The oxidation state of central atom in the anion of compound `NaH_(2)PO_(2)` will be . . . . . . . . .

Options

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

Explanation

To determine the oxidation state of the central atom (phosphorus, P) in the anion of the compound \( \text{NaH}_2\text{PO}_2 \), we can follow these steps:

### Step 1: Identify the components of the compound
The compound \( \text{NaH}_2\text{PO}_2 \) consists of sodium (Na), hydrogen (H), phosphorus (P), and oxygen (O). The anion in this case is \( \text{H}_2\text{PO}_2^- \).

### Step 2: Assign known oxidation states
- Sodium (Na) typically has an oxidation state of +1.
- 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 overall charge
Since the anion \( \text{H}_2\text{PO}_2^- \) has an overall charge of -1, we can set up the equation for the oxidation states. Let the oxidation state of phosphorus (P) be \( x \).

The equation based on the oxidation states is:
\[ x + 2(+1) + 2(-2) = -1 \]

### Step 4: Simplify the equation
Substituting the values into the equation:
\[ x + 2 - 4 = -1 \]
This simplifies to:
\[ x - 2 = -1 \]

### Step 5: Solve for \( x \)
Now, we can solve for \( x \):
\[ x = -1 + 2 \]
\[ x = +1 \]

### Conclusion
The oxidation state of the central atom (phosphorus) in the anion \( \text{H}_2\text{PO}_2^- \) is **+1**.

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