$I O_{3}^{-} - --- - + 3 H S O_{3}^{-} - ------- - ⟶ I^{-} + 3 H^{+} + 3 S O_{4}^{2 -}$
The equivalent weight of the underlined species by taking its molecular weight as $M$ is :

E (H S O_{3}^{-}) = \frac{M}{2}

E (I O_{3}^{-}) = \frac{M}{6}

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E (H S O_{3}^{-}) = \frac{M}{2}

E (I O_{3}^{-}) = \frac{M}{5}

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E (H S O_{3}^{-}) = \frac{M}{3}

E (I O_{3}^{-}) = \frac{M}{6}

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none of these

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Solution

The correct option is A $E (H S O_{3}^{-}) = \frac{M}{2}$ , $E (I O_{3}^{-}) = \frac{M}{6}$
The oxidation number of S in $H S O_{3}^{-}$ is +4.
The oxidation number of in $S O_{4}^{- 2}$ is +6.
Thus, the oxidation number of S changes by 2.
$Equivalent weight = \frac{Molecular weight}{Change in the oxidation number}$
Hence, $E = \frac{M}{2}$
The oxidation number of I in $I O_{3}^{-}$ is +5.
The oxidation number of I in $I^{-}$ is -1.
Thus, the oxidation number of S changes by 6.
$Equivalent weight = \frac{Molecular weight}{Change in the oxidation number}$
Hence, $E = \frac{M}{6}$

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