Oxidation number of oxygen in super oxides is

The general formula of superoxide is ${\mathrm{MO}}_2$ In super oxides, oxygen was combined with highly reactive alkali metals like Na ,Li and K . Here, M means alkali metal its valency is 1. So, the metal atom shows +1 oxidation state. So, $1(+1)+2\left(\mathrm{x}\right)=0$

2x=-1 

$\mathrm{x}=-\frac{1}{2}$

Therefore, the oxidation number of oxygen in super oxide ${\mathrm{MO}}_2 \mathrm{is}-\frac{1}{2}$