The geometry of XeOF_4 by VSEPR theory is | vedclass

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(C) In $XeOF_4$,the central atom $Xe$ has $8$ valence electrons. It forms $4$ single bonds with $F$ atoms and $1$ double bond with the $O$ atom,utiliz

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square pyramidal

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(C) In $XeOF_4$,the central atom $Xe$ has $8$ valence electrons. It forms $4$ single bonds with $F$ atoms and $1$ double bond with the $O$ atom,utilizing $6$ electrons for bonding. This leaves $2$ electrons,forming $1$ lone pair. Total electron pairs = $5$ bond pairs + $1$ lone pair = $6$ electron pairs. With $6$ electron pairs,the hybridization is $sp^3d^2$,which corresponds to an octahedral electron geometry. Due to the presence of $1$ lone pair,the molecular geometry is square pyramidal.

The geometry of $XeOF_4$ by $VSEPR$ theory is

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