In Xe `O_2F_2`, Xe has hybridization

Options

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

Answer

To determine the hybridization of xenon (Xe) in the compound XeO₂F₂, we can follow these steps:

### Step 1: Identify the central atom
The central atom in the molecule XeO₂F₂ is xenon (Xe).

### Step 2: Count the number of bonds
In XeO₂F₂:
- There are two oxygen (O) atoms and two fluorine (F) atoms bonded to xenon.
- Each O atom forms a double bond with Xe, which counts as one sigma bond (σ) and one pi bond (π) for each O. However, for hybridization, we only count the sigma bonds.
- Each F atom forms a single bond with Xe, which counts as one sigma bond each.

So, the total number of sigma bonds is:
- 2 (from O) + 2 (from F) = 4 sigma bonds.

### Step 3: Count the lone pairs
Xenon has 8 valence electrons. In XeO₂F₂:
- 4 electrons are used in bonding (2 for each double bond with O and 1 for each single bond with F).
- This leaves 4 valence electrons, which will form 2 lone pairs.

### Step 4: Calculate the steric number
The steric number (SN) is calculated as:
\[ \text{Steric Number} = \text{Number of Sigma Bonds} + \text{Number of Lone Pairs} \]
\[ \text{Steric Number} = 4 + 2 = 6 \]

### Step 5: Determine the hybridization
Based on the steric number:
- A steric number of 6 corresponds to sp³d² hybridization.

### Conclusion
Therefore, the hybridization of xenon in XeO₂F₂ is **sp³d²**.

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