Which of the following is non -existent

Options

A. `AlF_(6)^(3-)`
B. `COF_(6)^(3-)`
C. `BF_(6)^(3-)`
D. `SiF_(6)^(2-)`

Explanation

To determine which of the given species is non-existent, we need to analyze the options provided: ALF63-, BF63-, and SIF62-.

### Step-by-Step Solution:

1. **Identify the Elements and Their Electron Configurations:**
- **Aluminum (Al)**: Atomic number 13; electron configuration is 1s² 2s² 2p⁶ 3s² 3p¹.
- **Boron (B)**: Atomic number 5; electron configuration is 1s² 2s² 2p¹.
- **Silicon (Si)**: Atomic number 14; electron configuration is 1s² 2s² 2p⁶ 3s² 3p².
- **Fluorine (F)**: Atomic number 9; electron configuration is 1s² 2s² 2p⁵.

2. **Analyze Each Compound:**
- **ALF63-**: Aluminum can form a stable complex with fluorine, and the presence of 3 fluorine atoms is feasible.
- **SIF62-**: Silicon can also form a stable complex with fluorine, and the presence of 2 fluorine atoms is feasible.
- **BF63-**: Boron typically forms three bonds due to its three valence electrons. To form BF6, boron would need to utilize d-orbitals, which it does not have.

3. **Conclusion:**
- **BF63- is non-existent** because boron lacks d-orbitals necessary to accommodate six fluorine atoms. Boron can only form a maximum of four bonds (as in BF4-), making BF6 impossible.

### Final Answer:
The non-existent species among the options is **BF63-**.