In the following reaction correct change in phosphorus is explained by , `4P+3KOH+3H_2Orarr3KH_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 determine the correct change in phosphorus in the reaction \(4P + 3KOH + 3H_2O \rightarrow 3KH_2PO_2 + PH_3\), we need to analyze the oxidation states of phosphorus in the reactants and products.

### Step-by-Step Solution:

1. **Identify the oxidation state of phosphorus in the reactants:**
- In elemental form, phosphorus (P) has an oxidation state of 0. Therefore, for the reactant \(4P\), the oxidation state is 0.

2. **Determine the oxidation state of phosphorus in the products:**
- In \(KH_2PO_2\):
- The oxidation state of potassium (K) is +1.
- Each hydrogen (H) is +1, and there are 2 H atoms, contributing +2.
- Oxygen (O) has an oxidation state of -2, and there are 2 O atoms, contributing -4.
- Let the oxidation state of phosphorus in \(KH_2PO_2\) be \(x\).
- The sum of oxidation states in \(KH_2PO_2\) must equal 0:
\[ +1 + 2 + x - 4 = 0 \implies x - 1 = 0 \implies x = +1 \]
- Therefore, the oxidation state of phosphorus in \(KH_2PO_2\) is +1.

- In \(PH_3\):
- The oxidation state of hydrogen is +1, and there are 3 H atoms, contributing +3.
- Let the oxidation state of phosphorus in \(PH_3\) be \(y\).
- The sum of oxidation states in \(PH_3\) must equal 0:
\[ y + 3 = 0 \implies y = -3 \]
- Therefore, the oxidation state of phosphorus in \(PH_3\) is -3.

3. **Compare the changes in oxidation states:**
- Phosphorus changes from 0 in the reactants to +1 in \(KH_2PO_2\) (oxidation).
- Phosphorus changes from 0 in the reactants to -3 in \(PH_3\) (reduction).

4. **Conclusion:**
- Since phosphorus undergoes both oxidation (from 0 to +1) and reduction (from 0 to -3), it is both oxidized and reduced in the reaction.

### Final Answer:
The correct change in phosphorus is that it is both oxidized and reduced.