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Telangana TSBIE TS Inter 2nd Year Chemistry Study Material Lesson 6(c) Group-17 Elements Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material Lesson 6(c) Group-17 Elements

Very Short Answer Questions (2 Marks)

Question 1.
Which halogen produces O₂ and O₃ on passing through water?
Answer:
Fluorine liberates O₂ and O₃ with water.
2F₂ + 2H₂O → 4HF + O₂
3F₂ + 3H₂O → 6HF + O₃

Question 2.
Interhalogen compounds are more reactive than the constituent halogens except fluorine- explain.
Answer:
The interhalogen compounds are more reactive than the halogens (except F₂). This is because the A-X bond in interhalogens is weaker than the X – X bond in the halogens.

Question 3.
What is the use of ClF₃?
Answer:

1. ClF₃ is a covalent liquid. It is used to make gaseous UF₆.
   3ClF₃ + U → UF₆ + 3ClF
2. It acts as powerful fluorinating agent for inorganic substances.
   4ClF₃ + 6MgO → 6MgF₂ + 2Cl₂ + 3O₂

Question 4.
Write two uses of ClO₂.
Answer:

1. ClO₂ is used as a bleaching agent for paper pulp and textiles.
2. Water treatment.

Question 5.
Why are halogens coloured ?
Answer:
The colour of halogens is due to absorption of visible light by their molecules resulting in excitation of outer electrons to higher energy levels. The excitation energy required by small fluorine atoms is large and that required by the large iodine atoms is least. By absorbing different quanta of radiation, they display different colours. F₂ is yellow, Cl₂ is greenish yellow, Br₂ is red and I₂ is violet.

Question 6.
Write the reactions of Cl₂ and F₂ with water. [IPE ’14]
Answer:

1. Chlorine dissolves in water to give hydrochloric acid and hypochlorous acid.
   Cl₂ + H₂O → HCl + HOCl
2. Fluorine oxidises water to oxygen and ozone.
   2F₂ + 2H₂O → 4HF + O₂
   3F₂ + 3H₂O → 6HF + O₃

Question 7.
With which neutral molecule, ClO^– is iso- electronic? Is that molecule a Lewis base ?
Answer:
ClF is isoelectronic with ClO^–. ClF is a Lewis base.

Question 8.
Arrange the following in the order of the property indicated for each set.
a) F₂,Cl₂, Br₂, I₂ – increasing bond dissociation enthalpy.
b) HF, HCl, HBr, HI – increasing acidic strength.
c) HF, HCl, HBr, HI – increasing boiling points.
Answer:
a) I₂ < f₂ < Br₂ < Cl₂
b) HF < HCl < HBr < HI
c) HCl< HBr < HI < HF

Question 9.
Electron gain enthalpy of fluorine is less than that of chlorine. Explain.
Answer:
Fluorine has lower electron gain enthalpy than chlorine. The lower value is attributed to small size of fluorine. Because of its small size, there is repulsion between the electron pairs already present and the electrons that are added. Hence, electron adding up tendency of F atom is less.
Therefore, Electron gain enthalpy of fluorine is – 333 kJ / mol. is less than that of chlorine (-349 kJ/mol).

Question 10.
HF is a liquid while HCl is gas -explain.
Answer:
HF is an associated liquid because of hydrogen bonds.
H – F ……………… H – F ………… H – F
HCl is not an associated liquid. Hence its boiling point is less than that of HF. Hence it is a gas.

Question 11.
Bond dissociation enthalpy of F₂ is less than that of Cl_2^– explain.
Answer:
Due to small size of fluorine atoms, the F – F distance is also small. Hence internuclear repulsion is high. The large repulsions bet-ween lone pair of electrons present on the two fluorine atoms weaken the bond. This repulsion is less in other molecules, because the atoms are large in size.

Question 12.
Write the formulae of the compounds, in which oxygen has positive oxidation states and mention the oxidation states of oxygen in them.
Answer:

Question 13.
What is the use of O₂F₂ and I₂O₅?
Answer:
O₂F₂ oxidises plutonium to PuF₆ and the reaction is used in removing plutonium as PuF₆ from spent nuclear fuel.
I₂O₅ is a good oxidising agent and is used in the estimation of carbon monoxide.

Question 14.
Write two uses of hydrogen chloride.
Answer:

1. HCl is used for ‘pickling’ metals that is removing oxide layers from the surface.
2. To make metal chlorides.
3. In the manufacture of dyestuffs.

Question 15.
Explain the reactions of Cl₂ with NaOH.
Answer:

1. With cold, dilute NaOH, chloride and hypochlorite are formed.
   
2. With hot, cone. NaOH, chloride and chlorate are formed.
   

Question 16.
What happens when Cl2 reacts with dry slaked lime ? [Mar. 2018 AP & TS ; AP 17]
Answer:
Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O
Bleaching powder is formed when chlorine reacts with dry slaked lime.

Question 17.
Chlorine acts as an oxidising agent Explain with two examples. [AP ’16]
Answer:

1. Chlorine oxidises ferrous sulphate for ferric sulphate.
   2FeSO₄ + H₂SO₄ + Cl₂ → Fe₂ (SO₄)₃ + 2HCl
2. Chlorine oxidises SO₂ to SO₃ and in presence of water, H₂SO₄ is formed.
   SO₂ + 2H₂O + Cl₂ → H₂SO₄ + 2HCl

Question 18.
What is aqua-regia? Write its reaction with gold and platinum.
Answer:
When 3 parts cone. HC/ and 1 part cone. HN03 are mixed, aqua regia is formed.
\(\mathrm{Au}+4 \mathrm{H}^{+}+\mathrm{NO}_3^{-}+4 \mathrm{C} t \longrightarrow \mathrm{AuCl}_4^{-}+\mathrm{NO}+2 \mathrm{H}_2 \mathrm{O}\)
\(3 \mathrm{Pt}+16 \mathrm{H}^{+}+4 \mathrm{NO}_3^{-}+18 \mathrm{Cl} \longrightarrow 3 \mathrm{PtCl}_6^{2-}+4 \mathrm{NO}+8 \mathrm{H}_2 \mathrm{O}\)

Question 19.
How is Cl₂ manufactured by Deacon’s [AP & TS Mar. 19; (AP 17; TS 16)]
Answer:
Cl₂ is formed when HCl gas is oxidised by atmospheric oxygen in presence of CuCl₂ at 723K.

Question 20.
Chlorine acts as a bleaching agent only in the presence of moisture – Explain.
Answer:
Bleaching action of chlorine is due to oxidation. In presence of moisture, chlorine dis-solves in water to give nascent oxygen.
Cl₂ + H₂O → 2HCl + (O)
Coloured substance + (0) → Colourless substance

Question 21.
The decreasing order of acidic character among hypo halogen acids is HClO > HBrO > HIO. Give reason.
Answer:
The acidic character of HClO is due to its dissociation to give HCl.
HClO → HCl + (O)
Hypobromites and hypoiodides tend to disproportionate.
3BrO^– → 2Br^– + BrO₃^–
Hence stability of HBrO and HIO is less. Hence less acidic.

Question 22.
The acidic nature of oxoacids of chlorine is HOCl < HClO₂ < HClO₃ < HClO₄ – Explain. (Hint: HA + H₂O ⇌ H₃O⁺ + A^– conjugate base, greater the stability of A^–, lesser will be its basic strength or greater will be the tendency of HA to release H⁺. In other words, stronger will be the acid HA. Among the conjugate bases of oxoacids of chlorine, the stability order is OCl^– < ClO₂^– < ClO₃^– < ClO₄^–)
Answer:
Among the conjugate bases of oxoacids of chlorine, the stability order is
OCl^– < ClO₂^– < ClO₃^– < ClO₄^–
The more the stability of conjugate base, the stronger will be acid. Hence HClO₄ is a strong acid as ClO₄^– is stable. HOCl is weak as OCl^– is less stable.

Question 23.
What are interhalogen compounds? Give two examples.
Answer:
The binary compounds formed by the combination of halogens among themselves are called interhalogen compounds.
Ex : AX type ; ICl, Cl F
AX₃ type ; ClF₃, BrF₃
AX₅ type ; BrF₅, IF₅

Question 24.
Explain the structure of ClF₃.
Answer:
Excited chlorine

Chlorine undergoes sp³d hybridisation.
It assumes trigonal bipyramidal shape with two lone pairs of electrons. ClF₃ is planar, T- shaped with bond angle 90°. Lone pair distorts the angle slightly.

Question 25.
OF₂ should be called oxygen difluoride and not fluorine oxIde – Why?
(HInt : F Is more electronegative than Oxygen).
Answer:
As fluorine is more electronegative than oxygen, fluorine is in – 1 oxidation state. Hence the compounds of oxygen with fluorine are called fluorides. So, OF₂ is to be called oxygen difluoride.

Question 26.
Iodine is more soluble in KI than in water -Explain.
Answer:
Iodine combines with KI in aqueous solution to form KI₃.
KI + I₂ → KI₃
As complex formation increases solubility, solubility of I₂ is more in KI.

Question 27.
Among the hydrides of halogens (a) which is most stable? (b) which is most acidic? (c) which has lowest boiling point?
Answer:
a) HF is most stable,
b) HI is most acidic.
c) HCl has lowest boiling point.

Question 28.
Compare the bleaching action of Cl₂ and SO₂.
Answer:
Cl₂ acts as bleaching agent in presence of moisture. Cl₂ bleaches by oxidation.
Cl₂ + H₂O → HCl + (O)
Coloured substance + (0) → Colourless substance
SO₂ bleaches through reduction.
SO₂ + 2H₂O → H₂SO₄ + 2H
Coloured matter + 2[H] → Colourless matter

Question 29.
Give the oxidation states of halogens in the following. [TS ’15]
a) Cl₂O
b) ClO₂^–
c) KBrO₃
d) NaClO₄
Answer:
a) Let the oxidation state of Cl be x.
2x – 2 = 0
x = + 1
∴ O.S. of Cl in Cl₂O is + 1.

b) ClO₂^–; x – 4 = – 1
x = 4 – 1 = +3
∴ O.S. of Cl in ClO₂^– is + 3.

c) KBrO₃
O.S. of K = + 1; O.S. of O is -2
+1 + x – 6 = 0
x = + 5

d) NaClO₄
+1 + x – 8 = 0
x – 7 = 0
x = +7
O.S. of Cl in NaClO₄ is +7.

Question 30.
Describe the molecular shape of I₃^–. (Hint: Central iodine is of SP³d. – linear)
Answer:
Central Iodine atom is in sp³d hybridisation.

Short Answer Questions (4 Marks)

Question 31.
How can you prepare Cl₂ from HCl and HCl from Cl₂? Write the reactions.
Answer:

1. HCl is oxidised by MnO₂ to Cl₂.
   By heating manganese dioxide with concentrated HCl.
   MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O
2. Chlorine reacts with H₂ to form HCl.
   H₂ + Cl₂ → 2HCl

Question 32.
Write balanced equations for the following. [IPE ’14]
a) NaCl is heated with cone. H₂SO₄ in the presence of MnO₂
b) Chlorine is passed into a solution of NaI in water.
Answer:
a) Cl₂ gas liberated.
4NaCl + MnO₂ + 4H₂SO₄ → MnCl₂ + 4NaHSO₄ + 2H₂O + Cl₂
b) I₂ is liberated.
2NaI + Cl₂ → 2NaCl + I₂

Question 33.
Explain the structures of
a) BrF₅ and
b) IF₇.
Answer:
a) BrF₅: Central Bromine atom in BrF₅ undergoes sp³d² hybridisation. It has distorted octahedral shape.

b) IF₇ : Central Iodine atom in IF₇ undergoes sp³d³ hybridisation. It has pentagonal bipyramidal structure.

Question 34.
Write a short note on hydrogen halides.
Answer:
Halogens react with hydrogen forming hydrogen halides.
H₂ + X₂ → 2HX
Properties:

1. Stability decreases from HF to HI.
   HF > HCl > HBr > HI
2. The acidic strength increases in the order HF < HC/ < HBr < HI.
3. The boiling points HCl < HBr < HI < HF
   HF is associated liquid due to hydrogen bonding
   H – F …………. H – F …………. H – F

Uses:

1. HF is used in etching of glass.
2. HCl is used in the manufacture of chlorine, NH₄Cl and glucose.
3. HCl is used for extracting glue from bones and purifying bone black.

Question 35.
How is chlorine obtained in the laboratory ? How does it react with the following ? [(Mar. 18 AP) (AC & TS 16)]
a) cold dil. NaOH
b) excess NH₃
c) KI
Answer:
By heating concentrated hydrochloric acid with Manganese dioxide.
MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O
Reactions:
a) When chlorine is passed through cold dil. NaOH, chloride and hypochlorite are formed.
2NaOH + Cl₂ → NaCl + NaOCl + H₂O

b) Chlorine gives nitrogen and ammonium . chloride with excess ammonia.
8NH₃ + 3Cl₂ → 6NH₄Cl + N₂

c) When Cl₂ gas is passed through KI, I₂ is liberated.
2KI + Cl₂ → 2KCl + I₂

Question 36 .
What are interhalogen compounds? Give some examples to illustrate the definition. How are they classified?
Answer:
When two different halogens react with each other, interhalogen compounds are formed.
They are classified as AX, AX₃, AX₅ and AX₇ where A is halogen of larger size and X is smaller size.
They are prepared by direct combination or by the action of halogen on lower inter-halogen compounds.

Example:
AX Type : ClF, BrF, ICl
AX₃Type : ClF₃, BrF₃, ICl₃
AX₅ Type : BrF₅, IF₅
AX₇ Type: IF₇

Long Answer Questions (8 Marks)

Question 37.
How is ClF₃ prepared? How does it react with water? Explain its structure.
Answer:
ClF₃ is prepared by the reaction of Cl₂ with excess of Fluorine.

ClF₃ is a colourless gas. It has bent T-shape structure.

Reaction with water :
Chloric acid is formed when ClF₃ undergoes hydrolysis.
ClF₃ + 2H₂O → 3HF + HClO₂
Structure:

Excited state of chlorine
Chlorine undergoes sp³d hybridisation. There are two lone pairs and three bond pairs.

ClF₃ is T – shaped. The bond pairs and lone pairs occupy corners of trigonal bipyramid. The two lone pairs occupy the equatorial positions to minimise the repulsions.

Question 38.
How is chlorine prepared in the laboratory ? How does it react with the following ? [TS ’15]
a) Iron
b) hot cone. NaOH
c) acidified FeSO₄
d) Iodine
e) H₂S
f) Na₂S₂O₃
Answer:
Chlorine is prepared in the laboratory by heating concentrated Hydrochloric acid with Manganese dioxide.
MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O
It can also be prepared by heating a mixture of sodium chloride and concentrated H₂SO₄.
4NaCl + MnO₂ + 4H₂SO₄ → MnCl₂ + 4NaHSO₄ + 2H₂O + Cl₂
Reactions:
a) It oxidises Fe to Fe3+.
2Fe + 3Cl₂ → 2FeCl₃

b) When Cl₂ is passed through hot cone. NaOH sodium chlorate is formed.
6NaOH + 3Cl₂ → 5NaCl + NaClO₃ + 3H₂O

c) It oxidises acidified ferrous sulphate to ferric sulphate.
2FeSO₄ + H₂SO₄ + Cl₂ → Fe₂(SO₄)₃ + 2HCl

d) With I₂ in presence of water, it forms iodic acid.
I₂ + 6H₂O + 5Cl₂ → 2HIO₃+ 10HCl

e) It reacts with H₂S to form HCl.
H₂S + Cl₂ → 2HCl + S ↓

f) Hypo reacts with Cl₂ to form HCl and precipitates sulphur.
Na₂S₂O₃ + Cl₂ + H₂O → Na₂SO₄ + 2HCl + S ↓

Question 39.
Discuss anomalous behaviour of fluorine.
Answer:
Fluorine, the first element of halogen group shows anomalous properties.

1. Its high reactivity is due to small bond dissociation enthalpy of F – F bond and high electronegativity.
2. The electronegativity, ionisation enthalpy and electrode potentials are higher them expected.
3. Bond dissociation and electron gain enthalpy are quite lower than expected.
4. The anomalous behaviour of fluorine is due to its small size, highest electronega-tivity, low F- F bond dissociation enthalpy and non-availability of d-oribtals in the valence shell.
5. HF is a liquid while other hydrogen halides are gases.
6. Most of the reactions of fluorine are exothermic.
7. AgCl is insoluble in water whereas AgF is soluble.
8. CaCl₂ is soluble whereas CaF₂ insoluble.

Question 40.
How is chlorine prepared by electrolytic method? How does it react with a) NaOH and b) NH₃ under different conditions. [AP ’16, ’15]
Answer:
Chlorine is obtained by the electrolysis of brine solution. Chlorine is liberated at anode.

Brine solution is taken in a perforated U-shaped steel vessel. Graphite rod acts as anode, steel vessel acts as cathode. H₂ and NaOH are by products.

NaCl → Na⁺ + Cl^–
At Anode : 2Cl^– – 2e → Cl₂
2H₂O + 2e → 2OH^– + H₂
Na⁺ + OH^– → NaOH
Reactions:

1. When chlorine gas is passed through cold, dil. NaOH, NaCl and NaOCl are formed.
   2NaOH + Cl₂ → NaCl + NaOCl + H₂O
2. When Cl₂ gas is passed through hot concentrated NaOH, sodium chloride and sodium chlorate are formed.
   6NaOH + 3Cl₂ → 5NaCl + NaClO₃+ 3H₂O

b) With excess ammonia, Nitrogen gas is evolved.
8NH₃ + 3Cl₂ → 6NH₄Cl + N₂
When chlorine is in excess, NCl₃ and HCl are formed.
NH₃ + 3Cl₂ → NCl₃ + 3HCl

Question 41.
Write the names and formulae of oxoacids of chlorine. Explain their structures and acidic relative nature.
Answer:
Hypochlorous acid HOCl
Chlorous acid HClO₂
Chloric acid HClO₃
Perchloric acid HClO₄
Acidic character : Among the different oxoacids of chlorine, the acidic character follows the order.
HOCl < HClO₂ < HClO₃ < HClO₄
When these acids, dissolve in water their conjugate bases OCl^–, ClO₂^–, ClO₃^– and ClO₄^– are formed. As the oxidation of chlorine increases, number of n bonds increase and delocalisation increases. As delocalisation of n electrons increases, stability of conjugate base increases. Hence the ability of anion to accept proton decreases. Hence the basic character of anions increases in the order.
ClO₄^– < ClO₃^–– < ClO₂^– < ClO^–
ClO^– is strong base, its conjugate acid HOCl is weak.
ClO₄^– is weak base, its conjugate acid HClO₄ is strong.
Hence the acidic character increases and the order is
HOCl < HClO₂ < HClO₃ < HClO₄.

Structures: In all these acids, chlorine uses sp³ hybrid orbitals.
1) HOCl: It has linear shape no dπ – pπ bonds with three line pair of electrons.

2) HClO₂ : The anion of the acid is ClO₂^–. It has a V shape. There are two lone pairs in chlorine. The electron in 3d orbitals forms a π bond with oxygen.

HClO₃ : ClO₃^– is pyramidal. Two unpaired electrons in 3d form two dπ – pπ bonds with oxygen. There is one lone pair.

HClO₄ : Perchlorate ion has tetrahedral shape. Three dπ – pπ bonds are formed.

As there is more delocalisation the stability of ions increases in the order.
ClO^–< ClO₂^– < ClO₃^– < ClO₄^–

Intext Questions – Answers

Question 1.
Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F₂ and cl₂.
Answer:
Because of low enthalpy of dissociation of F – F bond and high hydration enthalpy, F₂ acts as strong oxidising agent than chlorine.

Question 2.
Give two examples to show the anoma¬lous behaviour of Flurine.
Answer:
HF is a liquid while others (HCl, HBr, Hr) are 4. gases. AgCl is insoluble in water whereas AgF is soluble.

Question 3.
Sea is the greatest source of some Halogens. Comment.
Answer:
Sea water contains chlorides, bromides and Iodides of sodium, potassium, magnesium and calcium. The deposits of dried up seas contain sodium chloride and carnalite. Certain forms of marine life and various seaweeds contain Iodine. Hence we can say that sea is the greatest source of some Halogens.

Question 4.
Give the reason for the bleaching action of Cl₂.
Answer:
Chlorine bleaches by oxidation. Coloured substance + (0) → Colourless substance

Question 5.
Name two poisonous gases which can be prepared from chlorine gas.
Answer:
Phosgene, Tear gas, Mustard gas.

Question 6.
Why is ICl more reactive than I₂?
The ICl bond is weaker than I – I bond.

Telangana TSBIE TS Inter 2nd Year Accountancy Study Material 3rd Lesson Accounting for Not-for-Profit Organisation Textbook Questions and Answers.

TS Inter 2nd Year Accountancy Study Material 3rd Lesson Accounting for Not-for-Profit Organisation

Very Short Answer Questions

Question 1.
Explain the meaning of Non-profit organisation.
Answer:
The organisations, whose main object is not to earn profit, but to render service to their members called as a non-profit organisation. Eg: Education Institutions, Hospitals clubs Religious Institutions, Co-operation societies.

Question 2.
Write the characteristics of Not-for-profit organisation.
Answer:

1. Not-for-profit organisations are to serve its members and the society in general.
2. These organisations will come into existence to promote Arts, Science, Religion and Spirituality.
3. These organisations do not declare and pay any dividend to its members.
4. These organisations are managed by elected members.

Question 3.
What is the differences between Capital expenditure and Revenue expenditure?
Answer:

Question 4.
What are the differences between Capital receipts and Revenue receipts?
Answer:

Question 5.
What is Deferred Revenue Expenditure?
Answer:

1. When heavy or unusual expenditure of revenue in nature is incurred in a particular year and the benefit of which extends beyond that year, it is treated as “Deferred Revenue Expenditure.
2. In other words the expenditure is revenue in nature, but its benefit is spread over a number of years.
3. For example Preliminary expenses, heavy expenditure on the advertisement.

Question 6.
Explain the differences between Payment and Expenditure.
Answer:

1. Expenditure includes all expenses incurred during particular period, it may be either paid or to be paid. Payment means actual cash paid. It may be for the current year or for a past period or for a future period.
2. Payment may be an expenditure, but an expenditure need not necessarily be a payment.

Question 7.
Explain the differences between Receipt and Income.
Answer:
1. Receipt means actual cash received. It may be for the current year or for a past period or for a future period. Whereas an income relates to a specific period, it may be accrued or earned during that period; it may be either received or receivable.

2. Receipt is different from income. An income may be receipt, but a receipt need not necessarily be an income.

Question 8.
Explain the following terms.
Answer:
1) Donations :

1. Any amount received by the organizations from the individuals and institutions voluntarily is called “Donations”.
2. Donations are divided into two types :
   A) General Donations. B) Specific Donations

2) Legacy :

1. Amount received by the organization as per the “will” of a person is called legacy. It is non-recurring nature and to be treated as capital income.
2. Accounting treatments: Legacy is treated as capital income and recorded on the liabilities side of the Balance sheet.

3) Entrance Free:

1. The amount/Fee, paid by the new members at the time of joining the organisations is known as Entrance fees.
2. It is treated as revenue income and it should be credited to income and expenditure account.

4) Subscriptions: It is an amount paid by the members regularly at periodical intervals. Subscriptions are regular source of income for the organization.

5) Capital fund : Excess of assets over liabilities is called capital fund. It is general fund which is similar to capital account of profit-making organisations.
Capital fund = Assets – Liabilities

6) Specific funds: Specific fund received for specific purposes such as building fund or tournament fund, prize fund etc.

Short Answer Questions

Question 1.
Write the features of receipts and payment Account.
Answer:

1. Receipts and Payment account is a real account that records all cash receipts and Payments.
2. It is a summary of cash and bank transactions.

Features :

1. Receipts and payments account is similar to cash book.
2. It is a real account.
3. This account reveals opening and closing balances of cash and bank.
4. It is maintained and cash basis of accounting.
5. Credit transactions are not recorded in receipts and payments account.
6. All cash receipts are shown on debit side, all cash payments are shown on credit side of the account.
7. This account depicts closing balance of cash and bank at the end of the year.

Question 2.
What is an Income and Expenditure account? Explain its features.
Answer:

1. The income and Expenditure account is similar to Profit and Loss account.
2. On debit side Revenue expenditure and on the credit side revenue income are recorded to ascertain the financial result of the organization.

Features :

1. It is similar to profit and loss account.
2. It is a nominal account.
3. All revenue expenses are recorded on the debit side and all revenue incomes on the credit side of the income and expenditure account.
4. Only revenue items are taken into consideration. Capital items are to be totally excluded.
5. There is no closing balance, but the difference between two sides of account shows either surplus or deficit.
6. It also records non-cash items like depreciation.
7. The surplus/deficit is transferred to capital fund in the balance sheet.

Question 3.
Explain the difference between Receipts and Payments account and Income and Expenditure account.
Answer:

Question 4.
What is donation? Explain the types of donations.
Answer:
Donations are the amount received by organizations from individuals and institutions as a gift.
Donations are two types :
a) General donations
b) Special donations (or) specific donations.

a) General donations: General donations is the amount given by the individuals without mentioning the purpose. So that the amount can be utilised for any purpose by the not-for- profit organizations.

b) Special donations (or) specific donations: A donation received by the organization for a specific purpose is called specific donations. For example: Donation for buildings.

Question 5.
Bring out the difference between Capital and Revenue items by giving suitable examples.
Answer:
I. Capital Items :
1) Capital Expenditure: Capital expenditure is the expenditure which incurred for the acquisition assets for the expenditure incurred to increase the earning capacity of the existing assets.
For Example Purchases of assets like buildings, machinery, furniture, investments, books, installation of assets etc.

2) Capital Receipts: Capital receipts of business consist of capital contributed by the members, special donations and sale of fixed assets.
Capital receipts is non-recurring nature.
For Example Legacies, Government grants, Life membership fees.

II. Revenue Items:
1) Revenue Expenses: These expenses are incurred in the day-to-day operations of the organizations. For Eg. Salaries, wages, transport, rent paid, postage, printing and stationery, newspapers, interest paid etc. and also includes expenditure on maintenance of fixed assets such as repairs, depreciations etc.

2) Revenue Receipts: These receipts are generally obtained from the day-to-day operations of the organizations. Eg: Subscriptions, rent received, interest received, proceeds from entertainment, lectures etc.

Textual Problems

Question 1.
From the following details, prepare Receipts and Payments account.
Opening balance of cash — 1,500
Opening bank balance — 4,500
Subscriptions collected — 8,000
Entertainment show receipts — 4,000
Entrance fees received — 2,000
Computer purchased — 3,000
Tournament expenses — 3,000
Entertainment show expenses — 1,800
Opening balance of cash — 1,500
Opening bank balance — 4,500
Subscriptions collected — 8,000
Entertainment show receipts — 4,000
Entrance fees received — 2,000
Computer purchased — 3,000
Tournament expenses 3,000
Entertainment show expenses 1,800
Answer:

Question 2.
From the following details, prepare the Receipts and Payments account of the Officers club for the year ending 31st March 2020.

Opening cash in hand — 1,400
Opening cash at bank — 3,400
Subscriptions received — 25,000
Donations collected — 7,000
Honorarium paid — 6,000
Rent paid — 3,000
Tournament expenses — 1,000
Shares purchased — 5,000
Entrance fee received — 2,600
Paid for internet connection — 1,500
Electricity charges — 500
Repairs and maintenance — 400
Stationery — 500
Postage — 1,000
Dinner expenses — 1,500
Cash in hand at the end — 4,000
Answer:

Question 3.
Prepare Receipts and Payments account of Sandeepani Educational society’ for the year ended 31-3-2020.
Balance of cash — 2,500
Bank balance — 3,000
Subscriptions received for 2018-19 — 4,000
Subscriptions received for 20 19-20 — 5,000
Donations received — 2,000
Salaries paid — 1,000
Life membership fees received — 3,000
Rent paid in advance — 1,500
Honorarium paid — 2,500
Govt. Securities purchased — 2,000
Rent paid for 2018-19 — 500
Rent paid for 20 19-20 — 2,500
Building fund received — 4,000
Tournament expenses — 800
Postage and stationery — 500
Purchase of books — 4,500
Answer:

Question 4.
Kamareddy Youth Club gives you their receipts and payments account and other information and requests you to prepare their income and expenditure account for the year ended 31-3-2019.

Additional Information :

1. Outstanding subscriptions on 31-3-2019 Rs. 1,500.
2. Subscriptions received in advance on 31-03-2019 Rs. 500.
3. Value of old furniture sold is Rs. 45,000.

Answer:

Question 5.
Following is the Receipts and Payments Account of a “Manasvi Library” for the year ended 31st March, 2020.

Additional Information :

1. The subscription amount includes Rs. 500 for the previous year and outstanding subscriptions for the current year are Rs. 1,500.
2. Subscription received in advance Rs. 500.
3. Capitalize half of the entrance fee.
4. Books are to be depreciated at 5% per annum.

You are required to prepare income and expenditure account.
Answer:

Question 6.
From the following receipts and payments account, prepare income and expenditure account of a Sports Club’, Karimnagar.

Additional Information :

1. Capitalize 50% of the donations, legacies, entrance fee and life membership fee.
2. Subscriptions still outstanding amount to Rs. 5,000.
3. Depreciate sports material and Furniture 10%.

Answer:

Question 7.
From the following receipts and payments account of Sri Kala Nilayam, Nirmal for the year ended 31-03-2019, prepare income and expenditure account.

Additional Information:

1. Outstanding salaries Rs. 500.
2. Subscriptions outstanding for 2018-19 Rs. 1,000.
3. Depreciate furniture by 10%.

Answer:

Question 8.
From the following receipts and payments account and other details of “Laharika Charitable Trust”. Prepare Income and Expenditure account for the year ended 31-03-2020.

Additional Information :

1. Subscriptions received in advance Rs. 1,500.
2. Outstanding rent Rs. 450.
3. The value of investments Rs. 40,000 and rate of interest is 3%.
4. Donations are received for a prizes to be awarded by the trust.

Answer:

Working Note :
Total Interest on Investment = 40,000 x 3/100 = 12,000
We already received = 1000
The remaining receivable interest on investment is 200

Question 9.
From the following information, prepare the income and expenditure account for the year ended 31-03-2020 of Kamareddy Cricket Club’.

Additional Information :

1. Subscriptions received included amount Rs. 2,200 related to the previous year.
2. Outstanding subscription Rs. 1,000.
3. Provide Rs. 300 depreciation on ground moving machine.
4. Sports material opening balance Rs. 4,000 and closing balance Rs. 7,500.

Answer:

Question 10.
The following is the receipts and payments account for the year ended 31st March, 2020.

Additional Information :

1. The subscription received amount includes Rs. 1,000 for the year 2020-2021.
2. Outstanding rent Rs. 600 and printing Rs. 300.
3. The value of investments Rs. 60,000 and rate of interest 8%.

You are required to prepare income and expenditure account.
Answer:

Working Note :
Total Interest on Investment = 60,000 x 8/100 = 4,800
We already received Interest = 4,000
Remaining interest receivable = 800.

Question 11.
Prepare income and expenditure account and balance sheet from the following Receipts and Payments account of “Supreetha” nursing home for the year ended 31-03-2019.

Additional Information :

1. Nursing home had investment value of Rs. 8,000 and Capital fund of the nursing home was Rs. 30,000.
2. Medicine opening balance as on 01.04.2018 is Rs. 2,000. Closing balance as on 31.03.2019 is Rs. 1,500.

Answer:

Balance Sheet of Supreetha Nursing home as on 31-3-2019

Question 12.
The following is the Receipts and Payments account of “Ashok Nagar Welfare Association’ for the year ended 31st December 2020.

Additional Information :

1. Capital fund of the association is Rs. 7,000.
2. Hall rent Rs. 1,000 related to last year and Rs. 600 received in advance.
3. Outstanding subscriptions for current year Rs. 3,600.

Prepare Income and expenditure account and Balance sheet from the above information.
Answer:

Balance Sheet of Ashok Nagar Welfare Association as on 31-12-2020

Question 13.
The ‘Sree Vedha Literary Society’ gives you the following Receipts and Payments account for the ended 31-03-2020.


Additional Information :

1. Outstanding subscriptions for current year Rs. 1,500.
2. Subscriptions received in advance Rs. 1,000.
3. Value of books on 31-03-2020 Rs. 1,800.
4. The furniture was purchased on 1.10.2019 and is to be depreciated by 20%.
5. Society had the inverter values of Rs. 3,200 and the capital fund of society Rs. 8,000.
6. Entrance Fee not to be capitalised.

Prepare Income and Expenditure account and Balance sheet from above information.
Answer:

Balance Sheet of Sree Vedha literary society as on 31-3-2020

Question 14.
From the following receipts and payments account and additional information prepare income and expenditure account and balance sheet of “Sreenidhi Library” as on 31.3.2020.

Additional Information :

1. Subscriptions outstanding Rs. 1,500.
2. Salaries outstanding Rs. 600.
3. Library had Bank Deposits Rs. 12,000 and Books Rs. 5,000 and Capital fund Rs. 41,000.
4. Provide Depreciation Books @ 10% p.a including purchase made during the year.

Answer:

Balance Sheet of Sreenidhi Library as on 31-3-2020

Question 15.
Sree Hanman Gym, Hyderabad gives you the following Receipts and Payments account for the ended 31-03-2020.

Additional Information :

1. Subscriptions received included Rs. 1,000 for the previous year.
2. Subscriptions outstanding for the current year Rs. 1,500.
3. Outstanding salaries t 1,000.
4. Half of the Entrance fee has to be capitalised.
5. Gym has investments worth 120,000 and total Capital fund of the Gym is Rs. 24,000. Prepare Income and Expenditure account and Balance sheet from the above information.

Answer:

Balance Sheet of Hanman Gym, Hyderabad as on 31-3-2015

Textual Examples

Question 1.
From the following information prepare the receipts and payments account of Hyderabad Cricket Club, for the year ending 31-03-2020.

Answer:
Receipt and Payments a/c of Hyderabad Cricket Club

Question 2.
From the following information prepare the receipts and payments account of “Nirmal Sports club for the year ended 31st March 2020.

Answer:
Receipt and payments A/c of Nirmal Sports Club

Question 3.
From the following particulars prepare the receipts and payments account year 31 March 2019.
Cash in hand ( 01-04-2019) — 2,000
Cash at bank (01-4-2019) — 28,000
Investments purchased — 4,000
Subscriptions received — 1,08,000
(including Rs. 5,000, Next financial year)
Subscriptions outstanding (for Last year) — 10,000
Furniture purchased — 6,000
Interest received on investments — 4,300
Books purchased — 9,000
Rent paid — 6,000
Rent due on 31-03-2019 — 400
Entrance fee — 2,700
Salaries paid — 20,000
Outstanding salaries — 3,000
Cash in Hand — 20,000
Answer:

Question 4.
From the following information prepare the income and expenditure account
Subscriptions received — 54,000
(including last year subscriptions Rs. 4,000)
Outstanding subscriptions — 20,000
Salaries (including last year salaries Rs. 2,400) — 17,200
Outstanding salaries — 3,200
Sundry expenses — 4,000
Tournament expenditure — 8,000
Meeting expenditure — 6,000
Traveling expenses — 6,400
Books purchased — 18,000
Newspapers — 4,000
Rent — 9,000
Postage and Telephone etc. — 3,600
Printing and Stationery — 4,400
Donations — 6,000
Sale of old newspaper — 400
Answer:

Note :

1. The purchase of books is capital expenditure and hence not considered in income and expenditure account.
2. Small amount of donations treated as a revenue income and taken as income on the credit side of income and expenditure account.

Question 5.
From the following receipts and payments account prepare the income and expenditure account for the year ended 31st March 2020.


Other Information :

1. Outstanding subscriptions for the year 2019-20 Rs. 2,500.
2. Prepaid Rent Rs. 1,300.
3. Outstanding Stationery bill Rs. 300.
4. Cost of Investments Rs. 10,000. Interest rate 10%
5. Capitalize the donations.

Answer:

Working Notes :
1) Purchase of furniture, Govt. Bonds are capital expenditures these are to be recorded in balance sheet as assets.

2) Current year 2019-20 subscriptions has to be treated as a revenue income.

3) Loss on sale of furniture :
Book value of furniture sold Rs. 2,000
Less: Sale value Rs. 1,650
Loss on sale of furniture Rs. 350
Loss on sale of furniture is revenue expenditure. It is shown in income and expenditure account debit side.

4) Outstanding interest on investments
Interest on Investments for the year = 10,000 x 10/100 = Rs. 1,000
Interest on Investment = Rs. 1,000
Less: Interest received = 800
Less: Outstanding interest on investments 200
Outstanding interest on Investments Rs. 200/- to be added to interest received credit side of income expenditure account.

5) Membership fees and entrance fees are small amounts treated as revenue incomes shown in credit side of income and expenditure account.

Question 6.
From the following receipts and payments account prepare the income and expenditure account of Kama Reddy Cricket Club for the year ending 31-03-2019.

Adjustments :

1. Capitalize the entrance fee and donations.
2. Closing stock sports material on 31-3-2019 t 5,000.
3. Outstanding subscriptions Rs. 2,135 for the year 2018-19.

Answer:


Working Notes :
1) Purchase of furniture Govt. Bonds are capital expenditure. These are to be recorded in the balance sheet as assets.

2) Calculation of depreciation on sports material :
Sports material purchased = Rs. 7,500
Less: Closing stock of sports material = Rs. 5,000
Depreciation on sports material = 2,500
It is shown in income and expenditure account debit side

3) Calculation of Interest on Investments:
Interest on Investments for the year = 12,000
Interest on Investments = Rs. 1,200
Less : Interest on Investments received = 900
Outstanding interest on Investments 300
It will be added credit side of income and expenditure a/c to interest on investments.

4) Profit on sale of old furniture:
Old furniture sale value = 1,685
Less: Old furniture book value = 1,500
Profit on sale old furniture = 185
Profit on sale of old furniture is treated as a revenue income. It is shown on credit side of income and expenditure account.

Question 7.
From the following receipts and payments account prepare the income and expenditure account and the balance sheet of Prime Sports Club, Hyderabad.

1. Capitalise the 50% of donations and life membership fee.
2. Outstanding subscriptions Rs. 5,000.
3. Provide for depreciation on furniture and buildings 5% and on sports material 10%.

Answer:

Working Notes :
Calculation of depreciation :
a) Building 5% = Rs. 40,000 x 5/100 = Rs. 2,000
b) Furniture 5% = Rs. 10,000 x 5/100 = Rs. 500
c) Sports material 10% = Rs. 5,000 x 10/100 = Rs. 500
Capitalisation of 50% of donations = Rs. 50,000 x 50/100 = Rs. 25,000
Capitalisation of 50% of life membership fee = Rs. 3,000 x 50/100 = Rs. 1,500

Balance Sheet of Prime Sports Club, Hyderabad

Question 8.
From the following receipts and payments account of Indoor Sports club and the sub joined information prepare income and expenditure account for the year ended 31 March, 2020.

Other Information :

1. The club has 50 members each paying an annual subscription of Rs. 500, subscriptions outstanding for the year 2018-19 were Rs. 2,500.
2. For the year 2019-20, salaries outstanding amounted Rs. 1,000, the salaries paid during current year include Rs. 3,000 for last year.
3. On 01-04-2019, the club owned building valued Rs. 1,00,000, Furniture Rs. 10,000. Books Rs. 10,000, capital fund Rs. 1,33,000.
4. Buildings are to be depreciated @ 10%.

Answer:

Indoor Sports Club Balance Sheet as on 31-03-2020

Question 9.
Prepare the income and expenditure account and balance sheet of Hyderabad Cricket Club from the following receipts and payments account for the year ended 31-03-2019.

Subscriptions due on 31-03-2018 — Rs. 3,200
Subscriptions due on 31-03-2019 — Rs. 3,800
Subscriptions received in Advance — Rs. 800
Salaries include Rs. 700 for the year 2017-18
Outstanding for the year 2018-19 — Rs. 900
Sports material on 31-3-2018 — Rs. 3,200
Sports material on 31-3-2019 — Rs. 3,600
Capital fund on 31-3-2018 — Rs. 24,300
Answer:


Balance Sheet of Hyderabad Cricket Club as on 31st March, 2019

Question 10.
Receipts and payments of the Railway Recreation Club for the year ended 31st March, 2020.

Additional Information :

1. Subscriptions outstanding for the 2019-20 Rs. 2,500.
2. Value of sports equipment on 01-04-2019 Rs. 1,000 and on 31-03-2020 Rs. 9,000.
3. Provide Rs. 1,000 depreciation on furniture.
4. Closing balance of postal stamps Rs. 200.
5. Capitalise half of the entrance fee
6. Capital fund on 01-04-2019 Rs. 5,000.

Prepare the Income and Expenditure account and Balance sheet of the club as on 31-03-2020.
Answer:

Railway Recreation Club Balance Sheet as on 31st March, 2020

Working Notes :

1. Calculation of Sports equipment used : (Depreciation)

2. Calculation of postal stamps used :

3. Donations received treated as capital income.

Telangana TSBIE TS Inter 2nd Year Chemistry Study Material Lesson 6(b) Group-16 Elements Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material Lesson 6(b) Group-16 Elements

Very Short Answer Questions (2 Marks)

Question 1.
Why is dioxygen a gas but sulphur a solid?
Answer:
Due to small size and high electronegativity oxygen forms multiple bond (=) and exists as O₂ molecule. These O₂ molecules are held together by weak van der Waals forces. Hence oxygen exists as gas.

Due to large size and less electronegativity Sulphur forms strong S – S bonds and exists as S₈ molecule. Hence Sulphur exists as solid.

Question 2.
What happens when
a) KClO₃ is heated with MnO₂,
b) O₂ is passed through KI solution.
Answer:
a) Oxygen is evolved.
2KClO₃ → 2KCl + 3O₂

b) Iodide is oxidised to I₂.
2KI + H₂O + O₃ → 2K0H + I₂ + O₂

Question 3.
Give two examples each for amphoteric oxides and neutral oxides.
Answer:
Examples of amphoteric oxides are Al₂O₃ and ZnO. Examples of neutral oxides are CO, NO and N₂O.

Question 4.
Oxygen generally exhibits an oxidation state of – 2 only while the other members of the group show oxidation states of +2, + 4 and + 6 also – Explain.
Answer:
The multiple oxidation states in the case of elements other than oxygen are due to the availability of d-orbitals in the valence shell of the atoms.

Oxygen exhibits -2 oxidation state only generally except with F, where it shows + I in F₂O₂ and + II in F₂O.

Question 5.
Write any two compounds in which oxygen shows an oxidation state different from -2. Give the oxidation states of oxygen in them.
Answer:

1. In peroxides oxidation state of oxygen is – 1. Ex: H₂O₂
2. In super oxides oxidation state of oxygen is -1/2. Ex : KO₂
3. In F₂O, oxidation state of oxygen is + 2.
4. In F₂O₂, oxidation state of oxygen is + 1.

Question 6.
Oxygen molecule has the formula O₂ while sulphur has S₈ – explain.
Answer:
Due to small atomic size and high electro-negativity in oxygen molecule, each oxygen atom is linked to other oxygen atom by a double bond. Hence its formula is O₂.
\(: \ddot{\mathrm{O}}=\ddot{\mathrm{O}}:\)
Due to large atomic size and less electro-negativity in sulphur molecule, eight ‘S’ atoms are linked together by single covalent bonds forming.
Puckered S₈ rings with crown configuration. Hence formula of sulphur is S₈.

Question 7.
Why is H₂O a liquid while H₂S is a gas? [IPE ’14]
Answer:
The O-H bond in H₂O is highly polar. There are hydrogen bonds among the molecules of H₂O. Hence it is present as a liquid.

There are no hydrogen bonds among H₂S molecules. So it exists as a gas at room temperature.

Question 8.
H₂O is neutral while H₂S is acidic-explain.
Answer:
Bond dissociation enthalpy of H-S bond is less than H – O bond. Hence H₂S is acidic.

Question 9.
Name the most abundant element present in earth’s crust.
Answer:
Oxygen

Question 10.
Which element of group-16 shows highest catenation?
Answer:
Sulphur

Question 11.
Among the hydrides of chalcogens, which is most acidic and which is most stable?
Answer:

1. Most acidic hydride of chalcogens is H₂Te.
2. Most stable hydride of chalcogens is H₂O.

Question 12.
Give the hybridisation of sulphur in the following.
a) SO₂
b) SO₃
c) SF₄
d) SF₆
Answer:
Hybridisation of S in SO₂ is sp².
Hybridisation of S in SO₃ is sp².
Hybridisation of S in SF₄ is sp³d.
Hybridisation of S in SF₂ is sp³d².

Question 13.
Write the names and formulae of any two oxyacids of sulphur. Indicate the oxidation state of sulphur in them.
Answer:

1. Sulphurous acid = H₂SO₃
   Oxidation state of sulphur is + 4.
2. Sulphuric acid = H₂SO₄
   Oxidation state of sulphur is + 6.
3. Pyrosulphuric acid = H₂S₂O₇
   Oxidation state of sulphur is + 6.

Question 14.
Explain the structures of SF₄ and SF₆.
Answer:
Structure of SF₄ :
Excited state configuration of S is

Hybridisation of Sulphur is sp³d.
SF₄ has distorted trigonal bipyramidal structure with one orbital being occupied by a lone pair of electrons.

Structure of SF₆:
Excited state configuration of Sulphur is

SF₆ has octahedral symmetry.

Question 15.
Give one example each for
a) a neutral oxide
b) a peroxide
c) a super oxide
Answer:
a) Nitric oxide NO is a neutral oxide.
b) Na₂O₂ is a peroxide. Peroxides contain O – O bond.
c) KO₂ is potassium super oxide.

Question 16.
What is tailing of mercury? How is it removed ? [AP & TS ’15]
Answer:
Ozone reacts with mercury to form Hg₂O. Due to dissolution of Hg₂O in Hg mercury loses its meniscus and sticks to the sides of glass. This is called tailing of mercury.
The menisus can be regained by shaking with water which dissolves Hg₂O.

Question 17.
Write the principle involved in the quantitative estimation of ozone gas.
Answer:
Ozone liberates I₂ from KI solution which can be titrated against a standard solution of Hypo using starch as an indicator.
2KI + H₂O + O₃ → 2KOH + I₂ + O₂
2Na₂S₂O₃ + I₂→ Na₂S₄O₆ + 2NaI

Question 18.
Write the structure of Ozone.
Answer:

O-O bond lengths in ozone are identical (128pm) and molecule is angular with a bond angle of 117°.

Question 19.
SO₂ can be used as anti-chlor. Explain.
Answer:
Sulphur dioxide reacts with chlorine in presence of charcoal to give sulphuryl chloride.
SO₂(g) + Cl₂ (g) → SO₂Cl₂ (l)
So it can remove chlorine and can be used as anti-chlor.

Question 20.
How is ozone detected?
Answer:

1. Ozone turns startch iodide paper blue.
2. It tails mercury.

Question 21.
How does ozone react with ethylene? [Mar. 18 – A.P.]
Answer:
When ozone is bubbled through the solution of ethylene in an inert solvent like CCl₄, at 195K, ethylene ozonide is formed.

NO + O₃ → NO₂ + O₂

Question 22.
Out of O₂ and O₃, which is paramagnetic?
Answer:
O₂ is paramagnetic. It contains two unpaired electrons in its molecular form O₂.

Question 23.
Between O₃ and O₂, ozone is a better oxidising agent – why?
Answer:
Due to the ease with which it liberates atoms of nascent oxygen O₃ → O₂ + (O). Ozone acts as a powerful oxidising agent.

Question 24.
Write any two uses each for O₃ and H₃SO₄.
Answer:
Uses of Ozone :

1. It is used as a germicide, disinfectant, and for sterilising water.
2. It is used for bleaching oils, ivory, flour, starch, etc.

Uses of H₂SO₄ :

1. H₂SO₄ is used in the manufacture of fertilisers e.g. ammonium sulphate, super phosphate.
2. Petroleum refining

Question 25.
Which form of sulphur shows paramagnetism?
Answer:
In vapour state sulphur partly exists as S₂ molecule which has two unpaired electrons in the antibonding π* orbitals. Hence exhibits paramagnetism.

Question 26.
How is the presence of SO₂ detected?
Answer:
SO₂ decolourises KMnO₄ solution in acid medium.
5SO₂ + 2\(\mathrm{MnO}_4^{-}\) + 2H₂O → 5\(\mathrm{SO}_4^{2-}\) + 4H⁺ + 2Mn⁺⁺

Question 27.
Why are group-16 elements called chal- cogens ?
Answer:
The elements Oxygen, Sulphur, Selenium, Tellurium of group 16- are collectively known as chalcogens meaning ‘ore forming’. Since many metals occur as oxides or sulphides in nature. The name is derived from the Greek word for brass. It indicates the association of sulphur and its congeners with copper.

Question 28.
Among chalcogens, which has highest electronegativity and which has highest electron gain enthalpy? ‘
Answer:
Oxygen has highest electronegativity (3.5 Pauling scale) among chalcogens. Sulphur has highest electron gain enthalpy among chalcogens.

Question 29.
Which hydride of group-16 has highest boiling point and weakest acidic character ?
Answer:
H₂O (water)

Short Answer Questions (4 Marks)

Question 30.
Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxida-tion states and hydride formation.
Answer:
Electronic configuration: All the elements of group 16 have six electrons in the outermost shell and have ns²np⁴ general electronic configuration.

Oxidation states:
General oxidation states of Group 16 elements are -2, +2, +4 and +6. Oxygen cannot show higher oxidation states +4, +6 because of non-availability of d-orbitals in the valence shell. Sulphur, Selenium, and Tellurium usually show +4 oxidation state in their compounds with oxygen and +6 with fluorine. The stability of +6 oxidation state decreases down the group and stability of +4 oxidation state increases. This is due to inert pair effect.

Hydride formation :
All the chalcogens form covalent hydrides of the formula H₂E. (E = S, Se, Te, Po)

Thermal stability of the hydrides decreases from H₂O to H₂Po. This is due to an increase in E – H bond length.

Water is a liquid while others are gases. Water exists as associated liquid due to hydrogen bonding.

The acidic character in aqueous solution increases from H₂O to H₂Te. This is due to decrease in charge density on conjugate bases OH^–, SH^–, SeH^–, TeH^–.

All the hydrides except water possess reducing property. The reducing property increases from H₂O to H₂Po. This trend can be attributed to decrease in thermal stability of hydrides from H₂O to H₂Po.

Thus there is a regular gradation in properties of these elements. Hence the inclusion of these elements in the same group is justified.

Question 31.
Describe the manufacture of H₂SO₄ by contact process. [TS ’16]
Answer:
Manufacture of H₂SO₄ by Contact process involves three steps.
i) Burning of sulphur or sulphide ores in air to generate SO₂.
S + O₂ → SO₂
4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂

ii) Conversion of SO₂ to SO₃ by the reaction with oxygen in the presence of a catalyst (V₂O₅).

iii) Absorption of SO₃ in H₂SO₄ to give oleum.
SO₃ + H₂SO₄ → H₂S₂O₇

iv) Dilution of oleum with water gives sul-phuric acid.
H₂S₂O₇ + H₂O → 2H₂SO₄

Question 32.
How is ozone prepared? How does it react with the following ? [Mar. 19, 18, 17, AP]
a) PbS
b)KI
c) Hg
d) Ag
Answer:
Preparation of Ozone :
When a slow dry stream of oxygen is passed through a silent electrical discharge, oxygen is converted to ozone. This product is called ozonised oxygen.
3O₂ → 2O₃, ∆H° (298K) = +142 kJ/mol.
If high concentration of ozone (> 10%) is required, a battery of ozonisers can be used. Pure ozone can be condensed in a vessel surrounded by liquid oxygen.
Properties:
a) Pbs is oxidised to PbSO₄.
PbS (s) + 4O₃ (g) → PbSO₄ + 4O₂
b) I₂ is liberated. I^– is oxidised to I₂.
2KI_(aq) + H₂O (l) + O₃(g) → 2KOH(aq) + I₂(s) + O₃(g)

c) Hg₂O formed sticks to the glass surface and mercury loses its meniscus due to dissolution of Hg₂O in Hg. This is called tailing of mercury.
2Hg + O₃ → Hg₂O + O₂

d) Silver metal is blackened. This is due to oxidation of the metal to its oxide
2Ag + O₃ → Ag₂O + O₂ (oxidation)

Question 33.
Write a short note on the allotropy of sulphur.
Answer:
Sulphur forms numerous allotropes. The two common crystalline forms are α (alpha) or rhombic sulphur and β (beta) or monoclinic sulphur. The stable form at room temperature is rhombic sulphur. It transforms to monoclinic sulphur when heated above 369K.

Rhombic sulphur:
This is yellow in colour. It is formed by evaporating the solution of roll sulphur in CS₂. It is soluble in CS₂. m.p. 385.8K.

Monoclinic sulphur:
This form of sulphur is prepared by melting rhombic sulphur in a dish and cooling the solution till crust is formed. Two holes are made in the crust and remaining liquid poured out.

α – sulphur transforms to β – sulphur above 369K. At 369K both are stable. This temperature is called transition temperature.
Both forms have S₈ puckered ring forms. It has a crown shape.

In cyclo – S₆, the ring adopts chair form. At 1000K, S₂ is the dominant species and is paramagnetic like O₂.

Question 34.
How does S02 react with the following?
a) Na₂SO₃ (aq)
b) Cl₂
c) Fe³⁺ ions
d) KMnO₄
Answer:
a) When SO₂ is passed into sodium sulphite solution, sodium hydrogen sulphite is formed.
Na₂SO₃ + H₂O + SO₂ → 2NaHSO₃

b) SO₂ reacts with chlorine in the presence of charcoal to give sulphuryl chloride.
SO₂(g) + Cl₂(g) → SO₂Cl₂ (l)

c) SO₂ reduces Fe³⁺ ions to Fe²⁺ ions.
2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO₄^2- + 4H⁺

d) It decolourises KMnO₄.
5SO₂ + 2MnO₄^– + 2H₂O → 5SO₄^2- + 4H⁺ + 2Mn⁺⁺

Question 35.
Starting from elemental sulphur, how is H2S04 prepared?
Answer:
Manufacture of H₂SO₄ by Contact process involves three steps.
i) Burning of sulphur or sulphide ores in air to generate SO₂.
S + O₂ → SO₂
4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂

ii) Conversion of SO₂ to SO₃ by the reaction with oxygen in the presence of a catalyst (V₂O₅).

iii) Absorption of SO₃ in H₂SO₄ to give oleum.
SO₃ + H₂SO₄ → H₂S₂O₇

iv) Dilution of oleum with water gives sul-phuric acid.
H₂S₂O₇ + H₂O → 2H₂SO₄

Question 36.
Describe the structures (shapes) of SO₄^-2 and SO₃.
Answer:
Structure of SO₃: SO₃ has planar triangular shape in the gas phase. S is in sp² hybridisation. Bond angle is 120°. S-O bond length 143pm,

In SO₃, sulphur is in second excited state.

In the solid state it may be cyclic or chain.

Structure of Sulphite ion SO₄^{– –}
SO₄^{– –} is tetrahedral in shape. Bond angle is 109°28′: S – O bond length 149pm.
Sulphur undergoes sp³ hybridisation.

Question 37.
Which oxide of sulphur can act as both oxidising and reducing agent? Give one example each.
Answer:
SO₂ acts both as oxidising and reducing agent.

1. H₂S is oxidised to sulphur.
   SO₂ + 2H₂S → 2H₂O + 3S
2. Fe³⁺ is reduced to Fe²⁺
   2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO₄^2- + 4H⁺

Question 38.
Explain the conditions favourable for the formation of SO₃ from SO₂ in the contact process of H₂SO₄.
Answer:
2SO₂ (g) + O₂(g) → 2SO₃ (g); ∆H = -196 kj/mol
The reaction is exothermic and reversible. Forward reaction leads to a decrease in volume. Therefore, low temperature and high pressure are favourable for maximum yield.

In practice, the plant is operated at a pressure of 2 bar and a temperature of 720K.

Question 39.
Complete the following.
a) KCl + H₂SO₄(conc.) →
b) Sucrose
c) Cu + H₂SO₄(conc.) →
d) C + H₂SO₄ (conc.) →
Answer:
a) HCl gas evolves.
2KCl + H₂SO₄ → K₂SO₄ + 2HCl

b) It removes water from carbohydrates.

c) SO₂ gas evolves. Cu is oxidised.
Cu . 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O

d) C is oxidised to CO₂.
C + 2H₂SO₄ →CO₂ + 2SO₂ + 2H₂O

Question 40.
Which is used for drying ammonia?
Answer:
Ammonia is dried using quick lime CaO.

Question 41.
Why cone. H₂SO₄, P₄O₁₀ and anhydrous CaCl₂ cannot be used for dry ammonia? (Hint: ammonia reacts with them forming (NH₄)₂SO₄ : (NH₄)₃ PO₄ and CaCl₂. 8 NH₃
Answer:
Ammonia forms (NH₄)₃ PO₄ with P₄O₁₀
Ammonia reacts with CaCl₂ forming CaCl₂ 8NH₃. Hence they cannot be used for drying ammonia.

Long Answer Questions (8 Marks)

Question 42.
Explain in detail the manufacture of sulphuric acid by contact process. [Mar. 2018 – TS]
Answer:
Contact process consists of three stages,
i) Burning of suphur or sulphide ores in air to generate SO₂.
S + O₂ → SO₂
4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂

ii) SO₂ is oxidised to SO₃ in presence of V₂O₅.

∆H = 196 kJ, mol^-1

iii) Absorption of SO₃ in H₂SO₄ to give oleum.
SO₃ + H₂SO₄ → H₂S₂O₇
Low temperature and high pressure are favourable for high yield of SO₃. In practice, a pressure of 2 bar and a temperature of 720K are employed.

The SO₃ gas from the catalytic converter is absorbed in concentrated H₂SO₄ to produce oleum. Dilution of oleum with water gives H₂SO₄ of the desired concentration.
SO₃ + H₂SO₄ → H₂S₂O₇
H₂S₂O₇ + H₂O → 2H₂SO₄

Question 43.
How is ozone prepared from oxygen? Explain its reaction with [AP Mar. ’18, 17, 16; 1PE ‘ 13,’ 14]
a) C₂H₄
b) KI
c) Hg
d) PbS
Answer:
Preparation of Ozone : When a slow dry stream of oxygen is passed through a silent electrical discharge, oxygen is converted to ozone. This product is called ozonised oxygen.
3O₂ → 2O₃, ∆H° (298K) = +142 kJ/mol.
If high concentration of ozone (> 10%) is required, a battery of ozonisers can be used. Pure ozone can be condensed in a vessel surrounded by liquid oxygen.

a) Ethylene reacts with ozone to form ethylene ozonide.

b) Iodide is oxidised to Iodine.
2KI + H₂O + O₃ → 2KOH + I₂ + O₂

c) Mercury is oxidised to Hg₂O by ozone.
As Hg₂O dissolves in Hg, Hg loses its meniscus and sticks to glass surface. It is called ‘tailing of mercury’.
2Hg + O₃ → Hg₂O + O₂

d) PbS is oxidised to PbSO₄ by O₃.
PbS + 4O₃ → PbSO₄ + 4O₂

Intext Questions – Answers

Question 1.
List the important sources of sulphur.
Answer:

1. Sulphur exists primarily such as gypsum CaSO₄ . 2H₂O.
2. Epsom salt MgSO₄ . 7H₂O
3. Sulphides such as
   galena PbS
   zinc blende ZnS
4. eggs, proteins, garlic, onion ……………

Question 2.
Write the order of thermal stability of the hydrides of Group 16 elements
Answer:
The thermal stability of hydrides of Group -16 decreases from H₂O to H₂PO.

Question 3.
Why is H₂O a liquid and H₂S a gas at room temperature and pressure ?
Answer:
There is molecular association through hydrogen bonds in H₂O. But, there is no such molecular association through hydrogen bond in H₂S. H and H₂O exists as liquid while H₂S exists as gas.

Question 4.
Which of the following does not react with oxygen directly ? Zn, Ti, Pt, Fe
Answer:
Platinum.

Question 5.
Complete the following reactions :
i) C₂H₄ + O₂ →
ii) 4 Al + 3O₂ →
Answer:
i) C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
ii) 4 Al + 3O₂ → 2 Al₂O₃

Question 6.
Why does O₃ act as a powerful oxidizing agent?
Answer:
O₃ can easily release nascent oxygen. Hence, it acts as powerful oxidizing agent.

Question 7.
How is O₃ estimated quantitatively ?
Answer:
O₃ reacts with excess of Kl solution and liberates Iodine. The liberated Iodine is titrated against a standard solution of sodium Thiosulphate.

Question 8.
What happens when SO₂ is passed through an aqueous solution of Fe(III) salt ?
Answer:
It reduces Fe(lII) salts to Fe(II) salts.
2Fe⁺³ + SO₂ + 2H₂O → 2Fe⁺² + SO₄^-2 + 4H⁺

Question 9.
Comment on the nature of two S-O bonds formed in S02 molecule. Are the two S-O bonds in this molecule equal ?
Answer:
The two S-O bonds in SO₂ molecule are covalent and they are not equal. One oxygen is linked by single bond and the other oxygen is linked by double bond. The S-O bond length is larger than S = O bond length.

Question 10.
How is the presence of SO₂ detected ?
Answer:
Due to the strong pungent smell, the presence of SO₂ can be detected. It decolourises acidified KMNO₄ solution.

Question 11.
Mention three areas in which H₂SO₄ plays an important role.
Answer:

1. H₂SO₄ is used for manufacture of fertilisers. e.g.: ammoium sulphate
2. Petroleum refining
3. Detergent industry
4. Metallurgical applications e.g.: electro-plating and galvanising.

Question 12.
Write the conditions to maximise the yield of H₂SO₄ by Contact process ?
Answer:
The yield of H₂SO₄ can be maximised by maintaining the following conditions.
a) low temperatures (720 k)
b) high pressures (2 bar).

Question 13.
Why is K_a2 < < K_a1 for in water?
Answer:
H₂SO₄ is a very strong acid in water largely because of its first ionisation to H₃O⁺ and \(\mathrm{HSO}_4^{-}\). The ionisation of \(\mathrm{HSO}_4^{-}\) to H₃O⁺ and SO₄^2- is very very small.

Hence , K_a2 < < K_a1

Students must practice these Maths 2B Important Questions TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type to help strengthen their preparations for exams.

TS Inter Second Year Maths 2B Definite Integrals Important Questions Long Answer Type

Question 1.
Evaluate \(\int_0^{\pi / 2} \frac{d x}{4+5 \cos x}\). [(TS) May ’18; (AP) Mar. ’16, ’15]
Solution:
Put tan \(\frac{x}{2}\) = t then \(\sec ^2 \frac{x}{2} \cdot \frac{1}{2} \mathrm{~d} x=\mathrm{dt}\)

Question 2.
Evaluate \(\int_0^{\pi / 4} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x\). [(AP) Mar. ’17; (TS) May ’15]
Solution:
Put sin x – cos x = t
⇒ (sin x – cos x)² = t²
(cos x + sin x ) dx = dt
sin²x + cos²x – 2 sin x cos x = t²
1 – sin 2x = t²
sin 2x = 1 – t²
Lower limit: x = 0 ⇒ t = -1
Upper limit: x = \(\frac{\pi}{4}\) ⇒ t = 0

Question 3.
Evaluate \(\int_0^1 \frac{\log (1+x)}{1+x^2} d x\). [(TS) Mar. ’20; (TS) May ’19, ’17; ’12]
Solution:
Put x = tan θ
then dx = sec²θ dθ
Lower limit: x = 0 ⇒ θ = 0
Upper limit: x = 1 ⇒ θ = \(\frac{\pi}{4}\)

Question 4.
Show that \(\int_0^{\pi / 2} \frac{x}{\sin x+\cos x} d x=\frac{\pi}{2 \sqrt{2}} \log (\sqrt{2}+1)\). [(AP) Mar. ’20, ’18; ’17 (TS)]
Solution:

Question 5.
Evaluate \(\int_0^{\pi / 2} \frac{\sin ^2 x}{\cos x+\sin x} d x\). [(TS) May ’15]
Solution:

Question 6.
Evaluate \(\int_0^\pi \frac{x}{1+\sin x} d x\)
Solution:

Question 7.
Evaluate \(\int_0^\pi \frac{x \sin x}{1+\sin x} d x\). [Mar. ’16 (TS); May; Mar. ’15 (AP); March ’13]
Solution:

Question 8.
Evaluate \(\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x\). [(AP) May ’18, ’16, ’14]
Solution:

Question 9.
Evaluate \(\frac{x \sin ^3 x}{1+\cos ^2 x} d x\). [(TS) May ’18; Mar. ’15]
Solution:

Question 10.
Evaluate \(\int_3^7 \sqrt{\frac{7-x}{x-3}} d x\)
Solution:
Put x = 3 cos²θ + 7 sin²θ then
dx = (3 . 2 cos θ (-sin θ) + 7 . 2 sin θ cos θ) dθ
= (-3 . 2 sin θ . cos θ + 7 . 2 sin θ cos θ) dθ
= 4 . 2 sin θ cos θ dθ
= 4 sin 2θ dθ
7 – x = 7 – 3 cos²θ – 7 sin²θ
= 7(1- sin²θ) – 3 cos²θ
= 7 cos²θ – 3 cos²θ
= 4 cos²θ
x – 3 = 3 cos²θ + 7 sin²θ – 3
= -3(1 – cos²θ) + 7 sin²θ
= -3 sin²θ + 7 sin²θ
= 4 sin²θ
Lower limit: x = 3 ⇒ θ = 0
Upper limit: x = 7 ⇒ θ = \(\frac{\pi}{2}\)

Question 11.
Show that the area enclosed between the curves y² = 12(x + 3) and y² = 20(5 – x) is \(64 \sqrt{\frac{5}{3}}\).
Solution:
Given curves are y² = 12(x + 3)
y = \(2 \sqrt{3} \sqrt{x+3}\) ……(1)
y² = 20(5 – x)
y = \(2 \sqrt{5} \sqrt{5-x}\) ……..(2)
Solving (1) and (2)
\(2 \sqrt{3} \sqrt{x+3}\) = \(2 \sqrt{5} \sqrt{5-x}\)
squaring on both sides
12(x + 3) = 20(5 – x)
⇒ 3x + 9 = 25 – 5x
⇒ 8x = 16
⇒ x = 2

Question 12.
Show that the area of the region bounded by \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) (ellipse) is πab. Also, deduce the area of the circle x² + y² = a². [(AP) May ’17; Mar. ’14]
Solution:
Given ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)


∴ Required area = πab sq. units
If b = a, the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) becomes circle as, x² + y² = a².
Area of circle = πa(a) = πa² (since b = a).

Question 13.
Let AOB be the positive quadrant of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) with OA = a, OB = b. Then show that the area bounded between the chord AB and the arc AB of the ellipse is \(\frac{(\pi-2) a b}{4}\).
Solution:
Given equation of ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
Since A = (a, 0) and B = (0, b).
We have the equation of chord AB is \(\frac{x}{a}+\frac{y}{b}=1\)

Question 14.
m Find the area bounded between the curves y² = 4ax, x² = 4by. [(AP) May ’19]
Solution:
Given curves are
y² = 4ax ………(1)
⇒ y = \(\sqrt{a} \sqrt{x} 2\)
x² = 4by ………(2)
⇒ y = \(\frac{x^2}{4 b}\)

Question 15.
Find the area bounded between the curves y² = 4ax, x² = 4ay.
Solution:
Given curves are y² = 4ax
⇒ y = \(2 \sqrt{a} \sqrt{x}\) …….(1)
x² = 4ay
⇒ y = \(\frac{x^2}{4 a}\) ………(2)

Solving (1) & (2)
\(2 \sqrt{a} \sqrt{x}\) = \(\frac{x^2}{4 a}\)
squaring on both sides
4ax = \(\frac{x^4}{16 a^2}\)
⇒ x⁴ = 64a³x
⇒ x⁴ – 64a³x = 0
⇒ x(x³ – 64a³) = 0
⇒ x = 0 (or) x³ – 64a³ = 0
⇒ x³ = 64a³
⇒ x = 4a
∴ x = 0, 4a

Question 16.
Evaluate \(\int_a^b \sqrt{(\mathbf{x}-\mathbf{a})(\mathbf{b}-\mathbf{x})} \mathbf{d x}\). [Mar. ’18 (TS)]
Solution:
Put x = a cos²θ + b sin²θ
dx = [a(2 cos θ) (-sin θ) + b 2 sin θ cos θ] dθ
= [-a sin 2θ + b sin 2θ] dθ
= (b – a) sin 2θ dθ
Now, x – a = a cos²θ + b sin²θ – a
= -a(1 – cos²θ) + b sin²θ
= -a sin²θ + b sin²θ
= (b – a) sin²θ
b – x = b – a cos²θ – b sin²θ
= b(1 – sin²θ) – a cos²θ
= b cos²θ – a cos²θ
= (b – a) cos²θ
Lower limit: x = a ⇒ θ = 0
Upper limit: x = b ⇒ θ = π/2

Question 17.
Find \(\int_{-a}^a\left(x^2+\sqrt{a^2-x^2}\right) d x\)
Solution:

Question 18.
Evaluate \(\int_0^\pi x \sin ^3 x d x\)
Solution:

Question 19.
Find \(\int_0^\pi x \sin ^7 x \cos ^6 x d x\). [Mar. ’19 (TS)]
Solution:

Question 20.
If I_n = \(\int_0^{\pi / 2} \sin ^n x d x\), then show that I_n = \(\frac{n-1}{n} I_{n-2}\). [Mar. ’15 (TS)]
Solution:

Question 21.
If I_n = \(\int_0^{\pi / 2} \cos ^n x d x\), then show that I_n = \(\frac{\mathbf{n}-1}{n} I_{n-2}\)
Solution:

Question 22.
The circle x² + y² = 8 is divided into two parts by the parabola 2y = x². Find the area of both parts.
Solution:
Given, the circle x² + y² = 8 ……..(1)
⇒ y = \(\sqrt{8-x^2}\)
The parabola is 2y = x² ……..(2)
⇒ y = \(\frac{x^2}{2}\)
Solving (1) and (2)
2y + y² = 8
⇒ y² + 2y – 8 = 0
⇒ (y + 4)(y – 2) = 0
⇒ y = -4 or y = 2
From (1) ⇒ x² + (-4)² = 8
x² = 8 – 16
-8 ≠ 8
From (2) ⇒ x² = 2(2)
∴ x = ±2

Question 23.
If I_n = \(\int_0^{\pi / 4} \tan ^n x d x\), then show that I_n+ I_n-2 = \(\frac{1}{n-1}\). [Mar. ’06, ’99, ’98]
Solution:

Put tan x = t
sec²x dx = dt
Lower limit: x = 0 ⇒ t = 0
Upper limit: x = \(\frac{\pi}{4}\) ⇒ t = 1

∴ I_n + I_n-2 = \(\frac{1}{n-1}\)

Question 24.
If \(I_{m, n}=\int_0^{\pi / 2} \sin ^m x \cos ^n x d x\), then show that \(I_{m, n}=\frac{m-1}{m+n} I_{m-2, n}\). [May ’99]
Solution:

Question 25.
Evaluate \(\int_2^6 \sqrt{(6-x)(x-2)} d x\)
Solution:
Put x = 2cos²θ + 6sin²θ then
dx = [2(2) cos θ . (-sin θ) + 6(2) sin θ cos θ] dθ
= [-(2) . 2 sin θ cos θ + 6(2) sin θ cos θ] dθ
= [-2 . sin 2θ + 6 sin 2θ] dθ
= 4 sin 2θ dθ
6 – x = 6 – 2 cos²θ – 6 sin²θ
= 6(1 – sin²θ) – 2 cos²θ
= 6 cos²θ – 2 cos²θ
= 4 cos²θ
x – 2 = 2 cos²θ + 6 sin²θ – 2
= -2(1 – cos²θ) + 6 sin²θ
= -2 sin²θ + 6 sin²θ
= 4 sin²θ
Lower limit: x = 2 ⇒ θ = 0
Upper limit: x = 6 ⇒ θ = \(\frac{\pi}{2}\)

Question 26.
Evaluate \(\int_4^9 \frac{d x}{\sqrt{(9-x)(x-4)}}\)
Solution:
Put x = 4 cos²θ + 9 sin²θ then
dx = [4(2) cos θ (-sin θ) + 9(2) sin θ . cos θ] dθ = 0
= (-4 sin 2θ + 9 sin 2θ) dθ
= 5 sin 2θ dθ
9 – x = 9 – 4 cos²θ – 9 sin²θ
= 9(1 – sin²θ) – 4 cos²θ
= 9 cos²θ – 4 cos²θ
= 5 cos²θ
x – 4 = 4 cos²θ + 9 sin²θ – 4
= -4 (1 – cos²θ) + 9 sin²θ
= -4 sin²θ + 9 sin²θ
= 5 sin²θ
Lower limit: x = 4 ⇒ θ = 0
Upper limit: x = 9 ⇒ θ = \(\frac{\pi}{2}\)

Telangana TSBIE TS Inter 2nd Year Chemistry Study Material Lesson 6(a) Group-15 Elements Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material Lesson 6(a) Group-15 Elements

Very Short Answer Questions (2 Marks)

Question 1.
Why does the reactivity of nitrogen differ from phosphorus?
Answer:
Nitrogen has unique ability to form pπ – pπ multiple bonds with itself and with other elements having small size and high electro-negativity. Nitrogen exists as a diatomic molecule with a triple bond (N ≡ N) between the two atoms. Consequently its bond enthalpy (941.4 kJ mol^-1) is very high. On the contrary phosphorus forms single bonds (p – p.) Hence, the reactivity of Nitrogen differ from phosphorus.

White phosphorus is less stable and therefore more reactive, because of angular strain in the P₄ molecule where the angles are only 60°. Hence phosphorous is much more reactive than nitrogen.

Question 2.
How is nitrogen prepared in the laboratory ? Write the chemical equations of the reactions involved.
Answer:
In the laboratory, dinitrogen is prepared by treating an aqueous solution of ammonium chloride with sodium nitrite.
NH₄Cl (aq) + NaNO₂(aq) → N₂(g) + 2H₂O(l) + NaCl (aq)
It can also be obtained by thermal decomposition of ammonium dichromate.

Question 3.
Nitrogen exists as diatomic molecule and phosphorus as P₄. Why? [TS 15]
Answer:
Nitrogen has unique ability to form pπ – pπ multiple bonds with itself and with other elements having small size and high electronegativity. Thus nitrogen exists as a diatomic molecule with a triple bond between the two atoms. Phosphorus forms p-p single bonds and exists as P₄ molecule.

As it has large size and lower eiectro- negavitity it does not form pπ – pπ multiple bonds with itself.

Question 4.
Why does nitrogen show catenation properties less them phosphorus ?
Answer:
Single N – N bond is weaker than the single P – P bond because of high electronic repulsion of the non-bonding electrons, owing to small bond length. Hence nitrogen shows less catenation property than phosphorus.

Question 5.
Nitrogen molecule is highly stable. Why? [IPE ’14]
Answer:
Nitrogen exists as a diatomic molecule with a triple bond between the two atoms (N ≡ N), The bond enthalpy of N₂ is very high (941.4 kj. mol^-1). Hence, nitrogen molecule is highly stable.

Question 6.
Why are the compounds of bismuth more stable in +3 oxidation state?
Answer:
The common oxidation states of Bismuth are +3 and +5. The stability of +5 oxidation state decreases and that of +3 state increases due to inert pair effect.

Question 7.
What is allotropy? Explain the different allotropic forms of phosphorus.
Answer:
Occurrence of an element in different physical forms with similar chemical properties but different physical properties is called allotropy.
Phosphorus is found in many allotropic forms, they are :
a) White or Yellow – P
b) Scarlet – P
c) Red – P
d) Violet – P
e) α – black – P and β – black – P

Question 8.
How do you account for the inert character of dinitrogen ?
Answer:
Nitrogen exists as a diatomic molecule with a triple bond N ≡ N.
The bond dissociation enthalpy of N ≡ N is very high. (941.4kJ mol^-1).
As, such a high energy is not available at room temperature nitrogen is inert.

Question 9.
Explain the difference in the structures of white and red phosphorus.
Answer:
White phosphorus is less stable and there fore more reactive because of angular strain in the P₄ molecule. The bond angle in P₄ is 60° only.
It consists of discrete tetrahedral P₄ molecules. These molecules are held by van der Waal’s forces.
Red phosphorus is polymeric consisting of chains of P₄ tetrahedra linked through covalent bonds. Hence less reactive.

Question 10.
How is a-black phosphorus prepared from red phosphorus ?
Answer:
When red phosphorus is heated in a sealed tube at 803K, a-black phosphorus is obtained.

Question 11.
Write the difference between the properties of white and red phosphorus.
Answer:

Question 12.
What is inert pair effect?
Answer:
The reluctance of electron pair in the outermost s-orbital to uncouple and take part in bonding is called inert pair effect.

Question 13.
Explain why is NH₃ basic while BiH₃ is only feebly basic.
Answer:
Ability to donate electron pair decreases as the atomic size increases from NH₃ to BiH₃. This is due to decrease in electron density. Hence NH₃ is basic while BiH₃ is feebly basic.

Question 14.
Arrange the hydrides of group-15 elements in the increasing order of basic strength and decreasing order of reducing character.
Answer:
Basic strength
NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃
Reducing character
NH₃ < PH₃ < AsH₃ < SbH₃ < BiH₃

Question 15.
PH₃ is a weaker base than NH₃ – Explain.
Answer:
Atomic size of phosphorous is more than that of Nitrogen. Ability to donate electron pair decreases from NH₃ to PH₃. Hence PH₃ is weaker base than NH₃.

Question 16.
A hydride of group-15 elements dissolves in water to form a basic solution. This solution dissolves the AgCl precipitate. Name the hydride. Write the chemical equations involved.
Answer:
The hydride is NH₃.
NH₃ + H₂O > NH₄OH
AgCl + 2NH₃ > [Ag(NH₃)⁺Cl^–

Question 17.
What happens when white phosphorus is heated with cone. NaOH solution in an inert atmospnere of CO₂ ? [AP Mar. ’19 ; AP ’15]
Answer:
Phosphine is formed.
P₄ + 3NaOH + 3H₂O → PH₃ ↑ + 3NaH₂PO₂ sodium hypophosphite

Question 18.
NH₃ forms hydrogen bonds but PH₃ does not. Why ? [TS ’15]
Answer:
N-H bond is more polar while P-H bond, is less polar. Nitrogen is a small atom with high electronegativity. Hence N-H bond is more polar. So, NH₃ can form hydrogen bonds.

Question 19.
The HNH angle is higher than HPH, HAsH and HSbH angles- Why?
Answer:
The NH₃ molecule is trigonal pyramidal. This shape results from the sp³ hybridisation of central atom. The decrease in the bond angle in HPH, HAsH and HSbH is due to increase in the size of the central atoms.

Question 20.
How do calcium phosphide and heavy water react?
Answer:
Calcium phosphide reacts with heavy water forming Calcium Deuteroxide and Deuterophosphine.
Ca₃P₂ + 6H₂O → 3Ca(OH)₂ + 2PD₃

Question 21.
Ammonia is a good complexing agent. Explain with an example. [IPE ’14]
Answer:
Ammonia can donate an electron pair and form dative bond. Hence ammonia can form complex compounds.

Question 22.
A mixture of Ca₃P₂ and CaC₂ is used in making Holme’s signal – Explain. [AP ’16]
Answer:
The spontaneous combustion of phosphine is technically used in Holme’s signals. Containers containing calcium phosphide and calcium carbide are pierced and thrown in the sea when the gases evolved burn and serve as a signal.

Question 23.
Which chemical compound is formed in the brown ring test of nitrate ions ?
Answer:
Fe²⁺ reduces \(\mathrm{NO}_3^{-}\) to NO.
No forms brown coloured complex with Fe²⁺
[Fe(H₂O)₆]²⁺ + NO > [Fe(H₂O)₅NO]²⁺ (brown) + H₂O

Question 24.
Give the resonating structures of NO, and N₂O₅.
Answer:

Question 25.
Why does R₃P = O exist but R₃N = O does not (R = alkyl group)?
Answer:
Nitrogen cannot form R₃N = O because of absence of d-orbitals in its valence shell. Nitrogen cannot form dπ – pπ bond as the heavier elements can e.g. R₃P = O.

Question 26.
How is nitric oxide (NO) prepared?
Answer:
Nitric oxide is obtained by the reduction of dilute HNO₃ with copper.
3Cu + 8HNO₃ → 2NO + SCu(NO₃)₂ + 4H₂O

Question 27.
Give one example each of normal oxide and mixed oxide of nitrogen.
Answer:
Normal oxide : NO₂, NO.
Mixed oxide : N₂O₃

Question 28.
NO is paramagnetic in gaseous state but diamagnetic in liquid and solid states – Why?
Answer:
NO contains odd number of valence electrons. It exists as a monomer in gaseous state and hence paramagnetic. It dimerises in liquid and solid states and is converted to N₂O₂ molecule with even number of electrons. Hence NO is diamagnetic in liquid and solid states.

Question 29.
Give an example of [Mar. 2018 – TS]
a) acidic oxide of phosphorus
b) neutral oxide of nitrogen.
Answer:
a) P₂O₃ and P₂O₅ are acidic oxides of phosphorus.
b) NO and N₂O are neutral oxides of nitrogen.

Question 30.
Explain the following.
a) reaction of alkali with red phosphorous.
b) reaction between PCl₃ and H₃PO₃.
Answer:
a) Red phosphorous is less reactive than white phosphorous, It does not react with alkali.
b) PCl₃ + 5H₃PO₃ → 3H₄P₂O₅ + 3HCl
Pyrophosphorous acid formed

Question 31.
How does PCl₃ react with
a) CH₃COOH
b)C₂H₅OH and
c) water
Answer:
a) PCl₃ reacts with CH₃COOH to give acetyl chloride.
3CH₃COOH + PCl₃ → 3CH₃COCl + H₃PO₃

b) Reacts with ethyl alcohol to give ethyl chloride.
3CH₃CH₂OH + PCl₃ → 3C₂H₅Cl + H₃PO₃

c) Hydrolysis of PCl₃ gives phosphorous acid.
PCl₃ + 3H₂O → H₃PO₃ + 3HCl

Question 32.
PCl₃ can act as an oxidising as well as a reducing agent – Justify.
Answer:
1) PCl₃ oxidises Sb to SbCl₃.
Sb + PCl₃ → SbCl₃ + P
2) It reduces PCl₃ to PCl₅.
PCl₃ + Cl₂ → PCl₅

Question 33.
Which of the following are not known ?
PCl₃, ASCl₃, SbCl₃, NCl₅, BiCl₅, PH₅
Answer:
NCl₅, BiCl₅, PH₅.

Question 34.
Which of the following is more covalent – SbCl₅ or SbCl₃?
Answer:
SbCl₅

Question 35.
Write the oxidation states of phosphorous in solid PCl₅.
Answer:
In the solid state PCl₅ exists as an ionic solid [PCl₄]⁺ [PCl₆]^– in which the cation is [PCl₄]⁺ and [PCl₆]^– is anion.
Oxidation state of phosphorous in PCl₄⁺ is + 5 and in PCl₆^– is + 5

Question 36.
Illustrate how copper metal can give different products on reaction with HNO₃.
Answer:
With dilute HNO₃, it gives NO.
3Cu + 8HNO₃ (dilute) → 3Cu (NO₃)₂ + 2NO + 4H₂O
With cone. HNO₃, NO₂ is obtained.
Cu + 4HNO₃ (conc) → Cu(NO₃)₂ + 2NO₂ + 2H₂O

Question 37.
Which oxide of nitrogen has oxidation number of N same as that in nitric acid?
Answer:
Nitrogen in N₂O₅ has same oxidation number as in Nitric acid.

Question 38.
Write the chemical reactions, that occur in the manufacture of nitric acid.
Answer:
Nitric acid is manufactured by Ostwald’s process and the following chemical reactions will occur.
4NH₃(g) + 5O₂(g) 4NO(g) + 6H₂O(g)
2NO (g) + O₂(g) ⇌ 2NO₂ (g)
3NO₂(g) + H₂O(l) → 2HNO₃(aq) + NO(g)

Question 39.
Iron becomes passive in cone HNO₃. Why ?
Answer:
Iron becomes passive in cone. HNO₃ because of formation of a passive film of oxide on the surface.

Question 40.
Give the uses of
a) nitric acid and
b) ammonia.
Answer:
a) Uses of nitric acid :

1. It is used in the manufacture of NH₄NO₃ for fertilisers and other nitrates for use in explosives,
2. It is used in the preparation of nitro-glycerin, trinitrotoluene and other nitro compounds.

b) Uses of Ammonia:

1. Ammonia is used to produce various nitrogeneous ferti- Users, (amonium nitrate, urea etc),
2. In the Ostwald process for preparation of HNO₃.
3. Liquid ammonia is used as refrigerant.

Question 41.
What are the oxidation states of phosphorus in the following ?
i) H₃PO₃
ii) PCl₃
iii) Ca₃P₂
iv)Na₃PO₄
v) POF₃
Answer:
i) H₃PO₃ ; 3 + x – 6 = 0; x = +3
ii) PCl₃ ; x – 3 = 0 ; x = +3
iii) + 6 + 2x = 0 ; 2x = -6 ; x = – 3
iv) + 3 + x – 8 = 0 ; x – 5 = 0 ; x = +5
v) x – 2 – 3 = 0 ; x = + 5

Question 42.
H₃PO₃ is diprotic while H₃PO₂ is monoprotic – Why?
Answer:

The H atoms which are attached with oxygen in P – OH form are ionisable and cause basicity. As per the structure, H₃PO₃ is diprotic while H₃PO₂ is monoprotic.

Question 43.
Give the disproportionation reaction of H₃PO₃.
Answer:
Orthophosphorus acid on heating disproportionates to give orthophosphoric acid and phosphine.

Question 44.
H₃PO₂ is a good reducing agent – Explain with an example.
Answer:
H₃PO₂ contains two P – H bonds and reduces AgNO₃ to metallic silver.
4AgNO₃ + 2H₂O + H₃PO₂ → 4Ag + 4HNO₃ + H₃PO₄

Question 45.
Draw the structures of
a) Hypo phosphoric acid
b) Cyclic meta phosphoric acid.
Answer:

Short Answer Questions (4 Marks)

Question 46.
Discuss the general characteristics of Group-15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity.
Answer:
Nitrogen, Phosphorous, Arsenic, Antimony and Bismuth are included in Group-15.
Metallic Nature: Nitrogen and Phosphorus are non-metals, Arsenic, Antimony are metalloids and Bismuth is a metal.
Electronic configuration : The valence shell electronic configuration of these elements is
Nitrogen 7 → 1s²2s²2p³
Phosphorus 15 → 1s²2s²2p⁶3s²3p³

Oxidation states:
The common oxidation states of Group 15 elements are -3, +3 and +5. The tendency to exhibit -3 oxidation state decreases down the group. This is due to increase in size and metallic character.

The stability of +5 oxidation state decreases down the group. The stability of +5 oxidation state decreases and that of +3 state increases down the group due to inert pair effect.

Nitrogen exhibits +1, +2, +4 oxidation states also. Phosphorus also shows +1 and +4 oxidation states in some oxoacids.

Nitrogen is restricted to maximum cova-lency of 4 since only four (one s and three p) orbitals are available for bonding. The heavier elements have vaccant d – orbitals which can be used for bonding.

Atomic size :
Covalent and ionic radii increase in size down the group. There is a considerable increase in covalent radius from N to P. However, from As to Bi only a small increase in covalent radius is observed. This is due to the presence of completely filled d and / or f – orbitals in heavier members.

Ionisation Enthalpy :
The ionisation enthalpy decreases down the group due to gradual increase in atomic size. The ionisation enthalpy of group 15 elements is much greater than that of group 14 elements in the corresponding periods. This is due to extra stable half-filled p-orbitals and small size of group 15 elements.

Electronegativity :
The electronegativity decreases down the group with increasing atomic size from N to Bi. However, amongst the heavier elements the difference is less.

Question 47.
Discuss the trends in chemical reactivity of group-15 elements.
Answer:
Nitrogen differs from the rest of the group 15 elements. This is due to its (i) small size (ii) high electronegativity (iii) high ionisation enthalpy and (iv) non-availability of d-orbitals.

The general oxidation states of group 15 elements are -3, +3 and +5. In Bismuth +3 is more stable than +5 oxidation state due to inert pair effect.

Hydrides: All the elements N to Bi form hydrides of the type EH₃ where E = N to Bi.
The stability decreases from NH₃ to BiH₃.
Ammonia is mild reducing agent while BiH3 is the strongest reducing agent. Reducing character increases from NH₃ to BiH₃.
Basic character decreases from NH₃ to BiH₃. NH₃ > PH₃ > ASH₃ > SbH₃ > BiH₃

Oxides:
These elements form two types of oxides E₂O₃ and E₂O₅. However nitrogen forms number of oxides due to pn – pn multiple bonding between nitrogen and oxygen atoms.

These oxides are acidic and acidic character decreases from N₂O₃ to Bi₂O₃. Similarly from N₂O₅ to Bi₃O₅ the acidic character decreases.

Pentoxides are more acidic than trioxides.

N₂O₃ and P₂O₃ are purely acidic. As₂O₃ and Sb₂O₃ are amphoteric. Bi₂O₃ is basic.

Halides :
Trihalides and pentahalidies are formed. Nitrogen does not form penta- halide due to non-availability of d – orbitals in its valence shell.

Pentahalides are more covalent than trihalides because the elements in the higher oxidation state exert more polarising power. All the trihalides of these elements except those of nitrogen are stable. Trihalides except BiF₃ are covalent in nature.

Reactivity towards metals : All these elements react with metals to form their binary compounds exhibiting – 3 oxidation state.
Example: Ca₃N₂ (Calcium nitride)
Ca₃P₂ (Calcium phosphide)

Question 48.
How does P₄ react with the following ?
a) SOCl₂
b) SO₂Cl₂
Answer:
a) P₄ reacts with thionyl chloride SOCl₂ to give PCl₃.
P₄ + 8 SOCl₂ → 4PCl₃ + 4SO₂ + 2S₂Cl₂

b) PCl₅ is formed when P₄ reacts with SO₂Cl₂.
P₄ + 10SO₂Cl₂ → 4PCl₅ + 10SO₂

Question 49.
Explain the anomalous nature of nitrogen in group -15.
Answer:
Nitrogen differs from the rest of the elements of Group-15 due to its small size, high electronegativity, high ionisation enthalpy and non-availability of d-orbitals.

Nitrogen has unique ability to form pπ – pπ multiple bonds with itself and with other elements (e.g.: C, O). Other elements cannot form pit – pn bonds as their atomic orbitals are large.

Thus nitrogen exists as N₂ with triple bond. Consequently bond enthalpy is high (225 kcal / mole, or 941.4 kJ/mole). Hence it is i~ert at room temperature.

Catenation tendency is weaker in Nitrogen as the N – N single bond is weaker because of high repulsion of the non-bonding electrons owing to small bond length.

Nitrogen cannot form NCl₅ because of the absence of d-orbitals in its valence shell.

Question 50.
Complete the following reactions:
a) Ca₃P₂ + H₂O →
b) P₄ + KOH →
c) CuSO₄ + NH₃ →
d) Mg + N₂ →
e) (NH₄)₂ Cr₂O₇
f) Decomposition of nitrous acid
Answer:
a) Phosphine is formed when calcium phosphide reacts with water.
Ca₃P₂ + H₂O → 3Ca(OH)₂ + 2PH₃ ↑

b) P₄ + KOH + 3H₂O → ↑ PH₃ T + 3KH₂PO₂ (Potassium hypophosphite)

c) CuSO₄(aq) + NH₃ (aq) ⇌ [Cu(NH₃)₄]²⁺(aq) + SO₄^-2 deep blue

d) At high temperatures N₂ directly combines with Mg to form Mg₃N₂.
3Mg + N₂ Mg₃N₂

e) Nitrogen is evolved when Ammonium dichromate decomposes on heating.

Question 51.
How does PCl₅ react with the following ?
a) Water
b) C₂H₅OH
c) CH₃COOH
d) Ag
Answer:
a) PCl₅ hydrolyses to give POCl₃ and H₃PO₄.
PCl₅ + H₂O → POCl₃ + 2HCl
POCl₃+ 3H₂O → H₃PO₄ + 3HCl

b) Ethyl chloride is formed.
C₂H₅OH + PCl₅ → C₂H₅Cl + POCl₃ + HCl

c) Acetyl chloride is formed.
CH₃COOH + PCl₅ → CH₃COCl + POCl₃ + HCl

d) Finely divided metals on heating with
PCl₅ to give corresponding chlorides.
2Ag + PCl₅ → 2AgCl + PCl₃

Question 52.
Complete the following:
a) NH₄NO₃
b) HNO₃ + P₄O₁₀ →
c) Pb(NO₃)₂
d) Zn + dil. HNO₃ →
e) P₄ + cone. HNO₃
f) HgCl₂ + PH₃ →
Answer:
a) Nitrous oxide forms.
NH₄NO₃ N₂O + 2H₂O
b) 4HNO₃ + P₄O₁₀ → 4HPO₃ + 2N₂O₅
c) 2Pb(NO₃)₂ 2PbO + 4NO₂ + O₂
d) 4Zn + 10 HNO₃ (dilute) → 4Zn(NO₃)₂ + 5H₂O + N₂O
e) P₄ + 20HNO₃ → 4H₃PO₄ + 20 NO₃ + 4H₂O
f) 3HgCl₂ + 2PH₃ → Hg₃P₂ + 6HCl

Long Answer Questions (8 Marks)

Question 53.
How is ammonia manufactured by Haber’s process ? Explain the reactions of ammonia with [Mar. 2018 . AP]
a) ZnSO_{4 (aq)}
b) CuSO_{4 (aq)}
c) AgCl_(s)
Answer:
On a large scale, ammonia is manufactured by Haber’s process.
N₂(g) + 3H₂(g) ⇌ 2NH₃(g); ∆_fH°= -46.1. kJmol^-1
The optimum conditions for production of ammonia are a pressure of about 2000 × 10⁵ Pa and a temperature of 700K. Iron oxide is the catalyst with small amounts of K₂O and Al₂O₃.

Compressed mixture of N₂ and H₂ in the volume ratio is heated to 700K at a pressure of 200 atm and passed over finely divided iron oxide mixed with small amounts of K₂O and Al₂O₃. Ammonia formed is liquified and unreacted mixture of N₂ and H₂ again pumped into catalytic chamber.

a) With ZnSO₄, Zinc hydroxide precipitate is formed.
ZnSO₄ (aq) + 2NH₄OH (aq) → Zn(OH)₂ (s) + (NH₄)₂SO₄ (aq)

b) When ammonia reacts with CuSO₄, deep blue colouration is obtained.
CuSO₄(aq) + 4NH₃(aq) ⇌ [Cu(NH₃)₄]²⁺ (aq) + SO₄^-2 deep blue

c) AgCl dissolves in NH₃.
AgCl + 2NH₃ (aq) → [Ag(NH₃)₂]Cl (aq)

Question 54.
How is nitric acid manufactured by Ostwald’s process ? How does it react with the following ? [TS Mar. 19 ; (AP ’17)]
a) Copper
b) Zn
c) S₈
d) P₄
Answer:
Ostwald’s process:
Step (1) : A mixture of ammonia and excess of atmospheric oxygen (1 : 10) is passed through a tower containing pt/Rh guaze catalyst maintained at 500 K temperature and 9 bar pressure. Then catalytic oxidation of NH₃ takes place, forming Nitric oxide.

Step (2) : The nitric oxide thus obtained is made to combine with oxygen. Then NO₂ is obtained.
2NO_(g) + O₂(g) ⇌ 2NO_2(g)

Step (3) :The nitrogen dioxide, is dissolved in water to give nitric acid
3NO₂(g) + H₂O_(l) → 2HNO₃(aq) + NO(g)
The NO formed is recycled.

Reactions:
a) Copper: N02 gas is liberated
Cu + 4 HNO₃ (conc) → Cu(NO₃)₂ + 2NO₂ + 2H₂O

b) Zinc : NO₂ gas is liberated
Zn + 4 HNO₃ (cone) → Zn(NO₃)₂ + 2H₂O + 2NO₂

c) Sulphur (S₈) : Oxidation takes place forming sulphuric acid
S₈ + 48 HNO₃(conc) → 8H₂SO₄ + 48 NO₂ + 16 H₂O

d) Phosphorous (P₄) : Oxidation takes place forming H₃PO₄
P₄ + 20 HNO₃ → 4H₃PO₄ + 20 NO₂ + 4H₂O

Intext Questions – Answers

Question 1.
Why are pentahalides more covalent than trihalides?
Answer:
Pentahalides are more covalent than trihalides because the elements in the higher oxidation state exert more polarising power.

Question 2.
Why is BiH₃ the strongest reducing agent amongst the hydrides of Group 15 elements?
Answer:
Because BiH₃ is least stable.

Question 3.
Why is N₂ less reactive at room temperature ?
Answer:
Because of high bond enthalpy of N ≡ N.

Question 4.
Mention the conditions required to maximise the yield of ammonia.
Answer:
In Haber’s process
N₂(g) + 3H₂ (g) ⇌ 2NH₃(g)
Conditions required for higher yield are

1. High pressure = 200 atm
2. Optimum temperature – 700 k
3. Catalyst: Iron oxide mixed with K₂O + Al₂O₃.

Question 5.
How does ammonia react with a solution of Cu⁺⁺ (aq)?
Answer:
Cu⁺⁺ (ag) + 4NH₃ (aq) ⇌ \(\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]_{(\mathrm{aq})}^{2+}\) deep blue
NH₃ forms complexion with Cu⁺⁺ by donating its lone pair of electrons, forming Tetraamine copper (II) ion.

Question 6.
What is the covalence of N in N₂O₅ structure ?
Answer:

From the structure of N₂O₅, it is evident that the covalence of N in N₂O₅ is four.

Question 7.
Bond angle in \(\mathrm{PH}_4^{+}\) is higher than that in PH₃. Why?
Answer:
In both PH₄⁺ and PH₃ the central P atom is sp3 hybridised. In \(\mathrm{PH}_4^{+}\) all the four orbitals are bonded whereas in PH₃ there is a lone pair on P. Due to lone pair-bond pair repulsion in PH₃ bond angle is less than 109°28′.

Question 8.
What happens when PCl₅ is heated?
Answer:
It decomposes to give Cl₂.
PCl₅ → PCl₃ + Cl₂

Question 9.
What is the basicity of H₃PO₄?
Answer:
3

Question 10.
What happens when H₃PO₃ is heated? (or write the disproportion of H₃PO₃)
Answer:
Orthphosphoric acid and Phosphine are formed.
4H₃PO₃ → 3H₃PO₄ + PH₃

Question 11.
What happen when white phosphorus is heated with conc. NaOH solution in an inert atmosphere of CO₂?
Answer:
Phosphine is formed.
P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂ sodium hypophosphite

Question 12.
Write a balanced equation for the hydrolytic reaction of PCl₅ in Heavy water ?
Answer:
PCl₅ + 4D₂O → D₃PO₄ + 5DCl

Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A Binomial Theorem Important Questions Long Answer Type to help strengthen their preparations for exams.

TS Inter Second Year Maths 2A Binomial Theorem Important Questions Long Answer Type

Question 1.
State and prove Binomial Theorem. [March ’09]
Solution:

Question 2.
Find the numerically greatest term in the expansion of (4 + 3x)¹⁵ where x = \(\frac{7}{2}\).
Solution:
Given
(4 + 3x)¹⁵
⇒ 4¹⁵ (1 + \(\frac{3 x}{4}\))¹⁵ = 4¹⁵ (1 + x)ⁿ
where n = 15; x = \(\frac{3 x}{4}\)
\(|x|=\left|\frac{3 x}{4}\right|=\left|\frac{3}{4} \cdot \frac{7}{2}\right|=\frac{21}{8}\)
Now, \(\frac{(n+1)|x|}{|x|+1}=\frac{(15+1)\left|\frac{3 x}{4}\right|}{\left|\frac{3 x}{4}\right|+1}\)
= \(\frac{16 \cdot \frac{21}{8}}{\frac{21}{8}+1}=\frac{336}{29}\) = 11.5
∴ T₁₂ is numerically greatest.
r + 1 = 12
⇒ r = 11
The general term in the expansion of (x + a)ⁿ is
T_{r + 1} = \({ }^n C_r\) x^{n – r} a^r
T_{11 + 1} = \({ }^{15} \mathrm{C}_{11}\) (4)^{15 – 11} (3x)¹¹
12th term in this expansion is

Question 3.
Find the numerically reatest tenu in the expansion of (4a – 6b)¹³ when a = 3, b = 5.
Solution:
Given (4a – 6b)¹³ = (4a)¹³ (1 – \(\frac{6 b}{4 a}\))¹³
= (4a)¹³ (1 + x)ⁿ
where, x = \(-\frac{6 b}{4 a}\), n = 13
|x| = \(\left|\frac{-6 b}{4 a}\right|=\left|\frac{-6 \cdot 5}{4 \cdot 3}\right|=\frac{5}{2}\)
Now, \(\frac{(n+1)|x|}{|x|+1}=\frac{(13+1) \frac{5}{2}}{\frac{5}{2}+1}\)
= \(\frac{14 \cdot 5}{5+2}=\frac{70}{7}\) = 10
∴ T₁₀ and T₁₁ are numerically greatest.
T₁₀:
r + 1 = 10
⇒ r = 9
The general term in the expansion of (1 + x)ⁿ is
T_{r + 1} = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) x^{n – r} a^r
T₁₀ in this expansion is
T₉₊₁ = \({ }^{13} \mathrm{C}_9\) (4a)^13-9 (- 6b)⁹
T10 = \({ }^{13} \mathrm{C}_9\) (4a)⁴ (- 6b)⁹
= \({ }^{13} \mathrm{C}_9\) (4⁴ . 3⁴) ((- 6)⁹ (5)⁹)
= \({ }^{13} \mathrm{C}_9\) 12⁴ . 30⁹
|T₁₀| = \({ }^{13} \mathrm{C}_9\) 12⁴ . 30⁹

T₁₁:
r + 1 = 11
⇒ r = 10.
The general term in the expansion of (1 – x)ⁿ is
T_{r + 1} = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) x^{n – r} a^r
T₁₁ in this expansion is
T₁₀₊₁ = \({ }^{13} \mathrm{C}_{10}\) (4a)^13-10 . (- 6b)¹⁰
= \({ }^{13} \mathrm{C}_{10}\) (4a)³ (6b)¹⁰
T₁₁ = \({ }^{13} \mathrm{C}_{10}\) (4 . 3)³ (6 . 5)¹⁰
= \({ }^{13} \mathrm{C}_{10}\) 12³ . 30¹⁰
T₁₁ = \({ }^{13} \mathrm{C}_{10}\) 12³ . 30¹⁰
∴ |T₁₀| = |T₁₁|.

Question 4.
If the coefficient of x¹⁰ in the expansion of \(\left(a x^2+\frac{1}{b x}\right)^{11}\) is equal to the coefficient of x^-10 in the expansion of \(\left(a x-\frac{1}{b x^2}\right)^{11}\) find the relation between a and b where a and b are real numbers. [AP – May 2015, AP – Mar. 2019]
Solution:
Case – I :
Given \(\left(a x^2+\frac{1}{b x}\right)^{11}\)
Here x = ax², a = \(\frac{1}{\mathrm{bx}}\); n = 11
Now, the general term in the expansion is
T_r+1 = \({ }^n C_r\) x^n-r a^r
= \({ }^{11} \mathrm{C}_{\mathrm{r}}\) a^11-r x^22-2r 1^r b^-r x^-r
= \({ }^{11} \mathrm{C}_{\mathrm{r}}\) a^11-r b^-r x^22-3r …………….(1)
To find the coefficient of x¹⁰,
Put 22 – 3r = 10
3r = 12
r = 4
Substituting r = 4 in equation (1) we get
T₄₊₁ = \({ }^{11} C_4\) a^11-4 b^-4 x^22-12
T₅ = \({ }^{11} C_4\) a⁷ b^-4 x¹⁰
T₅ = \({ }^{11} C_4\) a⁷ b^-4 x¹⁰
∴ The coeff. of x¹⁰ ¡n the expansion of \(\left(a x^2+\frac{1}{b x}\right)^{11}\) is \({ }^{11} C_4\) a⁷ b^-4

Case II:
Given \(\left(a x-\frac{1}{b x^2}\right)^{11}\)
Here x = ax, a = \(\left(\frac{-1}{b x^2}\right)\), n = 11
Now, the general term in the expansion is
T_{r + 1} = \({ }^n C_r\) x^n-r a^r
= \({ }^{11} C_r(a x)^{11-r}\left(\frac{-1}{b x^2}\right)^r\)
= \({ }^{11} \mathrm{C}_{\mathrm{r}}\) a^11-r x^11-r (- 1)^r b^-r x^-2r
= \({ }^{11} \mathrm{C}_{\mathrm{r}}\) (- 1)^r b^-r x^11-3r ……………(2)
To find the coefficient of x^-10
put 11 – 3r = – 10
⇒ 3r = 21
⇒ r = 7
Substitute r = 7 in equation (2) we get
T₇₊₁ = \({ }^{11} \mathrm{C}_7\) a^11-7 b⁷ (- 1)⁷ x^11-21
T₈ = – \({ }^{11} \mathrm{C}_7\) a⁴ b^-7 x^-10
∴ The coeff. of x^-10 in the expansion of \(\left(a x-\frac{1}{b x^2}\right)^{11}\) is – \({ }^{11} \mathrm{C}_7\) a⁴ b^-7
Given that, these coefficients are equal
\({ }^{11} \mathrm{C}_7\) a⁷ b^-4 = – \({ }^{11} \mathrm{C}_7\) a⁴ b^-7
\({ }^{11} C_4 \frac{a^7}{b^4}=-{ }^{11} C_4 \frac{a^4}{b^7}\)
\(\frac{a^7}{b^4}=\frac{-a^4}{b^7}\)
⇒ a³b³ = – 1
⇒ (ab)³ = (- 1)³
⇒ ab = – 1

Question 5.
If n is a positive integer, then show that
i) C₀ + C₁ + C₂ + ……………… + C_n = 2
ii) C₀ + C₂ + C₄ + ……………. = C₁ + C₃ + C₅ + ……….. = 2^n-1 [March ’97, ’90]
Solution:
We know that
(1 + x)ⁿ = C₀ + C₁x + C₂x² + ……………… + C_nxⁿ ………….(1)
i) Put x = 1 in equation (1)
⇒ (1 + 1)ⁿ = C₀ + C₁(1) + C₂(1)² + C₃(1)³ + C₄(1)⁴ + ……………… + C_n(1)ⁿ
2ⁿ = C₀ + C₁ + C₂ + C₃ + C₄ + …………. + C_n
∴ C₀ + C₁ + C₂ + C₃ + C₄ + …………. + C_n = 2ⁿ

ii) Put x = – 1 in equation (1), we get
(1 – 1)ⁿ = C₀ + C₁ (- 1) + C₂ (- 1)² + C₃ (- 1)³ + ……………. + C_n (- 1)ⁿ
o = C₀ – C₁ + C₂ – C₃ + C₄ – ………………
C₀ + C₂ + C₄ + …………………. = C₁ + C₃ + C₅ + ………….
Since a = b
⇒ a = b = \(\frac{a+b}{2}\)
C₀ + C₂ + C₄ + …………………. = C₁ + C₃ + C₅ + ………….
= \(\frac{\mathrm{C}_0+\mathrm{C}_2+\mathrm{C}_4+\ldots \ldots+\mathrm{C}_1+\mathrm{C}_3+\mathrm{C}_5+\ldots \ldots}{2}\)
= \(\frac{\mathrm{C}_0+\mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3+\mathrm{C}_4+\mathrm{C}_5+\ldots \ldots}{2}\)
= \(\frac{2^{\mathrm{n}}}{2}\)
= 2^{n – 1}
∴ C₀ + C₂ + C₄ + …………………. = C₁ + C₃ + C₅ + …………. = 2^{n – 1}
C₀ + C₂ + C₄ + …………………. = 2^{n – 1}
C₁ + C₃ + C₅ + …………………… = 2^{n – 1}

Question 6.
Prove that for any real numbers a, d, a . C₀ + (a + d) . C₁ + (a + 2d) . C₂ + ……………… + (a + nd) . C_n = (2a + nd) 2^{n – 1} [May ‘98]
Solution:
Let S = a . C₀ + (a + d) . C₁ + (a + 2d) . C₂ + ……………… + (a + nd) . C_n ………………(1)
By writing the terms in R.H.S of (1) in reverse order has done we get
S = (a + nd)C_n + (a + (n – 1)d)C_n-1 + (a + (n-2)d)C_n-2 + …………. + aC₀
S = (a + nd)C₀ + (a + (n – 1)d)C₁ + (a + (2n – 2)d)C₂ + ………….. + aC_n ……………(2)
Adding (1) and (2) we get

2s = (2a + nd) (C₀ + C₁ + C₂ + …………. + C_n)
2s = (2a + nd)²ⁿ
⇒ S = (2a + nd)^2n-1
∴ aC₀ + (a + d)C₁ + (a + 2d)C₂ + …………. + (a + nd)C_n = (2a + nd) 2^2n-1.

Question 7.
For r = 0, 1, 2, ….., n, prove that C₀ . C_r + C₁ . C_r+1 + C₂ . C_r+2 + ……………… + C_n-r . C_n = \({ }^{2 n} C_{(n+r)}\)
and hence deduce that
i) C₀² + C₁² + C₂² + ………….. + C_n² = \({ }^{2 n} C_n\)
ii) C₀ . C₁ + C₁ . C₂ + C₂ . C₃ + ………… + C_n-1 . C_n = \({ }^{2 n} C_{n+1}\)
[May ‘97, 95, ‘90, ‘94,’93, ‘92; March ‘98, ‘93, AP-Mar. 2018; TS – Mar. 2015]
Solution:
We know that
(1 + x)ⁿ = C₀ + C₁ . x + C₂ . x² + ………………….. + C_r x^r + C_n . xⁿ ………………(1)
(x + 1)ⁿ = C₀xⁿ + C₁x^n-1 + C₂x^n-2 + ………………… + C_rx^n-r + …………………. + C_n ………………(2)
Multiplying (2) and (1), we get
(C₀xⁿ + C₁x^n-1 + C₂x^n-2 + ………………… + C_rx^n-r + …………………. + C_n) (C₀ + C₁ . x + C₂ . x² + ………………….. + C_r x^r + C_n . xⁿ)
= (x + 1)ⁿ (1 + x)ⁿ = (1 + x)²ⁿ
Comparing the coefficient of x^n+r on bothsides we get,
C₀ . C_r + C₁ . C_r+1 + C₂ . C_r+2 + ……………… + C_n-r . C_n = \({ }^{2 n} C_{(n+r)}\) ……………….(3)

i) On substituting r = 0 in equation (3) we get
C₀ . C₀ + C₁ . C₁ + C₂ . C₂ + ……………….. + C_n . C_n = \({ }^{2 n} C_n\)
C₀² + C₁² + C₂² + ………….. + C_n² = \({ }^{2 n} C_n\)

ii) On substituting, r = 1 in equation (3) we get
C₀ . C₁ + C₁ . C₂ + C₂ . C₃ + ………… + C_n-1 . C_n = \({ }^{2 n} C_{n+1}\)

Question 8.
If n is a positive integer and x is any non-zero real number, then prove that
C₀ + C₁ . \(\frac{x}{2}\) + C₂ . \(\frac{x^2}{3}\) + C₃ . \(\frac{x^3}{4}\) + …………….. + C_n . \(\frac{\mathbf{x}^n}{n+1}\) = \(\frac{(1+x)^{n+1}-1}{(n+1) x}\). [May ’14, ’13, ’08, ’05, Mar. ’14, ’02, Board Paper, AP – May 2016]
Solution:
Let S = C₀ + C₁ . \(\frac{x}{2}\) + C₂ . \(\frac{x^2}{3}\) + C₃ . \(\frac{x^3}{4}\) + …………….. + C_n . \(\frac{\mathbf{x}^n}{n+1}\) = \(\frac{(1+x)^{n+1}-1}{(n+1) x}\)

Question 9.
If n is a positive integer, then prove that
C₀ + \(\frac{C_1}{2}+\frac{C_2}{3}+\ldots .+\frac{C_n}{n+1}=\frac{2^{n+1}-1}{n+1}\) [TS – Mar. 2017]
Solution:
Let S = C₀ + \(\frac{\mathrm{C}_1}{2}+\frac{\mathrm{C}_2}{3}+\ldots . .+\frac{\mathrm{C}_{\mathrm{n}}}{\mathrm{n}+1}\)
= \({ }^{\mathrm{n}} \mathrm{C}_0+\frac{1}{2} \cdot{ }^{\mathrm{n}} \mathrm{C}_1+\frac{1}{3} \cdot{ }^{\mathrm{n}} \mathrm{C}_2+\frac{1}{\mathrm{n}+1} \cdot{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}\)
Now multiplying on both sides with (n + 1), we get

Question 10.
Prove that \(\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\frac{C_7}{8}+\cdots \cdots=\frac{2^n-1}{n+1}\) [May 08].
Solution:

Question 11.
If (1 + x + x²)ⁿ = a₀ + a₁x + a₂x² + ……………… + a_2nx²ⁿ, then prove that
i) a₀ + a₁ + a₂ + …………….. + a_2n = 3ⁿ
ii) a₀ + a₂ + a₄ + ……………. + a_2n = \(\frac{3^n+1}{2}\)
iii) a₁ + a₃ + a₅ + ……………… + a_2n = \(\frac{3^n-1}{2}\)
iv) a₀ + a₃ + a₆ + a₉ + ………………. = 3^n-1 [TS May 2016; Board Paper]
Solution:
(1 + x + x²)ⁿ = a₀ + a₁x + a₂x² + ……………… + a_2nx²ⁿ
Put x = 1 in equation (1) we get,
(1 + 1 + 1²)ⁿ = a₀ + a₁ (1) + a₂ (1)² + ……………… + a_2n. (1)²ⁿ
a₀ + a₁ + a₂ + …………… + a_2n = 3ⁿ ……………(2)
Put x = – 1 in equation (1) we get,
(1 – 1 + 1²)ⁿ = a₀ + a₁ (- 1) + a₂ (- )² + a₃. (- 1)⁴ ……………… + a_2n. (- 1)²ⁿ
a₀ – a₁ + a₂ – a₃ + a₄ + …………… + a_2n = 1 ……………(3)
i) From (2),
a₀ + a₁ + a₂ + a₃ + …………… + a_2n = 3ⁿ

ii) Now, adding (2) and (3)
a₀ + a₁ + a₂ + a₃ + …………… + a_2n = 3ⁿ

iii) Now, subtracting (2) and (3)

iv) Put, x = 1 in equation (1) we get
(1 + 1 + 1²)ⁿ = a₀ + a₁ (1) + a₂(1)² + a₃ (1)³ + a₄ (1)⁴ + a₅ (1)⁵ + a₆ (1)⁶ + ……………… + a_2n (1)²ⁿ
= a₀ + a₁ + a₂ + a₃ + a₄ + a₅ + a₆ + …………….. + a_2n = 3ⁿ ………………(4)

Put x = ω in equation (1), we get
(1 + ω + ω²)ⁿ = a₀ + a₁ (ω) + a₂ (ω)² + a₃ (ω)³ + a₄ (ω)⁴ + a₅ (ω)⁵+ a₆ (ω)⁶ + ……………. + a_2n (ω)²ⁿ
= a₀ + a₁ . (ω) + a₂ (ω)² + a₃ (ω)³ + a₄ (ω)⁴ + a₅ (ω)⁵ + a₆ (ω)⁶ + ……………. + a_2n (ω)²ⁿ = 0 …………….(5)
put x = in equation (1) we get,
(1 + ω² + ω⁴)ⁿ = a₀ + a₁ . (ω)² + a₂ (ω)⁴ + a₃ (ω)⁶ + a₄ (ω)⁸ + a₅ (ω)¹⁰ + a₆ (ω)¹² + ……………. + a_2n (ω²)²ⁿ
= a₀ + a₁ . ω² + a₂ ω + a₃ + a₄ ω² + a₅ ω + a₆ + ……………. + a_2n (ω²)²ⁿ = 0
Now, adding (4), (5) and (6)

⇒ 3a₀ + 0 + 0 + 3a₃ + 0 + 0 + 3a₆ + ………….. = 3ⁿ
⇒ 3(a₀ + a₃ + a₆ + ………………..) = 3ⁿ
⇒ a₀ + a₃ + a₆ + ……………….. = 3^n-1.

Question 12.
If (1 + x + x² + x³)⁷ = b₀ + b₁x + b₂x² + ………………. + b₂₁x²¹, then find the value of
i) b₀ + b₂ + b₄ + …………. + b₂₀
ii) b₁ + b₃ + b₅ + …………….. + b₂₁
Solution:
Given
(1 + x + x² + x³)⁷ = b₀ + b₁x + b₂x² + ………………. + b₂₁x²¹ …………………….(1)
Substituting x = 1 in (1),
we get b₀ + b₁ + b₂ + …………. + b₂₁ = 47 ………………(2)
Substituting x = – 1 in (1),
we get b₀ – b₁ + b₂ + ……………… – b₂₁ = 0 ……………..(3)
i) (2) + (3)
⇒ 2b₀ + 2b₂ + 2b₄ + ………………….. + 2b₉₀ = 4⁷
⇒ b₀ + b₂ + b₄ + …………….. + b₂₀ = 2¹³

ii) (2) – (3)
⇒ 2b₁ + 2b₃ + 2b₅ + ……………. + 2b₂₁ = 4⁷
⇒ b₁ + b₃ + b₅ + ……………. + b₂₁ = 2¹³

Question 13.
the coefficients of x⁹, x¹⁰, x¹¹ in the expansion of (1 + x)ⁿ are in A.P., then prove that n² – 41n + 398 = 0.
Solution:
Coefficient of r in the expansion of (1 + x)ⁿ is \({ }^n C_r\).
Given coefficients of x⁹, x¹⁰, x¹¹ in the expansion of (1 + x)ⁿ are \({ }^n C_9,{ }^n C_{10},{ }^n C_{11}\) in AP., then 2\(\left({ }^n C_{10}\right)={ }^n C_9+{ }^n C_{11}\)
⇒ \(2 \frac{n !}{(n-10) ! 10 !}=\frac{n !}{(n-9) ! 9 !}+\frac{n !}{(n-11) !+11 !}\)
⇒ \(\frac{2}{10(n-10)}=\frac{1}{(n-9)(n-10)}+\frac{1}{11 \times 10}\)
⇒ \(\frac{2}{(n-10) 10}=\frac{110+(n-9)(n-10)}{110(n-9)(n-10)}\)
⇒ 22 (n – 9) = 110 + n² – 19n + 90
⇒ n² – 41n + 398 = 0.

Question 14.
1f 36, 84, 126 are three successive binomial coefficlenü* in the expansion of (1 + x)ⁿ, then find n. [May ’09, ’06] [TS – Mar. 2019]
Solution:
We know that
(1 + x)ⁿ = C₀ + C₁x + C₂x² + ………………. C_nxⁿ.
Let the three successive binomial coefficients in the expansion of (1 + x)ⁿ are \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1},{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}},{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}\)
Given that,
\({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}\) = 36 ………………(1)
\({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) = 84 ………………(2)
\({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}\) = 126 ……………..(3)

⇒ 2n – 2r = 3r + 3
⇒ 5r = 2n – 3 from (4)
⇒ 5 \(\left(\frac{3 n+3}{10}\right)\) = 2n – 3
⇒ 3n + 3 = 4n – 6
⇒ n = 9.

Question 15.
If the 2nd, 3rd and 4th terms in the expansion of (a + x)ⁿ are respectively 240, 720, 1080, find a, x, n. [TS – Mar. 2016; Ma ‘09, ’06]
Solution:
The general term in the expansion of (x + a)ⁿ is
T_r+1 = x^n-r . a^r
In the expansion, (a + x)ⁿ
T₂ = T₁₊₁
= \({ }^n C_1\) (a)^n-1 x¹ = 240 …………..(1)
T₃ = T₂₊₁
= \({ }^n C_2\) (a)^n-2 x² = 720 …………..(2)
T₄ = T₃₊₁
= \({ }^n C_3\) (a)^n-3 x³ = 1080 …………..(3)

⇒ 2(n – 2) x = 9a ………………(5)
\(\frac{(5)}{(4)}=\frac{2(n-2) x}{(n-1) x}=\frac{9 a}{6 a}\)
⇒ 4(n – 2) = 3(n – 1)
⇒ 4n – 8 = 3n – 3
⇒ n = 5
Substitute n = 5 in equation (4)
⇒ (5 – 1)x = 6a
⇒ 4x = 6a
⇒ 2x = 3a
⇒ x = \(\frac{3 a}{2}\)
Substitute n, x in equation (1).
\(5 C_1(a)^{5-1}\left(\frac{3 a}{2}\right)^1\) = 240
⇒ 5 . a⁴ . \(\frac{3}{2}\) . a = 240
⇒ a⁵ = 32
⇒ a = 2.
x = \(\frac{3 a}{2}\)
x = 3 . \(\frac{2}{2}\)
⇒ x = 3
∴ a = 2, x = 3, n = 5.

Question 16.
If the coefficients of rth, (r + 1)th and (r + 2)nd tenus in the expansion of (1 + x)ⁿ are in A.P then show that n² – (4r + 1) n + 4r² – 2 = 0.
Solution:
In the expansion (1 + x)ⁿ
The coeff. of T_r = \({ }^n \mathrm{C}_{\mathrm{r}-1}\);
The coeff. of T_r+1 = \({ }^n C_r\)
The coeÍf. of T_r+2 = \({ }^n \mathrm{C}_{\mathrm{r}+1}\)
Given that
\({ }^n \mathrm{C}_{\mathrm{r}-1}\), \({ }^n C_r\), \({ }^n \mathrm{C}_{\mathrm{r}+1}\) are in AP.
If a, b, c are in A.P.
Then 2b = a + c

2(n – r + 1) (r + 1) = (r + 1) r + (n – r + 1) (n – r)
⇒ 2nr + 2n – 2r² – 2r + 2r + 2 = r² + r + n² – nr – nr + r² + n – r
⇒ 2nr + 2n – 2r² + 2 = n² – 2nr + n + 2r²
⇒ n² – 2nr + n + 2r² – 2nr – 2n + 2r² – 2 = 0
⇒ n² – 4nr + 4r² – n – 2 = 0
⇒ n² – (4r + 1)n + 4r² – 2 = 0.

Question 17.
If ‘P’ and ‘Q’ are the sum of odd tenns and the sum of even terms respectively in the expansion of (x + a)ⁿ then prove that
i) P² – Q² = (x² – a²)ⁿ
ii) 4PQ = (x + a)²ⁿ – (x – a)²ⁿ [AP – March 2016, March ’10]
Solution:
Since P and Q are the sum of odd terms and sum of even terms respectively in the expansion of (x + a)ⁿ.
Now (x + a)ⁿ =

i) L.HS: P² – Q²
= (P + Q) (P – Q) = (x + a) (x – a)
= [(x + a) (x – a)]ⁿ
= (x² – a²)ⁿ = R.H.S.

ii) L.H.S : 4PQ = (P + Q)² – (P – Q)²
= ((x + a)ⁿ)² – ((x – a)ⁿ)²
= (x + a)²ⁿ – (x – a)²ⁿ = R.H.S.
∴ 4PQ = (x + a)²ⁿ – (x – a)²ⁿ.

Question 18.
If the coefficients of 4 consecutive term in the expansion of (1 + x)ⁿ are a₁, a₂, a₃, a₄ respectively then show that \(\frac{a_1}{a_1+a_2}+\frac{a_3}{a_3+a_4}=\frac{2 a_2}{a_2+a_3}\). [AP – Mar. 2017; May’11. ‘07, March ’11, ’95].
Solution:
Since a₁, a₂, a₃, a₄, are the coefficients of 4 consecutive terms in the expansion of (1 + x)ⁿ
Let a₁ = \({ }^n C_{r-1}\),
a₂ = \({ }^n C_r\),
a₃ = \({ }^n C_{r+1}\),
a₄ = \({ }^n C_{r+2}\)
L.H.S:

R.H.S:

Question 19.
Show that the middle term in the expansion of (1 + x)²ⁿ is \(\frac{1 \cdot 3 \cdot 5 \ldots \ldots(2 n-1)}{n !}(2 x)^n\).
Solution:
The expansion of (1 + x)²ⁿ contains 2n + 1 terms.
∴ Middle term is (n + 1)th term
∴ T_n+1 = \({ }^{2 n} C_n\) xⁿ

= \(\frac{2 n !}{n ! \cdot n !}\) xⁿ

= \(\frac{2 n(2 n-1)(2 n-2)(2 n-3) \ldots . .4 \times 3 \times 2 \times 1}{n ! n !} \cdot x^n\)

= \(\frac{1 \cdot 3 \cdot 5 \ldots .(2 n-1) \cdot 2 \times 4 \times 6 \times \ldots .2 n}{n ! n !} \cdot x^n\)

= \(\frac{1 \cdot 3 \cdot 5 \ldots .(2 n-1)(1 \times 2 \times 3 \times \ldots \ldots n) 2^n}{n ! n !} \cdot x^n\)

∴ Middle term = \(\frac{1 \cdot 3 \cdot 5 \ldots \ldots(2 n-1)}{n !}(2 x)^n\).

Question 20.
If (1 + 3x – 2x²)¹⁰ = a₀ + a₁x + a₂x² + ……………. + ax²⁰ then prove that
i) a₀ + a₁ + a₂ + ………….. + a₂₀ = 2¹⁰
ii) a₀ – a₁ + a₂ – a₃ + ……………. + a₂₀ = 4¹⁰
Solution:
Given
(1 + 3x – 2x²)¹⁰ = a₀ + a₁x + a₂x² + ……………. + ax²⁰ …………….(1)

i) Put x = 1, in (1), we get
(1 + 3 – 2)¹⁰ = a₀ + a₁ + a₂ + ………….. + a₂₀
⇒ a₀ + a₁ + a₂ + …………….. + a₂₀ = 2¹⁰.

ii) Put x = – 1 in (1), we get
(1 – 3 – 2)’°= a₀ – a₁ + a₂ – a₃ + ………………. + a₂₀
⇒ a₀ – a₁ + a₂ – a₃ + ……………. + a₂₀ = 4¹⁰.

Question 21.
If n is a positive integer, prove that \(\sum_{r=1}^n r^3 \cdot\left(\frac{{ }^n C_r}{{ }^n C_{r-1}}\right)^2=\frac{n(n+1)^2(n+2)}{12}\). [March ’13]
Solution:
Now, \(\frac{{ }^{n^n} C_r}{{ }^{n_C} C_{r-1}}=\frac{\frac{n !}{(n-r) ! r !}}{\frac{n !}{(n-r+1) !(r-1) !}}\)
= \(\frac{(n-r+1) !(r-1) !}{(n-r) ! r !}\)

Question 22.
Find the sum of the infinite series \(1+\frac{2}{3} \cdot \frac{1}{2}+\frac{2 \cdot 5}{3 \cdot 6}\left(\frac{1}{2}\right)^2+\frac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9}\left(\frac{1}{2}\right)^3+\ldots \infty\). [May ’91]
Solution:
Let the given series ¡s
S = 1 + \(\frac{2}{3} \cdot \frac{1}{2}+\frac{2 \cdot 5}{3 \cdot 6} \cdot \frac{1}{2^2}+\frac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9} \cdot \frac{1}{2^3}+\cdots \cdots\)
Comparing with 1 + nx + \(\frac{n(n-1)}{1 \cdot 2}\) x² + …………….

Question 23.
\(\frac{3 \cdot 5}{5 \cdot 10}+\frac{3 \cdot 5 \cdot 7}{5 \cdot 10 \cdot 15}+\frac{3 \cdot 5 \cdot 7 \cdot 9}{5 \cdot 10 \cdot 15 \cdot 20}+\cdots \cdots \cdots \infty\) [TS – Mar. 2017; May ’09]
Solution:
Let the given series is
S = \(\frac{3 \cdot 5}{5 \cdot 10}+\frac{3 \cdot 5 \cdot 7}{5 \cdot 10 \cdot 15}+\frac{3 \cdot 5 \cdot 7 \cdot 9}{5 \cdot 10 \cdot 15 \cdot 20}+\cdots \cdots \cdots \infty\)

S + \(\frac{3}{5}\) = \(\frac{3}{5}+\frac{3 \cdot 5}{5 \cdot 10}+\frac{3 \cdot 5 \cdot 7}{5 \cdot 10 \cdot 15}+\frac{3 \cdot 5 \cdot 7 \cdot 9}{5 \cdot 10 \cdot 15 \cdot 20}+\ldots \ldots\)

Question 24.
If x = \(\frac{1}{5}+\frac{1 \cdot 3}{5 \cdot 10}+\frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 15}+\ldots \ldots \cdots \infty\), then find 3x² + 6x. [TS – Mar. 2016, May ’14, ’11, ’08, Mar. ’14, ’12, ’06, TS – Mar. ’19]
Solution:
Given x = \(\frac{1}{5}+\frac{1 \cdot 3}{5 \cdot 10}+\frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 15}+\ldots \ldots \ldots\),
x + 1 = 1 + \(\frac{1}{5}+\frac{1 \cdot 3}{5 \cdot 10}+\frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 15}+\cdots \cdots\)
Now, comparing with

⇒ x² + 1 + 2x = \(\frac{5}{3}\)
⇒ 3x² + 6x + 3 = 5
⇒ 3x² + 6x = 2.

Question 25.
Find the sum of the infinite series \(\frac{3}{4}+\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}+\) [May ’10, March 11]
Solution:
Let the given series is
S = \(\frac{3}{4}+\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}+\ldots \ldots \ldots\)
S + 1 = 1 + \(\frac{3}{4}+\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}+\ldots \ldots \ldots\)
Now, comparing with

Question 26.
If x = \(\frac{1 \cdot 3}{3 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}+\frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12}+\ldots\) then prove that 9x² + 24x = 11. [AP – MAr. ’19, ’17, ’15, May ’16; TS – MAr. ’18, ’16, Mar. ’09, Board Paper, May ’15]
Solution:
Given

Question 27.
If x = \(\frac{5}{(2 !) \cdot 3}+\frac{5 \cdot 7}{(3 !) \cdot 3^2}+\frac{5 \cdot 7 \cdot 9}{(4 !) \cdot 3^3}+\ldots\) then find the value of x² + 4x. [TS – May; March ’13, May ’12]
Solution:
Given,

Question 28.
Find the sum to infinite terms of the series \(\frac{7}{5}\left(1+\frac{1}{10^2}+\frac{1 \cdot 3}{1 \cdot 2} \cdot \frac{1}{10^4}+\frac{1 \cdot 3 \cdot 5}{1 \cdot 2 \cdot 3} \cdot \frac{1}{10^6}+\ldots\right)\). [AP – Mar. ’18, ’16; May ’13, March ’05].
Solution:

Question 29.
Show that for any non-zero rational number x.
1 + \(\frac{x}{2}+\frac{x(x-1)}{2 \cdot 4}+\frac{x(x-1)(x-2)}{2 \cdot 4 \cdot 6}+\ldots \ldots\) = 1 + \(\frac{x}{3}+\frac{x(x+1)}{3 \cdot 6}+\frac{x(x+1)(x+2)}{3 \cdot 6 \cdot 9}+\ldots \ldots\) [March ’94]
Solution:
L.H.S = 1 + \(\frac{\mathrm{n}}{2}+\frac{\mathrm{n}(\mathrm{n}-1)}{2 \cdot 4}+\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{2 \cdot 4 \cdot 6}+\ldots\)

Some More Maths 2A Binomial Theorem Important Questions

Question 1.
Find the 7th term in the expansion of \(\left(\frac{4}{x^3}+\frac{x^2}{2}\right)^{14}\).
Solution:
Given \(\left(\frac{4}{x^3}+\frac{x^2}{2}\right)^{14}\)
Here, x = \(\frac{4}{x^3}\); a = \(\frac{x^2}{2}\); n = 14
r + 1 = 7
⇒ r = 6
The general term in the expansion of (x + a)ⁿ is
T_r+1 = \({ }^n C_r\) x^n-r a^r
The 7th term in the expansion of \(\left(\frac{4}{x^3}+\frac{x^2}{2}\right)^{14}\)

Question 2.
Find the 5th term in the expansion of (3x – 4y)⁷.
Solution:
Given (3x – 4y)⁷
Here, x = 3x; a = – 4y; n = 7
r + 1 = 5
r =4
The general term in the expansion of (x + a)ⁿ is
T_{r + 1} = \({ }^n C_r\) x^n-r a^r
The rth term in the expansion of (3x – 4y)⁷ is
T₄₊₁ = \({ }^7 \mathrm{C}_4\) (3x)^7-4 (- 4y)⁴
T₅ = \({ }^7 \mathrm{C}_4\) (3x)³ (4y)⁴
= 35 . 3³ . x³ 4⁴ . y⁴
= 241920 x³ . y⁴

Question 3.
Find the middle term(s) in \(\left(\frac{3}{a^3}+5 a^4\right)^{20}\).
Solution:
Given \(\left(\frac{3}{a^3}+5 a^4\right)^{20}\)
Here, x = \(\frac{3}{a^3}\), a = 5a⁴, n = 20
Since, n = 20 is even,
Middle term = \(\frac{n}{2}\) + 1
= \(\frac{20}{2}\) + 1
= 10 + 1 = 11th term
r + 1 = 11
⇒ r = 10
The general term in this expansion is
T_r+1 = \({ }^{{ }^n} C_r\) x^n-r a^r
11th term of given expansion is
T₁₀₊₁ = \({ }^{20} C_{10}\left(\frac{3}{a^3}\right)^{20-10}\left(5 a^4\right)^{10}\)
= \({ }^{20} \mathrm{c}_{10}\left(\frac{3}{a^3}\right)^{10}\left(5 a^4\right)^{10}\)

T₁₁ = \({ }^{20} C_{10}\) . a^-30 . 5¹⁰ . a⁴⁰
= \({ }^{20} C_{10}\) . 3¹⁰ . 5¹⁰ . a¹⁰

Question 4.
Find the middle term(s) in the expansion of (4x² + 5x³)¹⁷.
Solution:
Given (4x² + 5x³)¹⁷
Here, x = 4x², a = 5x³, n = 17
Since, n = 17 is odd, then
Middle terms \(\frac{\mathrm{n}+1}{2}, \frac{\mathrm{n}+3}{2}\) = 9, 10 terms
9’ term:
r + 1 = 9
⇒ r = 8
The general term in this expansion is
T_r+1 = \({ }^{{ }^n} C_r\) x^n-r a^r
9th term of given expansion is
T₈₊₁ = \({ }^{17} \mathrm{c}_9\) (4x²)^17-8 (5x³)⁸
= \({ }^{17} \mathrm{c}_9\) (4x²)⁹ (5x³)⁸
= \({ }^{17} \mathrm{c}_9\) . 4⁹ . x¹⁸ . 5⁸ . x²⁴
= \({ }^{17} \mathrm{c}_9\) . 4⁹ . 5⁸ . x⁴²

10th term:
r + 1 = 10
⇒ r = 9
The general term in this expansion is
T_r+1 = \({ }^{{ }^n} C_r\) x^n-r a^r
10th term in this expansion ¡s
T₉₊₁ = \({ }^{17} \mathrm{C}_9\)(4x²)^17-9 (5x³)⁹
T₁₀ = \({ }^{17} \mathrm{C}_9\) (4x²)⁸ (5x³)⁹
= \({ }^{17} \mathrm{C}_9\) . 4⁸ . 5⁹ . x¹⁶ . x²⁷
= \({ }^{17} \mathrm{C}_9\) . 4⁸ . 5⁹ . x⁴³

Question 5.
Find the coefficient of x¹¹ in \(\left(2 x^2+\frac{3}{x^3}\right)^{13}\).
Solution:
Given \(\left(2 x^2+\frac{3}{x^3}\right)^{13}\)
Here, x = 2x²,
a = \(\frac{3}{x^3}\), n = 13
Now, the general term in this expansion is
T_r+1 = \(\) x^n-r a^r
= \({ }^{13} \mathrm{C}_{\mathrm{r}}\) (2x^r)^13-r \(\left(\frac{3}{x^3}\right)^r\)
= \({ }^{13} \mathrm{C}_{\mathrm{r}}\) 2^13-r x^26-2r . 3² . x^-3r
= \({ }^{13} \mathrm{C}_{\mathrm{r}}\) 2^13-r 3^r x^26-5r ………………(1)
To find the coefficient of x¹¹
Put 26 – 5r = 11
5r = 15
⇒ r = 3
Now substituting r = 3 in equation (1) we get
T₃₊₁ = \({ }^{13} \mathrm{C}_3\) 2^13-3 3³ x^26-15
T₄ = \({ }^{13} \mathrm{C}_3\) 2¹⁰ 3³ x¹¹
The coefficient of x¹¹ in the expansion of \(\left(2 x^2+\frac{3}{x^3}\right)^{13}\) is \({ }^{13} \mathrm{C}_3\) 2¹⁰ 3³.

Question 6.
Find the term independent of x in the expansion of \(\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^2}\right)^{10}\).
Solution:
Given \(\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^2}\right)^{10}\)
Here, x = \(\sqrt{\frac{x}{3}}\), a = \(\frac{3}{2 x^2}\), n = 10
The general term in this expansion is
T_r+1 = \({ }^n C_r\) x^n-r a^r
T_r+1 = \({ }^{10} C_r\left(\sqrt{\frac{x}{3}}\right)^{10-r}\left(\frac{3}{2 x^2}\right)^r\)
= \({ }^{10} C_r \frac{x^{\frac{10-r}{2}}}{3^{\frac{10-r}{2}}} \cdot \frac{3^r}{2^r} \cdot x^{-2 r}\)
= \({ }^{10} \mathrm{C}_{\mathrm{r}} \frac{3^{\mathrm{r}}}{3^{\frac{10-\mathrm{r}}{2}} \cdot 2^{\mathrm{r}}} \mathrm{x}^{\frac{10-\mathrm{r}}{2}-2 \mathrm{r}}\) ……………….(1)
To find the term independent of x¹¹ (i.e., coeff. of x°)
Put \(\frac{10-\mathrm{r}}{2}\) – 2r = 0
10 – r – 4r = 0
5r = 10
⇒ r = 2
Substitute r = 2 in equation (1) we get
T = \({ }^{10} \mathrm{C}_2 \frac{3^2}{3^{\frac{10-2}{2}} \cdot 2^2} \cdot \mathrm{x}^{\frac{10-2}{2}-2^2}\)
= \({ }^{10} C_2 \frac{3^2}{3^3 \cdot 2^2} \cdot x^0\)
= \(\frac{10 \cdot 9}{1 \cdot 2} \times \frac{3^2}{3^4 \cdot 2^2}=\frac{10 \cdot 9}{1 \cdot 2} \cdot \frac{1}{9 \cdot 4} \cdot x^0=\frac{5}{4} x^0\)
The term independent of x in the given expansion is T₃ = \(\frac{5}{4}\).

Question 7.
Find the largest binomial coefficients In the expansion of (1 + x)¹⁹.
Solution:
Given, (1 + x)¹⁹
Here n = 19, an odd integer.
The largest binomial coefficients are \({ }^n C_{\frac{n-1}{2}}\) and \({ }^n C_{\frac{n+1}{2}}\)
\({ }^{19} \mathrm{C}_{\frac{19-1}{2}} \text { and }{ }^{19} \mathrm{C}_{\frac{19+1}{2}}\) = \({ }^{19} C_9 \text { and }{ }^{19} C_{10}\)
(Note that, \({ }^{19} C_{9}={ }^{19} C_{10}\)).

Question 8.
If the coefficients of (2r + 4)th, (r – 2)th terms in the expansion of (1 + x)²¹ equal find ‘r’.
Solution:
Given (1 – x)¹⁸
The general term in the expansion of (x + a)ⁿ is
T_r+1 = \({ }^n C_r\) x^n-r a^r
(2r + 4)th term in the expansion of (1 + x)¹⁸ is
T_(2r+3)+1 = \({ }^{18} \mathrm{C}_{2 \mathrm{r}+3}\) (1) x^2r+3
T_2r+4 = \({ }^{18} \mathrm{C}_{2 \mathrm{r}+3}\) x^2r+3
The coeff. of (2r + 4)th term is \({ }^{18} \mathrm{C}_{2 \mathrm{r}+3}\)
(r – 2) term in the expansion of (1 + x)¹⁸ is
T_(r-3)+1 = \({ }^{18} \mathrm{C}_{\mathrm{r}-3}\) 1^18-(r-3) x^r-3
T_r-2 = \({ }^{18} \mathrm{C}_{\mathrm{r}-3}\) . x^r-3
The coeff. of (r – 2)nd term is \({ }^{18} \mathrm{C}_{\mathrm{r}-3}\)
∴ Given that two coefficients are equal.
\({ }^{18} \mathrm{C}_{2 \mathrm{r}+3}={ }^{18} \mathrm{C}_{\mathrm{r}-3}\)
\({ }^n C_r={ }^n C_s\)
⇒ n = r + s (or) r = s
18 = 2r + 3 + r – 3
2r + 3 = r – 3
3r = 18
⇒ r = 6
⇒ r = – 6
Since, r is a positive integer, we get r = 6.

Question 9.
Find the set E of x for which the binomial expansion (7 + 3x)^-5.
Solution:
Given (7 + 3x)^-5
7^-5 (1 + \(\frac{3 x}{7}\))^-5
The binomial expansion of (7 + 3x)^-5 is valid when
\(\left|\frac{3 x}{7}\right|<1 \Rightarrow|x|<\frac{7}{3}\)
x ∈ \(\left(\frac{-7}{3}, \frac{7}{3}\right)\)

Question 10.
Find the set E of x for which the binomial expansion (7 – 4x)^-5 is valid.
Solution:
Given (7 – 4x)^-5 = 7^-5 (1 – \(\frac{4 x}{7}\))^-5
The binomial expansion of (7 – 4x)^-5 is valid when
\(\left|\frac{-4 x}{7}\right|<1 \Rightarrow\left|\frac{4 x}{7}\right|<1\)
⇒ x < \(\frac{7}{4}\)
⇒ x ∈ \(\left(\frac{-7}{4}, \frac{7}{4}\right)\)
∴ E = \(\left(\frac{-7}{4}, \frac{7}{4}\right)\)

Question 11.
Find the 6th term of (1 + \(\frac{x}{2}\))^-5
Solution:
Given, (1 + \(\frac{x}{2}\)))^-5
Comparing this with (1 + x)ⁿ, where
x = \(\frac{x}{2}\), n = – 5,
⇒ r + 1 = 6
⇒ r = 5
The general term in the binomial expansion of (1 + x)ⁿ is

Question 12.
Find the coefficient of x⁶ in (3x – \(\frac{4}{x}\))¹⁰.
Solution:
Given (3x – \(\frac{4}{x}\))¹⁰
Here, x = 3x; a = – \(\frac{4}{x}\), n = 10
The general term in this expansion is
T_r+1 = \({ }^n C_r\) x^n-r a^r
= \({ }^{10} \mathrm{C}_{\mathrm{r}}\) (3x)^10-r \(\left(\frac{-4}{x}\right)^{\mathbf{r}}\)
= \({ }^{10} \mathrm{C}_{\mathrm{r}}\) 3^10-r x^10-r (- 4)^r x^r
= \({ }^{10} \mathrm{C}_{\mathrm{r}}\) 3^10-r . (- 4)^10-r . x^10-2r …………..(1)
To find the coefficient of x^-6,
put 10 – 2r = – 6
⇒ 2r = 16
⇒ r = 8
Now, substituting r = 8 in equation (1) we get
T₈₊₁ = \({ }^{10} \mathrm{C}_8\) 3^10-8 (- 4)⁸ x^10-16
T₉ = \({ }^{10} \mathrm{C}_8\) 3² (4)⁸ x^-6
The coeff. of x^-6 in the expansion of (3x – \(\frac{4}{x}\))¹⁰ is \({ }^{10} \mathrm{C}_8\) . 3² . (4)⁸

Question 13.
Find the middle term (s) in the expansion of \(\left(4 a+\frac{3}{2} b\right)^{11}\).
Solution:
Given \(\left(4 a+\frac{3}{2} b\right)^{11}\)
Here, x = 4a; a = \(\frac{3}{2}\) b; n = 11
Since, n = 11 is odd,
Middle terms = \(\frac{\mathrm{n}+1}{2}, \frac{\mathrm{n}+3}{2}\) = 6, 7 terms.

6th term :
r + 1 = 6
⇒ r = 5
The general term in this expansion is
T_r+1 = \({ }^n C_r\) x^n-r a^r
6th term of given expansion ¡s
T₅₊₁ = \({ }^{11} C_5(4 a)^{11-5}\left(\frac{3}{2} b\right)^5\)
T₆ = \({ }^{11} \mathrm{C}_5(4 \mathrm{a})^6\left(\frac{3}{2} \mathrm{~b}\right)^5\)
= \(=11 C_5 \frac{4^6 \cdot 3^5}{2^5} \cdot a^6 \cdot b^5\)
T₆ = \({ }^{11} \mathrm{C}_6 \frac{2^{12} \cdot 3^5}{2^5}\) . a⁴b⁵
= \({ }^{11} \mathrm{C}_6\) . a⁶ . b⁵

7th term :
r + 1 = 7
⇒ r = 6
The general term in this expansion is
T_r+1 = \({ }^n C_r\) x^n-r a^r
7th term of given expansion is

Question 14.
Find the middle term (s) in the expansion of (4x² + 5x³)¹⁷.
Solution:
Given (4x² + 5x³)¹⁷.
Here, x = 4x², a = 5x³, n= 17.
Since, n = 17 is odd, then
middle terms = \(\frac{n+1}{2}, \frac{n+3}{2}\) = 9, 10 terms

9th term :
r + 1 = 9
⇒ r = 8
The general term in this expansion is
T_r+1 = \({ }^n C_r\) x^n-r a^r
∴ 9th term in this expansion is
T₉₊₁ = \({ }^{17} \mathrm{C}_9\)(4x²)^17-9 (5x³)⁹
T₁₀ = \({ }^{17} \mathrm{C}_9\) (4x²)⁸ (5x³)⁹
= \({ }^{17} \mathrm{C}_9\) 4⁸ . 5⁹ x¹⁶ . x²⁷
= \({ }^{17} \mathrm{C}_9\) 4⁸ . 5⁹ . x⁴³

Question 15.
Prove that 2 . C₀ + 5 . C₁ + 8 . C₂ + ……………… + (3n + 2) C_n = (3n + 4) 2^n-1
Solution:
Let S = 2 . C₀ + 5 . C₁ + 8 . C₂ + ……………… + (3n + 2) C_n = (3n + 4) 2^n-1 …………….(1)
By writing the terms in (1) in the reverse order
S = (3n + 2) . C_n + (3n – 1) . C_n-1 + (3n – 4) C_n-2 + ……………….. + 2. C₀
S = (3n + 2) C₀ + (3n – 1) C₁ + (3n – 4) C₂ + ……………… + 2 C_n ………….. (2)
Adding (1) and (2) we get,
S = 2 . C₀ + 5 . C₁ + 8 . C₂ + …………… + (3n + 2) . C_n
S = (3n + 2) . C₀ + (3n – 1) . C₁ + (3n – 4) . C₂ + ……………. + 2 . C_n
2S = (3n + 4)C₀ + (3n + 4) C₁ . (3n + 4) C₂ + …………… + (3n + 4) C_n
2S = (3n + 4) (C₀ + C₁ + C₂ + …………….. + C_n)
2S = (3n + 4) 2ⁿ
⇒ S = (3n + 4) 2^n-1
∴ 2 . C₀ + 5 . C₁ + 8. C₂ + …………… + (3n + 2) C_n = (3n + 4) . 2^n-1

Question 16.
Prove that
(C₀ + C₁) (C₁ + C₁) (C₂ + C₃) + ………………… + (C_n-1 + C_n) = \(\frac{(n+1)^n}{n !}\) . C₀ . C₁ . C₂ ……………. C_n

Question 17.
Write the first 3 terms in the expansion of \(\left(1+\frac{\mathrm{x}}{2}\right)^{-5}\).
Solution:
Given \(\left(1+\frac{\mathrm{x}}{2}\right)^{-5}\)
We know that,

Hence, the first three terms in the expansion of \(\left(1+\frac{\mathrm{x}}{2}\right)^{-5}\) are 1, \(\frac{-5 x}{2}, \frac{15 x^2}{4}\).

Question 18.
Find the coefficient of x⁶ in the expansion of (1 – 3x)^-2/5.
Solution:
Given that (1 + 3x)^-2/5
The general term of (1 – 3x)^-2/5 is
T_r+1 = \(\frac{n(n-1)(n-2) \ldots .(n-r+1)}{r !}\)
where x = – 3x, n = \(\frac{-2}{5}\)

T_r+1 = \(\frac{\frac{-2}{5}\left(\frac{-2}{5}-1\right)\left(\frac{-2}{5}-2\right) \ldots \ldots . .\left(\frac{-2}{5}-r+1\right)}{r !}(-3 x)^r\)

= \(\frac{\frac{-2}{5}\left(\frac{-7}{5}\right)\left(\frac{-12}{5}\right) \cdots \cdots \cdots\left(\frac{3}{5}-\mathrm{r}\right)}{\mathrm{r} !}(-3 \mathrm{r})^{\mathrm{r}}\)

= \(\frac{2 \cdot 7 \cdot 12 \ldots \ldots(5 \mathrm{r}-3)}{5^{\mathrm{r}} \cdot \mathrm{r} !}(3 \mathrm{x})^{\mathrm{r}}\)

Put r = 6
Then the coefficient of x^-6 is \(\frac{2 \cdot 7 \cdot 12 \ldots \ldots .27}{5^6 \cdot 6 !}(3)^6\)
= \(\frac{2 \cdot 7 \cdot 12 \ldots \ldots .27}{6 !} \cdot\left(\frac{3}{5}\right)^6\).

Question 19.
Eind the coefficient of x⁴ in the expansion of (1 – 4x)^3/5
Solution:
Given (1 – 4x)^3/5
The general term of (1 – 4x)^3/5 is
T_r+1 = \(\frac{n(n-1)(n-2) \ldots \ldots(n-r+1)}{r !}\) . x^r
where, x = – 4x, n = \(\frac{-3}{5}\)

Question 20.
Find the sum of the infinite series 1 + \(\frac{1}{3}+\frac{1 \cdot 3}{3 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}+\ldots \ldots\). [TS – Mar. 2015]
Solution:
Let the given series is
S = 1 + \(\frac{1}{3}+\frac{1 \cdot 3}{3 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}+\ldots \ldots\)
Now, comparing with

Question 21.
Find the sum of the infinite series 1 – \(\frac{4}{5}+\frac{4 \cdot 7}{5 \cdot 10}-\frac{4 \cdot 7 \cdot 10}{5 \cdot 10 \cdot 15}\) + …………..
Solution:
Let the given series is
S = 1 – \(\frac{4}{5}+\frac{4 \cdot 7}{5 \cdot 10}-\frac{4 \cdot 7 \cdot 10}{5 \cdot 10 \cdot 15}\) + …………..
Now comparing with

Question 22.
Find the sum of the infinite series \(\frac{3}{4 \cdot 8}-\frac{3 \cdot 5}{4 \cdot 8 \cdot 12}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16}\).
Solution:
Let the given series

Question 23.
If t = \(\frac{4}{5}+\frac{4 \cdot 6}{5 \cdot 10}+\frac{4 \cdot 6 \cdot 8}{5 \cdot 10 \cdot 15}\) + ……………….. ∞, then prove that 9t = 16.
Solution:
Given

Here p = 4; p + q = 6;
⇒ q = 2
⇒ x = \(\frac{2}{5}\)
∴ (1) ⇒ 1 + t = \(\left(1-\frac{2}{5}\right)^{\frac{-4}{2}}\)
⇒ 1 +t = \(\left(\frac{3}{5}\right)^{-2}\)
⇒ 1 + t = \(\frac{25}{9}\)
⇒ 9t =16.

Question 24.
If I, n are positive integers, 0 < f < 1 and if (7 + 4√3)ⁿ = 1 + f, then show that
i) I is an odd integer and
ii) (I + f) (1 – f) = 1.
Solution:
Sol. Since I, n are positive integers 0 < F < 1 and (7 + 4√3)ⁿ = 1 + F
Now, 36 < 48 < 49
⇒ 6 < 4√3 < 7 ⇒ – 6 > – 4√3 > – 7
⇒ 1 > 7 – 4√3 > 0
⇒ 0 < 7 – 4√3 < 1
⇒ 0< 7 – 4√3 < 1
⇒ o < (7 – 4√3)ⁿ < 1
Let (7 – 4√3)ⁿ = f
∴ 0 < f < 1
Now, 1 + f = (7 + 4√3)ⁿ

where k is an integer
⇒ I + F + f = Even integer.
Since, I is an integer, F + f is an integer.
But, 0 < F < 1, 0 < f < 1
⇒ 0 < F + f < 2.
∴ f + F = 1
From (1)
⇒ I + 1 = Even integer
⇒ Even integer – 1 = Odd Integer
∴ I is an odd integer.

ii) L.H.S: (I + f) (I – f) = (7 + 4√3)ⁿ . F
= (7 + 4√3)ⁿ (7 – 4√3)ⁿ
= (49 – 48)ⁿ = 1ⁿ = 1 = R.H.S
∴ LH.S = R.H.S.

Question 25.
Find the 8th term of \(\left(1-\frac{5 x}{2}\right)^{-3 / 5}\). [AP – Mar. ’18]
Solution:
Given \(\left(1-\frac{5 x}{2}\right)^{-3 / 5}\)
T₈ = \(\frac{\left(\frac{-3}{5}\right)\left(\frac{-3}{5}-1\right)\left(\frac{-3}{5}-2\right) \ldots\left(\frac{-3}{5}-6\right)}{7 !}\left(\frac{-5 x}{2}\right)^7\)
= \(\frac{\left(\frac{3}{5}\right)\left(\frac{8}{5}\right)\left(\frac{13}{5}\right) \ldots\left(\frac{33}{5}\right)}{7 !}\left(\frac{5 x}{2}\right)^7\)
= \(\frac{3 \times 8 \times 13 \times \ldots . \times 33}{7 !}\left(\frac{x}{2}\right)^7\)

Telangana TSBIE TS Inter 2nd Year Chemistry Study Material 7th Lesson d and f Block Elements & Coordination Compounds Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material 7th Lesson d and f Block Elements & Coordination Compounds

Very Short Answer Questions (2 Marks)

Question 1.
What are transition elements ? Give examples.
Answer:
Transition elements are defined as the elements having partially filled d-orbitals in the atoms in the elemental form or in the chemically significant stable oxidation states. Examples : Fe, Mn, Cr.

Question 2.
Which elements of 3d, 4d, and 5d series are not regarded as transition elements and why ?
Answer:
The elements Zn, Cd, Hg are not regarded as transition elements. They belong to 3d, 4d and 5d series respectively. Their general electronic configuration is (n – 1) d¹⁰ ns². The d-orbitals in these elements are completely filled. Therefore, they are not regarded as transition elements.

Question 3.
Why are d-block elements called transition elements ?
Answer:
The d-block elements are placed in between s and p-blocks in the periodic table. There is a transition in properties between the electropositive s-block elements and the electronegative p-block elements. Hence, d- block elements are named transition elements.

Question 4.
Write the general electronic configuration of transition elements.
Answer:
(n – 1) d¹⁰ ns^1-2
The (n – 1) stands for the penultimate shell. The (n -1) d orbitals may have one to ten electrons and the outermost ns orbital may have one or two electrons.

Question 5.
In what way is the electronic configuration of transition elements different from non-transition elements ?
Answer:
The transition elements have incompletely filled (n – 1) d orbitals.

In transition elements, the valence shell and penultimate shell are incompletely filled. Their general electronic configuration is (n- 1) d^{1 – 10} ns^{1 – 2}.

The non-transition elements i.e., the representative elements have the incompletely filled valence shell. The inner transition elements have the outer three energy levels incomplete.

Question 6.
Write the electronic configuration of chromium (Cr) and copper (Cu).
Answer:
Cr (24) – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d 4s¹ or [Ar] 4s¹ 3d⁵
Cu (29) – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ or [Ar] 4s¹ 3d¹⁰

Question 7.
Why do transition elements exhibits characteristic properties ?
Answer:
The transition elements are characterised by specific and special properties which can be explained on the basis of their chara-cteristic electronic configuration. The partly filled (n -1) d orbital is the cause of special properties like variable oxidation states, magnetic properties etc.

Question 8.
Scandium is a transition element. But zinc is not. Why ? [IPE 14]
Answer:
The atom of Sc has one unpaired d-electron. So it is a transition element. The electronic configuration of Zn is (Z = 30) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s².
The common oxidation state of Zn is +2.
The electronic configuration of Zn⁺⁺ is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰.
As neither Zn nor Zn⁺⁺ have partly filled 3d orbitals, Zinc is not considered as transition element.

Question 9.
Even though silver has d¹⁰ configuration, it is regarded as transition element Why ?
Answer:
Silver (Z = 47) can exhibit +2 oxidation state where in it will have incompletely filled d- orbitals (4d), hence it is a transition element.

Question 10.
Write the electronic configuration of Co²⁺ and Mn²⁺.
Answer:
Co²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷ Or [Ar] 3d⁷
Mn²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ Or [Ar] 3d⁵

Question 11.
Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to their +3 state ?
Answer:
Mn²⁺ has stable configuration because 3d orbitals are half-filled. Hence it is stable.

Fe²⁺ has the configuration [Ar] 3d⁶. As it gains half-filled electronic configuration jt! by losing an electron it is susceptible to oxidation.

Question 12.
Which metal in the first series of transition metals exhibits + 1 oxidation state most frequently and why ?
Answer:
Copper
Reason : The second ionisation enthalpy , is unusually high. Removal of second electron disrupts the stable configuration of Cu⁺.

Question 13.
Why do transition elements exhibit more than one oxidation state (variable oxidation states)?
Answer:
Transition elements exhibit general Oxidation state of + II by loosing the electrons in the outer “s” orbital. They also Show other oxidation states because of the participation of (n – 1) d electrons in bonding.

Both ns and (n -1) d electrons are available for bonding because there is very little difference between the energies of ns & (n – 1) d orbitals.

Question 14.
Though Sc is a transition element, it does not exhibit variable oxidation state. Why ?
Answer:
As there is only one d-electron Sc does not show vairable oxidation states. Thus Sc (II) is unknown. Further d¹ configuration is unstable.

Question 15.
Why is it difficult to obtain M³⁺ oxidation state in Ni, Cu and Zn ?
Answer:
As the third ionisation enthalpies are quite high, it is difficult to obtain M⁺³ state in Ni, Cu and Zn.

Question 16.
Why is Cr²⁺ reducing and Mn³⁺ oxidizing even though both have the same d4 electronic configuration ?
Answer:
Cr²⁺ is reducing as its configuration changes from d⁴ to d³, the latter having a half- filled d_xy, d_yz, d_zx orbitals. On the other hand, the change from Mn²⁺ to Mn³⁺ results in the half-filled d⁵ configuration which has extra stability.

Question 17.
Although Cr, Mo and W belong to the same group (group 6) Cr (VI) is a strong oxidizing agent while Mo (VI) and W (VI) are not. Why ?
Answer:
In Group 6, Mo(VI) and W(VI) are found to be more stable than Cr(VI). Thus Cr (VI) in the form of dichromate acidic medium is a strong oxidising agent, whereas MnO₃ and WO₃ are not.

Question 18.
What do you infer from the fact that M³⁺ / M²⁺ standard electrode potential of Mn is comparatively higher and that of Fe is comparitively lower ?
Answer:
The higher value for Mn shows that Mn²⁺ (d⁵) is particularly stable whereas comparatively low value for Fe shows the extra stability of Fe³⁺ (d⁵ configuration).

Question 19.
Transition elements have high melting points. Why ?
Answer:
The melting points of the metals are very high. The high values are attributed to the strong interactions present in the metals.

The involvement of greater number of electrons from (n – 1) d in addition to the ns electrons in the interatomic metallic bonding is the reason for high m.p.’s.

Question 20.
Among the first transition series (3d series) Chromium has highest melting point. Why ?
Answer:
In chromium, there is involvement of maximum number of (n -1) d electrons (d⁵) and one ns electron is interatomic metallic bonding. So the metallic bonding is strong. Hence the m.p is high.

Question 21.
Compared to s-block elements, the transition elements exhibit higher enthalpy of atomization. Why ?
Answer:
Because of large number of unpaired electrons in their atoms, they have stronger interatomic interaction. Hence stronger bonding between atoms resulting in higher enthalpies of atomisation.

Question 22.
Among the first transition series (3d series) zinc has lowest enthalpy of atomization. Why?
Answer:
As there are only two electrons available for metallic bond formation, the interatomic attractions are less. Hence the zinc has lowest enthalpy of atomisation.

Question 23.
How do you expect the density of transition element to vary in a given series and why ?
Answer:
Increase in density is expected from in the transition elements from Sc Z = 21 to copper Z = 29.
Reason :

1. Metallic bond strength increases due to increase in (n – 1) d electrons.
2. Metallic radius decreases.
3. Increase in atomic mass.
   As expected, the density increases from scandium to copper in the 3d series.

Question 24.
How do the atomic and ionic sizes vary among transition metals in a given series ?
Answer:
The atomic and ionic radii decrease slightly with the increase in a series of transition elements. This is because of poor screening by d-electrons.

New electron enters into a d orbital each time, nuclear charge increases by unity as we move from one element to the other in the transition series. As the shielding effect of a d-electron is not effective, the net attraction between the nuclear charge and the outermost electron increases and the ionic radius decreases.

Question 25.
Why do Mn, Ni and Zn exhibit more negative E^⊖ values than expected ?
Answer:
The stability of the half-filled d-subshell in Mn²⁺ and completely filled d¹⁰ configuration in Zn⁺⁺ are related to their more negative E^⊖ values. The value is more for Ni because of its highest negative \(\Delta \mathrm{H}_{\text {hyd }}^{\ominus}\) .

Question 26.
Among the first transition series (3d series) only copper has positive E^⊖ M²⁺/M value. Why?
Answer:
Positive E^⊖ value means less ability to act as reducing agent. Thus Cu cannot liberate H₂ from acids.
Reason:

1. Cu has high value for enthalpy of atomisation ∆_aH^⊖.
2. Low value for enthalpy of hydration ∆_hydH^⊖
3. The high energy to transform Cu(s) to Cu⁺⁺ (aq) is not balanced by its hydration enthalpy.

Question 27.
Cu¹¹ forms halides like CuF₂, CuCl₂ and CuBr₂ but not Cul₂. Why ?
Answer:
Cu²⁺ oxidises I^– to I₂. Hence CuI₂ is not known.
2CU²⁺ + 4I²⁺ → Cu₂I₂ (s) + I₂

Question 28.
The highest Mil fluoride is MnF₄ where as the highest oxide is Mn₂O₇. Why ?
Answer:
Oxygen can form multiple bonds with metals. This explains the ability of oxygen to stabilise high oxidation states. In the covalent oxide Mn₂O₇ each Mn is tetra-hedrally surrounded by O’s including a Mn – O – Mn bridge. The ability to stabilise higher oxidation state is less for fluorine when compared to oxygen. Hence it can form MnF₄ Only.

Question 29.
In its fluoride or Oxide, in which a transition metal exhibits highest oxidation state and why ?
Answer:
Mn exhibits its highest oxidation state in Mn₂O₇, because of the ability of oxygen to form multiple bonds with metals. In the covalent oxide, Mn₂O₇ each Mn is tetrahedrally surrounded by oxygen including a Mn – O – Mn bridge.

Question 30.
Why Zn²⁺ is diamagnetic whereas Mn²⁺ is paramagnetic ? [TS 15]
Answer:
Electronic configuration of Zn⁺⁺ is [Ar] 3d¹⁰.
As there are no unpaired electrons, Zn⁺⁺ is diamagnetic.
Electronic configuration of Mn⁺⁺ is [Ar] 3d⁵. As there are unpaired electrons, Mn⁺⁺ is paramagnetic.

Question 31.
Write ‘spin only’ formula to calculate the magnetic moment of transition metal ions.
Answer:
Magnetic moment μ = \(\sqrt{n(n+2)}\)
n = number of unpaired electrons
m = magnetic moment in Bohr magnetons

Question 32.
Calculate the ‘spin only’ magnetic moment of Fe²⁺ _(aq) ion. [AP 17]
Answer:
Fe (26) Electronic configuration [Ar] 3d⁶.4s² Fe²⁺. Electronic configuration is [Ar] 3d⁶
No. of unpaired electrons is 4.
Spin only formula μ = \(\sqrt{n(n+2)}\)
= \(\sqrt{4 \times 6}=\sqrt{24}\) = 4.9 BM

Question 33.
What is meant by ‘disproportionation’ ? Give an example of disproportionation reaction in aqueous solution.
Answer:
If an element undergoes both oxidation and reduction, it is called disproportionation reaction.
Ex : Many copper (I) compounds are un-stable in aqueous solution and undergo disproportionation.
2 Cu⁺ → Cu⁺⁺ + Cu

Question 34.
Aqueous Cu²⁺ ions are blue in colour, where as Aqueous Zn²⁺ ions are colourless. [TS 16]
Answer:
Cu⁺⁺ contains unpaired electron. It is coloured due to d-d transition. Zn⁺⁺ does not contain unpaired electron. Hence it is colourless.

Question 35.
What are complex compounds ? Give examples.
Answer:
Complex compounds are those in which the metal ions bind a number of anions or neutral molecules.
Ex: [Fe (CN)₆]^3-, [Fe(CN)₆]^4-, [Cu (NH₃)₄]²⁺

Question 36.
Why do the transition metals form a large number of complex compounds ?
Answer:
Transition metals form a large number of complex compounds due to

1. smaller sizes of the metal ions,
2. high ionic charges,
3. availability of d-orbitals for bond formation.

Question 37.
How do transition metals exhibit catalytic activity ?
Answer:
Transition metals possess good catalytic properties. This property is due to the the free valencies of the metals and also due to variable oxidation states.

Question 38.
Give two reactions in which transition metals or their compounds acts as catalysts.
Answer:

1. Vanadium pentoxide is used as catalyst in the manufacture of sulphuric acid by Contact process.
   
2. Pt is used as catalyst in Ostwald process.
   

Question 39.
What is an alloy ? Give example.
Answer:
When metals are mixed and the resulting liquid is allowed to solidify, the product formed is called an alloy. It may also contain non-metals.
Ex: German silver 25 – 50% Cu,
10 – 30% Ni
25 – 35% Zn
Nichrome 60% Ni, 25% Fe, 15% Cr.
Bell metal 80% Cu, 20% Sn

Question 40.
Why do the transition metals readily form alloys ?
Answer:
As the crystal structures of the transition metals are similar, alloys are formed among them readily.

Question 41.
How does the ionic character and acidic nature vary among the oxides of first transition series?
Answer:
The oxides dissolve in acids and bases to form oxometallic salts. Potassium dichromate and potassium permanganate are common examples.

Question 42.
What is the efffect of increasing pH on a solution of potassium dichromate ?
Answer:
It is converted into chromate by alkali. Chromate is yellow in colour.
K₂Cr₂O₇ + 2KOH → 2K₂CrO₄ + H₂O

Question 43.
Name the oxometal anions of the first series of transition metals in which the metal exhibits the oxidation state equal to its group number.
Answer:
6th group – K₂Cr₂O₇
Cr₂O₇^{– –} – is dichromate anion in which the oxidation number of Cr is + VI which is equal to its group number.
In permanganate anion \(\mathrm{MnO}_4^{-}\), Mn shows + VII state which is equal to its group number.

Question 44.
Permanganate titrations are carried out in the presence of sulphuric acid but not in presence of hydrochloric acid. Why ?
Answer:
The oxidation reactions of potassium per- mangante are generally carried out in presence of sulphuric acid. Hydrochloric acid should not be Used as the medium since it liberates chlorine by reaction with permanganate.
2KMnO₄ + 16HCl → 2MnCl₂ + 2KCl + 5Cl₂ ↑ + 8H₂O

Question 45.
What is lanthanoid contraction ? [T.S. Mar. 19]
Answer:
Gradual decrease in atomic and ionic sizes from La to Lu is called Lanthanoid con-traction.
Reason:

1. Imperfect shielding of one electron by another in the same set of orbitals (4f)
2. The shielding of one 4f electron by another is less than that of one d electron by another.
3. Nuclear charge increases along the series from La to Lu.
4. Due to poor shielding of 4f electrons, there is fairly regular decrease in the size of the entire 4f orbitals.

Question 46.
What are the different oxidation states exhibited by the lanthanoids ?
Answer:
Lanthanides react easily with water to give solutions giving + 3 ions. The principal oxidation state is + 3 although +4 and +2 oxidation states are also exhibited.

Question 47.
What is misch metal ? Give its composition and uses. [AP 16]
Answer:
Misch metal is an alloy of Lanthanoid metal.
Composition : 95% Lanthanoid metal and 5% iron and traces of S, C, Ca and Al.
It is used in making tracer bullets.

Question 48.
What is actinoid contraction ?
Answer:
The elements from Ac to Lr (Z = 89 to 103) are called actinides. In these elements the 5f orbital is gradually filled.

There is steady decrease in the size of M³⁺ and M⁴⁺ cations in actinide elements. The shielding of one electron in 5f orbital by another electron present in the same orbital is very poor. Due to this poor shielding effect, the increase in the nuclear charge by one unit brings the valence shell nearer to the nucleus and the size of the cation decreases.

Question 49.
What are coordination compounds ? Give two examples.
Answer:
Coordination compounds are those in which metal ions bind a number of anions or neutral molecules giving complex species with characteristic properties.

The molecules or ions containing one pair of electrons donate the electron pair to the central metal ion to form coordinate covalent bond.
Ex: [Fe(CN)₆]^4-
Hexacyanoferrate (II)

Question 50.
What is a coordination polyhedron ?
Answer:
The spacial arrangement of the ligand atoms which are directly attached to the central atom/ion defines a coordination polyhedron about the central atom. Most common coordination polyhedra are octahedral, square planar and tetrahedral.
For example, [Co(NH₃)₆]²⁺ is octahedral, Ni(CO)₄ is tetrahedral.

Question 51.
What is a double salt ? Give example.
Answer:
Those compounds which lose their identity in solution and break down into simple ions are called double salts. For example, an aqueous solution of potash alum K₂SO₄. Al₂ Al₂(SO₄)₃ . 24H₂O gives the test for K⁺, Al³⁺ and \(\mathrm{SO}_4^{2-}\) ions in its solution form.

Question 52.
What is the difference between a double salt and a complex compound ?
Answer:

Question 53.
What is a ligand ? [Mar. 2018, TS]
Answer:
A ligand is an ion or a molecule containing lone pairs of electrons which can be donated to a transition metal ion to form metal- ligand coordinate covalent bond.
Ex: NH₃, CN^–

Question 54.
Give one example each for ionic and neutral ligands.
Answer:
Ionic Ligands: Cyano CN^–; amino NH₂^–
Neutral Ligands : Carbonyl CO ; Ammine NH₃

Question 55.
How many moles of AgCl is precipitated when 1 mole of CoCl₃ is treated with AgNO₃ solution ?
Answer:
Three moles of AgCl is precipitated when 1 mole of COCl₃ is treated with AgNO₃ solution.

Question 56.
What is a chelate ligand ? Give example.
Answer:
A ligand which can form ring type complex is called chelate ligand.
Ethylenediamine H₂N – CH₂ – CH₂ – NH₂ is a chelate ligand. It can form ring type complex.

Question 57.
What is an ambidentate ligand ? Give example.
Answer:
Ligands which have two or more different donor sites but only one of these is attached to a single metal atom at a given time, are called ambidentate ligand.
Ex : NCS^– Thiocyanato M ← SCN isothio- cyanato M ← NSC

Question 58.
CuSO₄ . 5H₂O is blue in colour where as anhydrous CuSO₄ is colourless. Why ?
Answer:
A) Colour of a complex is due to splitting of d-orbitals and d-d transition of the electron.

In the absence of ligand, crystal field splitting does not occur and hence the substance is colourless. Due to the absence of ligand, the water molecules, anhydrous CuSO₄ is colourless. CuSO₄ . 5H₂O is blue in colour due to the water molecules and the splitting of d-orbitals.

Question 59.
FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1:1 molar ratio gives the test for Fe²⁺ ion but CuSO₄ mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu²⁺ ion. Why?
Answer:
FeSO₄ with (NH₄)₂SO₄ in 1 : 1 molar ratio forms a double salt FeSO₄. (NH₄)₂ SO₄ . 6H₂O which gives a test for all its constituent ions. On the other hand, CuSO₄ with aq. ammonia in 1 :4 molar ratio forms a coordination compound [Cu (NH₃)₄] SO₄. Hence it does not give test for Cu⁺⁺ ion.

Question 60.
How many geometrical isomers are poss-ible in the following coordination entities ?
i) [Cr(C₂O₄)₃]^3-
ii) [Co(NH₃)₃Cl₃]
Answer:
i) No geometric isomers are possible.
ii)

Another geometrical isomer occurs in [Ma₃b₃] type like [Co (NH₃)₃ Cl₃]. If three donor atoms of the same ligands occupy adjacent positions, at the corners of octa-hedral face we have the facial isomer (fac). When the positions occupied are around the meridian of the octahedron, we get the meridonial (mer) isomer.

Question 61.
What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution copper sulphate ? Why?
Answer:
When aq KCN is added to the aq. solution of CuSO₄, Octahedral complex of hexa cyan- idocuprate (10 is formed ,
CuSO₄ + 6KCN → K₄[CU(CN)₆] + K₂SO₄

Question 62.
[Cr(NH₃)₆]³⁺ is paramagnetic while [Ni(CN)₄]^2- is diamagnetic. Why ?
Answer:
In [Cr(NH₃)₆]³⁺, NH₃ is a weak field ligand and Cr has an electronic configuration of [Ar] 3d³ with three unpaired electrons. Hence [Cr(NH₃)₆]³ is paramagnetic. In [Ni(CN)₄]^2-, CN^– is a strong field ligand therefore Ni²⁺ ion has all the d-electrons paired. Hence, [Ni(CN)₄]^2- is diamagnetic.

Question 63.
A solution of [Ni(H₂O)₆]²⁺ is green but a solution of [Ni(CN)₂]^2- is colourless. Why ?
Answer:
In [Ni(H₂O)₆]²⁺, Ni²⁺ ion has two unpaired electron and it shows colour.
In [Ni(CN)₄]^2-, there is no unpaired electron. Hence it is colourless.
A solution of [Ni (H₂O)₆]²⁺ is green but a solution of [Ni (CN)₄]^2- colourless.

[Ni(H₂O)₆]²⁺ is an outer orbital complex.
As it contains two unpaired electrons, it is paramagnetic and coloured.
In the formation of [Ni(CN)₄]^2-, electron pairing takes place in Ni²⁺.

Ni²⁺ undergoes dsp² hybridisation and it is square planar. As it does not contain unpaired electron, it is colourless.

Question 64.
[Fe(CN)₄]^2- and [Fe(H₂O)₆]²⁺ are of different colours in dilute solutions. Why ?
Answer:

1. Different ligands produce different crystal field splittings.
2. Different geometric fields also produce different crystal field splittings.
   As d – d transitions require different energies, the colours exhibited by [Fe(CN)₄]^2- and [Fe(H₂O)₆]²⁺ are different.

Question 65.
What is the oxdiation state of cobalt in [AP 16]
i) K₃[CO(CO₄)₃] and
ii) [Co(NH₃)₆]³⁺
iii) K[Co(CO)₄] ?
Answer:
i) Ox. number of of Co in K₃ [Co(C₂O₄)₃] is + 3.
ii) Ox. no. of Co in [Co (NH₃)₆]³⁺ is + 3.
iii) Ox. on of cobalt in K[Co(CO)₄] is -1.

Short Answer Questions (4 Marks)

Question 66.
Compared to 3d series the corresponding transition metals of 4d and 5d transition series show high enthalpy of atomization. Explain.
Answer:
The melting points and boiling points of the transition elements are generally very high.

In the second and third row elements M – M bonds are much more common. The Metal – Metal (M – M) bonding occurs not only in the metals themselves but also in some compounds. M – M bonding is quite rare in the first row transition elements.

Hence because of increasing in atomic weight and also strong metallic bonding, the enthalpies of 4d and 5d series are higher than those of 3d series.

Question 67.
Compared to the changes in atomic and ionic sizes of elements of 3d and 4d series, the change in radii of elements of 4d and 5d series is virtually the same. Comment.
Answer:
When we compare the changes in ionic and atomic sizes of 3d series with those of 4d and 5d series, there is an increase from the 3d to 4d series of elements. But the radii of the third (5d) series are virtually the same as those of the corresponding members of 4d series (second series).

This is because of the filling of 4f before the 5d series of elements begin. The filling of 4f before 5d orbital results in a regular decrease in atomic radii called lanthanoid contraction. Because of lanthanide contraction, the expected increase in atomic size with increasing atomid number.

Question 68.
Account for the zero oxidation state of Ni and Fe in [Ni (CO)₄] and [Fe(CO)₅] respectively.
Answer:
Low oxidation are found when a complex compound has ligands capable of π-accep- tor character in addition to σ bonding. For example, Ni(CO)₄ and Fe(CO)₅, the oxidation of nickel and iron is zero. The metal- carbon-bond in metal carboyls possesses both σ and π character.

Question 69.
Why do the transition metal ions exhibit characteristic colours in aqueous solution. Explain giving examples.
Answer:
When an electron from a lower energy d- orbital of a metal ion in a complex is excited to a higher energy d-orbital of the same n value, the energy of excitation corresponds to the frequency of light absorbed. This frequency generally lies in the visible region. The colour absorbed corresponds to the complementary colour of the light absorbed. The frequency of light absorbed is determined by the nature of the ligand. In aqueous solutions, where water molecules are ligands
Ti³⁺ …………… purple
V⁴⁺ …………… blue
Cr³⁺ ……………. violet
Fe³⁺ ……………. yellow
Fe²⁺ ……………. green
To exhibit colour the ion should have at least unpaired (n -1) d electron. The colour is due to d – d transitions.

Question 70.
Explain the catalytic action of Iron (III) in the reaction between I^– and S₂O₈^2- ions.
Answer:
Transition metals and their compounds are known for their catalytic activity. This activity is ascribed to their ability to adopt multiple oxidation states to form complexes. For example, iron (III) catalyses the reaction between iodide and persulphate ions.
2I^– + S₂O₈^2- → I₂ + 2SO₄^2-
An explanation of this catalytic action can be given as :
2Fe³⁺ + 2I^– → 2Fe²⁺ + I₂
2Fe²⁺ + S₂O₈^2- → 2Fe²⁺ + 2SO₄^2-

Question 71.
What are interstitial compounds ? How are they formed ? Give two examples. [AP 15]
Answer:
Interstitial compounds are those which are formed when small atoms like H, C, or N are trapped inside the crystal lattices of metals. They are usually non-stoichiometric and are neither typically ionic nor covalent.
Example: TiC, Mn₄N, VH_0.56 and Ti H_1.7 etc.

Question 72.
Write the characteristics of interstitial compounds.
Answer:

1. Interstitial compounds are non-stoichiometric.
2. They are neither typically ionic nor covalent.
3. They have high melting points, higher than those of pure metals.
4. They are very hard.
5. They retain metallic conductivity.
6. They are chemically inert.

Question 73.
Write the characteristics properties of transition elements. [AP ’15]
Answer:
The transition elements are characterised by specific and special properties which are explained on the basis of their characteristic electronic configurations. They are

1. They exhibit variable oxidation states.
2. Most of the transition elements and their ions show paramagnetic property. This is due to presence unpaired d-electron.
3. They form coloured compounds.
4. They have alloy forming ability.
5. Complex forming ability.

Question 74.
Write down the electronic configuration of
(i) Cr³⁺
(ii) Cu⁺
(iii) Co²⁺
(iv) Mn²⁺
Answer:
i) Electronic configuration of Cr³⁺ = [Ar] 3d³
ii) Electronic configuration of Cu⁺ = [Ar] 3d¹⁰
iii) Electronic configuration of Co²⁺ = [Ar] 3d⁷ 4s³
iv) Electronic configuration of Mn²⁺ =[Ar] 3d⁵

Question 75.
What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: 3d³ 3d⁵ 3d⁸ and 3d⁴ ?
Answer:
Electronic configuration ………… Stable state
3d³ 4s² ………….. +2, and +5
3d⁵ 4s² ………….. +2, +7
3d⁸ 4s² …………. +2, +5
3d⁴ 4s² ………….. +3, +6

Question 76.
What is lanthanoid contraction? What are the consequences of lanthanoid contraction ?
Answer:
There is gradual decrease in the atomic and ionic sizes of the lanthanide elements. The filling of 4f before 5d orbital results in a regular decrease in atomic radii called lanthanoid contraction. This is due to the imperfect shielding of one electron by another in the same set of orbitals. However, the shielding of one 4f electron by another is less than that of one d electron by another, and as the nuclear charge increases along the series, there is fairly regular decrease in the size of the entire 4fⁿ orbitals.

Consequences:

1. The net result of the lanthanoid contraction is that the second and the third series exhibit similar radii (e.g.: Zr 160 pm Hf 159 pm) and have very similar physical and chemical properties much more than the expected on the basis of usual family relationship.
2. The decrease in metallic radius coupled with increase in atomic mass results in a general increase in the density of these elements.

Question 77.
How is the variability in oxidation states of transition metals different from that of the non transition metals ? Illustrate with examples.
Answer:
The variability of oxdiation states, a characteristic property of transition elements arises due to incomplete filling of d orbitals in such a way that their oxdiation states differ from each other by unity.
Ex: V^II, V^III, V^IV, V^V.

But in non-transition elements the oxidation states normally differ by a unit of two.
Although in the p-block elements the lower oxidation states are favoured by the heavier members (due to inert pair effect.)