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Question: Which of the following fluorides of xenon is impossible? (A) \(Xe{F_2}\) (B) \(Xe{F_3}\) (C) \...

Which of the following fluorides of xenon is impossible?
(A) XeF2Xe{F_2}
(B) XeF3Xe{F_3}
(C) XeF4Xe{F_4}
(D) XeF6Xe{F_6}

Explanation

Solution

Xenon is a chemical element with atomic number 5454. 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]4d105s25p6[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 XeF2Xe{F_2}, XeF4Xe{F_4} and XeF6Xe{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 2,4,62,4,6 half-filled orbitals.
It cannot combine with an odd number of F –atoms.
Thus, the formation of XeF3Xe{F_3} and XeF5Xe{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 4.54.5 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.