In the `XeF_4` molecule, the Xe atom is in the

Options

A. `sp^2` -hybridized state
B. `sp^3` -hybridized state
C. `sp^2d` -hybridized state
D. `sp^3d^2` -hybridized state

Answer

To determine the hybridization state of the xenon atom in the molecule \( \text{XeF}_4 \), we can follow these steps:

### Step 1: Determine the Valence Electrons
Xenon (Xe) is a noble gas and has 8 valence electrons. Each fluorine (F) atom has 7 valence electrons, but in the context of bonding, it will use only 1 electron to form a bond.

### Step 2: Count the Bonding Electrons
In \( \text{XeF}_4 \), xenon forms 4 bonds with 4 fluorine atoms. Therefore, 4 of xenon's valence electrons are used for bonding.

### Step 3: Calculate Remaining Electrons
After forming 4 bonds, xenon has \( 8 - 4 = 4 \) valence electrons left. These remaining electrons will exist as lone pairs.

### Step 4: Determine the Number of Lone Pairs
Since xenon has 4 remaining electrons, it will form 2 lone pairs (each lone pair consists of 2 electrons).

### Step 5: Calculate the Steric Number
The steric number is calculated by adding the number of bond pairs and lone pairs:
- Number of bond pairs = 4 (from the 4 Xe-F bonds)
- Number of lone pairs = 2

So, the steric number = \( 4 + 2 = 6 \).

### Step 6: Identify the Hybridization
A steric number of 6 corresponds to an \( \text{sp}^3\text{d}^2 \) hybridization.

### Step 7: Determine the Geometry and Shape
The geometry for \( \text{sp}^3\text{d}^2 \) hybridization is octahedral. However, due to the presence of 2 lone pairs, the molecular shape is square planar.

### Conclusion
Thus, the hybridization state of the xenon atom in \( \text{XeF}_4 \) is \( \text{sp}^3\text{d}^2 \).

### Final Answer
The xenon atom in \( \text{XeF}_4 \) is in the \( \text{sp}^3\text{d}^2 \) hybridized state.

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