The shape of XeOF_4 molecule is | vedclass

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(D) 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. This

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Square pyramidal

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(D) 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. This accounts for $6$ bonding electron pairs. Including $1$ lone pair,the total number of electron domains is $7$ (steric number $7$),which corresponds to $sp^3d^3$ hybridization. However,due to the presence of one lone pair and the double bond,the geometry is distorted. The molecule adopts a square pyramidal shape,where the $Xe$ atom is at the center of a square base formed by $4$ $F$ atoms,and the $O$ atom occupies the axial position. $	herefore$ $(d)$ is the correct answer.

The shape of $XeOF_4$ molecule is

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