- Chapter 1- Some Basic Concepts of Chemistry
- Chapter 2- Structure of The Atom
- Chapter 3- Classification of Elements and Periodicity in Properties
- Chapter 4- Chemical Bonding and Molecular Structure
- Chapter 5- States of Matter
- Chapter 6- Thermodynamics
- Chapter 7- Equilibrium
- Chapter 9- Hydrogen
- Chapter 10- The sBlock Elements
- Chapter 11- The p Block Elements
- Chapter 12- Organic Chemistry- Some Basic Principles and Techniques
- Chapter 13- Hydrocarbons
- Chapter 14- Environmental Chemistry

Chapter 1- Some Basic Concepts of Chemistry |
Chapter 2- Structure of The Atom |
Chapter 3- Classification of Elements and Periodicity in Properties |
Chapter 4- Chemical Bonding and Molecular Structure |
Chapter 5- States of Matter |
Chapter 6- Thermodynamics |
Chapter 7- Equilibrium |
Chapter 9- Hydrogen |
Chapter 10- The sBlock Elements |
Chapter 11- The p Block Elements |
Chapter 12- Organic Chemistry- Some Basic Principles and Techniques |
Chapter 13- Hydrocarbons |
Chapter 14- Environmental Chemistry |

(a) NaH

(b) NaHSO

(c) H

(d) K

(e) CaO

(f) NaBH

(g) H

(h) KAl(SO

**Answer
1** :

**Answer
2** :
**(a) **O.N. of iodine in KI_{3} **:** By conventional method, the oxidation number of iodine in the I_{3}^{_} ionmay be expressed as :**Explanation :** The oxidationnumber of iodine (I) comes out to be fractional which does not seem to bepossible. Let us consider the structure of I_{3}~ ion. In it, two atomsof iodine are linked to each other by covalent bond (I—I). The iodide ion (I~)is linked to the molecule by co-ordinate bond [I—I <—I]“. The molecule maybe represented as K^{+}[I—I <—1| . Now, in the anion, the oxidationnumber of the two I atoms is zero while I^{–} ion has – 1oxidation number. The overall oxidation number of L“ ion is :

(b) O.N. of S in H_{2}S_{4}O_{6} :By conventional method, the oxidation number of sulphur may be calculated as :

H^{+1}_{2 }S^{x}_{4 }O^{-2}_{6}2 + 4x + 6 (- 2) =0 or 4x = 12 – 2 = 10 or x = 5/2

Explanation : In order toaccount for the fractional value, let us assign oxidation numbers to differentsulphur atoms in the structural formula of the acid. The oxidation numbers ofthe two middle sulphur atoms is zero while the two atoms at the terminalpositions have + 5 oxidation number.

H

Explanation : In order toaccount for the fractional value, let us assign oxidation numbers to differentsulphur atoms in the structural formula of the acid. The oxidation numbers ofthe two middle sulphur atoms is zero while the two atoms at the terminalpositions have + 5 oxidation number.

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

(c) O. N. of Fe in Fe_{3}O_{4 }: Byconventional method, oxidation number of Fe may be calculated as :

Fe_{3}^{x} _{3} 0_{34}3x + 4(- 2) =0 or x = 8/3

Explanation : Fe_{3}O_{4} isa mixed oxide and is an equimolar mixture of Fe^{+2} O^{-2} andFe^{+3}_{2} O^{-2}_{3}Average O.N. of Fe = 1/3(2 + 2 x 3) =8/3

(d) O.N. of C in CH_{3}CH_{2}OH : Byconventional method, the oxidation number of carbon in ethanol molecule may becalculated as :

CH_{3}cH_{2}OH or C^{x}_{2} H^{+1}_{6} O^{-2}2x + 6(+ 1) + (- 2) = 0 or 2x + 6 – 2 = 0 or 2x = – 4 or x = – 2

Explanation : Let uscalculate the oxidation number of both C_{1} and C_{2} atoms.

C_{2} : Thecarbon atom is attached to three H atoms (less electronegative than carbon) andone CH_{2}OH group (more electronegative than carbon).

∴ O.N. of C_{2} = 3(+ 1) +x + (- 1) = O or x =-2

C1 : The carbon atom isattached to one OH group (O.N. = – 1), two H atoms (O.N. = + 1) and one CH_{3} group(O.N. = + 1)

O.N. of C1 = ( + 1) + x + 2 (+ 1) + 1(- 1) =0 or x = – 2 Average O.N. of C =l/2[ – 2 + (- 2)] = – 2

(e) O.N. of C in CH_{3}COOH : By conventionalmethod the O.N. of carbon may be calculated as:

CH3COOH or c_{2} H_{4} 62

2x + 4(+ 1) + 2(— 2) = 0 or x — 0

Explanation : Let uscalculate the oxidation number of both C_{2} and C1 atoms.

C1 : The carbon atom isattached to three H atoms (less electronegative than carbon) and one COOH group(more electronegative than carbon).

∴ O.N. of C_{2} =3 (+ 1) + x + 1(- 1) = 0 or x = – 2

C_{2} : Thecarbon atom is attached to one OH group (O.N. = – 1) one oxygen atom by doublebond

(O.N. = – 2) and one CH_{3} group (O.N. = + 1)

O.N. of C1 = 1( + 1) + Jt + 1 (-2) + 1(- 1) = O or x = + 2

∴ Average O.N. of C =1/2 [+ 2 + (- 2)] = 0

(b) Fe

(c) 2K(s) + F

(d) 4NH

**Answer
3** :
A chemical reaction may be regarded as redox reactionif one of the reacting species undergoes increase in O.N. (oxidation) and theother decrease in O.N. (reduction). Based upon this, let us try to justify thatthe reactions under study are redox reactions in nature.