/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 Why does xenon form stable compo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 22: Problem 31

Why does xenon form stable compounds with fluorine, whereas argon does not?

Short Answer

Expert verified
Xenon forms stable compounds with fluorine due to its relatively larger atomic size and lower ionization energy, which allows it to form bonds with highly electronegative elements such as fluorine. In contrast, argon remains unreactive with fluorine because its smaller atomic size and higher ionization energy prevent it from forming stable bonds.

Step by step solution

01

Noble Gases: Electronic Configuration

Noble gases have complete electron shells, making them unreactive under normal conditions. This is because they have achieved a stable electron configuration, and their low reactivity is attributed to a high ionization energy required to remove an electron from their outermost shell. The full electronic configuration of xenon (Xe) is \[1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \, 4s^2 \, 4p^6 \, 4d^{10} \, 5s^2 \, 5p^6\]. On the other hand, argon (Ar) has an electronic configuration of \[1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6\]. Their full electron shells give them stability and make them unreactive under normal conditions.
02

Xenon: Reactivity with Fluorine

Xenon, being a heavier noble gas, has a larger atomic size and weaker hold on its valence electrons. This leads to a relatively lower ionization energy, making it somewhat easier to remove an electron from its outermost shell, compared to lighter noble gases. In the presence of highly electronegative elements, such as fluorine (F), xenon can form compounds like XeF2, XeF4, and XeF6. Fluorine, due to its high electronegativity, attracts the valence electrons from xenon, making it possible for xenon to form bonds and ultimately create stable compounds.
03

Argon: Non-Reactivity with Fluorine

Argon, being a lighter noble gas, has a smaller atomic size, and a stronger hold on its valence electrons. This results in a higher ionization energy, making it very difficult to remove an electron from its outermost shell. As a result, argon remains unreactive with electronegative elements like fluorine, as it cannot form stable bonds. In conclusion, xenon can form stable compounds with fluorine due to its relatively larger atomic size and lower ionization energy, allowing it to form bonds with highly electronegative elements. On the other hand, argon remains unreactive with fluorine as its smaller atomic size and higher ionization energy prevent it from forming stable bonds.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The atomic and ionic radii of the first three group \(6 \mathrm{~A}\) elements are (a) Explain why the atomic radius increases in moving downward in the group. (b) Explain why the ionic radii are larger than the atomic radii. (c) Which of the three anjons would you expect to be the strongest base in water? Explain.

Write the formulas for the following compounds, and indicate the oxidation state of the group 4 A element or of boron in each: (a) boric acid, (b) silicon tetrabromide, (c) lead(II) chloride, (d) sodium tetraborate decahydrate \((\) borax \(),(e)\) boric oxide.

Boron nitride has a graphite-like structure with \(\mathrm{B}-\mathrm{N}\) bond distances of \(1.45 \AA\) within sheets and a separation of \(3.30 \AA\) between sheets. At high temperatures the \(B N\) assumes a diamondlike form that is almost as hard as diamond. Rationalize the similarity between \(\mathrm{BN}\) and elemental carbon.

Write complete balanced half-reactions for (a) reduction of nitrate ion to \(\mathrm{NO}\) in acidic solution, (b) oxidation of \(\mathrm{HNO}_{2}\) to \(\mathrm{NO}_{2}\) in acidic solution.

Write the chemical formula for each of the following compounds, and indicate the oxidation state of the halogen or noble-gas atom in each: (a) chlorate ion, (b) hydroiodic acid, (c) iodine trichloride, (d) sodium hypochlorite, (e) perchloric acid, (f) xenon tetrafluoride.
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.