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}

  • $\begingroup$ en.wikipedia.org/wiki/Peroxymonosulfuric_acid It has oxygen on higher ox state as all peroxyacids have $\endgroup$ – Mithoron Dec 13 '15 at 12:45
  • $\begingroup$ @hackwarewright Also called Caro's acid. $\endgroup$ – Pritt says Reinstate Monica May 12 '17 at 2:30
  • $\begingroup$ Structures are more important than chemical formulae while finding oxidation numbers of individual atoms. $\endgroup$ – Eashaan Godbole May 20 '19 at 14:34
  • $\begingroup$ Note that an oxidation number of $+8$ on $\ce{S}$ would require to open up the 2p shell, which does not happen for sulfur. $\endgroup$ – TAR86 May 20 '19 at 16:24

$\ce {H_2SO_5}$ 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 $-1$.


$\ce{H2SO5}$ has the Lewis structure shown below: enter image description here 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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