Question
The formation of molecular complex $$B{F_3} - N{H_3}$$ results in a change in hybridization of boron
A.
from $$s{p^2}$$ to $$ds{p^2}$$
B.
from $$s{p^2}$$ to $$s{p^3}$$
C.
from $$s{p^3}$$ to $$s{p^2}$$
D.
from $$s{p^3}$$ to $$s{p^3}d$$
Answer :
from $$s{p^2}$$ to $$s{p^3}$$
Solution :
In $$B{F_3},B$$ is $$s{p^2}$$ hybridized with one empty $${p_z}$$ orbital. The empty $${p_z}$$ orbital of $$B{F_3}$$ can be filled by lone pair of molecules such as $$N{H_3}.$$ When this occurs a tetrahedral molecule or ion is formed which is $$s{p^3}$$ hybridized.