# Question #b87ec

##### 1 Answer
Apr 18, 2015

The first thing you need to do when trying to figure out how the Lewis structure of a molecule looks like is to determine how many valence electrons that Lewis structure must account for.

In this case, xenon difluoride, $X e {F}_{2}$, will have a total of 22 valence electrons, 7 from each of the two fluorine atoms and 8 from the xenon atom.

Out of the 22 valence electrons the molecule has, you use 4 electrons to form the two single bonds between xenon and the fluorine, and 6 electrons as lone pairs for each of the two fluorine atoms, thus completing their octet.

This will leave you with another 6 electrons, which will be placed as lone pairs on the xenon atom.

As you can see, the xenon atom will have three lone pairs of electrons and form two single bonds, which means it will have an expanded octet (10 valence electrons, instead of 8).