What is the hybridization and structure of Xe | vedclass

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Question

(C) In $XeF_4$,the central atom $Xe$ has $8$ valence electrons.It forms $4$ bonds with $F$ atoms and has $2$ lone pairs.Total electron pairs = $4 	ex

Vedclass · Accepted Answer

$sp^3d^2$,Square planar

Vedclass · Answer

(C) In $XeF_4$,the central atom $Xe$ has $8$ valence electrons.It forms $4$ bonds with $F$ atoms and has $2$ lone pairs.Total electron pairs = $4 	ext{ (bond pairs)} + 2 	ext{ (lone pairs)} = 6$.Therefore,the hybridization is $sp^3d^2$.According to $VSEPR$ theory,with $4$ bond pairs and $2$ lone pairs,the geometry is octahedral,but the shape is square planar.

What is the hybridization and structure of $XeF_4$?

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