A) pentagonal bipyramidal
B) square planar
C) octahedral
D) distorted octahedral
Correct Answer: D
Solution :
In \[\text{Xe}{{\text{F}}_{\text{6}}}\text{,}\] Number of hybrid orbitals \[H=\frac{1}{2}[V+X-C+A]\] [where, V = Number of valence electrons X = Number of monovalent atom C and A = positive and negative charge] \[=\frac{1}{2}[8+6-0+0]\] \[=\frac{1}{2}[14]=7\] Thus, its hybridisation is \[\text{s}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{3}}}\]and on the basis of hybridisation, the structure should be pentagonal bipyramidal. But Xe also contains one lone pair of electrons, due to which the geometry of \[\text{Xe}{{\text{F}}_{6}}\]becomes distorted octahedral.
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