The shape of `XeF_(4)` is

Options

A. tetrahedral
B. square planar
C. pyramidal
D. linear

Explanation

To determine the shape of \( \text{XeF}_4 \), we can follow these steps:

### Step 1: Determine the Valence Electrons
- Xenon (Xe) has 8 valence electrons.
- Each fluorine (F) atom contributes 1 valence electron, and since there are 4 fluorine atoms, they contribute a total of 4 electrons.
- Total valence electrons = 8 (from Xe) + 4 (from 4 F) = 12 valence electrons.

### Step 2: Identify the Hybridization
- With 12 valence electrons, we can determine the hybridization:
- 12 valence electrons correspond to 6 electron pairs (since each pair consists of 2 electrons).
- This indicates \( \text{sp}^3\text{d}^2 \) hybridization.

### Step 3: Determine the Geometry
- The geometry associated with \( \text{sp}^3\text{d}^2 \) hybridization is octahedral.
- In an octahedral arrangement, there are 6 positions for bonding pairs.

### Step 4: Account for Lone Pairs
- In \( \text{XeF}_4 \), there are 4 bonding pairs (from the 4 Xe-F bonds) and 2 lone pairs on the xenon atom.
- The presence of lone pairs affects the shape but not the geometry.

### Step 5: Determine the Shape
- When considering the shape, we only account for the positions of the atoms and not the lone pairs.
- The arrangement of the 4 fluorine atoms, while ignoring the lone pairs, results in a square planar shape.

### Conclusion
- Therefore, the shape of \( \text{XeF}_4 \) is **square planar**.

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