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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

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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 (Ca虏鈦�) and negative fluoride ions (F鈦�). 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.

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Most popular questions from this chapter

Sodium chloride \((\mathrm{NaCl})\) exhibits predominantly ionic bonding. The \(\mathrm{Na}^{+}\)and \(\mathrm{Cl}^{-}\)ions have electron structures that are identical to which two inert gases?

Calculate the force of attraction between a \(\mathrm{K}^{+}\) and an \(\mathrm{O}^{2-}\) ion whose centers are separated by a distance of \(1.5 \mathrm{~nm}\).

The net potential energy between two adjacent ions, \(E_{N}\), may be represented by the sum of Equations \(2.8\) and \(2.9\); that is, $$ E_{N}=-\frac{A}{r}+\frac{B}{r^{n}} $$ Calculate the bonding energy \(E_{0}\) in terms of the parameters \(A, B\), and \(n\) using the following procedure: 1\. Differentiate \(E_{N}\) with respect to \(r\), and then set the resulting expression equal to zero, because the curve of \(E_{N}\) versus \(r\) is a minimum at \(E_{0 \text { - }}\) 2\. Solve for \(r\) in terms of \(A, B\), and \(n\), which yields \(r_{0}\), the equilibrium interionic spacing. 3\. Determine the expression for \(E_{0}\) by substituting \(r_{0}\) into Equation \(2.11\).

The net potential energy \(E_{N}\) between two adjacent ions is sometimes represented by the expression $$ E_{N}=-\frac{C}{r}+D \exp \left(-\frac{r}{\rho}\right) $$ in which \(r\) is the interionic separation and \(C\), \(D\), and \(\rho\) are constants whose values depend on the specific material. (a) Derive an expression for the bonding energy \(E_{0}\) in terms of the equilibrium interionic separation \(r_{0}\) and the constants \(D\) and \(\rho\) using the following procedure: 1\. Differentiate \(E_{N}\) with respect to \(r\) and set the resulting expression equal to zero. 2\. Solve for \(C\) in terms of \(D, \rho\), and \(r_{0}\) - 3\. Determine the expression for \(E_{0}\) by substitution for \(C\) in Equation \(2.12\). (b) Derive another expression for \(E_{0}\) in terms of \(r_{0}, C\), and \(\rho\) using a procedure analogous to the one outlined in part (a).

Compute the percent ionic character of the interatomic bonds for the following compounds: \(\mathrm{TiO}_{2}, \mathrm{ZnTe}, \mathrm{CsCl}, \mathrm{InSb}\), and \(\mathrm{MgCl}_{2}\).
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