The hybridisation of xenon atom in XeF_4 is | vedclass

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(C) The hybridisation of the central atom can be calculated using the formula: $X = \frac{1}{2} [V + M - C + A]$Where $V$ is the number of valence ele

Vedclass · Accepted Answer

$sp^3d^2$

Vedclass · Answer

(C) The hybridisation of the central atom can be calculated using the formula: $X = \frac{1}{2} [V + M - C + A]$Where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.For $XeF_4$:$V = 8$ (valence electrons of $Xe$)$M = 4$ (four $F$ atoms)$C = 0, A = 0$$X = \frac{1}{2} (8 + 4) = 6$$A$ value of $6$ corresponds to $sp^3d^2$ hybridisation.Therefore,the correct option is $(C)$.

Column $I$ (Molecule/ion)	Column $II$ (Hybridisation)
$A$. $H_3O^{+}$	$P$. $sp$
$B$. $NH_2^{-}$	$Q$. $sp^2$
$C$. $NO_3^{-}$	$R$. $sp^3$
$D$. $ClF_3$	$S$. $sp^3d$

The hybridisation of xenon atom in $XeF_4$ is

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Column $I$ (Molecule/ion) Column $II$ (Hybridisation)

$A$. $H_3O^{+}$ $P$. $sp$

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$D$. $ClF_3$ $S$. $sp^3d$

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