`P_(4)+NaOH to PH_(3) uarr+NaH_(2)PO_(2)`

Options

A. For disproportionation reaction.
B. For comproportionation reaction.
C. For either intermolecular redox reaction or displacement reaction
D. For either thermal combination redox reaction or thermal decomposition redox reaction.

Answer

To determine the type of reaction represented by the equation \( P_4 + NaOH \rightarrow PH_3 + NaH_2PO_2 \), we will analyze the oxidation states of the elements involved.

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
- Reactants: \( P_4 \) (white phosphorus) and \( NaOH \) (sodium hydroxide).
- Products: \( PH_3 \) (phosphine) and \( NaH_2PO_2 \) (sodium hypophosphite).

2. **Determine Oxidation States**:
- In \( P_4 \), phosphorus (P) is in the elemental state, so its oxidation state is 0.
- In \( NaOH \):
- Sodium (Na) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.
- Hydrogen (H) has an oxidation state of +1.
- In \( PH_3 \):
- Hydrogen (H) is +1, so for three hydrogens, the total is +3.
- Therefore, phosphorus must be -3 to balance the +3 from hydrogen (0 = -3 + 3).
- In \( NaH_2PO_2 \):
- Sodium (Na) is +1, and each hydrogen (H) is +1, contributing +2.
- Oxygen (O) is -2, contributing -4 for two oxygens.
- Let the oxidation state of phosphorus be \( x \).
- The equation becomes: \( 1 + 2 + x - 4 = 0 \) or \( x - 1 = 0 \), giving \( x = +1 \).

3. **Analyze Changes in Oxidation States**:
- Phosphorus in \( P_4 \) changes from 0 to -3 in \( PH_3 \) (reduction).
- Phosphorus in \( P_4 \) changes from 0 to +1 in \( NaH_2PO_2 \) (oxidation).

4. **Identify the Type of Reaction**:
- Since one species (phosphorus) is both oxidized and reduced in the same reaction, this is classified as a **disproportionation reaction**.

### Conclusion:
The reaction \( P_4 + NaOH \rightarrow PH_3 + NaH_2PO_2 \) is a **disproportionation reaction**.

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