Question

$XeF _{4}$ is square planar while $XeF _{6}$ has a distorted octahedrall structure. What is the correct explanation for this observation?

• A. Both molecules have one lone pair of electrons
• B. Both molecules have two lone pairs of electrons
• C. $XeF _{4}$ does not have any lone pair of electrons; $XeF _{6}$ has one lone pair of electrons on $Xe$
• D. $XeF _{4}$ has two lone pairs of electrons on $Xe ; XeF _{6}$ has one lone pair of electrons on $Xe$

Answer 1

Answer: (D)

$XeF _{4}$ is square planar and $s p^{3} d^{2}$ -hybridised.



$4 \sigma+2lp=6$ -hybrid orbital

$s p^{3} d^{2}$ -hybridisation

$XeF _{6}$ has distorted octahedral structure and $s p^{3} d^{3}$ -hybridised.



$6 \sigma+1 l p=7$ -hybrid orbital

$\left(s p^{3} d^{3}\right)$ (distorted octahedral)