Question

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

Answer 1

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