The non-existent species is

Options

A. `XeF_(5)`
B. `BrF_(5)`
C. `SbF_(5)`
D. `PF_(5)`

Explanation

To determine the non-existent species among the given options: XeF5, BrF5, SbF5, and PF5, we can analyze each compound based on the ability of the central atom to expand its valence shell and form the specified number of bonds.

### Step-by-Step Solution:

1. **Understanding the Valence Shell Expansion:**
- Elements in the p-block can expand their octet by utilizing available d-orbitals. This is crucial for forming compounds with more than four bonds.

2. **Analyzing Each Compound:**
- **BrF5 (Bromine Pentafluoride):**
- Bromine (Br) has the electronic configuration of [Ar] 4s² 4p⁵. It has vacant d-orbitals available for bonding. Thus, Br can form five bonds and BrF5 exists.

- **PF5 (Phosphorus Pentafluoride):**
- Phosphorus (P) has the electronic configuration of [Ne] 3s² 3p³. It can also expand its valence shell by utilizing its 3d orbitals. Therefore, PF5 can exist.

- **SbF5 (Antimony Pentafluoride):**
- Antimony (Sb) has the electronic configuration of [Kr] 5s² 5p³. Similar to Br and P, Sb can utilize its d-orbitals to form five bonds, making SbF5 a valid compound.

- **XeF5 (Xenon Pentafluoride):**
- Xenon (Xe) has the electronic configuration of [Kr] 5s² 5p⁶. While Xe can form compounds like XeF2 and XeF4, it does not have the capacity to form XeF5 because it cannot utilize d-orbitals (as it is a noble gas with a full valence shell). Therefore, XeF5 does not exist.

3. **Conclusion:**
- Among the given options, the non-existent species is **XeF5**.

### Final Answer:
The non-existent species is **XeF5**.

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