In which species the hybrid state of central atom is / are `sp^(3)` d ?

Options

A. `I_(3)^(-)`
B. `SF_(4)`
C. `PF_(5)`
D. `IF_(5)`

Answer

To determine the species in which the hybrid state of the central atom is \( sp^3d \), we need to analyze the hybridization based on the steric number of the central atom in various compounds. The steric number is calculated as the sum of the number of bond pairs and lone pairs around the central atom.

### Step-by-Step Solution:

1. **Understanding Hybridization**:
- Hybridization involves the mixing of atomic orbitals to form new hybrid orbitals. The notation \( sp^3d \) indicates that one s, three p, and one d orbital are involved in hybridization, leading to a total of five hybrid orbitals.

2. **Identifying the Compounds**:
- The question involves analyzing specific compounds to find those with a central atom that exhibits \( sp^3d \) hybridization. The compounds to consider are \( I_3^+ \), \( SF_4 \), \( PF_5 \), and \( IF_5 \).

3. **Analyzing \( I_3^+ \)**:
- Central atom: Iodine (I)
- Valence electrons: 7 (from group 17) - 1 (positive charge) = 6 electrons.
- Bond pairs: 2 (from two I atoms) and lone pairs: 2.
- Steric number = 2 (bond pairs) + 2 (lone pairs) = 4.
- Hybridization: \( sp^3 \) (not \( sp^3d \)).

4. **Analyzing \( SF_4 \)**:
- Central atom: Sulfur (S)
- Valence electrons: 6 (from group 16).
- Bond pairs: 4 (from four F atoms) and lone pairs: 1.
- Steric number = 4 (bond pairs) + 1 (lone pair) = 5.
- Hybridization: \( sp^3d \).

5. **Analyzing \( PF_5 \)**:
- Central atom: Phosphorus (P)
- Valence electrons: 5 (from group 15).
- Bond pairs: 5 (from five F atoms) and lone pairs: 0.
- Steric number = 5 (bond pairs) + 0 (lone pairs) = 5.
- Hybridization: \( sp^3d \).

6. **Analyzing \( IF_5 \)**:
- Central atom: Iodine (I)
- Valence electrons: 7 (from group 17).
- Bond pairs: 5 (from five F atoms) and lone pairs: 1.
- Steric number = 5 (bond pairs) + 1 (lone pair) = 6.
- Hybridization: \( sp^3d^2 \) (not \( sp^3d \)).

### Conclusion:
The species where the central atom has \( sp^3d \) hybridization are:
- \( SF_4 \)
- \( PF_5 \)

Thus, the correct answers are \( SF_4 \) and \( PF_5 \).