Chemistry - Complete s-block Question with Solution | TestHub

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 ${\mathrm{BaO}}_2$, $$\begin{array}{rl}& +2+(\mathrm{x}\times 2)=0\\ & \mathrm{x}=-1\end{array}$$ Thus, it is a peroxide. Only it gives ${\mathrm{H}}_2{\mathrm{O}}_2$ when reacts with dilute acids and has peroxide linkage as $$\begin{array}{rl}& {\mathrm{Ba}}^{2+}[\mathrm{O}-\mathrm{O}{]}^{2-}\\ & \text{ peroxide linkage }\end{array}$$ 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.