What is the change in hybridization (if any) of the Al atom in the following reaction.
`AlCl_3 + Cl^(-) to AlCl_4^(-)`

Explanation

To determine the change in hybridization of the aluminum (Al) atom in the reaction \( \text{AlCl}_3 + \text{Cl}^- \rightarrow \text{AlCl}_4^- \), we can follow these steps:

### Step 1: Identify the initial hybridization of Al in \( \text{AlCl}_3 \)
1. **Count the valence electrons of Al**: Aluminum is in group 13, so it has 3 valence electrons.
2. **Count the number of monovalent atoms (Cl)**: There are 3 chlorine atoms in \( \text{AlCl}_3 \).
3. **Consider the charge**: There is no charge on \( \text{AlCl}_3 \), so we do not add or subtract anything for charge.
4. **Apply the hybridization formula**:
\[
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