# What is the distinction between "oxides", "peroxides", and "superoxides"?

Mar 23, 2017

Oxygen generally assumes $- I I$ $\text{oxidation state}$.

#### Explanation:

While oxidation state is a formalism, oxygen generally is conceived to accept 2 electrons in its compounds to give a $- I I$ oxidation state. This is certainly true of water, $O {H}_{2}$, i.e. ${O}^{2 -} + 2 \times {H}^{+}$.

And in metal oxides: $N {a}_{2} O , M g O , F {e}_{2} {O}_{3}$, the metal oxidation states are $N a \left(+ I\right) , \text{i.e. } N {a}^{+}$, $M g \left(+ I I\right) , \text{i.e. } M {g}^{2 +}$, and $F e \left(+ I I I\right) , \text{i.e. } F {e}^{3 +}$ respectively.

Mar 26, 2017

I have been asked to expand this answer, and explain the difference between $\text{oxides}$, $\text{peroxides}$, and $\text{superoxides}$, and my attempt follows.

#### Explanation:

Now hydrogen peroxide is $H O - O H$, and it clearly contains an $O - O$ bond.  Because our definition of oxidation number is $\text{the charge left on the central atom when all the bonding pairs}$ $\text{of electrons are BROKEN, with the charge assigned}$ $\text{to the most electronegative atom,}$

this exercise results in the sharing of the electrons (because the oxygen atoms have equal electronegativity):

i.e. $H O - O H \rightarrow 2 \times \dot{O} H$

($\dot{O} H$ is the so-called hydroperoxyl radical, a NEUTRAL radical species.) The oxygens in hydrogen peroxide thus have a formal oxidation state of $- I$. Why? Because we consider the $O H$ species, where clearly hydrogen is LESS electronegative than oxygen.

And when we write peroxide salts, i.e. sodium peroxide, we use a formula of $N {a}_{2} {O}_{2}$; i.e. a salt of ""^(-)O-O^-.

And for $\text{superoxides}$, to continue the formalism, we write ${O}_{2}^{-}$, that is a mixed oxidation state dioxide of ${O}^{0}$ and ${O}^{-}$, to give an average oxidation number of ${O}^{- \frac{1}{2}}$.  In very old literature, this goes by the label $\text{hyperoxide}$