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

### Step 1: Identify the oxidation states of phosphorus in the reactants
In \( P_4 \), phosphorus is in its elemental form, so the oxidation state is 0.

### Step 2: Identify the oxidation states of phosphorus in the products
1. In \( PH_3 \) (phosphine), the oxidation state of phosphorus can be calculated as follows:
- Let the oxidation state of phosphorus be \( x \).
- The hydrogen has an oxidation state of +1.
- The equation becomes: \( x + 3(+1) = 0 \)
- Solving gives \( x + 3 = 0 \) → \( x = -3 \).

2. In \( NaH_2PO_2 \) (sodium dihydrogen phosphite), we can calculate the oxidation state of phosphorus:
- Let the oxidation state of phosphorus be \( y \).
- Sodium (Na) has an oxidation state of +1, hydrogen (H) has +1, and oxygen (O) has -2.
- The equation becomes: \( +1 + 2(+1) + y + 2(-2) = 0 \)
- Simplifying gives: \( 1 + 2 + y - 4 = 0 \) → \( y - 1 = 0 \) → \( y = +1 \).

### Step 3: Determine the changes in oxidation states
- In \( P_4 \) (0) to \( PH_3 \) (-3): Phosphorus is reduced (gains electrons).
- In \( P_4 \) (0) to \( NaH_2PO_2 \) (+1): Phosphorus is oxidized (loses electrons).

### Step 4: Conclusion about the reaction
Since the same element (phosphorus) undergoes both oxidation and reduction, this reaction is classified as a **disproportionation reaction**.

### Final Answer
The correct classification of the reaction is that it is a **disproportionation reaction**.

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