The hybrid orbital of the central atom in `AlF_(4)^(-)` is :

Options

A. `sp`
B. `sp^(2)`
C. `sp^(3)`
D. `sp^(3)d`

Answer

To determine the hybrid orbital of the central atom in the ion \( \text{AlF}_4^- \), we can follow these steps:

### Step 1: Identify the central atom and its electronic configuration
The central atom in \( \text{AlF}_4^- \) is aluminum (Al). The atomic number of aluminum is 13. Its electronic configuration is:
\[ \text{Al: } 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^1 \]

### Step 2: Determine the number of valence electrons
Aluminum has 3 valence electrons (2 from the 3s subshell and 1 from the 3p subshell). When it forms the \( \text{AlF}_4^- \) ion, it gains an additional electron, giving it a total of 4 valence electrons:
\[ \text{Valence electrons in } \text{AlF}_4^- = 3 + 1 = 4 \]

### Step 3: Determine the hybridization
To form bonds with the four fluorine atoms, we need to consider the hybridization. The 4 valence electrons will be used to form 4 bonds with the fluorine atoms.

1. **Excitation of electrons**: One of the 3s electrons is excited to the empty 3p orbital, resulting in:
- 1 electron in the 3s orbital
- 3 electrons in the 3p orbitals

2. **Formation of bonds**: This results in 4 unpaired electrons (1 from 3s and 3 from 3p) which can form 4 covalent bonds with the 4 fluorine atoms.

### Step 4: Determine the type of hybridization
Since aluminum is forming 4 equivalent bonds, the hybridization can be determined as follows:
- The 3s and 3p orbitals hybridize to form 4 equivalent sp³ hybrid orbitals.

### Conclusion
The hybrid orbital of the central atom in \( \text{AlF}_4^- \) is **sp³**.