Question

Find the number of lone pairs of electrons on $Xe$ in $XeO{F_4}$.

Answer 1

Hint: To find the number of lone pairs of electrons on $Xe$, we need to draw the structure of $XeO{F_4}$. We should also be aware of the oxidation states in which the central atom $Xe$ can exist, so as to know the number of electrons involved in bonding and number of lone pairs.

Complete step by step answer:
Xenon is found in multiple oxidation states such as $0$, $ + 2$, $ + 4$, $ + 6$ and $ + 8$. Due to this it forms a number of oxides, some of these oxides are stable while some are not.
Xenon also forms oxide fluorides which consists of both oxygen and fluorine atoms bonded to $Xe$. One of such oxide fluorides is xenon oxytetrafluoride, i.e., $XeO{F_4}$.
The O-atom is in $ - 2$ oxidation state and each F-atom is in $ - 1$ oxidation state. The overall charge on the molecule is zero, so the oxidation state of $Xe$ is,
$x + ( - 2) + ( - 1 \times 4) = 0$
$ \Rightarrow x - 2 - 4 = 0$
$ \Rightarrow x - 6 = 0$
$ \Rightarrow x = + 6$
$Xe$ is a noble gas and it has $8$ valence electrons. As it is in $ + 6$ oxidation state, it means it has shared $6$ electrons with the bonded atoms. Therefore, only $2$ more electrons are left in a non-bonded state. That is, one lone pair of electrons is present on $Xe$.
Structure of $XeO{F_4}$:

Note:
We can also find the number of lone pairs directly by calculating the number of electrons shared by $Xe$ to the bonded atoms, without the need to calculate the oxidation number. As $Xe$ has $8$ valence electrons, it is bonded to O-atom with a double bond, so $2$ electrons are shared; $4$ F-atoms are singly bonded to $Xe$, so $4$ electrons are shared with fluorine. The rest $2$ electrons are present as lone pairs of electrons.