1. Identify the species formed by $\mathrm{AlCl_3}$ in acidified aqueous solution

   In water, aluminium chloride does not remain as simple $\mathrm{AlCl_3}$ molecules. Since $\mathrm{Al^{3+}}$ is a small, highly charged ion, it becomes strongly hydrated.

   Thus, in aqueous solution, the aluminium ion exists mainly as the hexaaqua complex:

   $\mathrm{[Al(H_2O)_6]^{3+}}$

   Acidification prevents hydrolysis and precipitation of aluminium hydroxide, so this hydrated cation remains stable.

2. Determine the coordination number

   In $\mathrm{[Al(H_2O)_6]^{3+}}$, six water molecules are coordinated to $\mathrm{Al^{3+}}$.

   So, the coordination number is $6$.

3. Relate coordination number to geometry

   A coordination number of $6$ usually corresponds to octahedral geometry.

   Therefore, the ion formed is octahedral.

4. Check the options

   • A: Square planar $\rightarrow$ usually for coordination number $4$; incorrect.
   • B: Octahedral $\rightarrow$ correct for $\mathrm{[Al(H_2O)_6]^{3+}}$.
   • C: Trigonal bipyramidal $\rightarrow$ coordination number $5$; incorrect.
   • D: Tetrahedral $\rightarrow$ coordination number $4$; incorrect.

Hence, the correct answer is B.