The oxidation number of S in H₂SO₅ and oxidation number of Cr in CrO₅ is

In H₂SO₅, there is one peroxide linkage. Hence, two O atoms has -1 oxidation state.

Oxidation state of S (x) is calculated as:

$$\begin{gathered} 2(+1)+2(-1)+\mathrm{x}+3(-2)=0 \\ 2+(-2)+\mathrm{x}+(-6)=0 \\ \mathrm{x}=+6 \end{gathered}$$

In CrO₅, there are two peroxide linkages. Therefore, 4 O atoms have -1 oxidation state.

Oxidation state of Cr (x) is calculated as:

$$\begin{gathered} \mathrm{x}+1(-2)+4(-1)=0 \\ \mathrm{x}-4-2=0 \\ \mathrm{x}=+6 \end{gathered}$$