Geometry and hybridisation of `Xe` in `XeOF_(4)` molecule is

Options

A. square planar `sp^(3)d^(2)`
B. square pyramidal `sp^(3)d^(2)`
C. tetrahedral `sp^(3)`
D. None of the above

Explanation

To determine the geometry and hybridization of xenon in the molecule \( \text{XeOF}_4 \), we can follow these steps:

### Step 1: Determine the Valence Electrons
Xenon (Xe) has 8 valence electrons. Oxygen (O) contributes 6 valence electrons, and each fluorine (F) atom contributes 7 valence electrons. Since there are four fluorine atoms, the total contribution from fluorine is \( 4 \times 7 = 28 \) valence electrons.

**Total valence electrons in \( \text{XeOF}_4 \)**:
\[ \text{Total} = 8 (\text{Xe}) + 6 (\text{O}) + 28 (4 \times \text{F}) = 42 \text{ valence electrons} \]

### Step 2: Draw the Lewis Structure
In the Lewis structure of \( \text{XeOF}_4 \), xenon is the central atom. It forms:
- One double bond with oxygen (using 4 electrons).
- Four single bonds with the four fluorine atoms (using 8 electrons).

**Total electrons used in bonding**:
\[ 4 (\text{double bond with O}) + 4 \times 2 (\text{single bonds with F}) = 4 + 8 = 12 \text{ electrons} \]

### Step 3: Calculate Remaining Electrons
After forming the bonds, we can calculate the remaining electrons:
\[ \text{Remaining electrons} = 42 - 12 = 30 \text{ electrons} \]
Since each bond uses 2 electrons, we have 30 electrons left.

### Step 4: Determine Lone Pairs
Xenon will have 2 remaining electrons after forming bonds. These will form one lone pair on xenon.

### Step 5: Calculate Hybridization Index
The hybridization index is calculated as follows:
\[ \text{Hybridization Index} = \text{Number of sigma bonds} + \text{Number of lone pairs} \]
- **Number of sigma bonds**:
- 1 from the double bond with oxygen (1 sigma) + 4 from the single bonds with fluorine = 5 sigma bonds.
- **Number of lone pairs**: 1

Thus, the hybridization index is:
\[ \text{Hybridization Index} = 5 + 1 = 6 \]

### Step 6: Determine Hybridization
A hybridization index of 6 corresponds to \( \text{sp}^3\text{d}^2 \).

### Step 7: Determine Geometry
With \( \text{sp}^3\text{d}^2 \) hybridization and one lone pair, the geometry of the molecule is square pyramidal.

### Final Answer
- **Hybridization**: \( \text{sp}^3\text{d}^2 \)
- **Geometry**: Square pyramidal