In which of the following molecules, sigma bonds fomred by the overlap of `SP^(3)d` and P-orbitals.

Options

A. `PCl_(3)`
B. `CIF_(3)`
C. `SbCl_(3)`
D. `HClO_(4)`

Explanation

To determine which of the given molecules has sigma bonds formed by the overlap of sp³d hybrid orbitals and p orbitals, we will analyze the hybridization of the central atom in each molecule.

### Step-by-Step Solution:

1. **Identify the Molecules**: The molecules we need to analyze are PCl₃, ClF₃, SbF₃, and HClO₄.

2. **Determine Hybridization of Each Molecule**:
- **For PCl₃**:
- Phosphorus (P) is in Group 15 and has 5 valence electrons.
- It is bonded to 3 chlorine (Cl) atoms (each contributing 1 electron).
- Hybridization formula: (Valence electrons of P + Monovalent atoms - Charge) / 2
- Calculation: (5 + 3) / 2 = 4 → Hybridization is sp³.
- **Conclusion**: No sp³d hybridization, so this is not the correct answer.

- **For ClF₃**:
- Chlorine (Cl) is in Group 17 and has 7 valence electrons.
- It is bonded to 3 fluorine (F) atoms.
- Calculation: (7 + 3) / 2 = 5 → Hybridization is sp³d.
- **Conclusion**: Cl has sp³d hybridization, and F has p orbitals, so this is a potential candidate.

- **For SbF₃**:
- Antimony (Sb) is in Group 15 and has 5 valence electrons.
- It is bonded to 3 fluorine (F) atoms.
- Calculation: (5 + 3) / 2 = 4 → Hybridization is sp³.
- **Conclusion**: No sp³d hybridization, so this is not the correct answer.

- **For HClO₄**:
- Chlorine (Cl) is the central atom and has 7 valence electrons.
- It is bonded to 4 oxygens (O) and one hydrogen (H).
- Calculation: (7 + 4 + 1) / 2 = 6 → Hybridization is sp³d².
- **Conclusion**: This molecule does not have sp³d hybridization either.

3. **Final Conclusion**: The only molecule with sp³d hybridization where a sigma bond is formed by the overlap of sp³d and p orbitals is **ClF₃**.

### Answer: ClF₃