How to calculate the oxidation number of sulfur in H2SO5? I am in class 11th and I am having trouble solving the question.


  Calculate the oxidation number of sulphur in $\ce{H2SO5}$?
  (The answer is given as as $+6$.)


$\ce{H2SO5}$ exists it is named as "peroxy sulfuric acid".

How tried it:
\begin{align}
2(+1) + 1(x) + 5(-2) &amp;= 0\\
2 + x - 10 &amp;= 0\\
x&amp;=\pm8
\end{align}
$\ce {H_2SO_5}$ has an oxygen-oxygen bond. This means that two of the five oxygen atoms have an oxidation number of $-1$. Same case as in $\ce{H2O2}$.

$$2\cdot (+1) + 1\cdot(x) + 3\cdot(-2) + 2\cdot (-1) = 0\\2 + x - 6 - 2 = 0\\
x = +6$$

You have 3 oxygens with oxidation number $-2$ and 2 oxygen atoms with $-1$.