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 1

Answer: (B)

from $s{p}^2$ to $s{p}^3$

$B$ atom in $B{F}_3$ has 3 bond pairs of electrons and 0 lone pairs of electrons. It is $s{p}^2$ hybridised.

$B$ atom in $F_{3}B\leftarrow N{H}_3$ has 4 bond pairs of electrons and 0 lone pairs of electrons. It is $s{p}^3$ hybridised.

Hence, the formation of molecular complex $F_{3}B\leftarrow N{H}_3$ results in a change in hybridization of boron from $s{p}^2$ to $s{p}^3$.