Match List-I with List-II

	List-I		List-II
A.	$\text{XeO₃}$	(I)	$\text{sp³d; linear}$
B.	$\text{XeF₂}$	(II)	$\text{sp³; pyramidal}$
C.	$\text{XeOF₄}$	(III)	$\text{sp³d³; distorted octahedral}$
D.	$\text{XeF₆}$	(IV)	$\text{sp³d²; square pyramidal}$

Choose the correct answer from the options given below :

The correct answer is A-II, B-I, C-IV, D-III.

To match the xenon compounds in List-I with their hybridization and geometry in List-II, we can determine the steric number (number of bond pairs + number of lone pairs on the central Xenon atom) for each molecule:

A. ${\text{XeO}}_3$:
Xenon ($\text{Xe}$) has 8 valence electrons. In ${\text{XeO}}_3$, it forms three double bonds with three oxygen atoms (using 6 electrons). The remaining 2 valence electrons form 1 lone pair.
Steric number = 3 bond pairs ($\text{σ}$-bonds) + 1 lone pair = 4.
This corresponds to ${\text{sp}}^3$ hybridization. With one lone pair, the molecular geometry is pyramidal.
Therefore, A matches with (II).

B. ${\text{XeF}}_2$:
Xenon has 8 valence electrons. In ${\text{XeF}}_2$, it forms two single bonds with two fluorine atoms (using 2 electrons). The remaining 6 valence electrons form 3 lone pairs.
Steric number = 2 bond pairs + 3 lone pairs = 5.
This corresponds to ${\text{sp}}^3\text{d}$ hybridization. The three lone pairs occupy the equatorial positions to minimize repulsion, resulting in a linear geometry.
Therefore, B matches with (I).

C. ${\text{XeOF}}_4$:
Xenon has 8 valence electrons. In ${\text{XeOF}}_4$, it forms one double bond with oxygen (using 2 electrons) and four single bonds with fluorine atoms (using 4 electrons), using a total of 6 electrons. The remaining 2 valence electrons form 1 lone pair.
Steric number = 5 bond pairs ($\text{σ}$-bonds) + 1 lone pair = 6.
This corresponds to ${\text{sp}}^3{\text{d}}^2$ hybridization. With one lone pair, the molecular geometry is square pyramidal.
Therefore, C matches with (IV).

D. ${\text{XeF}}_6$:
Xenon has 8 valence electrons. In ${\text{XeF}}_6$, it forms six single bonds with six fluorine atoms (using 6 electrons). The remaining 2 valence electrons form 1 lone pair.
Steric number = 6 bond pairs + 1 lone pair = 7.
This corresponds to ${\text{sp}}^3{\text{d}}^3$ hybridization. Due to the presence of the lone pair, the geometry is a distorted octahedral.
Therefore, D matches with (III).

Combining all the matches, we get:
A-II, B-I, C-IV, D-III