The geometry of `XeF_6` molecule and the hybridization of Xe atom in the molecule is :

Options

A. Distorted octahedral and `sp^3 d^3`
B. Square planar and `sp^3d^2`
C. Pyramidal and `sp^3`
D. Octahedral and `sp^3 d^3`

Explanation

To determine the geometry of the `XeF_6` molecule and the hybridization of the xenon atom, we can follow these steps:

### Step 1: Determine the Valence Electrons of Xenon
Xenon (Xe) is a noble gas and has 8 valence electrons.

### Step 2: Count the Monovalent Atoms
In `XeF_6`, there are 6 fluorine (F) atoms. Each fluorine atom contributes 1 electron, so the total contribution from the monovalent atoms is 6.

### Step 3: Check for Charges
In this case, there are no positive or negative charges on the molecule. Therefore, both the positive charge (C) and negative charge (A) are 0.

### Step 4: Apply the Hybridization Formula
The formula for hybridization is given by:
\[ \text{Hybridization number} = \frac{1}{2} (V + M + C - A) \]
Substituting the values we have:
- \( V = 8 \) (valence electrons of Xe)
- \( M = 6 \) (monovalent atoms)
- \( C = 0 \) (positive charge)
- \( A = 0 \) (negative charge)

Now, substituting these values into the formula:
\[ \text{Hybridization number} = \frac{1}{2} (8 + 6 + 0 + 0) = \frac{1}{2} (14) = 7 \]

### Step 5: Determine the Hybridization
A hybridization number of 7 corresponds to \( sp^3d^3 \) hybridization.

### Step 6: Determine the Geometry
For a molecule with \( sp^3d^3 \) hybridization, the geometry is typically distorted octahedral. This is because the presence of lone pairs affects the spatial arrangement of the bonded atoms, leading to a distortion from the ideal octahedral shape.

### Conclusion
Thus, the hybridization of the xenon atom in `XeF_6` is \( sp^3d^3 \) and the geometry is distorted octahedral.

### Final Answer
- **Hybridization**: \( sp^3d^3 \)
- **Geometry**: Distorted octahedral

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