Question

$\underset{–}{P}{Cl}_5⟶P{Cl}_3+{Cl}_2$
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 P in $PC{l}_5$ is 5.
The oxidation number of P in $PC{l}_3$ is 3.
Thus, the oxidation number of P changes by 2.
$\text{Equivalent weight}=\frac{\text{Molecular weight}}{\text{Change in the oxidation number}}$
Hence, $E=\frac{M}{2}$