Chemistry - Complete s-block Question with Solution | TestHub

The oxidation state of $\mathrm{O}$in peroxide $\left({\text{O}}_2^{2-}\right)$ is $-1.$

The oxidation state of $\mathrm{O}$in:

${\text{KO}}_{\text{2}}=\frac{-1}{2}$

${\text{BaO}}_{\text{2}}= -1$

${\text{MnO}}_{\text{2}} = -2$

${\text{NO}}_{\text{2}}= -2$

Therefore, ${\mathrm{BaO}}_2$ is a peroxide.