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What is the atomic number (Z) of the noble gas that reacts with fluorine?
A) 54
B) 10
C) 18
D) 2

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Last updated date: 20th Jun 2024
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Answer
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Hint: Noble gas is very less reactive and enters in some of the chemical reactions only. To react with each other they should possess some similarities with other elements. And for the formation of a bond and involving in the chemical reactions the ionization enthalpy must be low.

Complete step by step solution:
So we know that the noble gases are also called as the inert gases since they do not react with the other elements as they do. And the cause of inertness toward other elements are because of their stable electronic configuration. The noble gases have the filled orbitals and hence less in energy and completely stable.
-The noble gases are
 Helium (He, Z=2)
Neon (Ne, Z=10),
Argon (Ar, Z=18),
Krypton (Kr, Z=30)
Xenon (Xe, Z=54)
Radon (Rn, Z=86)

The general electronic configuration of noble gas is $n{{s}^{2}}n{{p}^{6}}$ expect for He which is $n{{s}^{2}}$ configuration.
-Among these elements Xe possess similarities with oxygen like first ionization energy of xenon is almost similar to the oxygen both having values $1170 kJmo{{l}^{-1}}$ and $1180 kJmo{{l}^{-1}}$ respectively.
-The molecular diameters of both the oxygen and xenon is identical which gives a value of
400pm

- Neil Bartlett, a scientist studied these similarities and he put forward an argument that since oxygen combines with platinum hexafluoride, xenon should also form a similar compound with hexafluoride and he proved he was correct. He mixed deep-red vapors $Pt{{F}_{6}}$ with Xe which formed an orange-yellow solid at room temperature with a formula similar to the formulae of the compound formed by oxygen.
Hence the first stable noble gas compound was discovered and the reactivity study of noble gas gained the attention.
\[Pt{{F}_{6}}+{{O}_{2}}\to {{O}_{2}}^{+}{{\left[ Pt{{F}_{6}} \right]}^{-}}\]
\[Pt{{F}_{6}}+Xe\to X{{e}^{+}}{{\left[ Pt{{F}_{6}} \right]}^{-}}\]

-Another reason for the reactive is as the Xe atom is having a much greater size. It’s a larger molecule hence the impact or attraction of nucleus for the valence electrons will be less, hence very less amount of energy is required to remove the outer electrons, The first ionization enthalpy of the Xe is less and they react with the strong electron acceptor atom like fluorine which only needs an electron more to complete its shell. And as we move down the group in a periodic table first ionization enthalpy decreases.

So Xe reacts with Fluorine to gives xenon fluorides like $Xe{{F}_{2}},Xe{{F}_{4}},Xe{{F}_{6}}$
\[Xe+{{F}_{2}}\xrightarrow{{{400}^{\circ }}C}Xe{{F}_{2}}\] (In 2:1 ratio)
\[Xe+2{{F}_{2}}\xrightarrow[Ni]{{{400}^{\circ }}C}Xe{{F}_{4}}\] (In 1:5 ratio)

\[Xe+3{{F}_{2}}\xrightarrow[50atm]{{{300}^{\circ }}C}Xe{{F}_{6}}\] (In 1:20 ratio)
So Xe is the only noble gas known to react with fluorine having the atomic number 54.
So, the correct answer is “Option A”.

Note: As we have stated that the first ionization enthalpy decreases down the group then radon should be more reactive than xenon and radon is a larger molecule than xenon. But radon is a radioactive element, it has a very short half-life and hence its compounds are not studied or known yet.