Strong reducing behaviour of `H_(3)PO_(2)` is due to

Options

A. low oxidation state of phosphorus
B. presence of two ___OH groups and one P___H bond
C. presence of one ___OH group and two P___ H bonds
D. high electron gain enthalphy of phosphorus

Explanation

To determine the reason for the strong reducing behavior of \( H_3PO_2 \) (phosphorous acid), we can analyze the structure and the properties of the compound step by step.

### Step 1: Understand the Structure of \( H_3PO_2 \)

1. **Draw the Structure**: The molecular structure of \( H_3PO_2 \) consists of a phosphorus atom (P) bonded to one oxygen atom (O) with a double bond, one hydroxyl group (–OH), and two hydrogen atoms bonded to phosphorus (–P–H).

\[ \text{Structure: } H_2P(=O)(–OH) \]

### Step 2: Analyze the Oxidation State of Phosphorus

2. **Identify the Oxidation State**: In \( H_3PO_2 \), the oxidation state of phosphorus is +1. This low oxidation state is significant because elements in lower oxidation states tend to act as reducing agents.

### Step 3: Evaluate the Functional Groups

3. **Identify Functional Groups**: The presence of one hydroxyl (–OH) group and two phosphine (–P–H) bonds is crucial. The phosphine bonds can easily donate electrons, which contributes to the reducing behavior.

### Step 4: Determine the Reducing Behavior

4. **Understanding Reducing Behavior**: A reducing agent is a substance that can donate electrons or hydrogen ions in a chemical reaction. In \( H_3PO_2 \), the two P–H bonds and one –OH group facilitate the donation of hydrogen, enhancing its reducing capability.

### Step 5: Conclusion

5. **Final Reasoning**: The strong reducing behavior of \( H_3PO_2 \) is primarily due to the presence of one hydroxyl group and two phosphine groups. Therefore, the correct option is that it has one –OH group and two –P–H bonds.

### Final Answer:
The strong reducing behavior of \( H_3PO_2 \) is due to the presence of one hydroxyl group and two phosphine groups.