What will be the geometry of `XeOF_4` according to VSEPR theory ?

Options

A. Trigonal bidpyramidal
B. Square pyramidal
C. Pentagonal planar
D. Square planer

Answer

To determine the geometry of \( \text{XeOF}_4 \) according to VSEPR theory, we can follow these steps:

### Step 1: Count the Valence Electrons
- **Xenon (Xe)** has 8 valence electrons.
- **Oxygen (O)** has 6 valence electrons.
- Each **Fluorine (F)** has 7 valence electrons, and there are 4 fluorine atoms.

Calculating the total:
\[ \text{Total valence electrons} = 8 (\text{Xe}) + 6 (\text{O}) + 4 \times 7 (\text{F}) = 8 + 6 + 28 = 42 \text{ valence electrons} \]

### Step 2: Determine the Central Atom and Bond Pairs
- The central atom is **Xenon (Xe)**.
- It forms bonds with 4 fluorine atoms and 1 oxygen atom, resulting in 5 bond pairs.

### Step 3: Identify Lone Pairs
- After forming 5 bonds (4 with F and 1 with O), we need to account for the remaining valence electrons.
- Total electrons used in bonding = \( 5 \times 2 = 10 \) (since each bond consists of 2 electrons).
- Remaining electrons = \( 42 - 10 = 32 \) electrons.
- Each lone pair consists of 2 electrons, so we can form \( \frac{32}{2} = 16 \) lone pairs. However, since we already have 5 bond pairs, we need to check how many lone pairs are actually present on the central atom.

### Step 4: Calculate Lone Pairs on Xenon
- **Xenon** has 8 valence electrons initially.
- After using 10 electrons for bonding (5 bond pairs), it has \( 8 - 10 = -2 \), which means it does not have any lone pairs left.

### Step 5: Determine Hybridization
- With 5 bond pairs and 0 lone pairs, the hybridization of the central atom (Xe) can be determined.
- The hybridization corresponding to 5 bond pairs is \( \text{sp}^3\text{d} \).

### Step 6: Determine Geometry
- According to VSEPR theory, the geometry for a molecule with 5 bond pairs and no lone pairs is **Trigonal Bipyramidal**.

### Final Answer
The geometry of \( \text{XeOF}_4 \) according to VSEPR theory is **Trigonal Bipyramidal**.

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