What is the electron geometry of XeF4?

Answer 1

After taking a moment to appreciate seeing a compound with a noble gas in it, for once, we then start to consider some facts.

Xe has a valence of 8. Fluorine has a valence of 7.
Fluorine really wants 1 electron each to fill that octet. So we can imagine 4 covalent bonds to Xe and Fluorine is nice and happy.


So Xe has given away 4 of its electrons, it now has 4 left. That's two lone pairs.
Fluorines are going to repel each other equally and so must be equally spaced apart if at all possible. The same goes for the lone pairs. Lone pairs will repel fluorine and vice versa, and all by the same degree.


Sounds confusing but actually this one is really kind. We see a square planar molecule with two lone pairs, one above and one below the plane.
Each Fluorine is equidistant from another fluorine. And those lone pairs are equidistant from each other and all of the fluorines.
Practice this kind of thing, it'll soon become really easy.


For further reading, I suggest trying to look up VESPR theory. Also getting a model kit can help a lot when thinking about geometries. Think about the repulsions involved and use your imagination. You'll start seeing patterns when you practice it, and soon it will become second nature.