XeF_4 is square planar while XeF_6 has a dist | vedclass

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Question

(D) $XeF_4$ has $4$ bond pairs and $2$ lone pairs of electrons on $Xe$,resulting in $sp^3d^2$ hybridisation and a square planar geometry.$XeF_6$ has $

Vedclass · Accepted Answer

$XeF_4$ has two lone pairs of electrons on $Xe$; $XeF_6$ has one lone pair of electrons on $Xe$

Vedclass · Answer

(D) $XeF_4$ has $4$ bond pairs and $2$ lone pairs of electrons on $Xe$,resulting in $sp^3d^2$ hybridisation and a square planar geometry.$XeF_6$ has $6$ bond pairs and $1$ lone pair of electrons on $Xe$,resulting in $sp^3d^3$ hybridisation and a distorted octahedral geometry due to the presence of the lone pair.Thus,option $D$ is correct.

$XeF_4$ is square planar while $XeF_6$ has a distorted octahedral structure. What is the correct explanation for this observation?

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