Which of the following fluorides of xenon is impossible?
(A) $Xe{F_2}$
(B) $Xe{F_3}$
(C) $Xe{F_4}$
(D) $Xe{F_6}$

Hint: Xenon is a chemical element with atomic number $54$. It can only combine with an even number of F atoms to form xenon fluorides and not with odd numbers of F atoms.

Complete step by step solution:
Xenon is an inert gas. Its electronic configuration is $[Kr]4{d^{10}}5{s^2}5{p^6}$. All orbitals that are filled have paired electrons.
Xenon can combine with an even number of F atoms to form $Xe{F_2}$, $Xe{F_4}$ and $Xe{F_6}$.
This is because the promotion of 1, 2, or 3 electrons from the 5p filled orbitals to 5d vacant orbitals will give rise to $2,4,6$ half-filled orbitals.
It cannot combine with an odd number of F –atoms.
Thus, the formation of $Xe{F_3}$ and $Xe{F_5}$ is not possible.

Hence, option B is correct.

Note: Xenon is obtained commercially as a by-product of the separation of air into oxygen and nitrogen. It is $4.5$ times heavier than Earth’s atmosphere (which consists of a mixture of a number of gaseous elements and compounds). Its mass comes from its nucleus, which contains 54 protons and a varying (but similar) number of neutrons.