Anhydrous ${\mathrm{AlCl}}_3$ is covalent. From the data given below
Lattice energy $=5137\mathrm{KJ}/\mathrm{mol}.$

$\mathrm{\Delta }H$ hydration for  ${\mathrm{Al}}^{3+}=-4665\mathrm{KJ}/\mathrm{mol}$

$\mathrm{\Delta }H$ hydration for  ${\mathrm{Cl}}^{-}=-381\mathrm{KJ}/\mathrm{mol}$

Identify the correct statement.

${\mathrm{AlCl}}_3$ (Anhydrous) $\to {\mathrm{AlCl}}_3\left(\mathrm{aq}\right)$

$\mathrm{\Delta }H$ of reaction= Energy released during hydration + Energy used during ionization

$\begin{array}{l}\Rightarrow -4665-(3\times 381)+5137\\ =-671\end{array}$

Thus ${\mathrm{AlCl}}_3$ would become ionic in aqueous solution as the formation of ions will take place because

${\mathrm{\Delta H}}_{\text{hydration }}>{\mathrm{\Delta H}}_{\text{ionization }}$

Hence the correct option is c.