Hello Ushaben

Method to find oxidation number

• Oxidation number of Free elements (like ${\mathrm{O}}_2,{\mathrm{H}}_2$ , Na) is 0

Oxidation number of S in ${\mathrm{H}}_2{\mathrm{SO}}_4$

Let the oxidation number of S = x

and H = +1, O = −2

So, $\begin{array}{c}2(+1)+x+4(-2)=0\\ 2+x-8=0\\ x=+6\end{array}$

Oxidation number of S = +6