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# 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. $3,2,1$

Last updated date: 18th Jun 2024
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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 $Xe{F_6}$ .

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 = $\dfrac{{lp + bp}}{2}$ 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
$\dfrac{{8 + 2}}{2} = 5$
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
$\dfrac{{8 + 4}}{2} = 6$
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
$\dfrac{{8 + 6}}{2} = 7$
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 .