Question

${\left[Xe{O}_6\right]}^{4-}$ is octahedral whereas $Xe{F}_6$ is a disordered one, because

• A. fluorine is more electronegative than oxygen
• B. Xe has a lone pair in $Xe{F}_6$
• C. $Xe{F}_6$ is neutral whereas ${\left[Xe{O}_6\right]}^{4-}$ anionic
• D. $Xe-F$ bond has more ionic characters

Answer 1

Answer: (B)

Due to the presence of six fluoride ligands and one lone pair of electrons in $Xe{F}_6$ , the structure lacks perfect octahedral symmetry and is disordered. In ${\left[Xe{O}_6\right]}^{4-}$ all the valence shell electrons of xenon are involved in the bonding with oxygen and hence, no lone pair of electrons is present. So, it has perfect octahedral geometry.