Question

The hybridisations of $BC{l}_3$ and $NC{l}_3$ respectively are:

• A. $s{p}^2$ and $s{p}^3$
• B. $s{p}^3$ and $s{p}^2$
• C. $sp$ and $s{p}^3$
• D. $s{p}^2$ and $s{p}^2$

Answer 1

The correct option is A $s{p}^2$ and $s{p}^3$

The steric number of $BC{l}_3$
$=\frac{1}{2}(3+3+0)=3$
So, it is $s{p}^2$ hybridized and has three bond pairs and zero lone pairs. Thus, the shape is trigonal planar.

Similarly, $NC{l}_3$ has a steric number of
$=\frac{1}{2}(5+3)=4$
So, it is $s{p}^3$ hybridized. It has three bond pairs and one lone pair resulting in tetrahedral geometry.