Oxidation number of sulphur in oleum (H₂S₂O₇) is

Let the oxidation number of sulphur be x. The oxidation number of H and O are +1 and −2 respectively.

$$\begin{gathered} 2(+1)+2\mathrm{x}+7(-2)=0 \\ \mathrm{x}=+6 \end{gathered}$$