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Valency and Oxidation States

N2H4 reduces ...

Question

$N_{2} H_{4}$ reduces $I O_{3}^{-}$ to $I^{+}$ in acidic medium as :
$N_{2} H_{4} + I O_{3}^{-} + H^{+} ⟶ N_{2} + I^{+} + H_{2} O$ Equivalent mass of $K I O_{3}$ (molar mass = 214 g mol $^{- 1}$ ) is :

71.33

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107.0

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214.0

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53.5

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Solution

The correct option is D $53.5$
Change in oxidation state of I $= 5 - 1 = 4$
so eq. wt. $= 214 / 4 = 53.5$

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