Hybridisation of Xenon in `XeOF_4`

Options

A. `sp^3`
B. `sp^3d`
C. `sp^3d^2`
D. `sp^3d^3`

Answer

To determine the hybridization of xenon in \( \text{XeOF}_4 \), 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: Identify the Bonds Formed
In \( \text{XeOF}_4 \):
- Xenon forms 4 bonds with fluorine atoms (F).
- Xenon forms a double bond with oxygen (O).

### Step 3: Count the Electrons Used in Bonding
- Each bond with fluorine uses 1 electron, totaling 4 electrons for the 4 \( \text{Xe-F} \) bonds.
- The double bond with oxygen consists of 2 electrons (1 sigma and 1 pi bond).

Total electrons used in bonding:
\[ 4 \text{ (from Xe-F bonds)} + 2 \text{ (from Xe=O double bond)} = 6 \text{ electrons} \]

### Step 4: Calculate Remaining Electrons
Starting with 8 valence electrons and using 6 for bonding:
\[ 8 - 6 = 2 \text{ electrons} \]
These remaining 2 electrons will form a lone pair on xenon.

### Step 5: Determine the Steric Number
The steric number (SN) is calculated as follows:
- Count the number of bond pairs (sigma bonds).
- Count the number of lone pairs.

In \( \text{XeOF}_4 \):
- There are 5 bond pairs (4 from \( \text{Xe-F} \) bonds and 1 from the \( \text{Xe=O} \) bond).
- There is 1 lone pair.

Thus, the steric number is:
\[ \text{Steric Number} = \text{Number of Bond Pairs} + \text{Number of Lone Pairs} = 5 + 1 = 6 \]

### Step 6: Determine Hybridization
For a steric number of 6, the hybridization is \( \text{sp}^3\text{d}^2 \).

### Conclusion
The hybridization of xenon in \( \text{XeOF}_4 \) is \( \text{sp}^3\text{d}^2 \).

---