Question

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

Answer 1

Oxygen generally assumes `-II` `"oxidation state"`.

Explanation:

While oxidation state is a formalism, oxygen generally is conceived to accept 2 electrons in its compounds to give a `-II` oxidation state. This is certainly true of water, `OH_2`, i.e. `O^(2-) + 2xxH^+`.

And in metal oxides: `Na_2O, MgO, Fe_2O_3`, the metal oxidation states are `Na(+I), "i.e. " Na^+`, `Mg(+II), "i.e. " Mg^(2+)`, and `Fe(+III), "i.e. " Fe^(3+)` respectively.

Answer 2

I have been asked to expand this answer, and explain the difference between `"oxides"`, `"peroxides"`, and `"superoxides"`, and my attempt follows.

Explanation:

Now hydrogen peroxide is `HO-OH`, and it clearly contains an `O-O` bond.  Because our definition of oxidation number is `"the charge left on the central atom when all the bonding pairs"` `"of electrons are BROKEN, with the charge assigned"` `"to the most electronegative atom,"`

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

i.e. `HO-OHrarr2xxdotOH`

(`dotOH` 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 `OH` species, where clearly hydrogen is LESS electronegative than oxygen.

And when we write peroxide salts, i.e. sodium peroxide, we use a formula of `Na_2O_2`; i.e. a salt of `""^(-)O-O^-`.

And for `"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^(-1/2)`.  In very old literature, this goes by the label `"hyperoxide"`.