When $BCl_3$ is treated with water,it hydrolyses and forms $[B(OH)_4]^-$ only,whereas $AlCl_3$ in acidified aqueous solution forms $[Al(H_2O)_6]^{3+}$ ion. Explain what are the hybridizations of boron and aluminum in these species?

(N/A) In the trivalent state,most compounds are covalent and undergo hydrolysis in water. For example,trichlorides on hydrolysis in water form tetrahedral $[M(OH)_4]^-$ species; the hybridization state of element $M$ is $sp^3$.
$BCl_3 + 3H_2O \rightarrow B(OH)_3 + 3HCl$
$B(OH)_3 + H_2O \rightarrow [B(OH)_4]^- + H^+$
In $[B(OH)_4]^-$,boron is $sp^3$ hybridized.
Aluminum chloride in acidified aqueous solution forms the octahedral $[Al(H_2O)_6]^{3+}$ ion. In this complex ion,the $3d$-orbitals of $Al$ are involved and the hybridization state of $Al$ is $sp^3d^2$.
$AlCl_3 + 6H_2O \xrightarrow{HCl} [Al(H_2O)_6]^{3+} + 3Cl^-$