Which complex compound possesses `sp^(3)d^(2)` hybridization

Options

A. `SF_(6)`
B. `BrF_(5)`
C. `PF_(5)`
D. `[CrF_(6)]^(3-)`

Answer

To determine which complex compound possesses `sp^(3)d^(2)` hybridization, we need to analyze the given options based on their steric numbers. The steric number is calculated by adding the number of bond pairs and lone pairs around the central atom.

### Step-by-Step Solution:

1. **Understanding Hybridization and Steric Number**:
- Hybridization is determined by the steric number (SN).
- The hybridization types are as follows:
- SN = 2 → sp
- SN = 3 → sp²
- SN = 4 → sp³
- SN = 5 → sp³d
- SN = 6 → sp³d²

2. **Analyzing Each Compound**:
- **Option A: SF₆**
- Sulfur (S) has 6 valence electrons.
- It forms 6 bonds with fluorine (F).
- There are 0 lone pairs and 6 bond pairs.
- Steric Number = 0 (lone pairs) + 6 (bond pairs) = 6.
- Hybridization = sp³d².

- **Option B: BrF₅**
- Bromine (Br) has 7 valence electrons.
- It forms 5 bonds with fluorine and has 1 lone pair.
- Steric Number = 1 (lone pair) + 5 (bond pairs) = 6.
- Hybridization = sp³d².

- **Option C: PF₅**
- Phosphorus (P) has 5 valence electrons.
- It forms 5 bonds with fluorine and has 0 lone pairs.
- Steric Number = 0 (lone pairs) + 5 (bond pairs) = 5.
- Hybridization = sp³d.

- **Option D: CrF₆³⁻**
- Chromium (Cr) in the +3 oxidation state has 3d³ electronic configuration.
- It forms 6 bonds with fluorine and has 0 lone pairs.
- Steric Number = 0 (lone pairs) + 6 (bond pairs) = 6.
- Hybridization = sp³d².

3. **Conclusion**:
- The compounds that possess `sp^(3)d^(2)` hybridization are:
- SF₆ (Option A)
- BrF₅ (Option B)
- CrF₆³⁻ (Option D)
- PF₅ (Option C) does not possess `sp^(3)d^(2)` hybridization; it has `sp^(3)d` hybridization.

### Final Answer:
The correct options that possess `sp^(3)d^(2)` hybridization are **Option A (SF₆)**, **Option B (BrF₅)**, and **Option D (CrF₆³⁻)**.