Question

Which of the following has a bond angle of ${120}^{\circ }?$

• A. $Cl{F}_3$
• B. $NC{l}_3$
• C. $BC{l}_3$
• D. $N{H}_3$

Answer 1

The correct option is C $BC{l}_3$

The steric number in $Cl{F}_3$
$=\frac{1}{2}(7+3)=5$

So, $Cl{F}_3$ is $s{p}^{3}d$ hybridised. It has three bond pairs and two lone pairs. Thus, the geometry of $Cl{F}_3$ is trigonal bipyramidal and the shape is T-shape. Hence, the bond angle $(F-Cl-F)$ is ${90}^{\circ }$.

The steric number in $NC{l}_3$
$=\frac{1}{2}(5+3)=4$
So, $NC{l}_3$ is $s{p}^3$ hybridised. It has three bond pairs and one lone pair of electrons. Thus, the geometry of $NC{l}_3$ is tetrahedral and the shape is Pyramidal. Hence, the bond angle $(Cl-N-Cl)$ is around ${108}^{\circ }$.


The steric number in $BC{l}_3$
$=\frac{1}{2}(3+3)=3$
So, $BC{l}_3$ is $s{p}^2$ hybridised. It has three bond pairs. Thus, both the geometry and the shape of $BC{l}_3$ is trigonal planar. Hence, the bond angle (B-Cl-B) is ${120}^{\circ }$.


The steric number in $N{H}_3$ = $4$
So, $N{H}_3$ is $s{p}^3$ hybridised. It has three bond pairs and one lone pair. Thus, the geometry of $N{H}_3$ is tetrahedral and the shape of $N{H}_3$ is trigonal pyramidal. Hence, the bond angle (H-N-H) is lesser and is around ${107}^{\circ }$.