Chemistry - What is the oxidation state of oxygen in Na2O2?

Solution 1:

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.

Solution 2:

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.

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Solution 3:

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.