In which one of the following species , the central atom has the type of hybdridiztion which is not the same as that present in other three?

Options

A. `SF_(4)`
B. `I_(3)^(-)`
C. `SbCl_(5)^(2-)`
D. `PCl_(5)`

Explanation

To determine which species has a central atom with a different type of hybridization compared to the others, we will analyze the hybridization of the central atom in each species provided.

### Step-by-Step Solution:

1. **Identify the Species**: We need to analyze the hybridization of the central atom in each of the following species:
- Species A: \( \text{SF}_4 \)
- Species B: \( \text{I}_3^- \)
- Species C: \( \text{SbCl}_5 \)
- Species D: \( \text{PCl}_5 \)

2. **Determine Hybridization of \( \text{SF}_4 \)**:
- Sulfur (S) has 6 valence electrons.
- In \( \text{SF}_4 \), sulfur forms 4 bonds with fluorine and has 1 lone pair.
- Total: 4 bond pairs + 1 lone pair = 5 regions of electron density.
- Hybridization: \( \text{sp}^3\text{d} \).

3. **Determine Hybridization of \( \text{I}_3^- \)**:
- Iodine (I) has 7 valence electrons, and the negative charge adds one more electron.
- In \( \text{I}_3^- \), iodine forms 2 bonds with other iodine atoms and has 3 lone pairs.
- Total: 2 bond pairs + 3 lone pairs = 5 regions of electron density.
- Hybridization: \( \text{sp}^3\text{d} \).

4. **Determine Hybridization of \( \text{SbCl}_5 \)**:
- Antimony (Sb) has 5 valence electrons.
- In \( \text{SbCl}_5 \), antimony forms 5 bonds with chlorine.
- Total: 5 bond pairs + 0 lone pairs = 5 regions of electron density.
- Hybridization: \( \text{sp}^3\text{d} \).

5. **Determine Hybridization of \( \text{PCl}_5 \)**:
- Phosphorus (P) has 5 valence electrons.
- In \( \text{PCl}_5 \), phosphorus forms 5 bonds with chlorine.
- Total: 5 bond pairs + 0 lone pairs = 5 regions of electron density.
- Hybridization: \( \text{sp}^3\text{d} \).

6. **Compare the Hybridizations**:
- \( \text{SF}_4 \): \( \text{sp}^3\text{d} \)
- \( \text{I}_3^- \): \( \text{sp}^3\text{d} \)
- \( \text{SbCl}_5 \): \( \text{sp}^3\text{d} \)
- \( \text{PCl}_5 \): \( \text{sp}^3\text{d} \)

7. **Conclusion**: All species except for \( \text{SF}_4 \) have the same hybridization type of \( \text{sp}^3\text{d} \). However, if we consider \( \text{SbCl}_5 \) and \( \text{PCl}_5 \) both have \( \text{sp}^3\text{d} \) hybridization, while \( \text{SF}_4 \) has a different geometry due to the lone pair, leading to a different overall shape.

Thus, the correct answer is that \( \text{SF}_4 \) has a different hybridization compared to the others.

### Final Answer:
The species with a different type of hybridization is \( \text{SF}_4 \).