The hybridized state of `Al^(3+)` in the complex ion formed when `AlCl_3` is treated with aqueous acid is

Options

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

Explanation

To determine the hybridized state of \( \text{Al}^{3+} \) in the complex ion formed when \( \text{AlCl}_3 \) is treated with aqueous acid, we can follow these steps:

### Step 1: Identify the Complex Formed
When \( \text{AlCl}_3 \) is treated with aqueous acid, it forms a complex ion. The common complex formed is \( \text{[Al(H}_2\text{O)}_6]^{3+} \).

### Step 2: Determine the Coordination Number
In the complex \( \text{[Al(H}_2\text{O)}_6]^{3+} \), aluminum is surrounded by six water molecules. This indicates that the coordination number of aluminum in this complex is 6.

### Step 3: Identify the Geometry
With a coordination number of 6, the geometry of the complex is octahedral.

### Step 4: Determine the Hybridization
For an octahedral complex, the hybridization can be determined as follows:
- The central atom (in this case, \( \text{Al}^{3+} \)) undergoes hybridization to accommodate the six ligands.
- The hybridization corresponding to an octahedral geometry is \( \text{d}^2\text{s}^p \) or simply \( \text{sp}^3\text{d}^2 \).

However, since aluminum does not have d-orbitals available in its valence shell (as it is in the third period), it utilizes its s and p orbitals for hybridization. Thus, the hybridization can be simplified to \( \text{sp}^3 \).

### Conclusion
The hybridized state of \( \text{Al}^{3+} \) in the complex ion formed when \( \text{AlCl}_3 \) is treated with aqueous acid is \( \text{sp}^3 \).