Which of the following is a planar molecule ?

Options

A. `XeO_(2)F_(2)`
B. `XeOF_2`
C. `XeF_(4)`
D. `XeF_(6)`

Explanation

To determine which of the given molecules is planar, we need to analyze the hybridization and molecular geometry of each compound: \( \text{XeO}_2\text{F}_2 \), \( \text{XeOF}_2 \), \( \text{XeF}_4 \), and \( \text{XeF}_6 \).

### Step-by-Step Solution:

1. **Analyze \( \text{XeO}_2\text{F}_2 \)**:
- **Hybridization Calculation**:
\[ H = \frac{1}{2} \left( V + M - C + A \right) \]
Where:
- \( V = 8 \) (valence electrons of Xe)
- \( M = 2 \) (monovalent atoms, F)
- \( C = 0 \) (charge)
- \( A = 0 \) (additional atoms)

\[ H = \frac{1}{2} \left( 8 + 2 - 0 + 0 \right) = \frac{10}{2} = 5 \quad \text{(sp}^3\text{d)} \]
- **Geometry**: With 5 hybrid orbitals and 4 bonded atoms (2 O and 2 F), there is 1 lone pair. The arrangement is a seesaw shape, which is not planar.

2. **Analyze \( \text{XeOF}_2 \)**:
- **Hybridization Calculation**:
\[ H = \frac{1}{2} \left( 8 + 2 - 0 + 0 \right) = \frac{10}{2} = 5 \quad \text{(sp}^3\text{d)} \]
- **Geometry**: There are 3 bonded atoms (1 O and 2 F) and 2 lone pairs. The arrangement is trigonal bipyramidal with 2 lone pairs in equatorial positions, leading to a planar shape.

3. **Analyze \( \text{XeF}_4 \)**:
- **Hybridization Calculation**:
\[ H = \frac{1}{2} \left( 8 + 4 - 0 + 0 \right) = \frac{12}{2} = 6 \quad \text{(sp}^3\text{d}^2) \]
- **Geometry**: There are 4 bonded atoms (4 F) and 2 lone pairs. The arrangement is square planar, which is indeed planar.

4. **Analyze \( \text{XeF}_6 \)**:
- **Hybridization Calculation**:
\[ H = \frac{1}{2} \left( 8 + 6 - 0 + 0 \right) = \frac{14}{2} = 7 \quad \text{(sp}^3\text{d}^3) \]
- **Geometry**: There are 6 bonded atoms (6 F) and 1 lone pair. The arrangement is distorted octahedral due to the lone pair, which makes it non-planar.

### Conclusion:
The planar molecules among the given options are \( \text{XeOF}_2 \) and \( \text{XeF}_4 \).

### Final Answer:
**Planar Molecules**: \( \text{XeOF}_2 \) and \( \text{XeF}_4 \)