Question

What is oxidation state of sulphur in the following?

Peroxymonosulfuric acid

Answer 1

\[\ce{H2SO5}\] has the following structure:

\[\begin{array}{cc} \ce{O}\phantom{.....}\\ ||\phantom{.....}\\ \ce{H - O - S - O - O - H}\\ ||\phantom{.....}\\ \ce{O}\phantom{.....} \end{array}\]

the oxygens in peroxide linkage i.e. \[\ce{O - O}\] has oxidation number = -1 each

and in oxide linkage i.e = O and \[\ce{H - O - S}\] it is -2.

Let the oxidation number of sulphur be x.

Oxidation number of H = + 1

In \[\ce{H2SO5}\] ⇒ 2 × 1 + x + 3 × (- 2) + 2 × (-1) = 0

⇒ x + 2 - 6 - 2 = 0 

⇒ x - 6 = 0

⇒ x = + 6

Thus Oxidation number of sulphur in peroxomonosulphuric acid \[\ce{(H2SO5)}\] is + 6.