The oxidation number of sulphur in $\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8$ is

$\because$ 20-atoms are in peroxide linkage $\therefore$ Their oxidation state $=-1$ and, oxidation state of remaining 60 -atoms, $=-2$ Oxidation state of H -atoms $=+1$ Let oxidation state of S atom $=x$ $\begin{aligned} & \mathrm{S}_2 \quad \mathrm{H}_2 \quad \mathrm{O}_6 \quad \mathrm{O}_2 \\ & \therefore 2 x+2(+1)+6(-2)+2(-1)=0 \\ & 2 x+2-12-2=0 \\ & x=+6 \end{aligned}$