Question

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 1

`{:(Al_((g))^(3+)+ aq rarr Al_((aq)^(3+)) DeltaH_(1) = -4465 "kJ mole"^(-1),,,,),(3Cl_((g))^(Ө) + aq rarr 3Cl_((aq))^(Ө) DeltaH_(2) = 3 xx -381 "kJ mol"^(-1),,,,),(ulbar(Al_((g))^(3) + 3Cl_((g))^(Ө)+aq+toAl_((aq))^(3+)+3Cl_((aq))^(Theta)),,,,):}`
Therefore,
`(DeltaH_("hydration")) for AlCl_(3) = -4665 + 3(-381)`
=`-5808 kJ mol^(-1)`
Hydration energy is more than ionsiation energy of aluminium, so in aqueous solution, it exists in ionic form.
`AlCl_(3) +6H_(2)O rarr [Al(H_(2)O)_(6)]^(3-) + 3Cl^(Ө)`.