Question

When
BCl3
is
treated
with
water,
it
hydrolyses
and
forms
[B(OH)4]-
only
whereas


AlC13
in acidified
aqueous
solution
forms
[A1(H20)6]
3+
ion,
Explain
what
is
the


hybridisation
of
boron
and
aluminium
in
these
species?
​

Answer 1

Answer:

BCl3 + 3H2O→ B(OH)3 + 3HCl

B(OH)3 + H2O→[B(OH)4]- + H+

B(OH)3 due to its incomplete octet accepts an electron pair (as OH-) to give [B(OH)4]-. Boron in this ion involves one 2s orbital and three 2p orbitals. Thus, hybridisation of B in [B(OH)4]- is sp3.