Based on VSEPR model, match the xenon compounds given in List-I with the corresponding geometries and the number of lone pairs on xenon given in List-II and choose the correct option.

List-I	List-II
(P) XeF2	(1) Trigonal bipyramidal and two lone pair of electrons
(Q) XeF4	(2) Tetrahedral and one lone pair of electrons
(R) XeO3	(3) Octahedral and two lone pair of electrons
(S) XeO3F2	(4) Trigonal bipyramidal and no lone pair of electrons
	(5) Trigonal bipyramidal and three lone pair of electrons

The correct option is P-5, Q-3, R-2, S-4.

To determine the correct match between the xenon compounds (List-I) and their corresponding electron geometries and number of lone pairs (List-II), we can use the Valence Shell Electron Pair Repulsion (VSEPR) model.

Xenon (Xe) is a noble gas with 8 valence electrons in its outermost shell. The steric number (SN), which determines the electron-pair geometry, is calculated as:
$\text{Steric Number (SN)}=\text{Number of}\sigma \text{-bonds}+\text{Number of lone pairs on the central atom}$

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

(P) XeF₂ (Xenon difluoride):
- Fluorine (F) forms a single bond (monovalent).
- Number of bonding electrons = 2 (used for 2 Xe-F single bonds).
- Remaining valence electrons on Xe = 8 - 2 = 6 electrons (which form 3 lone pairs).
- Steric Number = 2 ($\sigma$-bonds) + 3 (lone pairs) = 5.
- For SN = 5, the electron-pair geometry is trigonal bipyramidal.
- Hence, XeF₂ has a trigonal bipyramidal geometry with three lone pairs.
Match: P-5

(Q) XeF₄ (Xenon tetrafluoride):
- Number of bonding electrons = 4 (used for 4 Xe-F single bonds).
- Remaining valence electrons on Xe = 8 - 4 = 4 electrons (which form 2 lone pairs).
- Steric Number = 4 ($\sigma$-bonds) + 2 (lone pairs) = 6.
- For SN = 6, the electron-pair geometry is octahedral.
- Hence, XeF₄ has an octahedral geometry with two lone pairs.
Match: Q-3

(R) XeO₃ (Xenon trioxide):
- Oxygen (O) forms double bonds (each oxygen atom forms 1 $\sigma$-bond and 1 $\pi$-bond).
- Number of bonding electrons = 6 (2 electrons shared per oxygen atom for 3 double bonds).
- Remaining valence electrons on Xe = 8 - 6 = 2 electrons (which form 1 lone pair).
- Steric Number = 3 ($\sigma$-bonds) + 1 (lone pair) = 4.
- For SN = 4, the electron-pair geometry is tetrahedral.
- Hence, XeO₃ has a tetrahedral geometry with one lone pair.
Match: R-2

(S) XeO₃F₂ (Xenon trioxide difluoride):
- Oxygen atoms form 3 double bonds (sharing 6 electrons).
- Fluorine atoms form 2 single bonds (sharing 2 electrons).
- Total bonding electrons = 6 + 2 = 8 electrons.
- Remaining valence electrons on Xe = 8 - 8 = 0 (no lone pairs).
- Steric Number = 5 (3 $\sigma$-bonds to oxygen + 2 $\sigma$-bonds to fluorine) + 0 (lone pairs) = 5.
- For SN = 5, the electron-pair geometry is trigonal bipyramidal.
- Hence, XeO₃F₂ has a trigonal bipyramidal geometry with no lone pairs.
Match: S-4

Combining the matches gives: P-5, Q-3, R-2, S-4.