Question

Question 8.2 What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?

(a) KI3 (b) H2S4O6 (c) Fe3O4 (d) CH3CH2OH (e) CH3COOH

Class XI Redox Reactions Page 272

Answer 1

(a) KI₃
let the oxidation number of I is x
+1 + 3x = 0
    x = -1/3
hence, the average oxidation number of I is -1/3 .but we know, oxidation number can't be fractional. actually,
       the structure of KI₃,     K⁺ (I⁰------I⁰<----I⁻)⁻
in this structure, a co-ordination bond is formed between I₂ molecule and I⁻ ion. hence, oxidation of I₃ is -1

(b)H₂S₄O₆
let the oxidation number of S = x
 +2+4x +6(-2)=0
  x = +5/2 
similarly, oxidation number of this compound depends upon structure of molecule. 
see structure,      O                       O
                              ||                        ||
          H---O---------S⁺⁵----S⁰-----S⁰-------S⁺⁵--------O-----H
                              ||                        || 
                              O                      O
after seeing the structure, it is clear that oxidation number of Satoms  are +5,0,0, +5 respectively.

(c)Fe₃O₄
let the oxidation number of Fe is x 
then, 3x + 4(-2) = 0
    x = + 8/3 
 hence, the average oxidation number of Fe is +8/3, but stoichiometrically, Fe₃O₄ is an equimolar mixture of FeO and Fe₂O₃. 
                 The oxidation number of Fe in FeO is +2 
and the oxidation number of Fe in Fe₂O₃ is +3 
then, 2x + x = +6 + 2 = +8 ⇒ x = +8/3
(d) CH₃CH₂OH
let the o.n of C is x 
  x + 3 + x + 2 - 2 + 1 =0
      x = -2 
therefore, average o.n of C is -2 
now, see the structure of given molecule.
              H                H
               |                  |
  H------- C² ------------- C¹ ---- O----H   [CH₃⁺ and CH₂OH⁻
              |                   |
             H                  H
    now, oxidation number of C₁ atom : x+ 2 -2 +1 = -1 ⇒ x = -2
    oxidation number of C₂ atom : x + 3 = +1 ⇒ x = - 2
thats why average oxidation number of C - atoms is -2 

(e) CH₃COOH
     let the oxidation number of C is x 
x + 3 + x -2 -2 + 1 = 0
  2x = 0  ⇒x = 0
therefore, average oxidation number of C is zero. 
                  let see the structure for understanding 
          H           O
           |            ||
H------C²----------C¹------O---H    [CH₃⁺ and COOH⁻ 
          |
         H
                now, oxidation number of C₁ atom : x - 2 -2 +1 =1 ⇒ x = +2
                        oxidation number of C₂ atom : x +3 = +1 ⇒ x = -2 
that's why average oxidation number of CH₃COOH is zero.
 
 

Answer 2

chemistry chapters 8 redox reaction