Question

The bond angle around B in $$BC{l}_3$$ and$$B{F}_3$$ is the same. Explain.

Answer 1

Hint: Boron is the element of group 13. It has 3 valence electrons in its outer shell. On the basis of hybridization, the structure is estimated. The structure helps in finding the bond angle. The number of lone pairs should also be considered.

Complete answer:
Boron is the first element of the group 13.
Its outer electronic configuration is $$2s^{2}2p^1$$
It means that it has 3 valence electrons in its outer shell. Hence, it can form 3 bonds.
Now, in $$BC{l}_3$$, boron is surrounded by 3 atoms of chlorine. Chlorine atom belongs to the group of halogens (group 17), which requires only one electron to complete its octet.
Hence, the chlorine forms a single bond with the boron atom.
Now, all the 3 valence electrons of boron form a bond with the chlorine atom hence there are no lone pairs present in the compound.
Since there are 3 bonds in the compound the hybridization will be $$s{p}^2$$
And we know that $$s{p}^2$$ hybridization forms trigonal planar geometry.
The trigonal planar has $${120}^{\circ }$$ bond angle around the central atom.

Now, in $$B{F}_3$$, boron is surrounded by 3 atoms of Fluorine. Fluorine atoms also belong to the group of halogens (group 17), which requires only one electron to complete its octet.
Hence, the chlorine forms a single bond with the boron atom.
Now, all the 3 valence electrons of boron form a bond with the Fluorine atom hence there is no lone pair present in the compound.
Since there are 3 bonds in the compound the hybridization will be $$s{p}^2$$
And we know that $$s{p}^2$$ hybridization forms trigonal planar geometry.
The trigonal planar has $${120}^{\circ }$$ bond angle around the central atom.

Hence the bond angle is the same in both cases.

Note: You may get confused that fluorine is a more electronegative atom so it would make a bond angle different from boron trichloride. The structure and bond angle depends on the number of bonds and the number of lone pairs around the central atom.