To determine the number of lone pairs on the central Xe atom in each compound, we first need to find the total number of valence electrons around Xe and then subtract the electrons used in bonding.

Step 1: Determine valence electrons for Xe and F.
Xenon (Xe) is a noble gas, so it has 8 valence electrons.
Fluorine (F) is a halogen, so it has 7 valence electrons.

Step 2: Calculate lone pairs for ${\mathrm{XeF}}_2$.
Number of bond pairs = 2 (for two Xe-F bonds).
Electrons used in bonding = $2\times 1=2$ electrons (each bond uses 1 electron from Xe).
Remaining valence electrons on Xe = $8-2=6$ electrons.
Number of lone pairs = $6/2=3$ lone pairs.

Step 3: Calculate lone pairs for ${\mathrm{XeF}}_4$.
Number of bond pairs = 4 (for four Xe-F bonds).
Electrons used in bonding = $4\times 1=4$ electrons.
Remaining valence electrons on Xe = $8-4=4$ electrons.
Number of lone pairs = $4/2=2$ lone pairs.

Step 4: Calculate lone pairs for ${\mathrm{XeF}}_6$.
Number of bond pairs = 6 (for six Xe-F bonds).
Electrons used in bonding = $6\times 1=6$ electrons.
Remaining valence electrons on Xe = $8-6=2$ electrons.
Number of lone pairs = $2/2=1$ lone pair.

Thus, the number of lone pairs on Xe in ${\mathrm{XeF}}_2,{\mathrm{XeF}}_4,{\mathrm{XeF}}_6$ is 3, 2, and 1 respectively.

Option Analysis:

• A) $2,3,1$: Incorrect. The number of lone pairs for ${\mathrm{XeF}}_2$ is 3, not 2, and for ${\mathrm{XeF}}_4$ is 2, not 3.
• B) $1,2,3$: Incorrect. The order is reversed for ${\mathrm{XeF}}_2$ and ${\mathrm{XeF}}_6$.
• C) $4,1,2$: Incorrect. All values are incorrect.
• D) $3,2,1$: Correct. This matches our calculated values.

Correct Answer: (D)