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

$2x=-1$  and $x=-\frac{1}{2}$

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