${MnO}_{2}$ when fused with KOH and oxidised in air gives a dark green compound $X$ .
In acidic solution, $X$ undergoes disproportionation to give an intense purple colour
compound $Y$ and ${MnO}_{2}$ . The compound $X and Y$ respectively, are

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$K_{2} {MnO}_{4} and {KMnO}_{4}$

$K_{2} {MnO}_{4} and {Mn}_{2} O_{7}$

${Mn}_{2} O_{7} and {KMnO}_{4}$

${KMnO}_{4} and K_{2} {MnO}_{4}$

answer is A.

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Detailed Solution

$2 {MnO}_{2} + 4 KOH + O_{2} \overset{Oxidised in air}{\to} \frac{2 K_{2} {MnO}_{4}}{X [Dark green compound]} + 2 H_{2} O$

$3 \overset{+ 6}{Mn} O_{4}^{- 2} + 4 H^{+} \overset{Disproportionation}{\to} \frac{2 \overset{+ 7}{Mn} O_{4}^{-}}{Y [Purple compound]} + \overset{+ 2}{Mn} O_{2} + 2 H_{2} O$

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