Among the following molecules how many are having `sp^(3)`d hybridization of this central atom ? `XeOf_(2), SF_(6), SF_(4), I_(3)^(-) , BrF_(3) , XeF_(2) , POCl_(3) , ClO_(4)^(-), XeF_(4) , PCl_(3) F_(2)`

Answer

To determine how many of the given molecules have `sp^3d` hybridization of the central atom, we will analyze each molecule step by step.

### Step-by-Step Solution:

1. **XeOF₂ (Xenon Oxyfluoride)**:
- Central atom: Xenon (Xe)
- Valence electrons: 8 (Group 18)
- Bonds: 2 (from 2 Fluorine atoms) + 1 (from Oxygen) = 3 bonds
- Lone pairs: 8 (total) - 3 (bonds) = 5 lone pairs
- Hybridization: Number of bond pairs + lone pairs = 3 + 2 = 5 (sp³d)

2. **SF₆ (Sulfur Hexafluoride)**:
- Central atom: Sulfur (S)
- Valence electrons: 6 (Group 16)
- Bonds: 6 (from 6 Fluorine atoms)
- Lone pairs: 6 (total) - 6 (bonds) = 0 lone pairs
- Hybridization: 6 (sp³d²)

3. **SF₄ (Sulfur Tetrafluoride)**:
- Central atom: Sulfur (S)
- Valence electrons: 6
- Bonds: 4 (from 4 Fluorine atoms)
- Lone pairs: 6 (total) - 4 (bonds) = 2 lone pairs
- Hybridization: 4 + 1 = 5 (sp³d)

4. **I₃⁻ (Triiodide Ion)**:
- Central atom: Iodine (I)
- Valence electrons: 7 (Group 17) + 1 (for the negative charge) = 8
- Bonds: 2 (from 2 Iodine atoms)
- Lone pairs: 8 (total) - 2 (bonds) = 6 lone pairs
- Hybridization: 2 + 3 = 5 (sp³d)

5. **BrF₃ (Bromine Trifluoride)**:
- Central atom: Bromine (Br)
- Valence electrons: 7
- Bonds: 3 (from 3 Fluorine atoms)
- Lone pairs: 7 (total) - 3 (bonds) = 4 lone pairs
- Hybridization: 3 + 2 = 5 (sp³d)

6. **XeF₂ (Xenon Difluoride)**:
- Central atom: Xenon (Xe)
- Valence electrons: 8
- Bonds: 2 (from 2 Fluorine atoms)
- Lone pairs: 8 (total) - 2 (bonds) = 6 lone pairs
- Hybridization: 2 + 3 = 5 (sp³d)

7. **POCl₃ (Phosphorus Oxychloride)**:
- Central atom: Phosphorus (P)
- Valence electrons: 5 (Group 15)
- Bonds: 1 (from Oxygen) + 3 (from 3 Chlorine atoms) = 4 bonds
- Lone pairs: 5 (total) - 4 (bonds) = 1 lone pair
- Hybridization: 4 + 1 = 5 (sp³d)

8. **ClO₄⁻ (Chlorate Ion)**:
- Central atom: Chlorine (Cl)
- Valence electrons: 7 + 1 (for the negative charge) = 8
- Bonds: 4 (from 4 Oxygen atoms)
- Lone pairs: 8 (total) - 4 (bonds) = 4 lone pairs
- Hybridization: 4 + 0 = 4 (sp³)

9. **XeF₄ (Xenon Tetrafluoride)**:
- Central atom: Xenon (Xe)
- Valence electrons: 8
- Bonds: 4 (from 4 Fluorine atoms)
- Lone pairs: 8 (total) - 4 (bonds) = 4 lone pairs
- Hybridization: 4 + 2 = 6 (sp³d²)

10. **PCl₃F₂ (Phosphorus Trichloride Difluoride)**:
- Central atom: Phosphorus (P)
- Valence electrons: 5
- Bonds: 3 (from 3 Chlorine atoms) + 2 (from 2 Fluorine atoms) = 5 bonds
- Lone pairs: 5 (total) - 5 (bonds) = 0 lone pairs
- Hybridization: 5 (sp³d)

### Summary of Hybridizations:
- XeOF₂: sp³d
- SF₆: sp³d²
- SF₄: sp³d
- I₃⁻: sp³d
- BrF₃: sp³d
- XeF₂: sp³d
- POCl₃: sp³d
- ClO₄⁻: sp³
- XeF₄: sp³d²
- PCl₃F₂: sp³d

### Conclusion:
The molecules with `sp³d` hybridization are:
1. XeOF₂
2. SF₄
3. I₃⁻
4. BrF₃
5. XeF₂
6. POCl₃
7. PCl₃F₂

**Total Count**: 7 molecules have `sp³d` hybridization.