Question

Oxidation number of sulphur in $H_2{S}_2{O}_8$ is

• A. +6
• B. +8
• C. +21
• D. +7

Answer 1

Answer: (A)

(i) Draw the structue of oxyacid $\left(H_2{S}_2{O}_8\right)$ to decide the oxidation state of oxygen. Peroxide linkage has $-1$ oxidation state and normally oxygen has $-2$ oxidation state.

$\therefore$ Two oxygens form peroxide linkage.
$\therefore$ Oxidation state of two oxygens is $-1$ each and rest of the six oxygens have $-2$ oxidation state each,
$H_2{S}_2{O}_8$
$+1\times 2+2x+(-1\times 2)+(-2\times 6)=0$
or $2+2x-2-12=0$
or $2x=12$
$\therefore x=+6$