# What is XeF_4's molecular shape?

Feb 7, 2017

$X e {F}_{4}$ is $\text{square planar...........}$
The molecular geometry is square planar, but the electronic geometry is octahedral. How? Well, the is a simple consequence of $\text{valence shell electron pair repulsion theory}$, which I will try to outline.
For $X e {F}_{4}$, we have 8 valence electrons from xenon, and $4 \times 7$ valence electrons from the bound fluorides; thus, we have 36 electrons, 18 electron pairs to distribute around 5 centres. Each fluorine atom has 4 electron pairs surrounding it; 3 lone pairs, and one $X e - F$ bond. And the central xenon atom has 4 bonding pairs, and 2 non-bonding lone pairs. The most thermodynamically stable state for these 6 electrons is as an OCTAHEDRON, with $4 \times X e - F$ bonds, and 2 non-bonding lone pairs on the central xenon atom.