Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 22

What type(s) of bonding would be expected for each of the following materials: solid xenon, calcium fluoride \(\left(\mathrm{CaF}_{2}\right),\) bronze, cadmium tel luride (CdTe), rubber, and tungsten?

Short Answer

Expert verified
Question: Identify the type of bonding expected in each material: solid xenon, calcium fluoride (CaF2), bronze, cadmium telluride (CdTe), rubber, and tungsten. Answer: The type of bonding in each material is as follows: - Solid xenon: van der Waals (dispersion) bonding - Calcium fluoride (CaF2): ionic bonding - Bronze: metallic bonding - Cadmium telluride (CdTe): mixed bonding (covalent and ionic) - Rubber: covalent bonding with partial hydrogen bonding - Tungsten: metallic bonding

Step by step solution

01

Solid Xenon

Xenon belongs to the noble gases group in the periodic table and exists as individual atoms. It forms a solid through weak London dispersion forces (Van der Waals forces) between xenon atoms. So, the type of bonding in solid xenon is van der Waals or dispersion bonding.
02

Calcium Fluoride (CaF2)

Calcium Fluoride consists of calcium (Ca) and fluorine (F) elements. Calcium, a metal, donates electrons to fluorine, a non-metal, to achieve a stable electronic configuration. This results in the transfer of electrons and formation of positive calcium ions (Ca²⁺) and negative fluoride ions (F⁻). The electrostatic force between these ions results in ionic bonding, so the type of bonding in Calcium Fluoride is ionic bonding.
03

Bronze

Bronze is an alloy primarily made from copper (Cu) and tin (Sn). Alloys are composed of two or more metallic elements, generally with metallic properties. In a metallic alloy, atoms of the different metals form a metallic bond by sharing a sea of free-moving electrons. Therefore, the type of bonding in bronze is metallic bonding.
04

Cadmium Telluride (CdTe)

Cadmium Telluride consists of cadmium (Cd) and tellurium (Te) elements. Both elements have semi-conducting features, and they form a solid crystal lattice in a 1:1 ratio. The bonding in CdTe is a combination of ionic and covalent character, so it exhibits mixed bonding (covalent and ionic).
05

Rubber

Rubber is an organic polymer made up of large chain molecules of carbon and hydrogen. The atoms in the molecule are bonded by sharing electrons, creating covalent bonding. Additionally, rubber exhibits some hydrogen bonding in its structure, so one can say the type of bonding in rubber is covalent bonding with partial hydrogen bonding.
06

Tungsten

Tungsten is a metallic element. In metals, atoms share electrons and form a "sea" of delocalized electrons, resulting in a metallic bond. Therefore, the type of bonding in tungsten is metallic bonding.

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

Relative to electrons and electron states, what does each of the four quantum numbers specify?

6 Allowed values for the quantum numbers of electrons are as follows: $$ \begin{aligned} n &=1,2,3, \ldots \\ l &=0,1,2,3, \ldots, n-1 \\ m_{l} &=0, \pm 1, \pm 2, \pm 3, \ldots, \pm l \\ m_{s} &=\pm \frac{1}{2} \end{aligned} $$ The relationships between \(n\) and the shell designations are noted in Table 2.1. Relative to the subshells, \(l=0\) corresponds to an \(s\) subshell \(l=1\) corresponds to a \(p\) subshell \(l=2\) corresponds to a \(d\) subshell \(l=3\) corresponds to an \(f\) subshell For the \(K\) shell, the four quantum numbers for each of the two electrons in the \(1 s\) state, in the order of \(n l m_{i} m_{s}\), are \(100 \frac{1}{2}\) and \(100\left(-\frac{1}{2}\right)\) Write the four quantum numbers for all of the electrons in the \(L\) and \(M\) shells, and note which correspond to the \(s, p\), and \(d\) subshells.

(a) Cite two important quantum-mechanical concepts associated with the Bohr model of the atom. (b) Cite two important additional refinements that resulted from the wave- mechanical atomic model.

With regard to electron configuration, what do all the elements in Group VIIA of the periodic table have in common?

For \(\mathrm{a} \mathrm{K}^{+}-\mathrm{Cl}^{-}\)ion pair, attractive and repulsive energies \(E_{A}\) and \(E_{R}\), respectively, depend on the distance between the ions \(r\), according to $$ \begin{aligned} E_{A} &=-\frac{1.436}{r} \\ E_{R} &=\frac{5.86 \times 10^{-6}}{r^{9}} \end{aligned} $$ For these expressions, energies are expressed in electron volts per \(\mathrm{K}^{+}-\mathrm{Cl}^{-}\)pair, and \(r\) is the distance in nanometers. The net energy \(E_{N}\) is just the sum of the preceding two expressions. (a) Superimpose on a single plot \(E_{N}, E_{R}\), and \(E_{A}\) versus \(r\) up to \(1.0 \mathrm{~nm}\). (b) On the basis of this plot, determine (i) the equilibrium spacing \(r_{0}\) between the \(\mathrm{K}^{+}\)and \(\mathrm{Cl}^{-}\)ions, and (ii) the magnitude of the bonding energy \(E_{0}\) between the two ions. (c) Mathematically determine the \(r_{0}\) and \(E_{0}\) values using the solutions to Problem \(2.14\) and compare these with the graphical results from part (b).
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Torque and Rotational Motion

Read Explanation

Classical Mechanics

Read Explanation

Fluids

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