In which of the following molecules, sigma bonds formed by the overlap of `sp^3 d and P` orbitals are absent ? 

Options

A. `PCl_5`
B. `ClF_4^(-)`
C. `SbCl_5`
D. `HClO_4`

Answer

To determine in which of the given molecules sigma bonds formed by the overlap of `sp^3 d` and `p` orbitals are absent, we will analyze each molecule's hybridization and bonding characteristics.

### Step-by-Step Solution:

1. **Identify the Molecules**: The molecules we need to analyze are PCl5, ClF4^-, SbCl5, and HCl4.

2. **Analyze PCl5**:
- **Central Atom**: Phosphorus (P)
- **Valence Electrons**: P has 5 valence electrons.
- **Monovalent Atoms**: There are 5 Cl atoms (each Cl is monovalent).
- **Hybridization Calculation**:
\[ \text{Hybridization number} = \frac{(5 + 5)}{2} = 5 \quad \text{(sp}^3\text{d)} \]
- **Sigma Bonds**: Sigma bonds are formed between the `sp^3 d` hybrid orbitals of phosphorus and the `p` orbitals of chlorine.
- **Conclusion**: Sigma bonds are present.

3. **Analyze ClF4^-**:
- **Central Atom**: Chlorine (Cl)
- **Valence Electrons**: Cl has 7 valence electrons.
- **Monovalent Atoms**: There are 4 F atoms (each F is monovalent).
- **Charge**: The negative charge adds 1 electron.
- **Hybridization Calculation**:
\[ \text{Hybridization number} = \frac{(7 + 4 + 1)}{2} = 6 \quad \text{(sp}^3\text{d}^2) \]
- **Sigma Bonds**: Sigma bonds are formed between the `sp^3 d^2` hybrid orbitals of chlorine and the `p` orbitals of fluorine.
- **Conclusion**: Sigma bonds are present.

4. **Analyze SbCl5**:
- **Central Atom**: Antimony (Sb)
- **Valence Electrons**: Sb has 5 valence electrons.
- **Monovalent Atoms**: There are 5 Cl atoms.
- **Hybridization Calculation**:
\[ \text{Hybridization number} = \frac{(5 + 5)}{2} = 5 \quad \text{(sp}^3\text{d)} \]
- **Sigma Bonds**: Sigma bonds are formed between the `sp^3 d` hybrid orbitals of antimony and the `p` orbitals of chlorine.
- **Conclusion**: Sigma bonds are present.

5. **Analyze HCl4**:
- **Central Atom**: Chlorine (Cl)
- **Valence Electrons**: Cl has 7 valence electrons.
- **Monovalent Atoms**: There are 4 H atoms (each H is monovalent).
- **Hybridization Calculation**:
\[ \text{Hybridization number} = \frac{(7 + 4)}{2} = 4 \quad \text{(sp}^3\text{d)} \]
- **Sigma Bonds**: Sigma bonds are formed between the `sp^3` hybrid orbitals of chlorine and the `p` orbitals of hydrogen.
- **Conclusion**: Sigma bonds are present.

### Final Conclusion:
The sigma bonds formed by the overlap of `sp^3 d` and `p` orbitals are absent in **ClF4^-** and **HCl4**.