The structure of `XeF_2` involves hybrdization of the type :

Options

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

Answer

To determine the type of hybridization in the structure of \( \text{XeF}_2 \), we can follow these steps:

### Step 1: Determine the Valence Electrons
- **Xenon (Xe)** is a noble gas with 8 valence electrons.
- **Fluorine (F)** has 7 valence electrons, but in \( \text{XeF}_2 \), each fluorine atom will form one bond with xenon, using 1 electron from xenon for each bond.

### Step 2: Calculate the Total Electrons Used in Bonding
- Since there are 2 fluorine atoms, xenon will use 2 of its valence electrons to form 2 bond pairs with fluorine.

### Step 3: Calculate Remaining Electrons
- After forming the 2 bonds, xenon will have \( 8 - 2 = 6 \) valence electrons left.
- These remaining 6 electrons will be arranged as lone pairs.

### Step 4: Determine the Number of Lone Pairs
- The 6 remaining electrons will form 3 lone pairs (since each lone pair consists of 2 electrons).

### Step 5: Calculate the Steric Number
- The steric number is calculated as the sum of the number of bond pairs and lone pairs.
- Here, we have 2 bond pairs (from the bonds with fluorine) and 3 lone pairs:
\[ \text{Steric Number} = \text{Number of Bond Pairs} + \text{Number of Lone Pairs} = 2 + 3 = 5 \]

### Step 6: Determine the Hybridization
- A steric number of 5 corresponds to \( \text{sp}^3\text{d} \) hybridization.

### Step 7: Determine the Geometry and Shape
- The geometry of \( \text{XeF}_2 \) is trigonal bipyramidal due to the \( \text{sp}^3\text{d} \) hybridization.
- However, the shape is linear because the three lone pairs occupy the equatorial positions, leaving the two fluorine atoms at the axial positions.

### Conclusion
The hybridization of \( \text{XeF}_2 \) is \( \text{sp}^3\text{d} \).

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