White `P` reacts with caustic soda, the products are `PH_(3)` and `NaH_(2)PO_(2)`. This reaction is an example of:

Options

A. Oxidation
B. Reduction
C. Disproportionation
D. Neutralisation

Explanation

To solve the question regarding the reaction of white phosphorus with caustic soda, we can follow these steps:

### Step 1: Identify the Reactants and Products
The reactants in the reaction are white phosphorus (P) and caustic soda (NaOH). The products formed are phosphine (PH₃) and sodium hypophosphite (NaH₂PO₂).

### Step 2: Write the Balanced Chemical Equation
The balanced chemical equation for the reaction can be written as:
\[ P_4 + 4 NaOH + 2 H_2O \rightarrow 2 PH_3 + 2 NaH_2PO_2 \]

### Step 3: Determine the Oxidation States
Next, we need to determine the oxidation states of phosphorus in the reactants and products:
- In white phosphorus (P₄), the oxidation state is 0.
- In phosphine (PH₃), the oxidation state of phosphorus is -3.
- In sodium hypophosphite (NaH₂PO₂), the oxidation state of phosphorus is +1.

### Step 4: Analyze the Changes in Oxidation States
Now, we analyze the changes in oxidation states:
- The phosphorus in P₄ (0) is reduced to PH₃ (-3), indicating a decrease in oxidation state.
- Simultaneously, another phosphorus in P₄ (0) is oxidized to NaH₂PO₂ (+1), indicating an increase in oxidation state.

### Step 5: Conclusion
Since there is both a decrease and an increase in the oxidation states of phosphorus in the same reaction, this reaction is classified as a disproportionation reaction.

### Final Answer
This reaction is an example of a disproportionation reaction.

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