Question

In which of the following sets of compounds does the bond angle remain constant for all members?
A.$N{H_3},P{H_3},As{H_3}$
B.$P{F_3},PC{l_3},PB{r_3}$
C.$O{H_2},O{F_2},OC{l_2}$
D.$B{F_3},BC{l_3},BB{r_3}$

Answer 1

The angle between two bonds in a covalent species which is present at the same atom is known as its bond angle. Hence, there should be at least two bonds in that covalent species. The lone pair repulsion mainly affects the bond angle. Therefore, if the central atom contains the lone pair of electrons, the angle of the compound will decrease.

Complete answer:
In these compounds, the surrounding atom is the same for three compounds but the central atom of each compound is varying, that is, N, P and As. Therefore, the bond angle is also varying. Hence, option (A) is incorrect.
Here, the central atom is the same for three compounds but the surrounding atom is different for all cases. Therefore, the bond angle is also different in each compound. Hence, the option (B) is incorrect.
In this case also, the central atom is the same and the surrounding atom is varying and there is a chance of varying the hybridization and bond angle. Hence, option (C) is incorrect.
In this case, the central atom is the same and it is boron. And the hybridization of all compounds is$s{p^2}$. And the sum of lone pairs and bond pairs of $B{F_3},BC{l_3},$ and $BB{r_3}$ is equal to the same.
Here, the bond pair is equal to three and the lone pair is equal to one. And the bond angle of $B{F_3},BC{l_3},$ and $BB{r_3}$ is equal to ${120^o}$.

Hence, the option (D) is correct.

Note:
We have to know that the bond angle mainly depends on the shape and hybridization of a compound. If the compounds have the same hybridization, then the sum of the lone pair and bond pair is equal to the same. Hence, the bond angle becomes equal. And the electronegativity and shape of molecules also affect the bond angle. The bond angle decreases with decreasing electronegativity.