Which among the following are isostructural ? .

Options

A. `XeO_(2)F_(2),SF_(4)`
B. `CO_(2),I_(3)^(Θ)`
C. `SO_(3)^(2-),CO_(3)^(2-)`
D. `CIF_(3),XeF_(2)`

Explanation

To determine which pairs of molecules are isostructural, we need to analyze the molecular geometry of each pair. Isostructural molecules have the same shape or geometry. Let's go through the options step by step.

### Step 1: Analyze the first pair (XCO2F2 and SF4)

1. **Calculate the valence electrons for XCO2F2:**
- X has 8 valence electrons (assuming it's a group 14 element).
- Oxygen (2 atoms) contributes: 2 × 6 = 12 electrons.
- Fluorine (2 atoms) contributes: 2 × 7 = 14 electrons.
- Total = 8 + 12 + 14 = 34 valence electrons.

2. **Determine the hybridization:**
- V = 34, so V/8 = 34/8 = 4.25 (approximately 4 bond pairs).
- Remainder = 34 - (4 × 8) = 2 (which corresponds to 1 lone pair).
- Hybridization = sp³d (1 lone pair).
- Geometry = Seesaw.

3. **Calculate the valence electrons for SF4:**
- Sulfur has 6 valence electrons.
- Fluorine (4 atoms) contributes: 4 × 7 = 28 electrons.
- Total = 6 + 28 = 34 valence electrons.

4. **Determine the hybridization:**
- V = 34, so V/8 = 34/8 = 4.25 (approximately 4 bond pairs).
- Remainder = 34 - (4 × 8) = 2 (which corresponds to 1 lone pair).
- Hybridization = sp³d (1 lone pair).
- Geometry = Seesaw.

**Conclusion for Pair 1:** Both XCO2F2 and SF4 have the same geometry (seesaw), thus they are isostructural.

### Step 2: Analyze the second pair (CO2 and I3-)

1. **Calculate the valence electrons for CO2:**
- Carbon has 4 valence electrons.
- Oxygen (2 atoms) contributes: 2 × 6 = 12 electrons.
- Total = 4 + 12 = 16 valence electrons.

2. **Determine the hybridization:**
- V = 16, so V/8 = 16/8 = 2 (2 bond pairs).
- Remainder = 16 - (2 × 8) = 0 (no lone pairs).
- Hybridization = sp.
- Geometry = Linear.

3. **Calculate the valence electrons for I3-:**
- Iodine (3 atoms) contributes: 3 × 7 = 21 electrons.
- Plus 1 for the negative charge = 22 valence electrons.

4. **Determine the hybridization:**
- V = 22, so V/8 = 22/8 = 2.75 (approximately 2 bond pairs).
- Remainder = 22 - (2 × 8) = 6 (which corresponds to 3 lone pairs).
- Hybridization = sp³d (3 lone pairs).
- Geometry = Linear.

**Conclusion for Pair 2:** Both CO2 and I3- are linear, thus they are isostructural.

### Step 3: Analyze the third pair (SO3^2- and CO3^2-)

1. **Calculate the valence electrons for CO3^2-:**
- Carbon has 4 valence electrons.
- Oxygen (3 atoms) contributes: 3 × 6 = 18 electrons.
- Plus 2 for the -2 charge = 24 valence electrons.

2. **Determine the hybridization:**
- V = 24, so V/8 = 24/8 = 3 (3 bond pairs).
- Remainder = 24 - (3 × 8) = 0 (no lone pairs).
- Hybridization = sp².
- Geometry = Trigonal planar.

3. **Calculate the valence electrons for SO3^2-:**
- Sulfur has 6 valence electrons.
- Oxygen (3 atoms) contributes: 3 × 6 = 18 electrons.
- Plus 2 for the -2 charge = 26 valence electrons.

4. **Determine the hybridization:**
- V = 26, so V/8 = 26/8 = 3.25 (approximately 3 bond pairs).
- Remainder = 26 - (3 × 8) = 2 (which corresponds to 1 lone pair).
- Hybridization = sp³ (1 lone pair).
- Geometry = Bent.

**Conclusion for Pair 3:** CO3^2- is trigonal planar while SO3^2- is bent, thus they are not isostructural.

### Step 4: Analyze the fourth pair (ClF3 and XCF2)

1. **Calculate the valence electrons for ClF3:**
- Chlorine has 7 valence electrons.
- Fluorine (3 atoms) contributes: 3 × 7 = 21 electrons.
- Total = 7 + 21 = 28 valence electrons.

2. **Determine the hybridization:**
- V = 28, so V/8 = 28/8 = 3.5 (approximately 3 bond pairs).
- Remainder = 28 - (3 × 8) = 4 (which corresponds to 2 lone pairs).
- Hybridization = sp³d (2 lone pairs).
- Geometry = T-shaped.

3. **Calculate the valence electrons for XCF2:**
- X has 8 valence electrons (assuming it's a group 14 element).
- Fluorine (2 atoms) contributes: 2 × 7 = 14 electrons.
- Total = 8 + 14 = 22 valence electrons.

4. **Determine the hybridization:**
- V = 22, so V/8 = 22/8 = 2.75 (approximately 2 bond pairs).
- Remainder = 22 - (2 × 8) = 6 (which corresponds to 3 lone pairs).
- Hybridization = sp³d (3 lone pairs).
- Geometry = Linear.

**Conclusion for Pair 4:** ClF3 is T-shaped while XCF2 is linear, thus they are not isostructural.

### Final Conclusion:
The pairs that are isostructural are:
- **Option A:** XCO2F2 and SF4
- **Option B:** CO2 and I3-