When boron trichloride reacts with water, it only forms `[B(OH)_4]^(-)` , whereas aluminium trichloride forms `[Al(H_2O)_6]^(3+)` in acidified aqueous solution. State the hybridisation of boron and aluminium in these species and explain your answer.

Explanation

Boron trichloricfe `(BCl_3)` undergoes hydrolysis to form orthoboric acid `[B(OH)_3]` at first. As atomic size of B is small and its electronegativity is high, `B(OH)_3` polarises `H_2O` molecule by accepting an `OH^(-)` ion thereby forming `[B(OH)_4]^(-)`and releasing a proton.
`BCl_3 + 3H_2O to B(OH)_3+3HCl`
`B(OH)_3+H_2O to [B(OH)_4]^(-)+ H^(+)`
There are no vacant d -orbitals in B as it lies in the second period and has only one s -and three p -orbitals. So, B can have four pairs of electrons in its valence shell, i.e., its maximum coordination number is 4. For this reason, `B(OH)_3` accepts one `OH^(-)` ion to form `[B(OH)_4]^(-)` in which B-atom is `sp^3` -hybridised. On the other hand, `AlCl_3` undergoes hydrolysis in acidic medium to form `[Al(H_2O)_6]^(3+)`
`AlCl_3 + 6H_2O to [Al(H_2O)_6]^(3+)+3Cl^(-)(aq)`
In acidic medium, concentration of `OH^(-)` ions is lower than `H^(+)` ions. Thus, `Al^(3+)` ions coordinate with `H_2O` molecules and not with `OH^(-)` ions. Due to availability of vacant d -orbitals in `Al^(3+)` ions, it can expand its coordination number from 4 to 6. Thus, it can form the complex ion, `[Al(H_2O)_6]^(3+)` in which hybridisation state of Al is `sp^3d^2`.