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

Options

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

Explanation

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

### Step 1: Identify the central atom and its valence electrons
The central atom in \( \text{AlF}_4^- \) is aluminum (Al). Aluminum is in group 13 of the periodic table and has 3 valence electrons.

### Step 2: Count the number of surrounding monovalent atoms
In \( \text{AlF}_4^- \), there are four fluorine (F) atoms surrounding the aluminum atom. Since fluorine is a monovalent atom (it has a valency of 1), we have 4 monovalent atoms.

### Step 3: Consider the charge on the ion
The charge on the \( \text{AlF}_4^- \) ion is -1. In our formula, we will add 1 for the negative charge.

### Step 4: Use the hybridization formula
The formula to calculate the number of hybrid orbitals (x) is:
\[ x = \frac{1}{2} \left( \text{Number of valence electrons of the central atom} + \text{Number of monovalent atoms} - \text{Charge on cation} + \text{Charge on anion} \right) \]

Substituting the values we have:
- Number of valence electrons of Al = 3
- Number of monovalent atoms = 4
- Charge on cation = 0 (since there is no cation)
- Charge on anion = -1 (which we will add as +1 in the formula)

So, we can plug these values into the formula:
\[ x = \frac{1}{2} \left( 3 + 4 - 0 + 1 \right) = \frac{1}{2} \left( 8 \right) = 4 \]

### Step 5: Determine the hybridization type
From the value of \( x \):
- If \( x = 2 \), hybridization is \( sp \)
- If \( x = 3 \), hybridization is \( sp^2 \)
- If \( x = 4 \), hybridization is \( sp^3 \)
- If \( x = 5 \), hybridization is \( sp^3d \)
- If \( x = 6 \), hybridization is \( sp^3d^2 \)

Since we calculated \( x = 4 \), the hybridization of the central atom (Al) in \( \text{AlF}_4^- \) is \( sp^3 \).

### Final Answer
The hybrid orbital of the central atom in \( \text{AlF}_4^- \) is \( sp^3 \).

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