Reaction Type and Oxidation States in Phosphorus Disproportionation

The reaction of white phosphorus ($\text{P}_4$) with an aqueous solution of sodium hydroxide ($\text{NaOH}$) under an inert atmosphere proceeds according to the balanced equation:

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

Here, phosphine ($\text{PH}_3$) is produced along with sodium hypophosphite ($\text{NaH}_2\text{PO}_2$).

1. Determination of Oxidation States:

1. In elemental white phosphorus ($\text{P}_4$): $\text{Oxidation state of P} = 0$

2. In phosphine ($\text{PH}_3$): Since hydrogen is less electronegative than phosphorus (or taking $\text{H}$ as $+1$): $\text{Oxidation state of P} + 3(+1) = 0 \implies \text{Oxidation state of P} = -3$

3. In sodium hypophosphite ($\text{NaH}_2\text{PO}_2$): Using the standard oxidation states for $\text{Na}$ ($+1$), $\text{H}$ ($+1$), and $\text{O}$ ($-2$): $(+1) + 2(+1) + x + 2(-2) = 0$ $1 + 2 + x - 4 = 0 \implies x = +1$ Thus, the oxidation state of phosphorus in $\text{NaH}_2\text{PO}_2$ is $+1$.

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2. Type of Reaction:

Phosphorus undergoes simultaneous:

• Reduction: from an oxidation state of $0$ in $\text{P}_4$ to $-3$ in $\text{PH}_3$.
• Oxidation: from an oxidation state of $0$ in $\text{P}_4$ to $+1$ in $\text{NaH}_2\text{PO}_2$.

A reaction in which the same element is simultaneously oxidized and reduced is classified as a disproportionation reaction (which is a specific class of redox reactions).

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Conclusion:

The correct description is:

• Reaction type: Disproportionation reaction
• Oxidation states of phosphorus: $-3$ (in $\text{PH}_3$) and $+1$ (in $\text{NaH}_2\text{PO}_2$)

Since none of the options list the correct oxidation state pair of $-3$ and $+1$, there is no correct option among the choices provided (hence, this question was awarded Marks to All / Zero Marks to all).