# Oxidation State of Oxygen in Na2O2?

I am asked to determine what is the O.S. of Oxygen in $\ce{Na2O2}$

I was under the impression that oxygen had a charge of -2 and I multiplied that charge by 2 giving me an Oxidation # of -4.

However, the computer program I am using states that I am wrong. Can someone clarify why?

• In the vast majority of cases this is true, but teachers like to assign substances that test the exceptions to rules, so be on the lookout for peroxides and superoxides. – Jason Patterson Sep 25 '14 at 13:41

Oxygen can take multiple oxidation states. This particular compound is sodium peroxide.

You're right that usually oxygen has a charge of -2, but in this case, there's no way that each $\ce{Na}$ can have an oxidation state of +2.

So you then work backwards, deciding if it's $\ce{Na+}$ then you have +2 from the sodium, and oxygen must have an average oxidation number of -1 per oxygen atom.

Besides peroxides ($\ce{O2^{2-}}$) there are also superoxides ($\ce{O2^{-}}$).

It's not surprising that these compounds are usually very reactive.

• I think this is a terminology difference. I meant to imply that there's no way to balance a -4 charge from two normal oxides with two sodium ions. You'd need +2 on each sodium, which clearly isn't the case. – Geoff Hutchison Sep 25 '14 at 2:52
• Do you really mean that "oxygen must have an average oxidation number of +1", rather than -1? I'm trying to understand more clearly the difference between oxidation number and oxidation state, but I must be missing something. – Cohen_the_Librarian Nov 16 '15 at 23:24
• @Cohen_the_Librarian Err yes, thanks. Fixed the typo. – Geoff Hutchison Nov 17 '15 at 0:23

There are "special rules" for determining the formal charge in some situations. Peroxides fall into one of these special cases, see here (see the section entitled "Oxygen in peroxides"). Each oxygen in a peroxide is assigned a formal charge of -1. Since sodium peroxide is neutral overall, each sodium atom is assigned an oxidation number of +1.

The structure of peroxides - $\ce{ROOR}$ - implies that the oxygens will generally exhibit -1 oxidation states. Oxidation states are assigned as if all bonding is ionic.

Given that oxygen is the second most electronegative element on the periodic table, we would expect that all the electrons in the $\ce{R-O}$ bond to be "taken" by oxygen (unless $\ce{R=F}$ as in the case of $\ce{FOOF}$).

In the $\ce{O-O}$ bond, however, the electrons are split 50/50 between the two oxygens (since both are of equal atomic electronegativities).

Therefore each oxygen has an oxidation state of -1 in peroxides (usually): each oxygen has 3 bonding electrons and 4 lone pair electrons for a total of 7 electrons, and oxygen by itself only has 6 electrons. 6 minus 7 is -1.