$K M n O_{4}$ acts as on oxidising agent in alkaline medium. When alkaline $K M n O_{4}$ is treated with KI, iodide ion is oxidised to :

I_{2}

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I O^{-}

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I O_{3}^{-}

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I O_{4}^{-}

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Solution

The correct option is C $I O_{3}^{-}$
$2 K M n O_{4} + H_{2} O + K I \to 2 M n O_{2} + 2 K O H + K I O_{3}$

$I + 6 O H^{-} \to I O_{3}^{-} + 3 H_{2} O + 6 e^{-}$

When alkaline KMnO4 is treated with KI, $I^{-}$ is oxidised to $I {O_{3}}^{-}$ .

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