In the reaction
`P_4 + 3 KOH + 3H_2 O to PH_3 + 3KH_2 PO_2`

Options

A. `P` is reduced only
B. `P` is oxidised only
C. oxidation state of `P` is 1
D. `P` is both oxidized and reduced

Explanation

To analyze the reaction \( P_4 + 3 KOH + 3 H_2 O \rightarrow PH_3 + 3 KH_2 PO_2 \) and determine the changes in oxidation states of phosphorus, we can follow these steps:

### Step 1: Identify the oxidation states of phosphorus in the reactants and products.
- In \( P_4 \) (elemental phosphorus), the oxidation state of phosphorus is **0**.
- In \( PH_3 \) (phosphine), we can calculate the oxidation state of phosphorus:
\[ X + 3(-1) = 0 \implies X - 3 = 0 \implies X = +3 \]
Therefore, the oxidation state of phosphorus in \( PH_3 \) is **-3**.

- In \( KH_2 PO_2 \) (potassium hypophosphite), we calculate the oxidation state of phosphorus:
\[ 1 + 2(-1) + X + 2(-2) = 0 \implies 1 - 2 + X - 4 = 0 \implies X - 5 = 0 \implies X = +5 \]
Therefore, the oxidation state of phosphorus in \( KH_2 PO_2 \) is **+1**.

### Step 2: Determine the changes in oxidation state.
- From \( P_4 \) (0) to \( PH_3 \) (-3): This indicates a **reduction** (decrease in oxidation state).
- From \( P_4 \) (0) to \( KH_2 PO_2 \) (+1): This indicates an **oxidation** (increase in oxidation state).

### Step 3: Conclude the overall changes.
Since phosphorus undergoes both reduction (in \( PH_3 \)) and oxidation (in \( KH_2 PO_2 \)), we conclude that phosphorus is both oxidized and reduced in this reaction.

### Final Answer:
The correct option is that phosphorus is both oxidized and reduced (Option 4).

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