d and f Block Elements Part 3

Concept:Disproportionation involves simultaneous oxidation and reduction of the same element.In acidic medium, manganate ion ($\mathrm{MnO}_4^{2-}$) disproportionates to give permanganate and manganese dioxide.Chemical Equation:$$3\mathrm{MnO}_4^{2-} + 4\mathrm{H}^+ \rightarrow 2\mathrm{MnO}_4^- + \mathrm{MnO}_2 + 2\mathrm{H}_2\mathrm{O}$$Explanation:The oxidation state of Mn in $\mathrm{MnO}_4^{2-}$ is $+6$ (since $x + 4(-2) = -2$ gives $x = +6$).In acidic medium, it disproportionates: one Mn atom is oxidized from $+6$ to $+7$ (forming $\mathrm{MnO}_4^-$), and another is reduced from $+6$ to $+4$ (forming $\mathrm{MnO}_2$).Thus, the products are $\mathrm{MnO}_4^-$ and $\mathrm{MnO}_2$.Shortcut:Manganate ion ($\mathrm{MnO}_4^{2-}$) is green; in acid it gives purple $\mathrm{MnO}_4^-$ and brown $\mathrm{MnO}_2$.So, the correct option is the one listing these two products.Answer:Option B: $\mathrm{MnO}_4^-$ and $\mathrm{MnO}_2$.