In which of the following pairs, the hybridisation of central atoms is same, but geometry is not the same?

Options

A. `SO_(3),CO_(3)^(2-)`
B. `SO_(3)^(2-),NH_(3)`
C. `PCl_(5),POCl_(3)`
D. `XeF_(2),Icl_(3)`

Answer

To solve the question, we need to analyze the hybridization and geometry of the central atoms in the given pairs. We will go through each option step by step.

### Step 1: Analyze Option A - SO3 and CO3^2-
1. **SO3 (Sulfur Trioxide)**:
- Valence electrons of Sulfur (S) = 6
- No monovalent atoms attached = 0
- No positive charge = 0
- Negative charge = 0
- Hybridization = (6 + 0 - 0 + 0) / 2 = 3 → sp²
- Geometry = Trigonal planar

2. **CO3^2- (Carbonate Ion)**:
- Valence electrons of Carbon (C) = 4
- No monovalent atoms attached = 0
- No positive charge = 0
- Negative charge = -2
- Hybridization = (4 + 0 - 0 - 2) / 2 = 3 → sp²
- Geometry = Trigonal planar

**Conclusion for Option A**: Both have sp² hybridization and trigonal planar geometry. Not our answer.

### Step 2: Analyze Option B - SO4^2- and NH3
1. **SO4^2- (Sulfate Ion)**:
- Valence electrons of Sulfur (S) = 6
- No monovalent atoms attached = 0
- No positive charge = 0
- Negative charge = -2
- Hybridization = (6 + 0 - 0 - 2) / 2 = 4 → sp³
- Geometry = Tetrahedral

2. **NH3 (Ammonia)**:
- Valence electrons of Nitrogen (N) = 5
- Monovalent atoms attached = 3 (H)
- No positive charge = 0
- No negative charge = 0
- Hybridization = (5 + 3 - 0 - 0) / 2 = 4 → sp³
- Geometry = Trigonal pyramidal

**Conclusion for Option B**: Both have sp³ hybridization but different geometries. This could be our answer.

### Step 3: Analyze Option C - PCl5 and POCl3
1. **PCl5 (Phosphorus Pentachloride)**:
- Valence electrons of Phosphorus (P) = 5
- Monovalent atoms attached = 5 (Cl)
- No positive charge = 0
- No negative charge = 0
- Hybridization = (5 + 5 - 0 - 0) / 2 = 5 → sp³d
- Geometry = Trigonal bipyramidal

2. **POCl3 (Phosphoryl Chloride)**:
- Valence electrons of Phosphorus (P) = 5
- Monovalent atoms attached = 3 (Cl)
- No positive charge = 0
- No negative charge = 0
- Hybridization = (5 + 3 - 0 - 0) / 2 = 4 → sp³
- Geometry = Tetrahedral

**Conclusion for Option C**: Different hybridization and geometry. Not our answer.

### Step 4: Analyze Option D - XeF2 and ICl3
1. **XeF2 (Xenon Difluoride)**:
- Valence electrons of Xenon (Xe) = 8
- Monovalent atoms attached = 2 (F)
- No positive charge = 0
- No negative charge = 0
- Hybridization = (8 + 2 - 0 - 0) / 2 = 5 → sp³d
- Geometry = Linear

2. **ICl3 (Iodine Trichloride)**:
- Valence electrons of Iodine (I) = 7
- Monovalent atoms attached = 3 (Cl)
- No positive charge = 0
- No negative charge = 0
- Hybridization = (7 + 3 - 0 - 0) / 2 = 5 → sp³d
- Geometry = T-shaped

**Conclusion for Option D**: Both have sp³d hybridization but different geometries. This is our answer.

### Final Answer:
The correct option is **D - XeF2 and ICl3** where the hybridization of the central atoms is the same (sp³d) but the geometry is different (linear vs T-shaped).