The hybridization state of the central atom in `AlI_3` is

Options

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

Explanation

To determine the hybridization state of the central atom in aluminum triiodide (AlI₃), we can follow these steps:

### Step 1: Determine the Valence Electrons
- **Aluminum (Al)** has an outer electronic configuration of 3s² 3p¹, which gives it a total of **3 valence electrons**.
- **Iodine (I)** has an outer electronic configuration of 5s² 5p⁵, which gives it a total of **7 valence electrons**.

### Step 2: Bond Formation
- Aluminum will bond with three iodine atoms. Each iodine atom needs one additional electron to complete its octet.
- Therefore, aluminum will share its three valence electrons with three iodine atoms, forming three Al-I bonds.

### Step 3: Calculate the Steric Number
- The steric number is calculated using the formula:
\[ \text{Steric Number} = \text{Number of Bonds} + \text{Number of Lone Pairs on the Central Atom} \]
- In AlI₃, there are **3 bonds** (one with each iodine) and **0 lone pairs** on aluminum since all its valence electrons are used in bonding.
- Thus, the steric number is:
\[ \text{Steric Number} = 3 + 0 = 3 \]

### Step 4: Determine Hybridization
- The hybridization can be determined based on the steric number:
- A steric number of **3** corresponds to **sp² hybridization**.

### Step 5: Determine the Molecular Geometry
- With sp² hybridization, the molecular geometry of AlI₃ is **trigonal planar**. This means that the three iodine atoms are arranged around the aluminum atom in a flat, triangular shape.

### Conclusion
- Therefore, the hybridization state of the central atom (aluminum) in AlI₃ is **sp²**.

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