Question

Which one of the following does not exist?
(A) $XeO{F}_4$
(B) $Ne{F}_2$
(C) $Xe{F}_2$
(D) $Xe{F}_6$

Answer 1

Hint: By observation, we can say that all the molecules are noble gas complexes; three of them are of xenon and one of them is of neon.
- Also, a number of xenon compounds with elements like fluorine and oxygen are known to exist in nature.

Complete Solution :
Let us discuss the compounds as stated above in detail;
Xenon forms various compounds with fluorine and oxygen in nature. Some of them can be stated with the reactions are:
$X{e}_{(g)}+F_{2(g)}\underset{1bar}{\overset{673K}{\to }}Xe{F}_{2(s)}$ ; here Xe is in excess.

$X{e}_{(g)}+3F_{2(g)}\underset{60-70bar}{\overset{573K}{\to }}Xe{F}_{6(s)}$ ; here Xe and fluorine is in the ratio as 1 : 20.

$Xe{F}_6+H_{2}O\to XeO{F}_4+2HF$ ; the above-mentioned compound reacts with water to give this reaction.
- From the above reactions, we can say that the compounds $XeO{F}_4$, $Xe{F}_2$ and $Xe{F}_6$ exists in nature. Whereas, $Ne{F}_2$ does not, as it would require d orbital for bonding and Ne does not have it presently as it belongs to the second period. And it cannot invest so much energy to excite the electrons for the formation of compounds as it would hence be unstable and return to the stable forms even if accidently formed.
So, the correct answer is “Option B”.

Note: In nature, compounds of xenon are known; but no true compounds of argon, helium and neon are known due to their extreme inert behaviour.
Also, $F_2$ can oxidise Xe to $X{e}^{2+}$ but cannot oxidise Ne to $N{e}^{2+}$ as, sum of first and second ionisation enthalpies of Ne are higher than that of Xe.