Question

Equivalent weight of $H_{3}P{O}_2$ (molecular weight = $M$) when it disproportionates into $P{H}_3$ and $H_{3}P{O}_3$ is:

• A. $M$
• B. $\frac{M}{2}$
• C. $\frac{M}{4}$
• D. $\frac{3M}{4}$

Answer 1

The correct option is C $\frac{M}{4}$

$3H_{3}P{O}_2\to 2H_{3}P{O}_3+P{H}_3$
The oxidation number of P changes from +1 to +3 and -3.
Thus, the change in the oxidation number is 4.
Hence, the equivalence mass of $H_{3}P{O}_2$ in the disproportionation reaction is, $\text{Equivalent mass}\frac{\text{Formula mass}}{\text{Change in the oxidation number}}=\frac{M}{4}$.