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) disproportionate 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 explain the reactions involved.

Answer 1

$IdentificationofAandB$

When an oxide of manganese (A) $Mn{O}_2$ is fused with $KOH$ in the presence of an oxidising agent and dissolves in water, it gives a dark green solution of compound (B) $K_{2}Mn{O}_4$.

The reaction is given below:

$2Mn{O}_2+4KOH+O_2\to 2K_{2}Mn{O}_4+2H_{2}O$
$\left(A\right)$ $\left(B\right)$

$IdentificationofC$

$K_{2}Mn{O}_4$ disproportionates in acidic solution to give purple compound (C) $Mn{O}_4^{-}$.

The reaction is given below:

$3Mn{O}_4^{2-}+4H^{+}\to Mn{O}_4^{-}+Mn{O}_2+2H_{2}O$
(C)

$IdentificationofD$

An alkaline solution of compound $Mn{O}_4^{-}$ (C) oxidises potassium iodide solution to a compound $KI{O}_3$ (D).

The reaction is given below:

$2Mn{O}_4^{-}+H_{2}O+KI$
$\downarrow$
$2Mn{O}_2+2O{H}^{-}+KI{O}_3$

Hence, $A=Mn{O}_2,B=K_{2}Mn{O}_4,C=Mn{O}_4^{-}andD=KI{O}_3$.