Equivalent weight of ${C r_{2} O_{7}}^{2 -}$ (mol. wt. = $M$ ) in the following reaction is :
${C r_{2} O_{7}}^{2 -} + 6 I^{-} + 14 H^{+} \to 2 C r^{3 +} + 3 I_{2} + 7 H_{2} O$

\frac{M}{6}

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\frac{M}{3}

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\frac{M}{2}

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M

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Solution

The correct option is A $\frac{M}{6}$
In acidic medium, the oxidation number of chromium atom decreases from $+ 6$ to $+ 3$ .
The net decrease in the oxidation number per Cr atom is $3$ .
The net decrease in the oxidation number for two Cr atoms is $6$ .
Hence, the oxidation number of ${C r_{2} O_{7}}^{2 -} = \frac{M}{6}$ .

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