Question

Equivalent weight of $H_{3}P{O}_2$ 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 D $\frac{3M}{4}$

Equivalent weight is defined as:
Molecular weight/ n-factor
$H_{3}P{O}_2\to P{H}_3+H_{3}P{O}_3$ (not balanced)
in $H_{3}P{O}_2\to P{H}_3$
Oxidation state of P decreases from +1 to -3 i.e. a loss of 4$e^{-}$ takes place.

$2H_{3}P{O}_2\to 2H_{3}P{O}_3$
Oxidation state of P increases from +1 to +3 i.e 2$e^{-}$ for a molecules, so 4$e^{-}$ for 2 molecules
n-factor is defined as the number of elelctrons transferred/number of moles
which will be 4/3
hence equivalent weight will be `M/(4/3)`
So answer is $\frac{3M}{4}$