Find the equivalent weight of ${{\text{P}}_{\text{4}}}$ in the following reaction:
\[{{\text{P}}_{\text{4}}} \to {\text{P}}{{\text{H}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}\]
A.40
B.$20.66$
C.60
D.31

Hint: The above reaction is a case of disproportionation reaction in which an element in a compound or in Free State gets both oxidized as well as reduced in the product. The equivalent weight of ${{\text{P}}_{\text{4}}}$ will be its molecular weight divided by its change in oxidation state.

Complete step by step solution:
The oxidation state of phosphorus in ${{\text{P}}_{\text{4}}}$ is zero since any element in its free elemental state has zero oxidation number. In phosphine the oxidation number of phosphorus is $ - 3$ to balance the three monovalent atoms of hydrogen. In phosphorous acid, the oxidation state of phosphorus is $ + 3$.
Hence, the phosphorus in the elemental state in ${{\text{P}}_{\text{4}}}$ converts to phosphine by accepting electrons while in phosphorous acid, it loses electrons to convert to the $ + 3$ oxidation state. Hence, the total change in the oxidation state is 3.
Ignoring the sign, change in oxidation state of ${{\text{P}}_{\text{4}}}$=$\dfrac{{3 \times 4}}{2} = 6$. So the change in oxidation state is 6 units.
Now, the equivalent weight = $\dfrac{{{\text{Molecular Weight}}}}{6}$ = $\dfrac{{31}}{6} = 20.66$.
Hence, the correct answer is $20.66$.

Notes: The oxidation state of any element can be defined as the number of electrons that can be accepted or given out by the element in the process of formation of a compound. The oxidation state of an element can be either positive or negative or even fractional, unlike the valency of the element which is always positive. The oxidation state of any element in its free state is always zero.