In the following reaction correct change in phosphorus is explained by , `4P+3KOH+3H_2Orarr3JH_2PO_2+PH_3`

Options

A. P is oxidized as well as reduced
B. P is oxidized only
C. P is reduced only
D. None of these

Answer

To solve the question regarding the change in oxidation state of phosphorus in the reaction:

\[ 4P + 3KOH + 3H_2O \rightarrow 3KH_2PO_2 + PH_3 \]

we will follow these steps:

### Step 1: Determine the oxidation states of phosphorus in the reactants
In the reactants, we have elemental phosphorus (P), which has an oxidation state of 0.

### Step 2: Determine the oxidation states of phosphorus in the products
1. In **PH₃** (phosphine), the oxidation state of phosphorus can be calculated as follows:
- Let the oxidation state of phosphorus be \( x \).
- The oxidation state of hydrogen is +1.
- Therefore, the equation becomes:
\[ x + 3(+1) = 0 \implies x + 3 = 0 \implies x = -3 \]
- So, the oxidation state of phosphorus in PH₃ is -3.

2. In **KH₂PO₂** (potassium dihydrogen phosphite), we can calculate the oxidation state of phosphorus:
- Let the oxidation state of phosphorus be \( y \).
- The oxidation state of potassium (K) is +1, and the oxidation state of hydrogen (H) is +1, and oxygen (O) is -2.
- The equation becomes:
\[ +1 + 2(+1) + y + 2(-2) = 0 \implies 1 + 2 + y - 4 = 0 \implies y - 1 = 0 \implies y = +1 \]
- So, the oxidation state of phosphorus in KH₂PO₂ is +1.

### Step 3: Analyze the changes in oxidation states
- In the reaction:
- Phosphorus changes from 0 in elemental form to -3 in PH₃ (reduction).
- Phosphorus changes from 0 in elemental form to +1 in KH₂PO₂ (oxidation).

### Step 4: Conclusion
The overall change in phosphorus is that it is both oxidized and reduced in the reaction. This is a redox reaction where phosphorus undergoes both oxidation and reduction.

### Final Answer
The correct change in phosphorus is explained by it being oxidized (to +1 in KH₂PO₂) and reduced (to -3 in PH₃).

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