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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 is the hybridizations of boron and aluminum in these species?
Solution
In trivalent state, most of compounds being covalent are hydrolyzed in water. For example, the trichlorides on hydrolysis in water form tetrahedral $\left[\mathrm{M}(\mathrm{OH})_{4}\right]$ species; the hybridization state of element $\mathrm{M}$ is $s p^{3}$.
$\mathrm{BCl}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{B}(\mathrm{OH})_{3}+3 \mathrm{HCl}$
$\mathrm{B}(\mathrm{OH})_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow\left[\mathrm{B}(\mathrm{OH})_{4}\right]^{-}+\mathrm{H}^{+}$
Aluminum chloride in acidified aqueous solution forms octahedral $\left[\mathrm{AI}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{+3}$ ion. In this complex ion, the $3 d$-orbitals of AI are involved and the hybridizations state of $\mathrm{AI}$ is $s p^{3} d^{2}$.
$\mathrm{AlCl}_{3}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{HCl}}{\longrightarrow}\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}+3 \mathrm{Cl}^{-}$
