When $I^{-}$ is oxidized by $M n O_{4}^{-}$ in alkaline medium, $I^{-}$ is converted into

I O_{3}^{-}

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I_{2}

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I O_{4}^{- 3}

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I O

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Solution

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

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$K M n O_{4}$ acts as an oxidizing agent in alkaline medium. When alkaline $K M n O_{4}$ is treated with KI, iodide ion is oxidized to

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