The oxidation number of P is `+3` in -

Options

A. `H_(3)PO_(3)`
B. `H_(3)PO_(4)`
C. `HPO_(3)-`
D. `H_(4)P_(2)O_(7)`

Answer

To determine the oxidation number of phosphorus (P) in the compounds provided, we will analyze each compound step by step.

### Step-by-Step Solution:

1. **Identify the Compounds**: The question asks for the oxidation number of phosphorus in various compounds. The first compound we will analyze is H₃PO₃.

2. **Analyze H₃PO₃**:
- Write the structure: H₃PO₃ consists of one phosphorus atom, three hydrogen atoms, and three oxygen atoms.
- Assign oxidation states:
- Hydrogen (H) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.
- Set up the equation based on the total charge:
\[ 3(+1) + x + 3(-2) = 0 \]
This simplifies to:
\[ 3 + x - 6 = 0 \]
\[ x - 3 = 0 \implies x = +3 \]
- Therefore, the oxidation number of phosphorus in H₃PO₃ is **+3**.

3. **Analyze H₃PO₄**:
- For H₃PO₄, the structure consists of one phosphorus atom, three hydrogen atoms, and four oxygen atoms.
- Set up the equation:
\[ 3(+1) + x + 4(-2) = 0 \]
This simplifies to:
\[ 3 + x - 8 = 0 \]
\[ x - 5 = 0 \implies x = +5 \]
- The oxidation number of phosphorus in H₃PO₄ is **+5**.

4. **Analyze HPO₃²⁻**:
- For HPO₃²⁻, we have one hydrogen atom, one phosphorus atom, and three oxygen atoms.
- Set up the equation:
\[ (+1) + x + 3(-2) = -1 \quad (\text{since the overall charge is -1}) \]
This simplifies to:
\[ 1 + x - 6 = -1 \]
\[ x - 5 = -1 \implies x = +4 \]
- The oxidation number of phosphorus in HPO₃²⁻ is **+4**.

5. **Analyze H₄P₂O₇**:
- For H₄P₂O₇, we have four hydrogen atoms, two phosphorus atoms, and seven oxygen atoms.
- Set up the equation:
\[ 4(+1) + 2x + 7(-2) = 0 \]
This simplifies to:
\[ 4 + 2x - 14 = 0 \]
\[ 2x - 10 = 0 \implies x = +5 \]
- The oxidation number of phosphorus in H₄P₂O₇ is **+5**.

6. **Conclusion**: The only compound where phosphorus has an oxidation number of **+3** is **H₃PO₃**.