Answer: Option A
:
A
Let us look at the ground state and the second excited state configurations:
These 5 orbitals may be assumed to hybridize to correspond to $s{p}^3{d}^2$ geometry. The 4 bond pairs (covalent bonds between Xe central atom and F atoms) lie in a plane symmetrically while the two lone pairs are distributed symmetrically above and below this plane.
Geometry is square bipyramidal or octahedral. For shape, we do not take the lone pairs into account. So shape will only be square planar.

What would be the shape of $Xe{F}_6?$ I will give you the answer but I encourage you to arrive at it by reason. You can assume that it has 6 bonding pairs and one lone pair:

Experimental evidence suggests that the structures of $Xe{F}_{2}andXe{F}_4$ are exactly as those predicted by VSEPR theory. For $Xe{F}_6$, given that the steric number = 6 bp + 1lp = 7, we can be tempted to conclude that it is similar to Iodine heptafluoride, thus having a pentagonal bipyramidal structure. But it has a face-capped octahedral structure. It is a structure where one lone pair is facing through one of the faces of the octahedron. In terms of the Valence bond theory, we could reason three of the ground state 5p electrons be promoted to the 5d orbitals giving the possibility of an $s{p}^3{d}^3$ hybridization. Molecular orbital theory fails to explain the actual structure of $Xe{F}_6$. According to MOT, the structure should be a completely symmetrical octahedral one. However, this is not the case. In fact, the structure of the molecule is “fluxional”. What do we mean by that? It simply means that the structure is constantly fluctuating between structures where the lone pair spreads across each of the 8 triangular faces.