How do you draw the Lewis structure for #"XeF"_4#?

1 Answer
Aug 4, 2016

Answer:

See explanation.

Explanation:

The first thing to do here is calculate how many valence electrons you have in one molecule of xenon tetrafluoride, #"XeF"_4#.

To do that, add the number of valence electrons that each atom brings to the table. You will have

#"Xe: " "8 e"^(-)#

#color(white)(a)"F: " "7 e"^(-)#

Since one molecule of xenon tetrafluoride contains one atom of xenon and four atoms of fluorine, the total number of valence electrons will be equal to

#"8 e"^(-) + 4 xx "7 e"^(-) = "36 e"^(-)#

Now, the xenon atom will act as your central atom. It will form four single bonds with the four fluorine atoms. Each single bond will account for #2# valence electrons, which means that you're left with

#"36 e"^(-) - 4 xx "2 e"^(-) = "28 e"^(-)#

Each atom of fluorine will complete its octet by adding #3# lone pairs of electrons. A lone pair of electrons contains of #2# **electrons, which means that you will be left with

#"28 e"^(-) - 4 xx (3 xx "2 e"^(-)) = "4 e"^(-)#

The remaining #4# valence electrons will be placed on the xenon atom as lone pairs.

The Lewis structure for xenon tetrafluoride will thus look like this

http://catalog.flatworldknowledge.com/bookhub/2273?e=ball-ch09_s05