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) &= 0\\
2 + x - 10 &= 0\\
x&=\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 $-1$.


 4

$\ce{H2SO5}$ has the Lewis structure shown below:
[![enter image description here](https://i.stack.imgur.com/ry30r.png)](https://i.stack.imgur.com/ry30r.png)
It is possible to assign the oxidation # of each atom by considering the electronegativities of the two atoms involved in each bond and assigning the bonding electrons to the more electronegative atom in each case. Oxygen atoms 3 and 4 are bonded to each other, so the bonding electrons are assigned one to each atom.


After assigning bonding electrons to the more electronegative atom in each bond, splitting the O-O bonding electrons and assigning lone pair electrons to the atom they are on, the oxidation # of each atom is found by the following formula:
$$ oxidation~number = group~number~of~element - assigned~electrons~in~the~structure$$
e.g. Oxygen 1: group # 6 (for oxygen) - assigned electrons 8 = -2 oxidation #


using these rules, the assigned oxidation #'s are:


Both H's: +1 each


Oxygens 1,2: -2 each


Oxygens 3,4: -1 each


S: +6


Note: there are no formal charges in the Lewis structure shown. There is another contributing form with single (dative) bonds to the two oxygens (# 2). This does not change the assigned oxidation numbers, but it does put 1- formal charges on the oxygens numbered 2, and a 2+ formal charge on the S.


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