Question

Find the sum of average of oxidation number of S in ${H_2}S{O_5}$ (peroxy mono sulphuric acid) and $N{a_2}{S_2}{O_3}$ (sodium thiosulphate).

Answer 1

Hint: First group elements generally have \[ + 1\] oxidation state and oxygen has $ - 2$ oxidation state.

Complete step by step answer:
In ${H_2}S{O_5}$, if we direct calculate the oxidation number of S, then
$
  2 \times 1 + x + 5 \times ( - 2) = 0 \\
  \Rightarrow x - 8 = 0 \\
  \Rightarrow x = + 8 \\
 $
But the oxidation state of S can’t be greater than\[ + 6\].
As shown in the structure of ${H_2}S{O_5}$
 $H - O - S{( = O)_2} - O - O - H$
2 Oxygen are in double bond with S, one OH group, one oxygen with single bond with S, and one bond with OH that is peroxide bond and in peroxide O has $ - 1$ oxidation state. So, two oxygen have $ - 1$ oxidation state.
Now, oxidation state of ${H_2}S{O_5}$will be
$ 2 \times 1 + x + 3 \times ( - 2) + 2 \times ( - 1) = 0 \\
  \Rightarrow 2 + x - 8 = 0 \\
  \Rightarrow x = + 6 \\
$
Now, the oxidation state of $N{a_2}{S_2}{O_3}$
$ 2 \times 1 + 2x + 3 \times ( - 2) = 0 \\
    \Rightarrow 2 + x - 6 = 0 \\
  \Rightarrow 2x = + 4 \\
  \Rightarrow x = + 2 \\
$

Hence, sum of average oxidation number is $6 + 2 = + 8$

Note:
When you find the oxidation number of an element in a molecule please check the oxidation state of the individual element as we did in${H_2}S{O_5}$. It can vary for an individual element.