JEE Main · 2024 · 30 Jan 2024 · Shift 2 · d and f Block Elements

Alkaline oxidative fusion of ${\text{MnO}}_2$ with atmospheric ${\text{O}}_2$ and molten KOH oxidises Mn(IV) to Mn(VI), giving the green manganate ion: $2{\text{MnO}}_2+4{\text{OH}}^{-}+{\text{O}}_2\to 2{\text{MnO}}_4^{2-}+2{\text{H}}_2\text{O}$; this manganate ion is A. Electrolytic oxidation of manganate in alkaline solution removes a further electron, ${\text{MnO}}_4^{2-}\to {\text{MnO}}_4^{-}+e^{-}$, giving the purple permanganate ion, B. A is ${\text{MnO}}_4^{2-}$ and B is ${\text{MnO}}_4^{-}$.