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  4. The structure of XeF6 is experimentally determined to be distorted octahedron. Its structure according to VSEPR theory is

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Q. The structure of $XeF_6$ is experimentally determined to be distorted octahedron. Its structure according to VSEPR theory is

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Solution:

In $XeF _{6}$, Xe, the central atom contains, 8 valence electrons. Out of which 6 are utilised with fluorine in bonding (i.e., 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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