Describe the change in hybridisation (if any) of the Al atom in the following reaction.
` AlCl_3 +Cl^(-) to AlCl_(4)^(-)`

Explanation

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

### Step 1: Determine the hybridization of Al in \( \text{AlCl}_3 \)

1. **Identify the valence electrons of Al**: Aluminum (Al) has an atomic number of 13, and its electron configuration is \( 1s^2, 2s^2, 2p^6, 3s^2, 3p^1 \). This means it has 3 valence electrons available for bonding.

2. **Draw the structure of \( \text{AlCl}_3 \)**: In \( \text{AlCl}_3 \), aluminum is bonded to three chlorine atoms. Each chlorine atom forms a single bond with aluminum, using one of its valence electrons.

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