p-Block Elements-I - Result Question 29
####29. Anhydrous $AlCl _3$ is covalent. From the data given below, predict whether it would remain covalent or become ionic in aqueous solution.
(Ionisation energy for $Al=5137 kJ mol^{-1}$
$ \begin{aligned} & \Delta H _{\text {hydration }} \text { for } Al^{3+}=-4665 kJ mol^{-1} \ & \Delta H _{\text {hydration }} \text { for } Cl^{-}=-381 kJ mol^{-1} \end{aligned}) $
(1997, 2M)
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Solution:
- The total hydration energy of $AlCl _3$
$=$ Hydration energy of $Al^{3+}+3 \times$ Hydration energy of $Cl^{-}$
$=-4665+3(-381) kJ / mol$
$=-5808 kJ / mol$
The above hydration energy is more than the energy required for ionisation of $AlCl _3$ into $Al^{3+}$ and $3 Cl^{-}$.
Due to this reason, $AlCl _3$ becomes ionic in aqueous solution. In aqueous solution, it is ionised completely as
$$ AlCl _3+6 H _2 O \longrightarrow\left[Al\left(H _2 O\right) _6\right]^{3+}+3 Cl^{-} $$
