Class 11

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 8 Redox Reactions Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 8 Redox Reactions

PSEB 11th Class Chemistry Guide Redox Reactions InText Questions and Answers

Question 1.
Assign oxidation numbers to the underlined elements in each of the following species:

Answer:
(a) Let the oxidation number of P be x.
Oxidation number of Na = +1
Oxidation number of H = +1
Oxidation number of O = – 2
In neutral compounds, the sum of the oxidation numbers of all the atoms is zero.
1 (+1) + 2 (+1) +1 00 + 4 (-2) = 0
1 + 2 + x – 8 = 0
3 + x + (-8) – 0
x = 8 – 3
⇒ x = + 5
Hence, the oxidation number of P is +5.

(b) Let the oxidation number of S be x.

1 (+1) +1 (+1) +1 (x) + 4 (-2) = 0
⇒ 1 + 1 + x – 8 = 0
⇒ x = + 6
Hence, the oxidation number of S is +6.

(c) Let the oxidation number of P be x.

4 (+1) + 2 (x) + 7 (-2) = 0
⇒ 4 + 2x – 14 = 0
⇒ 2x = +10
⇒ x = + 5
Hence, the oxidation number of P is + 5.

(d) Let the oxidation number of Mn is x.

2 (+1) + x + 4 (-2) = 0
⇒ 2 + x — 8 = 0
⇒ x = + 6
Hence, the oxidation number of Mn is + 6.

(e) Let the oxidation number of O be x.

1 (+ 2) + 2 (x) = 0
⇒ 2 + 2x = 0
⇒ x = – 1
Hence, the oxidation number of O is -1.

(f) Let the oxidation number of B be x.

1 (+1)+1 (x) + 4 (-1) = 0
⇒ 1 + x – 4 = 0
⇒ x = + 3
Hence, the oxidation number of B is + 3.

(g) Let the oxidation number of S is x.

2 (+1) + 2 (x) + 7 (-2) = 0
⇒ 2 + 2x -14 = 0
⇒ 2x = +12
x = +6
Hence, the oxidation number of S is + 6.

(h) Let the oxidation number of S be x.

1 (+1) +1 (+ 3) + 2 (x) + 8 (-2) +12 (2 x 1 + (-2)) = 0
⇒ 1 + 3 + 2x -16 + 24 – 24 = 0
⇒ 2x = 12
x = + 6
Hence, the oxidation number of S is + 6.

Question 2.
What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?


Answer:
(a) In

the oxidation number (O. N.) of K is +1. Hence, the average oxidation number of I is \(\frac{-1}{3}\). However, O.N. cannot be fractional.
Therefore, we will have to consider the structure of KI₃ to find the oxidation states.
In a KI₃ molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.

Hence, in a KI₃ molecule, the O.N. of the two I atoms forming the I₂ molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is -1.

(b) Let the oxidation number of S be x.

2 (+1) + 4 (x) + 6 (-2) = 0
=» 2 + 4x – 12 = 0
⇒ 4x = +10
⇒ x = + 2 \(\frac{1}{2}\)
However, O.N. cannot be fractional. Hence S must be present in different oxidation states in the molecule.

The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0.

(c) Let the oxidation number of Fe be x.

3(x) + 4(-2) = 0
3x – 8 = 0
x = \(+\frac{8}{3}\)
However O.N. cannot’be fractional. Here, one of the three Fe atoms exhibits the O.N. of +2 and the other two Fe atoms exhibit the O.N. of+3

(d) Let oxidation number of C be x.

2 (x) + 6 (+1) + 1 (-2) = 0
2x + 4 = 0
x = -2
Hence, the O.N. of C is – 2

(e) Let the oxidation number of C be x.

2 (x) + 4 (+1) + 2 (-2) = 0
2x = 0
x = 0
Therefore, the average oxidation number of C is zero.
Let us consider the structure of CH₃COOH

Oxidation number of atom = 1(+1) + x+1(-2) +1(-1) = 0
x = +2
Similarly, oxidation number of C₂ atom
3(+1) + x+ 1(-1) = 0
x = -2.

Question 3.
Justify that the following reactions are redox reactions:
(a) CuO (s) + H₂(g) → Cu(s) + H₂O(g)
(b) Fe₂O₃ (s) + 3CO(g) → 2Fe(s) + 3CO₂(g)
(c) 4BCl₃ (g) + 3LiAlH₄ (s) → 2B₂H₆(g) + 3LiCl (s) + 3 AlCl₃(S)
(d) 2K (s) + F₂ (g) → 2K⁺F^–(s)
(e) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O (g)
Answer:
(a) CuO (s) + H₂(g) → Cu(s) + H₂O(g)
Let us write the oxidation number of each element involved in the given reaction as :

Here, the oxidation number of Cu decreases from +2 in CuO to 0 in Cu i.e., CuO is reduced to Cu. Also, the oxidation number of H increases from 0 in H₂ to +1 in H₂O i.e., H₂ is oxidized to H₂O. Hence, this reaction is a redox reaction.

(b) Fe₂O₃ (s) + 3CO(g) → 2Fe(s) + 3CO₂(g)
Let us write the oxidation number of each element in the given reaction as:

Here, the oxidation number of Fe decreases from +3 in Fe₂O₃ to 0 in Fe i.e., Fe₂O₃ is reduced to Fe. On the other hand, the oxidation number of C increases from +2 in CO to +4 in CO₂ i.e., CO is oxidized to CO₂.
Hence, the given reaction is a redox reaction.

(c) 4BCl₃ (g) + 3LiAlH₄ (s) → 2B₂H₆(g) + 3LiCl (s) + 3 AlCl₃(S)
The oxidation number of each elements in the given reaction can be represented as:

In this reaction, the oxidation number of B decreases from +3 in BCl₃ to -3 in B₂H_6- i.e., BCl₃ is reduced to B₂H₆. Also, the oxidation number of H increases from -1 in LiAlH₄ to +1 in B₂H₆ i.e., LiAlH₄ is oxidized to B₂H₆. Hence, the given reaction is a redox reaction.

(d) 2K (s) + F₂ (g) → 2K⁺F^–(s)
The oxidation number of each element in the given reaction can be represented as:

In this reaction, the oxidation number of K increases from 0 in K to +1 in KF i.e., K is oxidized to KF. On the other hand, the oxidation number of F decreases from 0 in F₂ to -1 in KF i.e., F₂ is reduced to KF.
Hence, the above reaction is a redox reaction.

(e) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O (g)
The oxidation number of each elements in the given reaction can be represented as:

Here, the oxidation number of N increases from -3 in NH₃ to +2 in NO. On the other hand, the oxidation number of O₂ decreases from 0 in O₂ to -2 in NO and H₂O i.e.,O₂ is reduced. Hence, the given reaction is a redox reaction.

Question 4.
Fluorine reacts with ice and results in the change:
H₂O(S) + F₂(g) → HF(g) + HOF(g)
Justify that this reaction is a redox reaction.
Ans. Let us write the oxidation number of each atom involved in the given reaction above its symbol as:

Here, we have observed that the oxidation number of F increases from 0 in F₂ to +1 in HOF. Also, the oxidation number decreases from 0 in F₂ to -1 in HF. Thus, in the above reaction, F is both oxidized and reduced. Hence, the given reaction is a redox reaction.

Question 5.
Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) and \(\mathrm{NO}_{3}^{-}\). Suggest structure of these compounds. Count for the fallacy.
Answer:

2 (+1) +1 (x) + 5 (-2) = 0
⇒ 2 + x -10 = 0
⇒ x = + 8
However, the O.N. of S cannot be +8. S has six valence electrons. Therefore, the O.N. of S cannot be more than +6. The structure of H2S05 is shown as follows :

Now, 2 (+1) +1 (x) + 3 (-2) + 2 (-1) – 0
⇒ 2 + x – 6 – 2 = 0
⇒ x = + 6
Therefore, the O.N. of S is +6.

2 (x) + 7 (-2) = – 2 ⇒ 2x -14 = – 2
⇒ x = + 6
The structure of Cr2Oy is shown as follows:

Let the oxidation number of each Cr atom be
4(-2) + (-2) +1(-2) + 2x = 0
– 8 – 2 – 2 + 2x = 0
2x = +12
x = +6

Oxidation number of Cr in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) is same.
Hence, there is no fallacy.

1 (x) + 3(-2) = -1
= x – 6 = -1
x = +5
The structure of \(\mathrm{NO}_{3}^{-}\) is shown as follows:

Let the oxidation number of N be x.
1(-1) + x + 1(-2) + 1(-2) = 0
x = + 5
Oxidation number of N in \(\mathrm{NO}_{3}^{-}\) ion is same.
Hence, there is no fallacy.

Question 6.
Write formulas for 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:
(a) Mercury (II) chloride: Hg (II) Cl₂
(b) Nickel (II) sulphate: Ni (II) SO₄
(c) Tin (IV) oxide : Sn (IV) O₂
(d) Thallium (I) sulphate: Tl₂ (I) SO₄
(e) Iron (III) sulphate: Fe₂(III) (SO₄)₃
(f) Chromium (III) oxide: Cr₂(III)O₃

Question 7.
Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.
Answer:
The substances where carbon can exhibit oxidation states from -4 to +4 are listed in the following table :

Substance	Formula	Oxidation State of Carbon
Methane	CH4	-4
Ethane	C2H6	-3
Ethene	C2H4	-2
Ethyne	C2H2	-1
Dichloromethane	CH2Cl2	0
Hexachlorobenzone	C6Cl6	+1
Carbon monoxide	CO	+2
Oxalic acid	(COOH)2	+3
Carbon dioxide	CO2	+4

The substances where nitrogen can exhibit oxidation sates from -3 to +5 are listed in in the following table.

Substance	Formula	Oxidation State of Nitrogen
Ammonia	NH3	-3
Hydrazine	N2H4	-2
Hydride	N2H2	-1
Dinitrogen gas	N2	0
Nitrous oxide	N2O	+1
Nitric oxide	NO	+2
Dinitrogen trioxide	N2O3	+3
Nitrogen dioxide	NO2	+4
Nitrogen pentoxide	N2O5	+5

Question 8.
While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?
Answer:
(i) In sulphur dioxide (SO₂), the oxidation number (O.N.) of S is +4 and the range of the O.N. that S can have is from +6 to -2.
Therefore, SO₂ can act as an oxidising as well as reducing agent.
(ii) In hydrogen peroxide (H₂O₂), the O.N. of O is -1 and the range of the O.N. that O can have is from 0 to -2. O can sometimes also attain the oxidation numbers +1 and +2. Hence, H₂O₂ can act as an oxidising as well as reducing agent.
(iii) In ozone (O₃) the O.N. of O is zero and the range of the O.N. that O can have is from 0 to -2. Therefore, the O.N. of O can only decrease in this case. Hence,O₃ acts only as an oxidant.
(iv) In nitric acid (HNO₃) the O.N. of N is +5 and the range of the O.N. that N can have is from +5 to -3. Therefore, the O.N. of N can only decrease in this case. Hence, HNO₃ acts only as an oxidizing agent.

Question 9.
Consider the reactions:
(a) 6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆ (aq) + 6O₂(g)
(b) O₃(g) + H₂O₂(Z) → H₂O(1) + 2O₂(g)
Why it is more appropriate to write these reactions as:
(a) 6CO₂ (g) + 12H₂O(1) → C₆H₁₂O₆ (aq) + 6H₂O(l) + 6O₂ (g)
(b) O₃(g) + H₂O₂(Z) → + H₂O(Z) + O₂(g) + O₂(g)
Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.
Answer:
(a) The process of photosynthesis involves two steps :
Step 1: H₂O decomposes to give H₂ and O₂.
2H₂O(l) → 2H₂(g) + O₂(g)
Step 2: The H₂ produced in step 1 reduces CO₂ thereby producing glucose (C₆H₁₂O₆) and H₂O.
6CO₂(g) + 12H₂(g) → C₆H₁₂O₆(S) + 6H₂O(l)
Now, the net reaction of the process is given as:

It is more appropriate to write the reaction as given above because water molecules are also produced in the process of photosynthesis.
The path of this reaction can be investigated by using radioactive H₂O¹⁸ in place of H₂O

(b) O₂ is produced from each of the two reactants O₃ and H₂O₂. For this reason O₂ is written twice.
The given reaction involves two steps. First O₃ decomposes to form O₂ and O. In the second step H₂O₂ reacts with the O produced in the first step thereby producing H₂O and O₂.

The path of this reaction can be investigating by using \(\mathrm{H}_{2} \mathrm{O}_{2}^{18}\) or \(\mathrm{O}_{3}^{18}\).

Question 10.
The compound AgF₂ is unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why?
Answer:
AgF₂ → Ag + F₂
The oxidation state of Ag in Ag F₂ is +2. But +2 is an unstable oxidation state of Ag. Therefore, whenever Ag F₂ is formed, silver readily accepts an electron to form Ag⁺. This helps to bring the oxidation state of Ag down from +2 to a more stable state of +1. As a result, Ag F₂ acts as a very strong oxidizing agent.

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 two illustrations.
Answer:
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. This can be illustrated as follows:
(i) P₄ and F₂ are reducing and oxidising agents respectively.
If an excess of P₄ is treated with F₂ then PF₃ will be produced, where in the oxidation number (O.N.) of P is +3.

However, if P₄ is treated with an excess of F₂, then PF₅ will be produced, wherein the O.N. of P is +5.

(ii) K acts as a reducing agent, whereas O₂ is an oxidising agent.
If an excess of K reacts with O₂, then K₂O will be formed, wherein the O.N. of O is -2.

However, if K reacts with an excess of O₂, then K₂O₂ will be formed, wherein the O.N. of O is -1.

Question 12.
How do you account for the following observations?
(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 colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why?
Answer:
(a) Oxidation of toluene to benzoic acid in acidic medium.

Oxidation of toluene to benzoic acid in basic and neutral medium.

On industrial scale, alcoholic potassium permanganate is preferred to acidic or alkaline potassium permanganate because in the presence of alcohol, both the reactants KMnO₄ and C₆H₅CH₃ are mixed very well and form homogeneous solution and in homogeneous medium reaction takes place faster than in heterogeneous medium. Further more in neutral medium, OH^– ions are produced in the reaction itself.

(b)

HCl is a weak reducing agent. k cannot reduce H₂SO₄ to SO₂ that’s why pungent smelling gas HCl is obtained.

HBr is a strong reducing agent, it reduces H₂S0₄ to SO₂ and is itselfoxidised to Br₂. That’s why we get red vapours of bromine when conc.
H₂SO₄ reacts with inorganic mixture containing bromide salt.

Question 13.
Identify the substance oxidised, reduced, oxidising agent andreducing agent for each of the following reactions:
(a) 2AgBr(s) + C₆H₆O₂(aq) → 2Ag(s) + 2HBr(aq) + C₆H₄O₂(aq)
(b) HCHO(l) +2 [Ag (NH₃)₂]⁺ (aq) + 3OH^–(aq) → 2Ag(s) + HCOO^–(aq) + 4NH₃(aq) + 2H₂O (l)
(c) HCHO(l) + 2Cu²⁺ (aq) + 5 OH^– → Cu₂O (s) +HCOO^–(aq) + 3H₂O(l)
(d) N₂H₄ (Z) + 2H₂O₂ (1) → N₂ (g) + 4H₂O (l)
(e) Pb (s) +PbO₂ (s) + 2H₂SO₄ (ag) → 2PbSO₄(s) + 2H₂O(l)
Answer:
(a) Oxidised substance -C₆H₆O₂
Reduced substance – AgBr
Oxidising agent – AgBr
Reducing agent -C₆H₆O₂

(b) Oxidised substance – HCHO
Reduced substance – [Ag (NH₃)₂]⁺
Oxidising agent – [Ag (NH₃)₂]⁺
Reducing agent – HCHO

(c) Oxidised substance – HCHO
Reduced substance -Cu²⁺
Oxidising agent – Cu²⁺
Reducing agent – HCHO

(d) Oxidised substance – N₂H₄
Reduced substance -H₂O₂
Oxidising agent-H₂O₂
Reducing agent -N₂H₄

(e) Oxidised substance – Pb
Reduced substance – PbO₂
Oxidising agent – Pb O₂
Reducing agent – Pb

Question 14.
Consider the reactions:

Why does the same reductant, thiosulphate react differently with iodine and bromine?
Answer:
The average oxidation number (O.N.) of S in \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) is +2. Being a stronger oxidising agent than I₂, Br₂ oxidises \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-}\) in which the O.N. of S is +6. However I₂ is a weak oxidising agent. Therefore, it oxidises \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to \(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) in which the average O.N. of S is only +2.5. As a result, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) reacts differently with iodine and bromine. ,

Question 15.
Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.
Answer:
(i) F₂ can oxidize Cl^– to Cl₂, Br^– to Br₂ and I^– to I₂ as:
F₂(aq) + 2Cl^–(s) → 2F^–(aq) + Cl₂(g)
F₂(aq) + 2Br^–(aq) → 2F^–(aq) + Br₂(7)
F₂(aq) + 2I^–(aq) → 2F^–(aq) + I₂(s)

On the other hand, Cl₂, Br₂ and I₂ cannot oxidize F^– to F₂. The oxidizing power of halogens increases in the order of I₂ < Br₂ < Cl₂ < F₂. Hence fluorine is the best oxidant among halogens.

(ii) HI and HBr can reduce H₂SO₄ to SO₂, but HCl and HF cannot. Therefore HI and HBr are stronger reductants than HCl and HF.

2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O
2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O

Again, I- can reduce Cu²⁺ to Cu⁺ but Br^– cannot.

4I^–(aq) + 2Cu²⁺(aq) → Cu₂I₂(s) + I₂(aq)

Hence, hydroiodic acid is the best reductant among hydrohalic compounds.
Thus, the reducing power of hydrohalic acids increases in the order of
HF < HCl < HBr < HI.

Question 16.
Why does the following reaction occur?
\(\mathrm{XeO}_{6}^{4-}\)(aq) + 2F^– (aq) + 6H⁺(aq) → XeO₃(g) +F₂(g) + 3H₂O(i)
What conclusion about the compound Na₄XeO₆ (of which \(\mathrm{XeO}_{6}^{4-}\) is a part) can be drawn from the reaction.
Answer:
The given reaction occurs because \(\mathrm{XeO}_{6}^{4-}\) oxidizes being an oxidizing agent and F^– reduces \(\mathrm{XeO}_{6}^{4-}\)

In this reaction, the oxidation number (O.N.) of Xe decreases from +8 in \(\mathrm{XeO}_{6}^{4-}\) to +6 in XeO₃ and the O.N. of F increases from -1 in F^– to 0 in F₂.
Hence, we can conclude that Na₄XeO₆ is a strong oxidizing agent thanF^–.

Question17.
Consider the reactions:
(a) H₃PO₂ (aq) + 4 AgNO₃ (aq) + 2H₂O(l) → H₃PO₄ (aq) + 4Ag(s) + 4HNO₃ (aq)
(b) H₃PO₂ (oqr) + 2CuSO₄ (aq) + 2 H₂O(0 → H₃PO₄ (aq) + 2Cu(s) +H₂SO₄ (aq)
(c) C₆H₅CHO(l) + 2[Ag (NH₃)₂]+ (aq) + 3OH^– (aq) → C₆H₅COO^–(aq) + 2Ag(s) + 4NH₃ (aq) + 2H₂O(l)
(d) C₆H₅CHO(l) + 2Cu²⁺(aq) + 5OH^–(aqf) → No change observed.
What inference do you draw about the behaviour of Ag⁺ and Cu²⁺ from these reactions?
Answer:
Ag⁺ and Cu²⁺ act as an oxidizing agents in reactions (a) and (b) respectively.
In reaction (c), Ag⁺ oxidizes C₆H₅CHO to C₆H₅COO^–, but in reaction (d) Cu²⁺ cannot oxidize C₆H₅CHO. Hence, we can say that Ag⁺ is a stronger oxidising agent than Cu²⁺.

Question 18.
Balance the following redox reactions by ion-electron method:


Answer:
(a) Step 1 : The two half reactions involved in the given reaction are:
Oxidation half reaction :

Reduction half reaction:

Step 2 : Balancing I in the oxidation half reaction, we have
2I^– (aq) → I₂(s)
Now, to balance the charge, we add 2 e“ to the RHS of the reaction.
2I^–(aq) → I₂(s) + 2e^–

Step 3 : In the reduction half reaction, the oxidation state of Mn has reduced from +7 to +4. Thus, 3 electrons are added to the LHS of the reaction.
\(\mathrm{MnO}_{4}^{-}\) + 3e^– → MnO₂(aq)
Now, to balance the charge, we add 4 OH“ ions to the RHS of the reaction as the reaction is taking place in a basic medium.
\(\mathrm{MnO}_{4}^{-}(a q)^{-}\) + 3e^– → MnO₂(aq) + 4OH^–

Step 4 : In this equation, there are 6 O atoms on the RHS and 4 O atoms on the LHS. Therefore, two water molecules are added to the LHS.
\(\mathrm{MnO}_{4}^{-}\) (aq) + 2H₂O + 3e^– → MnO₂(aq) + 4OH^–

Step 5 : Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have:
6I^–(aq) → 3I₂(s) + 6e^–
2 \(\mathrm{MnO}_{4}^{-}\)(aq) + 4H₂O + 6e^– → 2MnO₂(s) + 8OH^–(aq)

Step 6 : Adding the two half reactions, we have the net balanced redox reaction as:
6I^–(aq) + 2\(\mathrm{MnO}_{4}^{-}\)(aq) + 4H₂O(l) → 3I₂(s) + 2MnO₂(s) + 8OH^–(aq)

(b) Following the steps as in part (a) we have the oxidation half reaction as:
SO₂(g) + 2H₂O(Z) → \(\mathrm{HSO}_{4}^{-}\)(aq) + 3H⁺ (aq) + 2e^– (aq)
And the reduction half reaction as:
\(\mathrm{MnO}_{4}^{-}\)(aq) + 8H⁺(aq) + 5e^– → Mn²⁺(aq) + 4H₂O(l)

Multiplying the oxidation half reaction by 5 and the reduction half reaction by 2, and then by adding them, we have the net balanced redox reaction as :

2Mno₄^–(aq) + 5SO₂(g) + 2H₂O(Z) + H⁺(aq) → 2Mn(aq) + 5HSO₄^–((aq)

(c) Following the steps as in part (a), we have the oxidation half reaction as:
Fe²⁺(aq) → Fe³⁺(aq) + e^–

And the reduction half reaction as:
H₂O₂(aq) + 2H⁺ (aq) + 2e^– → 2H₂O(l)

Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:
H₂O₂(aq) + 2Fe²⁺(aq) + 2H⁺(aq) → 2Fe³⁺ (aq) + 2H₂O(Z)

(d) Following the steps as in part (a), we have the oxidation half reaction as:

Question 19.
Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
(a) P₄(s) + OH^–(aq) → PH₃(g) + H₂PO₂^–(aq)
(b) N₂H₄(l) + CIO₃^–(aqr) → NO(g) + Cl^–(g)
(c) Cl₂O₇(g) + H₂O₂(aqr) → ClO^–₂(aq) + O₂(g) + H⁺
Answer:

O.N. increases by 1 per P atom.
P₄ acts both as an oxidising as well as a reducing agent.

Oxidation number method:
Total decrease in O.N. of P₄ in PH₃ = 3 x 4 = 12
Total increase in O.N. of P₄ in H₂PO₂ = 1 x 4 = 4
Therefore, to balance increase/decrease in O.N. multiply PH₃ by 1 and H₂PO₂^– by 3, we have,

P₄(s) + OH^–(aq) → PH₃(g) + 3H₂PO₂^–(aq)

To balance O atoms, multiply OH- by 6, we have,

P₄(s) + 6OH^–(aq) → PH₃(g) + 3H₂PO₂^–(aq)

To balance H atoms, add 3H₂O to L.H.S. and 3OH^– to the R.H.S. we have,

P₄ (s) + 6OH^–(aq) + 3H₂O(l) → PH₃(g) + 3H₂PO₂^–(aq) + 3OH^–(aq)
or P₄(s) + 3OH^–(aq) + 3H₂O(l) → PH₃(g) + 3H₂PO₂^–(aq) …(1)

Thus, Eq. (1) represents the correct balanced equation.

Ion-electron method : The two half reactions are:

Oxidation number method :
Total increase in O.N. of N = 2 x 4 = 8
Total decrease in O.N. of Cl = 1 x 6 = 6
Therefore, to balance increase/decrease in O.N. multiply N₂H₂ by 3 and \(\mathrm{ClO}_{3}^{-}\) by 4, we have,

3N₂H₄(l) + 4ClO₃^– (aq) → NO(g) + Cl^–(aq)

To balance N and Cl atoms, multiply NO by 6 and Cl^– by 4, we have,

3N₂H₄(l) + 4ClO₃^–(aq) → 6NO(g) + 4Cl^–(aq)

Balance O atoms by adding 6H₂O, in R.H.S.

3N₂H₄(l) + 4ClO₃^–(aq) → 6NO(g) + 4Cl^–(aq) + 6H₂O (l) …(1)

H atoms get automatically balanced and thus Eq. (1) represents the correct balanced equation.

Ion electron method :

Oxidation number method :
Total decrease in O.N. of Cl₂O₇ = 4 x 2 = 8
Total increase in O.N. of H₂O₂ = 2 x 1 = 2
∴ To balance increase/decrease in O.N. multiply H₂O₂ and O₂ by 4, we have,

Question 20.
What type of information can you draw from the following reaction?
(CN)₂(g) + 2OH^–(aq) → CN^–(aq) + CNO^–(aq) + H₂O(l)
Answer:
The oxidation number of carbon in (CN)₂ CN^– and CNO^– is +3, +2 and +4 respectively. These are obtained as shown below:
Let the oxidation number of C be x.
(i) (CN)₂
2 (x – 3) = 0
∴ x = +3

(ii) CN^–
x -3 = -1
∴ x = +2

(iii) CNO^–
x – 3 – 2 = -1
∴ x = + 4

The oxidation number of carbon in the various species is:

The following information we can drawn from the above reaction :
(i) Decomposition of cyanogen in the cyanide ion (CN^–) and cyanate ion (CNO^–) occurs in basic medium.
(ii) Cyanogen (CN)₂ acts as both reducing agent as well as oxidising agent.
(iii) The reaction is an example of disproportionation reaction.
(iv) Cyanogen (CN)₂ is called pseudohalogen while CN^–, CNO^– ions are called pseudohalide ions.

Question 21.
The Mn³⁺ ion is unstable in solution and undergoes disproportionation to give Mn²⁺, MnO₂ and H⁺ ion. Write a balanced ionic equation for the reaction.
Answer:
The given reaction can be represented as:

Mn³⁺ (aq) → Mn²⁺(aq) + MnO₂(s) + H⁺(aq)

The oxidation half equation is:

The oxidation number is balanced by adding one electron as:
Mn³⁺(aq) → MnO₂(s) + e^–

The charge is balanced by adding 4H+ ion as :
Mn³⁺(aq) → MnO₂(s) + 4H⁺ (aq) + e^–

The O atoms and H+ ions are balanced by adding 2H20 molecules as:
Mn³⁺(aq) + 2H₂O(l) → MnO₂(s) + 4H⁺ (aq) + e^– …(1)

The reduction half equation is:
Mn³⁺(aq) → Mn²⁺(aq)

The oxidation number is balanced by adding one electron as :
Mn³⁺ (aq) + e^– → Mn²⁺ (aq) … (2)

The balanced chemical equation can be obtained by adding equation (1) and (2) as :
2Mn³⁺(aq) + 2H₂O(l) → MnO₂(s) + Mn²⁺(aq) + 4H⁺(aq)

Question 22.
Consider the elements: Cs, Ne, I and F
(a) Identify the element that exhibits only negative oxidation state.
(b) Identify the element that exhibits only positive oxidation state.
(c) Identify the element that exhibits both positive and negative oxidation states.
(d) Identify the element which exhibits neither the negative nor the positive oxidation state.
Answer:
(a) F exhibits only negative oxidation state of-1.
(b) Cs exhibits only positive oxidation state of +1
(c) I exhibits both positive and negative oxidation states. It exhibits oxidation states of-1, 0, +1, + 3, + 5, and + 7.
(d) The oxidation state of Ne is zero. It exhibits neither negative nor positive oxidation states.

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 in water.
Answer:
The given redox reaction can be represented as:

Question 24.
Refer to the periodic table given in your book and now answer the following questions:
(a) Select the possible non-metals that can show disproportionation reaction.
(b) Select three metals that can show disproportionation reaction.
Answer:
In disproportionation reaction, one of the reacting substances always contains an element that can exist in at least three oxidation states.

Question 25.
In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with lO.OOg of ammonia and 20.00 g of oxygen?
Answer:
The balanced chemical equation for the given reaction is given as:

68 g of NH₃ reacts with 160 g of O₂
Therefore 10 g of NH₃ reacts with \(\frac{160 \times 10}{68}\). g of O₂ or 23.53 g of O₂
But the available amount of O₂ is 20 g which is less than the amount required to react with 10 g NH₃. So, O₂ is the limiting reagent and it limits the amount of NO produced. From the above balanced equation.
160 g of O₂; produces 120 g NO.
Therefore, 20 g of O₂; produces = \(\frac{120 \times 20}{160}\) = 15 g NO

Question 26.
Using the standard electrode potentials given in the table 8.1,
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 Fe³⁺ (aq)
(e) Br₂ (aq) and Fe²⁺(aqr)
Answer:

Question 27.
Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO₃ with silver electrodes.
(ii) An aqueous solution of AgNO₃ with platinum electrodes
(iii) A dilute solution of H₂SO₄ with platinum electrodes.
(iv) An aqueous solution of CuCl₂ with platinum electrodes,
Answer:
(i) AgNO₃ ionizes in aqueous solutions to form Ag⁺ and \(\mathrm{NO}_{3}^{-}\) ions.
On electrolysis, either Ag⁺ ions or H₂O molecules can be reduced at the cathode. But the reduction potential of Ag⁺ ions is higher than that of H₂O.

Hence, Ag+ ions are reduced at the cathode. Similarly, Ag metal or H20 molecules can be oxidized at the anode. But the oxidation potential of Ag is higher than that of H20 molecules.

Ag(s) → Ag⁺(aq) + e^– ;\(E^{\ominus}\) = -0.80V

2H₂O₂(g) → O₂(g) + 4H⁺(aq) + 4e^– ;\(E^{\ominus}\) = -1.23V
Therefore, Ag metal gets oxidized at the anode.

(ii) Pt cannot be oxidized easily. Hence, at the anode, oxidation of water occurs to liberate O₂. At the cathode, Ag⁺ ions are reduced and get deposited.

(iii) H₂SO₄ ionizes in aqueous solutions to give H⁺ and \(\mathrm{SO}_{4}^{2-}\) ions.

H₂SO₄ (aq) → 2H⁺(aq) + \(\mathrm{SO}_{4}^{2-}\) (aq)

On electrolysis, either of H⁺ ions or H₂O molecules can get reduced at the cathode. But the reduction potential of H⁺ ions is higher than that of H₂O molecules.

2H⁺(aq) + 2e^– → H₂(g); \(E^{\ominus}\) = 0.0 V
2H₂O(aq) + 2e^– → H₂(g) + 2OH^–(aq): \(E^{\ominus}\) = – 0.83V

Hence, at the cathode, H⁺ ions are reduced to liberate H₂ gas.
On the other hand, at the anode, either of \(\mathrm{SO}_{4}^{2-}\) ions or H₂O molecules can get oxidized. But the oxidation of \(\mathrm{SO}_{4}^{2-}\) involves breaking of more bonds than that of H₂O molecules. Hence, \(\mathrm{SO}_{4}^{2-}\) ions have a lower oxidation potential than H₂O. Thus, H₂O is oxidized at the anode to liberate O₂ molecules.

(iv) In aqueous solutions, CuCl₂ ionizes to give Cu²⁺ and Cl^– ions as
CuCl₂ (aq) → Cu²⁺ (aq) + 2Cl^– (aq)

On electrolysis either of Cu²⁺ ions or H₂O molecules can get reduced at the cathode. But the reduction potential of Cu²⁺ is more than that of H₂O molecules.

Cu²⁺(aq) + 2e^– → Cu(aq) ;\(E^{\ominus}\) = + 0.34V ;
H₂O(l) + 2e^– → H₂(g) + 2OH^– ;\(E^{\ominus}\) = – 0.83V

Hence, Cu²⁺ ions are reduced at the cathode and get deposited.
Similarly, at the anode, either of Cl^– or H₂O is oxidized. The oxidation , potential of H₂O is higher than that of Cl^–

2Cl^–(aq) → Cl₂(g) + 2e^– ;\(E^{\ominus}\) = -1.36V :K
2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e^– ;\(E^{\ominus}\) = -1.23V

But oxidation of H₂O molecules occurs at a lower electrode potential . than that of Cl^– ions because of over-voltage (extra voltage required to liberate gas). As a result, Cl^– ions are oxidized at the anode to liberate Cl₂ gas. :

Question 28.
Arrange the following metals in the order in which they displace each other from the solution of their salts.
Al, Cu, Fe, Mg and Zn. ‘
Answer:
A metal of stronger reducing power displaces another metal of weaker reducing power from its solution of salt.
The order of the increasing reducing power of the given metals is
Cu < Fe < Zn < Al < Mg. Hence, we can say that Mg can displace Al from its salt solution, but Al cannot displace Mg. Thus, the order in which the given metals displace each other from the solution of their salts is given below:
Mg > Al > Zn > Fe > Cu

Question 29.
Given the standard electrode potentials:
K⁺ / K = – 2.93V, Ag⁺ / Ag = 0.80 V, Hg²⁺ /Hg = 0.79 V
Mg²⁺/ Mg = -2.37 V, Cr³⁺/Cr = -0.74 V Arrange these metals in their increasing order of reducing power.
Answer:
The lower the electrode potential, the stronger is the reducing agent. Therefore, the increasing order of the reducing power of the given metal is
Ag < Hg < Cr < Mg < K.

Question 30.
Depict the galvanic cell in which the reaction Zn(s) + 2Ag⁺(aqr) → Zn²⁺(aqr) + 2Ag(s) takes place, further show:
(i) which of the electrode is negatively charged,
(ii) the carriers of the current in the cell, and
(iii) individual reaction at each electrode.
Answer:
The galvanic cell corresponding to the given redox reaction can be represented as:
Zn / Zn²⁺ (aq) | | Ag⁺(aq) / Ag
(i) Zn electrode is negatively charged because at this electrode, Zn oxidizes to Zn²⁺ and the leaving electrons accumulate on this electrode.
(ii) Ions are the carriers of current in the cell.
(iii) The reaction taking place at Zn electrode can be represented as:
Zn(s) → Zn²⁺ (aq) + 2e^–
and the reaction taking place at Ag electrode can be represented as:
Ag⁺(aq) + e^– → Ag(s)

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 9 Hydrogen Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 9 Hydrogen Important Questions

Very Short Answer Type Questions

Question 1.
Name the isotope of hydrogen which contains equal number of protons and neutrons.
Answer:
Deuterium (\({ }_{1}^{2} \mathrm{H}\))
Number of protons (p) = number of electrons
= atomic number = 1
Number of neutrons (n) = mass number – atomic number
= 2 – 1 = 1 .

Question 2.
Why is the ionisation enthalpy of hydrogen higher than that of sodium?
Answer:
Both H and Na contain one electron in the valence shell. But the size of H is much smaller as compare to that of Na and hence, the ionisation enthalpy of hydrogen is much higher (1312 kJ mol^-1) than that of Na (496 kJ mol^-1).

Question 3.
What do you mean by 15 volume H2O2 solution?
Answer:
‘15 volume H₂O₂’ means 1 mL of a 15 volume H₂O₂ solution gives 15 mL of O₂ at NTP.

Question 4.
Which isotope of hydrogen is radioactive?
Answer:
Tritium

Question 5.
Arrange H₂, D₂ and T₂ in the decreasing order of their
(i) boiling points
(ii) heat of fusion
Answer:
(i) T₂ > D₂ > H₂
(ii) T₂ > D₂ > H₂

Question 6.
Write the Lewis structure of hydrogen peroxide.
Answer:
The Lewis structure of hydrogen peroxide is :

Question 7.
Write one chemical reaction for the preparation of D₂O₂.
Answer:
D₂O₂ is prepared by distillation of potassium persulphate (K₂S₂O₈) with D₂O.

Question 8.
Suggest a method to show the electronegative nature of hydrogen.
Answer:
When sodium hydride is electrolysed, hydrogen is evolved at anode, which shows its electronegative nature.

Question 9.
What type of bonds are broken when water evaporates.
Answer:
Intermolecular hydrogen bonds are broken when water evaporates.

Short Answer Type Questions

Question 1.
Describe the industrial applications of hydrogen dependent on
(i) the heat liberated when its atoms are made to combine on the surface of a metal.
(ii) its effect on the unsaturated organic systems in the presence of a catalyst.
(iii) its ability to combine with nitrogen under specific conditions.
Answer:
(i) Due to this property, hydrogen is used in atomic hydrogen welding/cutting torch.
(ii) Due to this property hydrogen is used for the manufacture of vanaspati ghee from edible oils such as cotton-seed oil, soyabean oil, corn oil etc.

(iii) Due to this property dihydrogen is used for the manufacture of ammonia (Haber’s process).

Question 2.
Why does water show high boiling point as compared to hydrogen sulphide? Give reasons for your answer.
Answer:
Water show high boiling point as compared to hydrogen sulphide due to high electronegativity of oxygen (EN = 3.5), water undergoes extensive H-bonding as a result of which water exists as associated molecule.

For breaking these hydrogen bond, a large amount of energy is needed and hence the boiling point of H₂O is high. In other words, due to lower electronegativity of S (EN =2.5), hydrogen sulphide do not undergo H-bonding. Consequently, H₂S exists as discrete molecule and hence its boiling point is much lower than that of H₂O. That is why H₂S is a gas at room temperature.

Question 3.
If a given sample of water has degree of hardness equal to 46 ppm. If entire hardness is due to MgSO₄, how much MgSO₄ is present per kg of water?
Answer:
Given, degree of hardness = 46 ppm
Which means that 106 g of sample require 46 g of CaCO₃
∴ CaCO₃ present in 1000 g of water = \(\frac{46 \times 1000}{10^{6}}\) = 46 x 10^-3 g
1 mol (or 100 g) of CaC0₃ = 1 mol (or 120 g) of MgSO₄
∴ 46 x 10^-3 g of CaCO₃ = \(\frac{120 \times 46 \times 10^{-3}}{100}\)g = 0.055 g or 55 mg

Question 4.
What are the advantages in using hydrogen as a fuel?
Answer:
Hydrogen as a fuel has the following advantages :

1. It has high calorific value.
2. During combustion, it does not produce smoke or any unpleasant fumes.
3. It leaves no ash after burning. The only product of combustion is water.
4. It does not pollute the air because no pollutant is produced during its combustion.
5. It can be used in a fuel cell to generate electricity.
6. It can be used in the internal combustion engines with slight modifications.

Question 5.
Calculate the volume strength of a 3% solution of H₂O₂
Answer:
100 mL of H₂O₂ solution contains H₂O₂ = 3 g
∴ 1000 mL of H₂O₂ solution will contains
H₂O₂ = \(\frac{3}{100}\) x 1000 = 30 g
Consider the chemical equation,

Now 68 g of H₂O₂ gives O₂ at NTP = 22.7 L
∴ 30 g of H₂O₂ will give 02 at NTP = \(\frac{22.7}{68}\) x 30 = 10.014
But 30 g of H₂O₂ are present in 1000 mL of H₂O₂.
Hence, 1000 mL of H₂O₂ solution gives 02 at NTP = 1.0014 mL
∴ 1 mL of H₂O₂ solution will give O₂ at NTP = \(\frac{10014}{1000}\)= 10.01 mL
Hence, the volume strength of 3% H202 solution = 10.01

Long Answer Type Questions

Question 1.
(i) (a) How would you prepare dihydrogen from water by using a reducing agent?
(b) How would you prepare dihydrogen from a substance other than water?
(c) How would you prepare very pure dihydrogen in the laboratory?
(ii) Write a short note on hydrogenation of vegetable oils.
Answer:
(i) (a) Sodium metal is a good reducing agent. It reduces water to hydrogen (or dihydrogen).
2H₂O + 2Na → 2NaOH + H₂(g)
(b) Dihydrogen can be obtained by treating zinc with dilute HCl
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
(c) Highly pure dihydrogen (hydrogen gas) can be prepared by the following methods :
I. Fairly pure hydrogen can be obtained by treating pure magnesium or pure aluminium with chemically pure H₂SO₄ or HCl diluted with distilled water. The gas is passed over P₂O₅ and is collected by the displacement of mercury.
Mg(s) + H₂SO₄(aq) > MgSO₄(aq) + H₂(g)

II. Highly pure hydrogen gas can be obtained by electrolysing a warm solution of Ba(OH)2 in a U-tube using nickel electrodes. The gas is purified by passing it over heated platinum gauze when traces of oxygen combine with hydrogen forming water. The gas is then dried by passing it over caustic potash sticks and phosphorus pentoxide. Hydrogen is finally adsorbed in palladium and the impurities remain unadsorbed. On heating palladium under reduced pressure pure hydrogen is liberated.

(ii) When oils like groundnut oil or cotton seed oil (which are unsaturated compound i.e., have double bond) are treated with hydrogen in the presence of nickel as catalyst, they get converted into edible fats like margarine and vanaspati ghee (which are saturated compounds). This reaction is called hydrogenation of vegetable oils or hardening of oils.

Question 2.
Give ion electron equations for the following reactions :
(i) Oxidation of ferrous ions to ferric ions by hydrogen peroxide both in acidic and basic media.
(ii) Oxidation of iodide ion to iodine by hydrogen peroxide in acidic medium.
(iii) Reduction of acidified potassium dichromate solution.
(iv) Oxidation of sulphurous acid to sulphuric acid.
(v) Oxidation of ferrocyanide ions to ferricyanide ions in acidic medium.
Answer:

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 10 The s-Block Elements Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 10 The s-Block Elements Important Questions

Very Short Answer Type Questions

Question 1.
Complete the following reactions,
(i) \(\mathbf{O}_{2}^{2-}\) +H₂O →
(ii) O₂ +H₂O →
Answer:
(i) Peroxide ions react with water to form H202.
\(\mathbf{O}_{2}^{2-}\) + 2H₂O→ 20H’ + H₂O₂
(ii) Superoxides react with water to form H202 and 02.
\(2 \mathrm{O}_{2}^{-}\) + 2H₂O → 20H^– + H₂O₂ + O₂

Question 2.
(i) Predict giving reason, the outcome of the reaction

Answer:
(i) Large cation (K⁺) can stabilise large anion (I^–).
(ii) This is because the larger cation (K⁺) can stabilise larger anion (Cl^–).

Question 3.
Which colours are imparted to flame when the following elements are introduced in the flame one by one?
(i) Strontium (ii) Barium (iii) Calcium
Answer:
(i) Strontium — Brick red
(ii) Barium — Grassy green
(iii) Calcium — Crimson red

Question 4.
Sodium fire in the laboratory should not be extinguished by pouring water. Why?
Answer:
This is because sodium produces hydrogen gas with water which catches fire because of the exothermic nature of the reaction.

Question 5.
What is light soda ash? Wliy is it called so?
Answer:
Light soda ash is anhydrous Na₂CO₃. It is called so because it is fluffy solid with a low packing density of about 0.5 g cm^-3.

Question 6.
What is baryta water? Give its one use.
Answer:
Baryta is an aqueous solution of barium hydroxide. It can also be used for detection of CO₂.

Question 7.
Give the chemical formula of quick lime, slaked lime and lime water.
Answer:
Quick lime is CaO, slaked lime is Ca(OH)₂ and lime water is an aqueous solution of Ca(OH)₂.

Question 8.
Which magnesium compounds are the constituents of toothpaste?
Answer:
Mg(OH)₂ and MgCO₃ are the constituents of toothpaste.

Question 9.
What is the mixture of CaCN₂ and carbon known as?
Answer:
A mixture of calcium cyanamide (CaCN₂) and carbon is known as nitrolim. It is used as a fertiliser.

Question 10.
It is necessary to add gypsum in the final stages of preparation of cement. Explain why?
Answer:
Gypsum (CaSO₄ . 2H₂O) is added in the final stages of preparation of cement because it slows down the process of setting of cement so that it gets sufficiently hardened thereby imparting greater strength to it.

Short Answer Type Questions

Question 1.
How would you distinguish between
(i) Be(OH)₂ and Ba(OH)₂
(ii) BeSO₄ and BaSO₄
Answer:
(i) Be(OH)₂, beryllium hydroxide is soluble in aqueous sodium hydroxide solution whereas Ba(OH)₂, (barium hydroxide) does not, because Be(OH)₂ is amphoteric in nature and Ba(OH)₂ is basic in nature. Be(OH)₂ + 2NaOH → Na2[Be(OH)₄] (Sodium beryllate)
(ii) BeSO₄ is soluble in water whereas BaSO₄ is insoluble in water.

Question 2.
Element A bums in nitrogen to give an ionic compound B. Compound B reacts with water to give C and D. A solution of C becomes milky on bubbling carbon dioxide. Identify A, B, C and D.
Answer:
Element A is calcium

Compounds = Ca₃N₂ ; CompoundC = Ca(OH)₂ and Compound D = NH₃

Question 3.
What happens when
(i) chlorine gas is passed through a cold and dilute solution of NaOH?
(ii) yellow phosphorus is heated with NaOH solution?
(iii) carbon dioxide is passed through ammonical brine solution?
(iv) sodium hydrogen carbonate is heated?
Answer:
(i) Sodium hypochlorite and sodium chloride are obtained.
Cl2 + 2NaOH → NaCl + NaClO + H₂O
(ii) Phosphine gas is obtained.
P4 + 3NaOH + 3H₂O → 3NaH₂PO₂ + PH₃
(iii) Sodium hydrogen carbonate is precipitated.
NH₃ + H₂O + CO₂ + NaCl → NH₄Cl + NaHCO₃↓
(iv) Sodium ash is obtained.

Question 4.
Mention the various sources of sodium chloride and explain the preparation of sodium chloride from sea-water and salt mines.
Answer:
NaCl occurs abundantly in nature. Its major sources are (a) Sea water which contains 2.7 to 2.9 % NaCl.
(b) Water of inland lakes such as Sambhar Lake in Rajasthan.
(c) Salt-mines which contain rock salt are located in England, Australia, and Himachal Pradesh.
Preparation
(i) From sea water : Sea water is filled in big tanks where it slowly evaporates, leaving behind solid salt. In cold countries, where temperatures are very low, pure water get freeze. Ice formed is removed and concentration of NaCl in solution increases. The concentrated sodium can be separated and evaporated to get NaCl.
(ii) From salt-mines : Salt mines are located deep under the surface of the earth. Holes are made into these mines with the help of drillers and broken pieces of salt rocks are taken out by suitable means.

Question 5.
When water is added to compound (A) of calcium, solution of compound (B) is formed. When carhon dioxide is passed into the solution, it turns milky due to the formation of compound (C). If excess of carbon dioxide is passed into the solution, milkiness disappears due to the formation of compound (D). Identify the compound A, B, C and D. Explain why the milkiness disappears in the last step? [NCERT Exemplar]
Answer:
Appearance of milkiness on passing CO₂ in the solution of compound B indicates that compound B is lime water and compound C is CaCO₃. Since, compound B is obtained by adding H₂O to compound A, therefore compound A is quicklime, CaO. The reactions are as follows :

Long Answer Type Questions

Question 1.
Ions of an element of group 1 participate in the transmission of nerve signals and transport of sugars and amino acids into cells. This element imparts yellow colour to the flame in flame test and forms an oxide and a peroxide with oxygen. Identify the element and write chemical reaction to show the formation of
its peroxide. Why does the element impart colour to the flame?
Answer:
Yellow colour flame in flame test indicates that the alkali metal must be sodium. It reacts with O₂ to form a mixture of sodium peroxide, Na₂O₂ and sodium oxide, Na₂O.

Ionization enthalpy of sodium is low. When sodium metal or its salt is heated in Bunsen flame, the flame energy causes an excitation of the outermost electron which on reverting back to its initial position gives out the absorbed energy as visible light. That’s why sodium imparts yellow colour to the flame.

Question 2.
The stability of peroxide and superoxide of alkali metals increase as we go down the group. Explain giving reason.
Answer:
The stabilit-y of peroxides or superoxides increases as the size of metal ion increases, i.e., KO₂ < RbO₂ < CsO₂.
The reactivity of alkali metals towards oxygen to form different oxides is due to strong positive field around each alkali metal cation. Li⁺ is smallest, it does not allow 0^2- ion to react with O₂ further. Na⁺ is larger than Li, its positive field is weaker than Li⁺. It cannot prevent the conversion of O^2- into \(\mathrm{O}_{2}^{2-}\). The larger K⁺, Rb⁺ and Cs⁺ ions permit \(\mathrm{O}_{2}^{2-}\)ion to react with O₂ further forming superoxide ion (\(\mathrm{O}_{2}^{-}\)).

Further more, increased stability of the peroxide or superoxide with increase in the size of metal ion is due to the stabilisation of large anions by larger cations through lattice energy effect.

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 9 Hydrogen Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 9 Hydrogen

PSEB 11th Class Chemistry Guide Hydrogen InText Questions and Answers

Question 1.
Justify the position of hydrogen in the periodic table on the basis of its electronic configuration.
Answer:
Hydrogen is the first element in the periodic table. It has the electronic configuration 1s¹. It is similar to alkali metal (ns¹) of group I. It shows resemblance with alkali metals of group I of the periodic table. So it can be placed above the alkali metals in group I of the periodic table.
On the other hand, the electronic configuration of hydrogen shows that it is short of one electron to the nearest noble gas configuration (He) having the electronic configuration 1s². Like halogens it forms covalent bonds (H₂, Cl₂, Br₂, etc.) as well as ionic bonds (e.g. Na⁺ H^–). It forms H⁺ ion by giving one electron and hydride ion (H^–) by gaining one electron. On the basis of its electronic configuration (1s²) hydrogen is placed with other ns¹ elements namely alkali metals in the group I as well as in group 17 of the periodic table. Thus, the position of hydrogen in the periodic table is anomalous.
Hydrogen with so many unique characteristics is, therefore best placed separately in the periodic table of elements.

Question 2.
Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes?
Answer:
Hydrogen has following three isotopes :
1. Protium,\({ }_{1}^{1} \mathrm{H}\),
2. Deuterium, \({ }_{1}^{2} \mathrm{H}\) or D, and
3. Tritium, \({ }_{1}^{3} \mathrm{H}\) or T
The mass ratio of protium, deuterium and tritium is 1.008 : 2.014 : 3.016 or 1:2:3.

Question 3.
Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal conditions?
Answer:
The ionization enthalpy of hydrogen atom is very high (1312 kJ mol^-1). Hence, it is very hard to remove its electron. As a result, its tendency to exist in the monoatomic form is rather low. Instead, hydrogen forms a covalent bond with another hydrogen atom and exists as a diatomic (H₂) molecule.

Question 4.
How can the production of dihydrogen, obtained from ‘coal gasification’, be increased?
Answer:
Dihydrogen, produced by coal gasification method as :

The yield of dihydrogen (obtained from coal gasification) can be increased by reacting carbon monoxide (formed during the reaction) with steam in the presence of iron chromate as a catalyst.

This reaction is called the water-gas shift reaction. Carbon dioxide is removed by scrubbing it with a solution of sodium arsenite.

Question 5.
Describe the bulk preparation of dihydrogen by electrolytic method. What is the role of an electrolyte in this process?
Answer:
Electrolysis of acidified water using platinum electrodes gives dihydrogen.

The role of an electrolyte is to make water conducting.

Question 6.
Complete the following reactions :

Answer:

Question 7.
Discuss the consequences of high enthalpy of H-H bond in terms of chemical reactivity of dihydrogen.
Answer:
The ionization enthalpy of H-H bond is very high (1312 kJ mol^-1). This indicates that dihydrogen has a low tendency to form H+ ions. Its ionization enthalpy value is comparable to that of halogens. Hence, it forms diatomic molecules (H₂), hydrides with elements, and a large number of covalent bonds.
Since ionization enthalpy is very high, dihydrogen does not possess metallic characteristics (lustre, ductility, etc.) like metals.

Question 8.
What do you understand by (i) electron-deficient,
(ii) electron-precise, and (iii) electron-rich compounds of hydrogen? Provide justification with suitable examples.
Answer:
(i) An electron-deficient hydride : It has very few electrons, less than that required for representing its conventional Lewis structure e.g., diborane (B₂H₆). In B₂H₆, there are six bonds in all, out of which only four bonds are regular i.e., two electrons are shared by two atoms.
The remaining two bonds are three centered-two electron bonds i.e., two electrons are shared by three atoms. Hence, its conventional Lewis structure cannot be drawn.

(ii) An electron-precise hydride : It has sufficient number of electrons to be represented by its conventional Lewis structure e.g., CH₄. The Lewis structure can be written as :

Four regular bonds are formed where two electrons are shared by two atoms.

(iii) An electron-rich hybride : It contains excess electrons as lone pair e.g., NH₃

There are three regular bonds in all with a lone pair of electrons on the nitrogen atom.

Question 9.
What characteristics do you expect from an electron-deficient hydride with respect to its structure and chemical reactions?
Answer:
These hydrides do not have sufficient number of electrons to form normal covalent bonds, e.g., B in BF₃ has 6 electrons in its valence shell. These hydrides are trigonal planar in shape.

These hydrides act as Lewis acids, i.e., electron pair acceptor e.g.,

To make up the deficiency of electrons, these hydrides exist in polymeric forms e.g., B₂H₆, B₄H₁₀, etc. Electron deficient hydrides are very reactive. These reacts readily with metals and non-metals and their compounds, e.g.,

B₂H₆ + 3O₂(g) → B₂O₃(s) + 3H₂O(g)

Question 10.
Do you expect the carbon hydrides of the type (CraH2jt + 2) to act as ‘Lewis’ acid or base? Justify your answer.
Answer:
Carbon hydrides of the type C_nH_2n+2 are CH₄, C₂H₆ etc. in which number of electrons present are just sufficient to write down their conventional Lewis structures.

C has neither extra electrons nor less electrons. Such compounds of the formula C_nH_2n+2 are called ELECTRON-PRECISE compounds. They will act neither as Lewis acids nor as Lewis bases.

Question 11.
What do you understand by the term “non-stoichiometric hydrides”? Do you expect this type of the hydrides to be formed by alkali metals? Justify your answer.
Answer:
Non-stoichiometric hydrides are hydrogen-deficient compounds formed by the reaction of dihydrogen with d-block and f-block elements. These hydrides do not follow the law of constant composition.

For example :
LaH₂₈₇, YbH_2.55, TiH_{1.5 – 1.8} etc.
Alkali metals form stoichiometric hydrides. These hydrides are ionic in nature. Hydride ions have comparable sizes (208 pm) with alkali metal ions. Hence, strong binding forces exist between the constituting metal and hydride ion. As a result, stoichiometric hydrides are formed. Alkali metals will not form non-stoichiometric hydrides.

Question 12.
How do you expect the metallic hydrides to be useful for hydrogen storage? Explain.
Answer:
Metallic hydrides are hydrogen deficient, i.e., they do not hold the law of constant composition. It has been established that in the hydrides of Ni, Pd, Ce and Ac hydrogen occupies the interstitial position in lattices allowing further absorption of hydrogen on these metals. Metals like Pd, Pt, etc. have the capacity to accommodate a large volume of hydrogen. Therefore, they are used for the storage of hydrogen and serve as a source of energy.

Question 13.
How does the atomic hydrogen or oxy-hydrogen torch function for cutting and welding purposes? Explain.
Answer:
Atomic hydrogen atoms are produced by the dissociation of dihydrogen with the help of an electric arc. This releases a huge amount of energy (435.88 kJ mol^-1). This energy can be used to generate a temperature of 4000 K, which is ideal for welding and cutting metals. Hence, atomic k hydrogen or oxy-hydrogen torches are used for these purposes. For this reason, atomic hydrogen is allowed to recombine on the surface to be welded to generate the desired temperature.

Question 14.
Among NH₃, H₂O and HF, which would you expect to have highest magnitude of hydrogen bonding and why?
Answer:
The extent of hydrogen bonding depends upon electronegativity and the number of hydrogen atoms available for bonding. Among nitrogen, fluorine, and oxygen, the increasing order of their electronegativities are N < O < F. Hence, the order of the extent of hydrogen bonding is HF > H₂O > NH₃.

Question 15.
Saline hydrides are known to react with water violently producing fire. Can CO₂, a well known fire extinguisher, be used in this case? Explain.
Answer:
Saline hydrides (i.e., NaH, LiH, etc.) react with water to form a base and hydrogen gas. The chemical equation used to represent the reaction can be written as:
NaH(s) + H₂O(7) → NaOH(aq) + H₂(g)
This reaction is violent and produces fire.
This type of fire cannot be extinguished by CO₂ because it gets reduced by the hot metal hydride to form sodium format.
NaH + CO₂ → HCOONa

Question 16.
Arrange the following
(i) CaH₂, BeH₂ and TiH₂ in order of increasing electrical conductance.
(ii) LiH, NaH and CsH in order of increasing ionic character.
(iii) H-H, D-D and F-F in order of increasing bond dissociation enthalpy.
(iv) NaH, MgH₂ and H₂O in order of increasing reducing property.
Answer:
(i) BeH₂ < CaH₂ < TiH₂
(ii) LiH < NaH < CsH
(iii) F—F < H—H < D—D
(iv) H₂O < MgH₂ < NaH

Question 17.
Compare the structures of H₂O and H₂O₂.
Answer:
In gaseous phase, water molecule has a bent form with a bond angle of 104.5°. The O—H bond length is 95.7 pm. The structure can be shown as:

Hydrogen peroxide has a non-planar structure both in gas and solid phase. The dihedral angle in gas and solid phase is 111.5° and 90.2° respectively.

Question 18.
What do you understand by the term ‘auto-protolysis’ of water? What is its significance?
Answer:
Auto-protolysis (self-ionization) of water is a chemical reaction in which two water molecules react to produce a hydroxide ion (OH^–) and a hydronium ion (H₃O⁺).

The reaction involved can be represented as :

Auto-protolysis of water indicates its amphoteric nature i.e., its ability to act as an acid as well as a base. The acid-base reaction can be written as :

Question 19.
Consider the reaction of water with F₂ and suggest, in terms of oxidation and reduction, which species are oxidized/reduced.
Answer:

In these reactions, water acts as a reducing agent and hence itself gets oxidised to either oxygen or ozone. Fluorine acts as an oxidising agent and hence itself reduced to F^– ion.

Question 20.
Complete the following chemical reactions.
(i) PbS(s) + H₂O₂(ciq) →
(ii) MnO^–₄(aq) + H₂O₂(aq) →
(iii) CaO(s) + H₂O(g) →
(iv) AlCl₃(g) + H₂O(Z) →
(v) Ca₃N₂(s) + H₂O(l) →
Classify the above into (a) hydrolysis, (b) redox and (c) hydration reactions.
Answer:
(i) PbS(s) + 4H₂O₂(aq) → PbSO₄(s) + 4H₂O(l)
H₂O₂ is acting as an oxidizing agent in the reaction. Hence, it is a redox reaction.

(ii) 2MnO₄^– (aq) + 5H₂O₂(Z) + 6H⁺(aq) → 2Mn²⁺(aq) + 8H₂O(Z) + 5O₂(g)
H₂O₂ is acting as a reducing agent in the acidic medium, thereby oxidizing MnO^–₄(aq). Hence, the given reaction is a redox reaction.

(iii) CaO(s) + H₂O(g) → Ca(OH)₂(aq)
The reactions in which a compound reacts with water to produce other compounds are called hydrolysis reactions. The given reaction is hydrolysis reaction.

(iv) AlCl₃(g) + 6H₂O(Z) → [Al(OH₂)₆]³⁺ (aq) + 3Cl^–(aq)
It is a hydration reaction, because A1C13 is hydrated to [Al(OH₂)₆]³⁺.

(v) Ca₃N₂(s) + 6H₂O(Z) → 3Ca(OH)₂(aq) + 2NH₃(aq)
The reactions in which a compounds reacts with water to produce other compounds are called hydrolysis reactions. The given reaction represents hydrolysis of Ca₃N₂.

Question 21.
Describe the structure of the common form of ice.
Answer:
Ice is the crystalline form of water. It takes a hexagonal form if crystallized at atmospheric pressure, but condenses to cubic form if the temperature is very low.
The three-dimensional structure of ice is represented as :

The structure is highly ordered and has hydrogen bonding. Each oxygen atom is surrounded tetrahedrally by four other oxygen atoms at a distance of 276 pm. The structure also contains wide holes that can hold molecules of appropriate sizes interstitially.

Question 22.
What causes the temporary and permanent hardness of water?
Answer:
Temporary hardness of water is due to the presence of soluble salts of magnesium and calcium in the form of hydrogen carbonates (MHCO₃, where M = Mg, Ca) in water.
Permanent hardness of water is due to the presence of soluble salts of calcium and magnesium in the form of chlorides in water.

Question 23.
Discuss the principle and method of softening of hard water by synthetic ion-exchange resins.
Answer:
The process of treating permanent hardness of water using synthetic resins is based on the exchange of cations (e.g., Na⁺, Ca²⁺, Mg²⁺ etc.) and anions (e.g.,Cl^– , \(\mathrm{SO}_{4}^{2-}\), \(\mathrm{HCO}_{3}^{-}\) etc.) present in water by H⁺ and OH^– ions respectively.

Synthetic resins are of two types :
1. Cation exchange resins
2. Anion exchange resins

Cation exchange resins are large organic molecules that contain the -SO₃H group. The resin is firstly changed to RNa (from ROS₃H) by treating it with NaCl. This resin then exchanges Na⁺ ions with Ca²⁺ and Mg²⁺ ions, thereby making the water soft.

2RNa + M²⁺ (aq) → R₂M(s) + 2Na⁺ (aq)

There are cation exchange resins in H+ form. The resins exchange H⁺ ions for Na⁺, Ca²⁺ and Mg²⁺ ions.

2RH + M²⁺ (aq) ⇌ MR₂(s) + 2H⁺(aq)

Anion exchange resins exchange OH- ions for anions like Cl^–, \(\) and \(\) present in water.

During the complete process, water first passes through the cation exchange process. The water obtained after this process is free from mineral cations and is acidic in nature.
This acidic water is then passed through the anion exchange process where OH^– ions neutralize the H⁺ ions and de-ionize the water obtained.

Question 24.
Write chemical reactions to show the amphoteric nature of water.
Answer:
Water is amphoteric in character. It behaves both as an acid as well as a base. With acids stronger than itself, it behaves as a base and with bases stronger than itself, it acts as an acid.

Question 25.
Write chemical reactions to justify that hydrogen peroxide can function as an oxidizing as well as reducing agent.
Answer:
Hydrogen peroxide, H₂O₂ acts as an oxidizing as well as a reducing agent in both acidic and alkaline media. Reactions involving oxidizing actions are :

Question 26.
What is meant by ‘demineralised’ water and how can it be obtained?
Answer:
Water which does not contain cations and anions is called ‘demineralised’ water. It is soft water. Demineralised water is obtained the same way as soft water is obtained from hard water. Demineralised or deionised water is obtained by passing hard water first through a cation exchange resin (RCOOH or RSO₃H) which removes Ca²⁺ and Mg²⁺ ions from hard water by exchanging them with H⁺ ions and then passing through an anion exchange resin which removes Cl^– and \(\mathrm{SO}_{4}^{2-}\) ions present in hard water by exchanging them with OH^– ions.

Question 27.
Is demineralised or distilled water useful for drinking purposes? If not, how can it be made useful?
Answer:
Water is an important part of life. It contains several dissolved nutrients that are required by human beings, plants and animals for survival.
Demineralised water is free from all soluble minerals. Hence, it is not fit for drinking.
It can be made useful only after the addition of desired minerals in specific amounts, which are important for growth.

Question 28.
Describe the usefulness of water in biosphere and biological systems.
Answer:
Water is essential for all forms of life. It constitutes around 65% of the human body and 95% of plants. Water plays an important role in the biosphere owing to its high specific heat, thermal conductivity, surface tension, dipole moment and dielectric constant.
The high heat of vaporization and heat of capacity of water helps in moderating the climate and body temperature of all living beings.
It acts as a carrier of various nutrients required by plants and animals for various metabolic reactions.
Water is also required for photosynthesis in plants which releases O₂ into the atmosphere.

Question 29.
What properties of water make it useful as a solvent? What types of compound can it (i) dissolve, and (ii) hydrolyse?
Answer:
Water because of its high dielectric constant (78.39) has the ability to dissolve most of the inorganic (ionic) compounds and is, therefore, regarded as a universal solvent. Whereas the solubility of ionic compounds takes place due to ion-dipole interactions (i.e., solvation of ions) the solubility of covalent compounds such as alcohols, amines, urea, glucose, sugar etc. takes place due to tendency of these molecules to form hydrogen bonds with water.

(i) It can dissolve both ionic compounds as well as covalent compounds which can form hydrogen bonds with water.
Ionic compounds whose lattice energy is lower than hydration energy get dissolved in water.
(ii) Water can hydrolyse many oxides (metallic and non-metallic), hydrides, carbides, nitrides, phosphides and many other salts e.g.,

CaO(s) + H₂O(l) → Ca(OH)₂
SO₂(g) + H₂O(l) → H₂SO₃(aq)
CaH₂(s) + 2H₂O(l) → Ca(OH)₂(uq) + 2H₂(g)
SiCl₄(l) + 4H₂O(l) → SiO₂ -2H₂O(s) + 4HCl(aq)
Al₄C₃ + 12H₂O(l) → 4Al(OH)₃ + 3CH₄
Ca₃P₂(s) + 6H₂O → 3Ca(OH)₂ + 2PH₃

Question 30.
Knowing the properties of H20 and DaO, do you think that D20 can be used for drinking purposes?
Answer:
Heavy water (D₂O) acts as a moderator, i.e., it slows the rate of a reaction. Due to this property of D₂O, it cannot be used for drinking purposes because it will slow down anabolic and catabolic reactions takes place in the body and lead to a casualty.

Question 31.
What is the difference between the terms ‘hydrolysis’ and ‘hydration’?
Answer:
Hydrolysis is defined as a chemical reaction in which hydrogen and hydroxide ions (H+ and OH- ions) of water molecule react with a compound to form products. For example :
NaH + H₂O → NaOH + H₂
Hydration is defined as the addition of one or more water molecules to ions or molecules to form hydrated compounds. For example :
CUSO₄ + 5H₂O → CUSO₄ . 5H₂O

Question 32.
How can saline hydrides remove traces of water from organic compounds? ’
Answer:
Saline hydrides are ionic in nature. They react with water to form a metal hydroxide along with the liberation of hydrogen gas. The reaction of saline hydrides with water can be represented as :
AH(s) + H20(Z) → AOH(aq) + H2(g)
(where, A= Na, Ca, )

When added to an organic solvent, they react with water present in it. Hydrogen escapes into the atmosphere leaving behind the metallic hydroxide.Then, the dry organic solvent distills over.

Question 33.
What do you expect the nature of hydrides is, if formed by elements of atomic number 15, 19, 23 and 44 with dihydrogen? Compare their behaviour towards water,
Answer:
The elements of atomic number 15, 19, 23 and 44 are nitrogen, potassium, vanadium and ruthenium respectively.
1. Hydride of nitrogen
Hydride of nitrogen (NH₃) is a covalent molecule. It is an electron-rich hydride owing to the presence of excess electrons as a lone pair on nitrogen.

2. Hydride of potassium
Dihydrogen forms an ionic hydride with potassium owing to the high electropositive nature of potassium. It is crystalline and non-volatile in nature.

3. Hydrides of Vanadium and Ruthenium
Both vanadium and ruthenium belong to the d-block of the periodic table. The metals of d-block form metallic or non-stoichiometric hydrides. Hydrides of vanadium and ruthenium are therefore, metallic in nature having a deficiency of hydrogen.

4. Behaviour of hydrides towards water
Potassium hydride reacts violently with water as :
KH(s) + H₂O(aq) → KOH(aq) + H₂(g)
Ammonia (NH₃) behaves as a Lewis base and reacts with water as :
H₂O(Z) + NH₃(aq) ⇌ OH^–(aq) + \(\mathrm{NH}_{4}^{+}\)(aq)
Hydrides of vanadium and ruthenium do not react with water. Hence, the increasing order of reactivity of the hydrides is as: H < NH₃ < KH.

Question 34.
Do you expect different products in solution when aluminium (III) chloride and potassium chloride treated separately with
(i) normal water (ii) acidified water, and (iii) alkaline water? Write equations wherever necessary.
Answer:
Potassium chloride (KCl) is the salt of a strong acid (HCl) and strong base (KOH). Hence, it is neutral in nature and does not undergo hydrolysis in normal water. It dissociates into ions as follows :

In acidified and alkaline water, the ions do not react and remain as such. Aluminium (III) chloride is the salt of a strong acid (HCl) and weak base [Al(OH)₃]. Hence, it undergoes hydrolysis in normal water.

In acidified water, H⁺ ions react with Al(OH)₃ forming water and giving Al³⁺ ions. Hence, in acidified water, AlCl₃ will exist as Al³⁺(aq) and Cl^–(aq) ions.

In alkaline water the following reaction takes place :

Question 35.
How does H₂O₂ behave as a bleaching agent?
Answer:
H₂O₂ acts as a bleaching agent due to the nascent oxygen.
H₂O₂ → H₂O + O
Coloured matter + [O] → Colourless matter
It bleaches materials like silk, hair, ivory, cotton, wool, etc.

Question 36.
What do you understand by the terms :
(i) hydrogen economy (ii) hydrogenation (iii) ‘syngas’ (iv) water-gas shift reaction (v) fuel-cell?
Answer:
(i) Hydrogen economy : Hydrogen economy is a technique of using dihydrogen in an efficient way. It involves transportation and storage of dihydrogen in the form of liquid or gas.
Dihydrogen releases more energy than petrol and is more eco-friendly. Hence, it can be used in fuel cells to generate electric power. Hydrogen economy is about the transmission of this energy in the form of dihydrogen.

(ii) Hydrogenation : It refers to the addition of dihydrogen to another reactant. This process is used to reduce a compound in the presence of a suitable catalyst. For example, hydrogenation of vegetable oil using nickel as a catalyst gives edible fats such as vanaspati ghee etc.

(iii) Syngas : Syngas is a mixture of carbon monoxide and dihydrogen. Since the mixture of the two gases is used for the synthesis of methanol, it is called syngas, synthesis gas, or water gas.

Syngas is produced on the action of steam with hydrocarbons or coke at a high temperature in the presence of a catalyst.

(iv) Water gas shift reaction : The production of hydrogen by reacting carbon monoxide (CO) of syngas mixtures with steam in the presence of iron chromate as catalyst is called water-gas shift reaction.

CO₂ is removed by scrubbing with sodium arsenite solution.
(v) Fuel cell: It is a device which converts the energy produced during the combustion of a fuel directly into electrical energy. One such fuel cell is hydrogen-oxygen fuel cell. It does not cause any pollution. Fuel cells generated electricity with conversion efficiency of 70-85%.

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 10 The s-Block Elements Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 10 The s-Block Elements

PSEB 11th Class Chemistry Guide The s-Block Elements InText Questions and Answers

Question 1.
What are the common physical and chemical features of alkali metals?
Answer:
Physical properties of alkali metals are as follows :

1. They are quite soft and can be cut easily. Sodium metal can be easily cut using a knife.
2. They are light coloured and are mostly silvery white in appearance.
3. They have low density because of the large atomic sizes. The density increases down the group from Li to Cs. The only exception to this is K, which has lower density than Na.
4. The metallic bonding present in alkali metals is quite weak. Therefore they have low melting and boiling points.
5. Alkali metals and their salts impart a characteristic colour to flames. This is because the heat from the flame excites the electron present in the outermost orbital to a high energy level. When this excited electron reverts back to the ground state, it emits excess energy as radiation that falls in the visible region.
6. They also display photoelectric effect. When metals such as Cs and K are irradiated with light, they lose electrons.

Chemical properties of alkali metals are as follows :

Alkali metals are highly reactive due to their low ionization enthalpy. As we move down the group, the reactivity increases.
(1) They react with water to form respective oxides or hydroxides. As we move down the group, the reaction becomes more and more spontaneous.

(2) They react with water to form their respective hydroxides and dihydrogens. The general reaction for the same is given as :
2M + 2H²O → 2M⁺ + 2OH^– + H₂

(3) They react with dihydrogen to form metal hydrides. These hydrides are ionic solids and have high melting points.
2M + H₂ → 2M⁺ H^–

(4) Almost all alkali metals, except Li, react directly with halogens to form ionic halides.
2M + Cl₂ → 2MCl (M = Li, K, Rb, Cs)

Since Li⁺ ion is very small in size, it can easily distort the electron cloud around the negative halide ion. Therefore lithium halides are covalent in nature.

(5) They are strong reducing agents. The reducing power of alkali metals increases on moving down the group.
However, lithium is an exception. It is the strongest reducing agent among the alkali metals. It is because of its high hydration energy.

(6) They dissolve in liquid ammonia to form deep blue coloured solutions. Therefore, these solutions are conducting in nature.
M + (x + y) NH₃ → [M(NH₃)_x]⁺[M(NH₃)_y]^–

The ammoniated electrons cause the blue colour of the solution. These solutions are paramagnetic and if allowed to stand for some time, then they liberate hydrogen. This results in the formation of amides.
M⁺ + e^– + NH₃(l) → MNH + \(\frac{1}{2}\)H₂(g)

In a highly concentrated solution, the blue colour changes to bronze and the solution becomes diamagnetic.

Question 2.
Discuss the general characteristics and gradation in properties of alkaline earth metals.
Answer:
Physical and atomic properties of alkaline earth metals are as follows:

• The general electronic configuration of alkaline earth metals is (noble gas) ns².
• These metals lose two electrons to acquire the nearest noble gas configuration. Therefore, their oxidation state is +2.
• These metals have atomic and ionic radii smaller than that of alkali metals. Also when moved down the group, the effective nuclear charge decreases and this causes an increase in their atomic radii and ionic radii.
• Since the alkaline earth metals have large size, their ionization enthalpies are found to be fairly low. However, their first ionization enthalpies are higher than the corresponding group 1 metals.
• These metals are lustrous and silvery white in appearance. They are relatively less soft as compared to alkali metals.
• Atoms of alkaline earth metals are smaller than that of alkali metals. Also they have two valence electrons forming stronger metallic bonds. These two factors cause alkaline earth metals to have high melting and boiling points as compared to alkali metals.
• They are highly electropositive in nature. This is due to their low ionization enthalpies. Also the electropositive character increases on moving down the group from Be to Ba.
• Ca, Sr, and Ba impart characteristic colours to flames.
  Ca – Brick red Sr – Crimson red Ba – Apple green
  In Be and Mg, the electrons are too strongly bound to be excited. Hence, these do not impart any colour to the flame.

Chemical properties of alkaline earth metals are as follows :
The alkaline earth metals are less reactive than alkali metals and their reactivity increases on moving down the group.

(i) Reaction with air and water : Be and Mg are almost inert to air and water because of the formation of oxide layer on their surface.
(a) Powdered Be burns in air to form BeO and Be₃N₂.
(b) Mg, being more electropositive, burns in air with a dazzling sparkle to form MgO and Mg₃N₂.
(c) Ca, Sr, and Ba react readily with air to form respective oxides and nitrides.
(d) Ca, Ba, and Sr react vigorously even with cold water.

(ii) Alkaline earth metals react with halogens at high temperatures to form halides.
M + X₂ → MX₂ (X = F, Cl, Br, I)
(iii) All the alkaline earth metals, except Be, react with hydrogen to form hydrides.
(iv) They react readily with acids to form salts and liberate hydrogen gas.
M + 2HCl → MCl₂ + H₂(g) ↑
(v) They are strong reducing agents. However, their reducing power is less than that of alkali metals. As we move down the group, the reducing power increases.
(vi) Similar to alkali metals, the alkaline earth metals also dissolve in liquid ammonia to give deep blue coloured solutions.
M + (x – y) NH₃ → [M (NH₃)_x]²⁺ + 2 [e (NH₃)_y]^–

Question 3.
Why are alkali metals not found in nature?
Answer:
Alkali metals include lithium, sodium, potassium, rubidium, cesium, and francium. These metals have only one electron in their valence shell, which they lose easily, owing to their low ionization energies. Therefore, alkali metals are highly reactive and are not found in nature in their elemental state.

Question 4.
Find out the oxidation state of sodium in Na₂O₂.
Answer:
Let the oxidation state of Na be x. The oxidation state of oxygen, in case of peroxides, is -1.
Therefore, 2(x) + 2(-1) = 0 ⇒ 2x – 2 = 0 ⇒ 2x = 2 ⇒ x = ±1
Therefore, the oxidation state of sodium in Na₂O₂ is +1.

Question 5.
Explain why is sodium less reactive than potassium?
Answer:
In alkali metals, on moving down the group, the atomic size increases and the effective nuclear charge decreases.
Because of these factors, the outermost electron in potassium can be lost easily as compared to sodium. Hence, potassium is more reactive than sodium.

Question 6.
Compare the alkali metals and alkaline earth metals with respect to (i) ionization enthalpy (ii) basicity of oxides and (iii) solubility of hydroxides.
Answer:

• Ionization enthalpies : The first ionization enthalpies of the alkaline earth metals are higher than those of the corresponding alkali metals. This is due to their small size as compared to the corresponding alkali metals. But second ionization enthalpies of the alkaline earth metals are smaller than those of the corresponding alkali metals.
• Basicity of oxides : The oxides of the alkali and alkaline earth metals dissolves in water to form basic hydroxides. The alkaline earth metal hydroxides are however less basic and less stable than alkali metal hydroxides.
• Solubility of hydroxides : The solubility of hydroxides of alkaline earth metals is relatively less than their corresponding alkali metal hydroxides.

Question 7.
In what ways lithium shows similarities to magnesium in its chemical behaviour?
Answer:
Similarities between lithium and magnesium are as follows :
(i) Both Li and Mg react slowly with cold water.
(ii) The oxides of both Li and Mg are much less soluble in water and their hydroxides decompose at high temperature.
im 1

(iii) Both Li and Mg react withN2 to form nitrides.
im 2

(iv) Neither Li nor Mg form peroxides or superoxides.
(v) The carbonates of both are covalent in nature. Also, these decompose on heating.
im 3

(vi) Li and Mg do not form solid bicarbonates.
(vii) Both LiCl and MgCl₂ are soluble in ethanol owing to their covalent nature.
(viii) Both LiCl and MgCl₂ are deliquescent in nature. They crystallize from aqueous solutions as hydrates, for example, LiCl.₂H₂O and MgCl₂ . 8H₂O.

Question 8.
Explain why can alkali and alkaline earth metals not be obtained by chemical reduction methods?
Answer:
In the process of chemical reduction, oxides of metals are reduced using a stronger reducing agent. Alkali metals and alkaline earth metals are among the strongest reducing agents and the reducing agents that are stronger than them are not available. Therefore, they cannot be obtained by chemical reduction of their oxides.

Question 9.
Why are potassium and caesium, rather than lithium used in photoelectric cells?
Answer:
Potassium and caesium have much lower ionization enthalpy than that of lithium. Therefore, these metals on exposure to light emit electrons easily but lithium does not. That’s why K and Cs rather than Li are used in photoelectric cells.

Question 10.
When an alkali metal dissolves in liquid ammonia the solution can acquire different colours. Explain the reasons for this type of colour change.
Answer:
When an alkali metal is dissolved in liquid ammonia, it results in the formation of a deep blue coloured solution.
M + (x + y) NH₃ → [M (NH₃)_x]⁺ + [e^-1 (NH₃)_y] [Ammoniated electron]
The ammoniated electrons absorb energy corresponding to red region of visible light. Therefore, the transmitted light is blue in colour.
At a higher concentration (3 M), clusters of metal ions are formed. This causes the solution to attain a copper-bronze colour and a characteristic metallic lustre.

Question 11.
Beryllium and magnesium do not give colour to flame whereas other alkaline earth metals do so. Why?
Answer:
When an alkaline earth metal is heated, the valence electrons get excited to a higher energy level. When this excited electron comes back to its lower energy level, it radiates energy, which belongs to the visible region. Hence, the colour is observed. In Be and Mg, the electrons are strongly bound. The energy required to excite these electrons is very high. Therefore, when the electron reverts back to its original position the energy released does not fall in the visible region. Hence, no colour in the flame is seen.

Question 12.
Discuss the various reactions that occur in the Solvay process.
Answer:
In Solvay ammonia process, CO₂ is passed through brine, (a concentrated solution of NaCl) saturated with ammonia. The process involves the formation of a sparingly soluble sodium bicarbonate.
NaCl + NH₃ + CO₂ + H₂O → NaHCO₃↓ + NH₄Cl .
Sodium bicarbonate thus formed is filtered, dried and heated to obtain sodium carbonate.
im 4
CO₂ used in carbonating tower is prepared by heating calcium carbonate and the quicklime, CaO thus formed is dissolved in water to form slaked lime, Ca(OH)₂.
im 5
In ammonia recovery tower, NH3 is prepared by heating NH₄Cl with Ca(OH)₂.
im 6

Question 13.
Potassium carbonate cannot be prepared by Solvay process. Why?
Answer:
Solvay process cannot be used to prepare potassium carbonate. This is because unlike sodium bicarbonate, potassium bicarbonate is fairly soluble in water and does not precipitate out.

Question 14.
Why is Li₂CO₃ decomposed at a lower temperature whereas Na₂CO₃ at higher temperature?
Answer:
As we move down the alkali metal group, the electropositive character increases. This causes an increase in the stability of alkali carbonates.
However, lithium carbonate is not so stable to heat. This is because lithium carbonate is covalent. Lithium ion, being very small in size, polarizes a large carbonate ion, leading to the formation of more stable lithium oxide.
im 7
Therefore, lithium carbonate decomposes at a low temperature while a stable sodium carbonate decomposes at a high temperature.

Question 15.
Compare the solubility and thermal stability of the following compounds of the alkali metals with those of the alkaline earth metals (a) Nitrates (b) Carbonates (c) Sulphates.
Answer:
(a) Nitrates of alkali metals and alkaline earth metals :
(i) Nitrates of alkali metals are thermally not stable and decompose on heating to give MNO₂ and O₂ (except LiN03) whereas nitrates of alkaline earth metals decompose on heating give their oxides, nitrogen dioxide and oxygen gas.

im 8

(ii) Nitrates of alkali metals are highly soluble in water whereas alkaline earth metal nitrates are sparingly soluble and crystallise with six molecules of water.
(b) Carbonates of alkali metals and alkaline earth metals:
(i) Carbonates of alkali metals except Li are quite stable upto 1273 K and do not decompose, whereas carbonates of alkaline earth metals decompose at different temperatures, to give their oxides and carbon dioxide.

im 9

The thermal stability of carbonates of alkaline earth metals increase down the group.
BeCO₃ is least stable and BaCO₃ is most stable.

(ii) All the carbonates of alkali metals are generally soluble in water and their solubility increases rapidly on descending the group. This is due to the reason that their lattice energies decrease more rapidly than their hydration energies down the group. In the case of carbonates of alkaline earth metals they are sparingly soluble in water and their solubility decreases down the group from Be to Ba. For example, MgCO₃ is slightly soluble in water, but BaC03 is almost insoluble.

(c) Sulphates of alkali metals and alkaline earth metals :
(i) The sulphates of alkali metals are thermally quite stable whereas the sulphates of alkaline earth metals decompose on heating to give oxides and SO₃. The temperature of decomposition increases
down the group.
im 10

(ii) The sulphates of alkali metals Na and K are soluble in water. As far as the solubility of sulphates of alkaline earth metals in water is concerned, BeSO₄ and MgSO₄ are highly soluble, CaSO₄ is sparingly soluble, but the sulphates of Sr, Ba and Ra are virtually insoluble. Thus, the solubility of their sulphates in water decreases down the group.
BeSO₄ > MgSO₄ > CaSO₄ > SiSO₄ > BaSO₄

Question 16.
Starting with sodium chloride how would you proceed to prepare (i) sodium metal (ii) sodium hydroxide (iii) sodium peroxide (iv) sodium carbonate?
Answer:
(i) Sodium metal from sodium chloride: Sodium is prepared from fused (molten) sodium chloride. Sodium chloride is mixed with CaCl₂ and KF [to lower the M.Pt. of NaCl to 850-875 K] and subjected to electrolysis (in DOWN’S CELL) when the following reactions occur :

im 11

Sodium, liberated at the cathode, is collected in kerosene oil and chlorine gas is liberated at the anode.
(ii) Sodium hydroxide from sodium chloride: Sodium hydroxide (caustic soda) is generally prepared by the electrolysis of brine solution (NaCl solution in water) in Castner Kellner cell. A mercury cathode and carbon anode are used. Sodium metal discharged at the cathode combines with mercury to form sodium amalgam. Cl₂ gas is evolved at the anode.

im 12

The amalgam is treated with water to give sodium hydroxide and hydrogen gas.
2Na – amalgam + 2H₂O → 2NaOH + 2Hg + H₂↑

(iii) Sodium peroxide from sodium chloride: Sodium metal obtained, by the electrolysis of molten sodium chloride is heated with O₂ at about 575 K when sodium forms mainly sodium peroxide.

im 13

(iv) Sodium carbonate from sodium chloride: Sodium carbonate is prepared from an aqueous solution of NaCl by SOLVAY PROCESS. In this process, CO₂ is passed through NaCl solution saturated with ammonia, when following reactions occur :

2NH₃ + H₂O + CO₂ → (NH₄)₂ CO₃
(NH₄)₂ CO₃ + H₂O + CO₂ → 2NH₄HCO₃
NH₄HCO₃ + NaCl → NH₄Cl + NaHCO₃
im 14

Question 17.
What happens when (i) magnesium is burnt in air (ii) quick lime is heated with silica (iii) chlorine reacts with slaked lime (iv) calcium nitrate is heated?
Answer:
(i) Magnesium bums in air with a dazzling light to form MgO and Mg3N₂.
im 15

(ii) Quick lime (CaO) combines with silica (SiO₂) to form slag.
im 16

(iii) When chlorine is added to slaked lime, it gives bleaching powder.
im 17

(iv) Calcium nitrate, on heating decomposes to give calcium oxide.
im 18

Question 18.
Describe two important uses of each of the following (i) caustic soda (ii) sodium carbonate (iii) quicklime.
Answer:
(i) Uses of caustic soda
(a) It is used in soap industry.
(b) It is used as a reagent in laboratory.
(ii) Uses of sodium carbonate
(a) It is generally used in glass and soap industry.
(b) It is used as a water softener.
(iii) Uses of quick lime
(a) It is used as a starting material for obtaining slaked lime.
(b) It is used in the manufacture of glass and cement.

Question 19.
Draw the structure of (i) BeCl₂ (vapour) (ii) BeCl₂ (solid).
Answer:
(i) In the vapour state, BeCl₂ exists as a monomer with a linear structure.
im 19

(ii) In the solid state, BeCl₂ exists as a polymer in condensed phase.
im 20

Question 20.
The hydroxides and carbonates of sodium and potassium are easily soluble in water while the corresponding salts of magnesium and calcium are sparingly soluble in water. Explain.
Answer:
The atomic size of sodium and potassium is larger than that of magnesium and calcium. Thus, the lattice energies of carbonates and hydroxides formed by calcium and magnesium are much more than those of sodium and potassium.
Hence, carbonates and hydroxides of sodium and potassium dissolve readily in water whereas those of calcium and magnesium are only sparingly soluble.

Question 21.
Describe the importance of the following (i) limestone (ii) cement (iii) plaster of paris.
Answer:
(i) Importance of limestone
(a) It is used in the preparation of lime and cement.
(b) It is used as a flux during the smelting of iron ores.

(ii) Importance of cement
(a) It is used in plastering and in construction of bridges.
(b) It is used in concrete.

(iii) Importance of plaster of Paris
(a) It is used in surgical bandages.
(b) It is also used for making casts and moulds.

Question 22.
Why are lithium salts commonly hydrated and those of the other alkali ions usually anhydrous?
Answer:
Because of its smallest size among alkali metals, Li+ has the maximum degree of hydration. That’s why lithium salts are commonly hydrated and those of other alkali metal ions usually anhydrous.
im 21

Question 23.
Why is LiF almost insoluble in water whereas LiCl soluble not only in water but also in acetone?
Answer:
LiF is insoluble in water. On the contrary, LiCl is soluble not only in water, but also in acetone. This is mainly because of the greater ionic character of LiF as compared to LiCl. The solubility of a compound in water depends on the balance between lattice energy and hydration energy. Since fluoride ion is much smaller in size than chloride ion, the lattice energy of LiF is greater than that of LiCl. Also there is not much difference between the hydration energies of fluoride ion and chloride ion. Thus, the net energy change during the dissolution of LiCl in water is more exothermic than that during the dissolution of LiF in water. Hence, low lattice energy and greater covalent character are the factors making LiCl soluble not only in water, but also in acetone.

Question 24.
Explain the significance of sodium, potassium, magnesium and calcium in biological fluids.
Answer:
(i) Sodium (Na):
Sodium ions are found primarily in the blood plasma. They are also found in the interstitial fluids surrounding the cells.
(a) Sodium ions help in the transmission of nerve signals.
(b) They help in regulating the flow of water across the cell membranes.
(c) They also help in transporting sugars and amino acids into the cells.

(ii) Potassium (K):
Potassium ions are found in the highest quantity within the cell fluids.
(a) K⁺ ions help in activating many enzymes.
(b) They also participate in oxidising glucose to produce ATP.
(c) They also participate in transmitting nerve signals.

(iii) Magnesium (Mg) and calcium (Ca) :
Magnesium and calcium are referred to as macro-minerals. This term indicates their higher abundance in the human body system.
(a) Mg helps in relaxing nerves and muscles.
(b) Mg helps in building and strengthening bones.
(c) Mg maintains normal blood circulation in the human body system.
(d) Ca helps in the coagulation of blood.
(e) Ca also helps in maintaining homeostasis.

Question 25.
What happens when
(i) sodium metal is dropped in water?
(ii) sodium metal is heated in free supply of air?
(iii) sodium peroxide dissolves in water?
Answer:
(i) When sodium metal is dropped in water, it reacts violently to form sodium hydroxide and hydrogen gas. The chemical equation involved in the reaction is:
2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)
(ii) On being heated in air, sodium reacts vigorously with oxygen to form sodium peroxide. The chemical equation involved in the reaction is:
2Na(s) + O₂(g) → Na₂O₂(s)
(iii) When sodium peroxide is dissolved in water, it is readily hydrolysed to form sodium hydroxide and water. The chemical equation involved in the reaction is:
Na₂O₂(s) + 2H₂O(7) → 2NaOH(aq) + H₂O₂(aq)

Question 26.
Comment on each of the following observations:
(a) The mobilities of the alkali metal ions in aqueous solution are Li⁺ < Na⁺ < K⁺ < Rb⁺ < Cs⁺
(b) Lithium is the only alkali metal to form a nitride directly.
(c) \(\mathbf{H}^{\ominus}\) for M²⁺ (aq) + 2e^– → M(s) (where M = Ca, Sr or Ba) is nearly constant.
Answer:
(a) On moving down the alkali group, the ionic and atomic sizes of the metals increase. The given alkali metal ions can be arranged in the increasing order of their ionic sizes as:
Li⁺ < Na⁺ < K⁺ < Rb⁺ < Cs⁺

Smaller the size of an ion, the more highly is it hydrated. Since, Li⁺ is the smallest, it gets heavily hydrated in an aqueous solution. On the other hand, Cs+ is the largest and so it is the least hydrated. The given alkali metal ions can be arranged in the decreasing order of their hydrations as:
Li⁺ > Na⁺ > K⁺ > Rb⁺ > Cs⁺
Greater the mass of a hydrated ion, the lower is its ionic mobility. Therefore, hydrated Li⁺ is the least mobile and hydrated Cs⁺ is the most mobile. Thus, the given alkali metal ions can be arranged in the increasing order of their mobilities as:
Li⁺ < Na⁺ < K⁺ < Rb⁺ < Cs⁺

(b) Unlike the other elements of group 1, Li reacts directly with nitrogen to form lithium nitride. This is because Li⁺ is very small in size and so its size is the most compatible with the N^3- ion. Hence, the lattice energy released is very high. This energy also overcomes the high amount of energy required for the formation of the N^3- ion.

(c) Electrode potential (\(\mathbf{E}^{\ominus}\)) of any M²⁺/M electrode depends upon three factors :
(i) Ionisation enthalpy
(ii) Enthalpy of hydration
(iii) Enthalpy of vaporisation
The combined effect of these factors is approximately the same for Ca, Sr, and Ba. Hence, their electrode potentials are nearly constant.

Question 27.
State as to why
(a) a solution of Na₂CO₃ is alkaline?
(b) alkali metals are prepared by electrolysis of their fused chlorides?
(c) sodium is found to be more useful than potassium?
Answer:
(a) When sodium carbonate is added to water, it hydrolyses to give sodium bicarbonate and sodium hydroxide (a strong base). As a result, the solution becomes alkaline.
Na₂CO₃ + H₂O → NaHCO₃ + NaOH

(b) It is not possible to prepare alkali metals by the chemical reduction of their oxides as they themselves are very strong reducing agents. They cannot be prepared by displacement reactions either (wherein one element is displaced by another). This is because these elements are highly electropositive. Electrolysis of aqueous solutions can neither be used to extract these elements. This is because the liberated metals react with water. Hence, to overcome these difficulties, alkali metals are usually prepared by the electrolysis of their fused chlorides.

(c) Blood plasma and the interstitial fluids surrounding the cells are the regions where sodium ions are primarily found. Potassium ions are located withimthe cell fluids. Sodium ions are involved in the transmission of nerve signals, in regulating the flow of water across the cell membranes, and in transporting sugars and amino acids into the cells. Hence, sodium is found to be more useful than potassium.

Question 28.
Write balanced equations for reactions between
(a) Na₂O₂ and water
(b) KO₂ and water
(c) Na₂O and CO₂
Answer:
(a) The balanced chemical equation for the reaction between Na202 and water is:
Na₂O₂(s) + 2H₂O(l) → 2NaOH(aq) + H₂O₂(aq)

(b) The balanced chemical equation for the reaction between K02 and water is:
2KO₂(s) + 2H₂O(l) → 2KOH(aq) + H₂O₂(aq) + O₂(g)
or 4KO₂(s) + 2H₂O(l) → 4KOH(aq) + 3O₂(g)

(c) The balanced chemical equation for the reaction between Na₂O and CO₂ is:
Na₂O(s) + CO₂(g) + Na₂CO₃

Question 29.
How would you explain the following observations?
(i) BeO is almost insoluble but BeSO₄ is soluble in water, ‘
(ii) BaO is soluble but BaSO₄ is insoluble in water,
(iii) Lil is more soluble than KI in ethanol.
Answer:
(i) BeO is almost insoluble in water and BeSO₄ is soluble in water. Be²⁺ is a small cation with a high polarising power and O^2- is a small anion. The size compatibility of Be²⁺ and O^2- is high. Therefore, the lattice energy released during their formation is also very high. When BeO is dissolved in water, the hydration energy of its ions is not sufficient to overcome the high lattice energy. Therefore, BeO is insoluble in water. On the other hand, \(\mathrm{SO}_{4}^{2-}\) ion is a large anion. Hence, Be²⁺ can easily polarise \(\mathrm{SO}_{4}^{2-}\) ions, making BeSO₄ unstable. Thus, the lattice energy of BeSO₄ is not very high and so it is soluble in water.

(ii) BaOis soluble in water, but BaSO₄ is not. Ba²⁺ is a large cation and O^2- is a small anion. The size compatibility of Ba²⁺ and O^2- is not high. As a result, BaO is unstable. The lattice energy released during its formation is also not very large. It can easily be overcome by the hydration energy of the ions. Therefore, BaO is, soluble in water. In BaSO₄, Ba+ and \(\mathrm{SO}_{4}^{2-}\) are both large-sized. The lattice energy released is high. Hence, it is not soluble in water.

(iii) Lil is more soluble than KI in ethanol. As a result, due to its small size, the lithium ion has a higher polarising power than the potassium ion. It polarises the electron cloud of the iodide ion to a much greater extent than the potassium ion. This causes a greater covalent character in Li than in KI. Hence, Lil is more soluble in ethanol.

Question 30.
Which of the alkali metal is having least melting point?
(a) Na (b) K (c) Rb (d) Cs
Answer:
(d) Atomic size increases as we move down the alkali group. As a result, the binding energies of their atoms in the crystal lattice decrease. Also, the strength of metallic bonds decreases on moving down a group in the periodic table.
This causes a decrease in the melting point. Among the given metals, Cs is the largest and has the least melting point.

Question 31.
Which one of the following alkali metals gives hydrated salts?
(a) Li (b) Na (c) K (d) Cs
Answer:
(a) Smaller the size of an ion, the more highly is it hydrated. Among the given alkali metals, Li is the smallest in size. Also, it has the highest charge density and highest polarising power. Hence, it attracts water molecules more strongly than the other alkali metals. As a result, it forms hydrated salts such as LiCl • 2 H20. The other alkali metals are larger than Li and have weaker charge densities. Hence, they usually do not form hydrated salts.

Question 32.
Which one of the alkaline earth metal carbonates is thermally the most stable?
(a) MgCO₃
(b) CaCO₃
(c) SrCO₃
(d) BaCO₃
Answer:
(d) Thermal stability increases with the increase in the size of the cation present in the carbonate. The increasing order of the cationic size of the given alkaline earth metals is
Mg < Ca < Sr < Ba
Hence, the increasing order of the thermal stability of the given alkaline earth metal carbonates is
MgCO₃ < CaCO₃ < SrCO₃ < BaCO₃

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 1 Some Basic Concepts of Chemistry Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 1 Some Basic Concepts of Chemistry

Very Short Answer Type Questions

Question 1.
A liquid has a volume of 49.0 cm³ and a mass of 57.642 g. Find out the density of this liquid in SI unit.
Answer:
Density = \(\frac{\text { Mass }(\mathrm{kg})}{\text { Volume }\left(\mathrm{m}^{3}\right)}\)
Mass = 57.642 g = 57.642 x 10^-3 kg
Volume = 49.0 cm³ = 49.0 x (10^-3)³ = 49.0 x 10^-6 m³
∴ Density = \(\frac{57.642 \times 10^{-3}}{49.0 \times 10^{-6}}\) = 1.176 x 10³ kg/m³

Question 2.
Convert the following temperatures into degree Fahrenheit,
(i) 25°C, physiological (human body) temperature.
(ii) 35°C, the room temperature.
Answer:
(i) Given
C = 25 °C
°F = \(\frac{9}{5}\) °C + 32 = \(\frac{9}{5}\) x 25 + 32 = 45 + 32 = 77°F

(ii) Given, C = 35 °C
°F = \(\frac{9}{5}\) °C + 32 = \(\frac{9}{5}\) x 35 + 32 = 63 + 32 = 95°F

Question 3.
What is the mass (in grams) of a copper block whose dimensions are 5.0 inch x 6.0 inch and whose density is 8.96 g/cm³? Given that 1 inch = 2.54 cm
Answer:
Here, unit conversion factors are 1 = \(\frac{2.54 \mathrm{~cm}}{1 \text { inch }}=\frac{1 \text { inch }}{2.54 \mathrm{~cm}}\)
Hence, required mass (in g) = 5.0 inch x 6.0 inch x 4.0 inch
\(\) = 1.76 x 10⁴ g.

Question 4.
If 6.3 g of NaHCO₃ are added to 15.0 g of CH₃COOH solution, the residue is found to weigh 18.0 g. What is the mass of CO₂ released in the reaction?
Answer:

Sum of the masses of reactants = 6.3 + 15 = 21.3 g
Sum of the masses of products = x + 18
21.3 = x +18; x = 21.3 – 18 = 3.3 g
Thus, the mass of the C02 released is 3.3 g.

Question 5.
Why is the law of Gay Lussac’s not obeyed if any reactant or product is not a gas?
Answer:
If any reactant or product is a liquid or solid, the volume occupied by them is extremely small as compared to the gas and hence, the law is not obeyed.

Question 6.
Calculate the normality of solution containing 62.3 g of hydrated copper sulphate (CuSO₄ . 5H₂O) in 500 mL of solution.
Answer:
Mass of solute = 62.3 g
Equivalent mass of oxalic acid = \(\frac{249.5}{2}\) = 124.75 g
Gram equivalents of oxalic acid = \(\frac{62.3}{124.75}\) = 0.5
Volume of solution = 500 mL
Normality = \(\frac{0.5}{500}\) x 1000 = 1N

Question 7.
What is the difference between molality and molarity?
Answer:
Molality : It is defined as the number of moles of solute dissolved in 1 kg of solvent. It is independent of temperature. ,
Molarity : It is defined as the number of moles of solute dissolved in 1 L of solution. It depends upon temperature because, volume of solution °c temperature.

Question 8.
How is the term material different from matter?
Answer:
Anything which has mass and occupies space is called matter. However, material corresponds to the matter which has specific use.

Question 9.
Why are the atomic masses of most of the elements fractional?
Answer:
It is because most of the elements occur in nature as a mixture of isotopes and their atomic masses are the average relative atomic masses of the isotopes depending on their abundance.

Question 10.
Volume of a solution changes with change in temperature, then, will the molality of the solution be affected by temperature? Give reason for your answer.
Answer:
No, molality of solution does not change with temperature since mass remains unaffected with temperature.

Short Answer Type Questions

Question 1.
Express the following in the scientific notation.
(i) 0.000968
(ii) 157428
(iii) 90,000
(iv) (5.7 x 10⁶) x (4.2 x 10^-2)
(v) (6.8 x 10^-9) + (1.4 x 10^-6)
(vi) (456 x 10³ + 2.62 x 10²)
(vii) (9.87 x 10^-3 – 2.26 x 10^-4)
Answer:
(i) 0.000968 = 9.68 x 10^-4
(ii) 157428 = 1.57428 x 10⁵
(iii) 90,000 = 9 x 10⁴

(iv) (5.7 x 10⁶)x (4.2 x 10^-2) = (5.7 x 4.2)(10^6-2)
= 23.94 x 10⁴ =2.394 x 10⁵
(v) (6.8 x 10^-9) ÷ (1.4 x 10^-6) = x (10^-9-(-6)) = 4.857 x 10^-3
(vi) (4.56 x 10³ + 2.62 x 10²)= 45.6 x 10² + 2.62 x 10²
=(45.6 +2.62) x 10²
= 48.22 x 10²
= 4.822 x 10³
(vii) (9.87 x 10^-3 — 2.26 x 10^-4) = 9.87 x 10^-3 — 0.226 x 10^-3
= (9.87 – 0.226) x 10^-3 = 9.644 x 10^-3

Question 2.
The percentage composition of elements in NH3, H20, and N20 3 is given below. –
NH₃ → 82.35% N and 17.65% H
H₂O → 88.90% O and 11.10% H
N₂O₃ → 6a 15% O and 36.85% N
Show that these data are in accordance with the law of reciprocal proportion.
Answer:
(i) In NH₃, 1 part of H reacts with = \(\) = 4.67 part of N
(ii) In H₂O, 1 part of H reacts with = \(\) = 8.01 part of O
Thus, the ratio N : O :: 4.67 : 8.01 = 1 :1.72
(iii) In N₂O₃, N and O reacts with each other in the ratio N : O :: 36.85: 63.15 = 1 :1.71.
Thus, the two ratios are the same. Hence, it illustrates the law of reciprocal proportions.

Question 3.
A solution contains 25% water, 25% ethanol and 50% acetic acid by mass. Calculate the mole fraction of each component.
Answer:
Let the total mass of solution = 100 g
Mass of water = 25 g, Mass of ethanol = 25 g
Mass of acetic acid = 50 g
Moles of water = \(\frac{25}{18}\) = 1.388 (∵ Molar mass of H₂O= 18)
Moles of ethanol = \(\frac{25}{46}\) = 0.543 (∵ Molar mass of C₂H₅OH = 18)
Moles of acetic acid = \(\frac{50}{60}\) = 0.833 (∵ Molar mass of CH₃COOH = 18)
Total number of moles = 1.388 + 0.543 + 0.833 = 2.764
Mole fraction of water = \(\frac{1.388}{2.764}\) = 0.502
Mole fraction of ethanol = \(\frac{0.543}{2.764}\) = 0.196
Mole fraction of acetic acid = \(\frac{0.833}{2.764}\) = 0.302

Question 4.
Calculate the molality of a solution containing 20.7 g potassium carbonate dissolved in 500 mL of solution (assume density of solution = 1 g mL^-2).
Mass of K₂CO₃ = 20.7 g
Molar mass of K₂CO₃ = 2 x 39 + 12 + 3 x 16 = 138 mol^-1
Moles of K₂CO₃ = \(\frac{20.7}{138}\) = 0.15 .
Mass of solution = (500 mL) x (1 g mL^-1) = 500 g
Amount of water = 500 – 20.7 = 479.3 g
Molality = \(\frac{Moles of solute}
{Mass of solvent in gram}\) x 100
= \(\frac{0.15}{479.3}\) x 1000 = 0.313 mx`

Question 5.
45.4 L of dinitrogen reacted with 22.7 L of dioxygen and 45.4 L of nitrous oxide was formed. The reaction is given below 2N₂(g) + O₂(g) → 2N₂O(g)
Which law is being obeyed in this experiment? Write the statement of the law. [NCERT Exemplar]
Answer:

Hence, the ratio between the volumes of the reactants and the product in the given question is simple i.e., 2:1:2. It proves the Gay-Lussac’s law of gaseous volumes.
Gay-Lussac’s law of gaseous volumes : The law of combining volume states that when gases react together to form other gases, and when all volumes are measured at the same temperature and pressure, the ratio between the volumes of the gaseous reactants and products can be expressed in simple whole numbers.

Long Answer Type Questions

Question 1.
Copper oxide was prepared by the following methods.
(i) In first case, 1.75 g of the metal was dissolved in nitric acid and igniting the residual copper nitrate yielded 2.19 g of copper oxide.
(ii) In the second case, 1.14 g of metal dissolved in nitric acid were precipitated as copper hydroxide by adding caustic alkali solution. The precipitated copper hydroxide after washing, drying and heating yielded 1.43 g of copper oxide.
(iii) In the third case, 1.46 g of copper when strongly heated in a current of air yielded 1.83 g of copper oxide. Show that the given data illustrate the law of definite composition.
Answer:
Step I and II In the first experiment,
2.19 g of copper oxide contained 1.75 g of Cu.
∴ 100 g of copper oxide contained Cu
i.e., %ofCu = 79.91

In the second experiment,
1.43 g of copper oxide contained 1.14 g copper.
∴ 100 g of copper oxide contained Cu = \(\frac{1.14}{1.43}\) x 100 = 79.72 g,
i.e„ % of Cu = 79.72

In the third experiment,
1.83 g of copper oxide contained 1.46 g of copper
∴ 100 g of copper oxide contained Cu = \(\frac{1.46}{1.83}\) x 100 = 79.78g.
i.e., % of Cu = 79.78
Step III The percentage of copper in copper oxide derived from all the three experiments is nearly the same. Hence, the above data illustrate the law of definite composition.

Question 2.
Three oxides of nitrogen contained 63.6%, 46.7% and 30.4% nitrogen respectively. Show that these figures illustrate the law of multiple proportions.
Answer:
In case first,
Step I The oxide of nitrogen contains 63.6% N ‘ i.e., 63.6 g of N reacts with (100 – 63.6) g of O = 36.4 g of O.
Step II ∴ 1 g of N will react with \(\frac{36.4}{63.6}\) g of O = 0.57 g of O.

In case second,
Step I The oxide of nitrogen contains 46.7% N
i.e., 46.7 g of N reacts with (100 – 46.7) g of O = 53.3 g of O.

In case third,
Step I The oxide of nitrogen contains 30.4% N
i.e., 30.4 g of N reacts with (100 – 30.4) g of O = 69.6 g of O.
Step II ∴ 1 g of N will react with \(\frac{69.6}{30.4}\) of O = 2.26 g of O

Step III This means the ratio of the masses of oxygen which combine with 1 g of nitrogen is 0.57 : 1.14 : 2.26, i.e., 1: 2 : 4 is obviously in accordance with the law of multiple proportions.

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 2 Structure of Atom Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 2 Structure of Atom

Very Short Answer Type Questions

Question 1.
Which of the following will not show deflection from the path on passing through an electric field? Proton, cathode rays, electron, neutron
Answer:
Neutron is a neutral particle. Hence, it will not be deflected on passing through an electric field.

Question 2.
What is the nuclear radius of an atom whose mass number is 125?
Answer:
Nuclear radius, r = R₀A^1/3 where, R₀ = 1.4 x 10^-15 m,
∴ r = (1.4 x 10^-15 m) x (125)^1/3 = 7.0 x 10^-15 m.

Question 3.
The magnitude of charge on the electron is 4.8 x 10^-10 esu. What is the charge on the nucleus of a helium atom?
Answer:
Helium nucleus contains 2 protons and charge of a proton is same as that of an electron.
Therefore, the charge on the nucleus of a helium atom is (+2) x 4.8 x 10^-10 = + 9.6 x 10^-10 esu.

Question 4.
What is the difference in the origin of cathode rays and anode rays?
Answer:
Cathode rays originate from the cathode whereas anode rays do not originate from the anode. They are produced from the gaseous atoms by knock out of the electrons with high speed cathode rays.

Question 5.
What is the difference between atomic mass and mass number?
Answer:
Mass number is a whole number because it is the sum of number of protons and number of neutrons whereas atomic mass is fractional because it is the average relative mass of its atom as compared with mass of an atom of C-12 isotope taken as 12.

Question 6.
What is the difference between a quantum and a photon?
Answer:
The smallest packet of energy of any radiation is called a quantum whereas that of light is called photon.

Question 7.
Arrange s, p and rf-subshells of a shell in the increasing order of , effective nuclear charge (Zeff) experienced by the electron present in them. [NCERT Exemplar]
Ans. s-orbital is spherical in shape, it shields the electrons from the nucleus more effectively than p-orbital which in turn shields more effectively than d-orbital. Therefore, the effective nuclear charge (Zeff) experienced by electrons present in them is d < p < s.

Question 8.
Show the distribution of electrons in oxygen atom (atomic number 8) using orbital diagram.
Answer:

Question 9.
Nickel atom can lose two electrons to form Ni ion. The atomic number of Ni is 28. From which orbital will nickel lose two electrons? [NCERT Exemplar]
Answer:
₂₈Ni = 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁸, 4s²; Nickel will lose 2 electrons from 4s (outer most shell) to form Ni²⁺ ion.

Question 10.
Which of the following orbitals are degenerate?
3d_xy, 4d_xy, \(3 d_{z^{2}}\), 3d_yz, \(4 d_{z^{2}}\)
Answer:
The orbitals which belongs to same subshell and same shell are called degenerate orbitals. (3d_xy, \(3 d_{z} 2\), 3d_yz) and (4d_xy, 4dyz, 4d 2) are the two sets of degenerate orbitals.

Short Answer Type Questions

Question 1.
The Balmer series in the hydrogen spectrum corresponds to the transition from n₁ = 2 to n₂ = 3, 4 … . This series lies in the visible region. Calculate the wave number of line associated with the transition in Balmer series when the electron moves to n = 4 orbit. (R_H = 109677 cm1)
Answer:
From Rydberg formula,

Question 2.
Out of electron and proton which one will have, a higher velocity to produce matter waves of the same wavelength? Explain it.
Answer:
From de Broglie equation, wavelength, \(\lambda=\frac{h}{m v}\)
For same wavelength for two different particles, i.e., electron and proton, m₁v₁ = m₂v₂ (h is constant). Lesser the mass of the particle, greater will be the velocity. Hence, electron will have higher velocity.

Question 3.
Wavelengths of different radiations are given helow.
λ(A) = 300 nm, λ(B) = 300 pm, λ(C) = 3 nm, λ(D) = 30Å Arrange these radiations in the increasing order of their energies.
Answer:
(A) λ=3OOnm=3OO x 10^-9m
(B) λ =300µm=300 x 10^-6m
(C) λ =3nm = 3 x 10^-9 m
(D) λ = 30 = 30 x 10^-10m= 3 x 10^-9m
Energy, E = \(\frac{h c}{\lambda}\)
Therefore, E ∝ \(\frac{1}{\lambda}\)
Increasing order of energy is B

Question 4.
The electronic configuration of valence shell of Cu is 3d¹⁰4s¹ and not 3d⁹4s². How is this configuration explained?
Answer:
Configurations with completely filled and half-filled orbitals have extra stability. In 3d¹⁰4s¹, d-orbitals are completely filled and s-orbital is half-filled. Hence, it is a more stable configuration for Cu as compare to 3d⁹4s².

Question 5.
In each of the following pairs of salts, which one is more stable? (i) Ferrous and ferric salts (ii) Cuprous and cupric salts
Answer:
(i) Ferrous and ferric salts : In ferrous salts Fe²⁺, the configuration is 1s²,2s²,2p⁶,3s²,3p⁶,3d⁶. In ferric salts Fe³⁺, the configuration is 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁵. As half-filled 3d⁵ configuration is more stable therefore ferric salts are more stable than ferrous salts.
(ii) Cuprous and cupric salts : In cuprous salts, the configuration of Cu⁺ is 1s²,2s², 2p⁶, 3s², 3p⁶, 3d¹⁰. In cupric salts the configuration of Cu²⁺, is, 1s²,2s²,2p⁶,3s²,3p⁶,3d⁹. Although Cu+ has completely filled d-orbital, yet cuprous salts are less stable. This is because the nuclear charge is not sufficient enough to hold 18 electrons of Cu⁺ ion present in the outermost shell.

Long Answer Type Questions

Question 1.
When an electric discharge is passed through hydrogen gas, the hydrogen molecules dissociate to produce excited hydrogen atoms. These excited atoms emit electromagnetic radiation of discrete frequencies which can be given by the general formula \(\bar{v}=109677\left[\frac{1}{n_{i}^{2}}-\frac{1}{n_{f}^{2}}\right]\)
What points of Bohr’s model of an atom can he used to arrive at this formula? Based on these points derive the above formula giving description of each step and each term. [NCERT Exemplar]
Answer:
The two important points of Bohr’s model that can be used to derive the given formula are as follows :
(i) Electrons revolve around the nucleus in a circular path of fixed radius and energy. These paths are called orbits, stationary states or allowed energy states.
(ii) Energy is emitted or absorbed when an electron moves from higher stationary state to lower stationary state or from lower stationary state to higher stationary state respectively.
Derivation : The energy of the electron in the nth stationary state is given by the expression,
\(E_{n}=-R_{\mathrm{H}}\left(\frac{1}{n^{2}}\right)\)
n = 1,2,3 …. ……(i)
where R_H is called Rydberg constant and its value is 2.18 x 10^-18 J.
The Energy of the lowest state, also called the ground state, is
E₁ = -2.18 x 10^-18\(\left(\frac{1}{1^{2}}\right)\) = 2.18 x 10^-18J ……. (ii)
The energy gap between the two orbits is given by the equation,
ΔE = E_f – E_i … (iii)
On combining equations (i) and (iii)

Question 2.
Calculate the energy and frequency of the radiation emitted when an electron jumps from n = 3 to n = 2 in a hydrogen atom.
Answer:

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 1 Some Basic Concepts of Chemistry Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry

PSEB 11th Class Chemistry Guide Some Basic Concepts of Chemistry InText Questions and Answers

Question 1.
Calculate the molecular mass of the following:
(i) H₂O (ii)CO₂ (iii) CH₄
Answer:
(i) H₂O
The molecular mass of water (H20)
= (2 x Atomic mass of hydrogen) + (1 x Atomic mass of oxygen)
= [2(1.008 u) +1(16.00 u)]
= 2.016u +16.00u = 18.016 u=18.02u

(ii) CO₂
The molecular mass of carbon dioxide (CO₂)
= (1 x Atomic mass of carbon) + (2 x Atomic mass of oxygen)
= [1(12.011u) + 2(16.00u)]
= 12.011u + 32.00u = 44.01u

(iii) CH₄
The molecular mass of methane (CH₄)
= (1 x Atomic mass of carbon) + (4 x Atomic mass of hydrogen)
– [1(12.011u) + 4(1.008u)]
= 12.011u + 4.032u = 16.043u

Question 2.
Calculate the mass per cent of different elements present in sodium sulphate (Na₂SO₄).
Answer:
The molecular formula of sodium sulphate is Na₂SO₄
Molar mass of Na₂SO₄ = 2 x Atomic mass of Na + 1 x Atomic mass of S + 4 x Atomic mass of O
= [(2 x 23.0) + (1 x 32.066) + (4 x 16.00)]
= 46.0 + 32.066 + 64.00 = 142.066 g
Mass per cent of sodium

Question 3.
Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
Answer:

Question 4.
Calculate the amount of carbon dioxide that could be produced when
(i) 1 mol of carbon is burnt in air
(ii) 1 mole of carbon is burnt in 16g of dioxygen
(iii) 2 moles of carbon are burnt in 16g of dioxygen
Answer:

∵ 32.0g of O₂ produce 44.0 g of CO₂
∵ 16.0 g of O₂ produce = \(\frac{44}{32}\) x 16 = 22.0 g of CO₂
Amount of CO₂produced = 22.0g
(iii) Amount of CO₂produced when 2 moles (= 24 g) of C are burnt in 16.0g (limited amount ) of O₂ = 22.0g

Question 5.
Calculate the mass of sodium acetate (CH₃COONa) required to make 500mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0.245 g mol^-1.
Answer:
0.375M aqueous solution of sodium acetate = 1000mL of solution containing 0.375 moles of sodium actate
∴ Number of moles of sodium acetate in 500 mL.
= \(\frac{0.375}{1000}\) x 500 = 0.1875 mole.
Molar mass of sodium acetate = 82.0245 g mol^-1
∴ Required mass of sodium acetate
= 82.0245 g mol^-1 x 0.1875 mole
= 15.38 g

Question 6.
Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 gmLT1 and the mass per cent of nitric acid in it being 69%.
Answer:
Mass percent of nitric acid in the sample = 69%
Thus, 100 g of sample contains 69 g of nitric acid
Molar mass of nitric acid (HNO₃)
= {1 +14 + 3(16)} g mol^-1
= 1 +14 + 48 = 63 g mol^-1
∴ Number of moles in 69 g of HNO₃

Question 7.
How much copper can be obtained from 100 g of copper sulphate (CuSO₄)?
Answer:
1 mole of CuSO₄ contains 1 mole of copper.
Molar mass of CuSO₄
= (63.5) + (32.00) + 4(16.00)
= 63.5 + 32.00 + 64.00 = 159.5 g mol^-1
159.5 g of CuSO₄ contains = 63.5 g of copper
=> 100 g of CuSO₄ contains = \(\frac{63.5 \times 100 \mathrm{~g}}{159.5}\) = 39.81g
∴ Mass of copper that can be obtained from 100 g of CuSO₄ = 39.81 g

Question 8.
Determine the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are 69.9 and 30.1 respectively.
Answer:

Question 9.
Calculate the atomic mass (average) of chlorine using the following data:

	% Natural Abundance	Molar Mass
35cl	75.77	34.9689
37Cl	24.23	36.9659

Answer:
The average atomic mass of chlorine

= 26.4959 + 8.9568 = 35.4527
∴ The average atomic mass of chlorine = 35.4527

Question 10.
In three moles of ethane (C₂H₆), calculate the following:
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
(iii) Number of molecules of ethane.
Answer:
(i) 1 mole of C₂H₆ contains 2 moles of carbon atoms.
∴ 3 moles of C₂H₆ will contain = 2 x 3 = 6 moles of carbon atoms
(ii) 1 mole of C₂H₆ contains 6 moles of hydrogen atoms.
∴ 3 moles of C₂H₆ will contain =3 x 6 = 18 moles of hydrogen atoms
(iii) 1 mole of C₂H₆ contains 6.022 x 10²³ molecules of ethane.
∴ 3 moles of C₂H₆ will contain = 3 x 6.022 x 10²³
= 18.066 x 10²³ molecules of ethane

Question 11.
What is the concentration of sugar (C₁₂H₂₂O₁₁) in mol L^-1 if 20 g of sugar are dissolved in enough water to make a final volume up to 2 L?
Answer:
Molar mass of sugar (C₁₂H₂₂O₁₁)
= 12 x 12 + 22 x 1 +11 x 16 = 342 g mol^-1
Molar concentration = \(\frac{\text { Moles of solute }}{\text { Volume of solution in L }}=\frac{0.0585}{2 \mathrm{~L}}\)
= 0.0293 mol L^-1 = 0.0293 M

Question 12.
If the density of methanol is 0.793 kg L^-1, what is its volume needed for making 2.5 L of its 0.25 M solution?
Answer:
Final volume, V₂ = 2.5 L
Final molarity, M₂ = 0.25 M
Density of methanol = 0.793 kg L^-1
Molarity of initial solution M₁ = ?
Initial volume, V₁ = ?
Molar mass of methanol (CH₃OH) = (1 x 12) + (4 x 1) + (1 x 16)
= 32 g mol^-1 = 0.032 kg mol^-1
Molarity of methanol solution, M₁ = \(\frac{0.793 \mathrm{~kg} \mathrm{~L}^{-1}}{0.032 \mathrm{~kg} \mathrm{~mol}^{-1}}\) = 24.78 mol L^-1
Applying,
M₁V₁ = M₂V₂
(24.78 mol L^-1)V₁ = (2.5L) (0.25mol L^-1)
V₁ = 0.02522 L
V₁ = 25.22 mL

Question 13.
Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below :
1 Pa= IN m^-2
If mass of air at sea level is 1034 g cm-2, calculate the pressure in pascal.
Answer:
Pressure is defined as force acting per unit area of the surface.
i.e, P = F/A

= 101332.0 Nm^-2 [1N = 1kg ms^-2]
∴ Pressure = 1.01332 x 10⁵

Question 14.
What is the SI unit of mass? How is it defined?
Answer:
The SI unit of mass is kilogram (kg).
1 kilogram is defined as the mass equal to the mass of the international prototype of kilogram.

Question 15.
Match the following prefixes with their multiples :

	Prefixes	Multiples
(i)	micro	106
(ii)	deca	109
(iii)	mega	10-6
(iv)	giga	10-15
(v)	femto	10

Answer:

	Prefixes	Multiples
(i)	micro	10-6
(ii)	deca	10
(iii)	mega	106
(iv)	giga	109
(v)	femto	10-15

Question 16.
What do you mean by significant figures?
Answer:
The number of significant figures in a given data is the number of all certain digits plus one uncretain digit.
For example, if 15.6 mL is the result of an experiment, then 15 is certain while 6 is uncertain, and the total number of significant figures are 3.

Question 17.
A sample of drinking water was found to he severely contaminated with chloroform, CHC13, supposed to he carcinogenic in nature. The level of contamination was 15 ppm (by mass).
(i) Express this in per cent by mass.
(ii) Determine the molarity of chloroform in the water sample.
Answer:
(i) ∵ 10⁶ g of solution contains 15 g of CHCl₃
∴ 1 g of solution contains = \(\frac{15}{10^{6}}\) g of CHCl₃
100 g of solution contains = \(\frac{15}{10^{6}}\) x 10² = 15 x 10^-4 g of CHCl₃
∴ Percent by mass = 0.0015%

(ii) Molar mass of CHCl₃ = 12 +1 + 3 x 35.5 = 119.5 g mol^-1
0.0015% means 15 x 10^-4 g chloroform is present in 100 g sample.
Molarity = \(\frac{W \times 1000}{m \times \text { Volume of sample }}\)
[For water density = 1 g cm^-3; so mass = volume]
= \(\frac{15 \times 10^{-4} \times 1000}{119.5 \times 100}\)
= 1.25 x 10^-4 M

Question 18.
Express the following in the scientific notation :
(i) 0.0048
(ii) 234000
(iii) 8008
(iv) 500.0
(v) 6.0012
Answer:
(i) 0.0048 = 4.8 x 10^-3
(ii) 234000 = 2.34 x 10⁵
(iii) 8008 = 8.008 x 10³
(iv) 500.0 = 5.00 x 10²
(v) 6.0012 = 6.0012 x 10⁰

Question 19.
How many significant figures are present in the following?
(i) 0.0025
(ii) 208
(iii) 5005
(iv) 126000
(v) 500.0
(vi) 2.0034
Answer:
(i) 0.0025
There are 2 significant figures,
(ii) 208
There are 3 significant figures.
(iii) 5005
There are 4 significant figures.
(iv) 126000
There are 3 significant figures.
(v) 500.0
There are 4 significant figures.
(vi) 2.0034
There are 5 significant figures.

Question 20.
Round up the following upto three significant figures :
(i) 34.216 (ii) 10.4107
(iii) 0.04597 (iv) 2808
Answer:
(i) 34.2
(ii) 10.4
(iii) 0.0460
(iv) 2810

Question 21.
The following data are obtained when dinitrogen and dioxygen react together to form different compounds :

	Mass of dinitrogen	Mass of dioxygen
(i)	14 g	16 g
(ii)	14 g	32 g
(iii)	28 g	32 g
(iv)	28 g	80 g

(a) Which law of chemical combination is obeyed by the above experimental data? Give its statement.
(b) Fill in the blanks in the following conversions :
(i) 1 km = ………mm = ………pm
(ii) 1 mg = ………kg = ………ng
(iii) 1 mL = ………L = ………dm³
Answer:
On fixing the mass of dinitrogen as 14 g, then the masses of dioxygen which combines with the fixed mass (= 14 g) of dinitrogen will be 16, 32, 16, 40 which are in the simple whole number ratio of 1 : 2 : 1 : 2.5 or 2 : 4 : 2 : 5.

(a) Law of multiple proportions : This law was proposed by Dalton in 1803. According to this law, if tvo elements can combine to form two or more than two compounds, the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers.

Question 22.
If the speed of light is 3.0 x 10⁸ms^-1, calculate the distance covered by light in 2.00 ns.
Answer:
According to the question,
Time taken to cover the distance = 2.00 ns
2ns = 2.00 x 10^-9s [∵ 1ns = 10^-9s]
Speed of light = 3.0 x 10⁸ms^-1
Distance travelled by light in 2.00 ns
= Speed of light x Time taken
= (3.0 x 10⁸ms^-1)(2.00 x 10^-9s) = 6.00 x 10^-1 m = 0.6 m

Question 23.
In a reaction A + B₂ → AB₂
Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
Answer:
(i) According to the given reaction, 1 atom of A reacts with 1 molecule of B.
Thus, 300 atoms of A will react with 200 molecules of B thereby, leaving 100 atoms of A unused. Hence, B is the limiting reagent.
(ii) According to the given reaction, 1 mol of A reacts with 1 mol of B. Thus, 2 mol of A will react with 2 mol of B. As a result, 1 mol of B will not be consumed. Hence, A is the limiting reagent.
(iii) According to the given reaction, 1 atom of A combines with 1 molecule of B. Thus, all 100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric where no limiting reagent is present.
(iv) 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of B will combine only 2.5 mol of A. As a result, 2.5 mol of A will be left as such. Hence, B is the limiting reagent.
(v) 1 mol of atom A combines with 1 mql of molecule B. Thus, 2.5 mol of A will combine only 2.5 mol of B and the remaining 2.5 mol of B will be left as such. Hence, A is the limiting reagent.

Question 24.
Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: N2(g)+H2(g) → 2NH₃(g)
(i) Calculate the mass of ammonia produced if 2.00 x 10³ g dinitrogen reacts with 1.00 x 10³ g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?
On balancing the given chemical equation,
Answer:

⇒ 28 g N₂ reacts with 6 g H₂; 1 g N₂ reacts with \(\frac{6}{28}\) g H₂;
2.0 x 10³ g of N2 will react with \(\frac{2000 \times 6}{28}\) = 428.6 g of H₂.
Hence, N₂ is the limiting reagent.
∵ 28 g N2 produces 34 g NH3 34
∴ 1 g N2 produces \(\frac{34}{28}\) g NH₃
2.00 x 10³ f N₂ will produce = \(\frac{34 g}{28 g}\) x 2000g = 2428.57 g NH₃
(ii) N₂ is the limiting reagent and H₂ is the excess reagent. Hence, H₂ will remain unreacted.
(iii) Mass of H₂ remain unreacted = 1.00 x 10³ g – 428.6 g = 571.43 g

Question 25.
How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different?
Answer:
Molar mass of Na₂CO₃ = (2 x 23) +12.00 + (3 x 16) = 106 g mol^-1
Now, 1 mol Na₂CO₃ means 106 g Na₂CO₃
∴ 0.5 mol Na₂CO₃ = x °-5 mo1 Na₂C0₃ = 53 g Na₂C0₃
⇒ 0.50 M Na₂CO₃ = 0.50 mol/L Na₂CO₃
Hence, 0.50 mol Na₂CO₃ means 53 g Na₂CO₃ is present in 1 L of the solution.

Question 26.
If ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?
Answer:
Reaction of dihydrogen with dioxygen can be written as:

Now, 2 volumes of dihydrogen react with 1 volume of O₂ to produce 2 volumes of water vapour.
Hence, 10 volumes of dihydrogen will react with 5 volumes of dioxygen to produce 10 volumes of water vapour.

Question 27.
Convert the following into basic units :
(i) 28.7 pm
(ii) 15.15 pm
(iii) 25365 mg
Answer:
(i) 28.7 pm :
∵ 1 pm = 10^-12 m
∴ 28.7pm = 28.7 x 10^-12 m = 2.87 x 10^-11 m

(ii) 15.15pm
∵ 1 pm =10 12 m
∴ 15.15 pm = 15.15 x 10^-12 m = 1.515 x 10^-11 m

(iii) 25365 mg :
1 mg = 10^-3 g
25365 mg = 2.5365 x 10⁴ x 10^-3 g
= 2.5365 x 10¹ g
Since,
1 g = 10^-3 kg
2.5365 x 10¹ g = 2.5365 x 10¹ x 10^-3 kg.
∴ 25365 mg = 2.5365 x 10^-2 kg

Question 28.
Which one of the following will have largest number of atoms?
(i) 1 g Au(s)
(ii) 1 g Na(s)
(iii) 1g Li (s)
(iv) 1 g of Cl₂(g)
Answer:
(i) 1 g Au (s) = \(\frac{1}{197}\) mol atoms of Au (s)
= \(\frac{6.022 \times 10^{23}}{197}\) mol atoms of Au (s)
= 3.06 x 10²¹ atoms of Au (s)

(ii) 1 g Na (s) = \(\frac{1}{23}\) mol atoms of Na (s)
= \(\frac{6.022 \times 10^{23}}{23}\) atoms of Na (s)
= 0.262 x 10²³ atoms of Na (s)
= 26.2 x 10²¹ atoms of Na (s)

(iii) 1 g Li (s) = \(\frac{1}{7}\) mol atoms of Li (s)
= \(\frac{6.022 \times 10^{23}}{23}\) atoms of Li (s)
= 86.0 x 1021 atoms of Li (s)

(iv) 1 gCl₂ (g) = \(\frac{1}{71}\) mol molecules of Cl₂ (g)
= \(\frac{6.022 \times 10^{23}}{71}\) molecules of Cl? (g)
= 0.0848 x 10²³ molecules of Cl₂ (g)
= 8.48 x 10²¹ molecules of Cl₂ (g)
(one molecule of Cl₂ contains two atoms of Cl)
Number of atoms of Cl = 2 x 8.48 x 10²¹ = 16.96 x 10²¹ atoms of Cl
Hence, 1 g of Li (s) has largest number of atoms.

Question 29.
Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).
Answer:
Mole fraction of ethanol (C₂H₅OH)

Number of moles present in 1L water;
ⁿH₂O = \(\)
ⁿH₂O = 55.55 moles
Substituting the value of ⁿH₂O in eq (i).
\(\frac{n_{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}}{n_{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}+55.55}\) = 0.040
ⁿC₂H₅OH – 0.040ⁿC₂H₅OH = 2.222mol
0.96 ⁿC₂H₅OH = 2.222 mol ⁿC₂H₅OH = \(\frac{2.222}{0.96}\) mol.
ⁿC₂H₅OH = 2.314 mol
∴ Molarity of solution = \(\frac{2.314 \mathrm{~mol}}{1 \mathrm{~L}}\) = 2.314M

Question 30.
What will be the mass of one 12C atom in g?
Answer:
1 mol of carbon atoms = 6.022 x 10²³ atoms of carbon = 12 g of carbon
∴ Mass of one ¹²C atom = \(\frac{\text { Atomic mass of } \mathrm{C}}{\text { Avogadro’s number }}=\frac{12 \mathrm{~g}}{6.022 \times 10^{23}}\)
= 1.993 x 10^-23 g

Question 31.
How many significant figures should be present in the answer of the following calculations?
(i) \(\frac{0.02856 \times 298.15 \times 0.112}{0.5785}\)
(ii) 5 x 5.364
(iii) 0.0125 + 0.7864 + 0.0215
Answer:
\(\frac{0.02856 \times 298.15 \times 0.112}{0.5785}\)
Least precise number of calculation = 0.112
Number of significant figures in the answer = Number of significant figures in the least precise number = 3 /*
(ii) 5 x 5.364
Least precise number of calculation = 5.364
Number of significant figures in the answer = Number of significant figures in 5.364 = 4
(iii) 0.0125 + 0.7864 + 0.0215
Since the least number of decimal places in each term is four, the number of significant figures in the answer is also 4.

Question 32.
Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes :

Isotope	Isotopic molar mass	Abundance
36 Ar	35.96755 g mol-1	0.337%
38 Ar	37.96272 g mol-1	0.063%
40 Ar	39.9624 g mol-1	99.600%

Answer:
Molar mass of argon

= 0.121 + 0.024 + 39.802 g mol^-1 = 39.947 g mol^-1

Question 33.
Calculate the number of atoms in each of the following
(i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He.
Answer:
(i) ∵ 1 mol of Ar = 6.022 x 10²³ atoms of Ar
∴ 52 moles of Ar = 52 x 6.022 x 10²³ atoms of Ar
= 3.131 x 10²⁵ atoms

(ii) ∵ 1 atom of He = 4 u of He
Or, 4 u of He = 1 atom of He
∴ 1 u of He = 1/4 atom of He
∴ 52 u of He = 52/4 atoms of He = 13 atoms

(iii) ∵ 4 g of He = 6.022 x 10²³ atoms
6022 1023 52
∴ 52 g of He = \(\frac{6.022 \times 10^{23} \times 52}{4}\) = 7.8286 x 10²⁴ atoms

Question 34.
A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.
Answer:
(i) 44 g CO₂ = 12 g carbon
3.38 g CO₂ = 12/44 x 3.38g = 0.9218 g carbon
18 g H₂O = 2g hydrogen
0.690 g H₂O = 2/18 x 0.690 g = 0.0767 g hydrogen
Total mass of compound = 0.9218+0.0767 = 0.9985 g (because compound contains only carbon and hydrogen)
% of C in the compound = \(\frac{0.9218}{0.9985}\) x 100 = 92.32
% of H in the compound = \(\frac{0.0767}{0.9985}\) x 100 = 7.68
Calculation for Empirical Formula

Hence, empirical formula = CH
(ii) Calculation for molar mass of the gas
10.0 L of the given gas at STP weigh = 11.6 g
∴ 22.4 L of the given gas at STP will weigh = \(\frac{11.6 \times 22.4}{10}\) = 25.984 g
Molar mass = 25.984 = 26 g mol^-1
(iii) Empirical formula mass (CH) =12 + 1 = 13
∴ n = \(\frac{\text { Molecular mass }}{\text { Empirical formula mass }}=\frac{26}{13}\) = 2
Hence, molecular formula = n x CH = 2 x CH = C₂H₂

Question 35.
Calcium carbonate reacts with aqueous HC1 to give CaCl₂ and CO₂ according to the reaction,
CaC0₃(s) + 2 HCl(aq) -» CaCl₂(aq) + C0₂(g) + H₂O(l) What mass of CaC0₃ is required to react completely with 25 mL of 0.75 M HC1?
Answer:
The given reaction is
CaC0₃(s) + 2HCl(aq) → CaCl₂(aq) + C0₂(g) + H₂0(l)
Let us find out the weight of HCl present in 25 mL of 0.75 M HCl
1000 mL of 1.0 M HCl contains = 36.5 g
25 mL of 0.75 M HCl contains = \(\frac{36.5}{1000}\) x 25 x 0.75
= 0.6844 g of HCl
According to the equation;
73 g of HCl [2(1 + 35.5)] reacts with 100.0 g of CaC0₃
0.6844 g of HCl reacts with = \(\frac{100}{73}\) x 0.6844 = 0.94 g of CaC0₃

Question 36.
Chlorine is prepared in the laboratory by treating manganese dioxide (MnO₂) with aqueous hydrochloric acid according to the reaction
4HCl(oq) + MnO₂(s) > 2H₂O(l) + MnCl₂(aq) + Cl₂(g)
How many grams of HCl react with 5.0 g of manganese dioxide?
Answer:

According to the balanced chemical equation,
∵ 87 g of MnO₂ react with 4 x 36.5 g HCl
∴ 5 g of MnO₂ will react \(\frac{4 \times 36.5 \times 5}{87}\) = 8.39 g HCl

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 8 Redox Reactions Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 8 Redox Reactions

Very Short Answer Type Questions

Question 1.
What are spectator ions? Give one example.
Answer:
Spectator ions are ions that stay unaffected during a chemical reaction. They appear both as reactant and as product in an ionic equation. For example, in the following ionic equation, the sodium and nitrate ions are spectator ions.
Ag⁺ (aq) + NO₃^–(aq) + Na⁺ (aq) + Cl^– (aq) → AgCl(s) + Na⁺ (aq) + NO^–₃ (aq)

Question 2.
Why is anode called oxidation electrode, whereas cathode is called reduction electrode?
Answer:
At anode, loss of electrons takes place, i.e., oxidation takes place, whereas at cathode, gain of electrons takes place, i.e., reduction takes place.
Therefore, cathode is called reduction electrode and anode is called oxidation electrode.

Question 3.
Can we use KCl as electrolyte in the salt bridge of the cell?
Answer:
KCl cannot be used as electrolyte in the salt bridge because Cl^– ions will combine with Ag⁺ ions to form white precipitates of AgCl.

Question 4.
What would happen if no salt bridge were used in the electrochemical cell (e.g., Zn – Cu cell)?
Answer:
If no salt bridge is used, the positive ions (i.e., Zn²⁺ ) formed by loss of electrons will accumulate around the zinc electrode and negative ions (i.e., \(\mathrm{SO}_{4}^{2-}\)) left after reduction of Cu²⁺ ions will accumulate around the copper electrode. Thus, the solution will develop charges and the current stops flowing. Further, since the inner circuit is not complete, the current stops flowing.

Question 5.
Zn rod is immersed in CUSO₄ solution. What will you observe after an hour? Explain your observation in terms of redox reaction.
Answer:
The blue colour of CuSO₄ solution will get discharged and reddish brown copper metal will be deposited on Zn rod. This is because blue colour Cu²⁺ (in CuSO₄) gets reduced to Cu by accepting two electrons from Zn, which gets oxidised to colourless ZnSO₄.

Question 6.
What is the most essential conditions that must be satisfied in a redox reaction?
Answer:
In a redox reaction, the total number of electrons lost by the reducing agent must be equal to the number of electrons gained by the oxidising agent.

Question 7.
Find the value of n in \(\mathrm{MnO}_{4}^{-}\) + 8H⁺ + ne^– → Mn²⁺ + 4H₂O
Answer:
\(\mathrm{MnO}_{4}^{-}\) + 8H⁺ + ne^– → Mn²⁺ + 4H₂O
-1 + 8 + n = + 2
-1 – 2 + 8 + n = 0
n = – 5 or 5e^–

Question 8.
Can Fe³⁺ oxidise Br^– to Br₂ at 1 M concentrations?
\(\boldsymbol{E}^{\ominus}\)(Fe³⁺ /Fe²⁺) – 0.77 V and \(\boldsymbol{E}^{\ominus}\)(Br/Br^– ) = 1.09 V
Answer:
E^s ( Fe³⁺ / Fe²⁺) is lower than that of E^s(Br / Br^–).
Therefore, Fe²⁺ can reduce Br₂ but Br^– cannot reduce Fe³⁺. Thus, Fe³⁺ cannot oxidise Br^– to Br₂.

Question 9.
Identify the substance that get reduced in the following reaction:
Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)
Answer:
In the reaction, Fe₂O₃ loses oxygen and is reduced to Fe.

Question 10.
Can the following reaction, \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons 2 \mathrm{CrO}_{4}^{2-}+2 \mathrm{H}^{+}\) be regarded as a redox reaction?
Answer:
In this reaction, oxidation number of Cr in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) is +6 and oxidation number of Cr in \(\mathrm{CrO}_{4}^{2-}\) is +6. Since, during the reaction, the oxidation number of Cr has neither decreased nor increased, therefore, the above reaction is not a redox reaction.

Short Answer Type Questions

Question 1.
2Cu₂S + 3O₂ ⇌ 2Cu₂O + 2SO₂
In this reaction which substance is getting oxidised and which substance is getting reduced? Name the reducing agent and oxidising agent.
Answer:
Since, oxygen is being added to Cu, therefore, Cu₂S is oxidised to Cu₂O and the other reactant i.e., O₂ is getting reduced. Hence, Cu₂S is a reducing agent and O₂ is an oxidising agent.

Question 2.
One mole of N₂H₄ loses 10 moles electrons to form a new compound Y. Assuming that all the nitrogen appears in the new compound, what is the oxidation number of N in Y? There is no change in oxidation state of H.
Answer:
Suppose the oxidation number of N in Y is x
(N^2-)₂ → (2N)^x + 10e^–
(as N₂H₄ → Y +10e^–)
Therefore, 2x -10 = – 4, which gives x = + 3. Hence, oxidation number of N in Y = 3.

Question 3.
What are the net charges on the left and right side of the following equations? Add electrons as necessary to make each of them balanced half reactions.
(i) \(\mathrm{NO}_{3}^{-}+\mathbf{1 0 H}^{+} \longrightarrow \mathbf{N H}_{4}^{+}+3 \mathrm{H}_{2} \mathrm{O}\)
(ii) \(\mathrm{Cl}_{2}+4 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathbf{2 C l O}_{2}^{-}+8 \mathrm{H}^{+}\)
Answer:
(i) +9 charge on the left, +1 charge on the right; add 8 electrons to the left side.
(ii) 0 charge on the left, +6 charge on the right; add 6 electrons on the right side.

Question 4.
An iron rod is immersed in solution containing 1.0 M NiSO₄ and 1.0 M ZnSO₄. Predict giving reasons which of the following reactions is likely to proceed?
(i) Fe reduces Zn²⁺ ions,
(ii) Iron reduces Ni²⁺ ions. Given
\(E_{\mathbf{Z n}^{2+} / \mathbf{Z n}}^{\ominus}=-0.76 \mathrm{~V}, E_{\mathrm{Fe}^{2+} / \mathrm{Fe}^{=}}=-0.44 \mathrm{~V}\)
\(E_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{\ominus}=-0.25 \mathrm{~V}\)
Answer:
(i) Since \(E^{\ominus}\) of Zn is more negative than that of Fe, therefore, Zn will be oxidised to Zn²⁺ ions while Fe²⁺ ions will be reduced to Fe. In other words, Fe will not reduced Zn²⁺ ions.
(ii) Since, \(E^{\ominus}\) of Fe is more negative than that of Ni, therefore, Fe will be oxidised to Fe²⁺ ions while Ni²⁺ ions will be reduced to Ni. Thus, Fe reduces Ni²⁺ ions.

Question 5.
Copper dissolves in dilute nitric acid but not in dilute HC1. Explain.
Answer:
Since, \(E^{\ominus}\) of Cu²⁺/Cu electrode (+ 0.34 V) is higher than that of H⁺/H₂
electrode (0.0 V), therefore, H⁺ ions cannot oxidise Cu to Cu²⁺ ions and hence, Cu does not dissolve in dil. HCl.

In contrast, the electrode potential of \(\mathrm{NO}_{3}^{-}\) ion, i.e.\(\mathrm{NO}_{3}^{-}\) /NO electrode (+0.97 V) is higher than that of copper electrode and hence, it can oxidise Cu to Cu²⁺ ions and hence Cu dissolves in dil.HNO₃ due to oxidation of Cu by \(\mathrm{NO}_{3}^{-}\) ions and not by H⁺ ions.
Using standard electrode potential, the oxidative and reductive strength of a variety of substances can be composed.

Long Answer Type Questions

Question 1.
Why does fluorine doesn’t show disproportionation reaction?
Answer:
In a disproportionation reaction, the same species is simultaneously oxidised as
well as reduced. Therefore, for such a redox reaction to occur, the reacting species must contain an element which has atleast three oxidation states. The element, in reacting species, is present in an intermediate state while lower and higher oxidation states are available for reduction and oxidation to occur (respectively).
Fluorine is the strongest oxidising agent. It does not show positive oxidation state. That’s why fluorine does not show disproportionation reaction.

Question 2.
Which method can be used to find out strength of reductant/oxidant in a solution? Explain with an example.
Answer:
Measure the electrode potential of the given species by connecting the redox couple of the given species with standard hydrogen electrode. If it is positive, the electrode of the given species acts as reductant and if it is negative, it acts as an oxidant. Find the electrode potentials of the other given species in the same way, compare the values and determine their comparative strength as an reductant or oxidant.
Examples : Measurement of standard electrode potential of Zn⁺/Zn electrode using SHE as a reference electrode.

The EMF of the cell comes out to be 0.76 V. (reading of voltmeter is 0.76 V). Zn²⁺/Zn couple acts as anode and SHE acts as cathode.

Punjab State Board PSEB 11th Class Political Science Book Solutions Chapter 33 सर्वोच्च न्यायालय Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Political Science Chapter 33 सर्वोच्च न्यायालय

दीर्घ उत्तरीय प्रश्न-

प्रश्न 1.
भारत के उच्चतम न्यायालय का गठन किस प्रकार होता है ? भारत के उच्चतम न्यायालय के अधिकारों का वर्णन करो।
(How is the Supreme Court of India constituted ? Describe the powers of the Supreme Court.)
अथवा
भारत के सर्वोच्च न्यायालय की रचना, शक्तियों और कार्यों का वर्णन करो। (Discuss the composition, powers and functions of the Supreme Court of India.)
उत्तर-
केन्द्र तथा राज्यों के आपसी झगड़ों को निपटाने के लिए एक निष्पक्ष और स्वतन्त्र न्यायपालिका होना अनिवार्य है। इसके अतिरिक्त हमारे देश में संविधान को सर्वोच्च कानून माना गया है और नागरिकों को भी मौलिक अधिकार दिए गए हैं। संविधान की रक्षा और मौलिक अधिकारों की रक्षा के लिए स्वतन्त्र न्यायपालिका का और भी अधिक महत्त्व है। भारतीय संविधान के निर्माताओं ने निष्पक्ष और स्वतन्त्र न्यायपालिका के महत्त्व को समझते हुए भारतीय सुप्रीम कोर्ट की व्यवस्था की। सर्वोच्च न्यायालय की व्यवस्था संविधान के अनुच्छेद 124 में मिलती है।

रचना (Composition)-सर्वोच्च न्यायालय में पहले एक मुख्य न्यायाधीश और 7 अन्य न्यायाधीश होते थे। परन्तु दिसम्बर, 1977 में सर्वोच्च न्यायालय एक्ट में संशोधन करके सर्वोच्च न्यायालय की अधिकतम संख्या 17 निर्धारित की गई। अप्रैल, 1986 में सर्वोच्च न्यायालय के न्यायाधीशों की संख्या 17 से 25 कर दी गई। जनवरी, 2009 में सर्वोच्च न्यायालय के न्यायाधीशों की संख्या 25 से बढ़ाकर 30 कर दी गई। अतः अब सर्वोच्च न्यायालय में एक मुख्य न्यायाधीश और 30 अन्य न्यायाधीश हैं। वर्तमान समय में अल्तमस कबीर सर्वोच्च न्यायालय के मुख्य न्यायाधीश हैं। ।

न्यायाधीशों की नियुक्ति (Appointment) संविधान के अन्तर्गत सर्वोच्च न्यायालय के मुख्य न्यायाधीश तथा अन्य न्यायाधीशों की नियुक्ति करने का अधिकार राष्ट्रपति को दिया गया है। सर्वोच्च न्यायालय के मुख्य न्यायाधीश की नियुक्ति करते समय राष्ट्रपति सर्वोच्च न्यायालय तथा राज्यों के उच्च न्यायालयों के ऐसे न्यायाधीशों की सलाह लेता है जिन्हें वह उचित समझता है। अन्य न्यायाधीशों की नियुक्ति करते समय राष्ट्रपति के लिए मुख्य न्यायाधीश की सलाह लेना अनिवार्य है।

न्यायाधीशों की नियुक्ति सम्बन्धी सर्वोच्च न्यायालय का फैसला (Decision of the Supreme Court with regard to the appointment of the Judges)—सर्वोच्च न्यायालय ने 6 अक्तूबर, 1993 को एक महत्त्वपूर्ण निर्णय सुनाते हुए कहा कि सर्वोच्च न्यायालय के अन्य न्यायाधीशों की नियुक्ति करते समय कार्यपालिका के मुकाबले सर्वोच्च न्यायालय के मुख्य न्यायाधीश की सलाह को महत्त्व दिया जाएगा। 28 अक्तूबर, 1998 के सर्वोच्च न्यायालय की नौ सदस्यीय संविधान पीठ ने एक महत्त्वपूर्ण स्पष्टीकरण के तहत यह निर्धारित किया कि सर्वोच्च न्यायालय के न्यायाधीश की नियुक्ति में अपनी सिफ़ारिश देने से पूर्व मुख्य न्यायाधीश को सर्वोच्च न्यायालय के चार वरिष्ठतम न्यायाधीशों से विचार-विमर्श करना चाहिए। संविधान पीठ ने स्पष्ट किया है कि सलाहकार मण्डल की राय तथा मुख्य न्यायाधीश की राय जब तक एक न हो, तब तक सिफ़ारिश नहीं की जानी चाहिए। परामर्श किए गए चार न्यायाधीशों में से यदि दो न्यायाधीश भी विपरीत राय देते हैं तो मुख्य न्यायाधीश को सरकार को सिफ़ारिश नहीं भेजनी चाहिए। विचार-विमर्श प्रक्रिया का पालन किए बिना मुख्य न्यायाधीश द्वारा दी गई सिफ़ारिशें सरकार पर बाध्यकारी नहीं है।

योग्यताएं (Qualifications)-सर्वोच्च न्यायालय के न्यायाधीशों के लिए निम्नलिखित योग्यताएं निश्चित की गई हैं-

• वह भारत का नागरिक हो।
• वह कम-से-कम 5 वर्ष तक किसी एक, दो या अधिक उच्च न्यायालयों का न्यायाधीश रह चुका हो, या
• वह किसी एक, दो या उससे अधिक उच्च न्यायालयों में कम-से-कम 10 वर्ष तक वकालत कर चुका हो, या
• वह राष्ट्रपति की दृष्टि में कुशल विधिवेत्ता (Distinguished Jurist) हो।

कार्यकाल (Term of Office)–सर्वोच्च न्यायालय के न्यायाधीश 65 वर्ष की आयु तक अपने पद पर रहते हैं। उससे पहले वे स्वयं तो त्याग-पत्र दे सकते हैं, परन्तु राष्ट्रपति जब चाहे उन्हें उनके पद से नहीं हटा सकता। परन्तु इसका अर्थ यह नहीं कि 65 वर्ष से पहले किसी न्यायाधीश को अपदस्थ किया ही नहीं जा सकता। उन्हें अयोग्यता तथा कदाचार के आधार पर पदच्युत किया जा सकता है। इसके लिए यह व्यवस्था की गई है कि संसद् के दोनों सदन अलगअलग बैठकर अपने समस्त सदस्यों के स्पष्ट बहुमत तथा उपस्थित एवं मत डालने वाले सदस्यों के 2/3 बहुमत से किसी न्यायाधीश को पद से हटाए जाने का प्रस्ताव पास कर दें तो ऐसा प्रस्ताव पास होने पर राष्ट्रपति सम्बन्धित न्यायाधीश को उसके पद से हटा देगा। 11 मई, 1993 को लोकसभा में सर्वोच्च न्यायालय के न्यायाधीश वी० रामास्वामी के विरुद्ध महाभियोग का प्रस्ताव पास न हो सका क्योंकि कांग्रेस (इ) के सदस्यों ने मतदान में भाग नहीं लिया। प्रस्ताव के पक्ष में 196 मत पड़े जबकि प्रस्ताव पास होने के लिए 273 मतों की आवश्यकता थी।

वेतन तथा भत्ते (Salary and Allowances)—सर्वोच्च न्यायालय के मुख्य न्यायाधीश का मासिक वेतन 2,80,000 रुपए से और अन्य न्यायाधीशों का वेतन 2,50,000 रुपए है। इसके अतिरिक्त उन्हें एक नि:शुल्क निवासस्थान तथा यात्रा-भत्ता (जब कोई न्यायाधीश अपने कर्त्तव्य का पालन करते हुए यात्रा करता है) मिलता है। उनके वेतन तथा भत्ते भारत की संचित निधि से दिए जाते हैं जिस पर संसद् को मतदान का अधिकार नहीं। उनके वेतन और भत्ते वैसे तो समय-समय पर संसद् द्वारा निश्चित किए जाते हैं, परन्तु उनके कार्यकाल में घटाए नहीं जा सकते। सेवानिवृत्त होने पर न्यायाधीशों को पैन्शन मिलती है।

रिटायर होने पर वकालत पर पाबन्दी (Prohibition of Practice after Retirement)-सर्वोच्च न्यायालय के न्यायाधीश रिटायर होने के पश्चात् किसी न्यायालय में वकालत नहीं कर सकते, परन्तु आवश्यकता पड़ने पर राष्ट्रीय सर्वोच्च न्यायालय के किसी रिटायर न्यायाधीश को कोई विशेष कार्य सौंप सकता है और न्यायाधीश को इस कार्य का वेतन दिया जाता है।

शपथ (Oath) सर्वोच्च न्यायालय के प्रत्येक न्यायाधीश को अपना पद सम्भालने से पहले राष्ट्रपति या उसके द्वारा नियुक्त किसी अन्य अधिकारी के सामने अपने पद की शपथ लेनी पड़ती है।

स्थान (Seat of the Supreme Court)-संविधान के अनुसार इसका स्थान दिल्ली रखा गया है, परन्तु आवश्यकता पड़ने पर मुख्य न्यायाधीश राष्ट्रपति की स्वीकृति से अन्य स्थानों पर उसकी बैठक कर सकता है।

सर्वोच्च न्यायालय के क्षेत्राधिकार तथा कार्य (Jurisdiction and Function of the Supreme Court)-

सर्वोच्च न्यायालय भारत का सबसे बड़ा और अन्तिम न्यायालय है और इसीलिए इसके अधिकार तथा शक्तियां बड़ी व्यापक हैं। इसकी शक्तियों और कार्यों को कई श्रेणियों में बांटा जा सकता है, जोकि निम्नलिखित हैं :

1. प्रारम्भिक क्षेत्राधिकार (Original Jurisdiction)-भारतीय सर्वोच्च न्यायालय को कुछ मुकद्दमों में प्रारम्भिक क्षेत्राधिकार प्राप्त हैं अर्थात् कुछ मुकद्दमे ऐसे हैं जो सर्वोच्च न्यायालय में सीधे ले जाए जा सकते हैं। उन्हें इससे पहले किसी अन्य न्यायालय में ले जाने की आवश्यकता नहीं। अग्रलिखित मुकद्दमे सीधे सर्वोच्च न्यायालय में ले जाए जा सकते हैं :-

• भारत सरकार और एक या अधिक राज्यों के बीच उत्पन्न झगड़े।
• ऐसे झगड़े जिनमें भारत सरकार और राज्य एक तरफ हों तथा एक राज्य और कुछ राज्य दूसरी ओर।
• ऐसे झगड़े जो दो या दो से अधिक राज्यों के बीच हों और जिनका सम्बन्ध संविधान या किसी कानून की व्यवस्था आदि से हो।
• मौलिक अधिकारों के बारे में कोई भी मुकद्दमा सीधा सर्वोच्च न्यायालय में ले जाया जा सकता है।
• राष्ट्रपति और उपराष्ट्रपति के चुनाव सम्बन्धी विवादों का निर्णय सर्वोच्च न्यायालय ही करता है।

2. अपीलीय क्षेत्राधिकार (Appellate Jurisdiction)—जो मुकद्दमे सर्वोच्च न्यायालय में सीधे नहीं लाये जा सकते, वे अपील के रूप में सर्वोच्च न्यायालय के सामने लाए जा सकते हैं। सर्वोच्च न्यायालय को राज्य न्यायालयों के निर्णयों के विरुद्ध अपीलें सुनने का अधिकार है। ये अपीलें संवैधानिक, दीवानी और फ़ौजदारी तीनों प्रकार के मुकद्दमे में सुनी जा सकती हैं, परन्तु उच्च न्यायालय द्वारा दिए गए सभी मामलों के निर्णय के विरुद्ध अपील नहीं की जा सकती। सर्वोच्च न्यायालय में अपील करने के लिए कुछ शर्ते निश्चित हैं, जोकि निम्नलिखित हैं

(क) संवैधानिक मामलों में अपील (Appeal in Constitutional Cases)—संविधान के अनुसार यह व्यवस्था की गई है कि यदि उच्च न्यायालय यह प्रमाणित कर दे कि मुकद्दमे में संविधान को किसी धारा की व्यवस्था या कानून का कोई महत्त्वपूर्ण प्रश्न निहित है तो किसी भी मुकद्दमे में, चाहे वह दीवानी हो या फ़ौजदारी, छोटा हो या बड़ा उच्च न्यायालय के निर्णय के विरुद्ध अपील की जा सकती है। यदि उच्च न्यायालय किसी मुकद्दमे में ऐसा प्रमाणपत्र देने से इन्कार कर दे तो सर्वोच्च न्यायालय स्वयं ही इस प्रकार की अनुमति (Certificate of Leave) देकर अपील करने की विशेष आज्ञा प्रदान कर सकता है।

(ख) दीवानी मुकद्दमे में अपील (Appeal in Civil Cases)—संविधान के द्वारा दीवानी मामलों में सर्वोच्च न्यायालयों द्वारा अपील सुने जाने की व्यवस्था है। किसी भी राशि का मुकद्दमा सर्वोच्च न्यायालय के पास आ सकता है, जब उच्च न्यायालय यह प्रमाण-पत्र दे दे कि मुकद्दमा सर्वोच्च नयायालय के सुनने के योग्य है। यह ऐसे मुकद्दमे की अपीलें सुन सकता है जबकि उच्च न्यायालय यह प्रमाण-पत्र दे कि मुकद्दमे में कोई कानूनी प्रश्न विवादग्रस्त है। यदि उच्च न्यायालय किसी दीवानी मामले में इस प्रकार का प्रमाण-पत्र न दे तो सर्वोच्च न्यायालय स्वयं भी किसी व्यक्ति को अपील करने की विशेष आज्ञा दे सकता है। किसी दीवानी मुकद्दमे में संविधान की व्याख्या किसी महत्त्वपूर्ण प्रश्न के निहित होने की दशा में भी सर्वोच्च न्यायालय में अपील की जा सकती है।

(ग) फ़ौजदारी मुकद्दमे में अपील (Appeal in Criminal Cases) निम्नलिखित कई प्रकार के फ़ौजदारी मुकद्दमों में भी उच्च न्यायालय के निर्णय के विरुद्ध सर्वोच्च न्यायालय में अपील की जा सकती है

• यदि किसी फ़ौजदारी मुकद्दमे में निचले न्यायालय ने अभियुक्त को छोड़ दिया हो, परन्तु उच्च न्यायालय ने अपील में विमुक्ति के आदेश को रद्द करके अभियुक्त को मृत्यु-दण्ड दे दिया हो।
• उच्च न्यायालय ने किसी निचले न्यायालय में चल रहे किसी मुकद्दमे को अपने पास निरीक्षण के लिए मंगवा लिया हो और अभियुक्त को दोषी ठहराकर मृत्यु-दण्ड दे दिया हो।
• जिस मुकद्दमे में उच्च न्यायालय यह प्रमाणित कर दे कि मुकद्दमा सर्वोच्च न्यायालय में अपील किए जाने के योग्य है।
• सर्वोच्च न्यायालय किसी भी फ़ौजदारी मुकद्दमे में अपील करने की विशेष आज्ञा प्रदान कर सकता है।

3. संविधान की व्याख्या तथा रक्षा (Interpretation and Protection of the Constitution)-संविधान की व्याख्या तथा रक्षा करना भी सर्वोच्च न्यायालय का कार्य है। जब कभी संविधान की व्याख्या के बारे में कोई मतभेद उत्पन्न हो, तो सर्वोच्च न्यायालय द्वारा व्याख्या की जाती है और सर्वोच्च न्यायालय की व्याख्या को अन्तिम तथा सर्वोच्च माना जाता है। केवल संविधान की व्याख्या करना ही नहीं बल्कि इसकी रक्षा करना भी सर्वोच्च न्यायालय का कार्य है। यदि सर्वोच्च न्यायालय को यह विश्वास हो जाए कि संसद् द्वारा बनाया गया कानून या कार्यपालिका का आदेश संविधान का उल्लंघन करता है तो वह उस कानून या आदेश को असंवैधानिक घोषित करके रद्द कर सकता है।

4. मौलिक अधिकारों के विषय में क्षेत्राधिकार (Jurisdiction regarding Fundamental Rights)—यह न्यायालय स्वतन्त्रताओं और मौलिक अधिकारों का रक्षक है। 32वें अनुच्छेद के अनुसार इसे यह शक्ति दी गई है कि यह आदेश या लेख जैसे बन्दी प्रत्यक्षीकरण (Habeas Corpus), परमादेश (Mandamus), प्रतिषेध (Prohibition), अधिकार-पृच्छा (Quo-warranto) और उत्प्रेषण लेख (Certiorari) जारी करके मौलिक अधिकारों को लागू करवा सकता है। परन्तु संकटकाल के समय में मौलिक अधिकारों को लागू करवाने के लिए सर्वोच्च न्यायालय की सहायता लेने के अधिकार को राष्ट्रपति स्थगित कर सकता है।

5. सलाहकारी शक्तियां (Advisory Powers)-राष्ट्रपति किसी सार्वजनिक महत्त्व के विषय पर सर्वोच्च न्यायालय से परामर्श कर सकता है, परन्तु राष्ट्रपति के लिए यह जरूरी नहीं है कि वह सर्वोच्च न्यायालय के परामर्श के अनुसार चले। यह परामर्श कोई निर्णय नहीं होता।

6. अपने निर्णयों के पुनर्निरीक्षण का अधिकार (Power to Review its own Decisions)-सर्वोच्च न्यायालय अपने निर्णयों पर पुनर्विचार कर सकता है। सज्जन सिंह बनाम राजस्थान राज्य के मुकद्दमे में सर्वोच्च न्यायालय ने यह निर्णय दिया था कि संसद् मौलिक अधिकारों में संशोधन कर सकती है। परन्तु 1967 में गोलकनाथ के मुकद्दमे में सर्वोच्च न्यायालय ने यह निर्णय दिया था कि संसद् मौलिक अधिकारों में संशोधन नहीं कर सकती। 1973 के केशवानन्द भारती के मुकद्दमे में सर्वोच्च न्यायालय ने यह निर्णय दिया कि संसद् मौलिक अधिकारों में संशोधन कर सकती है।

7. एक अभिलेख न्यायालय (Court of Record)-सुप्रीम कोर्ट को एक अभिलेख न्यायालय माना गया है। इसकी समस्त कार्यवाही तथा निर्णय सदैव के लिए यादगार तथा प्रमाण के रूप में प्रकाशित किए जाते हैं तथा देश के समस्त न्यायालयों के लिए यह निर्णय न्यायिक दृष्टान्त (Judicial Precedents) के रूप में स्वीकार किए जाते हैं।

8. मुकद्दमे को स्थानान्तरित करने की शक्ति (Power Regarding Transference of Cases)-सर्वोच्च न्यायालय शीघ्र न्याय दिलाने के उद्देश्य से भी किसी भी मुकद्दमे को एक उच्च न्यायालय से दूसरे उच्च न्यायालय में भेज सकता है।

9. विविध कार्य (Miscellaneous Functions) सर्वोच्च न्यायालय के पास कई प्रकार की विविध शक्तियां हैं-

• यह अपना कार्य चलाने के लिए अपने पदाधिकारियों की नियुक्ति करता है।
• सर्वोच्च न्यायालय यह देखता है कि प्रत्येक न्यायालय में न्याय ठीक प्रकार हो रहा है अथवा नहीं।
• संघीय लोक सेवा आयोग के सदस्यों तथा सभापति को पदच्युत करने का अधिकार तो राष्ट्रपति के पास है परन्तु राष्ट्रपति ऐसा तभी कर सकेगा जब सर्वोच्च न्यायालय उसकी जांच-पड़ताल करके उसको अपराधी घोषित कर दे।

क्षेत्राधिकार में विस्तार (Enlargement of Jurisdiction)–संसद् को अपने अधिनियम द्वारा निम्नलिखित मामलों के सम्बन्ध में सर्वोच्च न्यायालय के क्षेत्राधिकार को विस्तृत करने का अधिकार है :

• संघ सूची में दिया गया कोई भी मामला।
• उच्च न्यायालयों के फैसलों के विरुद्ध फ़ौजदारी मामलों में अपीलीय क्षेत्राधिकार।
• कोई भी मामला जो भारत सरकार और किसी भी राज्य सरकार के समझौते द्वारा सर्वोच्च न्यायालय को दिया हो।
• मूल अधिकारों को लागू करने के अतिरिक्त किसी और उद्देश्यों के लिए निर्देश, आदेश तथा लेख जारी करना।
• सर्वोच्च न्यायालय को संविधान द्वारा सौंपे गए क्षेत्राधिकार को अच्छी प्रकार से प्रयोग करने के लिए ज़रूरी शक्ति।

सर्वोच्च न्यायालय की स्थिति (Position of the Supreme Court)—इस विवेचन के आधार पर यह कहा जा सकता है कि इस न्यायालय को संसार की सभी या संघात्मक सर्वोच्च न्यायालयों से अधिक शक्तियां प्राप्त हैं। श्री अल्लादी कृष्णा स्वामी अय्यर ने कहा है, “हमारी सुप्रीम कोर्ट को जितनी शक्तियां दी गई हैं, उतनी अधिक संसार में किसी अन्य सुप्रीम कोर्ट को प्राप्त नहीं हैं।”

एम० वी० पायली (M.V. Pylee) ने सर्वोच्च न्यायालय की स्थिति के बारे में लिखा है, “अपनी विभिन्न तथा व्यापक शक्तियों के कारण सर्वोच्च न्यायालय न्यायिक क्षेत्र में एक सर्वश्रेष्ठ संस्था ही नहीं, बल्कि वह देश के संविधान तथा कानून का भी रक्षक है।” (“The combination of such wide and varied powers in the Supreme Court of India makes it not only the supreme authority in the judicial field but also the guardian of constitution and law of the land.”)

लघु उत्तरीय प्रश्न

प्रश्न 1.
भारत में न्यायपालिका की स्वतन्त्रता की व्यवस्था के लिए किए गए तीन कार्यों का वर्णन करो।
उत्तर-
लोकतन्त्रात्मक तथा संघीय शासन प्रणाली की सफलता के लिए एक स्वतन्त्र तथा निष्पक्ष न्यायपालिका का होना अनिवार्य है। अतः संविधान निर्माताओं ने स्वतन्त्र न्यायपालिका की व्यवस्था के लिए संविधान में निम्नलिखित बातों का प्रबन्ध किया-

1. न्यायाधीशों की नियुक्ति-सर्वोच्च न्यायालय तथा राज्यों के उच्च न्यायालयों के न्यायाधीशों की नियुक्ति राष्ट्रपति करता है। परन्तु राष्ट्रपति न्यायाधीशों की नियुक्ति करने में स्वतन्त्र नहीं है। सर्वोच्च न्यायालय के मुख्य न्यायाधीश की नियुक्ति करते समय राष्ट्रपति सर्वोच्च न्यायालय तथा उच्च न्यायालयों के जिन न्यायाधीशों की सलाह ठीक समझे, परामर्श ले सकता है। सर्वोच्च न्यायालय के अन्य न्यायाधीशों की नियुक्ति करते समय राष्ट्रपति के लिए मुख्य न्यायाधीश की सलाह लेना अनिवार्य है। अक्तूबर, 1993 को सर्वोच्च न्यायालय ने एक निर्णय दिया कि सर्वोच्च न्यायालय के न्यायाधीशों की नियुक्ति करते समय कार्यपालिका सर्वोच्च न्यायालय के मुख्य न्यायाधीश की सलाह को महत्त्व देगी। 28 अक्तूबर, 1998 को सर्वोच्च न्यायालय की नौ सदस्यीय संविधान पीठ ने यह निर्णय दिया कि मुख्य न्यायाधीश को सलाह देने से पूर्व सर्वोच्च न्यायालय के चार वरिष्ठतम न्यायाधीशों से अवश्य विचार-विमर्श करना चाहिए। सलाहकार मण्डल की राय तथा मुख्य न्यायाधीश की राय जब तक एक न हो, तब तक सिफ़ारिश नहीं की जानी चाहिए।

2. न्यायाधीशों की योग्यताएं-संविधान में सर्वोच्च न्यायालय तथा अन्य न्यायालयों के न्यायाधीशों की योग्यताओं का वर्णन किया गया है। राष्ट्रपति उन्हीं व्यक्तियों को न्यायाधीश नियुक्त कर सकता है जिनके पास निश्चित योग्यताएं हों।

3. लम्बी अवधि-न्यायाधीशों को स्वतन्त्रता एवं निष्पक्षता के लिए लम्बी सेवा अवधि का होना अनिवार्य है। सर्वोच्च न्यायालय के न्यायाधीश 65 वर्ष की आयु तक अपने पद पर रह सकते हैं।

प्रश्न 2.
भारतीय सर्वोच्च न्यायालय का न्यायाधीश बनने के लिए आवश्यक योग्यताएं लिखें।
उत्तर-
राष्ट्रपति उसी व्यक्ति को सर्वोच्च न्यायालय का न्यायाधीश नियुक्त कर सकता है जिसमें निम्नलिखित योग्यताएं हों

• वह भारत का नागरिक हो।
• वह कम-से-कम पांच वर्ष तक एक या एक से अधिक उच्च न्यायालयों में न्यायाधीश के पद पर रह चुका हो।

अथवा

वह कम-से-कम 10 वर्ष तक उच्च न्यायालय का एडवोकेट रह चुका हो।

अथवा

वह राष्ट्रपति की दृष्टि में प्रसिद्ध कानून-विशेषज्ञ हो।

प्रश्न 3.
सर्वोच्च न्यायालय का प्रारम्भिक क्षेत्राधिकार बताएं।
उत्तर-
सर्वोच्च न्यायालय में कुछ मुकद्दमे सीधे ले जाए जा सकते हैं-

• यदि केन्द्र या किसी एक राज्य या कई राज्यों के बीच कोई झगड़ा उत्पन्न हो जाए तो उसका निर्णय सर्वोच्च न्यायालय करता है।
• यदि कुछ राज्यों के बीच किसी संवैधानिक विषय पर कोई झगड़ा उत्पन्न हो जाए तो वह झगड़ा भी सर्वोच्च न्यायालय द्वारा ही निपटाया जाता है।
• मौलिक अधिकारों के बारे में कोई भी मुकद्दमा सीधा सर्वोच्च न्यायालय के सामने ले जाया जा सकता है।
• यदि राष्ट्रपति या उप-राष्ट्रपति के चुनाव के बारे में कोई शंका या झगड़ा उत्पन्न हो जाए तो उसका निर्णय सर्वोच्च न्यायालय ही करता है।

प्रश्न 4.
सर्वोच्च न्यायालय के न्यायाधीशों का वेतन तथा अन्य सुविधाएं बताएं।
उत्तर-
सर्वोच्च न्यायालय के मुख्य न्यायाधीश को 2,80,000 रुपये मासिक तथा अन्य न्यायाधीशों को 2,50,000 रुपये मासिक वेतन मिलता है। वेतन के अतिरिक्त उन्हें कुछ भत्ते भी मिलते हैं। उन्हें रहने के लिए बिना किराये का निवास-स्थान भी मिलता है। उनके वेतन-भत्ते तथा दूसरी सुविधाओं में उनके कार्यकाल में किसी प्रकार की कटौती नहीं की जा सकती तथापि आर्थिक संकटकाल की उद्घोषणा के दौरान न्यायाधीशों के वेतन आदि घटाए जा सकते हैं। सेवा निवृत्त (Retire) होने पर उन्हें पेंशन भी मिलती है।

प्रश्न 5.
भारत में सर्वोच्च न्यायालय कैसे नागरिकों के मौलिक अधिकारों की रक्षा करता है ?
उत्तर-
यदि कोई व्यक्ति यह समझे कि सरकार ने उसके मौलिक अधिकारों में हस्तक्षेप किया है या कोई कानून मौलिक अधिकार के विरुद्ध बनाया गया है तो वह सर्वोच्च न्यायालय के पास जा सकता है और सर्वोच्च न्यायालय कई प्रकार के अभिलेख जारी कर सकता है और किसी कानून को अवैध भी घोषित कर सकता है।

प्रश्न 6.
सर्वोच्च न्यायालय भारतीय संविधान का संरक्षक है। टिप्पणी लिखिए।
उत्तर-
संविधान भारत की सर्वोच्च विधि है और किसी भी व्यक्ति, सरकारी कर्मचारी अधिकारी अथवा सरकार का कोई अंग इसके विरुद्ध आचरण नहीं कर सकता। इसकी रक्षा करना सर्वोच्च न्यायालय का कर्त्तव्य है। इसलिए सर्वोच्च न्यायालय को विधानमण्डलों द्वारा बनाए गए कानूनों तथा कार्यपालिका द्वारा जारी किए गए आदेशों पर न्यायिक निरीक्षण (Judicial Review) का अधिकार प्राप्त है। सर्वोच्च न्यायालय इस बात की जांच-पड़ताल तथा निर्णय कर सकता है कि कोई कानून या आदेश संविधान की धाराओं के अनुसार है या कि नहीं। यदि सर्वोच्च न्यायालय को यह विश्वास हो जाए कि किसी भी कानून से संविधान का उल्लंघन हुआ है तो वह उसे असंवैधानिक घोषित करके रद्द रक सकता है।

प्रश्न 7.
स्वतन्त्र न्यायपालिका क्या होती है ?
उत्तर-
न्यायपालिका की स्वतन्त्रता का अर्थ है कि न्यायपालिका, कार्यपालिका तथा विधानपालिका के अधीन अपना कार्य न करे। विधानपालिका तथा कार्यपालिका को न्यायपालिका के कार्यों में हस्तक्षेप करने का अधिकार नहीं होना चाहिए। न्यायाधीश तभी निष्पक्ष होकर न्याय कर सकते हैं जब उन पर किसी प्रकार का दबाव न हो। भारत में स्वतन्त्र न्यायपालिका की स्थापना की गई है।

प्रश्न 8.
सर्वोच्च न्यायालय के तीन प्रकार के क्षेत्राधिकारों के नाम लिखें। इसके सलाहकारी क्षेत्राधिकार का वर्णन करें।
उत्तर-
सर्वोच्च न्यायालय के मुख्य तीन क्षेत्राधिकार निम्नलिखित हैं-

1. प्रारम्भिक क्षेत्राधिकार।
2. अपीलीय क्षेत्राधिकार।
3. सलाहकारी क्षेत्राधिकार।

सलाहकारी क्षेत्राधिकार-राष्ट्रपति किसी सार्वजनिक महत्त्व के विषय पर सर्वोच्च न्यायालय से कानूनी सलाह ले सकता है। परन्तु राष्ट्रपति के लिए अनिवार्य नहीं है कि वह सर्वोच्च न्यायालय के परामर्श के अनुसार कार्य करे।

प्रश्न 9.
सर्वोच्च न्यायालय का संगठन क्या है ? न्यायाधीशों की नियुक्ति कौन करता है ?
उत्तर-
सर्वोच्च न्यायालय में एक मुख्य न्यायाधीश तथा कुछ अन्य न्यायाधीश होते हैं। आजकल सर्वोच्च न्यायालय में मुख्य न्यायाधीश के अतिरिक्त 30 अन्य न्यायाधीश हैं। सर्वोच्च न्यायालय के मुख्य न्यायाधीश तथा अन्य न्यायाधीशों की नियुक्ति राष्ट्रपति करता है। मुख्य न्यायाधीश की नियुक्ति करते समय राष्ट्रपति सर्वोच्च न्यायालय तथा राज्यों के उच्च न्यायालयों के ऐसे न्यायाधीशों की सलाह लेता है जिन्हें वह उचित समझता है। अन्य न्यायाधीशों की नियुक्ति करते समय राष्ट्रपति मुख्य न्यायाधीश की सलाह अवश्य लेता है। 28 अक्तूबर, 1998 को सर्वोच्च न्यायालय की नौ सदस्यीय संविधान पीठ ने यह निर्णय दिया कि मुख्य न्यायाधीश को सलाह देने से पूर्व सर्वोच्च न्यायालय के चार वरिष्ठतम न्यायाधीशों से विचार-विमर्श करना चाहिए। सलाहकार मण्डल की राय तथा मुख्य न्यायाधीश की राय जब तक एक न हो, तब तक सिफ़ारिश नहीं की जानी चाहिए।

प्रश्न 10.
सर्वोच्च न्यायालय के पांच अधिकारों के नाम लिखो।
उत्तर-

1. प्रारम्भिक क्षेत्राधिकार
2. अपीलीय क्षेत्राधिकार
3. सलाहकारी क्षेत्राधिकार
4. मौलिक अधिकारों का रक्षक
5. न्यायिक पुनर्निरीक्षण की शक्ति।

प्रश्न 11.
सर्वोच्च न्यायालय के न्यायाधीश को उसके पद से कैसे हटाया जा सकता है ?
उत्तर-
सर्वोच्च न्यायालय के न्यायाधीश को संसद् महाभियोग द्वारा हटा सकती है। न्यायाधीश को अयोग्यता तथा कदाचार आधार पर पदच्युत किया जा सकता है। यदि संसद् के दोनों सदन अलग-अलग बैठ कर अपने समस्त सदस्यों के स्पष्ट बहुमत तथा उपस्थित एवं मत डालने वाले सदस्यों के 2/3 बहुमत से किसी भी न्यायाधीश को पद से हटाए जाने का प्रस्ताव पास कर दे तो ऐसा प्रस्ताव पास होने पर राष्ट्रपति सम्बन्धित न्यायाधीश को उसके पद से हटा देगा।

प्रश्न 12.
न्यायिक पुनर्निरीक्षण का क्या अर्थ है ?
उत्तर-
न्यायिक पुनर्निरीक्षण न्यायालयों की वह शक्ति है जिसके द्वारा वह विधानमण्डल के कानूनों तथा कार्यपालिका के आदेशों की जांच कर सकता है और यदि वे कानून अथवा आदेश संविधान के विरुद्ध हों तो उनको असंवैधानिक एवं अवैध घोषित कर सकते हैं। न्यायालय कानून की उन्हीं धाराओं को अवैध घोषित करते हैं जो संविधान के विरुद्ध होती हैं न कि समस्त कानून को। न्यायालय उन्हीं कानूनों को अवैध घोषित कर सकता है जो उनके सामने मुकद्दमे के रूप में आते हैं।

प्रश्न 13.
सर्वोच्च न्यायालय की न्यायिक पुनर्निरीक्षण की शक्ति के बारे में आप क्या जानते हैं ?
उत्तर-
संविधान का संरक्षक तथा संविधान की अन्तिम व्याख्या करने वाली संस्था होने के कारण सर्वोच्च न्यायालय को न्यायिक पुनर्निरीक्षण की शक्ति प्राप्त है। यदि कोई नागरिक समझता है कि केन्द्र सरकार या राज्य विधानमण्डल द्वारा बनाया गया कोई कानून संविधान के विरुद्ध है तो वह नागरिक इस सम्बन्ध में सर्वोच्च न्यायालय को याचना कर सकता है। सर्वोच्च न्यायालय संसद् तथा राज्य विधानमण्डल द्वारा बनाए गए कानूनों की छानबीन करता है और यदि कोई कानून संविधान के विरुद्ध हो तो वह उसको असंवैधानिक घोषित कर सकता है। 9 मई, 1980 को सर्वोच्च न्यायालय ने 42वें संशोधन एक्ट के उस खण्ड 55 को रद्द किया, जिसमें संसद् को संविधान में संशोधन करने के असीमित अधिकार दिए गए थे। सर्वोच्च न्यायालय ने अधिकतर उन्हीं कानूनों को रद्द किया है जो अनुच्छेद 14, 19 तथा 31 (अनुच्छेद 31 को 44वें संशोधन द्वारा संविधान से निकाल दिया गया है) का उल्लंघन करते हैं।

अति लघु उत्तरीय प्रश्न

प्रश्न 1.
भारतीय सर्वोच्च न्यायालय का न्यायाधीश बनने के लिए आवश्यक योग्यताएं लिखें।
उत्तर-

• वह भारत का नागरिक हो।
• वह कम-से-कम पांच वर्ष तक एक या एक से अधिक उच्च न्यायालयों में न्यायाधीश के पद पर रह चुका हो।

अथवा
वह कम-से-कम 10 वर्ष तक उच्च न्यायालय का एडवोकेट रह चुका हो।

अथवा
वह राष्ट्रपति की दृष्टि में प्रसिद्ध कानून-विशेषज्ञ हो।

प्रश्न 3.
सर्वोच्च न्यायालय के किन्हीं दो प्रारम्भिक क्षेत्राधिकारों को बताएं।
उत्तर-
सर्वोच्च न्यायालय में कुछ मुकद्दमे सीधे ले जाए जा सकते हैं

• यदि केन्द्र या किसी एक राज्य या कई राज्यों के बीच कोई झगड़ा उत्पन्न हो जाए तो उसका निर्णय सर्वोच्च न्यायालय करता है।
• यदि कुछ राज्यों के बीच किसी संवैधानिक विषय पर कोई झगड़ा उत्पन्न हो जाए तो वह झगड़ा भी सर्वोच्च न्यायालय द्वारा ही निपटाया जाता है।

प्रश्न 4.
सर्वोच्च न्यायालय के न्यायाधीशों का वेतन तथा अन्य सुविधाएं बताएं।
उत्तर-
सर्वोच्च न्यायालय के मुख्य न्यायाधीश को 2,80,000 रुपये मासिक तथा अन्य न्यायाधीशों को 2,50,000 रुपये मासिक वेतन मिलता है। वेतन के अतिरिक्त उन्हें कुछ भत्ते भी मिलते हैं। उन्हें रहने के लिए बिना किराये का निवास-स्थान भी मिलता है।

प्रश्न 5.
सर्वोच्च न्यायालय के तीन प्रकार के क्षेत्राधिकारों के नाम लिखें।
उत्तर-

1. प्रारम्भिक क्षेत्राधिकार ।
2. अपीलीय क्षेत्राधिकार।
3. सलाहकारी क्षेत्राधिकार।

प्रश्न 6.
सर्वोच्च न्यायालय का संगठन क्या है ?
उत्तर-
सर्वोच्च न्यायालय में एक मुख्य न्यायाधीश तथा कुछ अन्य न्यायाधीश होते हैं। आजकल सर्वोच्च न्यायालय में मुख्य न्यायाधीश के अतिरिक्त 30 अन्य न्यायाधीश हैं। सर्वोच्च न्यायालय के मुख्य न्यायाधीश तथा अन्य न्यायाधीशों की नियुक्ति राष्ट्रपति करता है।

प्रश्न 7.
सर्वोच्च न्यायालय की न्यायिक पुनर्निरीक्षण की शक्ति के बारे में आप क्या जानते हैं ?
उत्तर-
संविधान का संरक्षक तथा संविधान की अन्तिम व्याख्या करने वाली संस्था होने के कारण सर्वोच्च न्यायालय को न्यायिक पुनर्निरीक्षण की शक्ति प्राप्त है। यदि कोई नागरिक समझता है कि केन्द्र सरकार या राज्य विधानमण्डल द्वारा बनाया गया कोई कानून संविधान के विरुद्ध है तो वह नागरिक इस सम्बन्ध में सर्वोच्च न्यायालय को याचना कर सकता है। सर्वोच्च न्यायालय संसद् तथा राज्य विधानमण्डल द्वारा बनाए गए कानूनों की छानबीन करता है और यदि कोई कानून संविधान के विरुद्ध हो तो वह उसको असंवैधानिक घोषित कर सकता है।

वस्तुनिष्ठ प्रश्न

प्रश्न I. एक शब्द/वाक्य वाले प्रश्न-उत्तर-

प्रश्न 1. सर्वोच्च न्यायालय का संगठन क्या है ?
उत्तर-सर्वोच्च न्यायालय का एक मुख्य न्यायाधीश तथा अन्य न्यायाधीश होते हैं। आजकल सर्वोच्च न्यायालय में मुख्य न्यायाधीश के अतिरिक्त 30 अन्य न्यायाधीश हैं।

प्रश्न 2. सर्वोच्च न्यायालय के न्यायाधीश और उच्च न्यायालय के न्यायाधीश कितनी आयु में सेवानिवृत्त होते हैं ?
उत्तर-सर्वोच्च न्यायालय के न्यायाधीश 65 वर्ष और उच्च न्यायालय के न्यायाधीश 62 वर्ष की आयु में सेवानिवृत्त होते हैं।

प्रश्न 3. सर्वोच्च न्यायालय का अध्यक्ष कौन होता है ?
उत्तर-सर्वोच्च न्यायालय का अध्यक्ष मुख्य न्यायाधीश होता है।

प्रश्न 4. सर्वोच्च न्यायालय की व्यवस्था किस अनुच्छेद के अधीन की गई है ?
उत्तर-सर्वोच्च न्यायालय की व्यवस्था संविधान के अनुच्छेद 124 में की गई है।

प्रश्न 5. सर्वोच्च न्यायालय की स्थापना कहां की गई है ?
उत्तर-सर्वोच्च न्यायालय की स्थापना नई दिल्ली में की गई है।

प्रश्न 6. सर्वोच्च न्यायालय के न्यायाधीशों की संख्या कौन निश्चित करती है ?
उत्तर-सर्वोच्च न्यायालय के न्यायाधीशों की संख्या संसद् निश्चित करती है।

प्रश्न 7. संविधान एवं मौलिक अधिकारों का संरक्षक किसे माना जाता है ?
उत्तर-संविधान एवं मौलिक अधिकारों का संरक्षक सर्वोच्च न्यायालय को माना जाता है।

प्रश्न 8. सर्वोच्च न्यायालय के अन्य न्यायाधीशों की नियुक्ति किसके परामर्श से की जाती है ?
उत्तर-मुख्य न्यायाधीश के।

प्रश्न 9. सर्वोच्च न्यायालय के मुख्य न्यायाधीश तथा अन्य न्यायाधीशों के वेतन बताओ।
उत्तर-सर्वोच्च न्यायालय के मुख्य न्यायाधीश को 2,80,000 रु० मासिक तथा अन्य न्यायाधीशों को 2,50,000 रु० मासिक वेतन मिलता है।

प्रश्न 10. सर्वोच्च न्यायालय का न्यायाधीश बनने के लिए कोई एक योग्यता लिखें।
उत्तर-सर्वोच्च न्यायालय का न्यायाधीश बनने के लिए आवश्यक है कि वह भारत का नागरिक हो।

प्रश्न 11. क्या सर्वोच्च न्यायालय के न्यायाधीश सेवानिवृत्त होने के पश्चात् वकालत कर सकते हैं ?
उत्तर-सर्वोच्च न्यायालय के न्यायाधीश सेवानिवृत्त होने के पश्चात् वकालत नहीं कर सकते हैं।

प्रश्न 12. उच्च न्यायालय के क्षेत्राधिकार में कौन वृद्धि कर सकता है ?
उत्तर-उच्च न्यायालय के क्षेत्राधिकार में संसद् वृद्धि कर सकती है।

प्रश्न 13. उच्च न्यायालयों के जजों के रिटायर होने की आयु कितनी है ?
उत्तर-62 वर्ष।

प्रश्न 14. उच्च न्यायालय का न्यायाधीश बनने के लिए कोई एक योग्यता लिखें।
उत्तर-उच्च न्यायालय का न्यायाधीश बनने के लिए आवश्यक है कि वह व्यक्ति भारत का नागरिक हो।

प्रश्न 15. उच्च न्यायालयों के न्यायाधीशों को कौन हटा सकता है ?
उत्तर-उच्च न्यायालयों के न्यायाधीशों को संसद् की सिफ़ारिश पर राष्ट्रपति हटा सकता है।

प्रश्न 16. उच्च न्यायालयों के न्यायाधीश की नियुक्ति कौन करता है ?
उत्तर-राष्ट्रपति।

प्रश्न 17. उच्च न्यायालयों के न्यायाधीशों को प्रति मास कितना वेतन मिलता है ?
उत्तर-2,25,000 रु०।।

प्रश्न 18. उच्च न्यायालयों के मुख्य न्यायाधीशों को प्रति मास कितना वेतन मिलता है ?
उत्तर-2,50,000 रु०।

प्रश्न II. खाली स्थान भरें-

1. सर्वोच्च न्यायालय के क्षेत्राधिकार में आरंभिक क्षेत्राधिकार,अपीलीय क्षेत्राधिकार तथा …………..
2. संविधान की व्याख्या तथा रक्षा करना ………….. का कार्य है।
3. सर्वोच्च न्यायालय संविधान के ……………… के अनुसार पांच प्रकार के लेख जारी कर सकता है।
4. सर्वोच्च न्यायालय की स्थापना …………… में की गई है।
उत्तर-

1. सलाहकारी क्षेत्राधिकार
2. सर्वोच्च न्यायालय
3. अनुच्छेद 32
4. नई दिल्ली।

प्रश्न III. निम्नलिखित में से सही एवं ग़लत का चुनाव करें-

1. सर्वोच्च न्यायालय की व्याख्या संविधान के अनुच्छेद 51 में की गई है।
2. वर्तमान समय में सर्वोच्च न्यायालय में मुख्य न्यायाधीश के अतिरिक्त 30 न्यायाधीश हैं।
3. सर्वोच्च न्यायालय के न्यायाधीशों की नियुक्ति राज्यपाल करता है।
4. सर्वोच्च न्यायालय के न्यायाधीश 65 वर्ष की आयु तक अपने पद पर रहते हैं।
5. संसद् सर्वोच्च न्यायालय के न्यायाधीशों को साधारण बहुमत से हटा सकती है।
उत्तर-

1. ग़लत
2. सही
3. ग़लत
4. सही
5. ग़लत ।

प्रश्न IV. बहुविकल्पीय प्रश्न-

प्रश्न 1.
निम्नलिखित में से किसको संविधान का संरक्षक माना जाता है ?
(क) राष्ट्रपति
(ख) सर्वोच्च न्यायालय
(ग) उच्च न्यायालय
(घ) संसद् ।
उत्तर-
(ख) सर्वोच्च न्यायालय ।

प्रश्न 2.
सर्वोच्च न्यायालय को कौन-सा अधिकार क्षेत्र प्राप्त नहीं है-
(क) प्रारम्भिक अधिकार क्षेत्र
(ख) अपीलीय
(ग) सलाहकारी
(घ) राजनीतिक।
उत्तर-
(घ) राजनीतिक।।

प्रश्न 3.
उच्च न्यायालय के न्यायाधीशों को कौन हटा सकता है-
(क) राष्ट्रपति
(ख) राज्यपाल
(ग) संसद्
(घ) प्रधानमन्त्री।
उत्तर-
(ग) संसद्।

प्रश्न 4.
न्यायिक पुनर्निरीक्षण की शक्ति है-
(क) जिला न्यायालयों के पास
(ख) केवल उच्च न्यायालयों के पास
(ग) केवल सर्वोच्च न्यायालय के पास
(घ) सर्वोच्च व उच्च न्यायालय दोनों के पास।
उत्तर-
(घ) सर्वोच्च व उच्च न्यायालय दोनों के पास।