When $I^{-}$ is oxidised by $M n O_{4}^{-}$ in alkaline medium, $I^{-}$ converts into:

I O_{3}^{-}

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I_{2}

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I O_{4}^{-}

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I O^{-}

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Solution

The correct option is A $I O_{3}^{-}$
When $I^{-}$ is oxidised by $M n O_{4}^{-}$ in alkaline medium, $I^{-}$ converts into $I O_{3}^{-}$ . The reaction occurs as follows.

$2 K M n O_{4} + 2 K O H \to 2 K_{2} M n O_{4} + H_{2} O + [O]$

$2 K_{2} M n O_{4} + 2 H_{2} O \to 2 M n O_{2} + 4 K O H + 2 [O] - ------------------------------------------------- -$

$2 K M n O_{4} + H_{2} O A l k a l i n e - ---- \to 2 M n O_{2} + 2 K O H + 3 [O]$

$K I + 3 [O] \to K I O_{3}$

i.e. $I O_{3}^{-} - i o n$ .

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