Question

In which of the following molecules the central atom has three lone pairs of electrons?
A) Ammonia.
B) Xenon difluoride.
C) Chlorine trifluoride.
D) Hydrogen sulphide.

Answer 1

Hint: We can calculate the steric number by totaling the number of atoms bonded to the central atom and the lone pairs of electrons present on the central metal atom. Let us know that if the steric number is 4, then we say the atom has $s{p^3}$ hybridization, if the steric number is 3, then atom has $s{p^2}$ hybridization, if the steric number is 2, then it has $sp$ hybridization. We can also predict the geometry of compounds using the steric number.

Complete answer:
Let us see the structure of species in options,
The Lewis structure of xenon difluoride is,

The steric number of central xenon atom is five which means that it is ${\text{s}}{{\text{p}}^{\text{3}}}{\text{d}}$ hybridized but it has only two bonding electrons thus it adopts the linear geometry with the bond angle of \[{\text{18}}{{\text{0}}^{\text{o}}}\] .
Hence option B is correct.
The structure of $N{H_3}$ is,

A ${\text{N}}{{\text{H}}_{\text{3}}}$ molecule has three covalent bonds and one lone pair of electrons. Therefore, it is appropriate to use ${\text{s}}{{\text{p}}_{\text{3}}}$ hybrid orbitals on the nitrogen atom. Three of these ${\text{s}}{{\text{p}}_{\text{3}}}$ orbitals form localized bond orbitals by combining with the fluorine p orbitals. Thus, the bonding in a ${\text{N}}{{\text{H}}_{\text{3}}}$ molecule in terms of three localized s-bond orbitals and one non-bonded lone pair in an $s{p_3}$ orbital on the nitrogen atom. There are eight valence electrons in a ${\text{N}}{{\text{H}}_{\text{3}}}$ molecule. Six of them occupy the three localized N (${\text{s}}{{\text{p}}_{\text{3}}}$) + (p) s-bond orbitals and two occupy the non-bonded N (${\text{s}}{{\text{p}}_{\text{3}}}$) orbital. The use of ${\text{s}}{{\text{p}}_{\text{3}}}$ orbitals implies that the H–N–H bond angles are ${\text{109}}{\text{.}}{{\text{5}}^{\text{o}}}$ . Hence option A is incorrect.
The structure of chlorine trifluoride is,

It is known that there are two lone pairs around the chlorine and thus option C is incorrect.
The structure of Hydrogen sulphide is,

It is known that there are two lone pairs around the oxygen and thus option D is incorrect.
And hence option B is correct.

Note:
We have to remember that hybridization means the mixing of two atomic orbitals of the same energy to give a new hybrid molecular orbital of the same energy. Based on the type of orbitals involved, hybridization is classified as $sp$ , $s{p^2}$ , $s{p^3}$ , \[ds{p^2}\] , \[ds{p^3}\] etc. Hybridization is often considered to have taken place with the assistance of an empty orbital. B contains 3 valence electrons, whereas 4 orbitals participate within the hybridization. Thus, one empty orbital participates within the hybridization.