`XeF_(2)` is isostructural with

Options

A. `SbCl_(3)`
B. `BaCl_(2)`
C. `TeF_(2)`
D. `ICl_(2)^(-)`

Explanation

To determine which molecule is isostructural with \( \text{XeF}_2 \), we need to analyze the structure and hybridization of \( \text{XeF}_2 \) and compare it with the given options.

### Step-by-Step Solution:

1. **Identify the Structure of \( \text{XeF}_2 \)**:
- \( \text{XeF}_2 \) consists of xenon (Xe) and two fluorine (F) atoms.
- Xenon has 8 valence electrons, and each fluorine atom contributes 1 electron. Thus, the total number of valence electrons is \( 8 + 2 = 10 \).

2. **Determine the Hybridization**:
- To find the hybridization, we use the formula:
\[ \text{Number of hybrid orbitals} = \frac{\text{Total valence electrons}}{2} = \frac{10}{2} = 5 \]
- This indicates that the hybridization of \( \text{XeF}_2 \) is \( \text{sp}^3\text{d} \).

3. **Geometry of \( \text{XeF}_2 \)**:
- With \( \text{sp}^3\text{d} \) hybridization and 3 lone pairs (since 2 fluorine atoms are bonded and 3 orbitals are occupied by lone pairs), the geometry is trigonal bipyramidal.
- The molecular shape, considering the lone pairs, is linear.

4. **Identify the Isostructural Molecule**:
- We need to find a molecule that also has \( \text{sp}^3\text{d} \) hybridization and a linear shape.
- The options provided are \( \text{SbCl}_3 \), \( \text{BaCl}_2 \), \( \text{TeF}_2 \), and \( \text{ICl}_2^- \).

5. **Analyze Each Option**:
- **\( \text{SbCl}_3 \)**: Hybridization is \( \text{sp}^3 \) (trigonal pyramidal shape).
- **\( \text{BaCl}_2 \)**: Hybridization is \( \text{s}^2\text{p}^2 \) (linear shape, but not \( \text{sp}^3\text{d} \)).
- **\( \text{TeF}_2 \)**: Hybridization is \( \text{sp}^3 \) (bent shape).
- **\( \text{ICl}_2^- \)**: Hybridization is \( \text{sp}^3\text{d} \) with a linear shape due to 3 lone pairs.

6. **Conclusion**:
- The molecule that is isostructural with \( \text{XeF}_2 \) is \( \text{ICl}_2^- \).

### Final Answer:
\( \text{ICl}_2^- \) is isostructural with \( \text{XeF}_2 \).