Statement-1: Aluminium chloride in acidified aqueous solution forms octahedral ${\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}$ ion. 

Statement-2: In ${\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}$ complex ion, the 3d orbitals of Al are involved and the hybridization state of Al is $s{p}^3{d}^2$

"Aluminum chloride in acidified aqueous solution forms octahedral ${\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}$ ion" is correct In Because in aqueous solution, aluminum chloride forms a complex with aluminum as a metal atom and water as a ligand. Six ligands are present. As a result, it is an octahedral complex, and the hybridization involved is [sp3d2], indicating that it is an outer orbital complex.

Both are valid, and Statement 2 explains Assertion correctly. ${\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}$.

So, Aluminum is in the +3 oxidation state in 

${\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}$