Question

Oxidation number of sulphur in peroxomonosulphuric acid $\left(H_{2}S{O}_5\right)$ is:

• A. 4
• B. 2
• C. 6
• D. -2

Answer 1

Answer: (C)

6

$H_{2}S{O}_5$ has the following structure:

$H-O-\stackrel{\stackrel{O}{\parallel }}{\underset{\underset{O}{\parallel }}{S}}-O-O-H$

the oxygens in peroxide linkage i.e. $O-O$ has oxidation number =-1 each

and in oxide linkage i.e $=O$ and $H-O-S$ it is -2

Let the oxidation number of sulphur be $x$.

Oxidation number of $H=+1$

In $H_{2}S{O}_5$: $2\times 1+x+3\times (-2)+2\times (-1)=0$

$x+2-6-2=0$ or $x=+6$

thus Oxidation number of sulphur in peroxomonosulphuric acid ($H_{2}S{O}_5$) is $+6$.