Question

The order of increasing bond angle in the molecules $BeC{l_2},BC{l_3},CC{l_4}{\text{ and S}}{{\text{F}}_{\text{6}}}$ is:
A. $S{F_6} < CC{l_4} < BC{l_3} < BeC{l_2}{\text{ }}$
B. $BeC{l_2}{\text{ < }}BC{l_3} < CC{l_4} < S{F_6}$
C. $S{F_6} < CC{l_4} < BeC{l_2}{\text{ }} < BC{l_3}$
D. $BC{l_3} < BeC{l_2}{\text{ }} < S{F_6}{\text{ < }}CC{l_4}$

Answer 1

Hint: The angle between the bonds of a covalent molecule is known as bond angle. The bond angle depends on the hybridization of the molecule because hybridization predicts the shape of the molecule and from the shape we can predict the bond angle.

Complete step by step answer:
The angle between the two bonds originating from the same atom is known as bond angle. And we know that bonds only exist in covalent bonding so bond angle is also possible in covalent compounds.
The bond angle of the compounds are determined by the hybridization of the central atom and the number of lone pairs present on the central and outer atom of the molecule. The atoms in a molecule position themselves in a way so that they feel minimum repulsion and attraction from the other atoms present in the covalent compound.
For the covalent compound $BeC{l_2}$
It has three atoms so these will try to come in linear positions so that they can maintain a maximum distance from each other. Due to linear geometry of molecules, the bond angle is $180^\circ $ .
For the covalent compound $BC{l_3}$
This compound contains four atoms, one of them is a central atom hence these atoms will try to maintain a maximum distance so the bond angle is approximately $120^\circ $ between each and every bond. The geometry with bond angle $120^\circ $ is known as trigonal.
For the covalent compound $CC{l_4}$
This covalent compound has five atoms and carbon is the central atom. The geometry of the compound is tetrahedral so the bond angle will be approximately $109^\circ $. The atoms will be present in a $3D$ plane to minimize the repulsion.
For the covalent compound $S{F_6}$
This compound has eight atoms, hence the geometry of this covalent product is octahedral. In octahedral geometry each and every bond has an angle of $90^\circ $ between them.
Hence from the above discussions we can conclude that $BeC{l_2}$has maximum and $S{F_6}$has minimum value of bond angle.
The order of bond angle is: $S{F_6} < CC{l_4} < BC{l_3} < BeC{l_2}{\text{ }}$

Hence option (A) is correct .

Note:
In some covalent compounds bond angles deviate from their ideal values, this happens due to the presence of lone pairs. The lone pairs repel the bonds, the repulsion from lone pairs is stronger than the repulsion from other bonds.