The shape of `XeF_(4)` molecule is

Options

A. Tetrahedral
B. Trigonal bipyramidal
C. Square pyramidal
D. Pentagonal bipyramidal

Answer

To determine the shape of the `XeF4` molecule, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Central Atom**:
The central atom in `XeF4` is xenon (Xe).

2. **Determine Valence Electrons**:
- Xenon (Xe) has 8 valence electrons.
- Each fluorine (F) atom has 7 valence electrons, and since there are 4 fluorine atoms, we have \(4 \times 7 = 28\) valence electrons from fluorine.
- Total valence electrons = \(8 + 28 = 36\).

3. **Draw the Lewis Structure**:
- Place Xe in the center and surround it with 4 F atoms.
- Form single bonds between Xe and each F atom. Each bond uses 2 electrons, so \(4 \times 2 = 8\) electrons are used for bonding.
- Remaining electrons = \(36 - 8 = 28\) electrons.
- Distribute the remaining electrons to the fluorine atoms to fulfill their octet (each F needs 8 electrons).

4. **Account for Lone Pairs**:
- After bonding, the xenon atom has 4 bond pairs (one for each Xe-F bond) and 2 lone pairs of electrons.
- This gives a total of 6 electron pairs around the xenon atom.

5. **Determine Hybridization**:
- The presence of 6 electron pairs (4 bond pairs + 2 lone pairs) indicates hybridization of `sp^3d^2`.

6. **Determine Molecular Geometry**:
- The electron pair geometry for `sp^3d^2` is octahedral.
- However, because there are 2 lone pairs, the molecular shape is affected.
- The lone pairs will occupy positions opposite each other to minimize repulsion, leading to a square planar arrangement of the fluorine atoms.

7. **Conclusion**:
The shape of the `XeF4` molecule is square planar.

### Final Answer:
The shape of `XeF4` is square planar.