Chemistry - Complete s-block Question with Solution | TestHub

Here statement D is wrong as${\mathrm{H}}_2{\mathrm{O}}_2$is oxidized to${\mathrm{O}}_2$by acidified$\mathrm{K}\mathrm{M}\mathrm{n}{\mathrm{O}}_4$.
$2\mathrm{M}\mathrm{n}{\mathrm{O}}_4^{-}+5{\mathrm{H}}_2{\mathrm{O}}_2+6{\mathrm{H}}^{+}\to 2\mathrm{M}{\mathrm{n}}^{2+}+8{\mathrm{H}}_2\mathrm{O}+5{\mathrm{O}}_2$
A. It is because rough surface catalyze decomposition of${\mathrm{H}}_2{\mathrm{O}}_2$.
B. Oxidation state of oxygen in${\mathrm{H}}_2{\mathrm{O}}_2$is -1. It can, therefore, be oxidized to${\mathrm{O}}_2$(zero oxidation state) or reduced to${\mathrm{H}}_2\mathrm{O}$or$\mathrm{O}{\mathrm{H}}^{-}$(-2 oxidation state).
C. It decomposes vigorously above 413 K. Its boiling point (423.2 K) is an extrapolated value.