In acidic medium, H₂O₂ changes ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-} \mathrm{to} {\mathrm{CrO}}_5$ which has two $\left(-\mathrm{O}-\mathrm{O}-\right)$ bonds. Oxidation state of Cr in CrO₅ is:

${\mathrm{CrO}}_5$has butterfly structure having two peroxo bonds

Peroxo oxygen has -1 oxidation state.

Let oxidation state of Cr be “X”

${\mathrm{CrO}}_5:\mathrm{x}+4(-1)+1(-2)=0\Rightarrow \mathrm{x}=+6$