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`

Answer

To find the oxidation state of the central atom (phosphorus) in the anion of the compound NaH₂PO₂, we can follow these steps:

### Step 1: Identify the Anion
The compound NaH₂PO₂ dissociates into Na⁺ and the anion H₂PO₂⁻. We need to focus on the anion H₂PO₂⁻.

### Step 2: Assign Oxidation States
In the anion H₂PO₂⁻:
- The oxidation state of hydrogen (H) is +1.
- The oxidation state of oxygen (O) is -2.

### Step 3: Set Up the Equation
Let the oxidation state of phosphorus (P) be x. The overall charge of the anion H₂PO₂⁻ is -1. Therefore, we can set up the following equation based on the sum of oxidation states:

\[ \text{Total oxidation state} = \text{Oxidation state of H} + \text{Oxidation state of P} + \text{Oxidation state of O} \]

This can be expressed as:

\[ 2(+1) + x + 2(-2) = -1 \]

### Step 4: Simplify the Equation
Now, simplify the equation:

\[ 2 + x - 4 = -1 \]

This simplifies to:

\[ x - 2 = -1 \]

### Step 5: Solve for x
Now, solve for x:

\[ x = -1 + 2 \]
\[ x = +1 \]

### Conclusion
The oxidation state of phosphorus (the central atom) in the anion H₂PO₂⁻ is +1.

Thus, the answer is **+1**.