The species that undergoes disproportionation in an alkaline medium are:

{C l}_{2}

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{M n O}_{4}^{2 -}

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{N O}_{2}

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{C l O}_{4}^{-}

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Solution

The correct options are
A ${C l}_{2}$
B ${N O}_{2}$
C ${M n O}_{4}^{2 -}$
In alkaline medium, chlorine disproportionates to form NaCl and sodium hypochlorite or sodium chlorate.
$2 N a O H (c o l d) + C l_{2} \to N a C l + N a C l O + H_{2} O$

$6 N a O H (h o t) + 3 C l_{2} \to 5 N a C l + N a C l O_{3} + 3 H_{2} O$

$M n O_{4}^{2 -}$ ion is stable in alkaline medium but disproportionates in acidic or neutral medium.

$3 M n O_{4}^{2 -} + 4 H^{+} \to 2 M n O_{4}^{-} + M n O_{2} + 2 H_{2} O$

$N O_{2}$ disproportionates in alkaline medium to form $K N O_{2} a n d K N O_{3}$
$2 N O_{2} + 2 K O H \to K N O_{2} + K N O_{3} + H_{2} O$

${C l O}_{4}^{-}$ ion does not undergo decomposition in alkaline medium.
Hence, options $A a n d C$ are correct.

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