Disproportionation Reaction of White Phosphorus with Sodium Hydroxide

The reaction of white phosphorus ($\text{P}_4$) with boiling aqueous sodium hydroxide ($\text{NaOH}$) in an inert atmosphere is represented by the chemical equation:

$\text{P}_4(\text{s}) + 3\text{NaOH}(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \longrightarrow \text{PH}_3(\text{g}) + 3\text{NaH}_2\text{PO}_2(\text{aq})$

To understand the nature of the reaction, let us determine the change in oxidation states of phosphorus:

1. In white phosphorus ($\text{P}_4$):
   Phosphorus is in its elemental state, so its oxidation state is $0$.

2. In phosphine ($\text{PH}_3$):
   Hydrogen has an oxidation state of $+1$.
   $x + 3(+1) = 0 \implies x = -3$
   The oxidation state of phosphorus decreases from $0$ to $-3$, which corresponds to a reduction process.

3. In sodium hypophosphite ($\text{NaH}_2\text{PO}_2$):
   Sodium has an oxidation state of $+1$, hydrogen is $+1$, and oxygen is $-2$.
   $(+1) + 2(+1) + x + 2(-2) = 0 \implies 3 + x - 4 = 0 \implies x = +1$
   The oxidation state of phosphorus increases from $0$ to $+1$, which corresponds to an oxidation process.

A redox reaction in which the same element simultaneously undergoes both oxidation and reduction from a single initial oxidation state is known as a disproportionation reaction.

Therefore, the given reaction is a disproportionation reaction.

Correct Option: (B)