Question

The number of lone pairs of electrons on \[Xe\] in $Xe{{F}_{2}},Xe{{F}_{4}}$ and $Xe{{F}_{6}}$ respectively are:
(A) 3,2,0
(B) 2,3,1
(C) 4,3,2
(D) 3,2,1

Answer 1

Hint: The valence shell of xenon has 8 electrons. Draw the structure of the above-mentioned compounds and find the number of lone pairs left in the central atom. A lone pair consists of 2 electrons.

Complete step by step answer:
-Xenon lies in the 18th group of the modern periodic table also known as the noble or inert gases.
-When the noble gases were first identified there were no compounds observed. However, Linus Pauling in 1933, predicted that heavier noble gases will be able to form compounds with oxygen and fluorine.
-In 1962, Neil Bartlett, noticed that the energy required for the formation of ${{O}_{2}}[Pt{{F}_{6}}]$ was almost equal to the formation of $Xe[Pt{{F}_{6}}]$. This is how the first noble gas compound was formed.
Based on these studies various compounds were synthesized of Xenon and Krypton.
We will now calculate the number of lone pairs present in $Xe{{F}_{2}},Xe{{F}_{4}}$ and $Xe{{F}_{6}}$.
Total number of electrons present in the valence shell of xenon is 8.

For the compound $Xe{{F}_{2}}$, 2 electrons are used to bond with the incoming F ions. So, the number of electrons remaining are 6 i.e. 3 lone pairs.

For the compound $Xe{{F}_{4}}$, 4 electrons are used to bond with the incoming F ions. So, the number of electrons remaining are 4 i.e. 2 lone pairs.

For the compound $Xe{{F}_{6}}$, 6 electrons are used to bond with the incoming F ions. So, the number of electrons remaining are 2 i.e. 1 lone pair.
So, the correct answer is “Option D”.

Note: The fluorides of Xenon undergo partial and complete hydrolysis to give oxides of xenon. Complete hydrolysis of $Xe{{F}_{6}}$ gives xenon trioxide i.e. $Xe{{O}_{3}}$. Partial hydrolysis of xenon fluoride gives $XeO{{F}_{4}}$ and $Xe{{O}_{2}}{{F}_{2}}$ as products.