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The structure of $\mathrm{XeF}_{6}$ is experimentally determined to be distorted octahedron. Its structure according to VSEPR theory is
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pentagonal bipyramid
In $\mathrm{XeF}_{6}$. Xe, the central atom $\mathrm{F}$ contains, 8 valence electrons. Out of which 6 are utilised with fluorine in bonding (ie., it contains six bond pairs of electrons) while one pair remains as lone pair. Thus, the total pairs $=6+1=7$ Hence, its shape is pentagonal bipyramid according to VSEPR theory.
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