Question

Among the following molecules (i) \[Xe{O_3}\](ii) \[XeO{F_4}\](iii) \[Xe{F_6}\] . Those having the same number of lone pairs on \[Xe\] are.
(a) (i) and (ii) only
(b) (i) and (iii) only
(c) (ii) and (iii) only
(d) (i), (ii) and (iii)

Answer 1

Hint: Xenon is a member of the noble gas family, and it has\[n{s^2}n{p^6}\]outer shell electronic configuration. Xenon can directly react with fluorine and oxygen and can form a variety of compounds such as., \[Xe{F_2}\], \[Xe{F_4}\], \[Xe{F_6}\], \[Xe{O_3}\] and \[XeO{F_2}\].

Complete Step by Step Solution:
In the given question we have to find out the compounds that have the same number of lone pairs of electrons. By using VSEPR theory we can easily find out the number of lone pairs of electrons in the given compounds of xenon.

The structure of \[Xe{O_3}\](xenon trioxide) has the \[Xe\] as a central atom and it is surrounded by three oxygen atoms via \[Xe = O\] bonds (covalent bonds). The exact structure of \[Xe{O_3}\]is represented below which represents the presence of one lone pair of electrons on the xenon centre.


Image: Structure of \[Xe{O_3}\].

Similarly, the structure of \[XeO{F_4}\] contain \[Xe\] as the central atom and it is surrounded by one oxygen and four fluorine atoms via \[Xe = O\] and \[Xe - F\] bond respectively. Like the structure of \[Xe{O_3}\], \[XeO{F_4}\] also has one lone pair of electrons on a xenon atom. The \[XeO{F_4}\]has a square pyramidal shape and \[s{p^3}{d^2}\]hybridization.


Image: Structure of \[XeO{F_4}\].

Similarly, the structure of \[XeO{F_4}\]contain \[Xe\]as the central atom and it is surrounded by one oxygen and four fluorine atoms via \[Xe = O\] and \[Xe - F\] bond respectively. Like the structure of \[Xe{O_3}\], \[XeO{F_4}\]also has one lone pair of electrons on a xenon atom. The \[XeO{F_4}\]has square pyramidal shape and \[s{p^3}{d^2}\]hybridization for \[Xe\] atoms.

Whereas the structure of \[Xe{F_6}\]has a different structure than the \[Xe{O_3}\]and \[XeO{F_4}\]. The \[Xe{F_6}\]molecule possesses six fluorine atoms which are bonded with the \[Xe\]atom. The central \[Xe\]atom has \[s{p^3}{d^3}\]hybridization and distorted octahedral or capped octahedral shape due to the presence of one lone pair of electrons on \[Xe\] centre.

Image: Structure of \[Xe{F_6}\].
Therefore from the above explanation we can also observe that the \[Xe{O_3}\]\[XeO{F_4}\]and \[Xe{F_6}\] have only one lone pair of electrons on xenon atom. Hence option (d) will be the correct option.

Note: The VSEPR theory also explains the lone pair and bond pair repulsion in the molecules. The VSEPR theory also explains the lone pair -lone pair repulsion is larger compared to the lone pair -bond pair and bond pair-bond pair repulsion. Therefore, lone pair-lone pair repulsion> lone pair -bond pair repulsion > bond pair -bond pair repulsion.