During the preparation of potassium permanganate, the intermediate green potassium manganate () is f — d and f Block Elements Chemistry Question

💡 Solution & Explanation

In acidic or neutral media, the manganate ion ($MnO_4^{2-}$, where $Mn$ is $+6$) is inherently unstable and undergoes a disproportionation redox reaction. It simultaneously oxidises itself to permanganate ($MnO_4^-$, where $Mn$ is $+7$) and reduces itself to solid manganese dioxide ($MnO_2$, where $Mn$ is $+4$). The balanced reaction is: $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$.