Step 1: Understanding the Topic
This question involves determining the molecular structure of various Xenon compounds to find how many lone pairs of electrons reside on the central Xenon atom. Xenon is a noble gas in Group 18, so it has 8 valence electrons, which it can use for bonding or retain as lone pairs.
Step 2: Key Approach - Electron Counting
For each compound, we will count the number of valence electrons Xenon uses for bonding. Each single bond uses one electron, and each double bond uses two. The remaining valence electrons on Xenon are then grouped into lone pairs (2 electrons per pair).
Step 3: Detailed Analysis of Each Compound
Xenon (Xe) starts with 8 valence electrons.
(i) $XeF_2$:
Xe forms 2 single bonds with 2 F atoms.
Electrons used in bonding = 2.
Electrons remaining on Xe = $8 - 2 = 6$.
Number of lone pairs = $6 / 2 = \mathbf{3}$.
(ii) $XeF_4$:
Xe forms 4 single bonds with 4 F atoms.
Electrons used in bonding = 4.
Electrons remaining on Xe = $8 - 4 = 4$.
Number of lone pairs = $4 / 2 = \mathbf{2}$.
(iii) $XeF_6$:
Xe forms 6 single bonds with 6 F atoms.
Electrons used in bonding = 6.
Electrons remaining on Xe = $8 - 6 = 2$.
Number of lone pairs = $2 / 2 = \mathbf{1}$.
(iv) $XeO_3$:
Oxygen is divalent, so Xe forms 3 double bonds with 3 O atoms.
Electrons used in bonding = $3 \times 2 = 6$.
Electrons remaining on Xe = $8 - 6 = 2$.
Number of lone pairs = $2 / 2 = \mathbf{1}$.
(v) $XeOF_4$:
Xe forms 1 double bond with O and 4 single bonds with F.
Electrons used in bonding = (1 $\times$ 2 for O) + (4 $\times$ 1 for F) = $2 + 4 = 6$.
Electrons remaining on Xe = $8 - 6 = 2$.
Number of lone pairs = $2 / 2 = \mathbf{1}$.
Step 4: Final Answer
By analyzing each compound, we find that the ones with exactly one lone pair on the central Xenon atom are:
\[
\boxed{XeF_6,\; XeO_3,\; \text{and } XeOF_4}
\]