The oxidation number of phosphorus do not involve oxidation reduction?

Options

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

Explanation

To determine whether the oxidation number of phosphorus involves oxidation-reduction reactions, we can analyze a specific compound containing phosphorus. Let's take the example of the compound \( \text{Ba}_2\text{PH}_4 \) (Barium phosphide).

### Step-by-step Solution:

1. **Identify the Compound**: We will analyze the compound \( \text{Ba}_2\text{PH}_4 \).

2. **Assign Oxidation States**:
- Barium (Ba) typically has an oxidation state of +2.
- Hydrogen (H) typically has an oxidation state of +1.
- Let the oxidation state of phosphorus (P) be \( x \).

3. **Set Up the Equation**:
- The compound is neutral, so the sum of the oxidation states must equal 0.
- The equation can be set up as follows:
\[ 2(+2) + 4(+1) + x = 0 \]

4. **Calculate the Contributions**:
- From barium: \( 2 \times +2 = +4 \)
- From hydrogen: \( 4 \times +1 = +4 \)

5. **Substitute and Solve for \( x \)**:
\[ +4 + +4 + x = 0 \]
\[ 8 + x = 0 \]
\[ x = -8 \]

6. **Conclusion**:
- The oxidation state of phosphorus in \( \text{Ba}_2\text{PH}_4 \) is -8.
- Since the oxidation state of phosphorus does not change in this reaction, it does not involve oxidation-reduction.

### Final Answer:
The oxidation number of phosphorus in the compound \( \text{Ba}_2\text{PH}_4 \) does not involve oxidation-reduction reactions.

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