How many lone pairs are there at xenon in `XeOF_(4)` ?

Explanation

To determine the number of lone pairs at xenon in the molecule \( \text{XeOF}_4 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Central Atom**:
- In \( \text{XeOF}_4 \), xenon (Xe) is the central atom.

2. **Determine the Valence Electrons**:
- Xenon is a noble gas and has 8 valence electrons.

3. **Count the Electrons Used in Bonding**:
- The molecule has 4 fluorine (F) atoms and 1 oxygen (O) atom.
- Each fluorine atom forms a single bond with xenon, using 1 electron per bond. Therefore, 4 fluorine atoms use \( 4 \times 1 = 4 \) electrons.
- Oxygen typically forms a double bond with xenon in this case, using 2 electrons.

4. **Calculate Total Electrons Used in Bonds**:
- Total electrons used in bonding = Electrons used for F bonds + Electrons used for O bond
- Total = \( 4 + 2 = 6 \) electrons.

5. **Determine Remaining Electrons**:
- Remaining electrons = Total valence electrons - Electrons used in bonding
- Remaining = \( 8 - 6 = 2 \) electrons.

6. **Identify Lone Pairs**:
- The remaining 2 electrons will be present as lone pairs on the xenon atom.
- Since each lone pair consists of 2 electrons, the number of lone pairs = \( \frac{2}{2} = 1 \).

7. **Final Structure**:
- The final structure of \( \text{XeOF}_4 \) will have 4 single bonds with fluorine, 1 double bond with oxygen, and 1 lone pair on xenon.

### Conclusion:
- Therefore, there is **1 lone pair** at xenon in \( \text{XeOF}_4 \).

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