Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 76

(a) Why does xenon react with fluorine, whereas neon does not? (b) Using appropriate reference sources, look up the bond lengths of \(\mathrm{Xe}-\mathrm{F}\) bonds in several molecules. How do these numbers compare to the bond lengths calculated from the atomic radii of the elements?

Short Answer

Expert verified
(a) Xenon reacts with fluorine due to its larger atomic size, which allows it to accommodate additional electrons and form stable compounds like XeF2, XeF4, and XeF6. In contrast, neon's full electron shell is firmly held by the nucleus, preventing it from reacting with fluorine. (b) Experimental bond lengths for Xe-F in several molecules, such as XeF2 (\(1.94~\mathrm{Å}\)), XeF4 (\(1.92~\mathrm{Å}\)), and XeF6 (\(1.98~\mathrm{Å}\)), are close to the theoretical bond length calculated from the atomic radii of the elements (\(1.95~\mathrm{Å}\)), indicating that the approximation using atomic radii is reasonably accurate in predicting the bond lengths of these xenon fluoride compounds.

Step by step solution

01

Discuss reactivity of xenon and neon

To begin, let's first discuss the reactivity of xenon (Xe) and neon (Ne) with fluorine (F). Both Xe and Ne are noble gases, known for their lack of reactivity due to their full outer electron shell. However, Xe is an exception to this general trend in noble gases because it has a larger atomic size which allows it to accommodate additional electrons and form stable compounds. Xe forms various compounds with fluorine, such as XeF2, XeF4, and XeF6. In contrast, neon's full electron shell is firmly held by the nucleus and doesn't provide sufficient energy to accommodate additional electrons, which means it remains unreactive with fluorine.
02

Research Xe-F bond lengths in molecules

Using appropriate reference sources, look up Xe-F bond lengths in several xenon fluoride compounds. Some examples are: 1. XeF2: \(1.94~\mathrm{Å}\) (Angstrom) 2. XeF4: \(1.92~\mathrm{Å}\) 3. XeF6: \(1.98~\mathrm{Å}\) These bond lengths may vary slightly depending on the source, but they should be in similar ranges.
03

Calculate bond lengths from atomic radii

Next, we will calculate the theoretical bond lengths for Xe-F bonds using atomic radii of Xe and F. The atomic radii for Xe and F are approximately: - Xe: \(1.31~\mathrm{Å}\) - F: \(0.64~\mathrm{Å}\) To calculate the bond length, add the atomic radii of the two atoms involved in the bond: Theoretical Xe-F bond length = Atomic radius of Xe + Atomic radius of F = \(1.31~\mathrm{Å} + 0.64~\mathrm{Å} = 1.95~\mathrm{Å}\)
04

Compare experimental values to theoretical bond lengths

Now, let's compare the bond lengths found in Step 2 to the theoretical bond length calculated in Step 3. 1. XeF2: Experimental \(1.94~\mathrm{Å}\), Theoretical \(1.95~\mathrm{Å}\) 2. XeF4: Experimental \(1.92~\mathrm{Å}\), Theoretical \(1.95~\mathrm{Å}\) 3. XeF6: Experimental \(1.98~\mathrm{Å}\), Theoretical \(1.95~\mathrm{Å}\) As observed, the experimental bond lengths for Xe-F bonds in different molecules are close to the theoretical bond length calculated from the atomic radii of the elements. This indicates that the approximation using atomic radii is reasonably accurate in predicting the bond lengths of these xenon fluoride compounds.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Discussing this chapter, a classmate says, "An element that commonly forms a cation is a metal." Do you agree or disagree?

Predict whether each of the following oxides is ionic or molecular: \(\operatorname{Sn} \mathrm{O}_{2}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{CO}_{2}, \mathrm{Li}_{2} \mathrm{O}, \mathrm{Fe}_{2} \mathrm{O}_{3}, \mathrm{H}_{2} \mathrm{O} .\)

Among elements 1-18, which element or elements have the smallest effective nuclear charge if we use Equation 7.1 to calculate \(Z_{\text { eff}}\)? Which element or elements have the largest effective nuclear charge?

The As\(-\)As bond length in elemental arsenic is 2.48 A. The \(\mathrm{Cl}-\mathrm{Cl}\) bond length in \(\mathrm{Cl}_{2}\) is 1.99 A. (a) Based on these data, what is the predicted \(\mathrm{As}-\mathrm{Cl}\) bond length in arsenic trichloride, \(A s C l_{3},\) in which each of the three Cl atoms is bonded to the As atom? (b) What bond length is predicted for \(A s C l_{3},\) , using the atomic radii in Figure 7.7?

Give three examples of ions that have an electron configuration of \(n d^{8}(n=3,4,5, \ldots).\)
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

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.

Sign-up for free