What is the hybridization of Xe in XeF_2? | vedclass

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(C) The hybridization of $Xe$ in $XeF_2$ is calculated using the formula: $H = \frac{1}{2} (V + M - C + A)$,where $V$ is the number of valence electro

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$sp^3d$

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(C) The hybridization of $Xe$ in $XeF_2$ is calculated using the formula: $H = \frac{1}{2} (V + M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.For $XeF_2$,$V = 8$ (valence electrons of $Xe$),$M = 2$ (two $F$ atoms).$H = \frac{1}{2} (8 + 2) = 5$.$A$ value of $5$ corresponds to $sp^3d$ hybridization.

What is the hybridization of $Xe$ in $XeF_2$?

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