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NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions.

Question 1.
Assign oxidation numbers to the underlined elements in each of the following species ;
(a) NaH₂PO₄
(b) NaHSO₄
(c) H₄P₂O₇
(d) K₂MnO₄
(e) CaO₂
(f) NaBH₄
(g) H₂S₂O₇
(h) KAl(SO₄)_2.12H₂O
Answer:

Question 2.
What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results ?|
(a) KI₃ (b) H₂S₄O_e (c) Fe₃O₄ (d) CH₃CH₂OH (e) CH₃COOH         (H P. Board 2015)
Answer:
(a) O.N. of iodine in KI₃ : By conventional method, the oxidation number of iodine in the I₃^_ ion may be expressed as :

Explanation : The oxidation number of iodine (I) comes out to be fractional which does not seem to be possible. Let us consider the structure of I₃~ ion. In it, two atoms of iodine are linked to each other by covalent bond (I—I). The iodide ion (I~) is linked to the molecule by co-ordinate bond [I—I <—I]“. The molecule may be represented as K⁺[I—I <—1| . Now, in the anion, the oxidation number of the two I atoms is zero while I^– ion has – 1 oxidation number. The overall oxidation number of L“ ion is :

(b) O.N. of S in H₂S₄O₆ : By conventional method, the oxidation number of sulphur may be calculated as :
H⁺¹₂  S^x₄ O^-2₆
2 + 4x + 6 (- 2) =0 or 4x = 12 – 2 = 10 or x = 5/2
Explanation : In order to account for the fractional value, let us assign oxidation numbers to different sulphur atoms in the structural formula of the acid. The oxidation numbers of the two middle sulphur atoms is zero while the two atoms at the terminal positions have + 5 oxidation number.

Average O.N. of S = —1/4 [5+ 0 + 0 +5] = 5/2

(c) O. N. of Fe in Fe₃O₄ : By conventional method, oxidation number of Fe may be calculated as :
Fe₃^x ₃ 0₃₄
3x + 4(- 2) =0 or x = 8/3
Explanation : Fe₃O₄ is a mixed oxide and is an equimolar mixture of Fe⁺² O^-2 and Fe⁺³₂ O^-2₃
Average O.N. of Fe = 1/3(2 + 2 x 3) =8/3

(d) O.N. of C in CH₃CH₂OH : By conventional method, the oxidation number of carbon in ethanol molecule may be calculated as :
CH₃cH₂OH or C^x₂ H⁺¹₆ O^-2
2x + 6(+ 1) + (- 2) = 0 or 2x + 6 – 2 = 0 or 2x = – 4 or x = – 2
Explanation : Let us calculate the oxidation number of both C₁ and C₂ atoms.
C₂ : The carbon atom is attached to three H atoms (less electronegative than carbon) and one CH₂OH group (more electronegative than carbon).
∴ O.N. of C₂ = 3(+ 1) +x + (- 1) = O or x = -2
C1 : The carbon atom is attached to one OH group (O.N. = – 1), two H atoms (O.N. = + 1) and one CH₃ group (O.N. = + 1)
O.N. of C1 = ( + 1) + x + 2 (+ 1) + 1(- 1) =0 or x = – 2 Average O.N. of C = l/2[ – 2 + (- 2)] = – 2

(e) O.N. of C in CH₃COOH : By conventional method the O.N. of carbon may be calculated as:
CH3COOH or c₂ H₄ 62
2x + 4(+ 1) + 2(— 2) = 0 or x — 0
Explanation : Let us calculate the oxidation number of both C₂ and C1 atoms.
C1 : The carbon atom is attached to three H atoms (less electronegative than carbon) and one COOH group (more electronegative than carbon).
∴ O.N. of C₂ = 3 (+ 1) + x + 1(- 1) = 0 or x = – 2
C₂ : The carbon atom is attached to one OH group (O.N. = – 1) one oxygen atom by double bond
(O.N. = – 2) and one CH₃ group (O.N. = + 1)
O.N. of C1 = 1( + 1) + Jt + 1 (-2) + 1(- 1) = O or x = + 2
∴ Average O.N. of C = 1/2 [+ 2 + (- 2)] = 0

Question 3.
How will you justify that the following reactions are redox reactions in nature ?
(a)  CuO(s) + H₂(g)  → Cu(s) + H₂O(g)
(b) Fe₂O₃(s) + 3CO(g) →2Fe(s) + 3CO₂(g)
(c) 2K(s) + F₂(g)→2K⁺F^–(s)
(d) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)
Answer:
A chemical reaction may be regarded as redox reaction if one of the reacting species undergoes increase in O.N. (oxidation) and the other decrease in O.N. (reduction). Based upon this, let us try to justify that the reactions under study are redox reactions in nature.

Question 4.
Fluorine reacts with ice as follows :
H₂O(s) + F₂(g) — HF(g) + HOF(g)
Justify that this reaction is a redox reaction.
Answer:

In this reaction, the F₂ molecule has undergone increase in O.N. in changing to HOF and decrease in O.N. in change to HF. It is therefore, a redox reaction. Since the same reacting species has undergone increase as well as decrease in O.N., it is also called disproportionation reaction.

Question 5.
Calculate the O.N. of sulphur, chromium and nitrogen in H₂SO₅, CrO₅ and NO^–₃ ion. Suggest structure of these compounds. Account for the fallacy if any.
Answer:
(1) Oxidation number of S in H₂SO₅
By conventional method, O.N. of sulphur may be calculated as :
2 x 1 + x + 5 X (-2)= 0 or 2 + x – 10 = 0 or x = + 8 (wrong)
But this cannot be true as maximum oxidation number for sulphur cannot exceed +6 as it has only six valence electrons. The oxidation number of the element can be calculated from the structure of the persulphuric acid.
The sulphur atom is linked one OH group (O.N. = – 1) two oxygen atoms (O.N. = – 2) and one peroxide (- O – O -) linkage (O.N. = – 1)
O.N. of S = 1 (- 1) + x + 2(- 2) + 1( – 1) = 0
    or 
x = + 6

(2) Oxidation number of Cr in CrO₅
By conventional method, O.N. of chromium may be calculated as :
x + 5 (-2) = 0 or x = + 10 (wrong)

This is wrong because maximum oxidation number of Cr cannot be more than +6, since it has only six electrons (3d⁵4.s¹) to take part in bond formation. The oxidation number of the metal atom can be calculated by taking into account its structure.
The Cr atom is linked with one oxygen atom by double bond (O. N. = – 2) and four oxygen atoms by peroxide linkages (O.N. = -1)
.’. O.N. of Cr = 4(- 1) \x + 1(- 2) = 0 or x = + 6

(3) Oxidation number N in NO^–₃ ion
By conventional method, O.N. of N atom in the ion may be calculated as : x + 3 (- 2) = – 1 or x = +5.This can be further made clear from the structure of the ion. In it, the nitrogen atom is linked to one single bonded oxygen atom (O.N. = – 2), one double bonded oxygen atom
(O.N.= – 2) and one oxygen atom by dative bond (O.N. = – 2).
O.N. of N = 1 (- 2) + x + 1( – 2) + 1 (- 2) = – 1 or x = + 5

Question 6.
Write the formulas of the following compounds :
(a) Mercury (II) chloride (b) Nickel (II) sulphate (c) Tin (IV) oxide (d) Thallium (I) sulphate (e) Iron (III) sulphate (f) Chromium (III) oxide.
Answer:
The formulas have been shown by stock notations. The Roman numerals given in parenthesis represent the oxidation state of the metal atom. Taking into account the oxidation state of the anions, the chemical formulas of the compounds may be written as :
(a) HgCl₂ (b) NiSO₄ (c) SnO₂ (d) T1₂SO₄ (e) Fe₂(SO₄)₃ (f) Cr₂O₃

Question 7.
Suggest a list of substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.
Answer:
Variable oxidation states of carbon :

Question 8.
While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid can act only as oxidising agents. Why ?
Answer:
In sulphur dioxide (SO₂) and hydrogen peroxide (H₂O₂), the oxidation states of sulphur and oxygen are +4 and -1 respectively. Since they can increase as well as decrease when these compounds take part in chemical reactions, they can act as oxidising as well as reducing agents. For example


In ozone (O₃), the oxidation state of oxygen is zero while in nitric acid (HNO₃), the oxidation state is nitrogen is +5. Since both them can undergo decrease in oxidation state and not an increase in its value, they can act only as oxidising agents and not as reducing agents.

Question 9.
Consider the reactions :
(a) 6CO ₂(g) + 6H₂O (l) → C₆H₁₂O₆(s) + 6O₂(g)
(b) O₃(g) + H₂O₂(l) → H₂O (l) + 2O₂(g)
Why it is more appropriate to write these reactions as :
(a) 6CO₂(g) + 12H₂O(l)→ C₆H₁₂O₆(s) + 6H₂O(l) + 6O₂(g)
(b) 6₃(g) + H₂O₂(l)→ H₂O(l) + O₂(g) + O₂(g)
Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.
Answer:
(a)  6CO₂(g) + 6H₂O(l) → C₆H₆O₆(s) + 6O₂(g)
Let us try to balance the equation by ion-electron method.


Note:
(1) In this reaction O₃ acts as an O.A. and H₂O₂ acts as a R.A.
(2) If a co-ordinate bond exist between two similar atoms, the donor atom acquires + 2 O.N. and the acceptor gets -2 O.N.
This explains the path of the reaction i.e., how electrons are lost and gained as well as and why it is more appropriate to write the equation as above.

Question 10.
The compound AgF₂ is an unstable compound. However if formed, the compound acts as a very strong oxidising agent. Why ?
Answer:
In AgFL the oxidation state of silver is + 2 i.e., it exists as Ag²⁺ ion. This is quite unstable since the normal stable oxidation state of the metal is + 1 or it exists as Ag⁺ ion. This means that if the compound AgF₂ be formed, it will undergo reduction by the gain of electrons. It is therefore, a very strong oxidising agent
Ag²⁺ + e^– — Ag⁺

Question 11.
Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations
Answer:

Question 12.
(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene, we use alcoholic potassium permanganate as an oxidant, why ? Write a balanced redox equation for the reaction.

(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get a colourless pungent smelling gas HC1. However, if mixture contains bromide, then red vapours of bromine are evolved. Why ?
Answer:

In the manufacture of benzoic acid from toluene, neutral medium is preferred over acidic or alkaline medium due to following reasons.

•  In the neutral medium, neither acid or base is to be added externally. This will definitely reduce the cost of the manufacturing process.
• Alcohol if used as solvent will help in the formation of a homogeneous mixture between toluene (non-polar) and KMnO₄ (ionic). Actually alcohol has non-polar alkyl group as well as polar OH group.

(b)
A chloride such as NaCl reacts with concentrated sulphuric acid to evolve hydrogen chloride gas upon heating. It has a pungent smell.

Hydrogen bromide is also expected to be formed in the same way when a bromide is heated with concentrated sulphuric acid. However, the acid being a strong oxidising agent will oxidise HBr to bromine (Br₂) which gets evolved  as red vapours.

Please note HC1 (g) is quite stable and is not oxidised to chlorine with concentrated sulphuric acid.

Question 13.
Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions :

1. 2AgBr(s) + C₆H₆O₂(aq) →2Ag(s) + 2HBr(aq) + C₆H₄O₂(aq)
2. HCHO(l) + 2[Ag(NH₃)₂]⁺(aq) + 3OH^–(aq) → 2Ag(s) + HCOO^– (aq) + 4NH₃(aq) + 2H₂O(l)
3. HCHO(l) + 2Cu²⁺(aq) + 5OH^–(aq)→ Cu₂(O(s) + HCOO-(aq) + 3H₂O(l)
4. N₂H₄(l) + 2H₂O₂(l) →N₂(g) + 4H₂O(l)
5. Pb(s) + PbO₂(s) + 2H₂SO₄(aq)→ 2PbSO₄(s) + 2H₂O(l)

Answer:


AgBr         : reduced ; acts as oxidising agent
C₆H₆O₂    : oxidised ; acts as reducing agent.

Question 14.
Consider the reactions :
2S₂O^2-₃(aq) + I₂(s) → S₄O^2-₆(aq) + 2I^–(aq)
S₂O² (aq) + 2Br₂(l) + 5H₂O (l) → 2SO^2-₄(aq) + 4Br-(ag) + 10H +(aq)
Why does the same reductant the osulphate react differently with iodine and bromine ?
Answer:
I₂ oxidises thiosulphate ion to tetrathionate ion i.e., from O.S. of + 2 for S (in S₂O^2-₃ ) to S.O. of + 5/2 for S (in S₄O^2-₄ ion).
Br₂ oxidises thiosulphate ion to sulphate ion i.e., from O.S. of + 2 for S (in S₂O^2-₃) to O.S. of + 6 for S
(in SO^2-₄ ion)
This is because Br₂ is a stronger

Question 15.
Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.
Answer:
The halogens (X₂) have strong electron accepting tendency. They are therefore, powerful oxidising agents. The relative order of oxidising powers of halogens is :

Fluorine is the strongest oxidising agent (oxidant). It can be justified by the fact that it can liberate the other halogens from their respective compounds. For example,
2KC1 + F_{2  →}  2KF + Cl₂
2KBr + F₂ → 2KF + Br₂
2KI + F₂ → 2KF + I₂
Among halogen acids (HX), the HI is the strongest reducing agent or reductant because its bond dissociation enthalpy is the minimum (299 kJ mol^-1)

(kJ mol^-1)
The iodination of methane for examples, is of reversible nature because HI formed in the reaction being a very strong reducing agent converts iodomethane back to methane

However, the halogenation with other halogen acids is not of reversible nature. This justifies that HI is the strongest reducing agent in the halogen acids.

Question 16.
Why does the following reaction occur ?
XeO^4-₆ (aq) + 2F^–(aq) + 6H⁺(aq)→ XeO₃(g) + 3H₂O(l)
What conclusion about the compound Na₄XeO₆ (of which XeO^4-₆ is a part can be drawn from the reaction ?
Answer:

In the reaction, O.N. of Xe decreases from + 8 to + 6. Therefore, XeO₆ is reduced.O.N. of F increases from – 1 to 0. Therefore, F^– is oxidised. The redox reaction occurs because Na₂XeO₆ (of which XeOg“ is a part) is a stronger oxidising agent than F₂.

Question 17.
Consider the reactions :
(a) H₃PO₂ (aq) + 4AgNO₃ (aq) + 2H₂O (l)→H₃PO₄ (aq) + 4Ag (s) + 4HNO₃ (aq)
(b) H₃PO₂ (aq) + 2CuSQ₄ (aq) + 2H₂O (l)→ H₃PO₄ (aq) + 2Cu (s) + 2H₂SO₄ (aq)
(c) C₆H₅CHO (I) + 2[Ag(NH₃)₂⁺ (aq) + 3OH^– (aq)→ C₆H₅COO (aq) + 2Ag (s) + 4NH₃ (aq) + 2H₂O (l)
(d) C₆H₅CHO (l) + 2Cu²⁺(aq) + SOH^–(aq)→No change observed
What inference do you draw about the behaviour of Ag⁺ and Cu²⁺ ions from these reactions ?
Answer:
(a) Hypophosphorus acid (H₃PO₂) reduces Ag⁺ ions to Ag which gets precipitated
(b) Hypophosphorus acid (H₃PO₂) reduces Cu²⁺ ions to Cu which gets precipitated
(C) Benzaldehyde (C₆H₅CHO) reduces Ag+ions to Ag which gets precipitated
(d) Benzaldehyde (C₆H₅CHO) is not in a position to reduce Cu²⁺ ions to Cu.
This shows that Ag⁺ ions are stronger oxidising agent than Cu²⁺ ions in all the reactions. This is quite evident from their relative positions in the reactivity series.

Question 18.
Balance the following equations by ion-electron method

Answer:

Question 19.
Balance the following equations in basic medium by ion-electron method and oxidation number method and identify the oxidising agent and the reducing agent.
(a) p₄(s) + OH⁺(aq) —> PH₃(g) + H₂PO^–₂ (aq)
(b) N₂H₄(7) + CIO^–₃ (aq) —> NOfe) + C l^–(g)
(c) C1₂O₇(g) + H₂O₂(aq) —> CIO^–₂(aq) + O₂(g) + H^–
Answer:

Question 20.
Write four informations about the reaction :
(CN)₂ (g) + 2OH^– (aq)—> CN^– (aq) + CNO (aq) + H₂O (l).
Answer:

1. The reaction involves the decomposition of cyanogen (CN)₂ in the alkaline medium.
2. Both (CN)₂ i.e. cyanogen and CN^– i.e. cyanide ions are pseudo halogens in nature i.e. they behave like halogens in characteristics.
3. Cyanogen undergoes disproportionation in the reaction. It undergoes simultaneous increase and decrease in O.N.of the species involved.
4. It is an example of redox reaction.

Question 21.
The Mn³⁺ ion is unstable in solution and under goes disproportionation to give Mn²⁺, MnO₂ and H⁺ ion. Write balanced ionic equation for the reaction.
Answer:

Question 22.
Consider the elements : Cs, Ne, I, F

1. Identify the element that exhibits only -ve oxidation state.
2. Identify the element that exhibits only +ve oxidation state.
3. Identify the element that exhibits +ve and -ve oxidation states.
4. Identify the element that neither exhibits + ve and -ve oxidation states.

Answer:

1. F (fluorine) exhibits only -ve oxidation state (- 1) in its compounds because it is the most electronegative element.
2. Cs (cesium) exhibits only +ve oxidation state (+ 1) its compounds because it is the most electropositive element.
3. Iodine has seven electrons in the valance shell as well as vacant 5d orbital to which electrons from 5p and 5s orbitals can be shifted. Therefore, the elements exhibits – 1 oxidation state as well as variable positive oxidation states of + 1, +3, +5 and +7 in its compounds.
4. Ne (neon) neither exhibits +ve nor -ve oxidation states because it is a noble gas element with completely filled orbitals (Is² 2s²2p⁶). Moreover, it has no vacant d-orbitals.

Question 23.
Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place.
(P.I.S.A. Based)
Answer:
Chlorine reacts with sulphur dioxide in the presence of water as follows
Cl₂ (aq) + SO₂ (aq) + H₂O (l) → Cl^– (aq) + SO₄^2- (aq)

Question 24.
From the periodic table, select three non-metals and three metals which can show disproportionation reactions.
Answer:
Non-metals and metals which show variable oxidation states can take part in the disproportionation reactions. For examples,
Non-metals : The non-metals showing disproportionation reactions states are : P₄, Cl₂ and S₈.

Question 25.
In the Ostwald process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.0 g of ammonia and reacting with 20.0 g of oxygen ?
Answer:



Question 26.
Using the standard electrode potentials given in the table, predict if the reaction between the following is feasible.
(a) Fe³⁺(aq) and I^–(aq)
(b) Ag⁺(aq) and Cu(s)
(c) Fe³⁺(aq) and Cu(s)
(d) Ag(s) and Fc³⁺(aq)
(f) Br₂(aq) and Fe²⁺(aq).
Answer:
A particular reaction can be feasible if e.m.f. of the cell based on the E° values is positive. Keeping this in mind, let us predict the feasibility of the reactions.

Question 27.
Predict the products of electrolysis of each of the following :

1. An aqueous solution of AgNO₃ using silver electrodes.
2. An aqueous solution of AgNO₃ using platinum electrodes.
3. A dilute solution of H₂SO₄ using platinum electrodes.
4. An aqueous solution of CuCl₂ using platinum electrodes.

Answer:
(1) An aqueous solution of AgNO₃ using platinum electrodes :     (P.I.S.A. Based)
Both AgNO₃ and water will ionise in aqueous solution

At cathode : Ag⁺ ions with less discharge potential are reduced in preference to H⁺ ions which will remain in solution. As a result, silver will be deposited at cathode.
Ag⁺ (aq) + e^– → Ag (deposited)
At anode : An equivalent amount of silver will be oxidised to Ag⁺ ions by releasing electrons.

As result of electrolysis, Ag from silver anode dissolves as Ag⁺(aq) ions while an equivalent amount of Ag⁺(aq) ions from the aqueous AgNO₃ solution get deposited on the cathode.

(2) An aqueous solution of AgNO₃ using platinum electrodes :

In the case, the platinum electrodes are the non-attackable electrodes. On passing current the following changes will occur at the electrodes.
At cathode : Ag⁺ ions will be reduced to Ag which will get deposited at the cathode.
At anode : Both NO^–₃ and OH^– ions will migrate. But OH^– ions with less discharge potential will be oxidised in preference to NO₃ ions which will remain in solution.
OH- (aq) → OH + e^–; 4OH → 2H₂O(l) + O₂ (g)
Thus, as a result of electrolysis, silver is deposited on the cathode while O₂ is evolved at the anode. The solution will be acidic due to the presence of HNO₃.

(3) A dilute solution of H₂SO₄ using platinum electrodes :
On passing current, both acid and water will ionise as follows :

At cathode :
H⁺ (aq) ions will migrate to the cathode and will be reduced to H₂.
H+ (aq) + e-→ H ; H + H→ H₂ (g)
Thus, H₂ (g) will evolve at cathode.
At anode : OH ions will be released in preference to SO^2-₄ ions because their discharge potential is less. They will be oxidised as follows :
OH^– (aq) → OH + e^– ; 4OH → 2H₂O(l) + O₂ (g)
Thus, O₂ (g) will be evolved at anode. The solution will be acidic and will contain H₂SO₄.

(4) An aqueous solution of CuCl₂ using platinum electrodes :

The electrolysis proceeds in the same manner as discussed in the case of AgNO₃ solution. Both CuCl₂ and H₂O will ionise as follows :

At cathode :  Cu²⁺ ions will be reduced in preference to H⁺ ions and copper will be deposited at cathode.
Cu²⁺ (aq) + 2e^– → Cu (deposited)
At anode : Cl^– ions will be discharged in preference to OH^– ions which will remain in solution.
Cl^–→Cl^–+ e^– ; Cl + Cl → Cl₂ (g) (evolved)
Thus, Cl₂ will evolve at anode.

Question 28.
Arrange the following metals in the order in which they displace each other from their salts. Al, Cu, Fe, Mg and Zn
Answer:
This is based upon the relative positions of these metals in the activity series. The metal placed lower in the series can displace the metals occupying a higher position present as its salt. Based upon this, the correct order is : Mg, Al, Zn, Fe, Cu.

Question 29.
Given the standard electrode potentials
K+/K = – 2.93 V, Ag+/Ag = 0.80V
Hg²+/Hg = 0.79 V ; Mg²⁺/Mg = – 2.37 V, Cr³⁺/Cr = – 0.74 Y
Arrange these metals in increasing order of their reducing power.
Answer:
If may be noted that lesser the E° value for an electrode, more will be reducing power. The increasing order of reducing power is :
Ag < Hg < Cr < Mg < K.

Question 30.
Depict the galvanic cell in which the reaction Zn(s) + 2Ag ⁺(aq)→ Zn²⁺(aq) + 2Ag(s) takes place. Further show :
(1) which electrode is negatively charged.
(2) the carriers of the current in the cell.
(3) individual reaction at each electrode.
Answer:
The galvanic cell for the reaction is

• Zinc electrode (anode) is negatively charged.
• Current flows from silver to zinc in outer circuit
• Anode : Zn (s) → Zn²⁺(aq) + 2e^–(oxidation)
  Cathode :   2Ag⁺ (ag) + 2e^– → 2Ag(s) (reduction)

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NCERT Solutions for Class 11 Chemistry Chapter 7 Equilibrium

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 7 Equilibrium.

Question 1.
A liquid is in equilibrium with its vapours in a sealed container at a fixed temperature. The volume of the container is suddenly increased,
(i) What is the initial effect of the change on the vapour pressure ?
(ii) How do the rates of evaporation and condensation change initially ?
(iii) What happens when equilibrium is restored finally and what will be the final vapour pressure ?
Answer:
(i) On increasing the volume of the container, the vapour pressure will initially decrease because the same amount of vapours are now distributed over a larger space.
(ii) On increasing the volume of the container, the rate of evaporation will increase initially because now more space is available. Since the amount of the vapours per unit volume decrease on increasing the volume, therefore, the rate of condensation will decrease initially.
(iii) Finally, equilibrium will be restored when the rates of the forward and backward processes become equal. However, the vapour pressure will remain unchanged because it depends upon the temperature and not upon the volume of the container.

Question 2.
What is Kc for the following reaction in a state of equilibrium :
2SO₂(g) + O₂(g) \(\rightleftharpoons \) 2SO₃(g) ?
Given : [SO₂] = 0.6 M ; [O₂]= 0.82 M ; and [SO₃] = 1.90 M
Answer:

Question 3.
At certain temperature and total pressure of 105Pa, iodine vapours contain 40t% by volume of iodine atoms in the equilibrium I₂ (g) \(\rightleftharpoons \) 2I (g). Calculate K_p for the equilibrium.
Answer:
According to available data :
Total pressure of equilibrium mixture = 10⁵ Pa
Partial pressure of iodine atoms (I) = \(\frac { 40 }{ 100 } \) × (10⁵ Pa) = 0.4 × 10⁵ Pa
Partial pressure iodine molecules (I₂) = \(\frac { 60 }{ 100 } \) × (10⁵ Pa) = 0.6 × 10⁵ Pa

Question 4.
Write the expression for the equilibrium constant for the following reactions
(i) 2NOCl(g)) \(\rightleftharpoons \) 2NO(g) + Cl₂(g)
(ii) 2CU(N0₃)₂(s) \(\rightleftharpoons \) 2CuO(s) + 4NO₂(g) + O₂(g)
(iii) CH₃COOC₂H₅(aq) + H₂O(l) \(\rightleftharpoons \) CH₃COOH(aq) + C₂H₅OH(aq)
(iv) Fe³⁺(aq) + 3OH^–(aq) \(\rightleftharpoons \) Fe(OH)₃(s)
(v) I₂(s) + 5F₂(g) \(\rightleftharpoons \) 2IF₅(l).
Answer:

Question 5.
Find the value of Kc for each of the following equilibria from the value of K_p:
(a) 2NOCl (g) \(\rightleftharpoons \) 2NO (g) + Cl₂ (g) ; K_p = 1.8 x 10^-2 atm at 500 K
(b) CaCO₃ (s) \(\rightleftharpoons \) CaO (s) + CO₂ (g) ; K_p = 167 atm at 1073 K.
Answer:
K_p and K_c are related to each other as K_p = K_c ( RT)^∆ng
The value of K_c can be calculated as follows :
(a) 2NOCl (g) \(\rightleftharpoons \) 2NO (g) + Cl₂ (g)
K_p = 1.8 × 10^-2 atm,
∆^ng = 3 – 2 = 1 ; R = 0.0821 litre atm K^-1 mol^-1 ; T = 500 K

(b) CaCO₃ (s) \(\rightleftharpoons \) CaO (s) + CO₂ (g)
K_p = 167 atm, ∆^ng = 1 – 0 = 1
R = 0.0821 liter atm K^-1 mol^-1 ; T = 1073 K

Question 6.
For the following equilibrium, K_c = 6.3 × 10¹⁴ at 1000 K
NO (g) + O₃ (g) \(\rightleftharpoons \) NO₂(g) + O₂ (g)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc for the reverse reaction ?
Answer:
For the reverse reaction K_c = \(\frac { 1 }{ { K }_{ c } } \) = \(\frac { 1 }{ 6.3\times { 10 }^{ 14 } } \) = 1.59 × 10^-15

Question 7.
Explain why pure liquids and solids can be ignored while writing the value of the equilibrium constant.
Answer:
In writing the expression for the equilibrium constant, molar concentrations of the species involved in the reaction are taken into account.

We all know that the molar concentration of a substance implies number of moles per unit volume (moles/volume) or (mass/volume).

Since mass divided by volume represents density of the substance therefore, molar concentration of a substance is directly related to its density.
Molar concentration = \(\frac { No.ofmoles }{ Volume } \) α \(\frac { Mass }{ Volume } \) (α density)
Density as we know, is an intensive property and does not depend upon the mass of the substance.

Consequently, molar concentration of a pure substance (solid or liquid) has always the same value and it can be ignored while writing the value of the equilibrium constant.

However, for substances in gaseous state or in aqueous solution, the amount in a given volume can vary and their molar concentration does not remain constant and cannot be ignored while writing the expression for the equilibrium constant.

Question 8.
Reaction between nitrogen and oxygen takes place as follows :
2N₂ (g) + O₂ (g) \(\rightleftharpoons \) 2N₂O (g)
If a mixture of 0.482 mol of N₂ and 0.933 mol of O₂ is placed in a reaction vessel of volume 10 L and allowed to form N₂0 at a temperature for which K_c = 2.0 × 10^-37, determine the composition of the equilibrium mixture.
Answer:
Let x moles of N₂ (g) take part in the reaction. According to the equation, \(\frac { x }{ 2 } \) moles of O₂ (g) will react to form x moles of N₂O (g). The molar concentration per litre of different species before the reaction and at the equilibrium point is :

The value of equilibrium constant (2.0 × 10^-37) is extremely small. This means that only small amounts of reactants have reacted. Therefore, x is extremely small and can be omitted as far as the reactants are concerned.

As x is extremely small, it can be neglected.
Thus, in the equilibrium mixture
Molar cone. of N₂ = 0.0482 mol L^-1
Molar cone. of O₂ = 0.0933 mol L^-1
Molar cone. of N₂O = 0.1 × x = 0.1 × 6.6 10^-20 mol L^-1
= 6.6 × 10^-21

Question 9.
Nitric oxide reacts with bromine and gives nitrosyl bromide as per reaction given below :
2NO (g) + Br₂ (g) \(\rightleftharpoons \) 2NOBr (g)
When 0.087 mole of NO and 0.0437 mole of Br₂ are mixed in a closed container at constant temperature, 0.0518 mole of NOBr is obtained at equilibrium. Determine the composition of the equilibrium mixture.
Answer:
The balanced chemical equation for the reaction is :
2NO (g) + Br₂ (g) \(\rightleftharpoons \) 2NOBr (g)
According to the equation, 2 moles of NO (g) react with 1 mole of Br₂ (g) to form 2 moles of NOBr (g). The composition of the equilibrium mixture can be calculated as follows :
No. of moles of NOBr (g) formed at equilibrium = 0.0518 mol (given)
No. of moles of NO (g) taking part in reaction = 0.0518 mol
No. of moles of NO (g) left at equilibrium = 0.087 – 0.0518 = 0.0352 mol
No. of moles of Br₂ (g) taking part in reaction = \(\frac { 1 }{ 2 } \) × 0.0518 = 0.0259 mol
No. moles of Br₂ (g) left at equilibrium = 0.0437 – 0.0259 = 0.0178 mol
The initial molar concentration and equilibrium molar concentration of different species may be represented as :

Question 10.
At 450 K, K_p = 2.0 × 10¹⁰ bar^-1 for the equilibrium reaction :
2SO₂ (g) = O₂ (g) \(\rightleftharpoons \) 2SO₃(g)
Answer:

Question 11.
A sample of HI (g) is placed in a flask at a pressure of 0.2 atm. At equilibrium partial pressure of HI (g) is 0.04 atm. What is K_p for the given equilibrium ?
2HI (g) \(\rightleftharpoons \) H₂(g) + I₂(g)
Answer:
pHI = 0.04 atm, pH₂ = 0.08 atm ; pI₂ = 0.08 atm
K_p = \(\frac { p{ H }_{ 2 }\times p{ I }_{ 2 } }{ { p }_{ HI }^{ 2 } } \) = \(\frac { (0.08atm)\times (0.08atm) }{ 0.04atm)\times (0.04atm) } \) = 4.0

Question 12.
A mixture of 1.57 mol of N₂, 1.92 mol of H₂ and 8.13 mol of NH₃ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant K_c for the reaction given as follows :
N₂(g) + 3H₂(g) \(\rightleftharpoons \) 2NH₃(g) is 1.7 × 10^-2.
Is this reaction at equilibrium ? If not, what is the direction of net reaction ?
Answer:
The reaction is : N₂(g) + 3H₂(g) \(\rightleftharpoons \) 2NH₃(g)

The equilibrium constant (Kc) for the reaction = 1.7 X 10^-2
As Q_c, ≠ K_c ; this means that the reaction is not in a state of equilibrium.
As Q_c > K_c ; the direction of the net reaction must be in the backward direction so that the molar concentration of NH₃ may decrease and equilibrium may be attained again.

Question 13.
The equilibrium constant expression for a gas reaction is,

Write the balanced chemical equation corresponding to this expression.
Answer:
Balanced chemical equation for the reaction is
4 NO(g) + 6 H₂O (g) \(\rightleftharpoons \) 4NH₃ (g) + 5 O₂(g)

Question 14.
If 1 mole of H₂O and 1 mole of CO are taken in a 10 litre vessel and heated to 725 K, at equilibrium point 40 percent of water (by mass) reacts with carbon monoxide according to equation!
H₂O(g) + CO(g) \(\rightleftharpoons \) H₂(g) + CO₂(g)
Calculate the equilibrium constant for the reaction.
Answer:
Number of moles of water originally present = 1 mol
Percentage of water reacted =40%
Number of moles of water reacted = \(\frac { 1\times 40 }{ 100 } \) = 0.4 mol
Number of moles of water left = (1 – 0.4) = 0.6 mol
According to the equation, 0.4 mole of water will react with 0.4 mole of carbon monoxide to form 0.4 mole of hydrogen and 0.4 mole of carbon dioxide.
Thus, the molar cone./litre per litre of the reactants and products before the reaction and at the equilibrium point are as follows :

Question 15.
At 700 K, the equilibrium constant for the reaction H₂(g) + I₂(g) \(\rightleftharpoons \) 2HI(g) is 54.8. If 0.5 mol L^-1 of HI (g) is present at equilibrium at 700 K, what are the concentration of H₂(g) and I₂(g) ? Assume that we initially started with HI(g) and allowed it to reach equilibrium at 700 K.
Answer:

Question 16.
What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M ?
2ICl(g) \(\rightleftharpoons \) I₂(g) + Cl₂(g) ; K_c = 0.14
Answer:
Suppose at equilibrium, the molar concentration of both I₂(g) and Cl₂(g) is x mol L^-1

Question 17.
K_p = 0.04 atm at 898 K for the equilibrium shown below. What is the equilibrium concentration of C₂H₆ when it is placed in a flask at 4 atm pressure and allowed to come to equilibrium.
C₂H₆(g) \(\rightleftharpoons \) C₂H₄(g) + H₂(g)
Answer:

Question 18.
The ester, ethyl acetate is formed by the reaction of ethanol and acetic acid and the equilibrium is represented as :
CH₃COOH(l) + C₂H₅OH(l) \(\rightleftharpoons \) CH₃COOC₂H₅(l) + H₂O(l)
(i) Write the concentration ratio (concentration quotient) Q for this reaction. Note that water is not in excess and is not a solvent in this reaction.
(ii) At 293 K, if one starts with 1.000 mol of acetic acid and 0.180 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii) Starting with 0.50 mol of ethanol and 1.000 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate
is found after some time. Has equilibrium been reached ?
Answer:

Question 19.
A sample of pure PCl₅ was introduced into an evacuated vessel at 473 K. After equilibrium was reached, the concentration of PCl₅ was found to be 0.5 x 10^-1 mol L^-1. If K_c is 8.3 × 10^-3, what are the concentrations of PCl₃ and Cl₂ at the equilibrium ?
Answer:
Let the initial molar concentration of PCl5 per litre = x mol
Molar concentration of PCl₅ at equilibrium = 0.05 mol
∴ Moles of PCl₅5 decomposed = (x – 0.05) mol
Moles of PCl₃ formed = (x – 0.05) mol
Moles of Cl₂ formed = (x – 0.05) mol
The molar cone. /litre of reactants and products before the reaction and at the equilibrium point are:

Question 20.
One of the reactions that takes place in producing steel from iron ore is the reduction of iron (II) oxide by
carbon monoxide to give iron metal and CO₂
FeO(s) + CO(g) \(\rightleftharpoons \) Fe(s) + CO₂(g); K_p = 0.265 atm at 1050 K
What are the equilibrium partial pressures of CO and CO₂ at 1050 K if the initial pressures are :
Pco = 1.4 atm and Pco₂ = 0.80 atm ?
Answer:

Question 21.
Equilibrium constant Kc for the reaction, N₂ (g) + 3H₂ (g) \(\rightleftharpoons \) 2NH₃ (g) at 500 K is 0.061. At particular time, the analysis shows that the composition of the reaction mixture is : 3.0 mol L^-1 of N₂ ; 2.0 mol L^-1 of H₂ ; 0.50 mol L^-1 of NH₃. Is the reaction at equilibrium ? If not, in which direction does the reaction tend to proceed to reach the equilibrium ?
Answer:

Question 22.
Bromine monochloride (BrCl) decomposes into bromine and chlorine and reaches the equilibrium :
2BrCl (g) \(\rightleftharpoons \) Br₂ (g) + Cl₂ (g)
The value of K_c is 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3× 10^-3 mol L^-1, what is its molar concentration in the mixture at equilibrium ?
Answer:

Question 23.
At 1127 K and 1 atmosphere pressure, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon has 90.55% by mass.
C(s) + CO₂(g) \(\rightleftharpoons \) 2CO(g)
Calculate K_c for the reaction at the above temperature.
Answer:

Question 24.

Answer:

Question 25.
Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume ?
(i) PCl₅(g) \(\rightleftharpoons \) PCl₃(g) + Cl₂
(ii) CaO(s) + CO₂(g) \(\rightleftharpoons \) CaCO₃(s)
3Fe(s) + 4H₂O(g) \(\rightleftharpoons \) Fe₃O₄(s) + 4H₂(g)
Answer:
(i) Pressure will increase in the forward reaction and number of moles of products will increase.
(ii) Pressure will increase in backward reaction and number of moles of products will decrease.
(iii) The change in pressure will have no effect on the equilibrium constant and there will be no change in the number of moles.

Question 26.
Which of the following reactions will get affected by increase in pressure ? Also mention whether the change will cause the reaction to go to the right or left direction.

Answer:
Only those reactions will be affected by increasing the pressure in which the number of moles of the gaseous reactants and products are different (np -/ nr) (gaseous). With the exception of the reaction (1) ; all the remaining five reactions will get affected by increasing the pressure. In general,
• The reaction will gq to the left if n_p > n_r.
• The reaction will go to the right if n_p < n_r.
Keeping this in mind,
(i) Increase in pressure will not affect equilibrium because n_p = n_r =3
(ii) Increase in pressure will favour backward reaction because n_p (2) > n_r (1)
(iii) Increase in pressure will favour backward reaction because n_p (10) > n_r (9)
(iv) Increase in pressure will favour forward reaction because n_p(1) < n_r (2)
(v) Increase in pressure will favour backward reaction because n_p (2) > n_r ( 1)
(vi) Increase in pressure will favour backward reaction because n_p (1) > n_r (0).

Question 27.
The equilibrium constant for the following reaction is 1.6 x 10⁵ at 1024 K.
H₂(g) + Br₂(g) \(\rightleftharpoons \) 2HBr(g)
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.
Answer:

Question 28.
Hydrogen gas is obtained from the natural gas by partial oxidation with steam as per following endothermic reaction :
CH₄ (g) + H₂O (g) \(\rightleftharpoons \) CO (g) + 3H₂ (g)
Write the expression for K_p for the above reaction
How will the value of K_p and composition of equilibrium mixture be affected by :
(i) increasing the pressure
(ii) increasing the temperature
(iii) using a catalyst ?
Answer:
The expression for K_p for the reaction is :

(i) By increasing the pressure, the no. of moles per unit volume will increase. In order to decrease the same, the equilibrium gets shifted to the left or in the backward direction. As a result, more of reactants will be formed and the value of K_p will decrease.
(ii) If the temperature is increased, according to Le Chatelier’s principle, the forward reaction will be favoured as it is endothermic. Therefore, the equilibrium gets shifted to the right and the value of K_p will increase.
(iii) The addition of catalyst will not change the equilibrium since it influences both the forward and the backward reactions to the same extent. But it will be attained more quickly.

Question 29.
What is the effect of :
(i) addition of H₂
(ii) addition of CH₃OH
(iii) removal of CO
(iv) removal of CH₃OH
On the equilibrium 2H₂ (g) + CO (g) \(\rightleftharpoons \) CH₃OH ?
Answer:
(i) Equilibrium will be shifted in the forward direction.
(ii) Equilibrium will be shifted in the backward direction.
(iii) Equilibrium will be shifted in the backward direction.
(iv) Equilibrium will be shifted in the forward direction.

Question 30.
At 473 K, the equilibrium constant Kc for the decomposition of phosphorus pentachloride (PCl₅) is 8.3 × 10^-3. If decomposition proceeds as :
PCl₅ (g) \(\rightleftharpoons \) PCl₃ (g) + Cl₂ (g) ; ∆H = + 124.0 kJ mol^-1
(a) Write an expression for Kc for the reaction.
(b) What is the value of Kc for the reverse reaction at the same temperature ?
(c) What would be the effect on Kc if
Answer:

Question 31.
Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H₂. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction.
CO(g) + H₂O(g) \(\rightleftharpoons \) CO₂(g) + H₂(g)
If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam so that pco = PH₂O = 4.0 bar, what will be the partial pressure of H₂ at equilibrium ? K_p = 0.1 at 400°C.
Answer:

Question 32.
Predict which of the following will have appreciable concentration of reactants and products :
(a) Cl₂ (g) \(\rightleftharpoons \) 2Cl (g) ; K_c = 5 ×10^-39
(b) Cl₂ (g) + 2NO (g) \(\rightleftharpoons \) 2NOCl (g) ; K_c = 3.7 × 10⁸
(c) Cl₂ (g) + 2NO₂ (g) \(\rightleftharpoons \) 2NO₂Cl (g) ; K_c = 1.8.
Answer:
Following conclusions can be drawn from the values of Kc.
(a) Since the value of K_c is very small, this means that the molar concentration of the products is very small as compared to that of the reactants.
(b) Since the value of K_c is quite large, this means that the molar concentration of the products is very large as compared to that of the reactants.
(c) Since the value of K_c is 1.8, this means that both the products and reactants have appreciable concentration.

Question 33.
The value of K_c for the reaction 3O₂(g) \(\rightleftharpoons \) 2O₃(g) is 2.0 × 1(10^-50 at 25°C. If equilibrium concentration of O₂ in air at 25°C is 1.6 × 10^-2, what is the concentration of O₂ ?
Answer:

Question 34.
The reaction CO(g) + 3H₂(g) \(\rightleftharpoons \) CH₄(g) + H₂O(g)
is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H₂ and 0.02 mol of H₂O and an unknown amount of CH₄ in the flask. Determine the concentration of CH₄ in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90.
Answer:

Question .35.
What is meant by conjugate acid-base pair ? Find the conjugate acid/base for the following species :
HNO₂,CN^–,HClO₄,OH^–,CO₃^2-,S^2-.
Answer:
An acid-base pair which differs by a proton only (HA \(\rightleftharpoons \) N A^– + H⁺ ) is known as conjugate acid-base pair. Conjugate acid : HCN, H₂O, HCO₃^–, HS^–.
Conjugate base : NO₂^–, ClO₄^–, O₂ .

Question 36.
Which of the following are Lewis Acids ?
H₂O, BF₃, H⁺ and NH₄⁺
Answer:
BF₃, H⁺ , NH₄⁺ ions are Lewis acids.

Question 37.
What will be the conjugate bases for the Bronsted acids ? HF, H₂SO₄ and H₂CO₃ ?
Answer:
Conjugate bases : F^–, HSO₄^– , HCO₃^–

Question 38.
Write the conjugate acids for the following Bronsted acids.
NH₂^–, NH₃ and HCOO^–
Answer:
NH₃, NH₄⁺ and H₂CO₃.

Question 39.
The species H₂O, HCO₃^–, HSO₄^– and NH₃ can act both as Bronsted acid and base. For each case, give the corresponding conjugate acid and base.
Answer:

Question 40.
Classify the following species into Lewis acids and Lewis bases and show how these can act as Lewis acid/Lewis base ?
(a) OH^– ions
(b) F^–
(c) H⁺
(d) BCl₃
Answer:
(a) OH^– ions can donate an electron pair and act as Lewis base.
(b) F^– ions can donate and electron pair and act as Lewis base.
(c) H⁺ ions can accept an electron pair and act as Lewis acid.
(d) BCl₃ can accept an electron pair since Boron atom is electron deficient. It is a Lewis acid.

Question 41.
The concentration of hydrogen ions in a sample of soft drink is 3.8 × 10^-3 M. What is its pH value ?
Answer:
pH = – log [H⁺] = – log (3.8 × 10^-3)
= (log 3 – log 3.8) = 3 – 0.5798 = 2.4202

Question 42.
The pH of soft drink is 3.76. Calculate the concentration of hydrogen ions in it.
Answer:
pH = – log [H⁺] or log [H⁺] = -pH = – 3.76 = \(\bar { 4 } \).24
[H⁺] = Antilog (\(\bar { 4 } \).24) = 1.74 x 10^-4M.

Question 43.
The ionisation constants of HF, HCOOH and HCN at 298 K are 6.8 x 10^-4, 1-8 × 10^-4 and 4.8 × 10^-9 respectively. Calculate the ionisation constant of the corresponding congugate bases.
Answer:

Question 44.
The ionisation constant of phenol is 1.0 x 10^-10. What is the concentration of phenate ion in 0.05 M solution of phenol and pH of solution ? What will be the degree of ionisation if the solution is also 0.01 M in sodium phenate ?
Answer:

Question 45.
The first dissociation constant H2S is 9.1 × 10^-8. Calculate the concentration of HS^– ions in its 0.1 M solution and how much will this concentration be affected if the solution is 0.1 M in HCl also. If the second dissociation constant of H₂S is 1.2 × 10^-13, calculate the concentration of S^2- ions under both the conditions.
Answer:

Question 46.
The ionization constant of acetic acid is 1.74 × 10^-5. Calculate the degree of dissociation of acetic acid in 0.05 M solution. Calculate the concentration of acetate ions in solution and the pH of the solution.
Answer:

Question 47.
It has been found that the pH of 0.01 M solution of organic acid is 4.15. Calculate the concentration of the anion, ionization constant of the acid and the pK_a.
Answer:

Question 48.
Assuming complete dissociation, calculate the pH of the following solutions :
(a) 0.003 M HCl
(b) 0.005 M NaOH
(c) 0.002 M HBr
(d) 0.002 M KOH.
Answer:

Question 49.
Calculate the pH of the following solutions :
(a) 2 g of TlOH dissolved in water to give 2 litre of solution.
(b) 0.3 g of Ca(OH)₂ dissolved in water to give 500 mL of solution.
(c) 0.3 g of NaOH dissolved in water to give 200 mL of solution.
(d) 1 mL of 13.6 M HCl diluted with water to give 1 litre of solution.
Answer:

Question 50.
The degree of ionization of 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and pKa of bromoacetic acid.
Answer:

 

Question 51.
The pH of 0-005 M codeine (C₁₈H₂₁NO₃) solution is 9.95. Calculate its ionizationconstant and pK_b.
Answer:

Question 52.
What is the pH of 0.001 M aniline solution ? The ionization constant of aniline is 4.27 × 10^-10. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
Answer:

Question 53.
Calculate the degree of ionization of 0.05 M acetic acid if its pK_a value is 4.74. How is degree of dissociation affected when the solution contains (a) 0.01 M HCl (b) 0.1 M HCl ?
Answer:

Question 54.
The ionization constant of dimethylamine is 5.4 × 10^-4. Calculate the degree of ionization of its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.01 M NaOH ?
Answer:

Question 55.
Calculate the hydrogen ion concentration in the following biological fluids whose pH values are as follows :
(a) Human muscles fluid 6.83
(b) Human stomach fluid 1.2.
(b) Human blood 7.38
(d) Human saliva 6.4
Answer:
(a) [H⁺] of human muscles fluid
pH = 6.83 or – log [H⁺] = 6.83 or log [H⁺] = – 6.83 = \(\bar { 7 } \).17
[H⁺] = Antilog [H⁺] = Antilog [\(\bar { 7 } \).17] = 1.479 ×10^-7 M

(b) [H⁺ ] of human stomach fluid
pH = 1.2 or – log [H⁺] = 1.2 or log [H⁺] = – 1.2
[H⁺] = Antilog (-1-2) = Antilog (2-8) = 6.309 × 10^-2 M

(c) [H⁺ ] of human blood
pH = 7.38 or – log [H⁺] = 7.38 or log [H⁺] = – 7.38
[H⁺] = Antilog (-7.38) = Antilog (\(\bar { 8 } \).62) = 4.168 × 10^-8 M

(d) [H⁺] of human saliva
pH = 6.4 or – log [H⁺] = 6.4 or log [H⁺] = – 6.4
[H⁺] = Antilog (-6.4) = Antilog (\(\bar { 7 } \).6) = 3.98 × 10^-7 M.

Question 56.
The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each case.
Answer:
(a) [H⁺] of milk
pH = 6.8 or – log [H⁺] = 6.8 or log [H⁺] = – 6.8
[H⁺] = Antilog (-6.80) = Antilog (7.20) = 1.58 × 10^-7 M

(b) [H⁺] of black coffee
pH = 5.0 or – log [H⁺] = 5.0 or log [H⁺] = – 5.0
[H⁺] = Antilog (-5.0) = Antilog (\(\bar { 5 } \)) = 10^-5 M

(c) [H⁺] of tomato juice
pH = 4.2 or – log [H⁺] = 4.2 or log [H⁺] = – 4.2
[H⁺] = Antilog (-4.2) = Antilog (\(\bar { 5 } \).80) = 6.31 × 10^-5 M

(d) [H⁺] of lemon juice
pH = 2.2 or – log [H⁺]= 2.2 or log [H⁺] = – 2.2
[H⁺] = Antilog (-2.2) = Antilog (\(\bar { 3 } \).80) = 6.310 × 10^-3 M

(e) [H⁺] of egg white
pH = 7.8 or – log [H⁺] = 7.8 or log [H+] = – 7.8
[H⁺] = Antilog (-7.8) = Antilog (\(\bar { 8 } \).20) = 1.58 × 10^-8 M

Question 57.
0.561 g KOH is dissolved in water to give 200 mL of solution at 298K. Calculate the concentration of potassium, hydrogen and hydroxyl ions. What is the pH of the solution ?
Answer:

Question 58.
The solubility of Sr(OH)₂ at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution. (Atomic mass of Sr = 87.6)
Answer:

Question 59.
The ionisation constant of propionic acid is 1.32 × 10^-5. Calculate the degree of ionisation of acid in its 0.05 M solution and also its pH. What will be its degree of ionisation if the solution is 0.01 M in HCl also ?
Answer:

Question 60.
The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate ionisation constant of the acid and also its degree of dissociation in the solution.
Answer:

Question 61.
The ionisation constant of nitrous acid is 4.5 × 10^-4. Calculate the pH value of 0.04 M NaNO₂ solution and also its degree of hydrolysis.
Answer:

Question 62.
A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionisation constant of pyridine.
Answer:

Question 63.
Predict if the solutions of the following salts are neutral, acidic or basic :
NaCl, KBr, NaCN, NH₄NO₃, NaNO₂, and KF
Answer:
Neutral :NaCl, KBr
Basic : NaCN, NaNO₂, KF
Acidic :NH₄NO₃

Question 64.
The ionisation constant of chloroacetic acid is 1.35 × 10^-3. What will be the pH of 0.1 M acid solution and of its 0.1 M sodium salt solution ?
Answer:

 

Question 65.
The ionic product of water at 310 K is 2.7 × 10^-14. What is the pH value of neutral water at this temperature ?
Answer:
Since water is neutral ; [H₃O⁺] = [OH^–]
We know that, [H₃O⁺] × [OH^–] = K_w = 2.7 × 10^-14
∴ [H₃O⁺]² = 2.7 × 10^-14
or [H₃O⁺] = (2.7 × 10^-14)^1/2 = 1.643 × 10^-7
pH of water = – log [H₃O⁺] = – log (1.643 × 10^-7)
= (-) (-7) + log (1.643) = (7 – log 1.643) = 7 – 0.2156 = 6.78.

Question 66.
Calculate the pH of the resultant mixtures
(a) 10 mL. of 0.2 M Ca(OH)₂ + 25 mL of 0.1 M HCl
(b) 10 mL of 0.01 M H₂SO₄ + 10 mL of 0.01 M Ca(OH)₂
(c) 10 mL of 0.1 M H₂SO₄ + 10 mL of 0.1 M KOH.
Answer:

Question 67.
Determine the solubilites of silver chromate, barium chromate and ferric hydroxide at 289 K from their solubility product constants. Determine also the molarities of the individual ions.
Given : K_spAg₂CrO₄) = 1.1 × 10^-12
K_sp(BaCrO₄) = 1.2 × 10^-10
K_sp[Fe(OH)₃] = 1.0 × 10^-38
Answer:

Question 68.
The solubility product constants of Ag₂CrO₄ and AgBr are 1.1 × 10^-12 and 5.0 × 10^-13 respectively. Calculate the ratio of the molarities of their saturated solutions.
Answer:

Question 69.
Equal volumes of 0.002 M solutions of sodium iodate and copper chlorate are mixed together. Will it lead to the precipitation of copper iodate ? (For copper iodate K_sp = 7.4 × 10^-8).
Answer:

Question 70.
The ionization constant of benzoic acid is 6.46 × 10^-5 and K_sp for silver benzoate is 2.5 × 10^-13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water ?
Answer:

Question 71.
What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide ? (For iron sulphide, K_sp = 6.3 × 10^-18).
Answer:

Question 72.
What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K. For calcium sulphate K_sp = 9.1 × 10^-6?
Answer:

Question 73.
The concentration of sulphide ion in 0.1 M HCl solution saturated with hydrogen sulphide is 1.0 × 10^-19 M. If 10 mL of this solution is added to 5 mL of 0.04 M solution of FeSO₄, MnCl₂, ZnCl₂ and CaCl₂, in which solutions precipitation will take place ? Given K_sp for FeS = 6.3 × 10^-18, MnS = 2.5 × 10^-13, ZnS = 1.6 × 10^-24 and CdS = 8.0 × 10^-27.
Answer:

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NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics.

Question 1.
Choose the correct answer.
A thermodynamic state function is a quantity
(i) used to determine heat changes
(ii) whose value is independent of path
(iii) used to determine pressure volume work
(iv) whose value depends on temperature only.
Answer:
(ii) whose value is independent of path

Question 2.
For the process to occur under adiabatic conditions, the correct condition is :
(i) ∆T = 0
(ii) ∆p = 0
(iii) q = 0
(iv) w = 0
Answer:
(iii) q = 0

Question 3.
The enthalpies of all elements in their standard states are :
(i) unity
(ii) zero
(iii) < 0
(iv) different for each element
Answer:
(ii) zero

Question 4.

Answer:

Question 5.
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are – 890.3 kJ mol^-1, -393.5 kJ mol^-1 and – 285.8 kJ mol^-1 respectively. Enthalpy of formation of CH₄ (g) will be
(i) -74.8 kJ mol^-1 (ii) -52.27 kJ mol^-1
(iii) + 74.8 kJ mol^-1 (iv) + 52.26 kJ mol^-1
Answer:

Question 6.
A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be
(i) possible at high temperature
(ii) possible only at low temperature
(iii) not possible at any temperature
(iv) possible at any temperature
Answer:
(iv) possible at any temperature

Question 7.
In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process ?
Answer:
Heat absorbed by the system, q = 701 J
Work done by the system = – 304 J
Change in internal energy (∆U) = q + w = 701 – 394 = 307 J.

Question 8.
The reaction of cyanamide, NH₂CN(s) with oxygen was affected in a bomb calorimeter and ∆U was found to be – 742.7 kJ mol^-1 of cyanamide at 298 K. Calculate the enthalpy change for the reaction at 298 K.
NH₂CN(y) + 3/2O₂(g) → N₂fe) + CO₂(g) + H₂O(l)
Answer:
∆U = – 742.7 kJ mol^-1 ; ∆^ng = 2 – \(\frac { 3 }{ 2 } \) = + \(\frac { 1 }{ 2 } \) mol.
R = 8.314 × 10^-3 kJ K^-1 mol^-1 ; T = 298 K
According to the relation, ∆H = ∆U + ∆^ng RT
AH = (- 742.7 kJ) + (\(\frac { 1 }{ 2 } \) mol) × (8.314 × 10^-3 kJ K^-1 mol^-1) × (298 K)
= -742.7 kJ + 1.239 kJ = -741.5 kJ.

Question 9.
Calculate the number of kJ necessary to raise the temperature of 60 g of aluminium from 35 to 55°C. Molar heat capacity of Al is 24 J mol^-1 K^-1.
Answer:
No. of moles of Al (m) = \(\frac { 60g }{ 27gmo{ l }^{ -1 } } \) = 2.22 mol
Molar heat capacity (C) = 24 J mol^-1 K^-1
Rise in temperature (∆T) = 55 – 35 = 20°C = 20 K
Heat evolved (q) = C × m × T = (24 J mol^-1 K^-1) × (2.22 mol) × (20 K)
= 1065.6 J = 1.067 kJ.

Question 10.
Calculate the enthalpy change on freezing of 1.0 mole of water at – 10.0°C to ice at – 10.0°C. ∆_fusH = 6.03 kJ mol^-1 at 0°C; C_p[H₂O(l)] = 75.3 J mol^-1 K^-1 ; C_p[H₂O(s)l = 36.8 J mol^-1 K^-1.
Answer:
Total change in enthalpy (AH) for the freezing process may be calculated as :
∆H = (1 male of water at 10°C → 1 mole of water at 0°C) + (1 mole of water at °C → 1 mole of ice at 0°C)
+ (1 mole of ice at °C → 1 mole of ice at – 10°C)
= C_p[H₂O(l)] × ∆T + ∆H_freezmg + C_p [H₂O(s)] × ∆T.
= (75.3 Jk^-1 mol^-1) (0 – 10 K) + (-6.03 kJ mol^-1) + (36.8Jk^-1 mol^-1) × (-10 K)
= (-753 J mol^-1) – (6.03 kJ mol^-1) – (368 J mol^-1)
= (-0.753 kJ mol^-1) – (6.03 kJ mol^-1) – (0.368 kJ mol^-1)
= – 7.151 kJ mol^-1

Question 11.
Enthalpy of combustion of carbon to carbon dioxide is – 393.5 kJ mol^-1. Calculate the heat released upon formation of 35.2 g of CO₂ from carbon and oxygen gas.
Answer:
The combustion equation is :
C(s) + O₂(g) → CO₂(g) ; ∆_cH = – 393.5 kJ mol^-1 (44 g)
(44g)
Heat released in the formation of 44g of CO₂ = 393.5 kJ
Heat released in the formation of 35.2 g of CO₂ = \(\frac { (393.5kJ)\times (35.2g) }{ (44g) } \) = 314.8 kJ.

Question 12.
Calculate the enthalpy of the reaction :
N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)
Given that;
∆_fH CO(g) = – 110 kJ mol^-1 ;
∆_fH CO₂(g) = – 393 kJ mol^-1
∆_fH. N₂O(g) = 81 kJ mol^-1 ;
∆_fH N₂O₄(g) = – 9.7 kJ mol^-1.
Answer:
Enthalpy of reaction (∆_rH) = [81 + 3(- 393)] – [9.7 + 3(- 110)]
= [81 – 1179] – [9.7 – 330] = – 778 kJ mol^-1.

Question 13.

Answer:

Question 14.

Answer:

Question 15.
Calculate the enthalpy change for the process
CCl₄ (g) → C (g) + 4 Cl (g) and calculate bond enthalpy of C—Cl in CC14 (g)
Given : ∆_vap H° (CCl₄) = 30.5 kJ mol^-1 ; ∆_fH°(CCl₄) = – 135.5 kJ mol^-1
∆_aH° (C) = 715.0 kJ mol^-1 where ∆_a H° is enthalpy of atomisation
∆_aH° (Cl₂) = 242 kJ mol^-1.
Answer:

Question 16.
For an isolated system ∆U = 0 ; what will be ∆S ?
Answer:
Change in internal energy (∆U) for an isolated system is zero because it does not exchange any energy with the surroundings. But entropy tends to increase in case of spontaneous reaction. Therefore, ∆S > 0 or positive.

Question 17.
For a reaction at 298 K
2 A + B → C
∆H = 400 kJ mol^-1 and ∆S = 0.2 kJ K’1 mol^-1.
At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range ?
Answer:
According to Gibbs-Helmholtz equation :
∆G = ∆H – T∆S
For ∆G = 0; ∆H = T∆S or T : = \(\frac { \triangle H }{ \triangle s } \)

T = \(\frac { (400kJmo{ l }^{ -1 }) }{ (0.2kJ{ K }^{ -1 }mo{ l }^{ -1 }) } \)
Thus, reaction will be in a state of equilibrium at 2000 K and will be spontaneous above this temperature.

Question 18.
For the reaction ; 2Cl (g) → Cl₂(g) ; what will be the signs of ∆H and ∆S ?
Answer:
∆H : negative (-ve) because energy is released in bond formation
∆S : negative (-ve) because entropy decreases when atoms combine to form molecules.

Question 19.
For a reaction ; 2A (g) + B (g) → 2D(g)

Calculate ∆U₂₉₈ for the reaction and predict whether the reaction is spontaneous or not.
Answer:
Let the mass of H₂ in the mixture = 20 g
The mass of O₂ in the mixture will be = 80 g

Question 20.

Answer:

Question 21.

Answer:

Question 22.

Answer:

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NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter Gases and Liquids

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter Solids, Liquids and Gases.

Question 1.
What will be the minimum pressure required to compress 500 dm³ of air at 1 bar to 200 dm³ at 30° C ?
Answer:
From the given data :
P₁ = 1 bar P₂ = ?
V₁ = 500 dm³ V₂ = 200 dm³
P₁V₁ = V₁P₂ or P₂ = \(\frac { { P }_{ 1 }{ V }_{ 1 } }{ { V }_{ 2 } } \)

Question 2.
A vessel of 120 mL capacity contains a certain amount of gas at 35°C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35°C. What would be its pressure ?
Answer:
From the given data :
P₁ = 1.2 bar P₂ = ?
V₁ = 120 mL V₂ = 180 mL
Since temperature is constant, Boyle’s Law can be applied.
P₁V₁ = V₁P₂ or P₂ = \(\frac { { P }_{ 1 }{ V }_{ 1 } }{ { V }_{ 2 } } \) ; P₂ = \(\frac { (1.2bar)\times (120mL) }{ 180mL } \) = 0.8 bar

Question 3.
Using the equation of state PV = nRT show that at a given temperature, the density of the gas is proportional to the gas pressure p.
Answer:

Question 4.
At 0°C, the density of a gaseous oxide at 2 bar is the same as that of nitrogen at 5 bar. What is the molecular mass of the oxide ?
Answer:

Question 5.
Pressure of lg of an ideal gas A at 27°C is found to be 2 bar. When 2g of another ideal gas B is introduced in the same flask at same temperature, the pressure becomes 3 bar. Find the relationship between their molecular masses.
Answer:
Let MA and MB be the molar masses of the two gases A and B. According to available data :

Question 6.
The drain cleaner, ‘Drainex’ contains small bits of aluminium which react with caustic soda to produce hydrogen. What volume of hydrogen at 20°C and one bar will be released when 0.15 g of aluminium reacts ?
Answer:
Step I. Calculation of the volume of hydrogen released under N. T.P. conditions.
The chemical equation for the reaction is :
2Al + 2NaOH + 2H₂O → 2NaAl0₂ + 3H₂
2×27=54 g Sod. meta aluminate 3×22400 mL
54g of Al at N.T.P. release H₂ gas = 3 × 22400 mL
0.15g of Al at N.T.P. release H₂ gas = \(\frac { (3\times 22400mL)\quad \times \quad (0.15g) }{ (54g) } \) =186.7mL.
Step II. Calculation of volume of hydrogen released at 20°C and 1 bar pressure.

Question 7.
What will be the pressure exerted (in pascal) by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm³ flask at 27°C ?
Answer:

Question 8.
What will be the pressure of the gaseous mixture when 0.5 L of H₂ at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1 L vessel at 27°C ?
Answer:
Step I. Calculation of partial pressure of H2 in 1 L vessel.
V₁, = 0-5 L V₂ = 1.0 L
P₁ = 0.8 bar P₂ = ?
According to Boyle’s Law, P₁V₁ = P₂V₂
P₂ = \(\frac { { P }_{ 1 }{ V }_{ 1 } }{ { V }_{ 2 } } \) = \(\frac { (0.8bar) \times  (0.5L) }{ (1.0L) } \) = 0.4bar
Step II. Calculation of partial pressure of O₂ in 1 L vessel.
V₁ = 2.0 L V₂ = 1.0L
P₁ = 0.7 bar P₂ = ?
According to Boyle’s Law, P₁V₁ = P₂V₂
P₂ = \(\frac { { P }_{ 1 }{ V }_{ 1 } }{ { V }_{ 2 } } \) = \(\frac { (0.7bar) \times  (2.0L) }{ (1.0L) } \) =1.4 bar
Step III. Calculation of total pressure of gaseous mixture.
P = P₁ +P₂ = (0.4 + 1.4 ) bar = 1.8 bar

Question 9.
The density of a gas is found to be 5.46 g/dm³ at 27°C and under 2 bar pressure. What will be its density at STP ?
Answer:

Question 10.
34.05 mL of phosphorus vapours weigh 0-0625 g at 546°C and 1.0 bar pressure. What is the molar mass of phosphorus ?
Answer:
According to ideal gas equation,
PV = n RT ; PV = \(\frac { WRT }{ M } \) or M = \(\frac { WRT }{ PV } \)
According to available data :
Mass of phosphorus vapours (W) = 0.0625 g
Volume of vapours (V) = 34.05 mL = 34.05 × 10³ L
Pressure of vapours (P) = 1.0 bar.
Gas constant (R) = 0.083 bar L K^-1 mol^-1
Temperature (T) = 546 + 273 = 819 K

Question 11.
A student forgot to add the reaction mixture to the round bottomed flask at 27°C but put it on the flame. After a lapse of time, he realised his mistake. By using a pyrometer, he found that the temperature of the flask was 477°C. What fraction of air would have been expelled out ?
Answer:
Since the student was working in the laboratory, there is no change in pressure. Thus, Charles’ Law is applicable.
According to given data : V₁ = V L (say) V₂ = ?
T₁ = 27+273 =300 K ; T₂ = 477 + 273 = 750K
\(\frac { { V }_{ 1 } }{ { T }_{ 1 } } \) = \(\frac { { V }_{ 2 } }{ { T }_{ 2 } } \) or V₂ = \(\frac { { V }_{ 1 }{ T }_{ 2 } }{ { T }_{ 1 } } \) = \frac { (VL)\times (750K) }{ 300K } = 2.5 VL
Thus, volume of air expelled = 2.5 V – V = 1.5 V 1.5V
Fraction of air expelled = \(\frac { 1.5V }{ 2.5V } \) = \(\frac { 3 }{ 5 } \) .

Question 12.
Calculate the temperature of 4.0 moles of a gas occupying 5 dm³ at 3.32 bar (R = 0.083 bar dm³ K^-1 mol^-1).
Answer:
According to idea gas equation, PV = n RT or T = [/latex] = \(\frac { PV }{ nR } \)
According to available data :
No. of moles of the gas (n) = 4.0 moles
Volume of the gas (V) = 5 dm³
Pressure of the gas (P) = 3.32 bar
Gas constant (R) = 0.083 bar dm³ K^-1 mol^-1

Question 13.
Calculate the total number of electrons present in 1.4 g of nitrogen gas.
Answer:
Molecular mass of N₂ = 28g
28g of N₂ represent molecules = 6.022 × 10²³
1.4 g of N₂ represent molecules = 6.022 × 10²³ × \(\frac { 1.4g }{ 28g } \) = 3.011 × 10²²
Atomic number of nitrogen (N) = 7
1 molecule of N₂ has electrones = 7 × 2 = 14
3.011 × 10²² molecules of N₂ have electrons = 4 × 3.011 × 10²² = 4.215 × 10²³ electrons

Question 14.
How much time would it take to distribute one Avogadro number of wheat grains if 10¹⁰ grains are distributed each second ?
Answer:
Time taken to distribute 10¹⁰ grains = 1 s

Question 15.
Calculate the total pressure of a mixture of 8 g of oxygen and 4 g of hydrogen confined in a vessel of 1 dm3 at 27°C (R = 0-083 bar dm³K^-1mol^-1)
Answer:

Question 16.
Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m and mass 100 kg is filled with helium at 1.66 bar at 27°C. (Density of air = 1.2 kg m^-3 and R = 0.083 bar dm³K^-1mol^-1).
Answer:
Step I Calculation of the mass of displaced air
Radius of balloon (r) = 10 m
Volume of balloon (V) = \(\frac { 4 }{ 3 } \) πr³ = \(\frac { 4 }{ 3 } \) × \(\frac { 22 }{ 7 } \) × (10 m)³ = 4190.5 m³
Mass of the displaced air = Volume of air (balloon) × Density of air
= (4190.5 m³) × (1.2 kg m^-3) = 5028.6 kg
Step II Calculation of the mass of the filled balloon.

= 279.37 × 10³ mol.
Mass of He present = Moles of He × Molar mass of He
= (279.37 × 10³ mol) × (4 g mol^-1)
= 1117.48 × 10³g = 1117.48 kg
Mass of filled balloon = 100 + 1117.48 = 1217.48 kg
Step III. Calcu lation of the pay load
Pay load = Mass of displaced air – Mass of filled balloon
= 5028.6 – 1217.48 =3811.12 kg.

Question 17.
Calculate the volume occupied by 8-8 g of Co₂ at 31.1 °C and 1 bar pressure (R = 0.083 bar LK^-1mol^-1).
Answer:

Question 18.
2.9 g of a gas at 95°C occupied the same volume as 0.184 g of hydrogen at 17°C and at the same pressure. What is the molar mass of the gas ?
Answer:

Question 19.
A mixture of hydrogen and oxygen at one bar pressure contains 20% by weight of hydrogen. Calculate the partial pressure of hydrogen.
Answer:
Let the mass of H₂ in the mixture = 20 g
The mass of O₂ in the mixture will be = 80 g

Question 20.
What would be SI units of a quantity pV²T²/n ?
Answer:

Question 21.
In term of Charles’ Law explain why -273 °C is the lowest possible temperature ?
Answer:
The temperature -273°C (or 0 K) is known as absolute zero temperature. Below this temperature, a substance cannot exist as a gas and changes to the liquid state. This means that the Charles’ Law can be applied only upto a temperature – 273°C since a substance fails to exist as gas below this temperature.

Question 22.
Critical temperature for carbon dioxide and methane are 31.1°C and -81.9°C respectively. Which of these has stronger intermolecular forces and why ?
Answer:
The values of critical temperatures suggest that the attractive forces in the molecules of carbon dioxide are more. In fact, both are non-polar gases but the van der Waals forces of attraction in carbon dioxide molecules are more because of greater molecular size.

Question 23.
Explain the physical significance of van der Waals parameters.
Answer:
Significance and Units of the van der Waals’ constants
The significance and the units of the constant ‘a’ and ‘b’ used in the van der Waals equation are discussed as van der Waals’ constant ‘a’ The constant ‘a’ is related to the magnitude of the attractive forces among the molecules in a particular gas. Thus, greater the value of ‘a’ more will be the attractive forces. It may be noted that ‘a’ is higher for gases like NH₃, HCl and SO₂ which can be easily liquefied than for gases like He, H₂ and N₂ which are liquefied with diffieculty. Actually the former are of polar nature and the attractive forces in their molecules are higher than the latter gases which are of non-polar nature. The units of the constant ‘a’ can be calculated as follows :

Units of ‘a’ are related to the units in which pressure ‘p’ and volume ‘V’ are expressed.
• Pressure is in atmospheres and volume in litres, a = atm. L² mol^-2
• Pressure is in atmospheres and volume in m³, a = atm m⁶ mol^-2
• Pressure is in Nm”2 and volume in m3, a = Nm^-2 m⁶ mol^-2
Van der Waals’ constant ‘b’. The constant ‘b’ represents the co-volume or excluded volume which is effective volume of the molecules in a gas. It is, in fact, four times the volume occupied by the gas molecules. The units of ‘h’ can be calculated as follows :
Units of ‘b’ may be L mol^-1 or m³ mol^-1.

NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter Solids

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 5(A) States of Matter Solids.

Question 1.
Define the term ‘amorphous’. Give a few examples of amorphous solids.
Answer:
Amorphous or amorphous solids imply those solids in which the constituent particles have short range order. These have irregular shapes and are isotropic in nature. Apart from that they do not have sharp melting points. A few examples of amorphous solids are : glass, rubber, plastic, celluose etc.

Question 2.
What makes glass different from a solid such as quartz ? Under what conditions could quartz be converted into glass ?
Answer:
Glass is a super cooled liquid and an amorphous substance. Quartz is the crystalline form of silica (SiO₂) ‘n which tetrahedral units SiO₄ are linked with each other in such a way that the oxygen atom of one tetrahedron is shared with another Si atom. Quartz cap be converted into glass by melting it and cooling the melt very rapidly. In glass, SiO₄ tetrahedra are joined in a random manner.

Question 3.
Classify each of the following solids as ionic, metallic, molecular, net work (covalent) or amorphous:
(a) Tetra phosphorus decoxide (P₄O₁₀)
(b) Graphite
(c) Brass
(d) Ammonium phosphate (NH₄)₃PO₄
(e) SiC
(f) Rb
(g)I₂
(h) LiBr
(i) P₄
(j) Si (k) Plastic.
Answer:
(a) Molecular solid
(b) Covalent (Net-work) solid
(c) Metallic solid
(d) Ionic solid
(e) Covalent solid (Network)
(f) Metallic solid
(g) Molecular solid
(h) Ionic solid
(i) Molecular solid
(j) Covalent solid
(k) Amorphous solid

Question 4.
(a)What is meant by the term coordination number ?
(b)What is the co-ordination number of atoms
(i) in a cubic close packed structure
(ii) in a body centred cubic structure ?
Answer:
(a)The co-ordination number of a constituent particle (atom, ion or molecule) is the number of nearest neighbours in its contact in the crystal lattice. „
(b)(i) Twelve, (ii) Eight

Question 5.
How can you determine the atomic mass of an unknown metal if you know its density and dimensions of its unit cell ? Explain your answer.
Answer:

We have learnt that different types of the cubic unit cells differ in their packing fractions i.e. with respect to the the space occupied by the constituting particles. These are expected to differ in their relative densities also. Let us derive a general relation between the edge length and the density in a unit cell.
Consider a unit cell with edge length = a pm = a × 10^-10 cm.
(∵ Because density is normally expressed in g cm^-3)
Volume of the unit cell = (a × 10^-10 cm)³ = a³ × 10^-30 cm³
Density of unit cell =\(\frac { Mass\quad of\quad unit\quad cell }{ Volume\quad of\quad unit\quad cell } \) ….(i)
Now, mass of unit cell = No. of atoms in a unit cell × Mass of each atom
Z = \(\frac { Atomic\quad mass\quad (M) }{ Avogadro’s\quad number(No) } \) ….(ii)
By substituting the value of mass of unit cell in equation (i)

The value of Z is different for different types of cubic unit cells.
It may be noted that the density of any crystalline solid is the same as that of its unit cell.

• While deriving the above relationship, the edge length is expressed in picometre (pm) units. In case, it is expressed in any other units, please convert the same into pm. There is no need to mention its units in the calculations.

Question 6.
(a) Stability of a crystal is reflected in the magnitude of the melting point. Comment.
(b) Collect the melting point of (i) Ice (ii) ethyl alcohol (iii) diethyl ether (iv) methane from a data book. What can you say about intermolecular forces between the molecules ?
Answer:
(a) The stability of a crystal depends upon the magnitude of force of interaction in the constituting particles. Greater the force of attraction present, more will be the stability of the crystal. For example, ionic solids such as NaCl, KC1 etc. have very high melting and boiling points while the molecular crystals such as naphthalene, iodine etc. have low values of melting and boiling points.

(b) The melting points of different substances are :
(i) Ice = 273 K
(ii) Ethyl alcohol = 155.7 K
(iii) Diethyl ether = 156.8 K
(iv) Methane = 90.5 K
The intermolecular forces in molecules of ice and ethyl alcohol are mainly hydrogen bonding. The magnitude is more in ice than in alcohol as is evident from the values of melting points.
The forces in the molecules of diethyl ether are dipolar forces while in methane, these are mainly the van der Waal’s forces which is quite evident from its melting point (least among the compounds listed).

Question 7.
How will you distinguish between the following pairs of terms :
(a) Hexagonal close packing and cubic close packing
(b) Crystal lattice and unit cell
(c) Tetrahedral void and octahedral void.
Answer:
(a) In hexagonal close packing (hep), the spheres of the third layer are vertically above the spheres of the first layer
(ABABAB type). On the other hand, in cubic close packing (ccp), the spheres of the fourth layer are present
above the spheres of the first layer (ABCABC type).
(b) Crystal Lattices and Unit Ceils
We have studied that in the crystalline solids, the constituent particles (atoms, ions or molecules) have a regular orderly arrangement throughout and it gets repeated again and again resulting in a definite pattern. These points are generally shown as dots (.) and the sites which are occupied by these points are known as lattice sites or crystal lattices. Crystal lattices may be either two dimensional or three dimensional in nature.
Two dimensional crystal lattices : A two dimensional lattice is planar and is represented in terms of two basic vectors (a and b) and the angle (L) between them. There are five different types of two dimensional lattices :

Three dimensional crystal lattices : A three dimensional lattice is a further extension of the two dimensional lattice. It depicts the actual shape as well as size of the constituent particles in the crystal.
It is therefore, called space lattice or cyrstal lattice. It may be defined as :
the regular three dimensional arrangement of the points in a crystal.
Please note that
• Each point in the lattice is known as lattice point or lattice site.
• Each point in the lattice represents the constituting particle. These particles may be atoms, ions or molecules depending upon the nature of the crystalline solid.
• Lattice points are joined by straight lines in order to depict the geometry of the solid. A typical space lattice has been shown in the fig. 5.37.

Unit Cell
If we carefully examine the space lattice given in the figure, we can find a group of lattice points (shown as shaded) which get repeated in different directions to form the complete lattice. This smallest repeating pattern is called unit cell which may be defined as :
the smallest but complete unit in the space lattice which when repeated over and again in the three dimensions, generates the crystal of the given substance.
In order to illustrate a unit cell, let us consider a block made of bricks of the same size. Each brick represents the unit cell while the block is similar to the space or crystal lattice. Thus, a unit cell is the fundamental building block of the space lattice.
It may be noted that crystals differ from each other with respect to the shape and size of the unit cells. A unit cell is completely characterised in terms of edge lengths and the angles between the edges.

(i) Edges or edge lengths. The edges a, b and c represent the dimensions of the unit cell in space along the three axes. The edges may or may not be mutually perpendicular.

(ii) Angles between the edges. There are three angles between the edges. These are denoted as \(\alpha \) (between b and c), \(\beta \) (between a and c) and \(\Upsilon \) (between a and b). Thus, a unit cell may be characterised by six parameters as shown in the Fig. 5.38. The various types
of crystal systems differ with respect to edge lengths as well as angles between the edges.

(c) Tetrahedral and Octahedral Voids or Interstitial Sites
In the discussion about the close packing of spheres in space, we have seen that even in the closest packed structure when the spheres (or particles) touch each other, certain space in between the particles is left unoccupied. For example, in both hep and ccp, 26% of the space is left unoccupied. In fact, the spheres can touch each other only at certain points. The space which is left in between the closest packed arrangement is known as void, hole or interstice. Two types of voids or interstitial sites are formed in close packed structures. These are tetrahedral and octahedral voids.

Tetrahedral Voids.
We have seen that a triangular void is formed in hexagonal two dimensional close packing . A tetrahedral void is formed when the triangular void made by three spheres of a particular layer and touching each other has a contact with one sphere either in the upper layer or the layer below it. Thus, a tetrahedral void is formed when four spheres placed at the comers of a tetrahedron touch each other as shown in the Figure 5.54.

It is interesting to note that the actual shape of the void is triangular but the arrangement of the spheres around the void is tetrahedral.
In a close packed structure, each sphere is in contact with three spheres in the layer above it and also three spheres in the layer below it. Thus, there are two tetrahedral voids associated with each sphere (or constituting particle). The radius of a tetrahedral void in a closest packed structure is 22.5% of the sphere which is involved in the arrangement. Thus,

Octahedral Voids
An octahedral void or site is formed when three spheres arranged at the corners of an equilateral triangle are placed over another set of spheres pointing in the opposite directions as shown in the Figure 5.56.
Thus, octahedral void is the empty space created at the centre of six spheres which are. placed octahedrally.

It has been found that there is only one octahedral site for each sphere. The radius of an octahedral void in a closest packed structure is 0.414 of the sphere which is involved in the arrangement.

Question 8.
How many lattice points are there in one unit cell of each of the following lattices
(a) face centred cubic (b) face centred tetragonal (c) body centred cubic ?
Answer:
(a) In face centred cubic arrangement,
Lattice points located at the corners of the cube = 8
Lattice points located at the centre of each face = 6
Total no. of lattice points = 8 + 6 = 14

(b) In face centred tetragonal, the number of lattice points is also the same i.e., 8 + 6 = 14.
(In both the cases, .particles per unit cell = 8 × \(\frac { 1 }{ 8 } \) + 6 × \(\frac { 1 }{ 2 } \) – 4)

(c) In body centred cubic arrangement,
Lattice points located at the corners of the cube = 8
Lattice points located in the centre of the body = 1
Total no. of lattice points =8+1=9
(the total number of particles per unit cell = 8 × \(\frac { 1 }{ 8 } \) + 1 = 2)

Question 9.
Explain :
(i) The basis of similarities and differences between metallic and ionic crystals.
(ii) Ionic solids are hard and brittle.
Answer:
(i) Basis of similarities. The basis of similarities between the metallic and ionic crystals are the presence of strong electrostatic forces of attraction. These are present among the ions in the ionic crystals and among the kernels and valence electrons in the metallic crystals. That is why both metals and ionic compounds are good conductors of electricity and have high melting points.
Basis of differences. The basis of differences is the absence of mobility of ions in the ionic crystals while the same is present in the valence electrons and kernels in case of metallic crystals. As a consequence, the ionic compounds conduct electricity only in the molten state while the metals can do so even in the solid state.
(ii) The ionic solids are hard and brittle because of strong electrostatic forces- of attraction which are present in the oppositely charged ions.
(iii) The ionic solids are hard because of the presence of strong inter ionic forces of attraction in the oppositely charged ions. These ions are arranged in three dimensional space. The ionic solids are brittle because the ionic bond is non-directional.

Question 10.
Calculate the efficiency of packing in case of metal crystal for :
(i) Simple cubic
(ii) Body centred cubic
(iii) Face centred cubic (with the assumption that the atoms are touching each other).
Answer:
(i) Simple Cubic Unit Cell
We have calculated the number of atoms or spheres (since an atom or particle may be regarded as the sphere) in different types of cubic unit cells. Now, let us calculate the relation between the edge length (a) and the radius (r) of the spheres present in these unit cells in case of pure elements i.e. elements composed of the same type of atoms.

It may be noted that :
(a) Distance between the centres of the spheres present on the corners of the edges of the cube is called edge length (a).
(b) Distance between the centres of the two nearest spheres is called nearest neighbour
distance (d) and is equal to r + r = 2r. (Here r is the radius of the spheres or the atomic radius.) .
We know that in a simple cubic unit cell, there is one atom (or one sphere) per unit cell. If r is the radius of the sphere, volume occupied by one sphere present in the unit cell = \(\frac { 4 }{ 3 } \) πr³.
Since the spheres at the corners touch each other ; Edge length (a) = r + r = 2r = d

(ii) Body Centred Cubic Unit Cell
We know that a body-centred cubic unit cell has 2 spheres (atoms) per unit cell. If r is the radius of the sphere Volume of one sphere = 4/3 nr3
Volume of two spheres present in the unit cell = \(\frac { 4 }{ 3 } \) πr³ × 2 = \(\frac { 8 }{ 3 } \) πr^{3</sup .
From the Fig. 5.62, it is evident that the sphere present in the centre of the cube touches a corner sphere so that the distance in their centres = 2r}

Length of the body diagonal (BD) in the cube = 4r.
By considering the triangle BCD and by applying Pythagorean theorem,
BD² = CD² + BC²
But BC² = AB² + AC²
∴ BD² = CD² + AB² + AC²
edge length = a, then
(4r)² = a² + a² + a²
16r² = 3a² or a = (16/3 r²)^1/2
a = \(\frac { 4r }{ \sqrt { 3 } } \) or r = \(\frac { \sqrt { 3 } a }{ 4 } \)
or d = 2r = \(\frac { 2\times \sqrt { 3 } a }{ 4 } \) = \(\frac { \sqrt { 3 } a }{ 2 } \)

(iii)Face centred cubic unit cell
We know that a face centred cubic unit cell (fcc) contains four spheres (or atoms) per unit cell.
The spheres at the corners are touching the sphere or atom present in the centre. However, they can not touch each other.
Figure 5.61, Faced centred cubic unit cell.

Volume occupied by one sphere of radius r = \(\frac { 4 }{ 3 } \) πr³
∴ Volume occupied by four spheres present in the unit cell = \(\frac { 4 }{ 3 } \) πr³ × 4 = \(\frac { 16 }{ 3 } \) πr³
Distance in their centres = r + r = 2r
Length of the face diagonal = r + 2r + r = 4r.
Now, according to Pythagorean theorem,
the square of the hypotenuse of a right triangle is equal to sum of the squares of the other two sides. If a is the edge length of the unit cell, then
a² + a² = (4r)² or 2a² = 16r²
a² = \(\frac { 16 }{ 2 } \) r² = 8r² or a = (8r²)^1/2 = \(\sqrt { 8 } \)r
a = 2r\(\sqrt { 2 } \) or r = \(\frac { a }{ 2\sqrt { 2 } } \)
d = 2r = \(\frac { 2a }{ 2\sqrt { 2 } } \) = \(\frac { a }{ \sqrt { 2 } } \) (d is the nearest neighbour distance.)

Question 11.
Silver crystallises in a face centred cubic lattice with all the atoms at the lattice points. The length of the edge of the unit cell as determined by X-ray diffraction studies is found to be 4.077 × 10^-8 cm. The density of silver is 10.5 g cm^-3. Calculate the atomic mass of silver.
Answer:

Question 12.
A cubic solid is made of two elements P and Q. Atoms Q are at the corners of the cube and P at the body centre. What is the formula of the compound ? What is the co-ordination number of P and Q ?
Answer:
Contribution by atoms Q present at the eight comers of the cube = \(\frac { 1 }{ 8 } \) × 8 = 1
Contribution by atom P present at the body centre = 1
Thus, P and Q are present in the ratio 1:1.
∴ Formula of the compound is PQ.
Co-ordination number of atoms P and Q = 8.

Question 13.
Niobium crystallises in body centred cubic structure. If the density is 8.55 g cm^-3, calculate atomic radius of niobium given that atomic mass of niobium is 93 g mol^-1.
Answer:

Question 14.
If the radius of octahedral void is r and the radius of the atom in close packing is R, derive the relation between r and R.
Answer:
A tetrahedral void has been shown in the Figure 1.54. Here three spheres form the triangle base while the fourth sphere lies at the top. The shaded sphere occupies the tetrahedral void. Let the radii of void and sphere be r and R respectively. Let the edge length of the cube be a.
From the right hand triangle ABC,
AB² = AC² + BC²
Here AB is the face diagonal while AC and BC are the edge lengths (a). Therefore,
AB² = a² + a²
or AB = (2a²) ^1/2= a\(\sqrt { 2 } \)

Question 15.
Copper crystallises into a fee lattice with edge length 3.61 × 10^-8 cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm^-3.
Answer:

Question 16.
Analysis shows that nickel oxide has formula Ni_0.98 O_1.00. What fraction of nickel exists as Ni²⁺ and as Ni³⁺ ions ?
Answer:
The ratio of Ni and O atoms in pure nickel oxide (NiO) = 1 : 1
Let x be the no. of Ni (II) atoms replaced by Ni (III) atoms in the oxide.
∴ No. of Ni (II) atoms present = (0.98 – x)
Since the oxide is neutral in nature,
Charge on Ni atoms = Charge on oxygen atoms
2(0.98 -x) + 3x =2
1.96 – 2x + 3x =2
x = 2 – 1.96 = 0.04
% of Ni (III) atoms in nickel oxide \(\frac { No.of\quad Ni(III)atoms }{ Total\quad no.Ni\quad atoms } \) × 100 = \(\frac { 0.04 }{ 0.98 } \) = 4.01%
% of Ni (II) atoms in nickel oxide = 100 – 4.01 = 95.99%

Question 17.
What are semi-conductors ? Describe the two main types of semiconductors and contrast their conduction mechanisms.
Answer:
Semi-conductors are the substances whose conductivity lies in between those of conductors and insulators. The two main types of semi-conductors are n-type and p-type.
Applications of n-type and p-type semi-conductors
Germanium and silicon are the elements which belong to group 14 and have four valence electrons in their atoms. These are both metalloids and also semi-conductors. A large variety of solid state materials have been prepared by the combination of elements belonging to group 13 and 15 or group 12 and 16. These have also four valence electrons. A few examples of compounds resulting from 13-15 combination are : InSb, A1P and GaAs. Similarly, compounds resulting from 12-16 combination are CdS, CdSe, HgTe. (Cd and Hg are the transition metals present in group 12). These have resulted in many sophisticated semi-conductors which have brought about a revolution in this field.
If a single crystal is doped with indium at one end and with arsenic at the other end, then one part is a p-type semi¬conductor while the other is «-type semi-conductor. In the middle, the region where the two sides meet represents n-p junction These junctions are used in making microcompressor devices and integrated circuits in the solid state electronics.
Some important applications of n- and p-type semi-conductors are listed.

1. Rectifiers : The role of rectifiers is to allow current from an outside source to flow in one direction and thus, converts an alternate current (AC) into direct current. For this, a circuit made from four diodes is needed. A diode is basically a transistor consisting of two zones ; one p-type and the other n-type along with a n-p junction in between.

2. Photovoltaic cells : A photovoltaic cell is a device which can convert light energy into electrical energy. In order to achieve this, a n-p junction is irridiated with photons of light having energy more than band gap. As a result, some n-p bonds break and generate electrons and positive holes. These electrons then get promoted from valence band to the conduction band. These extra electrons present in the conduction band make the n-type region more negative. At the same time in the p-type region, the electrons get trapped by some positive holes. By connecting the two regions in the external circuit, the electrons are in a position to flow from n-type region to the p-type region. The current flows in the opposite direction i.e., from p-type region to the n-type region. In this manner, photo-voltaic cells function. As pointed earlier, they work at the expense of light energy into electrical energy.

3. Transistors : These are basically single crystals of silicon which have been doped to give three zones i.e., p-n-p and n-p-n. These transistors have wide range of applications. They work as amplifiers and oscillators in radios, televisions and also in computers. In addition, these are used as photo-transistors, tunnel diodes, solar cells etc. Now-a-days, it has become possible to manufacture computer chips which are equivalent to many thousand single crystal transistors.

Question 18.
Non-stoichiometric cuprous oxide (Cu₂O) can be prepared in the laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semi-conductor ?
Answer:
The ratio less than 2 : 1 in Cu₂0 shows that some cuprous (Cu⁺) ions have been replaced by cupric (Cusup>2+) 10ns. In order to maintain the electrical neutrality, every two Cusup>+ ions will be replaced by one Cusup>2+ ion which results in creating cation vacancies leading to positive holes. Since the conduction is due to positive holes, it is a p-type semiconductor.

Question 19.
Ferric oxide crystallises in a hexagonal close packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of ferric oxide.
Answer:
There is one octahedral hole for each atom in hexagonal close packed arrangement.
If the number of oxide ions (O^2-) per unit cell is 1, then the number of Fe³⁺ ions = \(\frac { 2 }{ 3 } \) × octahedral holes = \(\frac { 2 }{ 3 } \) ×1 = 2/3. Thus, the formula of the compound = Fe_2/3O₁ or Fe₂C₃.

Question 21.
Gold (atomic radius = 0.14411m) crystallises in a face centred unit cell. What is the length of the side of the unit cell ?
Answer:
For a face centred cubic unit cell (fcc)
Edge length (a) = 2\(\sqrt { 2r } \) = 2 × 1.4142 × 0.144 mm = 0.407 nm

Question 22.
In terms of Band Theory, what is the difference (i) between a conductor and an insulator (ii) between a conductor and semi-conductor ?
Answer:
Explanation for Conductors, Insulators and Semi-conductors
The’variation in the electrical conductivity of the solids can be explained with the help of the band theory. In metals, the energy gap between the partially filled valence band and unoccupied conduction band is negligible. Rather they overlap and electron flow readily takes place. In semi-conductors, there is small energy gap between valence band and conduction band. However, some electrons may jump to the conduction band and these semi-conductors can exhibit a little electrical conductivity.

For example, in case of silicon and germanium, these are 111 kJ mol^-1 and 63 kJ mol^-1 respectively and the electron jump is feasible. The conductivity of the semi-conductors increases with the rise in temperature because more electrons are now in a position to jump from the valence band to the conduction band. Such type of semi-conductors are known as intrinsic semiconductors However, the conductivity of these semi-conductors is so low that it is hardly useful practically. In insulators, the energy gaps are very large and the no electron jump is feasible from the valence band to the conduction band. The energy gaps also called forbidden zones. The insulators therefore, do not conduct electricity. For example, in case of diamond the energy gap is of the order of 5 IT kJ mol^-1. Therefore, diamond is an insulator. From the above discussion, we conclude that the conductivity in the solids is primarily linked with the energy difference between the valence and conduction bands.

Question 23.
Explain the following terms with suitable examples.
(i) Schottky defect, (ii) Frenkel defect, (iii) Interstitials, (iv) F-centres.
Answer:
(i) Schottky defect. The defect was noticed by German scientist Schottky in 1930. This arises because certain ions are missing from the crystal lattice and vacancies or holes are created at their respective positions. Since a crystal is electrically neutral, the number of such missing cations (A⁺) and anions (B^–) must be the same. The crystal lattice of an ideal crystal and that sufferring from Schottky defect has been shown below. Only one cation and one anion have been indicated as missing.

These are only the planar representation of the crystal lattice and not the three dimensional or space lattice.

(ii) Frenkel Defect. This defect in the ionic crystals was discovered by a Russian scientist Frenkel in 1926. It results when certain ions leave their normal sites and occupy positions elsewhere in the crystal lattice. Holes are created at their respective positions. Since cations are smaller in size as compared to anions normally these are involved in Frenkel defect.
Frenkel defect is depicted in the Fig. 5.67 in which only one cation has been shown to be dislocated from its position.

(iii)Interstitial defect. This defect is noticed when certain constituent particles (atoms or molecules) occupy the interstitial sites in the crystal lattice.. As a result, the number of particles per unit volume increases and so is the density of the solid. These defects are very common in the interstitial compounds of transtion metals. For details, consult unit 8 (d and fblock elements).

Question 24.
Aluminium crystallises in a cubic close packed structure. Its metallic radius is 125 pm.
(a) What is the length of the side of the unit cell ?
(b) How many unit cells are there in 1-00 cm3 of aluminium ?
Answer:
Step I. Calculation of length of side of the unit cell
Forf.c.c. unit cell, a = 2\(\sqrt { 2r } \) = 2\(\sqrt { 2 } \)(125pm) = 2 × 1.4142 × (125 pm) = 354 pm.
Step II. Calculation of no. of unit cells in 1.00 cm³ of aluminium.
Volume of one unit cell = (354 pm)³ = (354 x 10^-10 cm)³ = 44174155 × 10^-30 cc.
No. of unit cells in 1.00 cc of Al metal = \(\frac { 1(cc) }{ 44174155\times { 10 }^{ -30 }(cc) } \) = 2.26 x 10²².

Question 25.
If NaCl is doped with 10^-3 mol % of SrCl₂, what is the concentration of cation vacancies ?
Answer:

Question 26.
Explain the following with suitable examples
(a) Ferromagnetism (b) Piezoelectric effect
(c) Paramagnetism (d) Ferromagnetism
(e) 12-16 and 13-15 compounds.
Answer:
(a)Ferromagnetic solids (Ferromagnetism). A few solids like iron, cobalt, nickel, gadolinium and Cr02 are attracted very strongly by magnetic fields. These are known as ferromagnetic solids. Apart from that, they can be even permanently magnetised or become permanent magnet. Actually, in the solid state, the metal ions of these substances are grouped together into small regions known as domains. Each domain behaves like a small magnet. When a ferromagnetic solid is placed in a magnetic field, all the domains get oriented in the direction of the magnetic field (Fig. 5.76 (a)). As a result, a strong magnetic effect is produced. The effect persists even when the magnetic field is removed. Thus, a ferromagnetic substance becomes a permanent magnet. CrO₂ is used in magnetic tapes for audio frequency.

(b) Piezoelectric effect.  It may be noted that all the three types of solids (ferromagnetic, anti-ferromagnetic and ferrimagnetic) listed above may become paramagnetic upon heating upto a certain temperature known as Curie temperature. The change in the magnetic character is because of the reallignment of the magnetic moments or domains in one direction upon heating.For example,
(i) Ferricmagnetic solid Fe₃O₄ becomes paramagnetic when heated to 850 K.
(ii) Antiferromagnetic solid V₂O₃ becomes paramagnetic when heated to 750 K.

(c) Paramagnetic solids (Paramagnetism). These are the solids attracted by a magnet and this property is known as paramagnetism. Actually, the atoms of the elements present have certain unpaired electrons. Their spins or magnetic moments may lead to magnetic character. However, the electron spins may mutually cancel due to random orientation under normal conditions and the substances may not have any magnetic character. Under the influence of an external magnetic field or magnet, the spins of certain electrons may allign to produce temporary magnetism i.e. they may be attracted by the magnet. However, they may lose their orientation and also magnetic character once the magnet is removed. This means that their magnetic character is temporary and is present as long as external magnetic field is present. Many transition metals such as Co, Ni, Fe, Cu, etc. and their ions are paramagnetic. Similarly, dioxygen (O₂) and nitric oxide (NO) have also unpaired electrons and they exhibit paramagnetism.

(d).Ferrimagnetic solids (Ferrimagnetism). In certain ferromagnetic solids, the magnetic moment of the domains allign in parallel and anti-parallel directions in unequal numbers as shown in the Fig. 5.76(c). As a result, they-‘have certain resultant magnetic moment or magnetic character which is of permanent nature. However, ferrimagnetic solids are less magnetic than ferromagnetic solids. For example, magnetic oxide of iron (Fe₃O₄) and ferrites with general formula MFe₂O₄ (where M = Mg, Cu, Zn etc.) are ferrimagnetic in nature.

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NCERT Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure.

Question 1.
Explain the formation of chemical bond.
Answer:
Inert gases or noble gas elements are present in group 18 also called zero group. This means that their valency is zero or their atoms can exist independently of their own. Let us have a close look at their electronic configurations given in the table

With the exception of first member helium which has only two electrons in its valence shell, the rest of the elements have eight electrons.

In 1916, G. N. Lewis and Kossel stated that the stability of noble gas elements is due to the presence of eight electrons in their valence shells (except helium) or due to the presence of complete octet. This is known as the octet rule. According to the rule, the atoms of different elements take part in chemical combination in order to complete their octet (to have eight electrons in the outermost or valence shell) or duplet (to have two valence electrons)in some cases such as H, Li, Re etc. Thus, the atoms take part in chemical combination or bond formation in order to complete their octet and to have the electronic configuration of nearest noble gas atoms.

Question 2.
Write the Lewis dot symbols of the following elements :
Be, Na, B, O, N, Br.
Answer:
Please remember that only the valence electrons are shown in the Lewis dot symbols of the atoms of the elements.

Question 3.
Write the Lewis dot symbols of the following atoms and ions :
S and S^2- ; P and P^3- ; Na and Na⁺; Al and Al³⁺ ; H and H^–
Answer:

Question 4.
Draw the Lewis structures of the following molecules and ions :
PH₃, H₂S, SiCl₄, BeF₂, AlI₃, C0₃^2-, HCOOH
Answer:

Question 5.
Define octet rule. Write its significance and limitations.
Answer:
the atoms of different elements take part in chemical combination in order to complete their octet (to have eight electrons in the outermost or valence shell) or duplet (to have two valence electrons)in some cases such as H, Li, Re etc. Thus, the atoms take part in chemical combination or bond formation in order to complete their octet and to have the electronic configuration of nearest noble gas atoms.

Significance : The octet rule i. e., the tendency of the atom of an element to acquire eight electrons in its valence shell is the basis for the formation of chemical bonds. This applies to both ionic and covalent bonds. For more details
Significance of Lewis Symbols
The Lewis symbols of the atoms have a lot of information to convey.
(a) They represent the number of valence or outermost electrons.
(b) No of valence electrons help in calculating the valency or valence of the element which is also called group valeme
i. e.„ the valence of the group to which a particular element belongs. The group valence is equal to either the number of dots in the Lewis symbol of the element or “Eight – no. of dots.” For example,

Limitations: The main limitation of the octet rule is that in certain molecules, the central atom may have either incomplete or expanded octet. But they are still quite stable. For more details,
Limitations of Octet Rule or Octet Theory
We have discussed the implication of octet rule which states that the atoms of same or different elements take part in chemica I combination to complete their octet either by transference or by sharing of electrons.
The octet rule can be applied to large number of covalent molecules, particularly those of organic nature. However, there are certain exceptions to the octet rule. These are illustrated as follows :
(a) Incomplete octet of the central atom. There are many stable covalent molecules in which the central atom has less tha t eight electrons after sharing i.e. it has an incomplete octet. For example,

(b) Old electron molecules. There are certain molecules in which even single or odd electrons may be present on th; atoms. For example,

(c)Expanded octet of the central atom. There are many examples in which the central atom may have even more that eight electrons after sharing i.e. it may have expanded octet. For example,

(d) Shapes of the molecules. The octet rule or octet theory gives no idea about the, shapes of the molecules which may be tri-atomic, tetra-atomic or polyatomic in nature.

Question 6.
Write the favourable conditions for the formation of ionic bond.
Answer:
The favourable conditions for the formation of ionic bond are :
(i) Low ionization enthalpy for the element to form positive ion or cation.
(ii) High electron gain enthalpy for the element to form negative ion or anion.
(iii) High magnitude of lattice energy or lattice enthalpy.
(i). Ionization enthalpy. In the formation of positive ion or cation, one of the atoms is to lose electrons and for this ionization enthalpy is needed. As already stated, it is the amount of energy required to remove the most loosely bound electron from an isolated gaseous atom. Thus, lesser the ionization enthalpy required, easier will be the formation of cation. For example,
A(g) → A⁺(g) + e^– ; ∆_iH₁ = + ve
The alkali metals and alkaline earth metals present in the s-block normally form cations since they have comparatively low ionization enthalpies.

(ii). Electron gain enthalpy. The electrons released in the formation of cation are to be accepted by the other atom . taking part in the ionic bond formation. The electron accepting tendency of an atom depends upon the electron gain
enthalpy. It may be defined as the energy released when an isolated gaseous atom takes up an electron to form anion. Greater the negative electron gain enthalpy, easier will be formation of anion. For example,
B(g) + e^– → B^–(g) ; ∆_egH The halogens present in group 17 have the maximum tendency to form anions as they have very high negative electron gain enthalpy.
The members of the oxygen family (group 16) such as oxygen also form anions but not so easily because energy is needed t( form divalent anion (^2- is comparatively high.
The details have been discussed in the unit 3 on Chemical families and periodic properties.

(iii). Lattice energy or enthalpy. The ionic compounds exist as crystalline solids and the arrangement is known as crysta lattice. Since the ions are charged species, energy known as lattice energy or enthalpy is released in the attraction o ‘ the ions. It may be defined as :
the energy released when one mole of crystalline solid is formed by the combination of oppositely charged ions.
It is denoted as U.
A⁺ (g) + B^– (g) → A⁺ B^– (s) ; U = – ve (Released)
Thus, greater the magnitude of lattice energy, more will be the stability of the ionic bond or ionic compound. The lattice energy depends upon the following factors :

(a) Size of the ions. The size of the ions influences the lattice energy. Smaller the size, lesser will be the internuclea distance and, thus, greater will be the lattice energy. For example, lattice energy of NaCl (769 8 kJ mol1) is mon than that of KC1 (698-7 kJ mol-1) because the radius of Na+ (95 pm) ion is smaller as compared to K+ (133 pm) ion.

(b) Charge on the ions. Greater the magnitude of charge on the ions, higher will be inter ionic attraction and thus greater will be the value of lattice energy.

Thus, we conclude that if the magnitude of lattice energy and negative electron gain enthalpy is greater than that of the ionization enthalpy required, a stable chemical bond will be formed. In case, it is less then the bond will not be formed.The Lattice Energy can be determined with the help of Born-Haber cycle. For the details, consult section 6.18 (Chapter 6 – Thermodynamics).

Question 7.
Predict the shapes of the following molecules using the V.S.E.P.R. model.
BeCl₂, SiCl₄, AsF₅, H₂S, PH₃,
Answer:
The shapes of the covalent molecules can be predicted with the help of V.S.E.P.R. theory.

Predicting the Shapes of Covalent Molecules or Ions
The shape of a covalent molecule or ion can be predicting theoretically by calculating the total number of electron pains (shared pairs as well as lone pairs) around the central atom in the molecule or ion. This can be done by the following relation.
= 1/2[No. of valence electrons on the central atom + No. of atoms linked to the central atom by covalent bonds]
(i) For positive ion (or cation), subtract the number of electrons equal to the positive charge from the valence electrors on the central atom.
(ii) For negative ion (or anion), add the number of equal to the negative charge to the valence electrons on the central atom.
The number of bond pairs or shared pairs = No. of atoms linked to the central atom by single bonds.
The number of lone pairs = Total number of electron pairs-No. of bond pairs or shared pairs.

Question 8.
Although geometries of NH₃ and H₂O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
Answer:
In \(\ddot { N }\)₃ and in H₂\(\ddddot { O } \) molecules the central atom is surrounded by four electron pairs. The geometries of these molecules are expected to be tetrahedral.
They however, get distorted on account of the presence of lone electron pairs. On comparison, the bond angle in NH₃ (107°) is more than in H₂O (104.5°).
This is on account of greater magnitude of force of lone pair : lone repulsion in H₂O molecule as compared to lone pair : shared pair repulsion in NH₃ molecule.
Ammonia molecule (NH₃)
In ammonia, the atomic number (Z) of nitrogen is 7 and its electronic configuration is 2, 5. Out of the five electrons present in the valence shell of nitrogen atom, three form shared pairs (bond pairs) with the electrons of three hydrogen atoms.
The remaining two electrons constitute one lone pair.

Thus, the nitrogen atom is surrounded by three shared pairs and one lone pair of electrons. According to VSEPR theory, the geometry of the molecule is irregular.
In order to have minimum force of repulsion in all the four electron pairs around the nitrogen atom, the shape is expected to be tetrahedral but the presence of one lone electron pair distorts the shape of the NH3 molecule.

Actually, Ip – bp repulsion is greater than the bp – bp repulsion. As a result, the two N-H bonds on both sides of the lone pair are stretched a little inwards. Thus, the bond angle in the molecule is 107° and is slightly less than the bond angle of a regular tetrahedron. The actual shape of the NH₃molecule is pyramidal .

Question 9.
How do we express bond strength in terms of bond order ?
Answer:
Bond strength is directly proportional to bond order, i.e., greater the bond order, more is the bond strength. For example,

Question 10.
Define bond length.
Answer:
Bond Length. Bond length may be defined as :
the average equilibrium distance between the centres of the two bonded atoms.
The bond length of different covalent bonds are determined by X-ray diffraction methods. For a covalent bond, it is the sum of covalent radii of the bonding atoms. For example, the bond length of C—Cl bond is rc + rC]. The values of the bond lengths are generally expressed in picometre (1 pm — 1(F12 m). The bond length is influenced by the following factors.

(i)Size of the atoms. As stated earlier, bond length is directly linked with the size of the bonding atoms. For example H—Br bond length (141 pm) is more than H—Cl bond (127 pm).
(ii)Multiplicity of bonds. The multiplicity of the bonds between two atoms brings them close to each other. As a result, the bond length decreases. For example, bond length for C—C bond is 154 pm and the value for C = C bond is 134 pm.

(iii)Types of hybridisation. The value of bond length is also influenced by the type of hybridisation. Since s-orbital is smaller in size as compared to the p-orbital, therefore, greater the s-character associated with a particular bond, smaller will be the bond length. For example,
sp³ C—H (111 pm) > sp² C—H bond (110 pm) > sp C—H bond (108 pm).

Question 11.
Explain the important aspects of resonance with respect to the \({ CO }_{ 3 }^{ 2- }\) ion
Answer:
Resonance
Covalent molecules are normally expressed as Lewis structures. But in some cases, a single Lewis structure fails to explain all the characteristics of the molecule. For example, the Lewis structure of ozone is represented as follows :

The two bonds in the molecule are expected to have different values. But the experimental studies have shown that both the bonds have the same lengths (128 pm). Moreover, the bond length is intermediate between the single and double bonds.

The problems can be solved if we consider that ozone is capable of existing in two forms in which the positions of the double bond and single bond are interchangeable.
Such structures for the molecule are known as resonating or contributing structures separated by (sign for resonance). None of these structures can independently explain all the properties of ozone.
It is believed that the actual structure is inbetween these two contributing structures and is known as resonance hybrid as

Thus, O to O bond length in the molecule is the same for both the bonds i.e. 128 pm. Resonance may be defined as :
the phenomenon as a result of which a molecule can be expressed in different forms none of which can explain all the properties of the molecule. The actual structure of the molecule is called resonance hybrid.

Question 12.
H₃PO₃ can be represented by the structures 1 and 2 as shown below. Can these structures be taken as the canonical forms of the resonance hybrid representing H₃PO₃ ? If not cite reasons for the same.
Answer:

The basic requirement for the resonance is that the canonical structures must differ in the arrangement of the electron pairs and not of the atoms. The structures (1) and (2) are not the canonical structures because in structure (1) phosphorus atom is linked with H (P—H bond) while in the other (2), it is linked with OH group (P—Oil bond).

Question 13.
Write the resonating structures for SO₃, No₂ and \({ NO }_{ 3 }^{ – }\)
Answer:


Note : Please note that in many cases, the co-ordinate bond is shown simply as covalent bond with positive (+) charge on donor atom and negative (-) charge on acceptor atom.

Question 14.
Using Lewis dot symbols, show electron transfer between the following atoms to form cations and anions :
(a) Na and Cl (b) K and S (c) Ca and O (d) Al and N.
Answer:

Question 15.
Although both CO₂ and H₂o are triatomic molecules, the shape of H₂o molecule is bent while that of Co₂ is linear. Explain this on the basis of dipole moment.
Answer:
Carbon dioxide is a linear molecule in which the two C = 0 bonds are oriented in the opposite directions at an angle of 180°. The dipole moment of C = O bond is 2.3D but due to linear structure of Co₂, the bond dipoles of two C = O bonds cancel with each other.

Therefore, the resultant dipole moment of the molecule is zero. On the other hand, H20 is a polar molecule having dipole moment 1.84 D. Actually H₂o molecule has a bent structure in which the O—H bonds are oriented at an angle of 104.5° and do not cancel the bond dipole moments of each other.

The molecular dipole moment of H₂o (p = 1.84 D) is the resultant of the individual values of the bond dipole moment of two O—H bonds.

Question 16.
Write the significance and applications of dipole moment.
Answer:
Applications of Dipole Moment
Some useful applications of dipole moment are given :
(i) In predicting the nature of the molecules. Molecules with specific dipole moments are polar in nature while those with zero value are non-polar. Thus, BeF₂ (μ= 0D) is non-polar while H₂0 (μ = 1.84 D) is polar.

(ii) In comparing the relative polarities of the molecules. The relative polarities of the molecules can be compared by their dipole moment values. Greater the value, more is polarity.

(iii) In determining the shapes of molecules. The dipole moment values are quite helpful in determining the general shapes of the molecules having three or more atoms. In case such molecules have dipole moments, then their shapes will not be symmetrical.

They will be either bent or angular, (e.g. H₂S with// = 0.90 D) or unsymmetrical (e.g. NH3 with// = 1.47 D). However, for molecules with zero dipole moment values, the shapes will be either linear (BeF₂, Co₂ etc.) or symmetrical (CH₄, CCl₄ etc.)

(iv) In calculating the percentage ionic character. The dipole moment values help in calculating the percentage ionic characters of polar bonds. It is the ratio of the observed dipole moment or experimentally determined dipole moment to the dipole moment for the complete transfer of electrons.

For example, the observed dipole moment of HCl molecule is 1.03 D. For complete transfer of electrons (100% ionic character) the charge on H⁺ and Cl^– ions would be equal to one unit (4.8 × 10^-10 esu) each. The bond length of H—Cl bond is 1.275 × 10^-8 cm. Therefore,
Dipole moment for complete electron transfer (μ_ionic) = q × d
= (4.8 × 10^-10esu) × (1.275 × 10^-8 cm)
= 6.12 × 10^-18 esu-cm = 6.12D
Observed dipole moment (μ_observed) = 1.03 D

In a similar manner, the percentage ionic character for other polar bonds can also be calculated.

(v) In distinguishing between cis and trans geometrical isomers. In organic compounds, dipole moment can halp in making a distinction between the cis and trans geometrical isomers of a substance. In general, cis isomer (less symmetrical) has more dipole moment value as compared to the trans isomer. For example, the dipole moment cis-but-2-ene is higher than that of trans-but-2-ene.

(vi) In identifying the position of the substituents in the disubstituted aromatic rings. The relative positions of the substituents such as methyl (- CH₃) groups present at ortho, meta and para positions can be identified from their dipole moments. The dipole moment of para-isomer is zero. Out of the ortho and meta isomers, the dipole moment of the ortho iosmer is higher.

Question 17.
Define electronegativity. How does it differ from electron gain enthalpy ?
Answer:
Difference between electron gain enthalpy and electronegativity.
The main points of distinction are given in a tabular form.

Question 18.
Explain with the help of suitable example polar covalent bond.
Answer:
We have studied that according to Lewis concept, a covalent bond represents a shared pair of electrons. If the atoms joined by covalent bond are the same then the shared electron pair is equidistant to both of them. In other words, it lies exactly in th: centre of the bonding atoms. As a result, no poles are developed and the bond is called non-polar covalent bond. Th: corresponding molecules are known as non-polar molecules. For example,

On the other hand, in case different atoms are linked to each other by covalent bond, then the shared electron pair will not lie in the centre because the bonding atoms differ in electronegativities.

For example, in hydrogen chloride molecule, chlorine with greater electronegativity (3.0) will have greater control over the electron pair as compared to hydrogen which is less electronegativ : (2.1). Since the electron pair is displaced more towards chlorine atom, the latter will acquire a partial negative charge (δ-).

At the same time, the hydrogen atom will have a partial positive charge (δ + ) with the magnitude of charge same as on the chlorine atom. Such a covalent bond is known as polar covalent bond or simply polar bond and is represented as follows :

The molecule with such a bond is called polar covalent molecule. It may be noted that greater the difference in the electronegativity of the bonding atoms, more will be the polarity of the bond. For example,

Question 19.
Arrange the bonds in order of increasing ionic character in the molecules :
LiF, K₂o, N₂, SO₂ and ClF₃
Answer:
N₂ < SO₂ < ClF₃v < K₂o < LiF.

Question 20.
The skeletal structure of CH3COOH as shown below is correct but some of the bonds are wrongly shown. Write the correct Lewis structure of acetic acid.

Answer:
Only the skeletal arrangement of the atoms in the above structure is correct. But it is not according to the Lewis concept as well as tetravalent nature of carbon. The correct structure of acetic acid is

Question 21.
Apart from tetrahedral geometry for methane (CH₄), another possible geometry is square planar with four ‘FT atoms at the comers of the square with the ‘C’ atom at its centre. Explain why CH₄ is not square planar.
Answer:
The tetrahedral and square planar structures of CH₄ are shown as follows :

According to VSEPR theory, the shared electron pairs around the central atom in a covalent molecule are so arranged in space that the force of repulsion in them is the minimum. Now, in a square planar geometry, the bond angle is 90° while it is 109°-28′ in tetrahedral geometry. This clearly shows that the electron repulsions are less in the tetrahedral geometry as compared to the square planar geometry. Thus, methane cannot be represented by square planar structure.

Question 22.
Explain why BeH₂ molecule has zero dipole moment although the Be—H bonds are polar.
Answer:
BeH₂ is a linear molecule (H—Be—H) with bond angle equal to 180°. Although the Be—H bonds are polar on account of electronegativity difference between Be (1.5) and FI (2.1) atoms, the bond polarities cancel with each other. The molecule has resultant dipole moment of zero.

Question 23.
Both NH₃ and NF₃ have identical shapes and same state of hybridisation. Both N—H and N—F bonds have almost the same electronegativity difference. But still, the two molecules have different dipole moment values. How will you account for it ?
Answer:
Both NH₃ and NF₃ have pyramidal geometries and the electronegativity difference of bonding atoms in both NH₃ (3.0 -2.1= 0.9) and NF₃ (4.0 – 3 0 = 1.0) molecules are also nearly the same. But the dipole moment of NH₃ (1.46D) is more than that of NF₃ (0.24D).

This is explained on the basis of the difference in the directions of the dipole moments. In NH₃, the dipole moments of the three N—H bonds are in the same direction as that of the lone electron pair.

But in NF3, the dipole moments of the three N—F bonds are in the direction opposite to that of the lone pair. Therefore, the resultant dipole moment in NH₃ is more than in NF₃.

Question 24.
What is meant by hybridisation of atomic orbitals ? Describe the shapes of sp, sp² and sp³ hybrid orbits.
Answer:
Hybridisation is of different types depending upon the nature and number of orbitals taking part in hybridisation. Let us discuss the hybridisation involving 5, p and d orbitals.
Hybridisation involving s and p-orbitals
The hybridisation involving 5 and p orbitals is of three types depending upon the orbitals that are involved. These are sp3, sp2 and sp types :

Question 25.
What is meant by the hybridisation of atomic orbitals ? What is the change in hybridisation (if any) of A1 atom in the reaction ?
AlCl₃ + Cl^– → AlC\({ l }_{ 4 }^{ – }\).
Answer:
Hybridisation is of different types depending upon the nature and number of orbitals taking part in hybridisation. Let us discuss the hybridisation involving 5, p and d orbitals.
In AlCl₃, the central Al atom is sp² hybridised while in the ion [AlCl₄]-, it has sp³ hybridisation.

Question 26.
Is there any change in the hybridisation of B and N atoms as a result of the following reaction ?
BF₃ + :NH₃ → F₃B.NH₃
Answer:
In BF₃, B atom is sp² hybridised. It forms an addition compound with \(\ddot { N }\)₃ (Where N atom is sp³ hybridised). In doing so, it takes up an electron pair from NH₃. This indicates that B atom undergoes a change in hybridisation from sp² to sp³. At the same time there is no change in hybridisation of N atom.

Question 27.
Draw diagrams showing the formation of a double and a triple bond between the carbon atoms in the (C₂H₄ and C₂H₄molecules.
Answer:
Formation of ethene (CH₂ = CH₂). In order to illustrate sp² hybridisation, let us discuss the orbital structure of ethene (H₂C = CH₂) molecule. Out of three hybridised orbitals belonging to each carbon atom, one is involved in the axial overlap with the similar orbital of the other carbon resulting in a sigma bond.

The remaining two hybridised orbitals of the carbon atoms form sigma bonds with the lx orbital of the hydrogen atoms. The unhybridised orbitals (2p^z) of both the carbon atoms participate in the sidewise overlap leading to a pi (π) bond as shown in the Fig. 4.27.

Question 28.
What is the total number of sigma and pi bonds in the following molecules ?
(a) C₂H₂ (b) C₂H₄
Answer:

Question 29.
Considering X-axis as the internuclear axis, which out of the following atomic orbitals will form a sigma bond ?
(a) 1s and 1s
(b) 1s and 2p_x
(c) 2p_y and 2p_y
(d) 1s and 2s.
Answer:
The sigma bond is formed by axial overlap along internuclear axis and is present in the following cases.
(a)1s and 1s
(b) 1s and 2p_x
(d) 1s and 2s.
2p_y and 2p_y atomic orbitals are involved in the sidewise overlap leading to the formation of π-bond.

Question 30.
Which hybrid orbitals are used by the carbon atoms in the following molecules ?
(a) H₃C—CH₃
(b) H₃C—CH=CH₂
(C) CH₃—CHO
(d) CH₃COOH
Answer:

Question 31.
What do you understand by bond pairs and lone pairs of electrons ? Illustrate by giving one example of each type.
Answer:
The electron pair involved in sharing between two bonding atoms is called shared pair or bond pair. At the same time, the electron pair which is not involved in sharing is called lone pair of electrons. For example, in BeH₂ molecule, the central Be atom is surrounded by two bond pairs. In the molecule of NH₃, the central B atom has three shared pairs and one lone pair around it.

Question 32.
Distinguish between a sigma and a pi bond.
Answer:
Types of Orbital Overlap (Sigma and Pi Bonds)
We have studied that the covalent bonds are formed by the overlap of the atomic orbitals. Depending upon the type of the overlapping, the covalent bonds are of two types known as sigma (\(\sigma \) ) and pi (π) bonds. These are described as below :
Sigma (\(\sigma \) ) bond. A sigma bond is formed when the two half filled atomic orbitals belonging to the bonding atoms overlap along their internuclear axes i.e. the line joining the centres of the nuclei of the two atoms.

Thus, the bond formed as a result of the axial overlap is known as \(\sigma \) bond. Since the axial overlap is quite large, the energy released during the overlap is also large. As a result, the sigma bond is a stable bond.
In most of the covalent molecules, s and p orbitals are involved in the bond formation. The axial overlap involving these orbitals is of three types :
(a) s-s overlap. This involves the axial overlap of the half filled v-atomic orbitals of the combining atoms. The cr-bond is
called s-s bond.
(b) s-p overlap. This involves the overlap of the s-atomic orbital belonging to one atom and p-atomic orbital to the other atom along their axes. The bond is known as s-p bond.
(c) p-p overlap. In this case, the atomic orbitals taking part in the axial overlap are both half filled p-orbitals. The \(\sigma \) -bond formed is called p-p bond.

Question 33.
Explain the formation of H? molecule on the basis of valence bond theory.
Answer:
Formation of hydrogen (H₂) molecule
In the light of the above discussion, let us consider the combination between atoms of hydrogen HA and HB. If eA and eB be their respective electrons, then attractive and repulsive forces may be shown in the Fig. 4.9.

Although the number of new attractive and repulsive forces is the same, but the magnitude of the attractive forces is more. Thus, when two hydrogen atoms approach each other the overall potential energy of the system decreases. Therefore, a stable molecule of hydrogen gets formed.

The potential energy changes which take place in the formation of a hydrogen molecule may also be shown graphically as follows: When the two hydrogen atoms are far separated (at infinite distance) the potential energy is zero. As they start coming closer to each other from infinite distance, energy is lost correspondingly.

Ultimately, a point of minimum energy is attained when the attractive and repulsive (or maximum stability) forces balance each other. At this stage, the two hydrogen atoms are bonded together to form a molecule of hydrogen.

M The distance between the centres of their nuclei is called bond length. In c case of hydrogen molecule, H—H bond length is 74 pm. It may be kept — in mind that the two hydrogen atoms can not be brought closer than 74 pm.

If it happens, then the repulsive forces will become more than the 3 attractive forces. As a result, the potential energy of the system will increase as shown by the dotted line.

The total decrease in the potential energy when one mole bonds of a particular type are formed between the atoms in the gaseous state is called bond energy. For example, the bond energy of H—H(g) is 433 kJ mol^-1.

It may be noted that the same energy is needed to break the molecules into the atoms and is known as bond dissociation energy. Thus, the bond dissociation energy of hydrogen is also 433 kJ mol^-1.

Question 34.
Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
Answer:
The molecular orbitals are formed by the linear combination of the wave functions of the participating atomic orbitals. They may combine either by addition or by subtraction. Let yA and yB represent the wave functions (or amplitude) of the two combining atomic orbitals A and B taking part in chemical combination.
(a) Combination by addition. When the two electron waves are in phase i.e. they have the same sign*, they will add up to give a new wave function expressed as φ (or \(\phi \)) = φ_A +φ_B.

(b) Combination by subtraction. When the two electron waves are out of phase i.e. they have opposite sign of the wave functions, then they will combine by subtraction and the resulting wave function φ* (or <()*) = φ_A – φ_B as shown below :

We know that the probable electron density is given by φ² and not by φ. On squaring we get :
φ² = φ_A² + φ_B² + 2φ_Aφ_B
φ*² = φ_A² + φ_B² – 2φ_Aφ_A
Here yA2 and yB2 represent the probable electron density in the two atomic orbitals whereas φ2 and φ*² indicate the electron density in the two molecular orbitals. These are called Bonding Molecular orbital (φ²) and Antibond ing Molecular orbital (φ*sup>2).

• Bonding Molecular Orbital (φ2) is formed by the linear combination of wave functions of the combining atomic orbitals [L.C.A.O) by addition.

• Antibonding Molecular Orbital (φ*sup>2) is formed by the linear combination of wave functions of the combining atomic orbitals (L.C. A. O) by subtraction. Let us illustrate both these orbitals by the combination of the atomic orbitals of hydrogen atoms

The formation of bonding molecular orbital when two atomic orbitals of hydrogen atoms (Is orbital) combine by the addition of their wave functions is shown below :

Similarly, the antibonding molecular orbital arising from the subtraction of the wave functions of the two participating atomic orbitals is shown as follows :

Thus we conclude that whenever two atomic orbitals combine, two molecular orbitals are formed. One of them is bonding molecular orbital (φ²M 0) while the other is antibonding molecular orbital (φ*²M.O)-

Question .35.
Use molecular orbital theory to explain why Be₂ molecule does not exist.
Answer:
The atomic no. (Z) of Be is 4. This means that 8 electrons are to be filled in the M.O of Be₂. The configuration is :

As the bond order comes out to be zero, the molecule of Be2 does not exist.

Question 36.
Compare the relative stability of the following species and indicate their magnetic properties :
O₂, \({ O }_{ 2 }^{ + }\),\({ O }_{ 2 }^{ – }\) (superoxide), \({ O }_{ 2 }^{ 2- }\) (peroxide)
Answer:
Bond orders of the different species are ;
O₂(2.0), \({ O }_{ 2 }^{ + }\)(2.5), \({ O }_{ 2 }^{ – }\)(l.5), \({ O }_{ 2 }^{ 2- }\)(1.0)
Relative stability : \({ O }_{ 2 }^{ + }\) > O₂ > \({ O }_{ 2 }^{ – }\) > \({ O }_{ 2 }^{ 2- }\)
For magnetic properties; consult section 4.32.

Question 37.
Write the significance of plus and minus signs in representing the orbitals.
Answer:
Plus and minus signs have been given to identify the nature of the electron waves. Plus (+ve) sign denotes crest while negative (-ve) sign denotes trough.

Question 38.
Describe the hybridisation in case of PCl₅. Why are the axial bonds longer as compared to equatorial bonds ?
Answer:
The hybridisation in the molecule of PCl₅ is same as in PF₅ because the central atom is the same in both the molecules and the two halogens chlorine (Cl) and fluorine (F) also belong to the same group (group 17). They are expected to have same hybridisation as well as same geometry. The P atom is sp²d hybridised and the molecule has trigonal bipyramidal geometry. The details of hybridisation of PF₅

Question 39.
Define hydrogen bond. Is it weaker or stronger than the van der Waals forces ?
Answer:
Hydrogen bond represents strong dipole dipole interactions between the H and the highly electronegative atoms (F, O and N) present in different bonds belonging to either different molecules or same molecule van der Waals forces are also dipole forces but are comparatively weak. For the details of hydrogen bond, consult section 4.34. For the details of van der Waals, forces, consult unit-5 on States of Matter.

Question 40.
What is meant by the term bond order ? Calculate the bond order of: N₂, O₂, \({ O }_{ 2 }^{ – }\) and \({ O }_{ 2 }^{ 2- }\); .
Answer:
Significance of Bond Order. The bond order conveys very important information about the molecules and ions. These are briefly discussed.
(a) Stability of molecules or ions. The stability of the molecules and their ions can be predicted in terms of bond order. If bond ‘ order is positive, (N_b > N_a), the molecule or ion will be stable. Incase, it is zero (N_b = N_a) or negative (N_b< N_a), the molecule or ion will be unstable. A fractional value of the bond order suggests that the molecule is unstable. However, it still exists.

(b) Bond dissociation enthalpy. Bond dissociation enthalpy is directly proportional to the bond order. Thus, more the value of bond order, greater will be the bond dissociation enthalpy and vice versa. This is further supported by the available data.

(c) Bond Length. Bond order is inversely proportional to bond length. This means that greater the bond order, smaller will be the bond length.

(d) Number of bonds. The value of bond order also predicts the number of covalent bonds in the molecule. Bond order 1, 2, 3 signifies the presence of single, double and triple bond respectively in a particular molecule, However, in some cases, the bond order is fractional (e.g. 0-5) and it is quite difficult to explain its existence.

We hope the NCERT Solutions for Class 11 Chemistry at Work Chapter 4 Chemical Bonding and Molecular Structure, help you. If you have any query regarding NCERT Solutions for Class 11 Chemistry at Work Chapter 4 Chemical Bonding and Molecular Structure, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties.

Question 1.
What is the basic theme of organisation in periodic table ?
Answer:
The basic theme of organisation in periodic table is to group the elements with same characteristics (physical and chemical) together so that it may become rather easy to follow these characteristics. Since these mainly depend upon the valence shell electronic configurations, the elements placed in a group have therefore same valence shell configuration of their atoms. In fact, the valence shell electronic configuration of the elements placed in a group gets repeated after definite gaps of atomic numbers (8, 8, 18, 18, 32).

Question 2.
Which important property did Mendeleev use to classify the elements in periodic table and did he stick to that ?
Answer:
The property used by Mendeleev is known as Mendeleev’s Periodic Law. According to the law,

Physical and chemical properties of the elements are a periodic function of their mass numbers.

Mendeleev did stick to it and classified elements into groups and periods. This classification was perhaps the first systematic way in order to study the characteristics of the elements. No doubt, later on many defects were pointed out and scientists challenged the basis of the classification.

Question 3.
What is the basic difference in approach between Mendeleev’s periodic law and Modern periodic law ?
Answer:
The atomic mass of the elements is the basis of periodicity according to Mendeleev’s periodic law. On the other hand, the atomic number of the elements constitutes the same according to Modern periodic law. For more details,

• Modern Periodic Law

Dmitri Mendeleev, a Russian Chemist, was the first to make a very significant contribution in the formation of the periodic table. He studied the chemical properties of the large number of the elements and in 1869 and came out with a statement that the chemical properties of the elements are a periodic function of their atomic masses. When Mendeleev came to know about the work of Lother Meyer, he modified his statement and gave a law called Mendeleev-Lother Meyer Periodic Law or simply Mendeleev’s Periodic Law.
Modern Periodic Law
It states that In the year 1913, an English physicist, Henry Moseley, a young physicist from England, studied the frequencies of the X-rays which were emitted when certain metals were bombarded with high speed electrons. He found that in all the cases, the square root of the frequency (Vv) was directly proportional to the atomic number of the atom of the metal. Upon investigation, he further stated that there was no co-relation between the frequency and the atomic mass.

Actually, when Moseley plotted a graph between and atomic number of the different metals, a straight line was obtained. But it was not the case when a graph was plotted between (4v) and atomic mass of the metals. These studies led Moseley to believe that atomic number and not the atomic mass is the fundamental property of an element. Therefore, atomic numbers must form the basis of the classification of the elements in the periodic table. In the light of the above observations. Moseley gave the modern periodic law which states that

Question 4.
On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
Answer:
The sixth period corresponds to sixth shell. The sub-shells present in this shell are 6s, 4f, 5p and 6d. The maximum number of electrons which can be present in these sub-shell are 2 + 14 + 6 + 10 = 32. Since the number of elements in a period correspond to the number of electrons in the shells, therefore, sixth period should have a maximum of 32 elements.

Question 5.
In terms of period and group where will you locate the element with Z = 114 ?
Answer:
We know that the gaps of atomic numbers in a group are 8, 8, 18, 18, 32. The element which proceeds the element with Z = 114 in the same group must have atomic number equal to 114 – 32 = 82. This represents lead (Pb) which is present in 6th period and group 14 of the p-block. This means that the element with Z = 114 must also belong to the group 14 of p-block and it must be a member of 7th period.

Question 6.
Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
Answer:
The element is chlorine (Cl) with atomic number (Z) = 17.

Question 7.
Which element do you think would have been named by :
(i) Lawrence Berkeley laboratory
(ii) Seaborg’s group ?
Answer:
(i) Lawrencium (Lr) with atomic number (Z) = 103
(ii) Seaborgium (Sg) with atomic number (Z) = 106.

Question 8.
Why do elements in a group have similar physical and chemical properties ?
Answer:
The elements in a group have the same valence shell electronic configurations of their atoms. However, they differ in the atomic sizes which tend to increase. Therefore, the elements in a group have similar chemical properties while their physical characteristics show a little variation.

Question 9.
What do atomic radius and ionic radius really mean to you ?
Answer:
For the definition and details of atomic radius,

Variation of Atomic Radius in the Periodic Table

We shall discuss the variation or change of the atomic radius along the period from left to the right and also down the group.
Variation in a period. Along a period, the atomic radii of the elements generally decrease from left to the right.
The atomic radii (metallic radii for Li, Be and B while covalent radii for the remaining elements) of the elements present in the second period are given :

The trend in atomic radii have also been shown in the figure 3.5. From the values, it is clear that the alkali metal (Li) placed on the extreme left has the maximum atomic radius while the halogen (F) on the extreme right has the minimum value.

Explanation. In moving from left to the right in a period, the nuclear charge gradually increases by one unit and at the same time one electron is also being added in the electron shell. Due to increased nuclear charge from left to the right, the electrons are also getting attracted more and more towards the nucleus. Consequently, the atomic size is expected to decrease as shown in case of the elements of second period.

Ionic Radii
Ionic compounds are crystalline solids and the ionic radii are related to the ions present in them. The ions formed by the loss of one or more electrons from the neutral atom is known as cation (positive ion) while the one formed by the gain of electrons by the neutral atom is called anion (negative ion). Ionic radius may be defined as :

the effective distance from the centre of the nucleus of the ion upto which it exerts its influence on the electron cloud.

Since the electron cloud may extend itself to a very large distance from the nucleus it may not be possible to determine the ionic radius experimentally. However, some theoretical methods have been given to calculate the radius of the ion and the values as given by Pauling are most acceptable.

But it is quite easy to determine the inter-nuclear distance between two oppositely charged ions (say Na⁺ and Cl^– ions) in the crystal lattice by X-ray studies. If the absolute radius of one of them is known, that of the other can be obtained by subtracting the same from the inter-nuclear distance.

Similarly for the details of ionic radius,  In a group, the atomic radii increase downwards.

For example, in sodium chloride crystals, the inter-nuclear distance is 276 pm and the radius of the Na⁺ ion is 95 pm (Pauling method). Therefore, the radius of CL^– ion is 276 – 95 = 181 pm.

It may be noted that the value of the ionic radius is linked with the atomic radius and it varies accordingly. Thus, the ionic radius always increases down the group and decreases along the period provided the ions involved have the same magnitude of charge.

This is on account of the increase in the number of electron shells and also due to increase in the magnitude of shielding or screening effect. In a period, the trend is the reverse.

The atomic radii decrease from left to the right because the electrons are filled in the same shell and no new shell is formed. The ionic radii of the elements also follow the same trend.

The atomic radii of the elements influence the other periodic properties such as ionisation enthalpy, electron gain enthalpy and electronegativity.

Question 10.
How do atomic radii vary in a period and in a group ? How do you explain the variation ?
Answer:
Variation in a group. The atomic radii of the elements in every group of the periodic table increase as we move downwards. Covalent Radii of the alkali metals present in group 1 are given. Since all of them are metallic in nature, their metallic radii have also been given for reference.

The trend in the atomic radii are also shown in the figure 3.6. From the values, it is quite clear that the atomic radius of lithium (Li) is the minimum while that of cesium (Cs) is the maximum. The value of the last element francium (Fr) is not known since being as radioactive in nature, it is unstable.

Explanation. On moving down a group, there is an increase in the principal quantum number and thus, increase in the number of electron shells. Therefore, the atomic size is expected to increase. But at the same time, there is an increase in the atomic number or nuclear charge also. As a result, the atomic size must decrease. However, the effect of increase in the electron shells is more pronounced than the effect of increase in nuclear charge. Consequently, the atomic size or atomic radius increases down a group. This is well supported by the values given in Table 3.5 for the alkali metals.

Question 11.
What do you understand by isoelectronic species ? Name a species that will be isoelectronic with each of the following atoms or ions.
(i) F^– (ii) Ar (in) Mg²⁺ (iv) Rb⁺.
Answer:
Isoelectronic species are those species (atoms/ions) which have same number of electrons (iso-same ; electronic- electrons). The isoelectronic species are listed :
(i) Na⁺
(ii) K⁺
(iii) Na⁺
(iv) Sr²⁺.

Question 12.
Consider the following species :
N^3-, O^2-, F^–, Na⁺, Mg²⁺, Al³⁺ .
(a) What is common in them ?
(b) Arrange them in order of increasing radii.
Answer:
(a) All of them are isoelectronic in nature and have 10 electrons each.

(b) In isoelectronic species, greater the nuclear charge, lesser will be the atomic or ionic radius. Based upon that, the increasing order of atomic radii are :
Al³⁺ < Mg²⁺ < Na⁺ < F^– < O^2- < N^3-.

Question 13.
Explain why are cations smaller and anions larger in radii than their parent atoms.
Answer:
Cations (positive ions) are formed by the loss of electrons from the parent atoms. Therefore, they have smaller radii than the parent elements. On the other hand, anions are formed when the parent atoms take up one or more electrons. They have therefore, bigger radii than the parent atoms. For illustration,

Question 14.
What is the significance of the terms— ‘isolated gaseous atom’ and ‘ground state’ while defining ionization enthalpy and electron gain enthalpy ?
Answer:
(a) Significance of term ‘isolated gaseous atom’ : The three states of matter as we know differ in the interparticle spaces. These are the maximum in the solid state while least in the gaseous state.

The atoms in the gaseous state are far separated in the sense that they do not have any mutual attractive and repulsive interactions. These are therefore, regarded as isolated atoms.

When an atom in the gaseous state is isolated, its electron releasing tendency and electron accepting tendency are both absolute in nature.

It means that their values of ionisation enthalpy and of electron gain enthalpy are not influenced by the presence of the other atoms. It is not possible to express these when the atoms are in the liquid or solid state due to the presence of inter atomic forces.

(b) Significance of ground state : The ground state means that in a particular atom, the electrons associated are in the lowest energy state i.e.. they neither lose nor gain electrons. This represents the normal energy state of an atom. Both ionisation enthalpy and electron gain enthalpy are generally expressed with respect to the ground state of an atom only.

Question 15.
The energy of an electron in the ground state of the hydrogen atom is – 2.18 × 10^-18 J. Calculate the ionisation enthalpy of atomic hydrogen in terms of J mol^-1.
Answer:
The ionisation enthalpy is for 1 mole atoms. Therefore, ground state energy of one mole atoms may be expressed as:
E_{(groun state)} = (-2.18 × 10^-18J) × (6.022 × lO²³ mol^-1) = – 1.312 × 10⁶Jmol^-1
Ionisation enthalpy = E_x – E_{groun state}
= 0 – (-1.312 × 10⁶ J mol^-1) = 1-312 × 10⁶ J mol^-1

Question 16.
Among the second period elements, the actual ionization enthalpies are in the order : Li<B<Be<C<O<N<F< Ne.
Explain why
(i) Be has higher ∆_iH₁ than B ?
(ii) O has lower ∆_i8H₁ than N and F ?
Answer:
(i)We know that the cations or positive ions are formed when the neutral atoms (generally metal atoms) lose electrons. They can do so only in the gaseous state since they are isolated and interatomic forces are the minimum. Now energy is needed to overcome the force of attraction between the nucleus and electrons so that the latter may be released. This is known as ionization enthalpy. It may be defined as :
The minimum amount of energy which is needed to remove the most loosely bound electron from a neutral isolated gaseous atom in its ground state to form a cation also in the gaseous state.
Atom (g) → Positive ion (g) + e^– ; IE (∆_iH₁)
(Cation)
Na (g) → Na⁺ (g) + e^– ; IE (∆_iH₁)
The energy required can also be expressed in the form of ionization potential which is the minimum amount of potential needed to remove the most loosly held electron from an isolated neutral gaseous atom in its ground state.

(ii) ∆_iH₁ values of the there elements present in second period are given :
N(1402 kJ mol^-1) ; O(1314 kJ mol^-1) ; F(1681 kJ mol^-1).
• ∆_iH₁ of O is expected to more than that of N but is actually lesser because the electronic configuration of N is more symmetrical as well as stable in comparison to O. For electronic configuration.
• ∆_iH₁ of O is less than that of F because the ionization enthalpy in general increases along a period.

Question 17.
∆_iH₁ value of Mg is more as compared to that of Na while its ∆_iH₂ value is less. Explain.
Answer:
The electronic configuration of the two elements and their first and second ionization enthalpies are.,given :

∆_iH₁ value of Mg is more than that of Na due to greater symmetry and smaller size. But ∆_iH₂ value of Na is higher because Na⁺ ion has the configuration of noble gas element neon while Mg⁺ ion does not have a symmetrical configuration.

Question 18.
What are the various factors due to which the ionisation enthalpy of the main group elements tends to decrease down the group ?
Answer:
Variation down a group. The ionization enthalpies of the elements decrease on moving from top to the bottom in any group. The trend is quite evident from the data given in the table 3.10. The ∆_iH₁ values of the members belonging to the alkali metals of group 1 have been given separately in the fig 3.10.

The decrease in ionization enthalpies down any group is because of the following factors :
(i) There is an increase in the number of the main energy shells (n) in moving from one element to the other.
(ii) There is also an increase in the magnitude of the screening effect due to the gradual increase in the number of inner electrons.
Although the nuclear charge qtlso increases down the group which is likely to result in increased ionization enthalpy but its effect is less pronounced than the two factors listed above. The net result is the decrease in ionization enthalpies in a group in the periodic table.

Question 19.
The first ionisation enthalpy values (kJ mol^-1) of group 13 elements are :

How would you account for their deviation from the general trend ?
Answer:
The decrease in ∆_iH₁ value from B to Al is quite expected because of the bigger size of Al atom. But the element Ga has ten 3d electrons present in the 3d sub-shell which do not screen as much as is done by s and p electrons. Therefore, there is an unexpected increase in the magnitude of effective nuclear charge resulting in increased 3 Classification of Elements and Periodicity in Properties value. The same explanation can be offered in moving from In to T1. The latter has fourteen 4f electrons with very poor shielding effect. This also results in unexpected increase in the effective nuclear charge resulting in inflated ∆_iH₁ value.

Question 20.
Which of the following pairs of elements would have a more negative electron gain enthalpy ?
(i) O or F
(ii) F or Cl
Answer:
(i) O or F. The negative electron gain enthalpy of F(∆_egH = – 328 kJ mol^-1) is more than that of O(∆_egH = – 141 kJ mol^-1). This is on account of smaller size of the F atom and its greater urge to have a noble gas configuration (only one electron is needed) as compared to oxygen (two electrons are needed).
(ii) F or Cl. The negative electron gain enthalpy of Cl (∆_egH = – 349 kJ mol^-1) is more than that of F(∆_egH = – 328 kJ mol^-1). Actually, the size of F atom (atomic radius = 72 pm) is quite small as compared to Cl atom (atomic radius = 99 pm). This results in greater crowding of electrons in small space around the nucleus in case of F and the attraction for outside electron is less as compared to Cl in which the atomic size is large and the crowding of electrons is less.

Question 21.
Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first ? Justify your answer.
Answer:
For oxygen atom ; O(g) + e^– → 0^–(g) ; (∆_egH) = – 141 kJ mol^-1
O^–(g) e^– → 0^2-(g) ; (∆_egH) = + 780 kJ mol^-1
The first electron gain enthalpy of oxygen is negative because energy is released when a gaseous atom accepts an electron to form monovalent anion. The second electron gain enthalpy is positive because energy is needed to overcome the force of repulsion between monovalent anion and the second incoming electron.

Question 22.
What is basic difference between the terms electron gain enthalpy and electronegativity ?
Answer:
Difference between electron gain enthalpy and electronegativity.
Both electronegativity and electron gain enthalpy are the electron attracting tendencies of the elements but they differ in many respects. The important points of distinction are listed.

Question 23.
How will you react to the statement that the electronegativity of N on Pauling scale is 3-0 in all the nitrogen compounds ?
Answer:
On Pauling scale, the electronegativity of nitrogen, (3-0) indicates that it is sufficiently electronegative. This is quite expected as well due to its very small size and its requirement to have only three electrons to achieve the noble gas configuration. But it is not correct to say that the electronegativity of nitrogen in all the compounds is 3. It depends upon its state of hydridisation in a particular compound. It may be noted that greater the percentage of ^-character, more will be the electronegativity of the element. Thus, the electronegativity of nitrogen increases in moving from sp3 hybridised orbitals to sp hydridised orbitals i.e., as sp³ < sp² < sp.

Question 24.
Describe the theory associated with the radius of an atom as it :
(a) gains an electron.
(b) loses an electron.
Answer:
(a) When an atom gains an electron. It changes into an anion (M + e^– → M^–). In
doing so, there is an increase in the size. Actually, there is no change in the atomic number or nuclear charge but there is an increase in the electrons by one. As a result, electrons experience less attraction towards the nucleus and the anionic radius or size increases. For example, the radius of F atom (72 pm) is less than that of F^– ion (136 pm).

(b) When an atom loses an electron. It changes into a cation (M → M⁺ + e^–).
In doing so, there is a decrease in size. In this case also, the nuclear charge is the same but the electrons have decreased by one. As a result, the electrons experience more attraction towards the nucleus and the cationic radius decreases. For example, radius of Na atom (157 pm) is less than that of Na⁺ ion (95 pm).

Question 25.
What are the major differences between metals and non-metals ?
Answer:
Metals and non-metals as the names suggest are quite different from one another. They differ in physical as well as chemical characteristics.
Distinction based upon physical properties. The main points of distinction are given :

Distinction based upon chemical properties. The main points of distinction are given :

Question 26.
Would you expect the ionization enthalpies of two isotopes of the same element to be same or different ? Justify your answer.
Answer:
The ionization enthalpies are related to the magnitude of nuclear charge as well as the electronic configuration of the elements. Since the isotopes of an element have same nuclear charge as well as electronic configuration, they have the same ionization enthalpies.

Question 27.
Use periodic table to answer the following questions :
(a) Identify the element with five electrons in the outer sub-shell.
(b) Identify the element that would tend to lose two electrons.
(c) Identify the element that would tend to gain two electrons.
Answer:
(a) Element belongs to nitrogen family (group 15) e.g. nitrogen
(b) Eletnent belongs to alkaline earth family (group 2) e.g. magnesium
(c) Element belongs to oxygen family (group 16) e.g. oxygen

Question 28.
The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas among the group 17 elements, it is F > Cl > Br > I. Explain.
Answer:
In group 1 elements (alkali metals) the reactivity of the metals is mainly due to the electron releasing tendency of their atoms, which is related to ionization enthalpy. Since ionization enthalpy decreases down the group, the reactivity of alkali metals increases.
In group 17 elements (halogens-non-metals), the reactivity is linked with electron accepting tendency of the members of the family. It is linked with electronegativity and electron gain enthalpy. Since both of them decrease down the group, the reactivity therefore decreases.
Thus, we conclude that the order of reactivity in two cases is different since in one case it is due to electron releasing tendency and it is because of electron accepting tendency in the other case.

Question 29.
Write the general electronic configuration of s, p, d and /block elements.
Answer:
s-Block elements : ns^1-2 (n varies from 2 to 7)
p-Block elements : ns² np^1-6 (n varies from 2 to 7)
d-Block elements : (n – 1) d^1-10 ns^1-2 (n varies from 4 to 7)
f-Block elements : (n – 2)f^1-14 (n – 1) d^0-1 ns² (n may have value either 6 or 7)

Question 30.
Assign the position of the elements having outer electronic configuration
(i) ns²p⁴ for n = 3
(ii) (n – 1 )d²ns² for n = 4 and
(iii) (n – 2) f⁷ (n – 1 )d¹ ns² for n = 6 in the periodic table.
Answer:
(i) The element is present in third period (n = 3) and belongs to group 16(10 + 2 + 4 = 16).
The element is sulphur : [1s² 2s² 2p⁶ 3s² 3p⁴].
(ii) The element is present in fourth period (.n = 4) and belongs to group 4 (2 + 2 = 4).
The element is titanium : [Ne] 3d² 4s².
(iii) The element is present in sixth period (n = 6) and belongs to group 3.
The element is gadolinium (Z = 64): [Xe] 4f⁷ 5d¹ 6s².

Question 31.
The first (∆_iH₁) and the second (∆_iH₂) ionization enthalpies (in kJ mol^-1) and the (∆_egH) electron gain enthalpy (in kJ mol¹) of a few elements are given below :

Which of the above elements is likely to be :
(a) the least reactive element.
(b) the most reactive metal.
(c) the most reactive non-metal.
(d) the least reactive non-metal.
(e) the metal which can form a stable binary halide of the formula MX₂ (X = halogen)
(f) the metal which can form a predominantly stable covalent halide of the formula MX (X = halogen) ?
Answer:
The nature and reactivity of the elements are guided by ionisation enthalpies and electron gain enthalpies. Based upon
the available data :
(a) The least reactive element is ‘V’. It has very high ∆_iH (2372 kJ mol^-1) and ∆_egH is (+ 48 kJ mol^-1) which indicate that it is a noble gas element. The value of ∆_egH suggests it to be helium (He).

(b) The most reactive metal is ‘II’. It has a very low ∆_iH equal to 419 kJ mol^-1 and its negative ∆_eg value is also quite small (- 48 kJ mol^-1). These values correspond to the alkali metal potassium (K).
In general, alkali metals are the most reactive metals in the periodic table.

(c) The most reactive non-metal is ‘III’. Since the element has very high ionization enthalpies (∆_iH₁ as well as ∆_iH₂) and very high negative electron gain enthalpy, it is expected to be a highly reactive non-metal. The values correspond to non-metal fluorine (F).

(d) The least reactive non-metal is ‘IV’. The character is revealed by the moderate values of ∆_iH₁ and ∆_iH₂. The value of electron gain enthalpy indicates that the element is iodine (I).

(e) The metal forming stable binary halide (MX₂) is ‘VI’. The formula MX₂ suggests that the metal belongs to the family of alkaline earth (group 2). The ∆_iH₁ and ∆_iH₂ values suggest it to be Mg (However, ∆_egH does not agree).

(f) The metal which can form a predominantly stable covalent halide of the formula. MX (halogen) is ‘I’. The metal is monovalent in nature and data suggests it to be lithium (Li) because ∆_iH₁ ] value is very small as compared to ∆_iH₂ value. In fact, in alkali metal lithium (Li), the monovalent cation (Li⁺) has the electronic configuration of the noble gas helium (He) The halide is LiX. The covalent character of LiX is due to the reason that the electron cloud of Li⁺ ion attracts the electron cloud of X-ion towards itself. As a result, the halide is of covalent nature.

Question 32.
Predict the formulas of the stable binary compounds that would be formed by the combination of the following
pairs of elements :
(a) Lithium and oxygen
(b) Magnesium and nitrogen
(c) Aluminium and iodine
(d) Silicon and oxy gen
(e) Phosphorus and fluorine
(f) Element 71 and fluorine
Answer:
(a) Li₂O (Lithium oxide)
(b) Mg₃N₂ (Magnesium nitride).
(c) AlI₃ (Aluminium iodide)
(d) SiO₂ (Silicon dioxide)
(e) PF₃ (Phosphorus trifluoride) or PF₅ (Phosphorus pentafluoride)
(f) The element with Z.
= 71 is lutetium (Lu) with electronic configuration [Xe] 4f¹⁴5d¹ 6s². With fluorine, it will form a binary compound of
formula LuF₃.

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NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom.

Question 1.
(i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.
Answer:
(i) Mass of an electron = 9.1 × 10^-28 g
9.1 × 10^-28 g is the mass of = 1 electron

(ii) One mole of electrons = 6.022 × 10²³ electrons
Mass of 1 electron = 9.1 × 10^-31 kg
Mass of 6.022 × 10²³ electrons = (9.1 × 10.31kg) × (6.022 × 10²³) = 5.48 × 10^-7 kg
Charge on one electron = 1.602 × 10^-19 coulomb
Charge on one mole electrons = 1.602 × 10^-19 × 6.022 × 10²³ = 9.65 × 10⁴ coulombs

Question 2.
(i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of ¹⁴C. (Assume that mass of a neutron = 1.675 × 10^-27kg).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH₃ at STP.
Will the answer change if the temperature and pressure are changed ?
Answer:
(i) One mole of methane (CH₄) has molecules = 6.022 × 10²³
No. of electrons present in one molecule of CH₄ = 6 + 4 = 10
No. of electrons present in 6.022 × 10²³ molecules of CH₄ = 6.022 × 10²³ × 10
= 6.022 × 10²⁴ electrons

(ii) Step I. Calculation of total number of carbon atoms
Gram atomic mass of carbon (C-14) = 14 g = 14 × 10³ mg
14 × 10³ mg of carbon (C-14) have atoms = 6.022 × 10²³

Step II. Calculation of total number and tatal mass of neutrons
No. of neutrons present in one atom (C-14) of carbon = 14 – 6 = 8
No. of neutrons present in 3-011 × 10²⁰ atoms (C-14) of carbon = 3.011 × 10²⁰ × 8
= 2.408 × 10²¹ neutrons
Mass of one neutron = 1.675 × 10^-27 kg
Mass of 2.408 × 10²¹ neutrons = (1.675X10^-27 kg) × 2.408 × 10²¹
= 4.033 × 10^-6 kg.

(iii) Step I. Calculation of total number ofNH₃ molecules
Gram molecular mass of ammonia (NH₃) = 17 g = 17 × 10³ mg
17 × 10³ mg of NH₃ have molecules = 6.022 × 10²³

Step II. Calculation of total number and mass of protons
No. of protons present in one molecule of NH₃ = 7 + 3 = 10 .
No. of protons present in 12.044 × 10²⁰ molecules of NH₃ = 12.044 × 10²⁰ × 10
= 1.2044 × 10²² protons
Mass of one proton = 1.67 × 10^-27 kg
Mass of 1.2044 × 10²² protons = (1.67 × 10^-27 kg) × 1.2044 × 10²²
= 2.01 × 10^-5 kg.
No, the answer will not change upon changing the temperature and pressure because only the number of protons and mass of protons are involved.

Question 3.
How many protons and neutrons are present in the following nuclei

Answer:

Question 4.
Write the complete symbol for the atom (X) with the given atomic number (Z) and atomic mass (A)
(i) Z = 17,A = 35
(ii) Z = 92, A = 233
(in) Z = 4, A = 9.
Answer:

Question 5.
Yellow light emitted from a sodium lamp has a wavelength (2) of 580 nm. Calculate the frequency (v) and wave number (v) of yellow light.
Answer:

Question 6.
Calculate the energy of each of the photons which
(i) correspond to light of frequency 3 × 1015 Hz
(ii) have wavelength of 0-50 A.
Answer:
(i) Energy of photon (E) = hv
h = 6.626 × 10^-34 J s ; v = 3 × 10¹⁵ Hz = 3 × 10¹⁵s^-1
∴ E = (6.626 × 10^-34 J s) × (3 × 10¹⁵ s^-1) = 1.986 × 10¹⁸ J
Energy of photon (E) = hv = \(\frac { hc }{ \lambda } \)
h = 6.626 × 10 34 J s; c = 3 × 10⁸ m s^-1 ;
λ= 0.50 Å = 0.5 × 10^-10 m.

Question 7.
Calculate the wavelength, frequency and wave number of light wave whose period is 2.0 × 10^-10 s.
Answer:

Question 8.
What is the number of photons of light with wavelength 4000 pm which provide 1 Joule of energy ?
Answer:
Energy of photon (E) = \(\frac { hc }{ \lambda } \)
h = 6.626 × 10^-34 Js, c = 3 × 10⁸ m s^-1, λ = 4000 pm = 4000 × 10^-12 = 4 × 10^-9 m

Question 9.
A photon of wavelength 4 × 10^-7 m strikes on metal surface ; the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon,
(ii) the kinetic energy of the emission
(iii) the velocity of the photoelectron. (Given 1 eV = 1.6020 × 10^-19 J).
Answer:

Question 10.
Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in k-J mol^-1.
Answer:

Question 11.
A 25 watt bulb emits monochromatic yellow light of wavelength 0.57 μm. Calculate the rate of emission of quanta per second.
Answer:

Question 12.
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 A. Calculate threshold frequency (v₀) and work function (W₀) of the metal.
Answer:

Question 13.
What is the wavelength of the light emitted when the electron in a hydrogen atom undergoes transition from the energy level with n = 4 to energy level n = 2 ? What is the colour corresponding to this wavelength ? (Given R_H = 109678 cm^-1)
Answer:

Question 14.
How much energy is required to ionise a hydrogen atom if an electron occupies n = 5 orbit ? Compare your answe r with the ionisation energy of H atom (energy required to remove the electron from n = 1 orbit)
Answer:

 

Question 15.
What is the maximum number of emission lines when the excited electron of a hydrogen atom in n = 6 drops to the ground state ?
Answer:
The maximum no. of emission lines = \(\frac { n(n – 1) }{ 2 }\) = \(\frac { 6(6 – 1) }{ 2 }\) =3 × 5 = 15

Question 16.
(i) The energy associated with first orbit in hydrogen atom is – 2.17 × 10^-18 J atom^-1. What is the energy associated with the fifth orbit ?
(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.
Answer:

(ii) For hydrogen atom ; r_n = 0.529 x n² Å
r₅ = 0.529 x (5)² = 13.225 Å = 1.3225 nm.

Question 17.
Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.
Answer:

Question 18.
What is the energy in joules required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of light emitted when the electron returns to the ground state ? The ground state electronic energy is – 2.18 × 11^-11 ergs.
Answer:

Question 19.
The electronic energy in hydrogen atom is given by E_n (-2.18 × 10^-18 s) / n²J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition ?
Answer:

Question 20.
Calculate the wavelength of an electron moving with a velocity of 2.05 × 10⁷ m s^-1.
Answer:

Question 21.
The mass of an electron is 9.1 × 10^-31 kg. If its kinetic energy is 3.0 × 10^-25 J, calculate its wavelength.
Answer:

Question 22.
Which of the following are iso-electronic species ?
Na⁺, K⁺, Mg²⁺, Ca²⁺, S^2-, Ar.
Answer:
Na⁺ and Mg²⁺ are iso-electronic species (have 10 electrons) K⁺, Ca²⁺ , S^2- are iso-electronic species (have 18 electrons)

Question 23.
(i) Write the electronic configuration of the following ions : (a) H (b) Na+ (c) 02~ (d) F^–.
(ii) What are the atomic numbers of the elements whose outermost electronic configurations are represented by :
(a) 3s¹ (b) Ip³ and (c) 3d⁶ ?
(iii) Which atoms are indicated by the following configurations ?
(a) [He]2s¹ (b) [Ne] 3s² 3p³ (c) [Ar] 4s² 3d¹.
Answer:
(i) (a) 1s²
(b) 1s² 2s² 2p⁶
(c) 1s²2s²2p⁶
(d) 1s²2s²2p⁶.
(ii) (a) Na (Z = 11) has outermost electronic configuration = 3s¹
(b) N (Z = 7) has outermost electronic configuration = 2p³
(c) Fe (Z = 26) has outermost electronic configuration = 3d⁶
(iii) (a) Li
(b) P
(c) Sc

Question 24.
What is the lowest value of n which allows ‘g’ orbital to exist ?
Answer:
The lowest value of l w’here ‘g’ orbital can be present = 4
The lowest value of n where ‘g’ orbital can be present = 4+1=5.

Question 25.
An electron is in one of the 3d orbitals. Give the possible values of n, l and nil for the electron.
Answer:
For electron in 3d orbital, n = 3, l = 2, mi = -2, -1, 0, +1, +2.

Question 26.
An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.
Answer:
No. of protons in a neutral atom = No. of electrons = 29
Electronic configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹.

Question 27.
Give the number of electrons in the species : H₂⁺, H₂ and 0₂⁺.
Answer:
H₂⁺ = one ; H₂ = two ; 0₂⁺ = 15

Question 28.
(i) An atomic orbital has n = 3. What are the possible values of l and m_l ?
(ii) List the quantum numbers m_l and l of electron in 3rd orbital.
(iii) Which of the following orbitals are possible ?
1p, 2s, 2p and 3f.
Answer:
(i) For n = 3; l = 0, 1 and 2.
For l = 0 ; m_l = 0
For l = 1; m_l = +1, 0, -1
For l = 2 ; m_l = +2, +1,0, +1, + 2
(ii) For an electron in 3rd orbital ; n = 3; l = 2 ; m_l can have any of the values -2, -1, 0,
+ 1, +2.
(iii) 1p and 3f orbitals are not possible.

Question 29.
Using s, p and d notations, describe the orbitals with follow ing quantum numbers :
(a) n = 1, l = 0
(b) n = 4, l = 3
(c) n = 3, l = 1
(d) n = 4, l = 2
Answer:
(a) 1s orbital
(b) 4f orbital
(c) 3p orbital
(d) 4d orbital

Question 30.
From the following sets of quantum numbers, state which are possible. Explain why the others are not possible.
(i) n = 0, l = 0, m_l = 0, m_s = +1/2
(ii) n = 1, l = 0, m_l = 0, m_s – – 1/2
(iii) n = 1, l = 1, m_l = 0, m_s= +1/2
(iv) n = 1, l = 0, m_l = +1, m_s= +1/2
(v) n = 3, l = 3, m_l = -3, m_s = +1/2
(vi) n = 3, l = 1, m_l = 0, m_s= +1/2
Answer:
(i) The set of quantum numbers is not possible because the minimum value of n can be 1 and not zero.
(ii) The set of quantum numbers is possible.
(iii) The set of quantum numbers is not possible because, for n = 1, l can not be equal to 1. It can have 0 value.
(iv) The set of quantum numbers is not possible because for l = 0. mt cannot be + 1. It must be zero.
(v) The set of quantum numbers is not possible because, for n = 3, l ≠ 3.
(vi) The set of quantum numbers is possible.

Question 31.
How many electrons in an atom may have the following quantum numbers ?
(a) n = 4 ; m_s = -1/2
(b) n = 3, l = 0.
Answer:
(a) For n = 4
Total number of electrons = 2n² = 2 × 16 = 32
Half out of these will have ms = —1/2
∴ Total electrons with ms (-1/2) = 16
(b) For n = 3
l= 0 ; m_l = 0, m_s +1/2, -1/2 (two e^–)

Question 32.
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.
Answer:

Thus, the circumference (2πr) of the Bohr orbit for hydrogen atom is an integral multiple of the de Broglie wavelength.

Question 33.
Calculate the number of atoms present in :
(i) 52 moles of He
(ii) 52 u of He
(iii) 52 g of He.
Answer:

Question 34.
Calculate the energy required for the process :
He⁺fe) → He²⁺(g) + e^–
The ionisation energy’ for the H atom in the ground state is 2.18 × 10^-18  J atom^-1
Answer:

For H atom (Z = 1), E_n =2.18 × 10^-18 × (l)² J atom^-1 (given)
For He⁺ ion (Z = 2), E_n =2.18 × 10^-18 × (2)² = 8.72 × 10^-18 J atom^-1 (one electron species)

Question .35.
If the diameter of carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across a length of a scale of length 20 cm long.
Answer:

Question 36.
2 × 10⁸ atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.
Answer:
The length of the arrangement = 2.4 cm
Total number of carbon atoms present = 2 ×10⁸

Radius of each carbon atom = \(\frac{ 1 }{ 2 }\)(1.2 × 10^-8) = 6.0 × 10^-9cm = 0.06 nm

Question 37.
The diameter of zinc atom is 2.6 Å. Calculate :
(a) the radius of zinc atom in pm
(b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side length wise.
Answer:
(a) Radius of zinc atom =\(\frac { 2.6\AA }{ 2 } \)= 1.3 Å = 1.3 × 10^-10m = 130 × 10^-12m = 130 pm
(b) Length of the scale = 1.6 cm = 1.6 × 10¹⁰ pm
Diameter of zinc atom = 260 pm

Question 38.
A certain particle carries 2.5 x 10^-16 C of static electric charge. Calculate the number of electrons present in it.
Answer:

Question 39.
In Millikan’s experiment, the charge on the oil droplets was found to be – 1.282 x 10^-18C. Calculate the number of electrons present in it.
Answer:

Question 40.
In Rutherford experiment, generally the thin foil of heavy atoms like gold, platinum etc. have been used to be bombarded by the a-particles. If a thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results ?
Answer:
We have studied that in the Rutherford’s experiment by using heavy metals like gold and platinum, a large number of a-particles sufferred deflection while a very few had to retrace their path.

If a thin foil of lighter atoms like aluminium etc. be used in the Rutherford experiment, this means that the obstruction offered to the path of the fast moving a-particles will be comparatively quite less.

As a result, the number of a-particles deflected will be quite less and the particles which are deflected back will be negligible.

Question 41.
Symbols \(_{ 35 }^{ 79 }{ Br }\) and ⁷⁹Br can be written whereas symbols \(_{ 79 }^{ 35 }{ Br }\) and ₃₅Br are not accepted. Answer in brief.
Answer:
In the symbol \(_{ A }^{ B }{ X }\) of an element :
A denotes the atomic number of the element
B denotes the mass number of the element.
The atomic number of the element can be identified from its symbol because no two elements can have the atomic number. However, the mass numbers have to be mentioned in order to identify the elements. Thus,
Symbols \(_{ 35 }^{ 79 }{ Br }\) and ⁷⁹Br are accepted because atomic number of Br will remain 35 even if not mentioned. Symbol \(_{ 79 }^{ 35 }{ Br }\) is not accepted because atomic number of Br cannot be 79 (more than the mass number = 35). Similarly, symbol 35Br cannot be accepted because mass number has to be mentioned. This is needed to differentiate the isotopes of an element.

Question 42.
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the symbol to the element.
Answer:
An element can be identified by its atomic number only. Let us find the atomic number.
Let the number of protons = x
Number of neutrons = x + [\(\frac { x\times 31.7 }{ 100 } \) = (x × 0.317x)
Now, Mass no. of element = no. of protons =no. neutrons
81 = x + x + 0-317 x = 2.317 x or x = \(\frac { 81 }{ 2.317 } \) = 35
∴ No. of protons = 35, No. of neutrons = 81 – 35 =46
Atomic number of element (Z) = No. of protons = 35
The element with atomic number (Z) 35 is bromine \(_{ 35 }^{ 81 }{ Br }\).

Question 43.
An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11 -1% more neutrons than the electrons, find the symbol of the ion.
Answer:
Let the no. of electron in the ion = x
∴ the no. of protons = x – 1 (as the ion has one unit negative charge)
and the no. of neutrons = x + \(\frac { x\times 11.1 }{ 100 } \) = 1.111 x
Mass no. or mass of the ion = No. of protons + No. of neutrons
(x – 1 + 1.111 x)
Given mass of the ion = 37
∴ x- 1 + 1.111 x = 37 or 2.111 x = 37 + 1 = 38
x = \(\frac { 38 }{ 2.111 } \) = 18
No. of electrons = 18 ; No. of protons = 18 – 1 = 17
Atomic no. of the ion = 17 ; Atom corresponding to ion = Cl
Symbol of the ion = \(_{ 17 }^{ 37 }{ Cl }\)^–

Question 44.
An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign symbol to the ion.
Answer:
Let the no. of electrons in the ion = x
∴ the no. of the protons = x + 3 (as the ion has three units positive charge)
and the no. of neutrons = x + \(\frac { x\times 31.7 }{ 100 } \) = xc + 0.304 x
Now, mass no. of ion = No. of protons + No. of neutrons
= (x + 3) + (x + 0.304x)
∴ 56 = (x + 3) + (x + 0.304x) or 2.304x = 56 – 3 = 53
x = \(\frac { 53 }{ 2.304 } \) = 23
Atomic no. of the ion (or element) = 23 + 3 = 26
The element with atomic number 26 is iron (Fe) and the corresponding ion is Fe³⁺.

Question 45.
Arrange the following type of radiations in increasing order of wavelength :
(a) radiation from microwave oven
(b) amber light from traffic signal
(c) radiation from FM radio
(d) cosmic rays from outer space and
(e) X-rays.
Answer:
Cosmic rays < X-rays < amber colour < microwave < FM

Question 46.
Nitrogen laser produces a radiation of wavelength of 337.1 nm. If the number of photons emitted is 5.6 x 10²⁴, calculate the power of this laser.
Answer:

Question 47.
Neon gas is generally used in sign boards. If it emits strongly at 616 nm, calculate :
(a) frequency of emission (b) the distance travelled by this radiation in 30s (c) energy of quantum (d) number of quanta present if it produces 2 J of energy.
Answer:

Question 48.
In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 x 10^-18 J from the radiations of 600 nm, calculate the number of photons received by the detector.
Answer:

Question 49.
Life times of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 x 10¹⁵, calculate the energy of the source.
Answer:
Time duration (t) = 2 ns = 2 x 10^-9 s

Energy of one photon, E = hv = (6.626 x 10^-34 Js) x (10⁹/2 s^-1) = 3.25 x 10^-25J
No. of photons = 2.5 x 10⁵
∴ Energy of source = 3.3125 x 10^-25 J x 2.5 x 10¹⁵ = 8.28 x 10^-10 J

Question 50.
The longest wavelength doublet absorption transition is observed at 589 nm and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.
Answer:

Question 51.
The work function for cesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the cesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron
Answer:

Question 52.
Following results are observed when sodium metal is irradiated with different wavelengths. Calculate threshold wavelength.

Answer:

Question 53.
The ejection of the photoelectrons from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.
Answer:

Question 54.
If the photon of the wavelength 150 pm strikes an atom, one of its inner bound electrons is ejected out with a velocity of 1.5 x 10⁷ m s^-1. Calculate the energy with which it is bound to the nucleus.
Answer:

Question 55.
Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 x 10¹⁵ (Hz) [1/3² – 1 /n²]
Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.
Answer:

Question 56.
Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
Answer:

Question 57.
Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 x 10⁶ m s^-1, calculate de Brogile wavelength associated with this electron.
Answer:

Question 58.
Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.
Answer:

Question 59.
If the velocity of the electron in Bohr’s first orbit is 2.19 x 10⁶ m s^-1, calculate the de Brogile wavelength associated with it.
Answer:

Question 60.
The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 x 10⁵ m s^-1. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.
Answer:

Question 61.
If the position of the electron is measured within an accuracy of ± 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is \(\frac { h }{ 4\pi } \) x 0.05 nm. Is there any problem in defining this value ?
Answer:

Since actual momentum is smaller than the uncertainty in measuring momentum, therefore, the momentum of electron can not be defined.

Question 62.
The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. List if any of these combination(s) has/have the same energy
(i) n = 4, l = 2, m_l = -2, m_s = -1/2
(ii) n = 3, l = 2, m_l = 1, m_s = +1/2
(iii) n = 4, l = 1, m_l = 0, m_s = +1/2
(iv) n = 3, l = 2, m_l = -2, m_s = -111
(v) n = 3, l = l, m_l = -1, m_s = +1/2
(vi) n = 4, l = 1, m_l = 0, m_s = +1/2
Answer:
The electrons may be assigned to the following orbitals :
(i) 4d
(ii) 3d
(iii) 4p
(iv) 3d
(v) 3p
(vi) 4p.
The increasing order of energy is :
(v) < (ii) = (iv) < (vi), = (iii) < (i)

Question 63.
The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electrons in 4p orbital. Which of these electrons experiences lowest effective nuclear charge ?
Answer:
4p electron experiences lowest effective nuclear charge because of the maximum magnitude of screening or shielding effect. It is farthest from the nucleus.

Question 64.
Among the following pairs of orbitals, which orbital will experience more effective nuclear charge (i) 2s and 3s (ii) 4d and 4f (iii) 3d and 3p ?
Answer:
Please note that greater the penetration of the electron present in a particular orbital towards the nucleus, more will be the magnitude of the effective nuclear charge. Based upon this,
(i) 2s electron will experience more effective nuclear charge.
(ii) 4d electron will experience more effective nuclear charge.
(iii) 3p electron will experience more effective nuclear charge.

Question 65.
The unpaired electrons in A1 and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus ?
Answer:
Configuration of the two elements are :
A1 (Z = 13) : [Ne]¹⁰3s²3p¹ ; Si (Z = 14) : [Ne] 103s23p2
The unpaired electrons in silicon (Si) will experience more effective nuclear charge because the atomic number of the element Si is more than that of A1.

Question 66.
Indicate the number of unpaired electrons in :
(a) P (b) Si (c) Cr (d) Fe and (e) Kr.
Answer:
(a) P (z=15) : [Ne]¹⁰3s²3p³ ; No. of unpaired electrons = 3
(b) Si (z=14) : [Ne]¹⁰3s²3p² ; No. of unpaired electrons = 2
(c) Cr (z=24): [Ar]¹⁸4s¹3d⁵ ; No. of unpaired electrons = 6
(d) Fe (z=26): [Ar]¹⁸4s²3d⁶ ; No. of unpaired electrons = 4
(e) Kr (z=36) : [Ar]¹⁸4s²3d¹⁰4p⁶ ; No. of unpaired electrons = Nil.

Question 67.
(a) How many sub-shells are associated with n = 4 ?
(b) How many electrons will be present in the sub-shells having ms value of -1/2 for n = 4 ?
Answer:
(a) For n = 4 ; No. of sub-shells = (l = 0, l = 1, l = 2, l = 3) = 4.
(b) Total number of orbitals which can be present = n² = 4² = 16.
Each orbital can have an electron with m_s = – 1/2 -‘. Total no. of electrons with m, = – 1/2 is 16.

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