In Xe{F_2} , Xe{F_4} and Xe{F_6} , the number of lone pairs on Xe are respectively :A. 2,3,1 B. 1,2,3 C. 4,1,2 D. 3,2,1

To determine the number of lone pairs on xenon (Xe) in each of the compounds XeF2, XeF4, and XeF6, we need to consider the valence electrons of xenon and the number of bonds it forms with fluorine atoms.

XeF2:
In XeF2, xenon forms 2 bonds with two fluorine atoms.
Xenon has 8 valence electrons (Group 18), and it uses 2 electrons for each bond with fluorine.
So, in XeF2, xenon uses 4 valence electrons for bonding (2 bonds * 2 electrons/bond = 4 electrons).
Therefore, the number of lone pairs on xenon in XeF2 is 8 - 4 = 4 lone pairs.

XeF4:
In XeF4, xenon forms 4 bonds with four fluorine atoms.
Using the same reasoning as above, xenon uses 8 valence electrons for bonding in XeF4.
Therefore, the number of lone pairs on xenon in XeF4 is 8 - 8 = 0 lone pairs.

XeF6:
In XeF6, xenon forms 6 bonds with six fluorine atoms.
Using the same reasoning as above, xenon uses 12 valence electrons for bonding in XeF6.
Therefore, the number of lone pairs on xenon in XeF6 is 8 - 12 = -4, which doesn't make sense in this context.

So, the correct answer is:
A. 4, 0, -4 (but -4 is not a valid option, so the answer should be A. 2, 3, 1)