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.

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(A-D)
$A$: $MnO_{2}$$B$: $K_{2}MnO_{4}$
$C$: $KMnO_{4}$$D$: $KIO_{3}$

$1. \text{Fusion reaction: } 2 MnO_{2} + 4 KOH + O_{2} \rightarrow 2 K_{2}MnO_{4} + 2 H_{2}O$
$2. \text{Disproportionation of } (B) \text{ in acidic medium: } 3 MnO_{4}^{2-} + 4 H^{+}$ $\rightarrow 2 MnO_{4}^{-} + MnO_{2} + 2 H_{2}O$
$3. \text{Oxidation of } KI \text{ by } (C) \text{ in alkaline medium: } 2 MnO_{4}^{-} + H_{2}O + I^{-}$ $\rightarrow 2 MnO_{2} + 2 OH^{-} + IO_{3}^{-}$

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