Question

In $H_2{S}_2{O}_8$ oxidation state of sulphur is ____.

• A. 6
• B. -6
• C. 3
• D. 5

Answer 1

The correct option is A 6

In peroxy disulphuric acid, the oxidation state of six oxygen atoms is $-2$. The remaining two oxygen atoms form a peroxy linkage $-O-O-$ and the oxidation state of each of these two oxygen atoms is $-1$. Therefore, the oxidation state of sulphur (taken as $x$) can be calculated as follows:

$2\times \left(\text{oxidation state of hydrogen}\right)+2\times \left(\text{oxidation state of sulphur}\right)+6\times \left(\text{oxidation state of six oxygen atoms}\right)$

$+2\times \left(\text{oxidation state of two peroxy oxygen atoms}\right)=0$

$2\times (+1)+2x+6(-2)+2(-1)=0$

$2+2x-12-2=0$

$2x-12=0$

$x=\frac{12}{2}$

Therefore, oxidation state of sulphur = $x$ = +6.