In `Ba(H_(2)PO_(2))_(2)` the oxidation number of phosphorous is

Options

A. `+5`
B. `+1`
C. `+3`
D. `+4`

Explanation

To determine the oxidation number of phosphorus 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). The formula indicates that there are two hydrogen phosphite groups (\( \text{H}_2\text{PO}_2 \)) in the compound.

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

### Step 3: Set up 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 \( \text{H}_2\text{PO}_2 \)) contributing \( 4 \times (+1) = +4 \)
- Let the oxidation state of phosphorus be \( p \).
- There are 4 oxygen atoms contributing \( 4 \times (-2) = -8 \)

### Step 4: Write the oxidation state equation
The sum of the oxidation states in a neutral compound must equal zero. Therefore, we can write the equation as follows:

\[ (+2) + (+4) + (2p) + (-8) = 0 \]

### Step 5: Simplify the equation
Combining the constants, we get:

\[ 2 + 4 + 2p - 8 = 0 \]
\[ 2p - 2 = 0 \]

### Step 6: Solve for \( p \)
Now, we can solve for \( p \):

\[ 2p = 2 \implies p = 1 \]

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

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