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Q.

$K M n O_{4}$ can be prepared from $K_{2} M n O_{4}$ as per the reaction.
$3 M n O_{4}^{2 -} + 2 H_{2} O ⇌ 2 M n O_{4}^{-} + M n O_{2} + 4 O H^{-}$
The reaction can go to completion by removing $O H^{-}$ ions on adding

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a

KOH

b

NaCl

c

$C O_{2}$

d

$S O_{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} C O_{2} + 2 O H^{-} \to C O_{3}^{2 -} + H_{2} O \\ S O_{2} + 2 O H^{-} \to S O_{3}^{2 -} + H_{2} O \end{array}$

$M n O_{4}^{-} + S O_{3}^{2 -} \to M n^{2 +} + S O_{4}^{2 -}$ Hence $S O_{2}$ can not be used

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