Match Xenon Compounds with Geometries and Lone Pair Numbers

To determine the correct matching according to the VSEPR (Valence Shell Electron Pair Repulsion) model, we calculate the steric number (number of $\sigma$-bonds $+$ number of lone pairs) and determine the electron pair geometry and the number of lone pairs on the central xenon ($\text{Xe}$) atom for each compound:

1. (P) $\text{XeF}_2$:

   • Xenon ($\text{Xe}$) has $8$ valence electrons.
   • It forms $2$ single $\sigma$-bonds with two fluorine atoms, using $2$ electrons.
   • Remaining electrons on $\text{Xe} = 8 - 2 = 6$ electrons, which equals $3$ lone pairs.
   • $\text{Steric Number} = 2\text{ ($\sigma$-bonds)} + 3\text{ (lone pairs)} = 5$.
   • An arrangement of $5$ electron pairs corresponds to a trigonal bipyramidal electron geometry with $3$ lone pairs of electrons.
   • Therefore, (P) matches with (5).

2. (Q) $\text{XeF}_4$:

   • Xenon ($\text{Xe}$) has $8$ valence electrons.
   • It forms $4$ single $\sigma$-bonds with four fluorine atoms, using $4$ electrons.
   • Remaining electrons on $\text{Xe} = 8 - 4 = 4$ electrons, which equals $2$ lone pairs.
   • $\text{Steric Number} = 4\text{ (}\sigma\text{-bonds)} + 2\text{ (lone pairs)} = 6$.
   • An arrangement of $6$ electron pairs corresponds to an octahedral electron geometry with $2$ lone pairs of electrons.
   • Therefore, (Q) matches with (3).

3. (R) $\text{XeO}_3$:

   • Xenon ($\text{Xe}$) has $8$ valence electrons.
   • It forms $3$ double bonds (each consisting of $1\ \sigma$-bond and $1\ \pi$-bond) with three oxygen atoms, using $3 \times 2 = 6$ electrons.
   • Remaining electrons on $\text{Xe} = 8 - 6 = 2$ electrons, which equals $1$ lone pair.
   • $\text{Steric Number} = 3\text{ (}\sigma\text{-bonds)} + 1\text{ (lone pair)} = 4$.
   • An arrangement of $4$ electron pairs corresponds to a tetrahedral electron geometry with $1$ lone pair of electrons.
   • Therefore, (R) matches with (2).

4. (S) $\text{XeO}_3\text{F}_2$:

   • Xenon ($\text{Xe}$) has $8$ valence electrons.
   • It forms $3$ double bonds with three oxygen atoms (using $6$ electrons) and $2$ single bonds with two fluorine atoms (using $2$ electrons).
   • Total valence electrons used = $6 + 2 = 8$ electrons.
   • Remaining electrons on $\text{Xe} = 0$, which equals $0$ lone pairs.
   • $\text{Steric Number} = 5\text{ (}\sigma\text{-bonds)} + 0\text{ (lone pairs)} = 5$.
   • An arrangement of $5$ electron pairs corresponds to a trigonal bipyramidal geometry with no lone pair of electrons.
   • Therefore, (S) matches with (4).

Thus, the correct mapping is: $\text{P} \rightarrow 5, \quad \text{Q} \rightarrow 3, \quad \text{R} \rightarrow 2, \quad \text{S} \rightarrow 4$

This corresponds to Option B.