The correct geometry and hybridization for `XeF_4` are

Options

A. Square planar, `sp^3 d^2`
B. Octahedral, `sp^3 d^2`
C. Trigonal bipyramidal, `sp^3 d`
D. Planar triangle, `sp^3 d^3`

Explanation

To determine the correct geometry and hybridization for \(XeF_4\), we can follow these steps:

### Step 1: Determine the Valence Electrons
- Xenon (Xe) is in Group 18 of the periodic table and has 8 valence electrons.
- Each fluorine (F) atom has 7 valence electrons, and since there are 4 fluorine atoms, they contribute \(4 \times 7 = 28\) valence electrons.
- However, we only need to consider the valence electrons of xenon for hybridization and geometry.

**Hint:** Count the valence electrons of the central atom (Xe) and the surrounding atoms (F) to understand the bonding situation.

### Step 2: Bonding and Lone Pairs
- In \(XeF_4\), xenon forms 4 bonds with the 4 fluorine atoms.
- This uses up 4 of the 8 valence electrons from xenon.
- The remaining 4 valence electrons will form 2 lone pairs on the xenon atom.

**Hint:** Remember that each bond uses 2 electrons, and lone pairs are the remaining electrons after bonding.

### Step 3: Calculate the Steric Number
- The steric number is calculated by adding the number of sigma bonds and lone pairs on the central atom.
- In \(XeF_4\), there are 4 sigma bonds (one for each \(Xe-F\) bond) and 2 lone pairs.
- Therefore, the steric number is \(4 + 2 = 6\).

**Hint:** The steric number helps determine the hybridization and geometry of the molecule.

### Step 4: Determine Hybridization
- A steric number of 6 corresponds to \(sp^3d^2\) hybridization.
- This involves the mixing of one s orbital, three p orbitals, and two d orbitals.

**Hint:** Hybridization can be deduced from the steric number, where each number corresponds to a specific type of hybridization.

### Step 5: Determine the Geometry
- The geometry associated with \(sp^3d^2\) hybridization is octahedral.
- However, due to the presence of lone pairs, the actual shape of the molecule is square planar.

**Hint:** While hybridization gives the geometry, the presence of lone pairs can alter the molecular shape.

### Final Answer
- The correct geometry of \(XeF_4\) is square planar, and the hybridization is \(sp^3d^2\).

### Summary
- **Hybridization:** \(sp^3d^2\)
- **Geometry:** Square planar (with an underlying octahedral arrangement due to hybridization).