Compound with the geometry square pyramidal and $s p^{3} d^{2}$ hybridisation is:

X e O F_{2}

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X e O F_{4}

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X e O_{4}

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X e O_{2} F_{2}

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Solution

The correct option is B $X e O F_{4}$
Hybridization of an atom is identified by calculating the sum of sigma bonds and lone pairs.
If a sum of sigma bonds and lone pairs is 2 $s p$ hybridization
If a sum of sigma bonds and lone pairs is 3 $s p^{2}$ hybridization
If a sum of sigma bonds and lone pairs is 4 $s p^{3}$ hybridization
If a sum of sigma bonds and lone pairs is 2 $s p^{3} d$ hybridization
If a sum of sigma bonds and lone pairs is 2 $s p^{3} d^{2}$ hybridization.
So the given molecule has 5 sigma bonds 1 lone pair so the sum is 6. So its hybridization is $s p^{3} d^{2}$ and the shape is square pyramidal.
Hence option B is correct.

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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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