Which one of the following compounds is a peroxide?

Key Idea:- In peroxides, the oxidation state of $O$ is - 1 and they give $\mathrm{H}_2 \mathrm{O}_2$, with dilute acids, and have peroxide linkage. In $\mathrm{KO}_2$, $$ +1+(\mathrm{x} \times 2)=0 $$ $x=-\frac{1}{2}$ (thus, it is a superoxide, not $a$ peroxide.) In $\mathrm{BaO}_2$, $$ \begin{aligned} & +2+(\mathrm{x} \times 2)=0 \\ & \mathrm{x}=-1 \end{aligned} $$ Thus, it is a peroxide. Only it gives $\mathrm{H}_2 \mathrm{O}_2$ when reacts with dilute acids and has peroxide linkage as $$ \begin{aligned} & \mathrm{Ba}^{2+}[\mathrm{O}-\mathrm{O}]^{2-} \\ & \text { peroxide linkage } \end{aligned} $$ In $\mathrm{MnO}_2$ and $\mathrm{NO}_2, \mathrm{Mn}$ and $\mathrm{N}$ exhibit variable oxidation states, thus, the oxidation state of $O$ in these is -2 . Hence, these are not peroxides. Thus, it is clear, that among the given molecules only $\mathrm{BaO}_2$ is a peroxide.