Question

The $ \text{C}{{\text{F}}_{\text{4}}} $ , $ \text{S}{{\text{F}}_{\text{4}}} $ , and $ \text{Xe}{{\text{F}}_{\text{4}}} $ contain the following electronic structure on their central atoms. Which one is the correct option?
(A) 1, 2 and 3 lone pairs of electrons respectively.
(B) 0, 1, and 2 lone pairs of electrons respectively.
(C) 1, 1 and 1 lone pairs of electrons respectively.
(D) No lone pairs of electrons on any molecule.

Answer 1

Hint: The lone pair of electrons refers to that electron pair which does not take part in covalent chemical bonding. It can be donated as a pair to any atom which has a dearth of electrons through coordinate covalent bonding. The electronic configuration of the central atom can help us find the number of lone pairs present.

Complete step by step solution:
To find out the number of the lone pairs, we need to see the electronic configuration of the central atom. In case of carbon tetrafluoride, the central atom is carbon which has four electrons in the valence shell and whose electronic configuration is $ \text{1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{\text{2}}} $ . Though the 2s orbital electrons in carbon are paired, they get unpaired due to energy released due to the two extra bonds formed in its unpairing. Hence there are no lone pair of electrons on the carbon atom.
In case of sulphur tetrafluoride, the central atom is sulphur which has six electrons in the valence shell and whose electronic configuration is $ \text{1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{6}}\text{3}{{\text{s}}^{\text{2}}}\text{3}{{\text{p}}^{4}} $ . There are 4 electrons in the 3p orbital, two of them are paired in the form of a lone pair while two are there as bond pairs. Also there is one lone pair as the 3s electron pair. Hence there is one lone pair of electrons on the sulphur atom.
In case of xenon tetrafluoride, the central atom is xenon which has eight electrons in the valence shell and whose electronic configuration is $ \left[ \text{Kr} \right]\text{4}{{\text{d}}^{\text{10}}}\text{5}{{\text{s}}^{\text{2}}}\text{5}{{\text{p}}^{\text{6}}} $ . There being eight electrons in the valence shell of the xenon atom, they fulfil the octet configuration and all the electrons are lone pairs in the beginning but due to the availability of the vacant 5d orbitals, the s, p, and d orbitals of the valence shell undergo hybridization and thus the xenon atom is left with only two lone pairs instead of four. Hence there are two lone pairs of electrons on the xenon atom.

Note:
The Valence Shell Electron Pair Repulsion Theory of the VSEPR theory guides the bonding pattern and the shape of the molecules according to the number of the lone pairs and the bond pairs present in the molecule and this is because, the order of repulsion among the electron pairs is:
Lone pair-lone pair > lone pair-bond pair > bond pair-bond pair.