Question

In the disproportionation reaction:
$H_{3}P{O}_2\to P{H}_3+H_{3}P{O}_3$
The $n-$factor of $H_{3}P{O}_2$ is:

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

Answer 1

The correct option is A $\frac{4}{3}$

$H_3\stackrel{+1}{P}{O}_2\to \stackrel{-3}{P}{H}_3+H_3\stackrel{+3}{P}{O}_3$
$H_3\stackrel{+1}{P}{O}_2\to H_3\stackrel{+3}{P}{O}_{3}oxidation$
$(n_f{)}_{\text{oxidation}}=|3-1|\times 1=2$
$H_3\stackrel{+1}{P}{O}_2\to \stackrel{-3}{P}{H}_{3}reduction$
$(n_f{)}_{\text{reduction}}=|-3-1|\times 1=4$
Valency factor in disproportionation reaction is given by
$\frac{1}{n_f}=\frac{1}{(n_f{)}_{\text{oxidation}}}+\frac{1}{(n_f{)}_{\text{reduction}}}$
$\frac{1}{n_f}=\frac{1}{2}+\frac{1}{4}$
$n_f=\frac{4}{3}$