Which of the following pairs of compounds is isoelectronic and isostructural?

Options

A. `BeCl_(2), XeF_(2)`
B. `TeI_(2), XeF_(2)`
C. `IBr_(2)^(-),XeF_(2)`
D. `IF_(3),XeF_(2)`

Explanation

To determine which pair of compounds is isoelectronic and isostructural, we need to follow these steps:

### Step 1: Understand the Definitions
- **Isoelectronic Species**: These are species that have the same number of total electrons.
- **Isostructural Species**: These are species that have the same molecular structure or geometry.

### Step 2: Analyze Each Option
We have four pairs of compounds to analyze.

#### Option 1: BE, CL2 and XCF2
- **Beryllium (Be)**: 2 valence electrons
- **Chlorine (Cl)**: 7 valence electrons × 2 = 14
- **Total for Cl2**: 2 + 14 = 16 electrons
- **Xenon (Xe)**: 8 valence electrons
- **Chlorine (Cl)**: 7 valence electrons × 2 = 14
- **Total for XCF2**: 8 + 14 = 22 electrons
- **Conclusion**: Not isoelectronic.

#### Option 2: TeI2 and XCF2
- **Tellurium (Te)**: 6 valence electrons
- **Iodine (I)**: 7 valence electrons × 2 = 14
- **Total for TeI2**: 6 + 14 = 20 electrons
- **Total for XCF2**: 22 electrons (as calculated before)
- **Conclusion**: Not isoelectronic.

#### Option 3: IBr2⁻ and XCF2
- **Iodine (I)**: 7 valence electrons
- **Bromine (Br)**: 7 valence electrons × 2 = 14
- **Charge**: -1 (adds 1 electron)
- **Total for IBr2⁻**: 7 + 14 + 1 = 22 electrons
- **Total for XCF2**: 22 electrons (as calculated before)
- **Conclusion**: Isoelectronic.

Now, we check if they are isostructural:
- For IBr2⁻, using the formula:
\[ \text{Steric Number} = \frac{(V + M + \text{Charge})}{2} \]
where V = 7 (I) + 14 (2 Br) + 1 (charge) = 22
\[ \text{Steric Number} = \frac{22}{2} = 11 \text{ (which is incorrect)} \]
Correct calculation:
\[ \text{Steric Number} = \frac{(7 + 14 + 1)}{2} = 11/2 = 5 \]
- **Geometry**: 2 bond pairs and 3 lone pairs → Trigonal bipyramidal.

- For XCF2:
- **Steric Number**: 8 (Xe) + 2 (2 F) = 10
\[ \text{Steric Number} = \frac{(8 + 2)}{2} = 5 \]
- **Geometry**: 2 bond pairs and 3 lone pairs → Trigonal bipyramidal.

Both have the same geometry.

#### Option 4: IF3 and XCF2
- **Iodine (I)**: 7 valence electrons
- **Fluorine (F)**: 7 valence electrons × 3 = 21
- **Total for IF3**: 7 + 21 = 28 electrons
- **Total for XCF2**: 22 electrons
- **Conclusion**: Not isoelectronic.

### Final Conclusion
The only pair that is both isoelectronic and isostructural is **IBr2⁻ and XCF2**.

### Answer
**The correct answer is: IBr2⁻ and XCF2.**

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