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_(2)O)_(6)]^(3+)` ion, Explain what is the hybridisation of boron and aluminium in these species?

Explanation

When `BCl_(3)` reacts with water, it first undergoes hydrolysis to form boric acid, `B(OH)_(3)`. Due to small size and high electronegativity of B, `B(OH)_(3)` polarizes `H_(2)O` molecule accepting an `OH^(-)` ion to form `[B(OH)_(4)]^(-)` species and releasing a proton :
`BCl_(3) + 3H_(2)O rarr B(OH)_(3) + 3HCl : B(OH)_(3) + H_(2)O rarr [B(OH)_(4)]^(-) + H^(+)`
Since B lies in the 2nd period, it has only ibe s-and three p-orbitals but no d-orbitals . In other words, at the maximum, it can have four pairs of electrons in the valence shell, ie., its maximum coordination number is In contrast, `AlCl_(3)` undergoes hydrolysis in acidfied aqueous solution to form `[Al(H_(2)O)_(6)]^(3+)`
`AlCl_(3) + "water" overset("HCl")rarr[Al(H_(2)O)_(6)]^(3+) + 3Cl^(-) (aq)`
This may be explained as follows : (i) In acidic medium, the conc. of `OH^(-)` ions is much lower than that of `H^(+)` atoms therefore, `Al^(3+)` c ions coordinate with `H_(2)O` molecules and not with `OH^(-)` ions.
(ii) Due to the presence of vacant d-orbital in `Al^(3+)` ions, it can expand its coordination number from 4 to 6 and hence forms `[Al(H_(2)O)_(6)]^(3+)` in which hybridization of `Al` is `sp^(3)d^(2)`.