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`.

Answer

IONIC

Explanation

Total hydration energy of `Al^(3+)` and `3 Cl^(-)` ions of `AlCl_(3). (Delta H_("Hydration")) = ` Hydration energy of `Al^(3+) + 3 xx` Hydration energy of `Cl^(-)`
`=[-4665 + 3 xx (-381) ] kJ mol^(-1) = - 5880 kJ mol^(-1)`
This amount of energy exceeds the energy needed for the isonisation of `Al` to `Al^(3+)` (i.e., `5808 gt 5137`). Beacuse of this `AlCL_(3)` becomes less ionic in aqueous solution. In aqueous solution `AlCl_(3)` exists in ionic form as `[Al (H_(2)O)_(6)]^(3+)` and `3 Cl^(-)`
`AlCl_(3) + 6H_(2)O rarr [Al (H_(2)O)_(6)]^(3+) + 3Cl^(-)`
`AlCl_(3) + aq rarr AlCl_(3) (aq) , Delta H = ?`
`Delta H =` (Energy released during hydration) - (Energy used during hydration)
`(-4665) - (3 xx 381) + 5137 = - 671 kJ mol^(-1)` Thus, formation of ions will take places.