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

Options

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

Explanation

To find the oxidation number of phosphorus (P) in the compound \( \text{Ba}(\text{H}_2\text{PO}_2)_2 \), we can follow these steps:

### Step 1: Identify the components of the compound
The compound consists of barium (Ba), hydrogen (H), phosphorus (P), and oxygen (O). We need to determine the oxidation states of each element in the compound.

### Step 2: Assign known oxidation states
- **Barium (Ba)**: As a Group 2 element, barium has an oxidation state of +2.
- **Hydrogen (H)**: Typically has an oxidation state of +1.
- **Oxygen (O)**: Typically has an oxidation state of -2.

### Step 3: Write the equation for the oxidation states
In the compound \( \text{Ba}(\text{H}_2\text{PO}_2)_2 \), we have:
- 1 barium atom contributing +2
- 4 hydrogen atoms (2 from each of the two \( \text{H}_2\text{PO}_2 \) units) contributing \( 4 \times (+1) = +4 \)
- 4 oxygen atoms (2 from each of the two \( \text{H}_2\text{PO}_2 \) units) contributing \( 4 \times (-2) = -8 \)
- Let the oxidation state of phosphorus be \( x \). Since there are 2 phosphorus atoms, their contribution will be \( 2x \).

### Step 4: Set up the equation
The sum of the oxidation states in a neutral compound must equal zero. Therefore, we can set up the following equation:
\[ +2 + 4 + 2x - 8 = 0 \]

### Step 5: Simplify the equation
Combine the constants:
\[ 2x - 2 = 0 \]

### Step 6: Solve for \( x \)
Add 2 to both sides:
\[ 2x = 2 \]
Now divide by 2:
\[ x = 1 \]

### Conclusion
The oxidation number of phosphorus in \( \text{Ba}(\text{H}_2\text{PO}_2)_2 \) is +1.