Anhydrous `AlCl_(3)` is covalent. From the date given below, predict whether it would remain covalent or become ionic in aqueous solution. (Ionisation energy for `Al` is `1537 kJ mol^(-1)`)
`Delta_("hydration") for Al^(3+) = - 4665 kJ mol^(-1)`
`Delta_("hydration") for Cl^(Ө) = - 381 kJ mol^-1`.

Explanation

To determine whether anhydrous \( \text{AlCl}_3 \) remains covalent or becomes ionic in aqueous solution, we need to analyze the provided data regarding ionization energy and enthalpy of hydration.

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Ionization energy of Aluminum (\( \text{Al} \)): \( 1537 \, \text{kJ/mol} \)
- Enthalpy of hydration for \( \text{Al}^{3+} \): \( \Delta H_{\text{hydration}}(\text{Al}^{3+}) = -4665 \, \text{kJ/mol} \)
- Enthalpy of hydration for \( \text{Cl}^- \): \( \Delta H_{\text{hydration}}(\text{Cl}^-) = -381 \, \text{kJ/mol} \)
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