Hybridisation state of $X e$ in $X e F_{2}$ , $X e F_{4}$ , and $X e F_{6}$ respectively are:

{sp}^{2}, {s p}^{3} d, {s p}^{3} d^{2}

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{sp}^{3} d, {s p}^{3} d^{2}, {s p}^{3} d^{3}

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{sp}^{3} d^{2}, {s p}^{3} d, {s p}^{3} d^{3}

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{sp}^{2}, {s p}^{3}, {s p}^{3} d

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Solution

The correct option is A ${sp}^{3} d, {s p}^{3} d^{2}, {s p}^{3} d^{3}$
Hybridisation is given as
$T o t a l n u m b e r o f b o n d p a i r s$ + $T o t a l n u m b e r o f l o n e p a i r s$

$X e F_{2}$ $2$ + $3$ = $5$ $S P^{3} d$
$X e F_{4}$ $4$ + $2$ = $6$ $S P^{3} d^{2}$
$X e F_{6}$ $6$ + $1$ = $7$ $S P^{3} d^{3}$

So the correct option is [B]

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Similar questions

X e F_{2}, X e F_{3}^{+}, X e F_{4}, X e F_{6}

What is the hybridization of P in $P C l_{5} (s)$ ?

B r F_{5}

	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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