluminium chloride is obtained by heating a mixture of alumina and coke in a current of chloride.

Question 20.
What happen when potash alum is treated with ammonium hydroxide?
Answer:
Potash alum forms aluminium hydroxide, when, treated with ammonium hydroxide
K₂SO_4. A₁₂(SO₄)₃. 24H₂O + 6NH₄OH → K₂SO₄ + 3(NH₄)2SO₄ + 24H₂O + 3Al(OH)₃

Question 21.
What is producer gas? How will you prepare producer gas?
Answer:
On industrial scale carbon monoxide is produced by the reaction of carbon with air. The carbon monoxide formed will contain nitrogen gas also and the mixture of nitrogen and carbon monoxide is called producer gas.

Question 22.
What is synthetic gas?
Answer:
A mixture of carbonmonoxide and hydrogen is called synthetic gas.

Question 23.
What is phosgene?
Answer:
When carbon monoxide is treated with chlorine in presence of light or charcoal, it forms a poisonous gas carbonyl chloride, when is also known as phosgene.

Question 24.
How will you prepare propanal by oxoprocess?
Answer:
In oxoprocess, ethene is mixed with carbon monoxide and hydrogen gas to produce propanal.

Question 25.
What are water gas equilibrium?
Answer:
The equilibrium involved in the reaction between carbon dioxide and hydrogen, has many industrial applications and is called water gas equilibrium.

Question 26.
Mention the uses of silicon tetrachloride.
Answer:

1. Silicon tetrachloride is used in the production of semiconducting silicon.
2. It is used as a starting material in the synthesis of silica gel, silicic esters, a binder for ceramic materials.

Question 27.
What are silicones?
Answer:
Silicones or poly siloxanes are organo silicon polymers with general empirical formula (R₂SiO). Since their empirical formula is similar to that of ketone (R₂CO), they were named silicones.

Question 28.
What are high-temperature polymers?
Answer:
Silicones are high-temperature polymers, because they have high thermal stability.

Question 29.
Why silicones are water repellent?
Answer:
All silicones are water repellent, this is due to the presence of organic side groups that surrounds the silicon which makes the molecule looks like an alkane. Therefore silicones are water repellent.

Samacheer Kalvi 12th Chemistry p-Block Elements – I 3 Marks Questions and Answers

Question 1.
Write a notes on ionisation enthalpy in p-block elements?
Answer:
1. As we move down a group, generally there is a steady decrease in ionisation enthalpy of elements due to increase in their atomic radius.

2. In p-block elements there are some minor deviations to this general trend. In group 13, from B to Al the ionisation enthalpy decreases as expected. But from Al to Tl there is only a marginal difference. This is due to the presence of inner d- and f-elements which has poor shielding effect compared to s and p electrons. As a result, the effective nuclear charge on the valance electrons increase.

3. A similar trend is also observed in group 14. The remaining groups (15-18) follows the general trend, in these groups the ionisation enthalpy decreases as we move down the group. Here poor shielding effect of d- and f-electrons are overcome by the increased shielding effect of the additional p-electrons.

4. The ionisation enthalpy of elements in successive groups is higher than the corresponding elements of the previous group as expected.

Question 2.
Give the uses of boron.
Answer:

1. Boron has the capacity to absorb neutrons. Hence, its isotope ¹⁰B₅ is used as moderator in nuclear reactors.
2. Amorphous boron is used as a rocket fuel igniter.
3. Boron is essential for the cell walls of plants.
4. Compounds of boron have many applications. For example eye drops, antiseptics, washing powders etc..contains boric acid and borax. In the manufacture of Pyrex glass , boric oxide is used.

Question 3.
How will be prepare borax from colemanite?
Answer:
Borax is a sodium salt of tetraboric acid. It is obtained from colemanite ore by boiling its solution with sodium carbonate.

Question 4.
Explain the extraction of boric acid from,

1. 1. Borax
   2. Colemanite

Answer:
1. Borax – Borax is treated with hydrochloric acid or sulphuric acid or nitric acid to give boric acid.

2. Colemanite – Colemanite is boiling with water, to give boric acid.

Question 5.
Explain the action of heat on boric acid.
Answer:
Boric acid when heated at 373 K gives metaboric acid and at 413 K, it gives tetraboric acid. When heated at red hot, it gives boric anhydride which is a glassy mass.

• 4 H₃BO₃ \(\underrightarrow { 373k }\) 4HBOz + H₂O
• 4HBO₂ \(\underrightarrow { 413k }\) H₂B₄O₇ + H₂O
• H₂B₄O₇ \(\underrightarrow { Red hot }\) 2B₂O₃ + H₂O

Question 6.
How will you convert, boric acid into,

1. Boron trifluoride
2. Borax

Answer:
1. Boron trifluoride:
Boric acid reacts with calcium fluoride in presence of cone. Sulphuric acid and gives boron trifluoride.

2. Borax:
Boric acid, when heated with soda ash it gives borax.

Question 7.
Mention the use of boric acid.
Answer:

1. Boric acid is used in the manufacture of pottery glazes, glass, enamels and pigments.
2. It is used as an antiseptic and as an eye lotion.
3. It is also used as a food preservative.

Question 8.
Explain the preparation of diborane.
Answer:
1. Diborane can also be obtained in small quantities by the reaction of iodine with sodium borohydride in diglyme.
2NaBH₄ + I₂ → B₂H₆ + 2NaI + H₂

2. On heating magnesium boride with HCl a mixture of volatile boranes are obtained.

• 2Mg₃B₂+ 12HCl → 6MgCl₂ + B₄H₁₀ + H₂
• B₄H₁₀+ H₂ → 2B₂Hg (Diborane)

Question 9.
What happen when diborane heated at various temperature?
Answer:
Diboranc is a gas at room temperature with sweet smell and it is extremely toxic. It is also highly reactive. At high temperatures it forms higher boranes liberating hydrogen.

Question 10.
Explain the reaction between diborane and ammonia?
Answer:
When treated with excess ammonia at low temperatures diborane gives diborane – diammonate on heating at higher temperature diborane gives borazole.

Question 11.
Why borontrifluoride is act as lewis acid? Explain the react between BF₃ and ammonia?
Answer:
BF₃ is an electron deficient compound and accepts electron pairs to form coordinate covalent bonds. Therefore, borontrifluoride is act as lewis acid. When BF₃ reacts with ammonia to form complex.

Question 12.
How does AlCl₃ reacts with following reagents?

1. H₂O
2. NH₄OH
3. NaOH

Answer:
1. When aluminium chloride reacts with water to form aluminium hydroxide.
AlCl₃ + 3H₂O → Al(OH)₃ + 3HCl

2. With ammonium hydroxide, aluminium chloride forms aluminium hydroxide.
AlCl₃ + 3NH₇OH → Al(OH)₃ + 3NH₄C₁

3. With excess of sodium hydroxide AlCl₃ produces sodium meta aluminate.
AlCl₃ + 4NaOH → NaAlO₂ + 2H₂O + 3NaCl

Question 13.
Give the uses of aluminium chloride.
Answer:

1. Anhydrous aluminium chloride is used as a catalyst in Friedels Crafts reactions.
2. It is used for the manufacture of petrol by cracking the mineral oils.
3. It is used as a catalyst in the manufacture on dyes, drugs and perfumes.

Question 14.
How will you prepare potash alum?
Answer:
The alunite the alum stone is the naturally occurring form and it is K₂SO₄. A₁₂(SO₄)₃. 4Al(OH)₃. When alum stone is treated with excess of sulphuric acid, the aluminium hydroxide is converted to aluminium sulphate. A calculated quality of potassium sulphate is added and the solution is crystallised to generate potash alum. It is purified by recrystallisation.

K₂SO₄. A₁₂(SO₄)₃. 4Al(OH)₃ + 6H₂SO₄ → K₂SO₄ + Al₂(SO ₄)₃ + 12H₂O
K₂SO₄ + A₁₂(SO₄)₂ + 24H₂O → K₂SO₄. A₁₂(SO₄)₃. 24H₂O

Question 15.
Mention the uses of the potash alum.
Answer:

1. It is used for purification of water
2. It is also used for water proofing and textiles
3. It is used in dyeing, paper and leather tanning industries
4. It is employed as a styptic agent to arrest bleeding.

Question 16.
What is burnt alum?
Answer:
When potash alum is heated at 365 K. It melts and form molten state. Again heated at 475 K they lose water of hydration and swells up. Thee swollen mass is known as burnt alum.

Question 17.
What are all the conditions are necessary for catenation?
Answer:
Essential condition for catenation:

1. The valency of elements is greater than or equal to two.
2. Element should have an ability to bond with itself.
3. The self bond must be as strong as its bond with other elements.
4. Kinetic inertness of catenated compound towards other molecules.

Question 18.
Give any three methods to prepare carbon monoxide.
Answer:
1. Carbon monoxide can be prepared by the reaction of carbon with limited amount of oxygen.
2C + O₂ → 2CO

2. On industrial scale carbon monoxide is produced by the reaction of carbon with air.

3. Pure carbon monoxide is prepared by warming methanoic acid with concentrated sulphuric acid which act as a dehydrating agent.
HCOOH + H₂SO₄ → CO + H₂O + H₂SO₄

Question 19.
Mention the uses the carbon monoxide.
Answer:

1. Equimolar mixture of hydrogen and carbon monoxide – water gas and the mixture of carbon monoxide and nitrogen – producer gas are important industrial fuels
2. Carbon monoxide is a good reducing agent and can reduce many metal oxides to metals.
3. Carbon monoxide is an important ligand and forms carbonyl compound with transition metals

Question 20.
Explain the methods to prepare carbon dioxide.
Answer:
1. On industrial scale it is produced by burning coke in excess of air.

2. Calcination of lime produces carbon dioxide as by product.
CaCO₃ \(\underrightarrow { \triangle }\) CaO + CO₂

3. Carbon dioxide is prepared in laboratory by the action of dilute hydrochloric acid on metal carbonates.
CaCO₂ + 2HCl → CaCl₂ + H₂O + CO₂

Question 21.
Give the uses of carbon dioxide.
Answer:

1. Carbon dioxide is used to produce an inert atomosphere for chemical processing.
2. Biologically, it is important for photosynthesis.
3. It is also used as fire extinguisher and as a propellent gas.
4. It is used in the production of carbonated beverages and in the production of foam.

Question 22.
Explain the preparation of silicon tetrachloride.
Answer:
1. Silicon tetrachloride can be prepared by passing dry chlorine over an intimate mixture of silica and carbon by heating to 1675 K in a porcelain tube.
SiO₂ + 2C + 2Cl₂ → SiCl₂ + 2CO

2. On commercial scale, reaction of silicon with hydrogen chloride gas occurs above 600 K.
Si + 4HCl → SiCl₄ + 2H₂

Question 23.
Complete the following reaction,

1. SiCl₄ + H₂O → ?
2. SiCl₄ + C₂HSOH → ?
3. 

Answer:

1. SiCl₄ + 4H₂O → Si(OH)₄ + 4HCl
2. SiCl₄ + C₂H₅OH → Si(OC₂H₅)₄+ 2C₁₂
3. 

Question 24.
Explain the types of silicones.
Answer:

1. Linear silicones – They are obtained by the hydrolysis and subsequent condensation of dialkyl or diaryl silicon chlorides.
   • Silicone rubbers – These silicones are bridged together by methylene or similar groups.
   • Silicone resins – They are obtained by blending silicones with organic resins such as acrylic esters.
2. Cyclic silicones – These are obtained by the hydrolysis of R₂SiCl₂.
3. Cross linked silicones – They are obtained by hydrolysis of RSiClr

Question 25.
Mention the properties of silicones.
Answer:

1. All silicones are water repellent.
2. They are thermal and electrical insulators.
3. Chemically they are inert.
4. Lower silicones are oily liquids whereas higher silicones with long chain structure are waxy solids.
5. The viscosity of silicon oil remains constant and doesn’t change with temperature and they don’t thicken during winter.

Question 26.
What are silicates and mention the types of silicates?
Answer:
Silicates:
The mineral which contains silicon and oxygen in tetrahedral [\({ { [S{ i }{ O }_{ 4 }] } }^{ 4- }\) units linked together in different patterns are called silicates.
types of Silicates:

1. Ortho silicates
2. Pyro silicates
3. Cyclic silicates
4. Ino silicates
5. Phyllo silicates
6. Tecto silicates

Samacheer Kalvi 12th Chemistry p-Block Elements – I 5 Mark Questions and Answers

Question 1.
An element (A) extracted from kernite. A reacts with nitrogen at high temperature gives B. A reacts with alkali to form C. Find out A, B and C. Give the chemical equations.
Answer:
1. An element (A) extracted form kemite is boron.

2. Boron reacts with nitrogen at high temperature gives Boron nitride (B)

3. Boron reacts with alkali (NaOH) to gives sodium borate (C).

Question 2.
Compound A is used in the manufacture of opticals. A on heating gives B. B further heating to form C. A reacts with hydrochloric acid to give D. Identify A, B and C. Explain the reaction.
Answer:
1. Compound (A) is borax, which is used in the manufacture of opticals.

2. Borax (A) on heating to give borax glass (B). Borax glass on further heating to give sodium metaborate (C).

3. Borax (A) reacts with hydrochloride acid to give boric acid.

Question 3.
Complete the reaction.

Answer:
Boric acid when heated at 373 K gives metaboric acid (A) and A heated at 413 K it gives tetraboric acid (B). B heated at red hot it gives boric anhydride (C).

Question 4.
Sodium borohydride reacts with iodine in the presence of diglyme to give A. A heated at 388 K give B. A heated at 373 K in sealed tube to form C. A further heated at red hot condition to give element D. Find out A, B, C and D. Give the reactions.
Answer:
1. Sodium borohydride reacts with iodine in the presence of diglyme to give diborane (A).

2. Diborane (A) heated at 388 K gives pentaborane (11) (B).

3. Diborane (A) heated at 373 K in the sealed tube to form decaborane

(14) (C).Diborane (A) heated at red hot conditions to give element boron (D) Diborane B₂H₆

Question 5.
How does carbonmonoxide reacts with following reagents,
(i) O₂
(ii) Cl₂
(iii) F₂O₃
(iv) H₂
(v) ethene
Answer:

Question 6.
Explain the preparation of silicones.
Answer:
Generally silicones are prepared by the hydrolysis of dialkyldichiorosilanes (R₂SiCl₂) or diaryldichlorosilanes Ar₂SiCl₂, which are prepared by passing vapours of RCl or ArCl over siliconat 570 K with copper as a catalyst.
2RCl + Si \(\underrightarrow { Cu/570k }\) R₂SiCl₂

The hydrolysis of dialkvlchloro silanes R₂SiCl₂ yields to a straight chain polymer which grown from both the sides

The hydrolysis of monoalkylchloro silanes RSiCl₃, yields to a very complex cross linked polymer Linear silicones can be converted into cyclic or ring silicones when water molecules is removed

Question 7.
Discuss the ortho and pyro silicates.
Answer:
Ortho silicates:
The simplest silicates which contain discrete [SiO₄]^4- tetrahedral units are called ortho silicates or neso silicates.
Examples:
Phenacite – Be₂SiO₄(Be²⁺) ions are tetrahedrally surrounded by O_2- ions). Olivine – (Fe/Mg)₂SiO₄( Fe²⁺ and Mg²⁺ cations are octahedrally surrounded by O^2-ions).

Pyro silicates:
Silicates which contain [Si₂O₇]^6- ions are called pyro silicates (or) Soro silicates They are formed by joining two [SiO₄]^4- tetrahedral units by sharing one oxygen atom at one comer, (one oxygen is removed while, joining).
Example:
Thortveitite – SC₂Si₂O₇.

Question 8.
Explain:

1. Cyclic silicates
2. Ino silicates

Answer:
1. Cyclic silicates:
Silicates which contain \(({ Si{ O }_{ 3 }) }_{ n }^{ 2n- }\) ions which are formed by linking three or more tetrahedral \(({ Si{ O }_{ 3 }) }_{ n }^{ 4- }\) units cyclically are called cyclic silicates. Each silicate unit shares two of its oxygen atoms with other units.
Example:
Beryl [Be₃Al₂(SiO₃)₆] (an aluminosilicate with each aluminium is surrounded by 6 Structure of Cyclic silicates

2. Ino silicones:
Silicates which contain V number of silicate units liked by sharing two or more oxygen atoms are called inosilicates. They are further classified as chain silicates and double chain silicates. Chain silicates (or pyroxenes) – These silicates contain \(({ Si{ O }_{ 3 }) }_{ n }^{ 2n- }\) ions formed by linking ‘n’ number of tetrahedral \(({ Si{ O }_{ 3 }) }_{ n }^{ 4- }\) units linearly. Each silicate unit shares two of its oxygen atoms with other Structure of Chain silicates units.

Example:
Spodumene – LiAl(SiO₃)₂ Double chain silicates (or amphiboles) – These silicates contains \([{ S{ i }_{ 4 }{ O }_{ 11 }] }_{ n }^{ 6n- }\) ions. In these silicates there are two different types of

3. Tetrahedra:

1. Those sharing 3 vertices
2. Those sharing only 2 vertices.

Examples.
Asbestos – These are fibrous and noncombustible silicates. Therefore they are used for thermal insulation material, brake linings, construction material and filters. Asbestos being carcinogenic silicates, their applications are restricted.

Question 9.
Write a notes on

1. Sheet silicates
2. Three dimensional silicate

Answer:
1. Sheet silicates:
Silicates which contain \(({ S{ i }_{ 2 }{ O }_{ 5 }) }_{ n }^{ 2n- }\) are called sheet or phyllo silicates. In these, Each \({ { [S{ i }{ O }_{ 4 }] } }^{ 4- }\) tetrahedron unit shares three oxygen atoms with others and thus by forming twodimensional sheets. These sheets, silicates form layered structures in which silicate sheets are stacked over each other. The attractive forces between these layers are very week, hence they can be cleaved easily just like graphite.

Example: Talc, Mica etc…

2. Three dimensional silicate:
Silicates in which all the oxygen atoms of \({ { [S{ i }{ O }_{ 4 }] } }^{ 4- }\) tetrahedra are shared with other tetrahedra to form three-dimensional network are called three dimensional or tecto silicates. They have general formula (SiO₂)_n.
Examples:
Quartz. These tecto silicates can be converted into Three dimentional aluminosilicates by replacing [\({ { [S{ i }{ O }_{ 4 }] } }^{ 4- }\) – units by [AlO₄]^5- units. E.g. Feldspar, Zeolites etc.,

Question 10.
Explain – Boron neutron capture therapy.
Answer:
1. The affinity of boron-10 for neutrons is the basis of a technique known as born neutron capture therapy (BNCT) for treating patients suffering from brain tumours.

2. It is based on the nuclear reaction that occurs when boron-10 is irradiated with low- energy thermal neutoms to give high linear energy a-particles and a Li-particles.

3. Boron compounds are injected into a patient with a brain tumour and the compounds collect preferentially in the tumour. The tumour area is then irradiated with thermal neutom and results in the release of an alpha-particle that damages the tissue in the tumour each time a boron-10 nucleus captures a neutron.

3. In this way damages can be limited preferentially to the tumour, leaving the normal brain tissue less affected.

4. BNCT has also been studied as a treatment for several other tumours of the head and neck, The breast, the prostate, the bladder and the liver.

Commn Errors and its Rectifications
Common Errors:

1. Ores formula may be difficult to remember.
2. Allotropes of elements may confuse the students in writing equation.

Rectifications:

1. Only chief ore from which the element is extracted can be easy to remember.
2. Crystalline forms are more chemically reactive.

Hope you love the Samacheer Kalvi 12th Chemistry Chapter Wise Material. Clearly understand the deep concept of Chemistry learning with the help of Tamilnadu State Board 12th Chemistry Chapter 2 p-Block Elements – I Questions and AnswersPDF. Refer your friends to and bookmark our website for instant updates. Also, keep in touch with us using the comment section.

Students can Download English Poem 4 Ulysses Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 12th English Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th English Solutions Poem Chapter 4 Ulysses

12th English Poem Ulysses Warm Up

Introduction:
The poem ‘Ulysses’ is a dramatic monologue that contains 70 lines of blank verse. Ulysses, the king of Ithaca, gathers his men together to prepare for the journey and exhorts them not to waste their time left on earth. Ulysses has grown old, having experienced many adventures at the battle of Troy and in the seas. After returning to Ithaca, he desires to embark upon his next voyage. His inquisitive spirit is always looking forward to more and more of such adventures.

The poem can be divided into three parts:

1. the thirst for adventure, which does not allow Ulysses to remain in his kingdom as a mere ruler
2. Ulysses handing over the responsibility to his son Telemachus, with total confidence in his abilities
3. Ulysses’ clarion call to his sailors, urging them to venture into unknown lands.

Samacheer Kalvi 12th Ulysses English Textual Questions

1. Complete the summary of the poem, choosing words from the list given below. Lines 1 to 32

It little profits that an idle king,
By this still hearth, among these barren crags,
Match’d with an aged wife, I mete and dole
Unequal laws unto a savage race,

That hoard, and sleep, and feed, and know not me.
I cannot rest from travel: I will drink
Life to the lees: All times I have enjoy’d
Greatly, have suffer’d greatly, both with those

That loved me, and alone, on shore, and when
Thro’ scudding drifts the rainy Hyades
Vext the dim sea: I am become a name;
For always roaming with a hungry heart

Much have I seen and known; cities of men
And manners, climates, councils, governments,
Myself not least, but honour’d of them all;
And drunk delight of battle with my peers,

Far on the ringing plains of windy Troy.
I am a part of all that I have met;
Yet all experience is an arch wherethro’
Gleams that untravell’d world whose margin fades

For ever and forever when I move.
How dull it is to pause, to make an end,
To rust unburnish’d, not to shine in use!
As tho’ to breathe were life! Life piled on life

Were all too little, and of one to me
Little remains: but every hour is saved
From that eternal silence, something more,
A bringer of new things; and vile it were

For some three suns to store and hoard myself,
And this gray spirit yearning in desire
To follow knowledge like a sinking star,
Beyond the utmost bound of human thought.

This is my son, mine own Telemachus,
To whom I leave the sceptre and the isle, –
Well-loved of me, discerning to fulfil
This labour, by slow prudence to make mild

A rugged people, and thro’ soft degrees
Subdue them to the useful and the good.
Most blameless is he, centred in the sphere
Of common duties, decent not to fail

offices of tenderness, and pay
Meet adoration to my household gods,
When I am gone. He works his work, I mine.
There lies the port; the vessel puffs her sail:

There gloom the dark, broad seas. My mariners,
Souls that have toil’d, and wrought, and thought with
me That ever with a frolic welcome took
The thunder and the sunshine, and opposed
Free hearts, free foreheads – you and I are old;

Old age hath yet his honour and his toil;
Death closes all: but something ere the end,
Some work of noble note, may yet be done,
Not unbecoming men that strove with Gods.

The lights begin to twinkle from the rocks:
The long day wanes: the slow moon climbs: the deep
Moans round with many voices. Come, my friends,
‘T is not too late to seek a newer world.

Push off, and sitting well in order smite
The sounding furrows; for my purpose holds
To sail beyond the sunset, and the baths
Of all the western stars, until I die.

It may be that the gulfs will wash us down:
It may be we shall touch the Happy Isles,
And see the great Achilles, whom we knew.
Tho’ much is taken, much abides; and tho’

We are not now that strength which in old days
Moved earth and heaven, that which we are, we are;
One equal temper of heroic hearts,
Made weak by time and fate, but strong in will
To strive, to seek, to find, and not to yield.

fullest	unquenchable	unattainable
experience	knowledge	king
matters	rust	adventure
unwilling	travel	breathing

Ulysses is (1) ______ to discharge his duties as a (2) ______ , as he longs for (3) ______ He is filled with an (4) ______ thirst for (5) ______ and wishes to live life to the (6) ______ He has travelled far and wide gaining (7) ______ of various places, cultures, men and (8) ______ He recalls with delight his experience at the battle of Troy. Enriched by his (9) ______ he longs for more and his quest seems endless. Like metal which would (10) ______ if unused, life without adventure is meaningless. According to him living is not merely (11) ______ to stay alive. Though old but zestful, Ulysses looks at every hour as a bringer of new things and yearns to follow knowledge even if it is (12) ______
Answers:

1. unwilling
2. king
3. adventure
4. unquenchable
5. travel
6. fullest
7. experience
8. matters
9. knowledge
10. rust
11. breathing
12. unattainable

Lines 33 to 42

prudence

kingdom

quest

tender

Ulysses desires to hand over his (1) _____ to his son Telemachus, who would fulfil his duties towards his subjects with care and (2) _____ Telemachus possesses patience and has the will to civilise the citizens of Ithaca in a (3) _____ way. Ulysses is happy that his son would do his work blamelessly and he would pursue his (4) _____ for travel and knowledge.
Answer:

1. kingdom
2. prudence
3. tender
4. quest

Lines 44 to 70

world	thunder	meaningful
gather	undaunted	heaven

Ulysses beckons his sailors to (1) _____ at the port where the ship is ready to sail. His companions who have faced both (2) _____ and sunshine with a smile, are united by their undying spirit of adventure. Though death would end everything, Ulysses urges his companions to join him and sail beyond the sunset and seek a newer (3) _____ , regardless of consequences. These brave hearts who had once moved (4) _____ and earth, may have grown old and weak . physically but their spirit is young and (5) _____ His call is an inspiration for all those who seek true knowledge and strive to lead (6) _____ lives.
Answers:

1. gather
2. thunder
3. world
4. heaven
5. undaunted
6. meaningful

2. Answer the following questions in one or two sentences each.

Ulysses Poem Questions And Answers Question (а)
‘Ulysses is not happy to perform his duties as a king.’ Why?
Answer:
Ulysses is not happy to perform the ordinary duties of a king mainly because his heart is in voyages beyond horizon. He is bored with the task of enforcing law and order and giving reward and punishment to a savage race.

Ulysses Questions And Answers Question (b)
What does he think of the people of his kingdom?
Answer:
The citizen of Ithaca hoard, sleep and feed. They don’t understand the aspirations of the dreamer Ulysses.

12th English Poem Ulysses Paragraph Question (c)
What has Ulysses gained from his travel experiences?
Answer:
Ulysses has met people hailing from different cultural backgrounds. He has learnt much from their manners, climates, councils and governments. He learnt strategies of warfare in battles.

Ulysses Poem Erc Question (d)
Pick out the lines which convey that his quest for travel is unending.
Answer:
“How dull it is to pause, to make an end,
To rust unburnished, not to shine in use!
The lines quoted above convey his quest for travel is unending.

12th English Ulysses Paragraph Question (e)
‘As tho’ to breathe were life!’ – From the given line what do you understand of Ulysses’ attitude to life?
Answer:
Ulysses strongly believes that just breathing is not life. Life has to be adventurous and full of action.

Ulysses Poem Questions And Answers Pdf Question (J)
What does Ulysses yearn for?
Answer:
Ulysses yearns for following knowledge like a sinking star beyond the utmost bound of human thought.

Ulysses Question Answer Question (g)
Who does the speaker address in the second part?
Answer:
The speaker addresses the readers in the second part explaining the difference between his roles and that of Telemachus.

Ulysses Poem Short Questions And Answers Question (h)
Why did Ulysses want to hand over the kingdom to his son?
Answer:
Ulysses finds Telemachus discerning and prudent. Besides, Ulysses is wearing the crown uneasily as the call for adventure and desire to sail beyond sunset is obsessing his mind. So, he wants to hand over the kingdom to Telemachus.

Ulysses Short Questions And Answers Question (i)
How would Telemachus transform the subjects?
Answer:
Ulysses believes that his son Telemachus is wise and kind enough to transform rugged citizens into mild and civilized subjects by his tenderness and love.

Question (j)
‘He works his work, I mine’ – How is the work distinguished?
Answer:
Telemachus would do the work of ruling Ithaca with prudence and tenderness. Ulysses will pursue his dream of adventure and try to meet great Achilles in the other world.

Question (i)
In what ways were Ulysses and his mariners alike?
Answer:
Both Ulysses and his fellow sailors are now old. They no more have the strength they possessed in olden days moving earth and heaven. They are made weak by time and fate but strong in will “to strive, to seek, to find and not to yield.” They share the heroic temper and undying quest for knowledge and adventure.

Question (j)
What could be the possible outcomes of their travel?
Answer:
The sailors and Ulysses may be washed down by the gulfs or they could touch Greek paradise and meet their hero Achilles. They may die happily braving the elements of nature.

3. Identify the figures of speech employed in the following lines.

Question (a)
“Thro” scudding drifts the rainy Hyades
Vext the dim sea…”
Answer:
The figure of speech Personification is employed in the above lines.

Question (b)
“For always roaming with a hungry heart”
Answer:
Metaphor

Question (c)
“And drunk delight of battle with my peers;”
Answer:
Metaphor

Question (d)
“…..the deep
Moans round with many voices.”
Answer:
Personification

Question (e)
“To follow knowledge like a sinking star.”
Answer:
Simile

Question (f)
“ There lies the port the vessel puffs her sai”
Answer:
Personification

Additional Questions

Question (a)
“I will drink life to the lees'”
Answer:
Metaphor

Question (b)
“Vext the dim sea:”
Answer:
Personification

Question (c)
“Yet all experience is an arch wherethro”
Answer:
Metaphor

Question (d)
“Gleams that untravell’d world whose margin fades”
Answer:
Metaphor

Question (e)
“To rust unburnish’d, not to shine in use!”
Answer:
Metaphor

Question (f)
“There gloom the dark, broad seas. My mariners,”
Answer:
Metaphor

Question (g)
“Souls that have toil’d, and wrought, and thought with me”
Answer:
Synecdoche (part of the whole)

Question (h)
“The thunder and the sunshine, and opposed”
Answer:
Metaphor

Question (i)
“T is not too late to seek a newer world.”
Answer:
Synecdoche

Question (j)
“…in order smite The sounding furrows;”
Answer:
Metaphor

Question (k)
“To sail beyond the sunset, and the baths”
Answer:
Metaphor

Question (l)
“It may be we shall touch the Happy Isles,”
Answer:
Allusion (in Greek mythology the place is known as Greek paradise)

Question (m)
“And see the great Achilles, whom we knew.”
Answer:
Allusion (Greek mythology)

Appreciate The Poem

4. Read the sets of lines from the poem and answer the questions that follow.

(a) “…I mete and dole
Unequal laws unto a savage race,
That hoard, and sleep, and feed, and know not me.”

Question (i)
What does Ulysses do?
Answer:
Ulysses, like a grocery shop owner, measures and delivers rewards and punishments unto a large number of uncivilized citizens.

Question (ii)
Did he enjoy what he was doing? Give reasons.
Answer:
No, he did not enjoy is work. He does not like the idea of ministering variable justice to people who like “drones” or animals just eat, sleep and multiply their kind. He wants to leave such work to his son.

Question (b)
“Yet all experience is an arch wherethrough
Gleams that untravelled world, whose margin fades
For ever and for ever when I move. ”

Question (i)
What is experience compared to?
Answer:
Experience is compared to an arch.

Question (ii)
How do the lines convey that the experience is endless?
Answer:
Through the arch of experience one can see the untravelled world. But the experience in the untravelled has a margin whose border fades as one moves forward. Thus experience is endless.

Question (c)
“Little remains: but every hour is saved
From that eternal silence, something more,
A bringer of new things; and vile it were”

Question (i)
How is every hour important to Ulysses?
Answer:
One lives in this world for a limited time. Every hour can provide new knowledge. So, every hour is very important.

Question (ii)
What does the term ‘Little remains’ convey?
Answer:
Ulysses realizes that he has become old. He has not much time left. He doesn’t want to die resting in his kingdom. He states that his remaining lifetime is very limited.

Question (d)
“This is my son, mine own Telemachus,
To whom I leave the sceptre and the isle
Well-loved of me,”

Question (i)
Who does Ulysses entrust his kingdom to, in his absence?
Answer:
Ulysses entrusts his kingdom to his beloved son Telemachus in his absence.

Question (ii)
Bring out the significance of the ‘sceptre’.
Answer:
Sceptre is an ornamental staff carried by a King on ceremonial occasions as a symbol of sovereignity. It symbolizes the power of a king.

(e) ‘‘That ever with a frolic welcome took
The thunder and the sunshine, and opposed”

Question (i)
What do ‘thunder’ and ‘sunshine’ refer to?
Answer:
Thunder and sunshine refers to misfortunes and happy days. Ulysses and his comrades had undergone both kinds of experiences.

Question (ii)
What do we infer about the attitude of the sailors?
Answer:
The sailors shared the undying quest for exploration, adventure and for seeking newer knowledge in the untravelled world. They even welcomed dangers in fighting with Gods. They enjoyed the thrill of action and never worried about the outcome of battles or quests. They have equal temper of heroic hearts.

(f) ‘‘Death closes all: but something ere the end,
Some work of noble note, may yet be done,
Not unbecoming men that strove with Gods.”

Question (i)
The above lines convey the undying spirit of Ulysses. Explain.
Answer:
Ulysses is aware of ageing and substantial decrease in his physical strength. He knows that will close in on him sooner or later. But before that happens, he wants to sail beyond the sunset/horizon and if possible meet warriors like Achilles. He wants to achieve something worthy of those who challenged and fought with Gods. Thus these lines show the undying spirit of Ulysses.

Question (ii)
Pick out the words in alliteration in the above lines,
Answer:
ere, end, noble, note are the words that alliterate.

(g) “……… for my purpose holds
To sail beyond the sunset, and the baths Of all the western stars, until I die.”

Question (i)
What was Ulysses’ purpose in life?
Answer:
Ulysses proposes to sail beyond sunset and baths. His goal is not death but is in death. He seeks life in death. Ordinary mortals can’t reach ‘Happy isles’ or baths while they are alive.Ulysses wants to find direct evidence of spiritual reality after death. He wants to venture into the unknown.

Question (ii)
How long would his venture last?
Answer:
His venture would last until he confronts his death.

Question (h)
“One equal temper of heroic hearts,
Made weak by time and fate, but strong in will
To strive, to seek, to find, and not to yield. ”

Question (i)
Though made weak by time and fate, the hearts are heroic. Explain.
Answer:
Ulysses and his compatriots have visited many strange places in their previous voyages and enjoyed misfortunes and glorious triumphs with the same heroic temperament. They might have become old and may not have the same strength they had in youth. But they still share the thirst for travel and pursuit of knowledge in the unexplored world. Their bravery and spiritual strength are intact.

Question (ii)
Pick out the words in alliteration in the above lines.
Answer:
Strive, seek, heroic, hearts are the words that alliterate.

5. Explain with reference to the context the following lines.

Question (a)
“I cannot rest from travel: I will drink
Life to the lees:”
Answer:
Reference: These lines are from the poem ‘Ulysses” written by Alfred Tennyson.
Context and Explanation: Ulysses, after spending many years in the seas returns to Ithaca and starts ruling his country. But his heart is not in the administration of his kingdom. He wants to sail again. At this context he says these words. He wishes to enjoy life to the fullest and so he can’t afford to idle away his remaining life as a king.

Question (b)
“I am become a name;
For always roaming with a hungry heart”
Answer:
Reference: These lines are from the poem ‘Ulysses” written by Alfred Tennyson.
Context and Explanation: Ulysses says these words while discussing the reputation he has earned among the common multitude due to his daring adventures. He has roamed the world like a hungry lion. This line has a biblical alusion as well “Blessed are they which do hunger and thirst after righteousness” – Matthew V. 6.

Though Ulysses is aware of his fame, it doesn’t motivate him to stay or settle down in the kingdom of Ithaca. His inquisitive spirit is always looking for newer knowledge through ‘the arch’ to the untravelled world.

Question (c)
“How dull it is to pause, to make an end,
To rust unburnished, not to shine in use!”
Answer:
Reference: These lines are from the poem ‘Ulysses” written by Alfred Tennyson.
Context and Explanation: The poet says these words while discussing the mental agony of Ulysses who is unable to settle down with his ageing wife Penelope and son Telemachus. Ulysses finds doling out justice to a savage people as ‘boring’. He does not want to settle down and die in Ithaca. He compares himself to a sword which may rust if left unused. He wants to lead an active and adventurous life till his death.

Question (d)
“To follow knowledge like a sinking star,
Beyond the utmost bound of human thought”
Answer:
Reference: These lines are from the poem ‘Ulysses” written by Alfred Tennyson.
Context and Explanation: The poet says these words while describing the quest Ulysses has for adventure and fulfillment. Similar to a sinking star, Ulysses wants to pursue in his failing old age to pursue knowledge like the goal of Goethe’s Faust, his quest is defined by the pursuit of new and unique knowledge “beyond the utmost bound of human thought”.

Question (e)
“He works his work, I mine.’’’’
Answer:
Reference: These lines are from the poem ‘Ulysses” written by Alfred Tennyson.
Context and Explanation: The poet says these words while justifying the decision of Ulysses to pass on his kingdom to Telemachus. Ulysses explains the polar difference between himself and his son Telemachus. His son will be a ‘fair’ and ‘decent’ ruler. Unlike Ulysses, Telemachus is rooted in regular political life. He enjoys leading “savage” population and the responsibility of showing the subject better moral codes of conduct and upholding justice. Whereas Ulysses finds this “slow” and intolerable. So, he wishes his son to rule Ithaca and for himself he wishes to set sail to the unknown.

Question (f)
“….you and I are old;
Old age hath yet his honour and his toil;”
Answer:
Reference: These lines are from the poem ‘Ulysses” written by Alfred Tennyson.
Context and Explanation: Poet, in this part of the monologue describes the address made by Ulysses to his compatriots who were with him during “thunders and sunshine”. He admits the fact that they are growing old. But he does not want to retire like ordinary mortals. He accepts gracefully the honour befitting old age as a result of varied cultural experiences. Yet, he does not want old people to bow out of the field of action. He sincerely believes there is more work to be done, lands to be explored and newer knowledge to be acquired in old age before death.

Question (g)
“The long day wanes: the slow moon climbs: the deep
Moans round with many voices.”
Answer:
Reference: These lines are from the poem ‘Ulysses” written by Alfred Tennyson.
Context and Explanation: The poet poet continues to discuss old age and the tantalising call of the oceans to conquer. Ulysses hints at the probable end of the cycle of life in the words “The long clay wanes”. The symbol of darkness or night is mostly associated with death. The lure of ocean to resume his voyages beyond the point of sunset is too tempting to resist. The dark unfathomable sea beckons him and his compatriots with mysterious voices.

Question (h)
“It may be we shall touch the Happy Isles,
And see the great Achilles, whom we knew.”
Answer:
Reference: These lines are from the poem ‘Ulysses” written by Alfred Tennyson.
Context and Explanation: The poet says these words through Ulysses as to the probable outcome of the daring voyage at the fag end of his life. Ulysses is uncertain about the probable outcome of his last voyage. But he infers that he might reach ‘Happy Isles’ and see the ‘great Achilles’ who was dipped in the river of life.

Question (i)
“We are not now that strength which in old days
Moved earth and heaven;”
Answer:
Reference: These lines are from the poem ‘Ulysses” written by Alfred Tennyson.
Context and Explanation: The poet says these words through Ulysses when he wants to justify the reasons for resuming the daring voyage. He admits the decline in the compatriots’ physical strength with which they were able to move heaven and earth in their youth. He asks his compatriots to ignore the infinity of age and draw on their inner spiritual strength to resume their voyage beyond sunset.

Question (j)
“To strive, to seek, to find, and not to yield.”
Answer:
Reference: These lines are from the poem ‘Ulysses” written by Alfred Tennyson.
Context and Explanation: Tennyson says these words through Ulysses who makes his motto loud and clear in these words. The final line of the poem is Ulysses’ enduring challenge to readers as well. The challenge if the aged ones could push ahead with vigour and strength of will no matter how fragile their bodies are. To yield to age or weakness is to be less than fully human. It might be honourable to live a peaceful settled life in the old age. But one would naturally miss out most exciting moments of life if one does not venture out, at least a little towards the unknown.

6. Answer the following questions in a paragraph of about 100 words each.

Question (a)
What makes Ulysses seek newer adventures?
Answer:
In the context of the poem, Ulysses has grown old. He has experienced all daring adventures. He has won the hearts of people during the battle at Troy. Back home, as per the prophecy of Tiresias, he rules Ithaca for a brief time. But he is fed up with a conventional duties of a king. He laments his own uselessness as a ruler of idle people who lead life like savages, just eating and sleeping. They don’t understand the over vaulting ambition of their adventurous king Ulysses who had moved earth and heavens in the past. He wishes to embark upon his next voyage. It might be his last. He is quite sensitive to the moans of the seas tantalising him and his compatriots to set sail quickly. He wants “to drink life to the lees”. Ulysses doesn’t want to bask on the glory he has earned in the past.

His inquisitive spirit is restless. He has seen much’ and acquired knowledge of various cultures of the world. But he considers all such experiences like an “arch” leading him to the unexplored or “untravelled world”. He wants to sail towards the area ‘beyond sunset’. He must shine in use like a sword but not “rust unbumished”. Yet at home, in the kingdom of Ithaca, he feels bored and yearns to truly engage with what is left of life. He is impatient for “new” experiences lamenting everyday and every hour to seek “something more”. His quest for adventure and fulfillment, like the goal of Goethe’s Faust is defined by the pursuit of new knowledge “beyond the utmost bound of human thought”.

Question (b)
List the roles and responsibilities Ulysses assigns to his son Telemachus, while he is away.
Answer:
The entire poem is a monologue. Yet the second part of the poem is an address to the readers justifying his decision to transfer the rule to his son Telemachus. The cloak of a king seems to be unfit for the temperament of Uly sses. He finds ruling Ithaca a boring thing. He finds Telemachus rooted to the political life of Ithaca. His role is merely to lead a ‘savage race’ to accept standard norms of behaviour in the society. He believes Telemachus fits well with the role of the ruler of “uninspired and imprudent citizens” and may discharge his duties with honour and grace. When he is away, he wants his son Telemachus to dispense variable justice to the subjects of Ithaca and guide them in the path of virtues and morals.

Question (c)
What is Ulysses’ clarion call to his sailors? How does he inspire them?
Answer:
In the third part of the poem, Ulysses makes a clarion call to his hearty compatriots (i.e.) mariners. They have been with him both during ‘thick and thin’ or thunders or sunshine. Similar to Ulysses they possess “free hearts and free foreheads” (i.e.) their hearts and brains are unburdened by domestic cares and resposibilities. They had frolicsome time fighting along with Ulysses against great warriors and Gods in the past. Ulysses does not want to live in memory of glory. He believes they need not waste away their precious time in nostalgic memories just recounting their escapades to younger generation. They can really do ‘ something of noble note’ before the end. He is conscious of the impending death in old age. But he tells it is not “too late to seek a newer world”.

The many “voices of the ocean” call out to the mariners to resume voyage. Ulysses is not content with having earned a name for himself. He has seen many countries and acquired knowledge of various cultures. Those experiences are not to be taken as accomplishments. They are just an ‘arch’leading them to “untravelled world” and constantly sailing to the ever expanding horizon. He does not want his compatriots to miss even an hour which could prrovide them novel experiences in their voyage. He persuades his compatriots to gather at the port as the sails are already puffing up welcoming them all. Their life would be one of fulfillment only when they venture out into the unkown on the seas. He uses an emotional bait to his mariners. He highlights the probable outcome of their voyage. They might reach the “Happy Isles” (i.e.) great paradise and meet Achilles, their war hero. No matter how much strength they have, they still have some “strength of will” left to strive, to seek, to find, and not to yield.

Listening Activity

Listen to the poem and fill in the blanks with appropriate words and phrases. If required listen to the poem again.

Wander-thirst

BEYOND the East the sunrise, beyond the West the sea, And East and West the wander-thirst that will not let me be; It works in me like madness, dear, to bid me say good-bye; For the seas call, and the stars call, and oh! the call of the sky!

I know not where the white road runs, nor what the blue hills are; But a man can have the sun for a friend, and for his guide a star; And there’s no end of voyaging when once the voice is heard, For the rivers call, and the roads call, and oh! the call of the bird!

Yonder the long horizon lies, and there by night and day The old ships draw to home again, the young ships sail away; And come I may, but go I must, and, if men ask you why, You may put the blame on the stars and the sun and the white road and the sky.

Choose the best option and complete the sentences:

Question 1
works like madness in the poet.
(a) Wander – Thirst
(b) Bidding Farewell
(c) Eastern Sunrise
(d) Western Seas
Answer:
(a) Wander – Thirst

Question 2.
A man could choose as his guide.
(a) the sun
(b) the hills
(c) a star
(d) a bird
Answer:
(c) a star

Question 3.
There is no end of once the voice is heard.
(a) walking
(b) roaming
(c) talking
(d) voyaging
Answer:
(d) voyaging

Question 4.
The old ships return, while the young ships
(a) drift
(b) move
(c) sail
(d) wander
Answer:
(c) sail

Question 5.
The blame is on the sun, stars, the road and the
(a) hills
(b) trees
(c) seas
(d) sky
Answer:
(d) sky

Ulysses About The Poet

Alfred Tennyson was the poet laureate of England and Ireland during Queen Victoria ‘s reign. Tennyson excelled in writing short lyrics such as “Break, Break, Break“, “ Tears Idle Tears“, “The Charge of the Light Brigade” and.‘Crossing the Bar’. Tennyson’s use of the musical qualities of words to emphasise his rhythms and meanings is sensitive.

Ulysses Summary in English

Introduction
‘Ulysses’ is a dramatic monologue in which Ulysses, the king of Ithaca expresses his undying thirst of adventure overseas. Tennyson has written this poem in memory of Arthur Henry Hallam who died young. Death is not the end for both Arthur and Ulysses.

Thirst for adventure
Ulysses addresses mariners in the third part. Ulysses does not see any worth in staying in the comfort of family life with his wife. As a king, he listens to the complaints of people and gives rewards and punishments for the citizens of his country. He discloses his inner nature, “cannot rest from travel!” He has become a living icon of travel overseas. He wants “to drink life to the lees”. He confesses that he has both enjoyed and suffered a lot during his travels. He has gained profound wisdom during his travels and battles. Enriched by his new gained cultural knowledge, he longs to resume his voyages. He believes that to rest is rust. Every hour has the •potential to bring new knowledge. So, it is meaningless to stay.

Passing on the inheritance
In the second part addresses the reader explaining why Ulysses doesn’t want to continue to rule. Ulysses wishes to hand over his kingdom to his son Telemachus. He believes he would rule the kingdom and render appropriate justice to his subjects. Telemachus, unlike Ulysses, is rooted to the soil. He is kind to the subjects and addresses their needs. Besides, Ulysses hopes that Penelope, his aging wife would be happy to spend her last years with her son and grandchildren. So, Ulysses can resume his voyage along with his old friends.

Call to set sail
Ulysses called his old companions to the port where the ship awaits them all. His old companions have fought many battles alongside him and share the undying quest for adventure. Ulysses persuades his friends to join him on his voyage to the edge of the world and beyond to find a new world. He knows the limitations of his companions. They may have grown old and weak due to age. Their physical powers may not be the same as when they had once moved heaven and earth. But he is confident that their spirit is young and undaunted. His clarion call would inspire invariably all those who seek knowledge and strive to lead meaningful lives.

Conclusion
People who are endowed with an unquenchable thirst for ever-expanding knowledge and incurable love for long distances Ulysses continues to be a source of inspiration.

Ulysses Summary in Tamil

முடிவுரை:
தன் அறிவு வேட்கையை அதிகப்படுத்துவதில் | ஆர்வம் கொண்டவர்களுக்கும் மற்றும் மாறாத தொலைதூரப் பயணத்தில் காதல் கொண்டவர்களுக்கும் | யுலிசஸ் எப்போதும் தூண்டுதலாக இருக்கிறார்.

சாகசத்தின் மேல் ஆவல்
யுலிசஸ் மூன்றாம் பாகத்தில் கப்பலோட்டிகளிடத்தில் சொற்பொழிவாற்றுகிறார். தன் குடும்பத்தாருடன் சொகுசு வாழ்க்கை வாழ்வதில் எந்தப் பயனும் இல்லை என எண்ணுகிறார். ஒரு அரசனாக குடிமக்களின் கூற்றுகளைக் கேட்டு அதற்கு ஏற்ற பரிசையும் மற்றும் தண்டனையும் வழங்குகிறார். கடற் பயணம் செய்யாமல் இருக்க முடியாது என்ற தன்னுடைய உள்ளுணர்வை வெளிப்படுத்துகிறார். கடல் கடந்த பயணங்களை மேற்கொள்ளும் உயிரோவியமாகத் திகழ்ந்தார். வாழ்க்கை முழுவதும் பயணிக்க விரும்புகிறார். அவர் பயணத்தில் மேற்கொண்ட இன்ப துன்பங்களை ஒப்புக் கொள்கிறார். அவர் பயணத்தின் போது ஏற்பட்ட அனுபவங்களாலும் மற்றும் எதிர்கொண்ட போர்களினாலும் ஆழ்ந்த ஞானத்தை அடைந்திருந்தார். புதிதாக பெறப்பெற்ற பண்புகளின் புலமையால், அவர் பயணங்களை மேற்கொள்வதில் ஆர்வத்தைக் காட்டினார். சோம்பலாக கிடந்தால் உடல் துருப்பிடித்து விடும் என எண்ணினார். ஒவ்வொரு கணமும் புதுப் பொலிவைக் கொண்டு வரும் ஆற்றல் மிக்கது. ஆதலால், ஓரிடத்தில் ஒண்டிக் கிடப்பதில் பயனில்லை என எண்ணினார்.

பரம்பரை சொத்துக்களை கைமாற்றம் செய்தல்:
இரண்டாம் பாக சொற்பொழிவில் அவருக்கு அரசாள்வதில் ஏன் விருப்பம் இல்லை என்பதைக் குறித்துக் கூறுகிறார். யுலிசஸ் தன் இராஜ்ஜியத்தை தன் புதல்வன் டெலிமேக்கஸ் (Tele Macus) இடம் கைமாற்றுகிறார். அவர் தன் புதல்வன் நல்ல முறையில் அரசாண்டு தன் குடிமக்களுக்கு தக்க நீதி வழங்குவான் என எண்ணினார். டெலிமாக்கஸ், யுலிசஸ் போல் அல்லாமல் ஒரே இடத்தில் ஊன்றி நிற்பவராய் இருந்தார். அவர்தம் குடிமக்களுக்கு அன்பார்ந்தவராகவும் அவர்கள் குறைகளைக் கேட்டறிபவராகவும் இருந்தார். அது தவிர யுலிசஸ் தன் வயது முதிர்ந்த மனைவி பெனைலோப்பிம், மகனுடனும் பேரப் பிள்ளைகளுடனும் மகிழ்ச்சியோடு இருப்பார் என நம்பினார். ஏனெனில், அது அவர் தன் தோழர்களுடன் கடல் பயணம் மேற்கொள்ளத் தக்கவாறு அமையும்.

கடல் பயணம் மேற்கொள்ளல்:
யுலிசஸ் தன் நண்பர்களைத் தமக்காக காத்துக் கொண்டிருக்கும் துறைமுகத்திற்கு வருமாறு அழைப்பு விடுகிறார். அவர் தோழர்களும் கடல் பயணத்தின் போது பல கடற்போர்களில் ஈடுபட்டு தனக்கு உள்ள கடல் பயணத்தின் மேல் உள்ள ஆவலைப் பகிர்ந்து கொண்டனர். உலகின் விளிம்பை அடைந்து அதற்கு அப்பாலும் செல்வதற்கு தன்னுடன் கடல் பயணம் மேற்கொள்ள நண்பர்களைத் தூண்டுகிறார். அவர் தம் நண்பர்களின் குறைகளை அறிந்திருந்தார். அவர்கள் வயது முதிர்ச்சியால் பலவீனமாகக் காணப்பட்டனர். அவர்கள், முன்னர் வானத்தையும், பூமியையும் அளந்தது போல் அவர்கள் உடல் வலிமை தற்போது இல்லை. ஆனால், அவர்கள் ஊக்கம் குன்றாமல், சளைக்கா வண்ணம் தொடரவேண்டும் என்பதில் உறுதியாக இருந்தார். அவரின் வீர முழக்கம் புதிய அனுபவத்தை நாடி மற்றும் அர்த்தமுள்ள வாழ்க்கையை வாழ முயற்சிக்கும் அனைவருக்கும் தூண்டுதலாக இருக்கும்.

முன்னுரை:
யூலிஸ் என்பது வியத்தகு மோனோலாலஜி ஆகும். இதில் இதாகா மன்னனான யுலிசஸ் வெளிநாடுகளில் சாகசத் தன்மையற்ற தாகத்தை வெளிப்படுத்துகிறார். டென்னிசன் இளம் கவிஞரான ஆர்தர் ஹென்றி ஹாலமின் நினைவாக இந்தக் கவிதை எழுதினார். ஆர்தர் மற்றும் யுலிசஸ் இருவருக்கும் முடிவே இல்லை.

Ulysses Glossary

Textual:

Additional:

Students can Download English Poem 3 All the World’s a Stage Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 12th English Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th English Solutions Poem Chapter 3 All the World’s a Stage

Warm Up

This is Life Cycle of butterfly.

All The World’s A Stage Poem Questions And Answers Question 1.
Discuss with your partner the different stages in the grow th of man from a new born to an adult.
Answer:
An infant pukes on the mother’s arms. As he is unable to articulate his needs, he keeps on crying like a kitten. Then he goes to school giving up his freedom. He is made to learn things he doesn’t want to learn. Then he becomes an adult hopelessly in love. He wastes his purple youth writing love letters or songs admiring the beauty of his love. Some join army or police force to serve the nation. At the peak of adulthood, they are quite touchy about honour and believe it to be more important than life itself.

Samacheer Kalvi 12th English All the World’s a Stage Textual Questions

1. Fill in the blanks using the words given in the box to complete the summary of the poem.

attention	treble	reluctantly
actors	maturity	reputation
serious	faculties	composing
enter	promises	dependent

Shakespeare considers the whole world a stage where men and women are only (1) _____ They (2) _____ the stage when they are borm and exit when they die. Every man, during his life time; plays seven roles based on age. In the first act, as an infant, he is wholly (3) _____ on the mother or a nurse. Later, emerging as a school child, he slings his bag over his shoulder and creeps most (4) _____ to school. His next act is that of a lover, busy (5) _____ ballads for his beloved and yearns for her (6) _____ In the fourth stage, he is aggressive and ambitious and seeks (7) _____ in all that he does. He (8) _____ solemnly to guard his country and becomes a soldier. As he grows older, with (9) _____ and wisdom, he becomes a fair judge. During this stage, he is firm and (10) _____ In the sixth act, he is seen with loose pantaloons and spectacles. His manly voice changes into a childish (11) _____ The last scene of all is his second childhood. Slowly, he loses his (12) _____ of sight, hearing, smell and taste and exits from the roles of his life.
Answer:

1. actors
2. enter
3. dependent
4. reluctantly
5. composing
6. attention
7. reputation
8. promises
9. maturity
10. serious
11. treble
12. faculties

2. From your understanding of the poem, answer the following questions briefly in a sentence or two.

12th English All The World’s A Stage Paragraph Question (a)
What is the world compared to?
Answer:
The world is compared to a stage.

12th English Poem All The World’s A Stage Question (b)
“And they have their exits and their entrances” – What do the words ‘exits’ and ‘entrances’ mean?
Answer:
‘Entrances’ means life. ‘Exits means death.

All The World’s A Stage Poem Appreciation Questions And Answers Question (c)
What is the first stage of a human’s life?
Answer:
The first stage of human life is “infant”. The babe on nurse’s arms pukes and mewls.

All The World’s A Stage Questions And Answers Question (d)
Describe the second stage of life as depicted by Shakespeare.
Answer:
The second stage is school boy. The boy goes to school with a heavy heart like a snail.

12th English Unit 3 Poem Question (e)
How does a man play a lover’s role?
Answer:
As a lover, man sings serenades seeking the attention of his lady love.

12th English All The World’s A Stage Question (f)
Bring out the features of the fourth stage of a man as described by the poet.
Answer:
In the fourth stage, man becomes aggressive and ambitious and seeks glory in all his pursuits. He is ready to enter the mouth of cannon for a moment of glory.

All The World’s A Stage Question And Answers Question (g)
When does a man become a judge? How?
Answer:
In the fifth stage, man grows mature and wise. He becomes an impartial judge. He is firm and serious about his opinions.

All The World’s A Stage Poem Questions And Answers Pdf Question (h)
Which stage of man’s life is associated with the ‘shrunk shank’?
Answer:
In the sixth stage, man becomes thin and weak. His fashionable dresses of youthful days have now become too lose to use for his shrunk shank (i.e.) legs that have become very lean with age.

All The World’s A Stage Question Answer Question (i)
Why is the last stage called second childhood?
Answer:
The last stage is called the second childhood. The old man slowly loses all his senses. He requirs the support of a nurse or wife to do anything. In this stage, he departs from the world.

3. Explain the following lines briefly with reference to the context.

All The World’s A Stage Question And Answer Question (a)
“They have their exits and their entrances;
And one man in his time plays many parts”
Answer:
Reference: These lines are from the poem ‘All the world’s a stage’ written by William Shakespeare.
Context and Explanation: The poet says this while hinting at the beginning and the end of life. The poet divides man’s life into seven stages. The first stage symbolises birth and the last stage death. So, he uses the words “entrances and exits”.

Question Answer Of All The World’s A Stage Question (b)
‘‘Jealous in honour, sudden and quick in quarrel,
Seeking the bubble reputation”.
Answer:
Reference: These lines are from the poem ‘All the world’s a stage’ written by William Shakespeare.
Context and Explanation: The poet says these words while describing the fourth stage when the young man becomes a soldier and runs after short-lived glory. He has inflated sense of honour and ready to insist on duels to settle matters touching his honour. He does not realise that the reputation he seeks is short-lived like a bubble.

All The World’s A Stage Summary Question (c)
“Is second childishness and mere oblivion;
Sans teeth, sans eyes, sans taste, sans everything.”
Answer:
Reference: These lines are from the poem ‘All the world’s a stage’ written by William Shakespeare.
Context and Explanation: The poet says this while man gets ready to leave this world (i.e.) the last stage of his life on this lonely planet. In this stage, man becomes totally forgetful. He loses his teeth, eyesight and taste. He loses all his senses of perception. Like a baby, he can’t do anything on his own. So, the poet calls this stage “second childhood” when the old man behaves in a childish manner.

All The World’s A Stage Poem Paragraph Additional Questions

Explain the following lines briefly with reference to the context.

All The World’s A Stage English Workshop Answers Question (a)
“His acts being seven ages. At first the infant,
Mewling and puking in the nurse’s arms;”
Answer:
Reference: These lines are from the poem ‘All the world’s a stage’ written by William . Shakespeare.
Context and Explanation: The poet says these words while describing the first seven stages of life on the stage (i.e.) earth. The first stage/Act is infancy. The babe vomits on the arms of the nurse and cries like a kitten.

All The World’s A Stage Reference To Context Question (b)
“All the world’s a stage,
And all the men and women merely players;”
Answer:
Reference: These lines are from the poem ‘All the world’s a stage’ written by William Shakespeare.
Context and Explanation: The poet says these words while philosophising and classifying stages of life. The poet compares the world to a stage. All men and women are simply actors playing different roles on the different stages of life.

All The World’s A Stage Poem Questions Question (c)
“Then the whining school-boy, with his satchel
And shining morning face, creeping like snail’
Answer:
Reference: These lines are from the poem ‘All the world’s a stage’ written by William Shakespeare.
Context’and Explanation: The poet says these words while describing the second stage of life. During boyhood, the school boy goes to school reluctantly in snail speed with a heavy heart. In ‘Romeo and Juliet’, Shakespeare compares a school boy going to school like a lover going away from his lady love with a heavy heart.

All The World’s A Stage Pictures Question (d)
“Sighing like furnace, with a woeful ballad
Made to his mistress’ eyebrow.”
Answer:
Reference: These lines are from the poem ‘All the world’s a stage’ written by William Shakespeare.
Context and Explanation: The poet says these words while describing the third stage of life when he becomes a lover. At this stage, he yearns for the attention of his lady love. He composes ballads expressing his agony caused by unrequitted love. He sings songs praising the beauty of his mistress trying to win her heart.

All The World’s A Stage Poem In Tamil Question (e)
“Seeking the bubble reputation
Even in the cannon’s mouth.”
Answer:
Reference: These lines are from the poem ‘All the world’s a stage” written by William Shakespeare.
Context and Explanation: The poet says these words while describing the fourth stage of life. In this stage, youngman becomes a soldier. He is quick to anger and attaches great importance to honour. He is ready to lay down his life for the fleeting bubble of reputation.

All The World’s A Stage Poem Reference To Context Question (f)
“…And then the justice,
In fair round belly with good capon lin’d,
With eyes severe and beard of formal cut,
Full of wise saws and modern instances;”
Answer:
Reference: These lines are from the poem ‘All the world’s a stage” written by William ‘ Shakespeare.
Context and Explanation: The poet says these words while describing the fifth stage of life. At this stage, he behaves like a judge pronouncing his decisive opinions with the modem instances. He quotes wise maxims from his own life experiences to influence other people. He is fond of eating delicacies unmindful of the protruding belly size.

Question (g)
“And so he plays his part The sixth age shifts
Into the lean and slipper ’dpantaloon,”
Answer:
Reference: These lines are from the poem ‘All the world’s a stage” written by William Shakespeare.
Context and Explanation: The poet says these words while describing the impact of ageing on the physical appearance. In the sixth stage, he becomes old, thin and unsteady.

Question (h)
“Hisyouthful hose, well said, a world too wide
For his shrunk shank; and his big manly voice,”
Answer:
Reference: These lines are from the poem ‘All the world’s a stage” written by William Shakespeare.
Context and Explanation: The poet says these words to describe the unsuitability of one’s own dress as one advances in years. As the young man turns old, his legs become thin and his trousers become very loose giving easy access to legs but tough to wear as the waistline has also thinned. His manly voice has become feeble. When he speaks, it looks like a child piping up his dreams.

Question (i)
“…Last scene of all,
That ends this strange eventful history,”
Answer:
Reference: These lines are from the poem ‘All the world’s a stage” written by William Shakespeare.
Context and Explanation: The poet says these words while describing the preparedness of the old man in the last stage of life to exit from this lonely planet. The poet beautifully says the “eventful history” (i.e.) life which was spiced up with many interesting things is now coming to a dramatic close. The eternal jewel of life, ‘the soul’, is going to depart the body which had kept it imprisoned for long. The soul celebrates the joy of freedom in death.

Appreciate The Poem

4. Read the poem once again carefully and identify the figure of speech that has been used in each of the following lines from the poem.

“All the world’s a stage,
And all the men and women merely players;
They have their exits and their entrances;
And one man in his time plays many parts,
His acts being seven ages. At first the infant,

Mewling and puking in the nurse’s arms;
Then the whining school-boy, with his satchel
And shining morning face, creeping like snail
Unwillingly to school. And then the lover,
Sighing like furnace, with a woeful ballad
Made to his mistress’ eyebrow. Then a soldier,
Full of strange oaths, and bearded like the pard,
Jealous in honour, sudden and quick in quarrel,

Seeking the bubble reputation
Even in the cannon’s mouth. And then the justice,
In fair round belly with good capon lin’d,
With eyes severe and beard of formal cut,
Full of wise saws and modern instances;
And so he plays his part. The sixth age shifts
Into the lean and slipper’d pantaloon,

With spectacles on nose and pouch on side;
His youthful hose, well sav’d, a world too wide
For his shrunk shank; and his big manly voice,
Turning again toward childish treble, pipes
And whistles in his sound. Last scene of all,
That ends this strange eventful history,
Is second childishness and mere oblivion;
Sans teeth, sans eyes, sans taste, sans everything.”

Question (a)
“All the world’s a stage”
Answer:
Metaphor

Question (b)
“And all the men and women merely players”
Metaphor

Question (c)
“And shining morning face, creeping like snail’
Simile

Question (d)
“Full of strange oaths, and bearded like the pard,”
Answer:
Simile

Question (e)
“Seeking the bubble reputation”
Answer:
Metaphor

Question (f)
“Hisyouthful hose, well sav’d, a world too wide”
Answer:
Alliteration

Question (g)
“and his big manly voice, turning again toward childish treble”
Answer:
Metaphor

5.Pick out the words in ‘alliteration’ in the following lines,

Question (a)
“and all the men and women merely players”
Answer:
and all the men and women merely players

Question (b)
“And one man in his time plays many parts”
Answer:
And one man in his time plays many parts

Question (c)
“Jealous in honour, sudden and quick in quarrel ”
Answer:
Jealous in honour, sudden and quick in quarrel.

6. Read the given lines and answer the questions that follow.

(a) “Then the whining school-boy, with his satchel
And shining morning face, creeping like snail
Unwillingly to school ”

Question (i)
Which stage of life is being referred to here by the poet?
Answer:
Boyhood is referred to here.

Question (ii)
What are the characteristics of this stage?
Answer:
Innocence, joy and care-free life are the characteristics of this stage in life.

Question (iii)
How does the boy go to school?
Answer:
The boy goes to school unwillingly. He is slow like a snail.

Question (iv)
Which figure of speech has been employed in the second line?
Answer:
Simile is employed in the second line.

(b) “Then a soldier,
full of strange oaths, and bearded like the pard,
Jealous in honour, sudden and quick in quarrel,
Seeking the bubble reputation Even in the cannon’s mouth.

Question (i)
What is the soldier ready to do?
Answer:
The soldier is ready to lay down his life.

Question (ii)
Explain ‘bubble reputation’.
Answer:
Reputation is a transitory thing. It doesn’t even last a minute like the life of a bubble.

Question (iii)
What are the distinguishing features of this stage?
In this stage, the youthful soldier attaches great value to honour. He is quick to temper and challenges people for fight for the sake of honour. He often swears to assert his valour.

(c) “And then the justice,
In fair round belly with good capon lin’d,
With eyes severe and beard of formal cut,
Full of wise saws and modern instances;’’’’

Question (i)
Whom does justice refer to?
Answer:
Justice refers to man in his fifth stage when he becomes critical of everyone else’s opinion in life.

Question (ii)
Describe his appearance.
Answer:
He has a pot belly and is fond of eating delicacies.

Question (iii)
How does he behave with the people around him?
Answer:
His eyes are severe. He often gives advice to people.

Question (iv)
What does he do to show his wisdom?
Answer:
To show of his wisdom, he often quotes modem examples and words of wisdom.

Additional Questions

Read the given lines and answer the questions that follow.

(a) “All the world’s a stage
And all the men and women merely players:
They have their exits and their entrances;
And one man in his time plays many parts,
His acts being seven ages.”

Question (i)
What are all the men and women of this world?
Answer:
The men and women of the world are just like players on the stage of life.

Question (ii)
Explain: ‘They have their exits and their entrances’.
Answer:
They take birth and enter the world. They die and depart from the world.

Question (iii)
How many parts does every man enact and play?
Answer:
Every man enacts and plays seven different roles in life.

Question (iv)
Why is this world compared to a stage?
Answer:
This world is like a big stage where men and women are ever busy in playing their respective roles.

(b) “At first the infant,
Mewling and puking in the nurse’s arms.
Then the whining schoolboy, with his satchel
And shining morning face, creeping like snail
Unwillingly to school.”

Question (i)
What does man do in the first stage of life?
Answer:
In the first stage of life man plays the role of an infant. He is always crying and vomiting in the nurse’s arms.

Question (ii)
Does the schoolboy show eagerness to go to school?
Answer:
No, the schoolboy doesn’t show any interest in going to school. Rather he is unwilling to go there.

Question (iii)
How does the schoolboy walk up to his school?
Answer:
He is inching slowly and unwillingly like a snail towards his school.

Question (iv)
Explain, ‘Mewling and pucking’.
Answer:
It means crying and vomiting.

(c) “And then the lover,
Sighing like furnace, with a woeful ballad
Made to his mistress’ eyebrow.”

Question (i)
What is the third stage of life?
The third stage of man’s life is that of a lover.

Question (ii)
What is the poetic device used in the second line?
Answer:
‘Simile’ is used as a poetic device in the second line.

Question (iii)
What does the lover do for his mistress?
Answer:
The lover is ahvays sighing and longing for his beloved. He writes a sad ballad describing the eyebrow of his mistress.

Question (iv)
Explain, ‘sighing like furnace’.
Answer:
It means moaning, breathing deeply and sadly like a fire place.

(d) “Then a soldier.
Full of strange oaths, and bearded like the pard,
Jealous in honour, sudden and quick in quarrel,
Seeking the bubble reputation.
Even in the cannon’s mouth.”

Question (i)
Describe the two traits of a soldier.
Answer:
A soldier is always ready to swear and is full of oaths. He is ever ready to compete for honour and glory.

Question (ii)
What is the poetic device used in : ‘bearded like a pard’?
Answer:
The poet uses a simile for comparison.

Question (iii)
Why does the soldier risk his life and what for?
Answer:
The soldier risks his life a momentary reputation and is ready even to enter the cannon’s mouth.

Question (iv)
How is the soldier bearded?
Answer:
He is bearded like a pard or a leopard.

(e) “The sixth age shifts
Into the lean and slippered pantaloon,
With spectacles on nose and pouch on side,
His youthful hose, well saved, a world too wide For his shrunk shank; and his big manly voice,
Turning again toward childish treble, pipes
And whistles in his sound.’’

Question (i)
What is a ‘lean and slippered pantaloon’?
Answer:
It means a thin old man wearing slippers and loose trousers.

Question (ii)
What does the phrase ‘a world too wide’ here mean?
Answer:
The stockings he bought in his youth have become too loose for his shrunk and thin legs.

Question (iii)
How does the ‘mainly voice’ turn into ‘childish’ in the sixth stage of life?
Answer:
His manly voice turns into childish trebles and whistles when he speaks as he has no teeth in his mouth.

Question (iv)
What is the sixth stage of man’s life?
Answer:
In the sixth stage of life man plays the role of a ‘lean and slippered pantaloon’.

(f) “Last scene of all,
That ends this strange eventful history,
Is second childishness and mere oblivion,
Sans teeth, sans eyes, sans taste, sans everything.”

Question (i)
What is the last scene of man’s life?
Answer:
The last scene that ends man’s eventual life is a ‘second-childishness’. In this stage he appears and behaves like a child.

Question (ii)
Why is the last stage of man has been called a ‘second childishness’?
Answer:
The last stage of man’s life has been called a ‘second childishness’ as man’s appearance and activities in this stage are quite similar to those of a child.

Question (iii)
How is the last stage of man’s life a ‘mere oblivion’?
Answer:
The last stage of life is a ‘mere oblivion’ as old age is another stage of forgetfulness.

Question (iv)
Explain ‘eventful history’.
Answer:
It means the life-long history of man full of interesting incidents and experiences.

7. Complete the table based on your understanding of the poem.

Stage	Characteristic
	crying
judge
soldier
	unhappy
second childhood
	whining
old man

Answer:

Stage	Characteristic
Baby (first stage)	crying
judge	Firm and serious
soldier	Aggressive and Ambitious
Lover	unhappy
second childhood	Loses senses
Boyhood (school)	whining
old man	Wise and judges others

8. Based on your understanding of the poem, answer the following questions in about 100 – 150 words each. You may add your own ideas if required, to present and justify your point of view.

Question (а)
Describe the various stages of a man’s life picturised in the poem “All the World’s a stage.”
Answer:
Shakespeare has beautifully portrayed this world as a huge open theatre where in all humans play seven acts/ages. In the first act, he is a helpless infant puking on the nurse’s arms mewling like a kitten. In the second stage, he is the grumbling/whining school student. He moves to school like a snal/unwllingly with his slate and bag. In the third Act, he is a lover sighing and yearning for the attention of his lady love.

He composes romantic ballads complaining his love that he needs a better deal. In the fourth Act, he becomes a quick-tempered soldier, aggressive and ambitious, ready to stake his life for the sake of bubble reputation. As he matures, he becomes a wise judge of contemporary life quoting wise maxims to endorse his opinion. He is firm and serious. In the sixth act, his stout legs become thin making his trousers of youth unsuitable. Thin and lean legs easily travel through them but are unable to stay due to a slimmed waist. His bass voice has become treble like that of a child. In the last act, he is sans teeth, sanys eyes, sans taste and sans everything (i.e.) loses all senses. He departs the world.

Question (b)
Shakespeare has skill fully brought out the parallels between the life of man and actors on stage. Elaborate this statement with reference to the poem.
Answer:
Shakespeare has beautifully compared the growth of humans by stages with his emergent role during that stage. In the first stage man plays the role of an infant. As an infant, he does represent characterisation of mewling and puking. In the second Act, he does the role of a school boy with the characteristics of unwillingness to go to schools and innocence shining in his face. In the third Act, he performs the role of a lover head over heels in love with a beautiful lady. He composes woeful romantic ballads and sings serenades to impress his love. In the fourth act, he plays the impressive role of a short-tempered, honour pursuing soldier.

He is ready to put his mouth in the Cannon’s mouth for conquering the bubble like honour in order to defend the territory of his country. In the fifth Act, he performs the role of a mature and fair judge criticising the ways of the world often spicing up his conversations with wise remarks and wit. His pot belly and well-cut beard shows the social status he enjoys in life. In the sixth act, he is old. He performs the role of a thin old man wearing ill-fitting loose garments with a changed treble in his voice. He is bespectacled and slow in walking. In the final act, he becomes a total invalid losing all senses of hearing, taste and sight. Then the performer leaves the stage (i.e.) the lonely planet.

Speaking Activity

Shakespeare describes the characteristics of the various stages of man. You are in the second stage of life. What do you think of your roles and responsibilities at this stage? Discuss with your partner and share your ideas with the class.
Answer:
At school age, imagination takes wings. Inquisitiveness is common among my peers. Parents, society and teachers want us only to study. But we need to explore the world around us. At home, it is our responsibility to keep our things in order. We need to assist the perennial worker, we mean, our moms in completing their domestic chores. Occasionally, we shall take care of siblings too not as a work but as a duty towards a family member who will be a life long companion to us.

Listening Activity

Listen to the poem and fill in the blanks with appropriate words and phrases. If required listen to the poem again.

The World Is Too Much with Us
The world is too much with us; late and soon, Getting and spending, we lay waste our powers; Little we see in Nature that is ours;We have given our hearts away, a sordid boon! This Sea that bares her bosom to the moon, The winds that will be howling at all hours,And are up- gathered now like sleeping flowers,

For this, for everything, we are out of tune; It moves us not. – Great God! I’d rather be A Pagan suckled in a creed outworn; So might I, standing on this pleasant lea, Have glimpses that would make me less forlorn; Have sight of Proteus rising from the sea; Or hear old Triton blow his wreathed horn.

The World Is Too Much with Us:
The world is too much with us; late and soon
Getting and spending, we lay waste our powers
Little we see in (1) ______ that is ours;
We have given (2) ______ away, a sordid boon!
This Sea that bares her bosom (3) ______
(4) ______ that will be howling at all hours,
And are up-gathered now like (5) ______
,For this, for everything, we are (6) ______ ;
It (7) ______ . us not. Great God! I’d rather be
A Pagan suckled in a creed outworn;
So might I, standing on this pleasant lea
Have glimpses that would make me less forlorn;
Have sight of Proteus rising (8) ______
Or hear old Triton blow his wreathed horn.
Answers:

1. Nature
2. our hearts
3. to the moon
4. The winds
5. sleeping flowers
6. out of tune
7. moves
8. from the sea

All the World’s a Stage About The Poet

William Shakespeare (1564-1616) was a prolific writer during the Elizabethan and Jacobean ages of British theatre (sometimes called the English Renaissance). Shakespeare’s plays are perhaps his most enduring legacy. Shakespeare’s poems remain popular to this day. Shakespeare’s rich and diverse works have spawned countless adaptations across multiple genres and cultures. His writings have been compiled in various iterations of The Complete Works of William Shakespeare. William Shakespeare continues to be one of the most important literary7 figures of the English language.

All the World’s a Stage Summary in english

Introduction
‘All the world’s a stage’ is an extract from the play ‘As you like it’, a romantic comedy by Shakespeare.

A metaphor defining world
Shakespeare claims this world as a stage in a theatre. All men and women are only actors. The stage has both exits and entrances. Similarly, men and women take birth and enter the world. They live their lives and go out of it when they die. Every man plays seven emergent roles and lives through seven stages of life.

Infancy and boyhood
With the birth of an infant begins the first stage of man’s life. The infant cries and vomits on the arms of his nurse. Then he grows into a school-going boy. He is unwilling to go to the school. He moves towards school at a snail’s speed.

Thirst for love and glory
In the third stage, man plays the role of a lover. He sighs like a fumace.He keeps on writing ballads praising the beauty of the eyes of his beloved. The fourth stage is that of a soldier. He keeps a beard like that of a leopard. He always runs after honour and fame. He is ready even to enter a cannon’s mouth just for momentary glory and bubble of reputation.

Wisdom and failing health
In the fifth stage, man plays the role of a justice. He is fond of eating chicken and develops a fat round belly. He is full of wise sayings and modem instances. He is a man of wisdom and knowledge. In the sixth stage, man becomes weak and thin in body. He wears slippers, spectacles and clothes that he bought when he was young. These pants and stockings have become loose for his shrunk and thin legs.

Second childhood
The seventh stage is the ‘second childhood’. In this stage, man becomes very old and starts behaving like a child. He is left with no teeth and becomes weak in eyesight. Actually, he loses taste and becomes a victim of forgetfulness. The poet describes this helpless state as “Sans teeth, sans eyes, sans taste and sans everything” nicely. Then the man departs from this world.

Conclusion
Shakespeare condenses the life of man beautifully and portrays it well. The revisit of childhood in old age proves his profound understanding of human life.

All the World’s a Stage Summary in Tamil

முன்னுரை
‘All the World’s a stage’ (‘உலகம் ஒரு நாடக மேடை’) என்ற கவிதை சேக்ஸ்பியரின் ‘As you like it’, என்ற | நகைச்சுவை கலந்த கற்பனை கதையின் ஒரு சாரம் ஆகும்.

உலகம் ஒரு நாடக மேடை:
சேக்ஸ்பியர் உலகத்தை ஒருநாடக மேடையாகக் கருதுகிறார். அதில் அனைத்து ஆணும், பெண்ணும் நடிகர்களே. இந்த நாடக மேடையின் உள்ளே வரவும் வெளியே செல்லவும் வழிகள் உள்ளன. அதே போல தான் மனிதன் பிறந்து இந்த உலகத்துக்கு வருகிறான். அவனது வாழ்நாளை வாழ்ந்துவிட்டு வெளியே போய்விடுகிறான். ஒவ்வொரு மனிதனும் ஏழு கதாபாத்திரங்களாக வாழ்க்கை மேடையில் நடிக்கிறான்.

குழந்தை பருவமும், விடலைப் பருவமும்:
குழந்தை பருவமே மனிதனின் முதல் பாகம் ஆகும். வாந்தியும், அழுகையுமாக முதல் பாகம் செவிலிப் பெண் தோளில் இருக்கிறான். பிறகு பள்ளிப் பருவம் அடைகிறான். பள்ளிக்கூடம் போக மனமில்லாது இருக்கிறான். பள்ளிக்கூடத்தை நோக்கி நத்தை போல் நகர்கிறான்.

காதல், புகழ் என ஈர்ப்புக்குள்ளாகிறான்:
மூன்றாம் பாகத்தில் காதலனாக கதாபாத்திரம் ஏற்கிறான். எரியும் அடுப்பைப் போன்று குமுறுகிறான். தன் காதலியின் கண்களைக் குறித்து கவிதை மழை பொழிகிறான். நான்காம் பாகத்தில் சிப்பாய் வேடம் ஏற்கிறான். சிறுத்தை போன்று மீசையை வளர்த்துக் கொள்கிறான். பேர் மற்றும் புகழின் பின்னால் ஓடுகிறான். தற்காலிக பேருக்கும், புகழுக்கும் ஆசைப்பட்டு பீரங்கி | வாயினுள் நுழையவும் தயாராக இருக்கிறான்.

அறிவு முதிர்ச்சியும், குன்றும் ஆரோக்கியமும்:
ஐந்தாம் பாகத்தில் தானே ஒரு நீதிபதி ஆகிறான். கோழி மாமிசத்தின் பால் ஆவல் கொண்டு அதை உண்டு பெரிய தொப்பையுடன் தோன்றுகிறான். அறிவு முதிர்ச்சியுடனும், புதுப் பொலிவுடனும் தோன்றுகிறான். அறிவும் ஆற்றலுமுடையவனாய்த் திகழ்கிறான். ஆறாம் பாகத்தில் உடல் வலுவிழந்து சோர்வடைகிறான், ஒல்லி வடிவமாய், காலில் செருப்புமாய் ஒரு சிரிப்பு நடிகனைப் போல் தோற்றமளிக்கிறான். இளம் வயதில் அணிந்த கண்ணாடியும், துணிகளும், செருப்பும் அணிந்து கொள்கிறான். சுருங்கிய தோல்களுக்கும், ஒல்லியான கால்களுக்கும் இந்த உடையும், செருப்பும் தொள தொளவென காணப்படுகின்றன.

இரண்டாம் குழந்தை பருவம்:
ஏழாம் பருவம் இரண்டாவது குழந்தை பருவம் எனலாம். இந்த பாகத்தில் மிகவும் வயது முதிர்ந்த ஒரு குழந்தையின் இயலாமைத் தனத்தை செயல்பாட்டில் காட்டுகிறான். பற்களை இழந்து, கண் பார்வைக் குன்றிப் போகிறான். குழந்தையின் குரல் போல் மாறி, குரல் ஒரு விசில் சத்தமாய் மாறுகிறது. இது கடைசி அத்தியாயம் எனலாம். அவனின் அதிசயமான பரப்பரப்பூட்டும் நிகழ்வுகள் நிறைந்த வரலாறானது ஒரு முடிவுக்கு வருகிறது. தன் இரண்டாம் குழந்தைப் பருவத்தில் பிறரைச் சார்ந்து வாழும் நிலையை அடைகிறான். பற்களை இழந்து, கண் பார்வையை இழந்து, நாவின் | சுவை இழந்து, பின் அனைத்தையும் இழக்கிறான். உலகத்தை விட்டு வெளியேறுகிறான்.

முடிவுரை:
சேக்ஸ்பியர் மனித வாழ்க்கையை சுருக்கி அழகாக அதை வர்ணித்து இருக்கிறார். வயோதிகத்தில் திரும்பும் குழந்தைத் தனம் என்பது அவர் மனித வாழ்க்கையை அவர் ஆழ்ந்து அறிந்து கொண்டதை உணர்த்துகிறது.

All the World’s a Stage Glossary

Textual:

Additional:

Students those who are preparing for the 12th Physics exam can download this Samacheer Kalvi 12th Physics Book Solutions Questions and Answers for Chapter 2 Current Electricity from here free of cost. These Tamilnadu State Board Solutions for Class 12th Physics PDF cover all Chapter 2 Current Electricity Questions and Answers. All these concepts of Samacheer Kalvi 12th Physics Book Solutions Questions and Answers are explained very conceptually as per the board prescribed Syllabus & guidelines.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity

Kickstart your preparation by using this Tamilnadu State Board Solutions for 12th Physics Chapter 2 Current Electricity Questions and Answers and get the max score in the exams. You can cover all the topics of Chapter 2 Current Electricity Questions and Answers pdf. Download the Tamilnadu State Board Solutions by accessing the links provided here and ace up your preparation.

Samacheer Kalvi 12th Physics Current Electricity Textual Evaluation Solved

Samacheer Kalvi 12th Physics Current Electricity Multiple Choice Questions   

Current Electricity Class 12 Problems With Solutions State Board Question 1.
The following graph shows current versus voltage values of some unknown conductor. What is the resistance of this conductor?

(a) 2 ohm
(b) 4 ohm
(c) 8 ohm
(d) 1 ohm
Answer:
(a) 2 ohm

Current Electricity Class 12 State Board Question 2.
A wire of resistance 2 ohms per meter is bent to form a circle of radius 1 m. The equivalent resistance between its two diametrically opposite points, A and B as shown in the figure is-

(a) π Ω
(b) \(\frac { π }{ 2 }\) Ω
(c) 2π Ω
(d) \(\frac { π }{ 4 }\) Ω
Answer:
(b) \(\frac { π }{ 2 }\) Ω

12th Physics Chapter 2 Book Back Answers Question 3.
A toaster operating at 240 V has a resistance of 120 Ω. The power is
(a) 400 W
(b) 2 W
(c) 480 W
(d) 240 W
Answer:
(c) 480 W

12th Physics Samacheer Kalvi Question 4.
A carbon resistor of (47 ± 4.7) k Ω to be marked with rings of different colours for its identification. The colour code sequence will be
(a) Yellow – Green – Violet – Gold
(b) Yellow – Violet – Orange – Silver
(c) Violet – Yellow – Orange – Silver
(d) Green – Orange – Violet – Gold
Answer:
(b) Yellow – Violet – Orange – Silver

Samacheer Kalvi 12th Physics Question 5.
What is the value of resistance of the following resistor?

(a) 100 k Ω
(b) 10 k Ω M
(c) 1 k Ω
(d) 1000 k Ω
Answer:
(a) 100 k Ω

Class 12 Physics Samacheer Kalvi Question 6.
Two wires of A and B with circular cross section made up of the same material with equal lengths. Suppose R_A = 3 R_B, then what is the ratio of radius of wire A to that of B?
(a) 3
(b) √3
(c) \(\frac { 1 }{ √3 }\)
(d) \(\frac { 1 }{ 3 }\)
Answer:
(c) \(\frac { 1 }{ √3 }\)

Current Electricity Short Notes Question 7.
A wire connected to a power supply of 230 V has power dissipation P₁ Suppose the wire is cut into two equal pieces and connected parallel to the same power supply. In this case power dissipation is P₂. The ratio \(\frac {{ p }_{2}}{{ p }_{1}}\) is.
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Samacheer Kalvi Guru 12th Physics Question 8.
In India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in USA. If the resistance of a 60W bulb for use in India is R, the resistance of a 60W bulb for use in USA will be
(a) R
(b) 2R
(c) \(\frac { R }{ 4 }\)
(d) \(\frac { R }{ 2 }\)
Answer:
(c) \(\frac { R }{ 4 }\)

Samacheer Kalvi 12th Physics Solutions Question 9.
In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1kW are connected. The voltage of electric mains is 220 V. The minimum capacity of the main fuse of the building will be (IIT-JEE 2014)
(a) 14 A
(b) 8 A
(c) 10 A
(d) 12 A
Answer:
(d) 12 A

Samacheerkalvi.Guru 12th Physics Question 10.
There is a current of 1.0 A in the circuit shown below. What is the resistance of P ?

(a) 1.5 Ω
(b) 2.5 Ω
(c) 3.5 Ω
(d) 4.5 Ω
Answer:
(c) 3.5 Ω

Samacheer Kalvi Physics Question 11.
What is the current out of the battery?

(а) 1 A
(b) 2 A
(c) 3 A
(d) 4 A
Answer:
(а) 1 A

12th Physics Solutions Samacheer Kalvi Question 12.
The temperature coefficient of resistance of a wire is 0.00125 per °C. At 300 K, its resistance is 1 Ω. The resistance of the wire will be 2 Ω. at
(a) 1154 K
(ft) 1100 K
(c) 1400 K
(d) 1127 K
Answer:
(d) 1127 K

Samacheer Kalvi 12th Physics Solution Book Question 13.
The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10Ω is
(a) 0.2 Ω
(b) 0.5 Ω
(c) 0.8 Ω
(d) 1.0 Ω
Answer:
(b) 0.5 Ω

Samacheer Kalvi Guru Physics Question 14.
A piece of copper and another of germanium are cooled from room temperature to 80 K. The resistance of
(a) each of them increases
(b) each of them decreases
(c) copper increases and germanium decreases
(d) copper decreases and germanium increases
Answer:
(d) copper decreases and germanium increases

Physics Class 12 Samacheer Kalvi Question 15.
In Joule’s heating law, when I and t are constant, if the H is taken along the y axis and I² along the x axis, the graph is
(a) straight line
(b) parabola
(c) circle
(d) ellipse
Answer:
(a) straight line

Samacheer Kalvi 12th Physics Current Electricity Short Answer Questions

Samacheer Kalvi.Guru 12th Physics Question 1.
Why current is a scalar?
Answer:
The current I is defined as the scalar product of current density and area vector in which the charges cross.
I = \(\vec { j } \) . \(\vec { A } \)
The dot product of two vector quantity is a scalar form. Hence, current is called as a scalar quantity.

12th Samacheer Physics Solutions Question 2.
Distinguish between drift velocity and mobility.
Answer:

Physics Solution Class 12 Samacheer Kalvi Question 3.
State microscopic form of Ohm’s law.
Answer:
Current density J at a point in a conductor is the amount of current flowing per unit area of the conductor around that point provided the area is held in a direction normal to the current.

Current Electricity Pdf Question 4.
State macroscopic form of Ohm’s law.
Answer:
The macroscopic form of Ohm’s Law relates voltage, current and resistance. Ohm’s Law states that the current through an object is proportional to the voltage across it and inversely proportional to the object’s resistance.
V = IR.

Current Electricity Class 12 Question 5.
What are ohmic and non-ohmic devices?
Answer:
Materials for which the current against voltage graph is a straight line through the origin, are said to obey Ohm’s law and their behaviour is said to be ohmic. Materials or devices that do not follow Ohm’s law are said to be non-ohmic.

Samacheer Kalvi Class 12 Physics Solutions Question 6.
Define electrical resistivity.
Answer:
Electrical resistivity of a material is defined as the resistance offered to current flow by a conductor of unit length having unit area of cross section.

Question 7.
Define temperature coefficient of resistance.
Answer:
It is defined as the ratio of increase in resistivity per degree rise in temperature to its resistivity at T₀

Question 8.
What is superconductivity?
Answer:
The ability of certain metals, their compounds and alloys to conduct electricity with zero resistance at very low temperatures is called superconductivity.

Question 9.
What is electric power and electric energy?
Answer:
1. Electric power:
It is the rate at which an electric appliance converts electric energy into other forms of energy. Or, it is the rate at which work is done by a source of emf in maintaining an electric current through a circuit.
P = \(\frac { W }{ t }\) = VI = I²R = \(\frac {{ V }_{2}}{ R }\)

2. Electric energy:
It is the total workdone in maintaining an electric current in an electric circuit for a given time.
W = Pt = VIt joule = I²Rt joule.

Question 10.
Define current density.
Answer:
The current density (J) is defined as the current per unit area of cross section of the conductor
J = \(\frac { 1 }{ A }\)
The S.I unit of current density.
\(\frac { A }{{ m }^{2}}\)
Or
Am²

Question 11.
Derive the expression for power P = VI in electrical circuit.
Answer:
The electrical power P is the rate at which the electrical potential energy is delivered,
P = \(\frac { dU }{ dt }\) = \(\frac { d }{ dt }\) (V.dQ) = V\(\frac { dQ }{ dt }\)
Since the electric current I = \(\frac { dQ }{ dt }\)
So the equation can be rewritten as P = VI.

Question 12.
Write down the various forms of expression for power in electrical circuit.
Answer:
The electric power P is the rate at which the electrical potential energy is delivered,
P = \(\frac { dU }{ dt }\) = \(\frac { 1 }{ dt }\) (V.dQ) = V.\(\frac { dQ }{ dt }\)
[dU = V.dQ]
The electric power delivered by the battery to any electrical system.
P = VI
The electric power delivered to the resistance R is expressed in other forms.
P = VI = I(IR) = I²R
P = IV = \((\frac { V }{ R })\) V = \(\frac {{ V }^{2}}{ R }\).

Question 13.
State Kirchhoff’s current rule.
Answer:
It states that the algebraic sum of the currents at any junction of a circuit is zero. It is a statement of conservation of electric charge.

Question 14.
State Kirchhoff’s voltage rule.
Answer:
It states that in a closed circuit the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit. This rule follows from the law of conservation of energy for an isolated system.

Question 15.
State the principle of potentiometer.
Answer:
The basic principle of a potentiometer is that when a constant current flows through a wire of uniform cross-sectional area and composition, the potential drop across any length of the wire is directly proportional to that length.

Question 16.
What do you mean by internal resistance of a cell?
Answer:
The resistance offered by the electrolyte of a cell to the flow of current between its electrodes is called internal resistance of the cell. An ideal battery has zero internal resistance and the potential difference across the battery equals to its emf. But a real battery is made of electrodes and electrolyte, there is resistance to the flow of charges within the battery. A freshly prepared cell has low internal resistance and it increases with ageing.

Question 17.
State Joule’s law of heating.
Answer:
It states that the heat developed in an electrical circuit due to the flow of current varies directly as:

1. the square of the current
2. the resistance of the circuit and
3. the time of flow.
   H = I²R?

Question 18.
What is Seebeck effect?
Answer:
Seebeck discovered that in a closed circuit consisting of two dissimilar metals, when the junctions are maintained at different temperatures, an emf (potential difference) is developed.

Question 19.
What is Thomson effect?
Answer:
Thomson showed that if two points in a conductor are at different temperatures, the density of electrons at these points will differ and as a result the potential difference is created between these points. Thomson effect is also reversible.

Question 20.
What is Peltier effect?
Answer:
When an electric current is passed through a circuit of a thermocouple, heat is evolved at one junction and absorbed at the other junction. This is known as Peltier effect.

Question 21.
State the applications of Seebeck effect.
Answer:
Applications of Seebeck effect:

1. Seebeck effect is used in thermoelectric generators (Seebeck generators). These thermoelectric generators are used in power plants to convert waste heat into electricity.
2. This effect is utilised in automobiles as automotive thermoelectric generators for increasing fuel efficiency.
3. Seebeck effect is used in thermocouples and thermopiles to measure the temperature difference between the two objects.

Samacheer Kalvi 12th Physics Current Electricity Long Answer Questions

Question 1.
Describe the microscopic model of current and obtain genera! form of Ohm’s Law.
Answer:
Microscopic model of current: Consider a conductor with area of cross-section A and an electric field E applied from right to left. Suppose there are n electrons per unit volume in the conductor and assume that all the electrons move with the same drift velocity \(\vec { v } \)_d.
The drift velocity of the electrons = v_d
The electrons move through a distance dx within a small interval of dt

v_d = \(\frac { dx }{ dt }\); dx = v_ddt ….. (1)
Since A is the area of cross section of the conductor, the electrons available in the volume or length dx is
= volume x number per unit volume
= A dx × n …… (2)
Substituting for dx from equation (1) in (2)
= (A v_d dt)n
Total charge in volume element dQ = (charge) x (number of electrons in the volume element)
dQ= (e)(A v_d dt)n
Hence the current, I = \(\frac { dQ }{ dt }\) = \(\frac{n e A v_{d} d t}{d t}\)
I = ne A v_d …….. (3)
Current denshy (J):
The current density (J) is defined as the current per unit area of cross section of the conductor
J = \(\frac { I }{ A }\)
The S.I. unit of current density,\(\frac { A }{{ m }^{2}}\) (or) Am^-2
J = \(\frac {{ neA v }_{d}}{ A }\) (from equation 3)
J = nev_d …….. (4)
The above expression is valid only when the direction of the current is perpendicular to the area A. In general, the current density is a vector quantity and it is given by
\(\vec { J } \) = ne\(\vec { v } \)_d
Substituting i from equation \(\vec { v } \)_d = \(\frac { -eτ }{ m }\) \(\vec { E } \)
\(\vec { J } \) = –\(\frac{n \cdot e^{2} \tau}{m}\)\(\vec { E } \) …… (5)
\(\vec { J } \) = -σ\(\vec { E } \)
But conventionally, we take the direction of (conventional) current density as the direction of electric field. So the above equation becomes
\(\vec { J } \) = σ\(\vec { E } \) …….. (6)
where σ = \(\frac{n \cdot e^{2} \tau}{m}\) is called conductivity.
The equation 6 is called microscopic form of ohm’s law.

Question 2.
Obtain the macroscopic form of Ohm’s law from its microscopic form and discuss its limitation.
Answer:
Ohm’s law: The Ohm’s law can be derived from the equation J = σE. Consider a segment of wire of length l and cross sectional area A.

When a potential difference V is applied across the wire, a net electric field is created in the wire which constitutes the current. For simplicity, we assume that the electric field is uniform in the entire length of the wire, the potential difference (voltage V) can be written as V = El
As we know, the magnitude of current density
J = σE = σ\(\frac { V }{ l }\) ……. (1)
But J = \(\frac { I }{ A }\), so we write the equation as
\(\frac { I }{ A }\) σ\(\frac { V }{ l }\)
By rearranging the above equation, we get
V = I \(\left( \frac { l }{ \sigma A } \right) \) ……… (2)
The quantity \(\frac { l }{ \sigma A } \)is called resistance of the conductor and it is denoted as R. Note that the oA resistance is directly proportional to the length of the conductor and inversely proportional to area of cross section.
Therefore, the macroscopic form of Ohm’s law can be stated as
V = IR.

Question 3.
Explain the equivalent resistance of a series and parallel resistor network.
Answer:
1. Resistors in series:
When two or more resistors are connected end to end, they are said to be in series. The resistors could be simple resistors or bulbs or heating elements or other devices. Fig. (a) shows three resistors R₁, R₂ and R₃ connected in series.

The amount of charge passing through resistor R₁ must also pass through resistors R₂ and R₃ since the charges cannot accumulate anywhere in the circuit. Due to this reason, the current I passing through all the three resistors is the same. According to Ohm’s law, if same current pass through different resistors of different values, then the potential difference across each resistor must be different. Let V₁, V₂ and V₃ be the potential difference (voltage) across each of the resistors R₁, R₂ and R₃ respectively, then we can write V₁ = IR_1, V₂ = IR₂ and V₃= IR₃. But the total voltage V is equal to the sum of voltages across each resistor.
V = V₁ + V₂ + V₃
= IR₁ + IR₂ + IR₃ ….. (1)
V = I(R₁ + R₂ +R₃)
V = I.R_S …… (2)
where R_s = R₁ + R₂ R₃ ……. (3)
When several resistances are connected in series, the total or equivalent resistance is the sum of the individual resistances as shown in fig. (b).

Note:
The value of equivalent resistance in series connection will be greater than each individual resistance.

2. Resistors in parallel:
Resistors are in parallel when they are connected across the same potential difference as shown in figer.

In this case, the total current I that leaves the battery in split into three separate paths. Let I₁, I₂ and I₃ be the current through the resistors R₁, R₂ and R₃ respectively. Due to the conservation of charge, total current in the circuit I is equal to sum of the currents through each of the three resistors.
I = I₁ + I₂ + I₃ ……. (1)
Since the voltage across each resistor is the same, applying Ohm’s law to each resistor, we have
I₁ = \(\frac { V }{{ R }_{1}}\) I₂ = \(\frac { V }{{ R }_{2}}\),I₁ = \(\frac { V }{{ R }_{3}}\)
Substituting these values in equation (1), we get
I = \(\frac { V }{{ R }_{1}}\) + \(\frac { V }{{ R }_{2}}\) +\(\frac { V }{{ R }_{3}}\) = V\(\left[ \frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } +\frac { 1 }{ { R }_{ 3 } } \right] \)
I = \(\frac { V }{{ R }_{p}}\)
\(\frac { 1 }{{ R }_{p}}\) = \(\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } +\frac { 1 }{ { R }_{ 3 } } \)
Here R_P is the equivalent resistance of the parallel combination of the resistors. Thus, when a number of resistors are connected in parallel, the sum of the reciprocal of the values of resistance of the individual resistor is equal to the reciprocal of the effective resistance of the combination as shown in the fig. (b).

Note:
The value of equivalent resistance in parallel connection will be lesser than each individual resistance.

Question 4.
Explain the determination of the internal resistance of a cell using voltmeter.
Answer:
Determination of internal resistance:
The emf of cell ξ is measured by connecting a high resistance voltmeter across it without connecting the external resistance R. Since the voltmeter draws very little current for deflection, the circuit may be considered as open. Hence, the voltmeter reading gives the emf of the cell. Then, external resistance R is included in the circuit and current I is established in the circuit. The potential difference across R is equal to the potential difference across the cell (V).

The potential drop across the resistor R is
V= IR …… (1)
Due to internal resistance r of the cell, the voltmeter reads a value V, which is less than the emf of cell ξ. It is because, certain amount of voltage (Ir) has dropped across the internal resistance r.

V = ξ – Ir
Ir = ξ – V …… (2)
Dividing equation (2) by equation (1), we get
\(\frac { Ir }{IR}\) = \(\frac { ξ – V }{V}\)
r = |\(\frac { ξ – V }{V}\)| R …… (3)
Since ξ, V and R are known, internal resistance r can be determined.

Question 5.
State and explain Kirchhoff’s rules.
Answer:
Kirchhoff’s first rule (current rule or junction rule):
Statement: It states that the algebraic sum of the currents at any junction of a circuit is zero. It is a statement of conservation of electric charge.

Explanation:
All charges that enter a given junction in a circuit must leave that junction since charge cannot build up or disappear at a junction. Current entering the junction is taken as positive and current leaving the junction is taken as negative.
Applying this law to the junction A,
I₁ + I₂ – I₃ – I₄ – I₅ = o
Or
I₁ + I₂ = + I₃ I₄ + I₅
Kirchhoff’s second rule (voltage rule or loop rule):
Statement: It states that in a closed circuit the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit. This rule follows from the law of conservation of energy for an isolated system. (The energy supplied by the emf sources is equal to the sum of the energy delivered to all resistors).

Explanation:
The product of current and resistance is taken as positive when the direction of the current is followed. Suppose if the direction of current is opposite to the direction of the loop, then product of current and voltage across the resistor is negative. It is shown in Fig. (a) and (b). The emf is considered positive when proceeding from the negative to the positive terminal of the cell. It is shown in Fig. (c) and (d).

Kirchhoff voltage rule has to be applied only when all currents in the circuit reach a steady state condition (the current in various branches are constant).

Question 6.
Obtain the condition for bridge balance in Wheatstone’s bridge.
Answer:
An important application of Kirchhoff’s rules is the Wheatstone’s bridge. It is used to compare resistances and also helps in determining the unknown resistance in electrical network. The – bridge consists of four resistances P, Q, R and S connected. A galvanometer G is connected between the points B and D. The battery is connected between the points A and C. The current through the galvanometer is I_G and its resistance is G.

Applying KirchhofFs current rule to junction B,
I₁ – I_G – I₃ = 0 …….. (1)
Applying Kirchhoff’s current rule to junction D,
I₂ – I_G – I₄ = 0 …….. (2)
Applying Kirchhoff’s voltage rule to loop ABDA,
I₁P + I_GG – I₂R = 0 …….. (3)
Applying Kirchhoff’s voltage rule to loop ABCDA,
I₁P + I₃Q – I₄S – I₂R = 0 …….. (4)
When the points B and D are at the same potential, the bridge is said to be balanced. As there is no potential difference between B and D, no current flows through galvanometer (I_G = 0). Substituting I_G = 0 in equation, (1), (2) and (3), we get
I₁ = I₃ …….. (5)
I₂ = I₄ …….. (6)
I₁P = I₂R …….. (7)
Substituting the equation (5) and (6) in equation (4)
I₁P + I₁Q – I₂R = 0
I₁(P + Q) = I₂ (R + S) …….. (8)
Dividing equation (8) by equation (7), we get
\(\frac { P + Q }{ P }\) = \(\frac { R + S }{ R }\)
1 + \(\frac { Q }{ P }\) = 1 + \(\frac { S }{ R }\)
⇒ \(\frac { Q }{ P }\) = \(\frac { S }{ R }\)
\(\frac { P }{ Q }\) = \(\frac { R }{ S }\) …….. (9)
This is the bridge balance condition. Only under this condition, galvanometer shows null deflection. Suppose we know the values of two adjacent resistances, the other two resistances can be compared. If three of the resistances are known, the value of unknown resistance (fourth one) can be determined.

Question 7.
Explain the determination of unknown resistance using meter bridge.
Answer:
The meter bridge is another form of Wheatstone’s bridge. It consists of a uniform manganin wire AB of one meter length. This wire is stretched along a meter scale on a wooden board between two copper strips C and D. Between these two copper strips another copper strip E is mounted to enclose two gaps G₁ and G₂ An unknown resistance P is connected in G₁ and a standard resistance Q is connected in G₂.

A jockey (conducting wire) is connected to the terminal E on the central copper strip through a galvanometer (G) and a high resistance (HR). The exact position of jockey on the wire can be read on the scale. A Lechlanche cell and a key (K) are connected across the ends of the bridge wire.

The position of the jockey on the wire is adjusted so that the galvanometer shows zero deflection. Let the point be J. The lengths AJ and JB of the bridge wire now replace the resistance R and S of the Wheatstone’s bridge. Then
\(\frac { P }{ Q }\) = \(\frac { R }{ S }\) = \(\frac { R’.AJ }{ R’.JB }\) …….. (1)
where R’ is the resistance per unit length of wire
\(\frac { P }{ Q }\) = \(\frac { AJ }{ JB }\) = \(\frac {{ l }_{1}}{ { l }_{2} }\) …….. (2)
P = Q \(\frac {{ l }_{1}}{ { l }_{2} }\) ……… (3)
The bridge wire is soldered at the ends of the copper strips. Due to imperfect contact, some resistance might be introduced at the contact. These are called end resistances. This error can be eliminated, if another set of readings are taken with P and Q interchanged and the average value of P is found.
To find the specific resistance of the material of the wire in the coil P, the radius r and length l of the wire is measured. The specific resistance or resistivity r can be calculated using the relation
Resistance = ρ-\(\frac { l }{ A }\)
By rearranging the above equation, we get
ρ = Resistance x \(\frac { A }{ l }\) …….. (4)
If P is the unknown resistance, equation (4) becomes
ρ = P\(\frac {{ πr }^{2}}{ l }\).

Question 8.
How the emf of two cells are compared using potentiometer?
Answer:
Comparison of emf of two cells with a potentiometer:
To compare the emf of two cells, the circuit connections are made as shown in figure. Potentiometer wire CD is connected to a battery Bt and a key K in series. This is the primary circuit. The end C of the wire is connected to the terminal M of a DPDT (Double Pole Double Throw) switch and the other terminal N is connected to a jockey through a galvanometer G and a high resistance HR. The cells whose emf ξ₁ and ξ₂ to be compared are connected to the terminals M_1, N₁ and M₂, N₂ of the DPDT switch.

The positive terminals of Bt, ξ₁ and ξ₂ should be connected to the same end C. The DPDT switch is pressed towards M₁, N₁ so that cell ξ₁ is included in the secondary circuit and the balancing length l₁ is found by adjusting the jockey for zero deflection, Then the second cells ξ₂ is included in the circuit and the balancing length l₂ is determined. Let r be the resistance per unit length of the potentiometer wire and I be the current flowing through the wire.
we have.
ξ₁ = Irl₁ …… (1)
ξ₂ = Irl₂ ……. (2)
By dividing (1) by (2)
\(\frac {{ ξ }^{1}}{ { ξ }^{2} }\) = \(\frac {{ l }^{1}}{ { l }^{2} }\) ……. (3)
By including a rheostat (Rh) in the primary circuit, the experiment can be repeated several times by changing the current flowing through it.

Samacheer Kalvi 12th Physics Current Electricity Numerical Problems

Question 1.
The following graphs represent the current versus voltage and voltage versus current for the six conductors A,B,C,D,E and F. Which conductor has least resistance and which has maximum resistance?
Solution:

According to ohm’s law, V = IR
Resistance of conductor, R = \(\frac { V }{ I }\)

Graph-I:
Conductor A, I = 4 A and V = 2 V
R = \(\frac { V }{ I }\) = \(\frac { 2 }{ 4 }\) = 0.5 Ω
Conductor B, I = 3 A and V = 4 V
R = \(\frac { V }{ I }\) = \(\frac { 4 }{ 3 }\) = 1.33 Ω
Conductor C, I = 2 A and V = 5 V
R= \(\frac { V }{ I }\) = \(\frac { 5 }{ 2 }\) = 2.5 Q Ω.

Graph-II:
Conductor D, I = 2 A and V = 4 V
R = \(\frac { V }{ I }\) = \(\frac { 4 }{ 2 }\) = 2 Ω d
Conductor E, I = 4A and V = 3 V
R = \(\frac { V }{ I }\) = \(\frac { 3 }{ 4 }\) = 1.75 Ω
Conductor F, I = 5A and V = 2 V
R = \(\frac { V }{ I }\) = \(\frac { 2 }{ 5 }\) = 0.4 Ω
Conductor F has least resistance, R_F = 0.4 Ω,
Conductor C has maximum resistance, R_C = 2.5 G.

Question 2.
Lightning is very good example of natural current. In typical lightning, there is 10⁹ J energy transfer across the potential difference of 5 x 10⁷ V during a time interval of 0.2 s.

Using this information, estimate the following quantities (a) total amount of charge transferred between cloud and ground (b) the current in the lightning bolt (c) the power delivered in 0.2 s.
Solution:
During the lightning energy, E = 10⁹ J
Potential energy, V = 5 x 10⁷ V
Time interval, t = 0.2 s
(a) Amount of charge transferred between cloud and ground,
q = It

(b) Current in the lighting bolt, E = VIt
I = \(\frac { E }{ Vt }\) = \(\frac{10^{9}}{5 \times 10^{7} \times 0.2}\) = 1 × 10⁹ × 1^-7
I = 1 × 10²
I = 100 A
∴ q = It = 100 × 0.2
q = 20 C.

(c) Power delivered, E = VIt
P = VI = 5 × 10⁷ × 100 = 500 × 10⁷
I = 5 × 10⁹ W
P = 5 GW.

Question 3.
A copper wire of 10⁶ m² area of cross section, carries a current of 2 A. If the number of electrons per cubic meter is 8 x 10²⁸, calculate the current density and average drift velocity.
Solution:
Cross-sections area of copper wire, A = 10⁶ m²
I = 2 A
Number of electron, n = 8 x 10²⁸
Current density, J = \(\frac { 1 }{ A }\) = \(\frac { 2 }{{ 10 }^{-6}}\)
J = 2 × 10⁶ Am^-2
Average drift velocity, V_d = \(\frac { 1 }{ neA }\)
e is the charge of electron = 1.6 × 10^-9 C
V_d = \(\frac{2}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 10^{-6}}\) = \(\frac { 1 }{{ 64. × 10 }{3}}\)
V_d = 0.15625 × 10^-3
V_d = 15.6 × 10^-5 ms^-1

Question 4.
The resistance of a nichrome wire at 0 °C is 10 Ω. If its temperature coefficient of resistance is 0.004/°C, find its resistance at boiling point of water. Comment on the result.
Solution:
Resistance of a nichrome wire at 0°C, R₀ = 10 Ω
Temperature co-efficient of resistance, α = 0.004/°C
Resistance at boiling point of water, RT = ?
Temperature of boiling point of water, T = 100 °C
R_T=R₀ ( 1 + αT) = 10[1 + (0.004 x 100)]
R_T= 10(1 +0.4) = 10 x 1.4
R_T = 14 Ω
As the temperature increases the resistance of the wire also increases.

Question 5.
The rod given in the figure is made up of two different materials.

Both have square cross sections of 3 mm side. The resistivity of the first material is 4 x 10^-3 Ω.m and it is 25 cm long while second material has resistivity of 5 x 10^-3 Ω.m and is of 70 cm long. What is the resistivity of rod between its ends?
Solution:
Square cross section of side, a = 3 mm = 3 x 10^-3 m
Cross section of side, A = a² = 9 x 10⁶ m
First material:
Resistivity of the material, ρ₁ = 4 x 10^-3 Ωm
length, l₁ = 25 cm = 25 x 10^-2 m
Resistance of the lord, R₁ = \(\frac{\rho_{l} l_{l}}{A}\) = \(\frac{4 \times 10^{-3} \times 25 \times 10^{-2}}{9 \times 10^{-6}}\) = \(\frac{100 \times 10^{-5} \times 10^{6}}{9}\)
R₁ = 11.11 x 10¹ Ω

Second material:
Resistivity of the material, ρ₂ = 5 x 10^-3 Ωm
length, l₂ = 70 cm = 70 x 10^-2 m
Resistance of the rod, R₂ = \(\frac{\rho_{2} l_{2}}{A}\) = \(\frac{5 \times 10^{-3} \times 70 \times 10^{-2}}{9 \times 10^{-6}}\) = \(\frac{350 \times 10^{-5} \times 10^{6}}{9}\)
R₂ = 38.88 x 10¹
R₂ = 389 Ω
Total resistance between the ends f the rods
R = R₁ + R₂ = 111 + 389
= 500 Ω

Question 6.
Three identical lamps each having a resistance R are connected to the battery of emf as shown in the figure.

Suddenly the switch S is closed, (a) Calculate the current in the circuit when S is open and closed (b)
What happens to the intensities of the bulbs A,B and C. (c) Calculate the voltage across the three bulbs when S is open and closed (d) Calculate the power delivered to the circuit when S is opened and closed (e) Does the power delivered to the circuit decreases, increases or remain same?
Solution:
Resistance of the identical lamp = R
Emf of the battery = ξ
According to Ohm’s Law, ξ = IR
(a) Current:
When Switch is open— The current in the circuit. Total resistance of the bulb,
R_s = R₁ + R₂ + R₃
R₁ = R₂ = R₃ = R
R_s = R + R + R = 3R
∴ Current, I = \(\frac { ξ }{{ R }_{s}}\)
⇒ I₀ = \(\frac { ξ }{ 3R }\)
Switch is closed— The current in the circuit. Total resistance of the bulb,
R_s = R + R = 2R
Current I = \(\frac { ξ }{{ R }_{s}}\)
I_c = \(\frac { ξ }{ 2R }\).

(b) Intensity:
When switch is open — All the bulbs glow with equal intensity.
When switch is closed — The intensities of the bulbs A and B equally increase. Bulb C will not glow since no current pass through it.

(c) Voltage across three bulbs:
When switch is open — Voltage across bulb A, V_A = I₀ R = \(\frac { ξ }{ 3R }\) x R = \(\frac { ξ }{ 3 }\)
similarly:
Voltage across bulb B, V_B = \(\frac { ξ }{ 3 }\)
Voltage across bulb C, V_C = \(\frac { ξ }{ 3 }\)
When switch is closed— Voltage across bulb A, V_A = I_cR = \(\frac { ξ }{ 2R }\) x \(\frac { ξ }{ 2 }\)
similarly:
Voltage across bulb B, V_B = I_cR \(\frac { ξ }{ 2 }\)
Voltage across bulb C, V_C = 0

(d) Power delivered to the circuit,
When switch is opened — Power P, = VI

When switch is closed — Power P, = VI

(e) Total power delivered to circuit increases.

Question 7.
The current through an element is shown in the figure. Determine the total charge that pass through the element at
(a) t = 0 s
(b) t = 2 s
(c) t = 5 s

Solution :
Rate of flow of charge is called current, I = \(\frac { dq }{ dt }\)
Total charge pass through element, dq = Idt
(a) At t= 0 s, I = 10 A
dq = Idt= 10 x 0 = 0 C.

(b) At t = 2 s, I = 5 A
dq = Idt = 5 x 2 = 10 C.

(c) At t = 5 s, I = 0
dq = Idt = 0 x 5 = 0 C.

Question 8.
An electronics hobbyist is building a radio which requires 150 Ω in her circuit, but she has only 220 Ω, 79 Ω and 92 Ω resistors available. How can she connect the available resistors to get desired value of resistance?
Solution:
Required effective resistance = 150 Ω
Given resistors of resistance, R₁ = 220 Ω, R₂ = 79 Ω, R₃ = 92 Ω
Parallel combination R₁ and R₂
\(\frac { 1 }{{ R }_{p}}\) = \(\frac { 1 }{{ R }_{1}}\) + \(\frac { 1 }{{ R }_{2}}\) = \(\frac { 1 }{ 220 }\) + \(\frac { 1 }{ 79 }\) = \(\frac { 79 + 220 }{ 220 × 79 }\)
R_p = 58 Ω

Parallel combination R_p and R₃
R_s = R_p + R₃ = 58 + 92
R_s = 150 Ω
Parallel combination of 220 Ω and 79 Ω in series with 92 Ω.

Question 9.
A cell supplies a current of 0.9 A through a 2 Ω resistor and a current of 0.3 A through a 7 Ω resistor. Calculate the internal resistance of the cell.
Solution:
Current from the cell, I₁ = 0.9 A
Resistor, R₁ = 2 Ω
Current from the cell, I₂ = 0.3 A
Resistor, R₂ 7 Ω
Internal resistance of the cell, r = ?
Current in the circuit I₁ = \(\frac { ξ }{{ r + R }_{1}}\)
ξ = I₁ (r + R₁) …… (1)
Current in the circuit, I₂ = \(\frac { ξ }{{ r + R }_{2}}\) …… (2)
Equating equation (1) and (2),
I₁r + I₁R₁ = I₂R₂ + I₂r
(I₁ – I₂)r = I₂R₂ – I₁R₁

r = 0.5 Ω.

Question 10.
Calculate the currents in the following circuit.

Solution:
Applying Kirchoff’s 1^st Law at junction B

I₁ – I₁ – I₃ = 0
I₃ = I₁ – I₂ …… (1)
Applying Kirchoff’s II^nd Law at junction in ABEFA
100 I₃ + 100 I₁ = 15
100 (I₃ + I₁) = 15
100 I₁ – 100 I₂ + 100 I₁ =15
200 I₁ – 100 I₂ = 15 …… (2)
Applying Kirchoff’s IInd Law at junction in BCDED
-100I₂ + 100 I₃ = 9
-100 I₂+ 100(I₁ – I₂) = 9
100I₁ – 200 I₂ = 9
Solving equating (2) and (3)

Substitute I₁ values in equ (2)
200(0.07) – 100 I₂ = 15
14 – 100 I₂ = 15
– 100 I₂ = 15 – 14
I₂ = \(\frac { -1 }{ 100 }\)
I₂ = -0.01A
Substitute I₁ and I₂ value in equ (1), we get
I₃ = I₁ – I₂ = 0.07 – (-0.01)
I₃ = 0.08 A.

Question 11
A potentiometer wire has a length of 4 m and resistance of 20 Ω. It is connected in series with resistance of 2980 Ω and a cell of emf 4 V. Calculate the potential along the wire.
Solution:
Length of the potential wire, l = 4 m
Resistance of the wire, r = 20 Ω
Resistance connected series with potentiometer wire, R = 2980 Ω
Emf of the cell, ξ = 4 V
Effective resistance, R_s = r + R = 20 + 2980 = 3000 Ω
Current flowing through the wire, I = \(\frac { ξ }{{ r + R }_{s}}\) = \(\frac { 4 }{ 3000 }\)
I = 1.33 x 10^-3 A
Potential drop across the wire, V = Ir
= 1.33 x 10^-33 x 20
V = 26.6 x 10^-3 V

= 6.65 x 10^-3
Potential garadient = 0.66 x 10^-2 Vm^-1

Question 12.
Determine the current flowing through the galvanometer (G) as shown in the figure.

Solution:
Current flowing through the circuit, I = 2A
Applying Kirchoff’s I^st law at junction P, I = I₁ + I₂
I₂ = I – I₁ …(1)
Applying Kirchoff’s I^nd law at junction PQSP
5 I₁ + 10 I_g – 15 I₂ = 0
5 I₁ + 10 I_g -15(I – I₁) = 0
20 I₁ + 10 I_g = 15 I
20 I₁ + 10 I_g = 15 x 2
÷ by 10 21 I₁ + I_g = 3 … (2)
Applying Kirchoff’s II^nd law at junction QRSQ
10(I₁ – I_g) – 20(I – I₁ – I_g) – 10 I_g = 0
10 I₁ – 10_g – 20(I – I₁ – I_g) – 10 I_g = 0
10 I₁ – 10_g – 20 I + 20 I₁ -20I_g – 10 I_g = 0
30 I ₁ – 40 I_g = 20 I
÷ by 10 ⇒ 3 I₁ – 4 I_g = 20 I
3 I₁ – 4 I_g = 2 I
3 I₁ – 4 I_g = 2 x 2 = 4

Question 13.
Two cells each of 5 V are connected in series across a 8 Ω resistor and three parallel resistors of 4 Ω, 6 SI and 12 Ω. Draw a circuit diagram for the above arrangement. Calculate (i) the current drawn from the cell (ii) current through each resistor.
Solution:
V₁ = 5 V; V₂ = 5 V
R₁ = 8 Ω; R₂ = 4 Ω; R₃ = 6 Ω; R₄ = 12 Ω
Three resistors R₂, R₃ and R₄ are connected parallel combination

Resistors R₁ and R_p are connected in series combination
R_s = R₁ +R_p = 8 + 2 = 10
R_s = 10Ω
Total voltage connected series to the circuit
V = V₁ + V₂
= 5 + 5 = 10
V = 10 V.
(i) Current through the circuit, I = \(\frac { V }{{ R }_{s}}\) = \(\frac { 10 }{ 10 }\)
I = 1 A
Potential drop across the parallel combination,
V’ = IR_p = 1 x 2
V’ 2 V

(ii) Current in 4 Ω resistor, I = \(\frac { V’ }{{ R }_{2}}\) = \(\frac { 2 }{ 4 }\) = 0.5 A
Current in 6 Ω resistor, I = \(\frac { V’ }{{ R }_{3}}\) = \(\frac { 2 }{ 6 }\) = 0.33 A
Current in 12 Ω resistor, I = \(\frac { V’ }{{ R }_{4}}\) = \(\frac { 2 }{ 12 }\) = 0.17 A

Question 14.
Four light bulbs P, Q, R, S are connected in a circuit of unknown arrangement. When each bulb is removed one at a time and replaced, the following behavior is observed.

Solution:

Question 15.
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63 cm, what is the emf of the second cell?
Solution:
Emf of the cell₁, ξ₁ = 1.25 V
Balancing length of the cell, l₁ = 35 cm = 35 x 10^-2 m
Balancing length after interchanged, l₂ = 63 cm = 63 x 10^-2 m
Emf of the cell₁, ξ₂ = ?
The ration of emf’s, \(\frac {{ ξ }_{1}}{{ ξ }_{2}}\) = \(\frac {{ l }_{1}}{{ l }_{2}}\)
The ration of emf’s, ξ₂ = ξ₁ = \(\left( \frac { { l }_{ 2 } }{ { l }_{ 1 } } \right) \)
= 1. 25 x \(\left( \frac { { 63×10 }^{ -2 } }{ { 35×10 }^{ -2 } } \right) \) = 12.5 x 1.8
ξ₂ = 2.25 V.

Samacheer Kalvi 12th Physics Current Electricity Additional Questions Solved

I. Choose the Correct Answer

Question 1.
When current I flows through a wire, the drift velocity of the electrons is v. When current 21 flows through another wire of the same material having double the length and area of cross-section, the drift velocity of the electrons will be-
(a) \(\frac { v }{ 4 }\)
(b) \(\frac { v }{ 2 }\)
(c) v
(d) 2 v
Answer:
(c) v
Hint:
V_d = \(\frac { 1 }{ nAe }\); v’_d = \(\frac { 2I }{ (2A)ne }\) = v_d

Question 2.
A copper wire of length 2 m and area of cross-section 1.7 x 10^-6 m² has a resistance of 2 x 10^-2 Ω. The resistivity of copper is
(a) 1.7 x 10^-8 Ωm
(b) 1.9 x ^-8 Ωm
(c) 2.1 x 10^-7 Ωm
(d) 2.3 x 10^-7 Ωm
Answer:
(a) 1.7 x 10^-8 Ωm
Hint:
Resistivity, ρ = \(\frac { RA }{ l }\) = \(\frac{2 \times 10^{-2} \times 1.7 \times 10^{-6}}{2}\) = 1.7 x 10^-8 Ωm.

Question 3.
If the length of a wire is doubled and its cross-section is also doubled, then its resistance will
(a) become 4 times
(b) become 1 / 4
(c) becomes 2 times
(d) remain unchanged
Answer:
(d) remain unchanged

Question 4.
A 10 m long wire of resistance 20 Ω is connected in series with a battery of emf 3 V and a resistance of 10 Ω. The potential gradient along the wire in volt per meter is
(a) 6.02
(b) 0.1
(c) 0.2
(d) 1.2
Answer:
(c) 0.2
Hint:
Potential difference across the wire = \(\frac { 20 }{ 3 }\) x 3 = 2 V
Potential gradient = \(\frac { v }{ l }\) = \(\frac { 2 }{ 10 }\) = 0.2 V/m

Question 5.
The resistivity of a wire
(a) varies with its length
(b) varies with its mass
(c) varies with its cross-section
(d) does not depend on its length, cross-section and mass.
Answer:
(d) does not depend on its length, cross-section and mass.

Question 6.
The electric intensity E, current density and conductivity a are related as
(a) j = σE
(b) j = \(\frac { E }{ σ }\)
(c) JE = s
(d) j = σ²E
Answer:
(a) j = σE

Question 7.
For which of the following dependences of drift velocity V_d on electric field E, is Ohm’s law obeyed?
(a) v_d ∝ E
(b) v_d ∝ E²
(c) v_d ∝ √E
(d) v_d = constant
Answer:
(a) v_d ∝ E
Hint:
v_d = \(\frac { 1 }{ nAe }\) = \(\frac { j }{ ne }\) = \(\left( \frac { σ }{ ne } \right) \)E ⇒ v_d ∝ E.

Question 8.
A cell has an emf of 1.5 V. When short circuited, it gives a current of 3A. The internal resistance of the cell is .
(a) 0.5 Ω
(b) 2.0 Ω
(c) 4.5 Ω
(d) \(\frac { 1 }{ 4.5 }\) Ω
Answer:
(a) 0.5 Ω
Hint:
r = \(\frac { ξ }{ I }\) = \(\frac { 1.5 }{ 3 }\) = 0.5 Ω.

Question 9.
The resistance, each of 1 Ω, are joined in parallel. Three such combinations are put in series. The resultant resistance is
(a) 9 Ω
(b) 3 Ω
(c) 1 Ω
(d) \(\frac { 1 }{ 3 }\) Ω
Answer:
(c) 1 Ω
Hint:
R_p = 1 + 1 + 1 = 3 Ω;
\(\frac { 1 }{{ R }_{s}}\) = \(\frac { 1 }{ 3 }\) + \(\frac { 1 }{ 3 }\) + \(\frac { 1 }{ 3 }\) = \(\frac { 3 }{ 3 }\) = 1
⇒ R_s = 1 Ω.

Question 10.
Constantan is used for making standard resistance because it has
(a) high resistivity
(b) low resistivity
(c) negligible temperature coefficient of resistance
(d) high melting point
Answer:
(c) negligible temperature coefficient of resistance

Question 11.
Kirchhoff’s two laws for electrical circuits are magnifestations of the conservation of
(a) charge only
(b) both energy and momentum
(c) energy only
(d) both charge and energy
Answer:
(d) both charge and energy

Question 12.
The resistance R₀ and R_t of a metallic wire at temperature 0° C and t° C are related as (a is the temperature co-efficient of resistance).
(a) R_t = R₀(1 + αt)
(b) R_t = R₀(1 – αt)
(c) R_t = R₀(1 + αt)²
(d) R_t = R₀(1 – αt) ²
Answer:
(a) R_t = R₀(1 + αt)

Question 13.
A cell of emf 2 V and internal resistance 0.1 Ω is connected with a resistance of 3.9 Ω. The voltage across the cell terminals will be
(a) 0.5 V
(b) 1.9 V
(c) 1.95 V
(d) 2 V
Answer:
(c) 1.95 V
Hint:
V = \(\frac { ER }{ R + r }\) = \(\frac { 2 x 3.9 }{ 3.9 + 0.1 }\) = 1.95 V.

Question 14.
A flow of 10⁷ electrons per second in a conduction wire constitutes a current of
(a) 1.6 x 10^-26 A
(b) 1.6 x 10¹² A
(c) 1.6 x 10^-12 A
(d) 1.6 x 10²⁶ A
Answer:
(c) 1.6 x 10^-12 A
Hint:
I = \(\frac { Q }{ t }\) = \(\frac{10^{7} \times 1.6 \times 10^{-19}}{1}\) = 1.6 x 10¹² A.

Question 15.
Sensitivity of a potentiometer can be increased by
(a) increasing the emf of the cell
(b) increasing the length of the wire
(c) decreasing the length of the wire
(d) none of the above
Answer:
(b) increasing the length of the wire

Question 16.
Potential gradient is defined as
(a) fall of potential per unit length of the wire.
(b) fall of potential per unit area of the wire.
(c) fall of potential between two ends of the wire.
(d) none of the above.
Answer:
(a) fall of potential per unit length of the wire.

Question 17.
n equal resistors are first connected in series and then in parallel. The ratio of the equivalent resistance in two cases is
(a) n
(b) \(\frac { 1 }{{ n }^{2}}\)
(c) n²
(d) \(\frac { 1 }{ n }\)
Answer:
(c) n²
Hint:
Required ratio = \(\frac{n \mathrm{R}}{\left(\frac{\mathrm{R}}{n}\right)}\) = n(c) n²

Question 18.
A galvanometer is converted into an ammeter when we connect a
(a) high resistance in series
(b) high resistance in parallel
(c) low resistance in series
(d) low resistance in parallel
Answer:
(d) low resistance in parallel

Question 19.
The reciprocal of resistance is
(a) conductance
(b) resistivity
(c) conductivity
(d) none of the above
Answer:
(a) conductance

Question 20.
A student has 10 resistors, each of resistance r. The minimum resistance that can be obtained by him using these resistors is
(a) 10r
(b) \(\frac { r }{ 10 }\)
(c) \(\frac { r }{ 100 }\)
(d) \(\frac { r }{ 5 }\)
Answer:
(b) \(\frac { r }{ 10 }\)

Question 21.
The drift velocity of electrons in a wire of radius r is proportional to
(a) r
(b) r²
(c) r³
(d) none of the above
Answer:
(d) none of the above

Question 22.
Kirchhoff’s first law, i.e. ∑I = 0 at a junction deals with conservation of
(a) charge
(b) energy
(c) momentum
Answer:
(a) charge

Question 23.
The resistance of a material increases with temperature. It is a
(a) metal
(b) insulator
(c) semiconductor
(d) semi-metal
Answer:
(a) metal

Question 24.
Five cells, each of emf E, are joined in parallel. The total emf of the combination is
(a) 5E
(b) \(\frac { E }{ 5 }\)
(c) E
(d) \(\frac { 5E }{ 2 }\)
Answer:
(c) E

Question 25.
A carbon resistance has colour bands in order yellow, brown, red. Its resistance is
(a) 41 Ω
(b) 41 x 10² Ω
(c) 41 x 10³ Ω
(d) 4.2 Ω
Answer:
(b) 41 x 10² Ω

Question 26.
The conductivity of a superconductor is
(a) infinite
(b) very large
(c) very small
(d) zero
Answer:
(a) infinite

Question 27.
The resistance of an ideal voltmeter is
(a) zero
(b) very high
(c) very low
(d) infinite
Answer:
(d) infinite

Question 28.
Carriers of electric current in superconductors are
(a) electrons
(b) photons
(c) holes
Answer:
(c) holes

Question 29.
Potentiometer measures potential more accurately because
(a) It measure potential in the open circuit.
(b) It uses sensitive galvanometer for null detection.
(c) It uses high resistance potentiometer wire.
(d) It measures potential in the closed circuit.
Answer:
(a) It measure potential in the open circuit.

Question 30.
Electromotive force is most closely related to
(a) electric field
(b) magnetic field
(c) potential difference
(d) mechanical force
Answer:
(c) potential difference

Question 31.
The capacitance of a pure capacitor is 1 farad. In DC circuit, the effective resistance will be
(a) zero
(b) infinite
(c) 1 Ω
(d) 0.5 Ω
Answer:
(b) infinite

Question 32.
The resistance of an ideal ammeter is
(a) zero
(b) small
(c) high
(d) infinite
Answer:
(a) zero

Question 33.
A milliammeter of range 10 mA has a coil of resistance 1 Ω. To use it as a voltmeter of range 10 V, the resistance that must be connected in series with it is
(a) 999 Ω
(b) 1000 Ω
(c) 9 Ω
(d) 99 Ω
Answer:
(a) 999 Ω
Hint:
R = \(\frac { V }{{ I }_{g}}\)-R_g = \(\frac { 10 }{{ 10 × 10 }^{-3}}\)-1 = 999 Ω.

Question 34.
A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. The current in the circuit is. 0.5 A. The terminal voltage of the battery when the circuit is closed is
(a) 10 V
(b) zero
(c) 8.5 V
(d) 1.5 V
Answer:
(c) 8.5 V
Hint:
V = ξ – Ir = 10 – (0.5 x 3) = 8.5 V.

Question 35.
Good resistance coils are made of
(a) copper
(b) manganin
(c) iron
(d) aluminium
Answer:
(b) manganin

Question 36.
A wire of resistance R is stretched to three times its original length. The new resistance is
(a) 3R
(b) 9R
(c) R/3
(d) R/9
Answer:
(b) 9R

Question 37.
n resistances, each of r Ω, when connected in parallel give an equivalent resistance of R Ω. If these resistances were connected in series, the combination would have a resistance in horns equal to
(a) n²R
(b) \(\frac { R }{{n}^{ 2 }}\)
(c) \(\frac { R }{ n }\)
(d) nR
Answer:
(a) n²R
Hint:
Resistance in parallel combination, R = \(\frac { r }{ n }\) ⇒ r = Rn
Resistance in series combination, R’ = nr = n²R

Question 38.
When a wire of uniform cross-section, having resistance R, is bent into a complete circle, the resistance between any two of diametrically opposite points will be
(a) \(\frac { R }{ 8 }\)
(b) \(\frac { R }{ 2 }\)
(c) 4R
(d) \(\frac { R }{ 4 }\)
Answer:
(d) \(\frac { R }{ 4 }\)
Hint:
It becomes two resistors each of (d) \(\frac { R }{ 2 }\), connected in parallel.

Question 39.
A steady current is set up in a metallic wire of non uniform cross-section. How is the rate of flow K of electrons related to the area of cross-section A?
(a) K is independent of A
(b) K ∝ A
(c) K ∝ A^-1
(d) K ∝ A²
Answer:
(c) K ∝ A^-1

Question 40.
Ohm’s Law is not obeyed by
(a) electrolytes
(b) discharge tubes
(c) vacuum tubes
(d) all of these
Answer:
(d) all of these

Question 41.
Which of the following has negative temperature coefficient of resistance?
(a) Copper
(b) Aluminium
(c) Germanium
(d) Iron
Answer:
(c) Germanium

II. Fill in the blanks

Question 1.
The material through which electric charge can flow easily is ……………….
Answer:
Copper.

Question 2.
A toaster operating at 240 V has a resistance of 120 Ω. The power is ……………….
Answer:
480 W.

Question 3.
In the case of insulators, as the temperature decreases, resistivity ……………….
Answer:
Increases.

Question 4.
When n resistors of equal resistance (R) are connected in series, the effective resistance is ……………….
Answer:
nR.

Question 5.
The net flow of charge at any point in the conductor is ……………….
Answer:
Zero.

Question 6.
The flow of free electrons in a conductor constitutes ……………….
Answer:
Electric current.

Question 7.
The rate of flow of charge through any wire is called ……………….
Answer:
Current.

Question 8.
The drift velocity acquired per unit electric field is the ……………….
Answer:
Mobility.

Question 9.
The reciprocal of resistance is ……………….
Answer:
Conductance.

Question 10.
The unit of specific resistance is ……………….
Answer:
Ohm meter.

Question 11.
The reciprocal of electrical resistivity is called ……………….
Answer:
Electrical conductivity.

Question 12.
With increase in temperature the resistivity of metals ……………….
Answer:
Increases.

Question 13.
The resistivity of insulators is of the order of ……………….
Answer:
10⁸ 10¹⁴ Ωm.

Question 14.
The resistivity of semiconductors is of the order of ……………….
Answer:
10^-2 -10² Ωm.

Question 15.
The materials which conduct electricity at zero resistance are called ……………….
Answer:
Superconductors.

Question 16.
Conductors turn into superconductors at ……………….
Answer:
Low temperatures.

Question 17.
The resistance of superconductors is ……………….
Answer:
Zero.

Question 18.
The phenomenon of superconductivity was discovered by ……………….
Answer:
Kammerlingh onnes.

Question 19.
Mercury becomes a superconductor at ……………….
Answer:
4.2 K.

Question 20.
With increase of temperature, resistance of conductors ……………….
Answer:
increases

Question 21.
In insulators and semiconductors, as temperature increases, resistance ……………….
Answer:
Decreases.

Question 22.
A material with a negative temperature coefficient is called a ……………….
Answer:
Thermistor.

Question 23.
The temperature coefficient for alloys is ……………….
Answer:
Low.

Question 24.
The electric current in an external circuit flows from the ……………….
Answer:
Positive to negative terminal.

Question 25.
In the electrolyte of the cell, current flows from ……………….
Answer:
Negative to positive terminal.

Question 26.
A freshly prepared cell has ………………. internal resistance.
Answer:
Low.

Question 27.
Kirchhoff’s first law is ……………….
Answer:
Current law.

Question 28.
The current law states that the algebraic sum of the currents meeting at any junction in a circuit is ……………….
Answer:
Zero.

Question 29.
Current law is a consequence of conservation of ……………….
Answer:
Charges.

Question 30.
Kirchhoff’s second law is ……………….
Answer:
Voltage law.

Question 31.
Kirchhoff’s second law is a consequence of conservation of ……………….
Answer:
Energy.

Question 32.
Wheatstone bridge is an application of ……………….
Answer:
Kirchhoff’s Law.

Question 33
………………. is a form of Wheatstone’s bridge.
Answer:
Metre bridge.

Question 34.
The temperature coefficient of manganin wire is ……………….
Answer:
Low.

Question 35
………………. is an instrument to measure potential difference.
Answer:
Potentiometer.

Question 36.
Unit of electrical energy is ……………….
Answer:
Joule.

Question 37.
An instrument to measure electrical power consumed is ……………….
Answer:
Watt meter.

Question 38.
………………. first introduced the electrochemical battery
Answer:
Volta.

Question 39.
Charging is a process of reproducing ……………….
Answer:
Active materials.

III. Match the following

Question 1.

Answer:
(i) → (b)
(ii) → (a)
(iii) → (d)
(iv) →(c)

Question 2.

Answer:
(i) → (b)
(ii) → (c)
(iii) → (d)
(iv) → (a)

Question 3.

Answer:
(i) → (d)
(ii) → (c)
(iii) → (b)
(iv) → (a)

Question 4.

Answer:
(i) → (b)
(ii)→ (d)
(iii) → (a)
(iv) → (c)

IV.Assertion and reason type

(a) If both assertion and reason are true and the reason in the correct explanation of the assertion.
(b) If both assertion and reason are true but the reason is not correct explanation of the assertion.
(c) If assertion is true but reason is false.
(d) If the assertion and reason both are false.
(e) If assertion is false but reason is true.

Question 1.
Assertion: Fuse wire must have high resistance and low melting point.
Reason: Fuse is used for small current flow only
Answer:
(c) If assertion is true but reason is false.

Question 2.
Assertion: In practical application, power rating of resistance is not important.
Reason: Property of resistance remains same even at high temperature
Answer:
(d) If the assertion and reason both are false.

Question 3.
Assertion: Electric appliances with metallic body e.g. heaters, presses, etc, have three pin connections, whereas an electric bulb has two pins.
Reason: Three pin connection reduce heating of connecting cables.
Answer:
(c) If assertion is true but reason is false.

Samacheer Kalvi 12th Physics Current Electricity Short Answer Questions

Question 1.
Define current?
Answer:
Current is defined as a net charge Q passes through any cross section of a conductor in time t
then, I = \(\frac { Q }{ t }\).

Question 2.
Define instantaneous current?
Answer:
The instantaneous current I is defined as the limit of the average current, as ∆t → 0.

Question 3.
What is resistance? Give its unit?
Answer:
The resistance is the ratio of potential difference across the given conductor to the current passing through the conductor V.
R = \(\frac { V }{ I }\).

Question 4.
What is meant by transition temperature?
Answer:
The resistance of certain materials become zero below certain temperature T_c. This temperature is known as critical temperature or transition temperature.

Question 5.
What is Joule’s heating effect?
Answer:
When current flows through a resistor, some of the electrical energy delivered to the resistor is converted into heat energy and it is dissipated. This heating effect of current is known as Joule’s heating effect.

Question 6.
What is meant by thermoelectric effect?
Answer:
Conversion of temperature differences into electrical voltage and vice versa is known as thermoelectric effect.

Question 7.
What is a thermopile? On what principle does it work?
Answer:
Thermopile is a device used to detect thermal radiation. It works on the principle of seebeck effect.

Question 8.
What is a thermistor?
Answer:
A material with a negative temperature coefficient is called a thermistor.
Eg:

1. Insulator
2. Semiconductor.

Question 9.
State principle of potentiometer?
Answer:
The principle of potentiometer states that the emf of the cell is directly proportional to its balancing length.
ξ ∝ l
ξ = Irl.
Samacheer Kalvi 12th Physics Current Electricity Long Answer Questions

Question 1.
Explain the concept of colour code for carbon resistors.
Answer:
Color code for Carbon resistors:
Carbon resistors consists of a ceramic core, on which a thin layer of crystalline carbon is deposited. These resistors are inexpensive, stable and compact in size. Color rings are used to indicate the value of the resistance according to the rules.

Three coloured rings are used to indicate the values of a resistor: the first two rings are significant figures of resistances, the third ring indicates the decimal multiplier after them. The fourth color, silver or gold, shows the tolerance of the resistor at 10% or 5%. If there is no fourth ring, the tolerance is 20%. For the resistor, the first digit = 5 (green), the second digit = 6 (blue), decimal multiplier = 10³ (orange) and tolerance = 5% (gold). The value of resistance = 56 x 10³ Q or 56 kΩ with the tolerance value 5%.

Question 2.
Explain in details of temperature dependence of resistivity.
Answer:
Temperature dependence of resistivity:
The resistivity of a material is dependent on temperature. It is experimentally found that for a wide range of temperatures, the resistivity of a conductor increases with increase in temperature according to the expression,
ρ_T = ρ₀ [1 + α(T -T₀)] ……. (1)
where ρ_T is the resistivity of a conductor at T₀C, ρ₀ is the resistivity of the conductor at some reference temperature To (usually at 20°C) and a is the temperature coefficient of resistivity. It is defined as the ratio of increase in resistivity per degree rise in temperature to its resistivity atT₀.
From the equation (1), we can write
ρ_T – ρ₀ = αρ₀ (T -T₀)
∴ α = \(\frac{\rho_{\mathrm{T}}-\rho_{0}}{\rho_{0}\left(\mathrm{T}-\mathrm{T}_{0}\right)}\) = \(\frac{\Delta p}{\rho_{0} \Delta T}\)
where ∆_ρ = ρ_T – ρ₀ is change in resistivity for a change in temperature ∆T = T – T₀. Its unit is per °C.

1. α of conductors:
For conductors a is positive. If the temperature of a conductor increases, the average kinetic energy of electrons in the conductor increases. This results in more frequent collisions and hence the resistivity increases. Even though, the resistivity of conductors like metals varies linearly for wide range of temperatures, there also exists a nonlinear region at very low temperatures. The resistivity approaches some finite value as the temperature approaches absolute zero. As the resistance is directly proportional to resistivity of the material, we can also write the resistance of a conductor at temperature T °C as
R_T -R₀ = [1 + α(T -T₀)] ……. (2)
The temperature coefficient can be also be obtained from the equation (2), +

where ∆R = R_T -R₀ is change in resistance during the change in temperature ∆T = T – T₀

2. α of semiconductors:
For semiconductors, the resistivity decreases with increase in temperature. As the temperature increases, more electrons will be liberated from their atoms (Refer unit 9 for conduction in semi conductors). Hence the current increases and therefore the resistivity decreases. A semiconductor with a negative temperature coefficient of resistance is called a thermistor.
We can understand the temperature dependence of resistivity in the following way. The electrical conductivity, σ = \(\frac{n e^{2} \tau}{m}\) \(\frac{m}{n e^{2} \tau}\). As the resistivity is inverse of σ, it can be written as,
σ = \(\frac{n e^{2} \tau}{m}\) \(\frac{m}{n e^{2} \tau}\) …… (4)
The resistivity of materials is

1. inversely proportional to the number density (n) of the electrons
2. inversely proportional to the average time between the collisions (τ).

In metals, if the temperature increases, the average time between the collision (τ) decreases and n is independent of temperature. In semiconductors when temperature increases, n increases and τ decreases, but increase in n is dominant than decreasing x, so that overall resistivity decreases.

Question 3.
Explain the effective internal resistance of cells connected in series combination. Compare the results to the external resistance.
Answer:
Cells in series Several cells can be connected to form a battery. In series connection, the negative terminal of one cell is connected to the positive terminal of the second cell, the negative terminal of second cell is connected to the positive terminal of the third cell and so on. The free positive terminal of the first cell and the free negative terminal of the last cell become the terminals of the battery.

Suppose n cells, each of emf ξ volts and internal resistance r ohms are connected in series with an external resistance R.
The total emf of the battery = nξ
The total resistance in the circuit = nr + R
By Ohm’s law, the current in the circuit is

Case (a) If r << R, then,
I = \(\frac { nξ }{ R }\)nI₁ ≈ nI₁ ……. (2)
where, I, is the current due to a single cell \(\left(\mathrm{I}_{1}=\frac{\xi}{\mathrm{R}}\right)\)
Thus, if r is negligible when compared to R the current supplied by the battery is n times that supplied by a single cell.
Case (b) If r >> R, I = \(\frac { nξ }{ nr }\) ≈ \(\frac { ξ }{ R }\) …….. (3)
It is the current due to a single cell. That is, current due to the whole battery is the same as that due to a single cell and hence there is no advantage in connecting several cells. Thus series connection of cells is advantageous only when the effective internal resistance of the cells is negligibly small compared with R.

Question 4.
Explain the effective internal resistance of cells connected in parallel combination. Compare the results to the external resistance.
Answer:
Cells in parallel: In parallel connection all the positive terminals of the cells are connected to one point and all the negative terminals to a second point. These two points form the positive and negative terminals of the battery.
Let n cells be connected in parallel between the points A and B and a resistance R is connected between the points A and B. Let ξ, be the emf and r the internal resistance of each cell.

The equivalent internal resistance of the battery is \(\frac { 1 }{{ r }_{eq}}\) = \(\frac { 1 }{ r }\) + \(\frac { 1 }{ r }\) + ….. \(\frac { 1 }{ r }\) (n terms) = \(\frac { n }{ r }\).
So r_eg = \(\frac { r }{ n }\) and the total resistance in the circuit = R + \(\frac { r }{ n }\). The total emf is the potential difference between the points A and B, which is equal to ξ. The current in the circuit is given by

Case (a) If r << R, then,
I = \(\frac { nξ }{ R }\) = nI₁ …….. (2)
where II is the current due to a single cell and is equal to \(\frac { ξ }{ R }\) when R is negligible. Thus, the
current through the external resistance due to the whole battery is n times the current due to a single cell.

Case (b) If r << R. I = \(\frac { ξ }{ R }\) …… (3)
The above equation implies that current due to the whole battery is the same as that due to a single cell. Hence it is advantageous to connect cells in parallel when the external resistance is very small compared to the internal resistance of the cells.

Samacheer Kalvi 12th Physics Current Electricity Numerical Problems

Question 1.
Show that one ampere is equivalent to a flow of 6.25 x 10¹⁸ elementary charges per second.
Solution:
Here I = 1 A, t= 1s, e = 1.6 x 10^-19 C
As I = \(\frac { q }{ t }\) = \(\frac { ne }{ t }\)
Number of electrons, n = \(\frac { It }{ e }\) = \(\frac { 1 × 1 }{{ 1.6 × 10 }^{19}}\) = 6.25 × 10¹⁸

Question 2.
Calculate the resistivity of a material of a wire 10 m long. 0.4 mm in diameter and having a resistance of 2.0 Ω.
Solution:
Here l = 10 cm, r = 0.2 mm = 0.2 x 10^-3 m, R = 2 Ω
\(\left[ r\quad =\quad \frac { d }{ 2 } \right] \)

= 2.513 x 10^-8 Ω.

Question 3.
A wire of 10 ohm resistance is stretched to thrice its original length. What will be its (i) new resistivity and (ii) new resistance?
Solution:
(i) Resistivity p remains unchanged because it is the property of the material of the wire.
(ii) In both cases, volume of wire is same, so

Question 4.
A copper wire has a resistance of 10 Ω. and an area of cross-section 1 mm². A potential difference of 10 V exists across the wire. Calculate the drift speed of electrons if the number of electrons per cubic metre in copper is 8 x 10²⁸ electrons.
Solution:
Here, R = 10 Ω, A = 1 mm² = 10^-6 m², V = 10 V, n = 8 x 10²⁸ electrons/m³
Now,
I = enAv_d
∴ \(\frac { V }{ R }\) = enAv_d (or) v_d = \(\frac { V }{ enAR }\)
= \(\frac{10}{1.6 \times 10^{-19} \times 8 \times 10^{28} \times 10^{-6} \times 10}\) = 0.078 x 10^-3 ms^-1
= 0.078 x ms^-1

Question 5.
(i) At what temperature would the resistance of a copper conductor be double its resistance at 0°C.
(ii) Does this temperature hold for all copper conductors regardless of shape and size? Given a for Cu = 3.9 x 10^-3 °C^-1.
Solution:

Thus the resistance of copper conductor becomes double at 256 °C.
(ii) Since a does not depend on size and shape of the conductor. So the above result holds for all copper conductors.

Question 6.
Find the value of current I in the circuit shown in figure.
Solution:

In the circuit, the resistance of arm ACB (30 + 30 = 60 Ω) is the parallel with the resistance of arm AB (= 30 Ω).
Hence, the effective resistance of the circuit is
R = \(\frac { 30 × 60 }{ 30 + 60 }\) = 20 Ω
Current, I = \(\frac { V }{ R }\) = \(\frac { 2 }{ 20 }\) = 0.1 A.

Common Errors and Its Rectifications:

Common Errors:

1. Sometimes students think that charge and current are same.
2. In doing calculation part students can’t give the importance to mention the units.
3. They may confuse the parallel and series network of the resistance.

Rectifications:

1. Charge q = ne Current I = q/t
2. Unit is very importance to the every physical quantities.
3. If the resistors are series, their resultant is sum of the all the reciprocal of individual resistance. If the resistors are parallel their resultant is sum of the individual resistance.

We believe that the shared knowledge about Tamilnadu State Board Solutions for 12th Physics Chapter 2 Current Electricity Questions and Answers learning resource will definitely guide you at the time of preparation. For more details visit Tamilnadu State Board Solutions for 12th Physics Solutions and for more information ask us in the below comments and we’ll revert back to you asap.

Are you searching for the Samacheer Kalvi 12th Chemistry Chapter Wise Solutions PDF? Then, get your Samacheer Kalvi 12th Chapter Wise Solutions PDF for free on our website. Students can Download Chemistry Chapter 3 p-Block Elements – II Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Chemistry Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements – II

All concepts are explained in an easy way on our website. So, students can easily learn and practice Tamilnadu State Board 12th Chemistry Chapter 3 p-Block Elements – II Question and Answers. You can enjoy the digitized learning with the help of the Tamilnadu State Board Chemistry Online Material.

Samacheer Kalvi 12th Chemistry p-Block Elements – II TextBook Evalution

I. Choose the correct answer.

12th Chemistry Chapter 3 Book Back Answers Question 1.
In which of the following, NH₃ is not used?
(a) Nessler’s reagent
(b) Reagent for the analysis of IV group basic radical
(c) Reagent for the analysis of III group basic radical
(d) Tollen’s reagent
Answer:
(a) Nessler’s reagent

Samacheer Kalvi Guru 12th Chemistry Question 2.
Which is time regarding nitrogen?
(a) least electronegative element
(b) has low ionisation enthalpy than oxygen
(c) d-orbitals available
(d) ability to form pπ – pπ bonds with itself
Answer:
(d) ability to form pπ – pπ bonds with itself

12th Chemistry 3rd Lesson Book Back Answers Question 3.
An element belongs to group 15 and 3 rd period of the periodic table, its electronic configuration would be …………
(a) 1s²2s²2p⁴
(b) 1s²2s²2p³
(c) 1s²2s²2p⁶3s²3p²
(d) 1s²2s²2p⁶3s²3p³
Answer:
(d) 1s²2s²2p⁶3s²3p³

Samacheerkalvi.Guru 12th Chemistry Question 4.
Solid (A) reacts with strong aqueous NaOH liberating a foul smelling gas(B) which spontaneously bum in air giving smoky rings. A and B are respectively …………
(a) P₄(red) and PH₃
(b) P₄(white) and PH₃
(c) S₈ and H₂S
(d) P₄(white) and H₂S
Answer:
(b) P₄(white) and PH₃

Samacheer Kalvi Guru Class 12 Chemistry Question 5.
In the brown ring test, brown colour of the ring is due to …………
(a) a mixture of NO and NO₂
(b) Nitroso ferrous sulphate
(c) Ferrous nitrate
(d) Ferric nitrate
Answer:
(b) Nitroso ferrous sulphate

Samacheer Kalvi 12th Chemistry Question 6.
On hydrolysis, PCl₃ gives …………
(a) H₃PO₃
(b) PH₃
(c) H₃PO₄
(d) POOL
Answer:
(a) H₃PO₃

Chemistry Class 12 Samacheer Kalvi Question 7.
P₄O₆ reacts with cold water to give …………
(a) H₃PO₃
(b) H₄P₂O₇
(c) HPO₃
(d) H₃PO₄
Answer:
(a) H₃PO₃

Samacheer Kalvi Class 12 Chemistry Solutions Question 8.
The basicity of pyrophosphorous acid ( H₄P₂O₅) is …………
(a) 4
(b) 2
(c) 3
(d) 5
Answer:
(b) 2

Samacheer Kalvi Guru Chemistry Question 9.
The molarity of given orthophosphoric acid solution is 2M. its normality is …………
(a) 6N
(b) 4N
(c) 2N
(d) none of these
Answer:
(a) 6N

Chapter 3 Chemistry Class 12 Notes Question 10.
Assertion – bond dissociation energy of fluorine is greater than chlorine gas
Reason – chlorine has more electronic repulsion than fluorine
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(d) Both assertion and reason are false. The converse is true.

Samacheer Kalvi 12th Chemistry Solutions Question 11.
Among the following, which is the strongest oxidizing agent?
(a) Cl₂
(b) F₂
(c) Br₂
(d) I₂
Answer:
(b) F₂

Class 12 Chemistry Samacheer Kalvi Question 12.
The correct order of the thermal stability of hydrogen halide is …………
(a) HI > HBr > HCl > HF
(b) HF > HCl > HBr > HI
(c) HCl > HF > HBr > HI
(d) HI > HCl > HF > HBr
Answer:
(b) HF > HCl > HBr > HI

Samacheer Kalvi 12 Chemistry Solutions Question 13.
Which one of the following compounds is not formed?
(a) XeOF₄
(b) XeO₃
(c) XeF₂
(d) NeF₂
Answer:
(d) NeF₂

Class 12 Chemistry Chapter 3 Notes Pdf Download Question 14.
Most easily liquefiable gas is …………
(a) Ar
(b) Ne
(c) He
(d) Kr
Answer:
(c) He

12th Chemistry Chapter 7 Book Back Answers Question 15.
XeF₆ on complete hydrolysis produces …………
(a) XeOF₄
(b) XeO₂F₄
(c) XeO₃
(d) XeO₂
Answer:
(c) XeO₃

Question 16.
On oxidation with iodine, sulphite ion is transformed to …………
(a) S₄\({ O }_{ 6 }^{ 2- }\)
(b) S₂\({ O }_{ 6 }^{ 2- }\)
(c) S\({ O }_{ 4 }^{ 2- }\)
(d) S\({ O }_{ 3 }^{ 2- }\)
Answer:
(c) S\({ O }_{ 4 }^{ 2- }\)

Question 17.
Which of the following is strongest acid among all?
(a) HI
(b) HF
(c) HBr
(d) HCl
Answer:
(a) HI

Question 18.
Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules?
(a) Br₂ > I₂ > F₂ > Cl₂
(b) F₂ > Cl₂ > Br₂ > I₂
(c) I₂ > Br₂ > Cl₂ > F₂
(d) Cl₂ > Br₂ > F₂ > I₂
Answer:
(d) Cl₂ > Br₂ > F₂ > I₂

Question 19.
Among the following the correct order of acidity is …………
(a) HClO₂ < HCIO < HClO₃ < HClO₄
(b) HClO₄ < HClO₂ < HCIO < HClO₃
(c) HClO₃ < HClO₄ < HClO₂ < HCIO
(d) HCIO < HClO₂ < HClO₃ < HClO₄
Answer:
(d) HCIO < HClO₂ < HClO₃ < HClO₄

Question 20.
When copper is heated with cone HNO₃ it produces …………
(a) CU(NO₃)₂ , NO and NO₂
(b) Cu(NO₃)₂ and N₂O
(c) CU(NO₃)₂ and NO₂
(d) Cu(NO₃)₂ and NO
Answer:
(c) CU(NO₃)₂ and NO₂

II. Answer the following questions:

Question 1.
What is inert pair effect?
Answer:
In p-block elements, as we go down the group, two electrons present in the valence s-orbital become inert and are not available for bonding (only p-orbital involves chemical bonding). This is called inert pair effect.

Question 2.
Chalcogens belongs to p-block. Give reason.
Answer:
Chalcogens are ore forming elements. Most of the ores are oxides and sulphides, therefore oxygen, sulphur and other group 16 elements are called Chalcogens. In O, S, Se, Te and Po last electron enters to p-orbital. Therefore Chalcogens belongs to p-block.

Question 3.
Explain why fluorine always exhibit an oxidation state of -1?
Answer:
Fluorine the most electronegative element than other halogens and cannot exhibit any positive oxidation state. Fluorine does not have d-orbital while other halogens have d-orbitals. Therefore fluorine always exhibit an oxidation state of-1 and others in halogen family shows +1, +3, +5 and +7 oxidation states.

Question 4.
Give the oxidation state of halogen in the following.

1. OF₂
2. O₂F₂
3. Cl₂O₃
4. I₂O₄

Answer:
1. OF₂
+2 + 2(x) = 0
+2 = -2x
2x = -2
x =-1

2. O₂F₂
2(+1) + 2x = 0
2x = – 2
x = – 1

3. Cl₂O₃
2(x) + 3(-2) =0
2x = +6
x = +3

4. I₂O₄
2(x) + 4(-2) =0
2x = +8
x = +4

Question 5.
What are interhalogen compounds? Give examples.
Answer:
Each halogen combines with other halogens to form a series of compounds called interhalogen compounds. For example, Fluorine reacts readily with oxygen and forms difluorine oxide (F₂O) and difluorine dioxide (F₂O₂).

Question 6.
Why fluorine is more reactive than other halogens?
Answer:
Fluorine is the most reactive element among halogen. This is due to the minimum value of F – F bond dissociation energy. Hence fluorine is more reactive than other halogens.

Question 7.
Give the uses of helium.
Answer:

1. Helium and oxygen mixture is used by divers in place of air oxygen mixture. This prevents the painful dangerous condition called bends.
2. Helium is used to provide inert atmosphere in electric arc welding of metals
3. Helium has lowest boiling point hence used in cryogenics (low temperature science).
4. It is much less denser than air and hence used for filling air balloons

Question 8.
What is the hybridisation of iodine in IF₇? Give its structure.
Answer:
Hybridisation of iodine in IF₇ is sp³d³ Structure of IF₇ is pentagonal bipyramidal.

Question 9.
Give the balanced equation for the reaction between chlorine with cold NaOH and hot NaOH.
Answer:
1. Reaction between chlorine with cold NaOH:
Cl₂+ H₂O → HCl + HOCl
HCl + NaOH → NaCl + H₂O
HOCl + NaOH → NaOCl + H₂O
Overall reaction

Chlorine reacts with cold NaOH to give sodium chloride and sodium hypochlorite.

2. Reaction between chlorine with hot NaOH:
Cl₂ + H₂O → HCl + HOCl
HCl + NaOH → NaCl + H₂O
HOCl + NaOH → NaOCl + H₂O
3NaOCl → NaClO₃+ 2NaCl
Overall reaction

Chlorine reacts with hot NaOH to give sodium chlorate and sodium chloride.

Question 10.
How will you prepare chlorine in the laboratory?
Answer:
1. Chlorine is prepared by the action of cone, sulphuric acid on chlorides in presence of manganese dioxide.
4NaCl + MnO₂ + 4H₂SO₄ → Cl₂ + MnCl₂ + 4NaHSO₄ + 2H₂O

2. It can also be prepared by oxidising hydrochloric acid using various oxidising agents such as manganese dioxide, lead dioxide, potassium permanganate or dichromate.
PbO₂ + 4HCl → PbCl₂ + 2H₂O + Cl₂
MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂
2KMnO₄ + 16HCl → 2KCl + 2MnCl + 8H₂O + 5Cl₂
K₂Cr₂O₇ + 14HCl → 2KCl + 2CrCl₃ + 7H₂O + 3Cl₂

3. When bleaching powder is treated with mineral acids chlorine is liberated
CaOCl₂ + 2HCl → CaCl₂ + H₂O + Cl₂
CaOCl₂ + H₂SO₄ → CaSO₄ + H₂O + Cl₂

Question 11.
Give the uses of sulphuric acid.
Answer:

1. Sulphuric acid is used in the manufacture of fertilisers, ammonium sulphate and super phosphates and other chemicals such as hydrochloric acid, nitric acid etc.
2. It is used as a drying agent and also used in the preparation of pigments, explosives etc.

Question 12.
Give a reason to support that sulphuric acid is a dehydrating agent.
Answer:
Sulphuric acid is highly soluble in water and has strong affinity towards water and hence it can be used as a dehydrating agent. When dissolved in water it forms mono ( H₂SO₄. H₂O ) and di ( H₂SO₄. 2H₂O ) hydrates and the reaction is exothermic. The dehydration property can also be illustrated by its reaction with organic compounds such as sugar, oxalic acid and formic acid.

Question 13.
Write the reason for the anamolous behaviour of Nitrogen.
Answer:
1. Due to its small size, high electro negativity, high ionisation enthalpy and absence of d-orbitals.

2. N, has a unique ability to form pπ – pπ multiple bond whereas the heavier members of this group (15) do not form pπ – pπ bond, because their atomic orbitals are so large and diffused that they cannot have effective overlapping.

3. Nitrogen exists a diatomic molecule with triple bond between the two atoms whereas other elements form single bond in the elemental state.

4. N cannot form dπ – pπ bond due to the absence of d – orbitals whereas other elements can.

Question 14.
Write the molecular formula and structural formula for the following molecules.
(a) Nitric acid
(b) dinitrogen pentoxide
(c) phosphoric acid
(d) phosphine
Answer:

Question 15.
Give the uses of argon.
Answer:
Argon prevents the oxidation of hot filament and prolongs the life in filament bulbs.

Question 16.
Write the valence shell electronic configuration of group-15 elements.
Answer:
General electronic configuration of group 15 elements are ns²np³.

1. Nitrogen – [He] 2s² 2p³
2. Phosphorous – [Ne] 3s² 3p³
3. Arsenic – [Ar] 3d¹⁰ 4s² 4p³
4. Antimony – [Kr] 4d¹⁰ 5s² 5p³
5. Bismuth – [Ne] 4f¹⁴ 5s¹⁰ 6s² 6p³

Question 17.
Give two equations to illustrate the chemical behaviour of phosphine.
Answer:
1. Phosphine reacts with halogens to give phosphorous penta halides.
PH₃ + 4Cl₂ → PCl₅ + 3HCl

2. Phosphine forms coordination compound with lewis acids such as boron trichloride.

3. Phosphine precipitates some metal from their salt solutions.
3AgNO₃ + PH₃ → Ag₃P + 3HNO₃

Question 18.
Give a reaction between nitric acid and a basic oxide.
Answer:
Nitric acid reacts with bases and basic oxides to form salts and water.

1. ZnO + 2HNO₃ → Zn(NO₃)₂ + H₂O
2. 3FeO + 10HNO₃ → 3Fe(NO₃)₃ + NO + 5H₂O

Question 19.
What happens when PCl₅ is heated?
Answer:
On heating phosphorous pentachloride, it decomposes into phosphorus trichloride and chlorine.
PCl₅ \(\underrightarrow { \triangle }\) PCl₃ + Cl₂

Question 20.
Suggest a reason why HF is a weak acid, whereas binary acids of the all other halogens are strong acids.
Answer:
The hydrogen halides are extremely soluble in water due to the ionisation.
X + H₂O → H₃O⁺ + X^–
( X = F, Cl, Br or I )

Solutions of hydrogen halides are therefore acidic and known as hydrohalic acids. Hydrochloric, hydrobromic and hydroiodic acids are almost completely ionised and are therefore strong acids but HF is a weak acid. For HF,

• HF + H₂O \(\rightleftharpoons\) H₃O⁺ + F^–
• HF + F^– → \({ HF }_{ 2 }^{ – }\)

At high concentration, the equilibrium involves the removal of fluoride ions is important. Since it affects the dissociation of hydrogen fluoride, therefore it is a weak acid.

Question 21.
Deduce the oxidation number of oxygen in hypofluorous acid – HOF.
Answer:
In case of O – F bond is HOF, fluorine is most electronegative element. So its oxidation number is -1. Thereby oxidation number of O is +1. Similarly in case of O – H bond is HOF. O is highly electronegative than H. So its oxidation number is -1 and oxidation number of H is +1. So, Net oxidation of oxygen is – 1 + 1 = 0.

Question 22.
What type of hybridisation occur in

1. BrF₅
2. BrF₃

Answer:
1. BrF₅
BrF₅ is a AX₅ type. Therefore is has sp³d² hybridisation. Hence, BrF₅ molecule has square pyramidal shape.

2. BrF₃
BrF₃ is a AX₃ type. Therefore it has sp³d hybridisation. Hence, BrF₃ molecule has T-shape.

Question 23.
Complete the following reactions.
Answer:

Samacheer Kalvi 12th Chemistry p-Block Elements – II Evaluate yourself

Question 1.
Write the products formed in the reaction of nitric acid (both dilute and concentrated) with zinc.
Answer:

Samacheer Kalvi 12th Chemistry p-Block Elements – II Additional Questions

Samacheer Kalvi 12th Chemistry p-Block Elements – II 1 Mark Questions and Answers

I. Choose the best answer.

Question 1.
About 78% of earth atmosphere contains,…………
(a) P
(b) As
(c) N
(d) Bi
Answer:
(c) N

Question 2.
Which one of the following is not a pnictogens?
(a) Nitrogen
(b) Oxygen
(c) Phosphorous
(d) Antimony
Answer:
(b) Oxygen

Question 3.
Which one of the following shows isotopes?
(a) Nitrogen
(b) Arsenic
(c) Antimony
(d) Bismuth
Answer:
(a) Nitrogen

Question 4.
Nitrogen gas in atmosphere is separated industrially from liquid air by …………
(a) simple distillation
(b) Fractional distillation
(c) Sublimation
(d) Distillation under reduced pressure
Answer:
(b) Fractional distillation

Question 5.
Bond order for nitrogen molecule is …………
(a) 1
(b) 2
(c) 3
(d) 0
Answer:
(c) 3

Question 6.
Nitrogen gas is …………
(a) Inert
(b) Noble
(c) More reactive
(d) Less reactive
Answer:
(a) Inert

Question 7.
Which one of the following is used in cryosurgery?
(a) Liq N₂
(b) Liq NH₃
(c) Liq Na
(d) Liq H₂
Answer:
(a) Liq N₂

Question 8.
The dielectric constant of ammonia is (K) …………
(a) 10^-30
(b) 10^-14
(c) 10³⁰
(d) 10¹⁴
Answer:
(a) 10^-30

Question 9.
When ammonia reacts with copper sulphate solution to give complex, the colour of complex is …………
(a) violet
(b) deep blue
(c) blue
(d) Red
Answer:
(b) deep blue

Question 10.
H – N – H bond angle in NH₃ is …………
(a) 109° 28’
(b) 107° 28’
(c) 104°
(d) 107°
Answer:
(d) 107°

Question 11.
Shape of ammonia is …………
(a) Planar
(b) Square planar
(c) Pyramidal
(d) Square pyramidal
Answer:
(c) Pyramidal

Question 12.
Nitric acid prepared in large scales using …………
(a) Ostwald’s process
(b) Haber’s process
(c) Contact process
(d) Deacon’s process
Answer:
(a) Ostwald’s process

Question 13.
Benzene undergoes nitration reaction to form nitrobenzene in this reaction takes place due to the formation of …………
(a) Hydronium ion
(b) Hydride ion
(c) Nitronium ion
(d) Nitrasonium ion
Answer:
(c) Nitronium ion

Question 14.
Oxidation state of N m FINO₃ is…………
(a) ±2
(b) +3
(c) +4
(d) +5
Answer:
(d) +5

Question 15.
Compound used in photography is …………
(a) AgNO₃
(b) AgBr
(c) AgCl
(d) AgI
Answer:
(a) AgNO₃

Question 16.
Sodium nitrate
(a) Photography
(b) Firearms
(c) Royal water Sosurgerv …………
(d) Cryosurgery
Answer:
(b) Firearms

Question 17.
Nitrogen sesquoxide colour is …………
(a) colourless
(b) Brown
(c) Blue
(d) Red
Answer:
(c) Blue

Question 18.
White phosphorous is also called as …………
(a) Red phosphorous
(b) Black phosphorous
(c) Scarlet phosphorous
(d) Yellow phosphorous
Answer:
(d) Yellow phosphorous

Question 19.
White (Yellow) phosphorous glows in the dark due to oxidation which is called …………
(a) phosphorescence
(b) phosphorus
(c) Fluorescence
(d) Liminoscence
Answer:
(a) phosphorescence

Question 20.
Yellow phosphorous reacts with alkali on boiling in an inert atmosphere liberates …………
(a) Phosphorous acid
(b) Phosphoric acid
(c) Phosphine
(d) Pyrophosphoric acid
Answer:
(c) Phosphine

Question 21.
Consider the following statements.
(i) phosphine is the most important hydride of phosphorous
(ii) phosphine is a poisonous gas with rotten egg smell.
(iii) phosphine is a powerful reducing agent

Which of the above statement(s) is / are correct?
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (ii) only
Ansswer:
(c) (i) and (iii)

Question 22.
When phosphine is heated with air it bums to gives …………
(a) Orthophosphoric acid
(b) Metaphosphoric acid
(c) Pyrophosphoric acid
(d) Phosphoroustrioxide
Answer:
(b) Metaphosphoric acid

Question 23.
Hybridisation of P in phosphine is …………
(a) sp³d
(b) sp³d²
(c) sp3d3
(d) sp3
Answer:
(d) sp3

Question 24.
Compounds used in Holme’s signal are …………
(a) Phosphine + Acetylene
(b) H₃PO₃+H₃PO₃
(c) Calcium carbide + calcium phosphide
(d) Calcium carbonate + calcium phosphate
Answer:
(c) Calcium carbide + calcium phosphide

Question 25.
Chalgogens are also called as …………
(a) Ore forming elements
(b) Group-16 elements
(c) group 17 elements
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)

Question 26.
Element present in the volcanic ashes is …………
(a) Oxygen
(b) Sulphur
(c) Selenium
(d) Tellurium
Answer:
(b) Sulphur

Question 27.
The decomposition of potassium chlorate speed up in the presence of …………
(a) MnO₂
(h) Mn₃O₄
(c) MnSO₄
(d) KMnO₄
Answer:
(a) MnO₂

Question 28.
Pure ozone is …………
(a) yellow gas
(b) blue gas
(c) Pale blue gas
(d) bright blue gas
Answer:
(c) Pale blue gas

Question 29.
Shape of ozone …………
(a) V-shape
(b) Linear shape
(c) bent shape
(d) spherical shape
Answer:
(c) bent shape

Question 30.
The rate of decompositon of ozone drops sharply in …………
(a) acidic medium
(b) alkaline medium
(c) neutral medium
(d) Ether medium
Answer:
(b) alkaline medium

Question 31.
Which one of the following used as fuel in rockets?
(a) Liq O₂
(b) Liq CO₂
(c) Liq N₂
(d) Liq He – O₂
Answer:
(a) Liq O₂

Question 32.
Find out crystalline allotrophic form of sulphur?
(a) γ – sulphur
(b) λ – sulphur
(c) α – sulphur
(d) milk of sulphur
Answer:
(c) α – sulphur

Question 33.
Consider the following statements
(i) α – sulphur is the only thermodynamically stable allotrophic form.
(ii) At 140 ° C the mono clinic sulphur melts to form mobile pale yellow liquid called γ – sulphur
(iii) Monoclinic sulphur is stable between 96°C-119°C and slowly changes into λ- sulphur

Which of the above statement(s) is / are not correct?
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (ii) and (iii)
Answer:
(d) (ii) and (iii)

Question 34.
Sulphur di oxide, how many times heavier than air?
(a) 2 times
(b) 2.5 times
(c) 2.2 times
(d) 2.3 times
Answer:
(c) 2.2 times

Question 35.
Which one of following has temporary bleaching action?
(a) Chlorine
(b) SO₃
(c) H₃SO₄
(d) SO₂
Answer:
(d) SO₂

Question 36.
Sulphuric acid can be manufactured by …………
(a) Ostwald’s process
(b) Lead chamber process
(c) Deacon’s process
(d) Haber’s process
Answer:
(b) Lead chamber process

Question 37.
Sulphuric acid is manufactured by contact process, catalyst used in contact process is …………
(a) V₂O₅
(b) TiCl₄
(c) Fe
(d) Mo
Answer:
(c) V₂O₅

Question 38.
Benzene reacts with sulphuric acid to gives …………
(a) sulphate
(b) sulphide
(c) sulphonic acid
(d) sulphite
Answer:
(c) sulphonic acid

Question 39.
Reagent used to detect sulphate ion is …………
(a) BaCl₂
(b) BaSO₃
(c) (CH,COO),Pb
(d) both (a) and (c)
Answwer:
(d) both (a) and (c)

Question 40.
Deacon’s process is used to manufacture …………
(a) Cl₂
(b) F₂
(C) Br
(d) I₂
Answer:
(a) Cl₂

Question 41.
Catalyst used in Deacon’s process is …………
(a) CuCl₂
(b) Cu₂Cl₂
(c) CuBr
(d) Cu₂Br₂
Answer:
(b) Cu₂Cl₂

Question 42.
C₁₀H₁₆ + 8C₁₂ \(\underrightarrow { \triangle }\) A. Identify A?
(a) Methane
(b) Ethane
(c) Carbon
(d) Propane
Answer:
(c) Carbon

Question 43.
Passing chlorine gas through dry slaked lime to produce …………
(a) CaOCl
(b) CaOCl₂
(c) CaO
(d) CaCl₂
Answer:
(b) CaOCl₂

Question 44.
Which one of the following is used for purification of drinking water?
(a) SO₃
(b) SO₂
(c) Br₂ / H₂O
(d) Cl₂
Answer:
(d) Cl₂

Question 45.
Which one of the following is a weak acid?
(a) HF
(b) HCl
(c) HBr
(d) HI
Answer:
(a) HF

Question 46.
Reagent not stored in glass bottles?
(a) HCI
(b) HBr
(c) HF
(d) HI
Answer:
(c) HF

Question 47.
More reactive element is
(a) Fluorine
(b) Chlorine
(c) Bromine
(d) Iodine
Answer:
(a) Fluorine

Question 48.
The correct order of the acidity of hydrohalic acids?
(a) HF > HCI > HBr > HI
(b) HCI >HF >HBr >HI
(c) HBr > HCI >HF > HI
(d) HI > HBr > HCI > HF
Answer:
(d) HI > HBr > HCI > HF

Question 49.
Consider the following statements
(i) In interhalogen compounds, the central atom will be the smaller one.
(ii) It can be formed only between two halogen and not more than two halogens.
(iii) They are strong reducing agents.

Which of the above statement(s) is / are not correct?
(a) (i) only
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (iii) only
Answer:
(c) (i) and (iii)

Question 50.
Shape of ClF₃ is …………
(a) Linear
(b) T-shape
(c) Pyrimidal
(d) Square planar
Answer:
(b) T-shape

Question 51.
Which one of the following is more acidic?
(a) HOCl
(b) HCIO₂
(c) HClO₃
(d) HClO₄
Answer:
(d) HClO₄

Question 52.
When XeF₆ reacts with 2.5 M NaOH gives …………
(a) Na₄XeO₆
(b) Na₂XeO₃
(c) XeO₂F₂
(d) XeO₃
Answer:
(a) Na₄XeO₆

Question 53.
Shape of XeF₆ is …………
(a) Octahedron
(b) Distorted octahedron
(c) Pyramidal
(d) Tetrahedron
Answer:
(b) Distorted octahedron

Question 54.
Which one of the following can penetrate through dense fog?
(a) He
(b) Ne
(c) Kr
(d) Rn
Answer:
(c) Kr

Question 55.
Find out radioactive element?
(a) He
(b) Rn
(c) Xe
(d) Ar
Answer:
(b) Rn

II. Fill in the blanks:

1. The 11th most abundant element is ………….
2. …………. is the principle gas of atmosphere.
3. Nitrogen is chemically ………….
4. …………. process for the synthesis of ammonia.
5. …………. is used for the manufacture of calcium cyanamide
6. …………. is a pungent smelling gas.
7. Ammonia acts as a …………. agent.
8. With excess of chlorine, ammonia reacts to give …………. an explosive substance.
9. When excess ammonia is added to aqueous solution of copper sulphate …………. colour compound is formed.
10. Pure nitric acid becomes …………. on standing.
11. …………. is used in gunpower for firearms.
12. The decomposition of ammonium nitrate gives ………….
13. White phosphorous is colourless but becomes pale yellow due to formation of a …………. upon standing.
14. The white phosphorous can be changed into …………. by heated it to 420°C in the absence of air and light.
15. …………. reacts with alkali on boiling in an inert atmosphere liberating phosphine.
16. …………. is used in the match boxes.
17. Phosphine is …………. smelling gas.
18. Phosphine has a …………. shape.
19. …………. is used for producing smoke screen.
20. When phosphorous trichloridie is hydrolysed with cold water it gives ………….
21. Elements belonging group 16 are called ………….
22. Under ordinary condition oxygen exists as a …………. gas.
23. Allotrophic form of oxygen is …………. and ………….
24. Pure ozone is …………. gas.
25. …………. is used in welding purpose.
26. Monoclinic sulphur is stable between 96° and 119°C and slowly changes into ………….
27. …………. gas is found in volcanic eruptions.
28. A large amount of …………. gas is released into atmosphere from plants using coal and oil and copper melting plants.
29. Sulphurdioxide gas has …………. odour.
30. Sulphurdioxide can be used for …………. and …………. in agriculture.
31. In SO₃ S-atom undergoes …………. hybridisation.
32. In SO₃ a double bond arises between S and O is due to …………. overlapping.
33. High boiling point and viscosity of sulphuric acid is due to ………….
34. …………. is used as a drying agent.
35. The main source of fluorine is ………….
36. The main source of chlorine is ………….
37. Chlorine is a …………. gas.
38. Chlorine is soluble in water and its solution is referred as ………….
39. …………. is produced by passing chlorine gas through dry slaked lime.
40. …………. is used in extraction of gold and platinum.
41. …………. is used for extraction of glue from bone.
42. At room temperature, hydrogen halides are gases but …………. can be readily liquefied.
43. Liberation of iodine which gives a …………. colouration with starch.
44. Each halogen combines with other halogens to form a series of compounds called ………….
45. Structure of AX₇ type is ………….
46. Oxidation state of Cl in HClO₄ is ………….
47. Noble gases have the …………. ionisation energy.
48. Xenon reacts with PtF₆ and gave an ………….
49. Kr and fluorine gases are irradiated with SbF it forms ………….
50. Shape of XeOF₄ is ………….
51. Helium used for filling air ………….
52. …………. is used in fluorescent bulbs.
53. …………. is used in high speed electronic flash bulbs.
54. Radon is a source of …………. rays.
55. …………. is formed by the hydrolysis of urea.
56. …………. element subtimes at 889 K.
57. Yellow phosphorus has a characteristics …………. smell.

Answers:

1. phosporous
2. Nitrogen
3. Inert
4. Haber’s
5. Nitrogen
6. Ammonia
7. Reducing
8. Nitrogen trichloride
9. Deep blue
10. Yellow
11. Nitric acid / NaNO
12. Nitrous oxide
13. Layer of red phosphorous
14. Red phosphorous
15. White phosphorous
16. Red phosporous
17. Rotten fish
18. Pyramidal
19. Phosphine
20. phosporous acid
21. Chalgogens
22. diatomic
23. dioxygen and ozone
24. pale blue
25. Oxyacetylene
26. Rhombic sulphur
27. Sulphurdioxide
28. SO₂
29. Suffocating
30. disinfecting crops and plants
31. sp²
32. pπ – dπ
33. Hydrogen bonding
34. Sulphuric acid
35. Fluorite
36. Sodium chloride
37. Green yellow
38. Chlorine water
39. Bleaching powder
40. Chlorine
41. Hydrochloric acid
42. hydrogen fluroide
43. blue-black
44. Inter-halogen compounds
45. pentagonal bipyramidal
46. +7
47. largest
48. orange yellow solid [XePtF₆]
49. KrF₂. 2SbF₃
50. Square pyramidal
51. Balloons
52. Krypton
53. Xenon
54. Gamma
55. Ammonia
56. Arsenic
57. Garlic

II. Match the following:

Question 1.
(i) Haber’s process – (a) HNO₃
(ii) Deacon’s process – (b) Ammonia
(iii) Contact process – (c) Chlorine
(iv) Ostwald’s process – (d) H₂SO₄
Answer:
(i) b
(ii) c
(iii) d
(iv) a

Question 2.
(i) Nitric acid – (a) Purification of bone black
(ii) HCl – (b) Photography
(iii) White (yellow) phosphorous – (c) Rotten fish smell
(iv) Phosphine – (d) Phosphorescence
Answer:
(i) b
(ii) a
(iii) d
(iv) c

Question 3.
(i) Nitrogen sesquoxide – (a) H₂N₂O₂
(ii) Nitrous oxide – (b) H₄N₂O₄
(iii) Hyponitrous acid – (c) N₂O
(iv) Hydronitrous acid – (d) N₂O₃
Answer:
(i) d
(ii) c
(iii) a
(iv) b

Question 4.
(i) N₂O – (a) +5
(ii) N₂O₄ – (b) +3
(iii) N₂O₅ – (c) +1
(iv) N₂O₃ – (d) +4
Answer:
(i) c
(ii) d
(iii) a
(iv) b

Question 5.
(i) White phosphorous – (a) Volcanic eruptions
(ii) Red phosphorous – (b) Yellow phosphorous
(iii) Phosphine – (c) Match boxes
(iv) SO₂ – (d) smoke screen
Answer:
(i) b
(ii) c
(iii) d
(iv) a

Question 6.
(i) ammonia – (a) suffocating odour
(ii) SO₂ – (b) Rotten fish smell
(iii) PH₃ – (c) Greenish yellow gas
(iv) Cl₂ – (d) pungent smelling gas
Answer:
(i) d
(ii) a
(iii) b
(iv) c

Question 7.
(i) XeF₄ – (a) sp³
(ii) XeOF₂ – (b) sp³d²
(iii) XeO₃ – (c) sp³d³
(iv) XeF – (d) sp³d
Answer:
(i) b
(ii) d
(iii) a
(iv) c

Question 8.
(i) Helium – (a) flash bulbs
(ii) Neon – (b) radioactive
(iii) Krypion – (c) air balloons
(iv) Radon – (d) Brilliant red
Answer:
(i) c
(ii) d
(iii) a
(iv) b

IV. Assertion and Reason

Question 1.
Assertion (A) – Xenon is used in high speed electronic flash bulbs used by photographers.
Reason (R) – Xenon emits an intense light in discharge tubes instantly.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) A and R are correct and R explains A

Question 2.
Assertion (A) – Noble gases have the largest ionisation energy compared to any other elements.
Reason (R) – Noble gases have incomplete filled orbital.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(c) A is correct but R is wrong

Question 3.
Assertion (A) – Hydrogen iodide decomposes at 400⁰C while hydrogen fluoride and hydrogen chloride are stable at this temperature.
Reason (R) – Thermal stability of hydrogen halides decreases from fluoride to iodide.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) A and R are correct and R explains A

Question 4.
Assertion (A) – The bleaching of chlorine is temporary.
Reason (R) – Chlorine oxidises ferrous salts to ferric salts.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(d) A is wrong but R is correct

Question 5.
Assertion (A) – Sulphuric acid is highly reactive.
Reason (R) – Sulphuric acid can act as strong acid and an oxidising agent.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) A and R are correct but doesn’t explains A

Question 6.
Assertion (A) – Sulphuric acid is a high boiling point and viscous liquid.
Reason (R) – This is due to the association of molecules together through hydrogen bonding.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) A and R are correct and R explains A

Question 7.
Assertion (A) – Monoclinic sulphur is less stable than rhomobic sulphur.
Reason (R) – Monoclinic sulphur is stable between 96°C – 119°C and slowly changes into rhombic sulphur.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) A and R are correct and R explains A

Question 8.
Assertion (A) – Nitrogen gas is chemically inert.
Reason (R) – Nitrogen has low bonding energy.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(c) A is correct but R is wrong

V. Find the odd one out
Question 1.
(a) NO
(b) HNO₃
(C) NO₂
(d) N₂O
Answer:
(b) HNO₃
Hint: HNO₃ is acid and other are oxides.

Question 2.
(a) Nitrous acid
(b) Nitric acid
(c) Hyponitrous acid
(d) Pemitrous acid
Answer:
(d) Pemitrous acid
Hint: Pemitrous acid contains peroxide linkage others doesn’t have peroxide linkage.

Question 3.
(a) White phosphorous
(b) Red phosphorous
(c) phosphorous pentaoxide
(d) black phosphorous
Answer:
(c) phosphorous pentaoxide
Hint: P₂O₅ is a compound of phosphorous and others are allotropic form of phosphorous.

Question 4.
(a) PH₃
(b) HPO₃
(c) H₃PO₃
(d) H₃PO₄
Answer:
(a) PH₃
Hint: PH₃ is hydrides of phosphorous and others are oxo acids of phosphorous.

Question 5.
(a) He
(b) Ne
(c) Ar
(d) Xe
Answer:
(d) Xe
Hint: Xe forms several chemical compounds than others.

VI. Find out the correct pair.

Question 1.
(a) Helium – filament bulbs
(b) Krypton – prevent bonds
(c) Xenon – Lasers
(d) Radon – flash bulbs
Answer:
(c) Xenon – Lasers

Question 2.
(a) Ra – gamma rays
(b) Xe – cancer growth
(c) Ne – balloons
(d) Kr – advertisement bulb
Answer:
(a) Ra – gamma rays

Question 3.
(a) ClF₃ – Linear
(b) BrF₅ – T Shaped
(c) IF₄ – square pyrimidal
(d) BrF₅ – square pyrimidal
Answer:
(d) BrF₅ – square pyrimidal

Question 4.
(a) XeOF₂ – sp³
(b) XeF₆ – sp³d³
(c) XeF₄ – sp³
(d) XeOF₄ – sp³d
Answer:
(b) XeF₆ – sp³d³

Question 5.
(a) OF₂ = -1
(b) Cl₄O₄ = -1
(c) I₂O₄ = -1
(d) I₂O₉ = -1
Answer:
(a) OF₂ = -1

Question 6.
(a) Royal water – Dissolving gold
(b) Chlorine – Entraction of glue from bone
(c) HCl- Extraction of gold
(d) Chlorine – Temporary bleaching
Answer:
(a) Royal water – Dissolving gold

Question 7.
(a) O – 2s²2p³
(b) S – 3s²3p⁴
(c) Se – 4d¹⁰5s²5p⁴
(d) Te – 3d¹⁰4s²4p⁴
Answer:
(b) S – 3s²3p⁴

VI.Find out the incorrect pair.

Question 1.
(a) Nitrogen gas – inert
(b) Ammonia – pungent smelling gas
(c) Nitric acid – oxidizing agent
(d) Phosphine – rotten egg smell
Answer:
(d) Phosphine – rotten egg smell

Question 2.
(a) Liquid nitrogen – biological preservation
(b) Nitric acid – photography
(c) white phosphorous – yellow phosphorous
(d) phosphorous – welding
Answer:
(d) phosphorous – welding

Question 3.
(a) N₂O = +1
(b) N₂O = +2
(c) N₂O₃ = +5
(d) NO₂ = +4
Answer:
(c) N₂O₃ = +5

Question 4.
(a) Hvponitrous acid – N₂O
(b) Nitrous acid – HNO₂
(c) pernitric acid – HNO₄
(d) pernitrous acid – HOONO
Answer:
(a) Hvponitrous acid – N₂O

Question 5.
(a) PH₃ – Holme’s signal
(b) O₂ – welding
(c) H₂SO₄ – Disinfecting crops
(d) SO₃ – Bleaching hair
Answer:
(c) H₂SO₄ – Disinfecting crops

Question 6.
(a) H₂SO₄ drying agent
(b) Chlorine – Deacon’s process
(c) HCI – purification of bone black
(d) Helium – flash bulbs
Answer:
(d) Helium – flash bulbs

Question 7.
(a) ICl – Linear
(b) CIF₃ – T shape
(c) IF₅ – pentagonal bipyramidal
(d) IF₇ – pentagonal bipyramidal
Answer:
(c) IF₅ – pentagonal bipyramidal

Question 8.
(a) HOCl = +2
(b) HOCl = +3
(e) HOCI = +5
(d) HOCl₄ = +7
Answer:
(a) HOCl = +2

Question 9.
(a) XeF₄ – sp³d³
(b) XeF₆ – sp³d³
(e) XeOF₄ – sp³d²
(d) XeO₃ – sp³d
Answer:
(d) XeO₃ – sp³d

Question 10.
(a) He – cryogenics
(b) Ne – advertisement
(c) Kr – flurescent bulbs
(d) Ra – Lasers.
Answer:
(d) Ra – Lasers.

Samacheer Kalvi 12th Chemistry p-Block Elements – II 2 Mark Questions and Answers

Question  1.
What are pnictogens?
Answer:
The group – 15 elements like nitrogen, phosphorous Arsenic, Antimony and Bismuth are collectively called as pnictogens. Their general outer electronic configuration is ns²np³.

Question 2.
What happen when sodium azide undergoes thermal decomposition?
Answer:
Pure nitrogen gas can be obtained by the thermal decomposition of sodium azide about 575K
2NaN₃ \(\underrightarrow { 575K }\) 2Na + 3N₂

Question 3.
How will you prepare ammonia from nitrogen? and mention the name of process?
Answer:
Nitrogen directly reacts with hydrogen gives ammonia. This reaction is favoured by high pressures and at optimum temperature in the presence of iron catalyst,
N₂ + 3H₃ \(\rightleftharpoons\) 2NH₃
This process is called as Haber’s process.

Question 4.
Why nitrogen gas is chemically inert?
Answer:
The chemically inert character of nitrogen is largely due to high bonding energy of the molecules 225 cal mol^-1 .Interestingly the triply bonded species is notable for its less reactivity in comparison with other iso-electronic triply bonded systems such as -C \(\equiv\) C – C \(\equiv\) Q, X- C \(\equiv\) N, etc.

Question 5.
Mention the uses of nitrogen?
Answer:

1. Nitrogen is used for the manufacture of ammonia, nitric acid and calcium cyanamide etc.
2. Liquid nitrogen is used for producing low temperature required in cryosurgery, and so in biological preservation

Question 6.
Explain the action of heat on ammonia.
Answer:
Above 500°C ammonia decomposes into its elements. The decomposition may be accelerated by metallic catalysts like Nickel, Iron. Almost complete dissociation occurs on continuous sparking.
2NH₃ \(\underrightarrow { { 500 }^{ 0 }C } \) N₂ + 3H₂

Question 7.
Prove ammonia act as a reducing agent?
Answer:
Ammonia reduces the metal oxides to metal when passed over heated metallic oxide.
3PbO + 2NH₃ → 3Pb + N₂ + 3H₂O

Question 8.
What happen when copper sulphate reacts with ammonia?
Answer:
When excess ammonia is added to aqueous solution copper sulphate a deep blue colour Complex [Cu(NH₃)₄ ]²⁺ is formed.

Question 9.
Pure nitric acid is colourless, on standing it becomes yellow. Justify your answer.
Answer:
Nitric acid decomposes on exposure to sunlight or on being heated, into nitrogen dioxide, water and oxygen.
4HNO₄ → 4NO₂ + 2H₂O + O₂
Due to this reaction pure acid or its concentrated solution becomes yellow on standing.

Question 10.
Write the products formed in the reaction of nitric acid with dilute and concentrated with magnesium.
Answer:
1. Magnesium with cone. HNO₃:
4Mg + 10HNO₃ → 4Mg(NO₃)₂ + NH₄NO₃ + 3H₂O
Magnesium reacts with concentraated nitric acid to gives both magnesium nitrate and ammonium nitrate.

2. Magnesium with dil HNO₃:
Magnesium reacts with dilute nitric acid to gives both magnesium nitrate and nitrous oxide.
4Mg + 10HNO₃ → 4 Mg(NO₃)₂ + N₂O + 5H₂O

Question 11.
Give the uses of nitric acid.
Answer:

1. Nitric acid is used as a oxidising agent and in the preparation of aquaregia.
2. 4. Salts of nitric acid are used in photography (AgNO₃) and gunpowder for firearms. (NaNO₃) .

Question 12.
How will you prepare nitric oxide from sodium nitrite?
Answer:
When sodium nitrite reacts with ferrous sulphate in the presence of sulphuric acid to gives nitric oxide.
2NaNO₂ + 2FeSO₄ + 3H₂S → Fe₂(SO₄)₃ + 2NaHSO₄ + 2H₂O + 2NO

Question 13.
How will you prepare nitrogen pentoxide?
Answer:
Nitric acid reacts with phosphorous pentaoxide to give nitrogen pentoxide.
2HNO₃ + P₂O₅ → N₂O₅ + 2HPO₃

Question 14.
Draw the structure of

1. N₂O
2. N₂O₃

Answer:
1. N₂O (Nitrous oxide)

2. N₂O₃ (Nitrogen sesquoxide)

Question 15.
Draw- the structure of

1. N₂O₄
2. N₂O₃

Answer:
1. N₂O₄ (Nitrogen tetraoxide)

2. N₂O₅

Question 16.
Draw the structure of

1. Hyponitrous acid
2. Hydronitrous acid.

Answer:
(a) Hypon trous acid – H₂N₂O₂
HO – N = H – OH

(b) Hydronitrous acid:

Question 17.
Mention the allotropic forms of phosphorous?
Answer:
The most common allotropic forms of phosphorous

1. white phosphorous
2. Red phosphorous
3. Black phosphorous

Question 18.
Why white phosphorous is also known as yellow phosphorous?
Answer:
The freshly prepared white phosphorus is colourless but becomes pale yellow due to formation of a layer of red phosphorus upon standing. Hence it is also known as yellow phosphorus.

Question 19.
What is phosphorescence?
Answer:
White (yellow) phosphorous glows in the dark due to oxidation which is called phosphorescence.

Question 20.
Why white phosphorous undergoes spontaneous combustion in air?
Answer:
White phosphorous ignition temperature is very low and hence it undergoes spontaneous combustion in air at room temperature and during in the combusition process, white phosphorous produces P₂O₅.

Question 21.
Draw the structure of

1. white phosphorous
2. red phosphorous

Answer:
1. White phosphorous

2. Red phosphorous

Question 22.
How will you convert red phosphorous into P₂O₃ and P₂Q₅?
Answer:
Red phosphorus reacts with oxygen on heating to give phosphorus trioxide or phosphorus pentoxide.
P₄ + 3O₂ \(\underrightarrow { \triangle }\) P₄O₆(phosphorous trioxide)
P₄ + 5O₂ \(\underrightarrow { \triangle }\) P₄Q₁₀ (phosphorous pentaoxide)

Question 23.
How will you prepare orthophosphoric acid from phosphorous?
Answer:
When phosphorous is treated with cone.nitric acid it is oxidised to orthophosphoric acid. This reaction is catalysed by iodine crystals.

Question 24.
Mention the uses of phosphorous?
Answer:

1. The red phosphorus is used in the match boxes.
2. It is also used for the production of certain alloys such as phosphor bronze.

Question 25.
Show that phosphine is weakly basic?
Answer:
Phosphine is weakly basic and forms phosphonium salts with halogen acids.
PH₃ +HI → PH₄I
PH₄I + H₂O \(\underrightarrow { \triangle }\) PH₃+ H₃O^– + I^–

Question 26.
Draw the structure of PCl₃
Answer:

Question 27.
How will you prepare PCl₃ from white phosphorous?

1. When a slow stream of chlorine is passed over white phosphorous, PCl₃ is formed
2. It can also be obtained by treating white phosphorous with thionyl chloride.

Question 28.
What happen when PCl₃ is treated with cold water?
Answer:
When PCl₃ is hydrolysed with cold water it gives phosphorous acid.

Question 29.
How will you prepare PCl₃ ?
Answer:
When PCl₃ is treated with excess chlorine, phosphorous pentachloride is obtained.
PCl₃+ Cl₂ → PCl₅

Question 30.
Mention the uses of PCl₃ and PCl₅?
answer:

1. Phosphorus trichloride is used as a chlorinating agent and for the preparation of H₂PO₃.
2. Phosphorous pentachloride is a chlorinating agent and is useful for replacing hydroxyl groups by chlorine atom.

Question 31.
Draw the structure of H₃PO₂ and H₃PO₃?
Answer:

Question 32.
Draw the structure of H₄P₂O₆ and H₄P₂O₃?
Answwer:
(i) H₄P₂O₆
(ii) H₄P₂O₃

Question 33.
Give the method to prepare hypophosphorous acid and pyrophosphoric acid?
Answer:
1. Hypophosphorous acid – Phosphorous reacts with water to give hypophosphorous acid.

2. Pyrophosphoric acid – Phosphorous acid is heated they produce pyrophosphoric acid.

Question 34.
Complete the reactions

1. HgO \(\underrightarrow { \triangle }\) ?
2. BaO₂ \(\underrightarrow { \triangle }\) ?

Answer:

1. HgO \(\underrightarrow { \triangle }\) 2Hg + O₂
2. 2BaO₂ \(\underrightarrow { \triangle }\) 2BaO + O₂

Question 35.
Give and explain the reaction used to estimation of ozone.
Answer:
Ozone oxidises potassium iodide to iodine.
O₃ + 2KI + H₂O → 2KOH + O₂ + I₂
This reaction is quantitative and can be used for estimation of ozone.

Question 36.
Write a short notes on Rhombic sulphur?
Answer:
Rhombic sulphur also known as a sulphur, is the only thermodynamically stable allotropic form at ordinary temperature and pressure. The crystals have a characteristic yellow colour and composed of S₈ molecules. When heated slowly above 96°C, it converts into monoclinic sulphur. Upon cooling below 96°C the β form converts back to a form.

Question 37.
Write a notes on monoclinic sulphur?
Answer:
Monoclinic sulphur also contains S₈ molecules in addition to small amount of S₆ molecules. It exists as a long needle like prism and is also called as prismatic sulphur. It is stable between 96° – 119°C and slowly changes into rhombic sulphur.

Question 38.
What are λ – sulphur?
Answer:
Sulphur also exists in liquid and gaseous states. At around 140° C the monoclinic sulphur melts to form mobile pale yellow liquid called λ sulphur.

Question 39.
How will you prepare sulphurdioxide by laboratory method?
Answer:
Sulphur dioxide is prepared in the laboratory treating a metal or metal sulphite with sulphuric acid.
Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O
\({ SO }_{ 3 }^{ – }\) + 2H⁺ → H₂O + SO₂

Question 40.
Complete the reactions,
Answer:

1. Zns + O₂ \(\underrightarrow { \triangle }\) ?
2. FeS₂ + O₂ \(\underrightarrow { \triangle }\) ?

Answer:

1. 2Zns + 3O₂ \(\underrightarrow { \triangle }\) 2ZnO + 2SO₂
2. 4FeS₂ + 11O₂ \(\underrightarrow { \triangle }\) 2Fe₂O₃ + 8SO₂

Question 41.
What happen when sulphurdioxdie reacts with sodium hydroxide and sodium carbonate?
Answer:
Sulphur dioxide reacts with sodium hydroxide and sodium carbonate to form sodium bisulphite and sodium sulphite respectively.

Question 42.
Mention the uses of sulphurdioxide?
Answer:

1. Sulphur dioxide is used in bleaching hair, silk, wool etc…
2. It can be used for disinfecting crops and plants in agriculture.

Question 43.
Why sulphuric acid is high boiling and viscous liquid ?
Answer:
Sulphuric acid is high boiling and viscous liquid this is due to the association of molecules together through hydrogen bonding.

Question 44.
Explain the reaction between benzene and sulphuric acid.
Answer:
Sulphuric acid reacts with benzene to give benzene sulphuric acid.

Question 45.
Give the uses of sulphuric acid?
Answer:

1. Sulphuric acid is used in the manufacture of fertilisers, ammonium sulphate and super phosphates and other chemicals such as hydrochloric acid, nitric acid etc…
2. It is used as a drying agent and also used in the preparation of pigments, explosives etc..

Question 46.
Draw the structure of H₂SO₃ and H₂SO₄ ?
Answer:
1. H₂SO₃

2. H₂SO₄

Question 48.
Draw the structure of

1. Pyrosulphuric acid
2. Peroromonosulphuric acid?

Answer:
1. Pyrosulphuric acid H₂S₂O₇

2. Peroromonosulphuric acid H₂SO₅

Question 49.
What happen when chlorine reacts with trupentine?
Answer:
When chlorine burnt with turpentine it forms carbon and hydrochloric acid.

Question 50.
How will you prepare hydrochloric acid by laboratory method?
Answer:
It is prepared by the action of sodium chloride and concentrated sulphuric acid

• NaCl + H₂SO₄ → NaHSO₄ + HCl
• NaHSO₄ + NaCl → Na₂SO₄ + HCl

Dry hydrochloric acid is obtained by passing the gas through conc. sulphuric acid

Question 51.
Mention the uses of hydrochloric acid?
Answer:

1. Hydrochloric acid is used for the manufacture of chlorine, ammonium chloride, glucose from com starch etc.,
2. It is used in the extraction of glue from bone and also for purification of bone black.

Question 52.
Complet the following reaction?
Answer:

Question 53.
Why noble gases have the largest ionisation energy?
Answer:
Noble gases have the largest ionisation energy compared to any other elements in a given row as they have completely filled orbital (ns²np⁶) in their outer most shell. They are extremely stable and have a small tendency to gain or lose electrons. Hence noble gases have the largest ionisation energy.

Question 54.
Mention the uses of Neon?
Answer:
Neon is used in advertisement as neon sign and the brilliant red glow is caused by passing electric current through neon gas under low
pressure.

Question 55.
Give the uses of Krypton?
Answer:

1. Krypton is used in fluorescent bulbs, flash bulbs etc…
2. Lamps filed with krypton are used in airports as approaching lights as they can penetrate through dense fog.

Question 56.
Mention the application of Xenon?
Answer:

1. Xenon is used in fluorescent bulbs, flash bulbs and lasers.
2. Xenon emits an intense light in discharge tubes instantly. Due to this it is used in high speed electronic flash bulbs used by photographers.

Question 57.
Give the uses of Radon?
Answer:

1. Radon is radioactive and used as a source of gamma rays
2. Radon gas is sealed as small capsules and implanted in the body to destroy malignant i.e. cancer growth.

Question 58.
Why are pentahalides more covalent than trihalides?
Answer:
Since elements in the +5 oxidation state have less tendency to lose electrons than in the +3 oxidation state, therefore elements in the +5 oxidation state are more covalent than in the +3 oxidation state. In other words, pentahalides are more covalent than trihalides.

59. Why is N₂ less reactive at room temperature?
Answer:
Due to the presence of triple bond between the two nitrogen atoms, the bond dissociation energy of N₂ (941.4 kJ mol^–) is very high. Therefore, N₂ is less reactive at room temperature.

Question 60.
Why is ICl more reactive than I₂ ?
Answer:
ICl bond is weaker than I – I bond. Therefore ICl can break easily to form halogen atoms which readily brings about the reaction.

Question 61.
Why is helium used in diving apparatus?
Answer:
It is used as a diluent for oxygen in modern diving apparatus because of its very low solubility in blood.

Question 62.
Give the disproportionation reaction of H₃PO₃?
Answer:
H₃PO₃ on heating undergoes self-oxidation reduction as:

Question 63.
Why is red Phosphorus less reactive than white Phosphorus?
Answer:
White Phosphorus is more reactive than red Phosphorus under normal conditions because of angular strain in the P₄ molecule where the angles are only 60°.

Question 64.
Nitrogen does not form any pentahalide like Phosphorus. Why?
Answer:
Nitrogen does not form pentahalide due to non-availability of the d-orbitals in its valence shell.

Question 65.
Give reason for the following. Among the noble gases only Xenon is well known to form chemical compounds?
Answer:
Xe is largest in size and has the highest polarising power.

Question 66.
Why is hydrogen sulphide, with greater molar mass a gas, while water a liquid at room temperature?
Answer:
H₂O molecules are associated with intermolecular H-bonding, H₂S is not because oxygen is more electronegative and smaller in size than sulphur. That is why H₂O is a liquid and H₂S is a gas.

Question 67.
Noble gases are chemically inert. Give reasons.
Answer:
Noble gases are chemically inert because they have their octet complete except Helium, i.e. they have a stable electronic configurations.

Question 68.
Why do noble gases exist as monoatomic?
Answer:
Noble gases have stable electronic configuration, that is why they have no tendency to lose or gain electrons.Therefore they do not form covalent bond.

Question 69.
Nitrogen exists as diatomic molecule and Phosphrus as P⁴. Why?
Answer:
Nitrogen has a triple bond between its two atoms because of its small size and high electro negativity. Phosphorus P⁴ has single bond, that is why it is tetra-atomic.

Question 70.
Why is H₂S more acidic than water?
Answer:
It is because bond dissociation energy of S – H bond is less than O – H bond due to longer bond length.

Samacheer Kalvi 12th Chemistry p-Block Elements – II 3 Marks Questions and Answers

Question 1.
Complete the following reactions.
(i) 6Li + N₂ → ?
(ii) 3Ca + N₂ → ?
(iii) 2B + N₂ → ?
Answer:

Qustion 2.
How is ammonia prepared?
Answer:
1. Ammonia is formed by the hydrolysis of urea.
NH₂CONH₂ + H₂O → 2NH₃ + CO₂

2. Ammonia is prepared in the laboratory by heating an ammonium salt with a base.
\({ 2NH }_{ 4 }^{ + }\) + OH^– → 2NH₃+ H₂O
2NH₄Cl + CaO → CaCl₂ + 2NH₃ + H₂O

3. It can also be prepared by heating a metal nitrides such as magnesium nitride with water.
Mg₃N₂+ 6H₂O → 3Mg(OH)₂ + 2NH₃

Question 3.
Explain the structure of ammonia.
Answer:
Ammonia molecule is pyramidal in shape N – H bond distance is 1.016 A and H – H bond distance is 1.645 A with a bond angle 107°. The structure of ammonia may be regarded as a tetrahedral with one lone pair of electrons in one tetrahedral position hence it has a pyramidal shape as shown in the figure.

Question 4.
How will you prepare nitric acid?
Answer:
Nitric acid is prepared by heating equal amounts of potassium or sodium nitrate with concentrated sulphuric acid.
KNO₃ + H₂SO₄ → KHSO₄ + HNO₃

The temperature is kept as low as possible to avoid decomposition of nitric acid. The acid condenses to a fuming liquids which is coloured brown by the presence of a little nitrogen dioxide which is formed due to the decomposition of nitric acid.
4HNO₃ → 4NO₂ + 2H₂O + O₂

Question 5.
Discuss the Commercial method to prepare Nitric acid.
(Or)
How will you prepare nitric acid by Ostwald’s process?
Answer:
Nitric acid prepared in large scales using Ostwald’s process. In this method ammonia from Haber’s process is mixed about 10 times of air. This mixture is preheated and passed into the catalyst chamber where they come in contact with platinum gauze. The temperature rises to about 1275 K and the metallic gauze brings about the rapid catalytic oxidation of ammonia resulting in the formation of NO, which then oxidised to nitrogen dioxide.
4NH₃ + 5O₂ → 4NO + 6H₃O + 120kJ
2NO + O₂ → 2NO₃

The nitrogen dioxide produced is passed through a series of adsorption towers. It reacts with water to give nitric acid. Nitric acid formed is bleached by blowing air.
6NO₂ + 3H₂O → 4HNO₃ + 2NO + H₂O

Question 6.
Draw the structure of the following compounds.
(a) Nitrous acid
(b) Nitric acid
(c) Pernitric acid
Answer:

Question 7.
Identify A, B and C from the following reactions.

1. P₄ + Mg → A
2. P₄+ Ca → B
3. P₄ + Na → C

Answer:

1. P₄ + 6Mg → 2Mg₃P₂ (Magnesium phosphide)
2. P₄ + 6Ca → 2Ca₃P₂ (Calcium phosphide)
3. P₄+ 12Na → 2Na₃P (Sodium phosphide)

Question 8.
How will you prepare phosphine and explain the purification of phosphine?
Answer:
Phosphine is prepared by action of sodium hydroxide with white phosphorous in an inert atmosphere of carbon dioxide or hydrogen.

Phosphine is freed from phosphine dihydride (P₂H₄) by passing through a freezing mixture. The dihydride condenses while phosphine does not. Phosphine can also prepared by the hydrolysis of metallic phosphides with water or dilute mineral acids.

Phosphine is prepared in pure form by heating phosphorous acid.

A pure sample of phosphine is prepared by heating phosphonium iodide with caustic soda solution.

Question 9.
What happens when PH₃ reacts with oxygen or air?
Answer:
When phosphine air or oxygen it bums to give meta phosphoric acid.

Question 10.
Explain the structure of phosphine.
Answer:
In phosphine, phosphorus shows sp₃ hybridisation. Three orbitals are occupied by bond pair and fourth comer is occupied by lone pair of electrons. Hence, bond angle is reduced to 94°. Phosphine has a pyramidal shape.

Question 11.
Discuss the uses of phosphine.
Answer:
Phosphine is used for producing smoke screen as it gives large smoke. In a ship, a pierced container with a mixture of calcium carbide and calcium phosphide, liberates phosphine and acetylene when thrown into sea. The liberated phosphine catches fire and ignites acetylene. These burning gases serves as a signal to the approaching ships. This is known as Holmes signal.

Question 12.
How does PCl₃ reacts with the following reagents?

1. C₂H₅OH
2. C₂H₅COOH

Answer:

1. C₂H₅OH + PCl₃ → C₂H₅Cl + H₃PO₃
2. C₂H₅COOH + PCl₃ → C₂H₅COCl + H₃PO₃

Question 13.
Explain the reaction between PCl₅ and water.
Answer:
Phosphorous pentachloride reacts with water to give phosphoryl chloride and orthophosphoric acid.

Question 14.
Explain the structure of phosphorous trioxide (P₂O₃).
Answer:
In phosphorous trioxide four phosphorous atoms lie at the comers of a tetrahedron and six oxygen atoms along the edges. The P – O bond distance is 165.6 pm which is shorter than the single bond distance of P – O (184 pm) due to pπ – dπ bonding and results in considerable double bond character.

Question 15.
Discuss the structure of phosphorous pentaoxide (P₂O₃).
Answer:
In P₄O₁₀ each P atoms form three bonds to oxygen atom and also an additional coordinate bond with an oxygen atom. Terminal coordinate P – O bond length is 143 pm, which is less than the expected single bond distance. This may be due to lateral overlap of filled p orbitals of an oxygen atom with empty d-orbital on phosphorous.

Question 16.
Mention the uses of oxygen.
Answer:

1. Oxygen is one of the essential component for the survival of living organisms.
2. It is used in welding (oxyacetylene welding)
3. Liquid oxygen is used as fuel in rockets etc.

Question 17.
Give and explain reducing behaviour of sulphur dioxide.
Answer:
As Sulphur dioxide can readily be oxidised, it acts as a reducing agent. It reduces chlorine into hydrochloric acid.
SO₂ + 2H₂O + Cl₂ → H₂SO₄ + 2HCl
It also reduces potassium permanganate and dichromate to Mn²⁺ and Cr³⁺ respectively.

• 2KMnO₄ + 5SO₂ + 2H₂O → K₂SO₄ + 2MnSO₄ + 2H₂SO₄
• K₂Cr₂O₇ + 3SO₂ + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + H₂O

Question 18.
Why bleaching action of sulphur dioxide is temporary?
Answer:
In presence of water, sulphur dioxide bleaches coloured wool, silk, sponges and straw into colourless due to its reducing property.
SO₂ + 2H₂O → 2H₂SO₄ + 2(H)

However, the bleached product (colourless) is allowed to stand in air, it is reoxidised by atmospheric oxygen to its original colour. Hence bleaching action of sulphur dioxide is temporary.

Question 19.
Explain the structure of sulphur dioxide.
Answer:
In sulphur dioxide, sulphur atom undergoes sp² hybridisation. A double bond arises between S and O is due to pπ- dπ overlapping.

Question 20.
How will you manufacture sulphuric acid by contact process?
Answer:
Manufacture of sulphuric acid by contact process involves the following steps:
1. Initially sulphur dioxide is produced by burning sulphur or iron pyrites in oxygen/air.
S + O₂ → SO₂
4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂

2. Sulphur dioxide formed is oxidised to sulphur trioxide by air in the presence of a catalyst such as V₂O₅ or platinised asbestos.

3. The sulphur trioxide is absorbed in concentrated sulphuric acid and produces oleum (H₂S₂O₇). The oleum is converted into sulphuric acid by diluting it with water.

To maximise the yield the plant is operated at 2 bar pressure and 720 K. The sulphuric acid obtained in this process is over 96 % pure.

Question 21.
Prove that H₂SO₄ is a strong dibasic acid.
Answer:
Sulphuric acid forms two types of salts namely sulphates and bisulphates.

Question 22.
Draw the structure of
(a) Marshall’s acid
(b) polythionic acid
(c) Dithionic acid
Answer:

Question 23.
What happens when chlorine reacts with ammonia?
Answer:
1. If chlorine reacts with excess ammonia, it gives nitrogen and ammonium chloride.
8NH₃ + 3Cl₂ → N₂ + 6NH₄Cl

2. If ammonia reacts with excess chlorine, it gives nitrogen trichloride and ammonium chloride.
4NH₂ + 3Cl₂ → NCl₂ + 3NH₄Cl

Question 24.
Bleaching action of chlorine is permanent. Justify this statement.
Answer:
Chlorine is a strong oxidising and bleaching agent because of the nascent oxygen.

Colouring matter + Nascent oxygen → Colourless oxidation product
Therefore, bleaching of chlorine is permanent. It oxidises ferrous salts to ferric, sulphites to sulphates and hydrogen sulphide to sulphur.

Question 25.
Give the uses of chlorine.
Answer:

1. Purification of drinking water
2. Bleaching of cotton textiles, paper and rayon
3. It is used in extraction of gold and platinum

Question 26.
What is Royal water?
Answer:
When three parts of concentrated hydrochloric acid and one part of concentrated nitric acid are mixed, aquaregia is obtained. This is also known as Royal water. This is used for dissolving gold, platinum etc.
Au + 4H⁺ + \({ 4NO }_{ 3 }^{ – }\) + 4 Cl^– → \({ AuCl }_{ 4 }^{ – }\) + NO + 2H₂O
3Pt + 16H⁺ + \({ 4NO }_{ 3 }^{ – }\) + 18Cl^– → \({ 3PtCl }_{ 3 }^{ 2- }\) + NO + 8H₂O

Question 27.
Why HF is not stored in glass bottles?
Answer:
HF rapidly reacts with silica and glass to form soluble salt and that process leads to break the glass bottles. Hence HF is not stored in glass bottles.
SiO₂ + 4HF → SiF₄ + 2H₂O
Na₂SiO₃ + 6HF → Na₂SiF₆ + 3H₂O

Question 28.
Give the properties of inter halogen compounds.
Answer:
Properties of inter halogen compounds:

1. The central atom will be the larger one.
2. It can be formed only between two halogen and not more than two halogens.
3. Fluorine can’t act as a central metal atom being the smallest one.
4. Due to high electronegativity with small size fluorine helps the central atom to attain high coordination number
5. They can undergo the auto ionization.
6. They are strong oxidizing agents.

Question 29.
How will you prepare Xenon fluoride?
Answer:
Xenon fluorides are prepare by direct reaction of xenon and fluorine under different conditions as shown below.

Question 30.
Complete the following reaction.
(i) XeOF₄ + SiO₂ → A + SiF₆
(ii) A + SiO₂ → B + SiF₆
Answer:

Question 31.
What happens when XeF₆ reacts 2.5 M solution of NaOH?
Answer:
When XeF₆ reacts with 2.5 M of NaOH, sodium per xenate is obtained.

Question 32.
Give two examples to show the anomalous behaviour of fluorine.
Answer:
The anomalous behaviour of fluorine is due to its:

1. Small size
2. highest electronegativity
3. low F-F bond dissociation enethalpy
4. non-availability of d-orbitals in its valence shell.

The two examples are:
1. Due to non-availability of d-orbitals in its valence shell, fluorine cannot expand its octet, therefore, shows only -1 oxidation state while all other halogens due to the presence of d-orbitals shows positive oxidation states of +1, +3, +5 and +7 besides oxidation state of -1.

2. Due to its small size, the three lone pair of electrons on each F atom in F – F molecule, repel the bond pair. As a result, F – F bond dissociation energy is lower than that of Cl – Cl bond.

Question 33.
Why does the reactivity of nitrogen differ from Phosphorus?
Answer:
Nitrogen exists as a diatomic molecule (N \(\equiv\) N). Due to the presence of a triple bond between the two N – atoms the bond dissociation energy is large (941.4 kJ mol^-1). As a result nitrogen is said to be chemically inert in its elemental state. In contrast, P – P single bond is much weaker (213 kJ mol^-1) than N \(\equiv\) N triple bond Therefore, phosphorus is much more reactive than nitrogen.

Question 34.
Why does NH₃ form hydrogen bond but PH₃ does not?
Answer:
The electronegativity of N (3.0) is much higher than that of FI (2.1). As a result, N – H bond is quite polar and hence NH₃ undergoes intermolecular H-bonding. In contrast, both P and H have an electronegativity of 2.1. Therefore P – H bond is non-polar and hence PH₃ does not undergo H – bonding.

Question 35.
Can PCl₅ act as an oxidising as well as a reducing agent? Justify.
Answer:
Oxidation state of P in PCl₅ is +5. Since P has five valence electrons in its valence shell, therefore it cannot increase its oxidation state beyond +5 by donating its electrons, therefore PCl₅ cannot act as a reducing agent. However, it can decrease its oxidation number from +5 to +3 or some lower value, therefore PCl₅ acts as an oxidising agent. For example, it oxidises Ag to AgCl.

Samacheer Kalvi 12th Chemistry p-Block Elements – II 5 Mark Questions and Answers

Question 1.
How does ammonia react with

1. Excess Cl₂
2. Na
3. CuSO₄
4. O₂ / ∆

Answer:
1. Reaction with Excess Cl₂

2. Reaction with Na:

3. Reaction with CuSO₄

4. Reaction with O₂

Question 2.
Explain the reaction of metals with nitric acid.
Answer:
The reactions of metals with nitric acid are explained in 3 steps as follows:
Primary reaction:
Metal nitrate is formed with the release of nascent hydrogen
M + HNO₃ → MNO₃ + (H)

Secondary reaction:
Nascent hydrogen produces the reduction products of nitric acid.

Tertiary reaction:
The secondary products either decompose or react to give final products

For examples:
Copper reacts with nitric acid in the following manner
3Cu + 6HNO₃ → 3Cu(NO₃)₂ + 6(H)
6(H) + 3HNO₃ → 3HNO₂ + 3H₂O
3HNO₂ → HNO₃ + 2NO + H₂O

Overall reaction
3Cu + 8HNO₃ → 3CU(NO₃)₂ + 2NO + 4H₂O

The concentrated acid has a tendency to form nitrogen dioxide
Cu + 4HNO₃ → 3CU(NO₃)₂ + 2NO₂ + 2H₂O

Question 3.
How will you prepare ozone by laboratory method? Explain the structure of ozone.
Answer:
In the laboratory ozone is prepared by passing electrical discharge through oxygen. At a potential of 20,000 V about 10% of oxygen is converted into ozone it gives a mixture known as ozonised oxygen. Pure ozone is obtained as a pale blue gas by the fractional distillation of liquefied ozonised oxygen.


Structure of ozone:
The ozone molecule have a bent shape and symmetrical with delocalised bonding between the oxygen atoms.

Question 4.
A is a king of acid. A reacts with HBr to give B and Bromine. A reacts with Na₂CO₃ to give C and carbon dioxide. Identify A, B and C. Give the reaction.
Answer:
1. King of acid is sulphuric acid (A).

2. Sulphuric acid (A) reacts with HBr to give SO₂ (B) and Bromine.

3. Sulphuric acid (A) reacts with Na2CO₃ to give sodium sulphate (C) and CO₂.

Question 5.
How does Sulphuric acid react with the following:
(a) Al
(b) KNO₃
(c) NaBr
(d) C₆H₆
Answer:

Question 6.

1. Explain the test for sulphate (or) sulphuric acid.
2. What happens when sulphuric acid reacts with oxalic acid?

Answer:
1. Dilute solution of sulphuric acid or aqueous solution of sulphates gives white precipitate with barium chloride solution. It can also be detected using lead acetate solution. Here a white precipitate of lead sulphate is obtained.

2. Sulphuric acid reacts with oxalic acid to give CO and CO₂

Question 7.
Discuss the manufacture of chlorine.
Answer:
Electrolytic process:
When a solution of brine (NaCl) is electrolysed, Na+ and CP ions are formed. Na⁺ ion reacts with OH^– ions of water and forms sodium hydroxide. Hydrogen and chlorine are liberated as gases.

Deacons process:
In this process a mixture of air and hydrochloric acid is passed up a chamber containing a number of shelves, pumice stones soaked in cuprous chloride are placed. Hot gases at about 723 K are passed through a jacket that surrounds the chamber.

The chlorine obtained by this method is dilute and is employed for the manufacture of bleaching powder. The catalysed reaction is given below,

Question 8.

1. How is bleaching powder prepared?
2. What happens when benzene reacts with chlorine?

Answer:
1. Bleaching powder is prepared by passing chlorine gas through dry slaked lime (Calcium hydroxide).

2. Benzene reacts with chlorine in the presence of ferric chloride to give chlorobenzene.

Question 9.
Cone. H₂SO₄ is added followed by heating to each of the following test tubes labelled (I) to (IV). Identify in which of the above test tube the following change will be observed. Support your answer with the help of chemical equation:
(a) formation of black substance
(b) evolution of brown gas
(c) evolution of colourless gas
(d) formation of a brown substance which on dilution becomes blue.
(e) disappearance of yellow powder along with evolution of colourless gas.
Answer:

Question 10.
Give reasons for each of the following:

1. Bleaching of flowers by Cl₂ is permanent while by SO₂ is temporary.
2. Molten aluminium bromide is a poor conductor of electricity.
3. Nitric oxide becomes brown when released in air.
4. PCl₅ is ionic in nature in the solid state.
5. Ammonia is a good complexing agent.

Answer:
1. Cl₂ bleaches by oxidation, while SO₂ does it by reduction. The reduced product gets oxidise again and the colour is regained back.

2. Aluminium bromide exists as a dimer, Al₂Br₆. In this structure, each aluminium atom forms one coordinate bond by accepting a lone pair of electrons from the bromine atom of another aluminium bromide molecule and thus complete the octet of electrons. Due to lack of free electrons, molten aluminium bromide is a poor conductor of electricity.

3. Nitric oxide reacts with air and oxidised into NO₂ which is brown in colour.
2NO + O₂ → 2NO₂

4. In solid state PCl₅ exists as [PCl₄]⁺ [PCl₆]^– and hence it is ionic in nature. Due to its ionic nature, it conducts current on fusion.

5. N atom in ammonia has lone pair of electrons which can coordinate with other atoms or cations required for the stability of electron pair.

Question 11.
How will you prepare the following compounds.
(a) Hyponitrous acid
(b) Nitrous acid
(c) Pernitrous acid
(d) Pernitric acid
Answer:

Common Errors And Its Rectificaations:

Common Errors:

1. Oxidation number rules may be confused.
2. Oxidation number of oxygen may get confused.
3. Molecular formula and compound name may get confused.

Rectifications:

1. Oxidation number of fluorine is always -1.
2. Always oxygen is -2 but in peroxide it is -1. When it comes as a first element its oxidation number is positive..
3. Student should memorise the formula using short cut method, e.g. Phosphorous acid – H₃PO₃ Phosphoric acid – H₃PO₄ e.g. Phosphorous acid – H₃PO₃ By pronunciation, ‘O’ indicates oxygen is more.

Hope you love the Samacheer Kalvi 12th Chemistry Chapter Wise Material. Clearly understand the deep concept of Chemistry learning with the help of Tamilnadu State Board 12th Chemistry Chapter 3 p-Block Elements – II Questions and AnswersPDF. Refer your friends to and bookmark our website for instant updates. Also, keep in touch with us using the comment section.