Oxidation number of Xe in `XeF_(5)^(-)` is :

Options

A. `+1`
B. `+2`
C. `+3`
D. `+4`

Answer

To find the oxidation number of xenon (Xe) in the ion \( \text{XeF}_5^{-} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the oxidation states of the elements involved**:
- Fluorine (F) typically has an oxidation state of -1.

2. **Set up the equation based on the overall charge of the ion**:
- The ion \( \text{XeF}_5^{-} \) has a total charge of -1.
- Let the oxidation state of xenon (Xe) be \( x \).

3. **Write the equation for the total charge**:
- The contribution to the charge from the five fluorine atoms is \( 5 \times (-1) = -5 \).
- Therefore, the equation can be set up as:
\[ x + 5(-1) = -1 \]

4. **Simplify the equation**:
- This simplifies to:
\[ x - 5 = -1 \]

5. **Solve for \( x \)**:
- Adding 5 to both sides gives:
\[ x = -1 + 5 \]
- Thus,
\[ x = 4 \]

6. **Conclusion**:
- The oxidation number of xenon in \( \text{XeF}_5^{-} \) is +4.

### Final Answer:
The oxidation number of Xe in \( \text{XeF}_5^{-} \) is +4.

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