# Question #aaab2

Jan 30, 2017

London dispersion forces.

Xenon tetrafluoride is ${\text{XeF}}_{4}$, a covalent compound with a square planar geometry. When you draw out the Lewis structure using the electron-counting method:

• $\text{Xe}$ $\implies$ $\text{8 valence}$
• $4 \times \text{F}$ $\implies$ $4 \times \text{7 valence}$

With $36$ valence electrons, and fluorine tending to make single bonds, we have $\boldsymbol{8}$ bonding valence electrons surrounding $\text{Xe}$ for the four single bonds total, and the $\boldsymbol{4 \times 6}$ nonbonding valence electrons surrounding the fluorine atoms total.

That leaves $4$ to be distributed onto xenon, which symmetrically distribute themselves to minimize repulsions.

Since ${\text{XeF}}_{4}$ is symmetrical, any polarities in its bonds cancel out, leaving it nonpolar. Therefore, it has london dispersion, and that's pretty much it.