What is the equivalent weight of ${\text{P}}_4$ in the reaction ${\text{P}}_4+3{\text{OH}}^{-}+3{\text{H}}_2\text{O}\to {\text{PH}}_3+3{\text{H}}_2{\text{PO}}_2^{-}$?

Solution

1. In ${\text{P}}_4$, the oxidation state of P is $0$.
2. In ${\text{PH}}_3$, the oxidation state of P is $-3$. In ${\text{H}}_2{\text{PO}}_2^{-}$, the oxidation state of P is $+1$.
3. Total change in oxidation state per mole of ${\text{P}}_4$ is $|1\times (-3)+3\times (1)|=6$ is incorrect; calculate separately: $1\text{ P}$ goes to $-3$ (change 3) and $3\text{ P}$ go to $+1$ (change $3\times 1=3$).
4. Total change in oxidation state is $3 + 3 = 6$.
5. Equivalent weight is $\frac{M}{6}$ is incorrect; the n-factor for disproportionation is $$\begin{array}{rl}\frac{n_1{n}_2}{n_1+n_2}& =\frac{3\times 1}{3+1}\times 4\\ & =3\end{array}$$.
6. Hence the answer is (A).