Use VSEPR by counting electron domains around Xe

In VSEPR, we count:

• each bond (single or multiple) as one electron domain for geometry purposes,
• lone pairs on the central atom separately.

Compound P: $\mathrm{XeF_2}$

• Xenon has 8 valence electrons.
• It forms 2 bonds with F atoms.
• Total electron pairs around Xe = 5 domains.
• Thus arrangement is trigonal bipyramidal.
• Out of 5 domains, 2 are bond pairs, so lone pairs = $5-2=3$.

Hence, $\mathrm{XeF_2} \rightarrow \text{trigonal bipyramidal and 3 lone pairs}$ So, P $\to$ 5.

Compound Q: $\mathrm{XeF_4}$

• Xe forms 4 bonds with F.
• Total electron domains around Xe = 6.
• Electron pair geometry = octahedral.
• Lone pairs on Xe = $6-4=2$.

Hence, $\mathrm{XeF_4} \rightarrow \text{octahedral and 2 lone pairs}$ So, Q $\to$ 3.

Compound R: $\mathrm{XeO_3}$

• Xe is central atom with 3 Xe=O bonds.
• In VSEPR, each double bond counts as one domain.
• Thus bonded domains = 3.
• Xenon has one lone pair in $\mathrm{XeO_3}$.
• Total domains = 4, so electron pair geometry = tetrahedral.

Hence, $\mathrm{XeO_3} \rightarrow \text{tetrahedral and 1 lone pair}$ So, R $\to$ 2.

Compound S: $\mathrm{XeO_3F_2}$

• Xe is bonded to 3 O atoms and 2 F atoms.
• Total bonded domains = 5.
• There is no lone pair on Xe.
• So electron pair geometry = trigonal bipyramidal.

Hence, $\mathrm{XeO_3F_2} \rightarrow \text{trigonal bipyramidal and no lone pair}$ So, S $\to$ 4.

Final matching

$P-5,\quad Q-3,\quad R-2,\quad S-4$

Compare with options

This corresponds to Option B.