The species that cannot exist is

Options

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

Answer

To determine which species cannot exist among the given options, we need to analyze the electronic configurations and bonding capabilities of each species. Let's break down the analysis step by step.

### Step 1: Analyze SiF6^2-
- Silicon (Si) has the electronic configuration of [Ne] 3s² 3p².
- To form SiF6^2-, silicon would need to accommodate 6 fluorine atoms.
- Silicon can use its 3s and 3p orbitals for bonding, but it does not have enough available orbitals to accommodate 6 fluorine atoms without exceeding its valence shell capacity.
- Therefore, SiF6^2- cannot exist.

### Step 2: Analyze BF6^3-
- Boron (B) has the electronic configuration of [He] 2s² 2p¹.
- To form BF6^3-, boron would need to accommodate 6 fluorine atoms.
- Boron has only 3 valence electrons (2 from 2s and 1 from 2p). It can form a maximum of 4 bonds (due to the empty 2p orbital), but it cannot accommodate 6 fluorine atoms.
- Therefore, BF6^3- cannot exist.

### Step 3: Analyze SF6
- Sulfur (S) has the electronic configuration of [Ne] 3s² 3p⁴.
- Sulfur can utilize its 3s, 3p, and 3d orbitals for bonding.
- It can form 6 bonds with fluorine atoms, as it can promote one of its 3p electrons to the empty 3d orbital.
- Therefore, SF6 can exist.

### Step 4: Analyze AlF6^3-
- Aluminum (Al) has the electronic configuration of [Ne] 3s² 3p¹.
- To form AlF6^3-, aluminum would need to accommodate 6 fluorine atoms.
- Aluminum can use its 3s and 3p orbitals, but it would also need to utilize the empty 3d orbital to accommodate 6 bonds.
- Therefore, AlF6^3- can exist.

### Conclusion
Based on the analysis, the species that cannot exist is **SiF6^2-** and **BF6^3-** due to the inability of silicon and boron to accommodate 6 fluorine atoms based on their electronic configurations.