Hybridization and structure of $X e F_{4}$ is:

s p^{3} d

, triogonal bipyramidal

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s p^{3}

, tetrahedral

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s p^{3} d^{2}

, square planar

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s p^{3} d^{2}

, hexagonal

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Solution

The correct option is B $s p^{3} d^{2}$ , square planar
In $X e F_{4}$ out of 8 valencies of Xe 4 are satisfied by F. Thus, it Xe has 4 bond pair and 2 lone pairs. Thus its hybridization is $s p^{3} d^{2}$ and its geometry is square planar.

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	Molecule	Hybridization	Shape	Number of lone pairs of electrons
$(1)$	$X e O_{3}$	$s p^{3}$	pyramidal	$1$
$(2)$	$X e F_{4}$	$s p^{3} d^{2}$	planar	$3$
$(3)$	$X e F_{2}$	$s p^{3} d$	linear	$2$
$(4)$	$X e O_{4}$	$s p^{3} d^{2}$	pyramidal	$1$

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