Question

$4{Fe}^{2+}+\underset{–}{2{NH}_{2}OH}⟶4{Fe}^{3+}+N_{2}O+4H^{+}$
The equivalent weight of the underlined compound by taking its molecular weight as $M$ is :

• A. $E=\frac{M}{2}$
• B. $E=\frac{M}{4}$
• C. $E=\frac{M}{3}$
• D. none of these

Answer 1

The correct option is A $E=\frac{M}{2}$

The oxidation number of N in $N{H}_{2}OH$ is -1.
The oxidation number of N in $N_{2}O$ is +1.
Thus, the oxidation number of S changes by 2.
$\text{Equivalent weight}=\frac{\text{Molecular weight}}{\text{Change in the oxidation number}}$
Hence, $E=\frac{M}{2}$