Question

# Draw the structure of $Xe{{F}_{4}}$.

Hint: The structure of $Xe{{F}_{4}}$ is governed by the VSEPR theory of electron repulsion. With this idea in mind, try to predict the structure of the given compound.
Before we move onto the structure of $Xe{{F}_{4}}$ , let us first look into the VSEPR theory of electron repulsion.
Now, with the specifics of the VSEPR theory done and accounted for, let us now move on to the specific structure of $Xe{{F}_{4}}$.
VSEPR theory predicts that $Xe{{F}_{4}}$ is square planar. There are six electron pairs around the central atom (Xe), four of which are bonding, and two are lone pairs. Thus, we have octahedral electron pair geometry, and square planar molecular geometry.
It’s possible that someone might tell you that xenon exhibits $sp{}^\text{3}d{}^\text{2}$ hybridization. Don’t believe that. It’s been pretty well established that for the representative elements there is no d-orbital participation in the bonding in hypervalent molecules.
Note: While a traditional compound unlike $Xe{{F}_{4}}$ such as xenon hexafluoride ($Xe{{F}_{6}}$) without the presence of lone pairs in its structure would exhibit octahedral geometry with different bond angles, the presence of lone pairs increases electron repulsions leading to a square planar structure with 900 bond angles.