White phosphorous on reaction with `NaOH` gives `PH_(3)` and

Options

A. `Na_(2)HPO_(3)`
B. `NaH_(2)PO_(2)`
C. `NaH_(2)PO_(3)`
D. `Na_(3)PO_(4)`

Answer

To solve the question about the reaction of white phosphorus with sodium hydroxide (NaOH), we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants are white phosphorus (P₄) and sodium hydroxide (NaOH).

2. **Write the Reaction**:
The reaction of white phosphorus with sodium hydroxide in the presence of water can be represented as:
\[ P_4 + NaOH + H_2O \rightarrow PH_3 + NaH_2PO_2 \]
Here, phosphine (PH₃) and sodium hypophosphite (NaH₂PO₂) are the products.

3. **Balancing the Reaction**:
To balance the reaction, we need to ensure that the number of atoms of each element is the same on both sides. The balanced equation is:
\[ P_4 + 4 NaOH + 2 H_2O \rightarrow 2 PH_3 + 2 NaH_2PO_2 \]
This indicates that 1 mole of P₄ reacts with 4 moles of NaOH and 2 moles of water to produce 2 moles of PH₃ and 2 moles of NaH₂PO₂.

4. **Determine the Oxidation States**:
- In P₄, phosphorus has an oxidation state of 0.
- In PH₃, let the oxidation state of phosphorus be \(y\):
\[ y + 3(+1) = 0 \implies y = -3 \]
- In NaH₂PO₂, let the oxidation state of phosphorus be \(x\):
\[ +1 + 2(+1) + x + 2(-2) = 0 \implies x + 1 - 4 = 0 \implies x = +3 \]

5. **Identify the Type of Reaction**:
The reaction involves the same element (phosphorus) being oxidized and reduced, which characterizes it as a disproportionation reaction. In this case, phosphorus goes from an oxidation state of 0 in P₄ to -3 in PH₃ (reduction) and +3 in NaH₂PO₂ (oxidation).

6. **Final Conclusion**:
The products of the reaction are phosphine (PH₃) and sodium hypophosphite (NaH₂PO₂). Therefore, the correct answer to the question is that white phosphorus on reaction with NaOH gives PH₃ and NaH₂PO₂.