In the given reaction: `P_4 + NaOH + H_2O rarr PH_3 + NaH_2PO_2`

Options

A. P is reduced only
B. P undergoes disproportionation reaction
C. P is oxidised only
D. O is reduced

Answer

To analyze the reaction \( P_4 + NaOH + H_2O \rightarrow PH_3 + NaH_2PO_2 \), we need to determine the changes in oxidation states of phosphorus in the reactants and products.

### Step 1: Write down the oxidation states of phosphorus in the reactants and products.
- In \( P_4 \) (elemental phosphorus), the oxidation state of phosphorus is 0.
- In \( PH_3 \) (phosphine), the oxidation state of phosphorus is -3.
- In \( NaH_2PO_2 \) (sodium hypophosphite), we need to determine the oxidation state of phosphorus. The oxidation states of sodium (Na) is +1, hydrogen (H) is +1, and oxygen (O) is -2. The overall charge of the compound is neutral, so we can set up the equation:
\[ +1 + 2(+1) + x + 2(-2) = 0 \]
Simplifying this gives:
\[ 1 + 2 + x - 4 = 0 \implies x - 1 = 0 \implies x = +1 \]
Therefore, the oxidation state of phosphorus in \( NaH_2PO_2 \) is +1.

### Step 2: Identify the changes in oxidation states.
- Phosphorus changes from 0 in \( P_4 \) to -3 in \( PH_3 \) (reduction).
- Phosphorus changes from 0 in \( P_4 \) to +1 in \( NaH_2PO_2 \) (oxidation).

### Step 3: Determine if the reaction is a disproportionation reaction.
A disproportionation reaction is one in which the same element is both oxidized and reduced. Here, phosphorus is being reduced from 0 to -3 in \( PH_3 \) and oxidized from 0 to +1 in \( NaH_2PO_2 \).

### Conclusion:
Since phosphorus undergoes both oxidation and reduction, this reaction is indeed a disproportionation reaction.

The answer to the question is:
**P undergoes disproportionation reaction.**

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