Match the List-I with List-II.

List-I	List-II
A. XeO3 B. XeF2 C. XeOF4 D. XeF6	I. sp3d; linear II. sp3; pyramidal III. sp3d3; distorted octahedral IV. sp3d2; square pyramidal

Choose the correct answer from the options given below.

(1) A-II,   B-I,   C-IV,   D-III
(2) A-II,   B-I,   C-III,   D-IV
(3) A-IV,   B-II,   C-III,   D-I
(4) A-IV,   B-II,   C-I,   D-III

The correct option is (1) A-II, B-I, C-IV, D-III.

To determine the correct matching between the xenon compounds in List-I and their hybridization and shape in List-II, we can determine the steric number (number of valence shell electron pairs around the central xenon atom) for each molecule:

The formula for the steric number (SN) is given by:
$\mathrm{SN}=\frac{1}{2}\left[\right(\text{Valence electrons of Xe})+(\text{Number of monovalent atoms})-(\text{Charge on cation})+(\text{Charge on anion}\left)\right]$

Let's analyze each compound step-by-step:

A. XeO₃ (Xenon trioxide):
Xenon has 8 valence electrons. Oxygen is a divalent atom, so it does not count in the number of monovalent atoms.
$\mathrm{SN}=\frac{1}{2}[8+0]=4$
A steric number of 4 corresponds to sp³ hybridization.
Since there are 3 Xe-O double bonds, there are 3 bonding pairs and 1 lone pair on the xenon atom.
The geometry with 3 bond pairs and 1 lone pair in sp³ hybridization is pyramidal.
Therefore, A matches with II.

B. XeF₂ (Xenon difluoride):
Xenon has 8 valence electrons, and there are 2 monovalent fluorine atoms.
$\mathrm{SN}=\frac{1}{2}[8+2]=5$
A steric number of 5 corresponds to sp³d hybridization.
With 2 bonding pairs (Xe-F bonds), there are 3 lone pairs on the xenon atom.
To minimize repulsion, the 3 lone pairs occupy the equatorial positions of the trigonal bipyramid, and the 2 fluorine atoms occupy the axial positions.
Thus, the molecular shape is linear.
Therefore, B matches with I.

C. XeOF₄ (Xenon oxytetrafluoride):
Xenon has 8 valence electrons, oxygen is divalent (0), and there are 4 monovalent fluorine atoms.
$\mathrm{SN}=\frac{1}{2}[8+4]=6$
A steric number of 6 corresponds to sp³d² hybridization.
There are 5 bonding groups (1 Xe=O and 4 Xe-F bonds) and 1 lone pair on the xenon atom.
The arrangement of 5 bond pairs and 1 lone pair in an octahedral geometry results in a square pyramidal shape.
Therefore, C matches with IV.

D. XeF₆ (Xenon hexafluoride):
Xenon has 8 valence electrons, and there are 6 monovalent fluorine atoms.
$\mathrm{SN}=\frac{1}{2}[8+6]=7$
A steric number of 7 corresponds to sp³d³ hybridization.
There are 6 bonding pairs and 1 lone pair.
The presence of 6 bond pairs and 1 lone pair leads to a distorted octahedral shape (also referred to as capped octahedral).
Therefore, D matches with III.

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