`XeF_(2)` is isostructural with

Options

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

Explanation

To determine which compound is isostructural with \( \text{XeF}_2 \), we need to analyze the molecular structure of \( \text{XeF}_2 \) and compare it with the given options: \( \text{TeF}_2 \), \( \text{ICl}_2^- \), \( \text{SbCl}_3 \), and \( \text{BaCl}_2 \).

### Step-by-Step Solution:

1. **Calculate the Total Valence Electrons for \( \text{XeF}_2 \)**:
- Xenon (Xe) has 8 valence electrons.
- Each fluorine (F) has 7 valence electrons, and there are 2 fluorine atoms.
- Total valence electrons = \( 8 + (7 \times 2) = 8 + 14 = 22 \).

2. **Determine Bond Pairs and Lone Pairs**:
- To find the number of bond pairs and lone pairs, we can use the following method:
- Divide the total valence electrons by 8 to find the number of bond pairs.
- The remainder, when divided by 2, gives the number of lone pairs.
- Calculation:
- \( 22 \div 8 = 2 \) (which means 2 bond pairs, since \( 8 \times 2 = 16 \)).
- Remainder = \( 22 - 16 = 6 \).
- \( 6 \div 2 = 3 \) (which means 3 lone pairs).
- Therefore, \( \text{XeF}_2 \) has **2 bond pairs** and **3 lone pairs**.

3. **Analyze the Given Options**:
- **Option 1: \( \text{TeF}_2 \)**:
- Tellurium (Te) has 6 valence electrons.
- Total valence electrons = \( 6 + (7 \times 2) = 6 + 14 = 20 \).
- \( 20 \div 8 = 2 \) (2 bond pairs).
- Remainder = \( 20 - 16 = 4 \).
- \( 4 \div 2 = 2 \) (2 lone pairs).
- **Result**: \( \text{TeF}_2 \) has 2 bond pairs and 2 lone pairs (not isostructural).

- **Option 2: \( \text{ICl}_2^- \)**:
- Iodine (I) has 7 valence electrons, and each chlorine (Cl) has 7 valence electrons.
- With the negative charge, total valence electrons = \( 7 + (7 \times 2) + 1 = 7 + 14 + 1 = 22 \).
- \( 22 \div 8 = 2 \) (2 bond pairs).
- Remainder = \( 22 - 16 = 6 \).
- \( 6 \div 2 = 3 \) (3 lone pairs).
- **Result**: \( \text{ICl}_2^- \) has 2 bond pairs and 3 lone pairs (isostructural).

- **Option 3: \( \text{SbCl}_3 \)**:
- Antimony (Sb) has 5 valence electrons.
- Total valence electrons = \( 5 + (7 \times 3) = 5 + 21 = 26 \).
- \( 26 \div 8 = 3 \) (3 bond pairs).
- Remainder = \( 26 - 24 = 2 \).
- \( 2 \div 2 = 1 \) (1 lone pair).
- **Result**: \( \text{SbCl}_3 \) has 3 bond pairs and 1 lone pair (not isostructural).

- **Option 4: \( \text{BaCl}_2 \)**:
- Barium (Ba) has 2 valence electrons.
- Total valence electrons = \( 2 + (7 \times 2) = 2 + 14 = 16 \).
- \( 16 \div 8 = 2 \) (2 bond pairs).
- Remainder = \( 16 - 16 = 0 \).
- \( 0 \div 2 = 0 \) (0 lone pairs).
- **Result**: \( \text{BaCl}_2 \) has 2 bond pairs and 0 lone pairs (not isostructural).

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

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