Strong reducing behaviour of `H_(3)PO_(4)` is due to :

Options

A. High oxidation state of phosphorus
B. Presence of two`-OH` group and one `P-H` bond
C. Presence of one `-OH` group and two `P-H` bond
D. High electron gain enthalpy of phosphorus

Explanation

To determine the strong reducing behavior of \( H_3PO_4 \) (phosphoric acid), we need to analyze the structure and bonding of the compound. However, it seems there is a correction needed in the question, as the strong reducing behavior is actually associated with \( H_3PO_2 \) (phosphorous acid), not \( H_3PO_4 \).

### Step-by-Step Solution:

1. **Identify the Correct Compound**:
The question mentions \( H_3PO_4 \), but the strong reducing behavior is actually attributed to \( H_3PO_2 \). Therefore, we will focus on \( H_3PO_2 \).

2. **Analyze the Structure of \( H_3PO_2 \)**:
- The structure of \( H_3PO_2 \) includes:
- One phosphorus atom (P)
- One double bond to an oxygen atom (O)
- One hydroxyl group (–OH)
- Two phosphorus-hydrogen (P–H) bonds
- This can be represented as:
\[ H_2P(=O)(–OH)(–H) \]

3. **Identify the Reducing Agents**:
- The presence of P–H bonds in \( H_3PO_2 \) is crucial. The P–H bonds can donate electrons, which is a characteristic of reducing agents.
- The P–H bonds can easily be oxidized, leading to the reduction of other species.

4. **Oxidation State of Phosphorus**:
- In \( H_3PO_2 \), the oxidation state of phosphorus can be calculated as follows:
- Let the oxidation state of phosphorus be \( x \).
- The equation is:
\[ 3(1) + x + 2(-2) = 0 \implies 3 + x - 4 = 0 \implies x = +1 \]
- The oxidation state of phosphorus in \( H_3PO_2 \) is +1, which is relatively low compared to other phosphorus oxoacids.

5. **Conclusion**:
- The strong reducing behavior of \( H_3PO_2 \) is due to the presence of P–H bonds, which can easily donate electrons, making it a strong reducing agent.

### Final Answer:
The strong reducing behavior of \( H_3PO_2 \) is due to the presence of P–H bonds that can donate electrons.