© 2012 pearson education, inc. chapter 12 solids and modern materials

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© 2012 Pearson Education, Inc. Chapter 12 Solids and Modern Materials

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Page 1: © 2012 Pearson Education, Inc. Chapter 12 Solids and Modern Materials

© 2012 Pearson Education, Inc.

Chapter 12

Solids and Modern Materials

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Bonding in Solids• There are four general types of solids.

• Metallic solids share a network of highly delocalized electrons.

• Ionic solids are sets of cations and anions mutually attracted to one another.

• Covalent-network solids are joined by an extensive network of covalent bonds.

• Molecular solids are discrete molecules that are linked to one another only by van der Waals forces.

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Bonding in Solids

• In crystalline solids atoms are arranged in a very regular pattern.

• Amorphous solids are characterized by a distinct lack of order in the arrangement of atoms.

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Crystal Lattices

One can deduce the pattern in a crystalline solid by thinking of the substance as a lattice of repeating shapes formed by the atoms in the crystal.

The individual shapes of the lattice, then, form "tiles," or unit cells, that must fill the entire space of the substance.

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Crystal Lattices

• There are seven basic three-dimensional lattices:– Cubic– Tetragonal– Orthorhombic– Rhombohedral– Hexagonal– Monoclinic– Triclinic

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A. Cubic B. Monoclinic

C. Orthorhombic D. Tetragonal

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Crystal Lattices

Within each major lattice type, additional types are generated by placing lattice points in the center of the unit cell or on the faces of the unit cell.

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Crystal LatticesOnce one places atoms within a unit cell, the structure of the compound can be seen by bonding the atoms to one another across unit cells.

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Metallic Structure

The structures of many metals conform to one of the cubic unit cells.

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Cubic StructuresOne can determine how many atoms are within each unit cell which lattice points the atoms occupy.

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Close Packing

The atoms in a crystal pack as close together as they can based on the respective sizes of the atoms.

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A. Cubic latticeB. Trigonal latticeC. Rhombic latticeD. Hexagonal lattice

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Sample Exercise 12.1 Calculating Packing Efficiency

It is not possible to pack spheres together without leaving some void spaces between the spheres. Packing efficiency is the fraction of space in a crystal that is actually occupied by atoms. Determine the packing efficiency of a face-centered cubic metal.

Practice ExerciseDetermine the packing efficiency by calculating the fraction of space occupied by atoms in abody-centered cubic metal.

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A. Packing efficiency increases as the number of nearest neighbors decreases.

B. Packing efficiency increases as the the volume occupied by atoms decreases compared to the volume of the unit cell.

C. Packing efficiency decreases as the number of nearest neighbors decreases.

D. Packing efficiency decreases as the size of the nearest neighbors decreases.

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Alloys• Alloys are combinations of two or more

elements, the majority of which are metals.• Adding a second (or third) element changes

the properties of the mixture to suit different purposes.

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Alloys

• In substitutional alloys, a second element takes the place of a metal atom.

• In interstitial alloys, a second element fills a space in the lattice of metal atoms.

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A. The solvent in a solid solution is the one with the majoritycomponent and the solute is the one with a minority component.

B. The solvent in a solid solution is the one with the minoritycomponent and the solute is the one with a majority component.

C. The solute in a solid solution is the one of smaller size and thesolvent is the one of larger size.

D. The solute in a solid solution is the one of larger size and thesolvent is the one of smaller size.

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A. Interstitial alloy, because B has a small size and can occupy “holes” between Pd atoms.

B. Substitutional alloy, because B has a size that is sufficiently large to occupy positions held by Pd atoms.

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A. A unit cell does not show the empirical or molecular formula.B. Some atoms in a unit cell may be shared with adjoining unit cells reducing the net number of atoms, which results in the empirical formula. In this case all atoms are not shared with other units cells.C. Some atoms in a unit cell may be shared with adjoining unit cells reducing the net number of atoms, which results in the empirical formula. In this case all atoms are shared equally with other unit cells and there is one Co atom at the center.D. Some atoms in a unit cell may be shared with adjoining unit cells reducing the net number of atoms, which results in the empirical formula. In this case the Sm atoms are shared at corners, eight of the Co atoms are shared at faces, and one Co atom sits at the center, unshared.

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Metallic Bonding• In elemental samples of nonmetals and metalloids,

atoms generally bond to each other covalently.• Metals, however, have a dearth of valence

electrons; instead, they form large groups of atoms that share electrons among them.

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A. Red goldB. Purple gold

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Nonbonding electron pairs: Cl S P SiA. 2 1 1 1B. 2 2 2 0C. 3 2 1 0D. 3 3 2 1

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Metallic Bonding

• One can think of a metal, therefore, as a group of cations suspended in a sea of electrons.

• The electrical and thermal conductivity, ductility, and malleability of metals is explained by this model.

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A Molecular-Orbital Approach

As the number of atoms in a chain increases, the energy gap between molecular orbitals (MOs) essentially disappears, and continuous bands of energy states result.

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A. Fourth period: V; Fifth period: Mo; Sixth period: W. All are located near the middle of the period.B. Fourth period: Cr; Fifth period: Tc; Sixth period: Os. All are located near the middle of the period.C. Fourth period: V and Cr; Fifth period: Mo; Sixth period: W. All arelocated near the middle of the period.D. Fourth period: V and Cr; Fifth period: Ru; Sixth period: Ir. All arelocated near the middle of the period.

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A. No change in spacing of energy between MOs.B. MO energy spacing decreases as the number of atoms in the chain increases.C. MO energy spacing increases as the number of atoms in the chain increases.D. A large energy gap between bonding MOs and antibonding MOs occurs.

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A. 4s, 4p, and 3d (4s electrons distribute into all three orbitals)B. 4s and 3d (some leakage of 4s electron density into 3d)C. 4s and 4p (some leakage of 4s electron density into 4p)D. 4s only (no leakage of 4s electron density into 3d ever occurs)

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A. Tungsten (W) has the greater number of electrons in antibonding orbitals and gold (Au) has the higher melting point.

B. Au has the greater number of electrons in antibonding orbitals and W has the higher melting point.

C. W has both the greater number of electrons in antibonding orbitals and higher melting point.

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Ionic Solids• In ionic solids, the

lattice comprises alternately charged ions.

• Ionic solids have very high melting and boiling points and are quintessential crystals.

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A. Metals cleave more readily because the atoms have zero charge. Ions in ionic substances are tightly bound to others of opposite charge and this limits cleavage.B. Metals don’t cleave readily as the atoms that have zero charge are strongly attracted to one another through metallic bonding. Nearest neighbor ions in ionic substances can move and rearrange themselves during cleavage.C. Metals contain atoms of the same size and thus the principles in Figure 12.25 do not apply.D. Metal atoms are more electropositive than cations in ionic solids and thus only move with great force because of their greater attraction to other metal atoms.

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Ionic Solids

The different-sized ions in an ionic compound minimize the distance between oppositely charged ions while keeping like-charged ions away from each other.

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A. Yes, all lattice points are equivalent.B. Yes, because although an ionic substance has at least two different atoms, the different

atoms can lie on the same lattice point.C. No, the lattice points in the metallic structure are different from those in an ionic one.D. No, an ionic substance has at least two different atoms and different atoms cannot lie on

the same lattice points.

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A. Yes, if the anions are sufficiently large.B. Yes, if the anions are sufficiently small.C. No, because anions repel one another.D. No, because each anion adopts different lattice points in the unit cell.

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NaF MgF2 ScF3 Na+ F– Mg2+ F– Sc3+ F–

A. 4 4 2 4 1 3B. 1 1 1 2 2 3C. 2 1 1 1 1 3D. 2 2 2 4 1 2

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A. 2

B. 4

C. 6

D. 8

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Sample Exercise 12.2 Calculating the Empirical Formula and Density of an Ionic Solid

The unit cell of a binary compound of copper and oxygen is shown here. Given this image and the ionic radii rCu

+ = 0.74 Å and rO2 = 1.26 Å, (a)

determine the empirical formula of this compound, (b) determine the coordination numbers of copper and oxygen, (c) estimate the length of the edge of the cubic unit cell, and (d) estimate the density of the compound.

Practice ExerciseEstimate the length of the cubic unit cell edge and the density ofCsCl (Figure 12.26) from the ionic radii of cesium, 1.81 Å, and chloride, 1.67 Å.

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Molecular Solids

• The physical properties of molecular solids are governed by van der Waals forces.

• The individual units of these solids are discrete molecules.

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A. Benzene possesses stronger intermolecular forces and hasmolecules packed more efficiently.

B. Benzene possesses stronger intermolecular forces and toluene has

molecules packed more efficiently.C. Toluene possesses stronger intermolecular forces and has

molecules packed more efficiently.D. Toluene possesses stronger intermolecular forces and benzene

has molecules packed more efficiently.

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Covalent-Network andMolecular Solids

• Diamonds are an example of a covalent-network solid, in which atoms are covalently bonded to each other.– They tend to be hard

and have high melting points.

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Covalent-Network andMolecular Solids

• Graphite is an example of a molecular solid, in which atoms are held together with van der Waals forces.– They tend to be

softer and have lower melting points.

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Semiconductors

In the closely packed molecular orbitals in these compounds, there is a gap between the occupied MOs (valence band) and the unoccupied ones (conduction band).

These Group IVA elements have gaps between their valence and conducting bands of 0.08 to 3.05 eV (7 to 300 kJ/mol).

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Semiconductors• Among elements, only Group IVA, all of which have 4

valence electrons, are semiconductors.• Inorganic semiconductors (like GaAs) tend to have an

average of 4 valence electrons (3 for Ga, 5 for As).

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Sample Exercise 12.3 Qualitative Comparison of Semiconductor Band Gaps

Will GaP have a larger or smaller band gap than ZnS? Will it have a larger or smaller band gap than GaN?

Practice ExerciseWill ZnSe have a larger or smaller band gap than ZnS?

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Doping

By introducing very small amounts of impurities that have more (n-type) or fewer (p-type) valence electrons, one can increase the conductivity of a semiconductor.

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Sample Exercise 12.4 Identifying Types of Semiconductors

Practice ExerciseSuggest an element that could be used to dope silicon to yield a p-type material.

Which of the following elements, if doped into silicon, would yield an n-type semiconductor: Ga, As, or C?

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PolymersPolymers are molecules of high molecular mass made by sequentially bonding repeating units called monomers.

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Some Common Polymers

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Addition Polymers

Addition polymers are made by coupling the monomers by converting bonds within each monomer to bonds between monomers.

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Condensation Polymers• Condensation polymers are made by

joining two subunits through a reaction in which a smaller molecule (often water) is also formed as a by-product.

• These are also called copolymers.

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Synthesis of Nylon

Nylon is one example of a condensation polymer.

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A. An addition polymer, because it has no C=C bond available for a polymer reaction.

B. An addition polymer, because it has a C=C bond available for a polymer reaction.

C. A condensation polymer, because it has one reactive site H2N– that can react with H2N on another molecule.

D. A condensation polymer because it has a reactive site H2N– that can react with a –COOH site on another molecule with water also forming.

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Properties of Polymers

Interactions between chains of a polymer lend elements of order to the structure of polymers.

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A. Increasing the amount of vinyl acetate increases the extent of side chain branching and inhibits crystallinity. Decreasing the number of crystalline regions results in a decreasing melting point.

B. Increasing the amount of vinyl acetate increases the extent of side chain branching and increases crystallinity. Increasing the the number of crystalline regions results in a decreasing melting point.

C. Vinyl acetate is more flexible than ethylene and this decreases the degree of crystallinity.

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Properties of Polymers

Stretching the polymer chains as they form can increase the amount of order, leading to a degree of crystallinity of the polymer.

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Cross-Linking

Chemically bonding chains of polymers to each other can stiffen and strengthen the substance.

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Cross-Linking

Naturally occurring rubber is too soft and pliable for many applications.

In vulcanization, chains are cross-linked by short chains of sulfur atoms, making the rubber stronger and less susceptible to degradation.

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Nanoparticles

Different size particles of a semiconductor (like Cd3P2) can emit different wavelengths of light, depending on the size of the energy gap between bands.

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A. Wave length decreases and band gap increases.B. Wave length decreases and band gap decreases.C. Wave length increases and band gap increases.D. Wave length increases and band gap increases.

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A. Yes, by decreasing size of the particles to form nanocrystals the band gap and wavelength will decrease resulting in a shift of emission to the visible region.

B. Yes, by increasing size of the particles to form nanocrystals the band gap and wavelength will decrease resulting in a shift of emission to visible region.

C. No, by decreasing the size of the particles to form nanocrystals the band gap and wavelength will decrease resulting in a shift of emission deeper into UV region.

D. No, by increasing size of the particles to form nanocrystals the band gap and wavelength will decrease resulting in a shift of emission deeper into UV region.

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NanoparticlesFinely divided metals can have quite different properties than larger samples of metals.

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Carbon Nanotubes

Carbon nanotubes can be made with metallic or semiconducting properties without doping.

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A. Four bonds and bonding is like that in diamond.B. Four bonds and bonding is like that in graphite.C. Three bonds and bonding is like that in diamond.D. Three bonds and bonding is like that in graphite.