Question

The correct increasing bond angle among $B{F_3}$, $P{F_3}$ and $Cl{F_3}$ follows the order.
A.$B{F_3} < {\text{ }}P{F_3} < {\text{ }}Cl{F_3}$
B.$P{F_3} < {\text{ }}B{F_3} < {\text{ }}Cl{F_3}$
C.$Cl{F_3} < {\text{ }}P{F_3} < {\text{ }}B{F_3}$
D.$B{F_3} < {\text{ }}P{F_3} < {\text{ }}Cl{F_3}$

Answer 1

Here we have to choose the correct option for increasing bond angle among the given. Bond angle is the angle between two bonds originating from a common atom. Bond angle is the concept inside the molecular geometry of the molecule.

Complete step by step answer:

Let’s start with discussing the bond angle for better understanding of the question. Bond angle is the angle between two bonds originating from a common atom. Bond angle is the concept inside the molecular geometry of the molecule. Molecule geometry is the three dimensional arrangement of atoms in a molecule. It includes the shape of the molecule, bond length, bond angle and any other geometric parameter.
Coming to the question, the geometry of $B{F_3}$ is trigonal planar in which the fluorine atoms lies in the same plane and having bond angle of $B{F_3}$ is ${\text{12}}{{\text{0}}^{\text{o}}}$ . The bond angle of $P{F_3}$ is ${\text{10}}{{\text{9}}^{\text{o}}}$ due to the lone pair at the top, the shape of the molecule is trigonal pyramidal. The bond angle of $Cl{F_3}$ is ${\text{9}}{{\text{0}}^{\text{o}}}$ due to lone pair the shape of the molecule is T shape.
Hence,
$\therefore$ The answer to this question will C. $ClF3{\text{ }} < {\text{ }}PF3{\text{ }} < {\text{ }}BF3$.

Note:
We must remember that the best bond angle is ${\text{109}}{\text{.}}{{\text{5}}^{\text{o}}}$ and it is always called as an ideal bond angle. The reason behind it is that a perfect tetrahedral is being formed at this angle. This angle is found in H-C-H and in this the hydrogen atoms repel each other equally which leads to the formation of this angle. Also this geometry got the greatest distance between all the four lone pairs.