How many lone pairs of electrons are on the #Xe# atom in #XeF_6#? A) 0 B) 3 C) 2 D) 1

Question

12550 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-25T03:25:40

D) #1# lone pair.

First, let's draw the Lewis Structure for #XeF_6#:

The total number of electrons in #XeF_6# is #8+(7xx6)=50# electrons.
#Xe# will be the central atom. So, let's just connect #Xe# and #F# atoms with a single bond each.
We have #50-12=38# electrons left to draw on.
This means that we have enough electrons to not need double bonds and complete the octet for #F# using lone pairs.
However, we still have #38-36=2# electrons left to use.
These electrons will need to be placed on #Xe#, since #Xe#, having its valence electrons in the #5p# orbital, have empty #5d#, #4f#, and #5f# orbitals—allowing it have more electrons than an octet.
(For more information on exceeding octets, check out this article on Wikibooks!)

So, there are #2# electrons on #Xe#—which is #1# pair.