Question

In $Xe{F}_2$ , $Xe{F}_4$ and $Xe{F}_6$ , the number of lone pairs on $Xe$ are respectively :
A. $2,3,1$
B. $1,2,3$
C. $4,1,2$
D. $3,2,1$

Answer 1

Hint: Xenon belongs to the noble gas family and so has eight electrons in its outermost shell . It reacts directly with fluorine under appropriate conditions to form three binary fluorides , which are $Xe{F}_2$ , $Xe{F}_4$ and $Xe{F}_6$ .

Complete step by step answer:
Our aim is to find out the lone pair of electrons on $Xe$ in each of the three binary compounds .
We can find out the number of lone pairs of electrons according to VSEPR theory .
According to VSEPR theory the total number of electron pairs in a molecule is given by
total number of electron pairs = ${\displaystyle \frac{lp+bp}{2}}$ where , lp=lone pair and bp = bond pair
In case of $Xe{F}_2$ there are two $Xe-F$ bonds , also the number of electron pairs are
${\displaystyle \frac{8+2}{2}}=5$
Out of this two are bond pairs , therefore the number of lone pairs of electrons on Xenon is 3 .
In case of $Xe{F}_4$ there are four $Xe-F$ bonds , also the number of electron pairs are
${\displaystyle \frac{8+4}{2}}=6$
Out of this four are bond pairs , so the number of lone pairs of electrons on Xenon is 2 .
In case of $Xe{F}_6$ there are six $Xe-F$ bonds , also the number of electron pairs are
${\displaystyle \frac{8+6}{2}}=7$
Out of this six are bond pairs so only one lone pair is left .
So in $Xe{F}_2$ , $Xe{F}_4$ and $Xe{F}_6$ , the number of lone pairs on $Xe$ are 3,2,1

So, the correct answer is Option D .

Note:
All the Xenon fluorides are colourless , crystalline solids and sublime readily at $298K$ . All the fluorides are extremely strong oxidising and fluorinating agents . Also the lower fluorides form higher fluorides when heated with fluorine gas under pressure .