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 $I O_{3}^{-}$
$2 K M n O_{4} + 2 K O H \to 2 K_{2} M n O_{4} + H_{2} O + O$
$\frac{2 K_{2} M n O_{4} + 2 H_{2} O \to 2 M n O_{2} + 4 K O H + 2 O}{2 K M n O_{4} + H_{2} O a l k a l m e - ----- \to 2 M n O_{2} + 2 K O H + 3 [O]}$
$\frac{K I + [O] \to K I O_{3}}{2 K M n O_{4} + K I + H_{2} O \to 2 K O H + 2 M n O_{2} + K I O_{3}}$
$K I + 3 [O] \to K I O_{3}^{-}$
Hence option A is correct.

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