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Chapter 22: Problem 12

Provide an explanation for the inability of \(\mathrm{O}_{2}\) to react directly with xenon.

Short Answer

Expert verified
Xenon, being a noble gas, has a fully filled outermost electron shell, making it stable and less reactive. Oxygen, particularly \(\mathrm{O}_{2}\), is also stable and doesn't readily react. Because of the stability of both, \(\mathrm{O}_{2}\) cannot react directly with xenon under standard conditions.

Step by step solution

01

Understanding the nature of xenon

Xenon is a noble gas. The noble gases, are located in group 18 of the periodic table. These elements are characterized by having their outermost electron shells fully occupied. Specifically for xenon, its outermost shell has a stable electron configuration with 8 electrons.
02

Explaining the non-reactivity of xenon

Because xenon, like other noble gases, has a full set of electrons in its outermost shell, it is generally unreactive. The electron shell structure of xenon makes it stable and not eager to accept or donate electrons, impeding bond formation. This contributes to its inability to directly react with other elements, including oxygen.
03

Understanding the nature of oxygen

Elemental oxygen exists as \(\mathrm{O}_{2}\), a diatomic molecule. This molecule is stable and does not readily break apart to form other compounds.
04

Explaining the inability of \(\mathrm{O}_{2}\) to react directly with xenon

Given the highly stable nature of both xenon and oxygen, neither has a tendency to break their current stable state and form a new bond. Therefore, \(\mathrm{O}_{2}\) doesn't react directly with xenon under standard conditions.

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Most popular questions from this chapter

Write an appropriate formula for each of the following: (a) hydroxylamine; (b) calcium hydrogen phosphate; (c) lithium nitride.

Despite the fact that it has the higher molecular mass, XeO \(_{4}\) exists as a gas at \(298 \mathrm{K},\) whereas \(\mathrm{XeO}_{3}\) is a solid. Give a plausible explanation for this observation.

Write plausible half-equations and a balanced oxidation-reduction equation for the disproportionation of \(\mathrm{XeF}_{4}\) to \(\mathrm{Xe}\) and \(\mathrm{XeO}_{3}\) in aqueous acidic solution. Xe and \(\mathrm{XeO}_{3}\) are produced in a 2: 1 mol ratio, and \(\mathrm{O}_{2}(\mathrm{g})\) is also produced.

Explain why the volumes of \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g})\) obtained in the electrolysis of water are not the same.

The following bond energies are given for \(298 \mathrm{K}: \mathrm{O}_{2}\) \(498 ; \mathrm{N}_{2}, 946 ; \mathrm{F}_{2}, 159 ; \mathrm{Cl}_{2}, 243 ;\) ClF, \(251 ; \mathrm{OF}\left(\text { in } \mathrm{OF}_{2}\right)\) \(213 ; \mathrm{ClO}\left(\operatorname{in} \mathrm{Cl}_{2} \mathrm{O}\right), 205 ;\) and \(\mathrm{NF}\left(\mathrm{in} \mathrm{NF}_{3}\right), 280 \mathrm{kJmol}^{-14}\) Calculate \(\Delta H_{f}\) at \(298 \mathrm{K}\) for \(1 \mathrm{mol}\) of \((\mathrm{a}) \mathrm{ClF}(\mathrm{g})\) (b) \(\mathrm{OF}_{2}(\mathrm{g}) ;(\mathrm{c}) \mathrm{Cl}_{2} \mathrm{O}(\mathrm{g}) ;(\mathrm{d}) \mathrm{NF}_{3}(\mathrm{g})\).
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