Question

When an oxide of manganese (A) is fused with KOH in the presence of an oxidising agent and dissolved in water, it gives a dark green solution of compound (B). Compound (B) disproportionates in neutral or acidic solution to give purple compound (C). An alkaline solution of compound (C) oxidises potassium iodide solution to a compound (D) and compound (A) is also formed. Identify compounds A to D and also explain the reactions involved.

Answer 1

The compounds (A), (B), (C) and (D) are given as under:

A = \[\ce{MnO2}\]

B = \[\ce{K2MnO4}\]

C = \[\ce{KMnO4}\]

D = \[\ce{KlO3}\]

The reactions are explained as under:

\[\ce{\underset{(A)}{2MnO2} + 4KOH + O2 -> \underset{(B)}{2K2MnO4} + 2H2O}\]

\[\ce{3MnO^{2-}4 + 4H^+ -> \underset{(C)}{2MnO^{-}4} + MnO2 + 2H2O}\]

\[\ce{2MnO^{-}4 + H2O + Kl -> \underset{(A)}{2MnO2} + 2OH^{-} + \underset{(D)}{KlO3}}\]