Oxidation number of `P` in `Ba(H_(2)PO_(2))_(2)` is

Options

A. `-1`
B. `+1`
C. `+2`
D. `+3`

Answer

To find the oxidation number of phosphorus (P) in the compound \( Ba(H_2PO_2)_2 \), we can follow these steps:

### Step-by-Step Solution

1. **Identify the oxidation states of known elements**:
- Barium (Ba) is an alkaline earth metal and typically has an oxidation state of +2.
- Hydrogen (H) generally has an oxidation state of +1.
- Oxygen (O) usually has an oxidation state of -2.

2. **Write the formula for the compound**:
- The compound is \( Ba(H_2PO_2)_2 \). This means there are two \( H_2PO_2 \) groups in the formula.

3. **Calculate the total contribution from barium**:
- Since there are two barium ions, the total contribution from barium is:
\[ 2 \times (+2) = +4 \]

4. **Calculate the contribution from hydrogen**:
- Each \( H_2PO_2 \) group contains 2 hydrogen atoms, and since there are 2 groups, the total number of hydrogen atoms is:
\[ 2 \times 2 = 4 \]
- The total contribution from hydrogen is:
\[ 4 \times (+1) = +4 \]

5. **Calculate the contribution from oxygen**:
- Each \( H_2PO_2 \) group contains 2 oxygen atoms, and since there are 2 groups, the total number of oxygen atoms is:
\[ 2 \times 2 = 4 \]
- The total contribution from oxygen is:
\[ 4 \times (-2) = -8 \]

6. **Set up the equation for the oxidation state of phosphorus (P)**:
- Let the oxidation state of phosphorus be \( x \).
- The total charge of the compound must equal zero. Therefore, we can set up the equation:
\[ (+4) + (+4) + (x \text{ from P}) + (-8) = 0 \]

7. **Simplify the equation**:
- Combine the known contributions:
\[ 4 + 4 - 8 + x = 0 \]
\[ 0 + x = 0 \]

8. **Solve for \( x \)**:
- This simplifies to:
\[ x = +1 \]

### Conclusion
The oxidation number of phosphorus (P) in \( Ba(H_2PO_2)_2 \) is **+1**.