What is the oxidation state of P in `Ba(H_(2)PO_(2))_(2)` ?

Options

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

Explanation

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

### Step 1: Identify the oxidation states of known elements
- **Barium (Ba)**: The oxidation state of barium is +2 because it is an alkaline earth metal.
- **Hydrogen (H)**: The oxidation state of hydrogen is +1.
- **Oxygen (O)**: The oxidation state of oxygen is -2.

### Step 2: Write the formula for the compound
The compound can be written as:
\[ \text{Ba}(\text{H}_2\text{PO}_2)_2 \]
This indicates there are 2 units of \( \text{H}_2\text{PO}_2 \).

### Step 3: Set up the equation for the overall charge
Since the compound is neutral, the sum of the oxidation states must equal zero. Let \( x \) be the oxidation state of phosphorus (P).

### Step 4: Calculate the contributions from each component
- **Barium**: There are 2 barium atoms contributing \( 2 \times (+2) = +4 \).
- **Hydrogen**: Each \( \text{H}_2\text{PO}_2 \) has 2 hydrogen atoms, and there are 2 such units, contributing \( 2 \times 2 \times (+1) = +4 \).
- **Oxygen**: Each \( \text{H}_2\text{PO}_2 \) has 2 oxygen atoms, and there are 2 such units, contributing \( 2 \times 2 \times (-2) = -8 \).
- **Phosphorus**: Each \( \text{H}_2\text{PO}_2 \) has 1 phosphorus atom, and there are 2 such units, contributing \( 2x \).

### Step 5: Write the equation
Now, we can write the equation for the total oxidation states:
\[ (+4) + (+4) + (-8) + (2x) = 0 \]

### Step 6: Simplify the equation
Combining the constants:
\[ 4 + 4 - 8 + 2x = 0 \]
This simplifies to:
\[ 0 + 2x = 0 \]

### Step 7: Solve for \( x \)
From the equation:
\[ 2x = 0 \implies x = +1 \]

### Conclusion
Thus, the oxidation state of phosphorus (P) in \( \text{Ba}(\text{H}_2\text{PO}_2)_2 \) is +1.

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