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

Aug 4, 2016

See explanation.

#### Explanation:

The first thing to do here is calculate how many valence electrons you have in one molecule of xenon tetrafluoride, ${\text{XeF}}_{4}$.

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

${\text{Xe: " "8 e}}^{-}$

$\textcolor{w h i t e}{a} {\text{F: " "7 e}}^{-}$

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

${\text{8 e"^(-) + 4 xx "7 e"^(-) = "36 e}}^{-}$

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

${\text{36 e"^(-) - 4 xx "2 e"^(-) = "28 e}}^{-}$

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

${\text{28 e"^(-) - 4 xx (3 xx "2 e"^(-)) = "4 e}}^{-}$

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