Question

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

• A. Octahedron
• B. Trigonal bipyramid
• C. Pentagonal bipyramid
• D. Tetragonal bipyramid

Answer 1

Answer: (C)

Pentagonal bipyramid

Answer 2

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.