C h a p t e r 1 / Introduction

A familiar item that is fabricated from three different material types is the beverage

container. Beverages are marketed in aluminum (metal) cans (top), glass (ceramic) bot-

tles (center), and plastic (polymer) bottles (bottom). (Permission to use these photo-

graphs was granted by the Coca-Cola Company.)

1

L e a r n i n g O b j e c t i v e sAfter careful study of this chapter you should be able to do the following:

1. List six different property classifications of mate-rials that determine their applicability.

2. Cite the four components that are involved in thedesign, production, and utilization of materials,and briefly describe the interrelationships be-tween these components.

3. Cite three criteria that are important in the mate-rials selection process.

1.1 HISTORICAL PERSPECTIVE

Materials are probably more deep-seated in our culture than most of us realize.Transportation, housing, clothing, communication, recreation, and food produc-tion—virtually every segment of our everyday lives is influenced to one degree oranother by materials. Historically, the development and advancement of societieshave been intimately tied to the members’ ability to produce and manipulate materi-als to fill their needs. In fact, early civilizations have been designated by the levelof their materials development (i.e., Stone Age, Bronze Age).

The earliest humans had access to only a very limited number of materials,those that occur naturally: stone, wood, clay, skins, and so on. With time theydiscovered techniques for producing materials that had properties superior to thoseof the natural ones; these new materials included pottery and various metals. Fur-thermore, it was discovered that the properties of a material could be altered byheat treatments and by the addition of other substances. At this point, materialsutilization was totally a selection process, that is, deciding from a given, ratherlimited set of materials the one that was best suited for an application by virtue ofits characteristics. It was not until relatively recent times that scientists came tounderstand the relationships between the structural elements of materials and theirproperties. This knowledge, acquired in the past 60 years or so, has empoweredthem to fashion, to a large degree, the characteristics of materials. Thus, tens ofthousands of different materials have evolved with rather specialized characteristicsthat meet the needs of our modern and complex society; these include metals,plastics, glasses, and fibers.

The development of many technologies that make our existence so comfortablehas been intimately associated with the accessibility of suitable materials. An ad-vancement in the understanding of a material type is often the forerunner to thestepwise progression of a technology. For example, automobiles would not havebeen possible without the availability of inexpensive steel or some other comparablesubstitute. In our contemporary era, sophisticated electronic devices rely on compo-nents that are made from what are called semiconducting materials.

1.2 MATERIALS SCIENCE AND ENGINEERING

The discipline of materials science involves investigating the relationships that existbetween the structures and properties of materials. In contrast, materials engineeringis, on the basis of these structure–property correlations, designing or engineeringthe structure of a material to produce a predetermined set of properties. Throughoutthis text we draw attention to the relationships between material properties andstructural elements.

2

4. (a) List the three primary classifications of solidmaterials, and then cite the distinctive chemi-cal feature of each.

(b) Note the other three types of materials and,for each, its distinctive feature(s).

1.2 Materials Science and Engineering ● 3

‘‘Structure’’ is at this point a nebulous term that deserves some explanation.In brief, the structure of a material usually relates to the arrangement of its internalcomponents. Subatomic structure involves electrons within the individual atomsand interactions with their nuclei. On an atomic level, structure encompasses theorganization of atoms or molecules relative to one another. The next larger struc-tural realm, which contains large groups of atoms that are normally agglomeratedtogether, is termed ‘‘microscopic,’’ meaning that which is subject to direct observa-tion using some type of microscope. Finally, structural elements that may be viewedwith the naked eye are termed ‘‘macroscopic.’’

The notion of ‘‘property’’ deserves elaboration. While in service use, all materi-als are exposed to external stimuli that evoke some type of response. For example,a specimen subjected to forces will experience deformation; or a polished metalsurface will reflect light. Property is a material trait in terms of the kind andmagnitude of response to a specific imposed stimulus. Generally, definitions ofproperties are made independent of material shape and size.

Virtually all important properties of solid materials may be grouped into sixdifferent categories: mechanical, electrical, thermal, magnetic, optical, and deterio-rative. For each there is a characteristic type of stimulus capable of provokingdifferent responses. Mechanical properties relate deformation to an applied loador force; examples include elastic modulus and strength. For electrical properties,such as electrical conductivity and dielectric constant, the stimulus is an electricfield. The thermal behavior of solids can be represented in terms of heat capacity andthermal conductivity. Magnetic properties demonstrate the response of a material tothe application of a magnetic field. For optical properties, the stimulus is electromag-netic or light radiation; index of refraction and reflectivity are representative opticalproperties. Finally, deteriorative characteristics indicate the chemical reactivity ofmaterials. The chapters that follow discuss properties that fall within each of thesesix classifications.

In addition to structure and properties, two other important components areinvolved in the science and engineering of materials, viz. ‘‘processing’’ and ‘‘perfor-mance.’’ With regard to the relationships of these four components, the structureof a material will depend on how it is processed. Furthermore, a material’s perfor-mance will be a function of its properties. Thus, the interrelationship betweenprocessing, structure, properties, and performance is linear, as depicted in theschematic illustration shown in Figure 1.1. Throughout this text we draw attentionto the relationships among these four components in terms of the design, production,and utilization of materials.

We now present an example of these processing-structure-properties-perfor-mance principles with Figure 1.2, a photograph showing three thin disk specimensplaced over some printed matter. It is obvious that the optical properties (i.e., thelight transmittance) of each of the three materials are different; the one on the leftis transparent (i.e., virtually all of the reflected light passes through it), whereasthe disks in the center and on the right are, respectively, translucent and opaque.All of these specimens are of the same material, aluminum oxide, but the leftmostone is what we call a single crystal—that is, it is highly perfect—which gives riseto its transparency. The center one is composed of numerous and very small single

Processing Structure Properties Performance

FIGURE 1.1 The four components of the discipline of materialsscience and engineering and their linear interrelationship.

crystals that are all connected; the boundaries between these small crystals scattera portion of the light reflected from the printed page, which makes this materialoptically translucent. And finally, the specimen on the right is composed not onlyof many small, interconnected crystals, but also of a large number of very smallpores or void spaces. These pores also effectively scatter the reflected light andrender this material opaque.

Thus, the structures of these three specimens are different in terms of crystalboundaries and pores, which affect the optical transmittance properties. Further-more, each material was produced using a different processing technique. And, ofcourse, if optical transmittance is an important parameter relative to the ultimatein-service application, the performance of each material will be different.

1.3 WHY STUDY MATERIALS SCIENCE AND ENGINEERING?Why do we study materials? Many an applied scientist or engineer, whether mechan-ical, civil, chemical, or electrical, will at one time or another be exposed to adesign problem involving materials. Examples might include a transmission gear,the superstructure for a building, an oil refinery component, or an integrated circuitchip. Of course, materials scientists and engineers are specialists who are totallyinvolved in the investigation and design of materials.

Many times, a materials problem is one of selecting the right material from themany thousands that are available. There are several criteria on which the finaldecision is normally based. First of all, the in-service conditions must be character-ized, for these will dictate the properties required of the material. On only rareoccasions does a material possess the maximum or ideal combination of properties.Thus, it may be necessary to trade off one characteristic for another. The classicexample involves strength and ductility; normally, a material having a high strengthwill have only a limited ductility. In such cases a reasonable compromise betweentwo or more properties may be necessary.

A second selection consideration is any deterioration of material propertiesthat may occur during service operation. For example, significant reductions inmechanical strength may result from exposure to elevated temperatures or corrosiveenvironments.

Finally, probably the overriding consideration is that of economics: What willthe finished product cost? A material may be found that has the ideal set of

4 ● Chapter 1 / Introduction

FIGURE 1.2Photograph showing the light

transmittance of three aluminum oxidespecimens. From left to right: single-crystal material (sapphire), which is

transparent; a polycrystalline and fullydense (nonporous) material, which is

translucent; and a polycrystallinematerial that contains approximately 5%

porosity, which is opaque. (Specimenpreparation, P. A. Lessing; photography

by J. Telford.)

1.4 Classification of Materials ● 5

properties but is prohibitively expensive. Here again, some compromise is inevitable.The cost of a finished piece also includes any expense incurred during fabricationto produce the desired shape.

The more familiar an engineer or scientist is with the various characteristicsand structure–property relationships, as well as processing techniques of materials,the more proficient and confident he or she will be to make judicious materialschoices based on these criteria.

1.4 CLASSIFICATION OF MATERIALS

Solid materials have been conveniently grouped into three basic classifications:metals, ceramics, and polymers. This scheme is based primarily on chemical makeupand atomic structure, and most materials fall into one distinct grouping or another,although there are some intermediates. In addition, there are three other groupsof important engineering materials—composites, semiconductors, and biomaterials.Composites consist of combinations of two or more different materials, whereassemiconductors are utilized because of their unusual electrical characteristics; bio-materials are implanted into the human body. A brief explanation of the materialtypes and representative characteristics is offered next.

METALSMetallic materials are normally combinations of metallic elements. They have largenumbers of nonlocalized electrons; that is, these electrons are not bound to particularatoms. Many properties of metals are directly attributable to these electrons. Metalsare extremely good conductors of electricity and heat and are not transparent tovisible light; a polished metal surface has a lustrous appearance. Furthermore,metals are quite strong, yet deformable, which accounts for their extensive use instructural applications.

CERAMICSCeramics are compounds between metallic and nonmetallic elements; they are mostfrequently oxides, nitrides, and carbides. The wide range of materials that fallswithin this classification includes ceramics that are composed of clay minerals,cement, and glass. These materials are typically insulative to the passage of electricityand heat, and are more resistant to high temperatures and harsh environments thanmetals and polymers. With regard to mechanical behavior, ceramics are hard butvery brittle.

POLYMERSPolymers include the familiar plastic and rubber materials. Many of them are organiccompounds that are chemically based on carbon, hydrogen, and other nonmetallicelements; furthermore, they have very large molecular structures. These materialstypically have low densities and may be extremely flexible.

COMPOSITESA number of composite materials have been engineered that consist of more thanone material type. Fiberglass is a familiar example, in which glass fibers are embed-ded within a polymeric material. A composite is designed to display a combinationof the best characteristics of each of the component materials. Fiberglass acquiresstrength from the glass and flexibility from the polymer. Many of the recent materialdevelopments have involved composite materials.

SEMICONDUCTORSSemiconductors have electrical properties that are intermediate between the electri-cal conductors and insulators. Furthermore, the electrical characteristics of thesematerials are extremely sensitive to the presence of minute concentrations of impu-rity atoms, which concentrations may be controlled over very small spatial regions.The semiconductors have made possible the advent of integrated circuitry that hastotally revolutionized the electronics and computer industries (not to mention ourlives) over the past two decades.

BIOMATERIALSBiomaterials are employed in components implanted into the human body forreplacement of diseased or damaged body parts. These materials must not producetoxic substances and must be compatible with body tissues (i.e., must not causeadverse biological reactions). All of the above materials—metals, ceramics, poly-mers, composites, and semiconductors—may be used as biomaterials. �For example,in Section 20.8 are discussed some of the biomaterials that are utilized in artificialhip replacements.�

1.5 ADVANCED MATERIALS

Materials that are utilized in high-technology (or high-tech) applications are some-times termed advanced materials. By high technology we mean a device or productthat operates or functions using relatively intricate and sophisticated principles;examples include electronic equipment (VCRs, CD players, etc.), computers, fiber-optic systems, spacecraft, aircraft, and military rocketry. These advanced materialsare typically either traditional materials whose properties have been enhanced ornewly developed, high-performance materials. Furthermore, they may be of allmaterial types (e.g., metals, ceramics, polymers), and are normally relatively expen-sive. In subsequent chapters are discussed the properties and applications of anumber of advanced materials—for example, materials that are used for lasers,integrated circuits, magnetic information storage, liquid crystal displays (LCDs),fiber optics, and the thermal protection system for the Space Shuttle Orbiter.

1.6 MODERN MATERIALS’ NEEDS

In spite of the tremendous progress that has been made in the discipline of materialsscience and engineering within the past few years, there still remain technologicalchallenges, including the development of even more sophisticated and specializedmaterials, as well as consideration of the environmental impact of materials produc-tion. Some comment is appropriate relative to these issues so as to round outthis perspective.

Nuclear energy holds some promise, but the solutions to the many problemsthat remain will necessarily involve materials, from fuels to containment structuresto facilities for the disposal of radioactive waste.

Significant quantities of energy are involved in transportation. Reducing theweight of transportation vehicles (automobiles, aircraft, trains, etc.), as well asincreasing engine operating temperatures, will enhance fuel efficiency. New high-strength, low-density structural materials remain to be developed, as well as materi-als that have higher-temperature capabilities, for use in engine components.

6 ● Chapter 1 / Introduction

References ● 7

Furthermore, there is a recognized need to find new, economical sources ofenergy, and to use the present resources more efficiently. Materials will undoub-tedly play a significant role in these developments. For example, the direct con-version of solar into electrical energy has been demonstrated. Solar cells employsome rather complex and expensive materials. To ensure a viable technology,materials that are highly efficient in this conversion process yet less costly mustbe developed.

Furthermore, environmental quality depends on our ability to control air andwater pollution. Pollution control techniques employ various materials. In addition,materials processing and refinement methods need to be improved so that theyproduce less environmental degradation, that is, less pollution and less despoilageof the landscape from the mining of raw materials. Also, in some materials manufac-turing processes, toxic substances are produced, and the ecological impact of theirdisposal must be considered.

Many materials that we use are derived from resources that are nonrenewable,that is, not capable of being regenerated. These include polymers, for which theprime raw material is oil, and some metals. These nonrenewable resources aregradually becoming depleted, which necessitates: 1) the discovery of additionalreserves, 2) the development of new materials having comparable properties withless adverse environmental impact, and/or 3) increased recycling efforts and thedevelopment of new recycling technologies. As a consequence of the economics ofnot only production but also environmental impact and ecological factors, it isbecoming increasingly important to consider the ‘‘cradle-to-grave’’ life cycle ofmaterials relative to the overall manufacturing process.

�The roles that materials scientists and engineers play relative to these, aswell as other environmental and societal issues, are discussed in more detail inChapter 21.�

R E F E R E N C E S

The October 1986 issue of Scientific American, Vol.255, No. 4, is devoted entirely to various advancedmaterials and their uses. Other references forChapter 1 are textbooks that cover the basic funda-mentals of the field of materials science and engi-neering.Ashby, M. F. and D. R. H. Jones, Engineering Mate-

rials 1, An Introduction to Their Properties andApplications, 2nd edition, Pergamon Press, Ox-ford, 1996.

Ashby, M. F. and D. R. H. Jones, Engineering Mate-rials 2, An Introduction to Microstructures, Pro-cessing and Design, Pergamon Press, Oxford,1986.

Askeland, D. R., The Science and Engineering ofMaterials, 3rd edition, Brooks/Cole PublishingCo., Pacific Grove, CA, 1994.

Barrett, C. R., W. D. Nix, and A. S. Tetelman, ThePrinciples of Engineering Materials, PrenticeHall, Inc., Englewood Cliffs, NJ, 1973.

Flinn, R. A. and P. K. Trojan, Engineering Ma-terials and Their Applications, 4th edition, JohnWiley & Sons, New York, 1990.

Jacobs, J. A. and T. F. Kilduff, Engineering Materi-als Technology, 3rd edition, Prentice Hall, Up-per Saddle River, NJ, 1996.

McMahon, C. J., Jr. and C. D. Graham, Jr., Intro-duction to Engineering Materials: The Bicycleand the Walkman, Merion Books, Philadel-phia, 1992.

Murray, G. T., Introduction to Engineering Materi-als—Behavior, Properties, and Selection, Mar-cel Dekker, Inc., New York, 1993.

Ohring, M., Engineering Materials Science, Aca-demic Press, San Diego, CA, 1995.

Ralls, K. M., T. H. Courtney, and J. Wulff, Intro-duction to Materials Science and Engineering,John Wiley & Sons, New York, 1976.

Schaffer, J. P., A. Saxena, S. D. Antolovich, T. H.Sanders, Jr., and S. B. Warner, The Science and

Design of Engineering Materials, 2nd edition,WCB/McGraw-Hill, New York, 1999.

Shackelford, J. F., Introduction to Materials Sciencefor Engineers, 5th edition, Prentice Hall, Inc.,Upper Saddle River, NJ, 2000.

Smith, W. F., Principles of Materials Science and

Engineering, 3rd edition, McGraw-Hill BookCompany, New York, 1995.

Van Vlack, L. H., Elements of Materials Scienceand Engineering, 6th edition, Addison-WesleyPublishing Co., Reading, MA, 1989.

8 ● Chapter 1 / Introduction

C h a p t e r 2 / Atomic Structure andInteratomic Bonding

This micrograph, which

represents the surface of a

gold specimen, was taken

with a sophisticated atomic

force microscope (AFM). In-

dividual atoms for this (111)

crystallographic surface

plane are resolved. Also

note the dimensional scale

(in the nanometer range) be-

low the micrograph. (Image

courtesy of Dr. Michael

Green, TopoMetrix Corpo-

ration.)

Why Study Atomic Structure and Interatomic Bonding?

An important reason to have an understandingof interatomic bonding in solids is that, in someinstances, the type of bond allows us to explain amaterial’s properties. For example, consider car-bon, which may exist as both graphite anddiamond. Whereas graphite is relatively soft and

9

has a ‘‘greasy’’ feel to it, diamond is the hardestknown material. This dramatic disparity in proper-ties is directly attributable to a type of interatomicbonding found in graphite that does not exist indiamond (see Section 3.9).

L e a r n i n g O b j e c t i v e sAfter careful study of this chapter you should be able to do the following:

1. Name the two atomic models cited, and note thedifferences between them.

2. Describe the important quantum-mechanicalprinciple that relates to electron energies.

3. (a) Schematically plot attractive, repulsive, andnet energies versus interatomic separationfor two atoms or ions.

(b) Note on this plot the equilibrium separationand the bonding energy.

4. (a) Briefly describe ionic, covalent, metallic, hy-drogen, and van der Waals bonds.

(b) Note what materials exhibit each of thesebonding types.

2.1 INTRODUCTION

Some of the important properties of solid materials depend on geometrical atomicarrangements, and also the interactions that exist among constituent atoms ormolecules. This chapter, by way of preparation for subsequent discussions, considersseveral fundamental and important concepts, namely: atomic structure, electronconfigurations in atoms and the periodic table, and the various types of primaryand secondary interatomic bonds that hold together the atoms comprising a solid.These topics are reviewed briefly, under the assumption that some of the materialis familiar to the reader.

A T O M I C S T R U C T U R E2.2 FUNDAMENTAL CONCEPTS

Each atom consists of a very small nucleus composed of protons and neutrons,which is encircled by moving electrons. Both electrons and protons are electricallycharged, the charge magnitude being 1.60 � 10�19 C, which is negative in sign forelectrons and positive for protons; neutrons are electrically neutral. Masses forthese subatomic particles are infinitesimally small; protons and neutrons have ap-proximately the same mass, 1.67 � 10�27 kg, which is significantly larger than thatof an electron, 9.11 � 10�31 kg.

Each chemical element is characterized by the number of protons in the nucleus,or the atomic number (Z).1 For an electrically neutral or complete atom, the atomicnumber also equals the number of electrons. This atomic number ranges in integralunits from 1 for hydrogen to 92 for uranium, the highest of the naturally oc-curring elements.

The atomic mass (A) of a specific atom may be expressed as the sum of themasses of protons and neutrons within the nucleus. Although the number of protonsis the same for all atoms of a given element, the number of neutrons (N) may bevariable. Thus atoms of some elements have two or more different atomic masses,which are called isotopes. The atomic weight of an element corresponds to theweighted average of the atomic masses of the atom’s naturally occurring isotopes.2

The atomic mass unit (amu) may be used for computations of atomic weight. Ascale has been established whereby 1 amu is defined as ��� of the atomic mass of

10

1 Terms appearing in boldface type are defined in the Glossary, which follows Appendix E.2 The term ‘‘atomic mass’’ is really more accurate than ‘‘atomic weight’’ inasmuch as, in thiscontext, we are dealing with masses and not weights. However, atomic weight is, by conven-tion, the preferred terminology, and will be used throughout this book. The reader shouldnote that it is not necessary to divide molecular weight by the gravitational constant.

2.3 Electrons in Atoms ● 11

the most common isotope of carbon, carbon 12 (12C) (A � 12.00000). Within thisscheme, the masses of protons and neutrons are slightly greater than unity, and

A � Z � N (2.1)

The atomic weight of an element or the molecular weight of a compound may bespecified on the basis of amu per atom (molecule) or mass per mole of material.In one mole of a substance there are 6.023 � 1023 (Avogadro’s number) atoms ormolecules. These two atomic weight schemes are related through the followingequation:

1 amu/atom (or molecule) � 1 g/mol

For example, the atomic weight of iron is 55.85 amu/atom, or 55.85 g/mol. Sometimesuse of amu per atom or molecule is convenient; on other occasions g (or kg)/molis preferred; the latter is used in this book.

2.3 ELECTRONS IN ATOMS

ATOMIC MODELSDuring the latter part of the nineteenth century it was realized that many phenomenainvolving electrons in solids could not be explained in terms of classical mechanics.What followed was the establishment of a set of principles and laws that governsystems of atomic and subatomic entities that came to be known as quantummechanics. An understanding of the behavior of electrons in atoms and crystallinesolids necessarily involves the discussion of quantum-mechanical concepts. How-ever, a detailed exploration of these principles is beyond the scope of this book,and only a very superficial and simplified treatment is given.

One early outgrowth of quantum mechanics was the simplified Bohr atomicmodel, in which electrons are assumed to revolve around the atomic nucleus indiscrete orbitals, and the position of any particular electron is more or less welldefined in terms of its orbital. This model of the atom is represented in Figure 2.1.

Another important quantum-mechanical principle stipulates that the energiesof electrons are quantized; that is, electrons are permitted to have only specificvalues of energy. An electron may change energy, but in doing so it must make aquantum jump either to an allowed higher energy (with absorption of energy) orto a lower energy (with emission of energy). Often, it is convenient to think ofthese allowed electron energies as being associated with energy levels or states.

Orbital electron

Nucleus

FIGURE 2.1 Schematic representation ofthe Bohr atom.

12 ● Chapter 2 / Atomic Structure and Interatomic Bonding

These states do not vary continuously with energy; that is, adjacent states areseparated by finite energies. For example, allowed states for the Bohr hydrogenatom are represented in Figure 2.2a. These energies are taken to be negative,whereas the zero reference is the unbound or free electron. Of course, the singleelectron associated with the hydrogen atom will fill only one of these states.

Thus, the Bohr model represents an early attempt to describe electrons in atoms,in terms of both position (electron orbitals) and energy (quantized energy levels).

This Bohr model was eventually found to have some significant limitationsbecause of its inability to explain several phenomena involving electrons. A resolu-tion was reached with a wave-mechanical model, in which the electron is consideredto exhibit both wavelike and particle-like characteristics. With this model, an elec-tron is no longer treated as a particle moving in a discrete orbital; but rather,position is considered to be the probability of an electron’s being at various locationsaround the nucleus. In other words, position is described by a probability distributionor electron cloud. Figure 2.3 compares Bohr and wave-mechanical models for thehydrogen atom. Both these models are used throughout the course of this book;the choice depends on which model allows the more simple explanation.

QUANTUM NUMBERSUsing wave mechanics, every electron in an atom is characterized by four parameterscalled quantum numbers. The size, shape, and spatial orientation of an electron’sprobability density are specified by three of these quantum numbers. Furthermore,Bohr energy levels separate into electron subshells, and quantum numbers dictatethe number of states within each subshell. Shells are specified by a principal quantumnumber n, which may take on integral values beginning with unity; sometimes theseshells are designated by the letters K, L, M, N, O, and so on, which correspond,respectively, to n � 1, 2, 3, 4, 5, . . . , as indicated in Table 2.1. It should also be

0 0

�1 ´ 10�18

�2 ´ 10�18

(a) (b)

�5

�10

�15

n = 1 1s

Ene

rgy

(J)

Ene

rgy

(eV)

n = 2

n = 3

2s

3s

2p

3p3d

�1.5

�3.4

�13.6

FIGURE 2.2 (a) Thefirst three electron

energy states for theBohr hydrogen atom.

(b) Electron energystates for the first three

shells of the wave-mechanical hydrogenatom. (Adapted fromW. G. Moffatt, G. W.Pearsall, and J. Wulff,

The Structure andProperties of Materials,Vol. I, Structure, p. 10.

Copyright 1964 byJohn Wiley & Sons,

New York. Reprintedby permission of John

Wiley & Sons, Inc.)

2.3 Electrons in Atoms ● 13

1.0

0

(a) (b)

Orbital electron Nucleus

Prob

abili

ty

Distance from nucleus

FIGURE 2.3 Comparison ofthe (a) Bohr and (b) wave-mechanical atom models interms of electrondistribution. (Adapted fromZ. D. Jastrzebski, TheNature and Properties ofEngineering Materials, 3rdedition, p. 4. Copyright 1987 by John Wiley & Sons,New York. Reprinted bypermission of John Wiley &Sons, Inc.)

Table 2.1 The Number of Available Electron States in Some of theElectron Shells and Subshells

PrincipalNumber of ElectronsShellQuantum Number

of StatesSubshellsNumber n Designation Per Subshell Per Shell1 K s 1 2 2

2 Ls 1 2

8p 3 6

s 1 23 M p 3 6 18

d 5 10

s 1 2

4 Np 3 6

32d 5 10f 7 14

14 ● Chapter 2 / Atomic Structure and Interatomic Bonding

noted that this quantum number, and it only, is also associated with the Bohr model.This quantum number is related to the distance of an electron from the nucleus,or its position.

The second quantum number, l, signifies the subshell, which is denoted by alowercase letter—an s, p, d, or f ; it is related to the shape of the electron subshell.In addition, the number of these subshells is restricted by the magnitude of n.Allowable subshells for the several n values are also presented in Table 2.1. Thenumber of energy states for each subshell is determined by the third quantumnumber, ml . For an s subshell, there is a single energy state, whereas for p, d, andf subshells, three, five, and seven states exist, respectively (Table 2.1). In the absenceof an external magnetic field, the states within each subshell are identical. However,when a magnetic field is applied these subshell states split, each state assuming aslightly different energy.

Associated with each electron is a spin moment, which must be oriented eitherup or down. Related to this spin moment is the fourth quantum number, ms , forwhich two values are possible (� �� and � ��), one for each of the spin orientations.

Thus, the Bohr model was further refined by wave mechanics, in which theintroduction of three new quantum numbers gives rise to electron subshells withineach shell. A comparison of these two models on this basis is illustrated, for thehydrogen atom, in Figures 2.2a and 2.2b.

A complete energy level diagram for the various shells and subshells using thewave-mechanical model is shown in Figure 2.4. Several features of the diagram areworth noting. First, the smaller the principal quantum number, the lower the energylevel; for example, the energy of a 1s state is less than that of a 2s state, which inturn is lower than the 3s. Second, within each shell, the energy of a subshell levelincreases with the value of the l quantum number. For example, the energy of a3d state is greater than a 3p, which is larger than 3s. Finally, there may be overlapin energy of a state in one shell with states in an adjacent shell, which is especiallytrue of d and f states; for example, the energy of a 3d state is greater than that fora 4s.

Principal quantum number, n

Ene

rgy

1

s

sp

sp

sp

sp

df sp

sp

df

d

d

d

f

2 3 4 5 6 7

FIGURE 2.4 Schematicrepresentation of the relativeenergies of the electrons for thevarious shells and subshells. (FromK. M. Ralls, T. H. Courtney, and J.Wulff, Introduction to MaterialsScience and Engineering, p. 22.Copyright 1976 by John Wiley &Sons, New York. Reprinted bypermission of John Wiley & Sons,Inc.)

2.3 Electrons in Atoms ● 15

ELECTRON CONFIGURATIONSThe preceding discussion has dealt primarily with electron states—values of energythat are permitted for electrons. To determine the manner in which these statesare filled with electrons, we use the Pauli exclusion principle, another quantum-mechanical concept. This principle stipulates that each electron state can hold nomore than two electrons, which must have opposite spins. Thus, s, p, d, and fsubshells may each accommodate, respectively, a total of 2, 6, 10, and 14 electrons;Table 2.1 summarizes the maximum number of electrons that may occupy each ofthe first four shells.

Of course, not all possible states in an atom are filled with electrons. For mostatoms, the electrons fill up the lowest possible energy states in the electron shellsand subshells, two electrons (having opposite spins) per state. The energy structurefor a sodium atom is represented schematically in Figure 2.5. When all the electronsoccupy the lowest possible energies in accord with the foregoing restrictions, anatom is said to be in its ground state. However, electron transitions to higher energystates are possible, as discussed in Chapters 12 �and 19.� The electron configurationor structure of an atom represents the manner in which these states are occupied.In the conventional notation the number of electrons in each subshell is indicatedby a superscript after the shell–subshell designation. For example, the electronconfigurations for hydrogen, helium, and sodium are, respectively, 1s1, 1s2, and1s22s22p63s1. Electron configurations for some of the more common elements arelisted in Table 2.2.

At this point, comments regarding these electron configurations are necessary.First, the valence electrons are those that occupy the outermost filled shell. Theseelectrons are extremely important; as will be seen, they participate in the bondingbetween atoms to form atomic and molecular aggregates. Furthermore, many ofthe physical and chemical properties of solids are based on these valence electrons.

In addition, some atoms have what are termed ‘‘stable electron configurations’’;that is, the states within the outermost or valence electron shell are completelyfilled. Normally this corresponds to the occupation of just the s and p states forthe outermost shell by a total of eight electrons, as in neon, argon, and krypton;one exception is helium, which contains only two 1s electrons. These elements (Ne,Ar, Kr, and He) are the inert, or noble, gases, which are virtually unreactivechemically. Some atoms of the elements that have unfilled valence shells assume

Incr

easi

ng e

nerg

y

3p3s

2s

1s

2p

FIGURE 2.5 Schematic representation of thefilled energy states for a sodium atom.

16 ● Chapter 2 / Atomic Structure and Interatomic Bonding

stable electron configurations by gaining or losing electrons to form charged ions,or by sharing electrons with other atoms. This is the basis for some chemicalreactions, and also for atomic bonding in solids, as explained in Section 2.6.

Under special circumstances, the s and p orbitals combine to form hybrid spn

orbitals, where n indicates the number of p orbitals involved, which may have avalue of 1, 2, or 3. The 3A, 4A, and 5A group elements of the periodic table (Figure2.6) are those which most often form these hybrids. The driving force for theformation of hybrid orbitals is a lower energy state for the valence electrons. Forcarbon the sp3 hybrid is of primary importance in organic and polymer chemistries.

Table 2.2 A Listing of the Expected Electron Configurationsfor Some of the Common Elementsa

AtomicElement Symbol Number Electron ConfigurationHydrogen H 1 1s1

Helium He 2 1s2

Lithium Li 3 1s22s1

Beryllium Be 4 1s22s2

Boron B 5 1s22s22p1

Carbon C 6 1s22s22p2

Nitrogen N 7 1s22s22p3

Oxygen O 8 1s22s22p4

Fluorine F 9 1s22s22p5

Neon Ne 10 1s22s22p6

Sodium Na 11 1s22s22p63s1

Magnesium Mg 12 1s22s22p63s2

Aluminum Al 13 1s22s22p63s23p1

Silicon Si 14 1s22s22p63s23p2

Phosphorus P 15 1s22s22p63s23p3

Sulfur S 16 1s22s22p63s23p4

Chlorine Cl 17 1s22s22p63s23p5

Argon Ar 18 1s22s22p63s23p6

Potassium K 19 1s22s22p63s23p64s1

Calcium Ca 20 1s22s22p63s23p64s2

Scandium Sc 21 1s22s22p63s23p63d 14s2

Titanium Ti 22 1s22s22p63s23p63d 24s2

Vanadium V 23 1s22s22p63s23p63d 34s2

Chromium Cr 24 1s22s22p63s23p63d 54s1

Manganese Mn 25 1s22s22p63s23p63d 54s2

Iron Fe 26 1s22s22p63s23p63d 64s2

Cobalt Co 27 1s22s22p63s23p63d 74s2

Nickel Ni 28 1s22s22p63s23p63d 84s2

Copper Cu 29 1s22s22p63s23p63d 104s1

Zinc Zn 30 1s22s22p63s23p63d 104s2

Gallium Ga 31 1s22s22p63s23p63d 104s24p1

Germanium Ge 32 1s22s22p63s23p63d 104s24p2

Arsenic As 33 1s22s22p63s23p63d 104s24p3

Selenium Se 34 1s22s22p63s23p63d 104s24p4

Bromine Br 35 1s22s22p63s23p63d 104s24p5

Krypton Kr 36 1s22s22p63s23p63d 104s24p6

a When some elements covalently bond, they form sp hybrid bonds. This is espe-cially true for C, Si, and Ge.

1

H1.0080

3

Li6.939

4

Be9.0122

11

Na22.990

12

Mg24.312

19

K39.102

20

Ca40.08

37

Rb85.47

38

Sr

21

Sc44.956

39

Y87.62

55

Cs132.91

56

Ba137.34

5

B10.811

13

Al26.982

31

Ga69.72

49

In114.82

81

Tl204.37

6

C12.011

14

Si28.086

32

Ge72.59

50

Sn118.69

82

Pb207.19

7

N14.007

15

P30.974

33

As74.922

51

Sb121.75

83

Bi208.98

8

O15.999

16

S32.064

34

Se78.96

52

Te127.60

84

Po(210)

9

F18.998

17

Cl35.453

35

Br79.91

53

I126.90

85

At(210)

10

Ne20.183

2

He4.0026

18

Ar39.948

36

Kr83.80

54

Xe131.30

86

Rn(222)

22

Ti47.90

40

Zr91.2288.91

72

Hf178.49

23

V50.942

41

Nb92.91

73

Ta180.95

24

Cr51.996

42

Mo95.94

74

W183.85

25

Mn54.938

43

Tc(99)

75

Re186.2

26

Fe55.847

44

Ru101.07

76

Os190.2

27

Co58.933

45

Rh102.91

77

Ir192.2

28

Ni58.71

46

Pd106.4

78

Pt195.09

29

Cu63.54

29

Cu63.54

47

Ag107.87

79

Au196.97

30

Zn65.37

48

Cd112.40

80

Hg200.59

66Dy

162.50

98

Cf(249)

67Ho

164.93

99

Es(254)

68Er

167.26

100

Fm(253)

69Tm

168.93

101

Md(256)

70Yb

173.04

102

No(254)

71Lu

174.97

103

Lw(257)

57La

138.91

89

Ac(227)

58Ce

140.12

90

Th232.04

59Pr

140.91

91

Pa(231)

60Nd

144.24

92

U238.03

61Pm

(145)

93

Np(237)

62Sm

150.35

94

Pu(242)

63Eu

151.96

95

Am(243)

64Gd

157.25

96

Cm(247)

65Tb

158.92

97

Bk(247)

87

Fr(223)

88

Ra(226)

Atomic number

Symbol

Metal

Nonmetal

Intermediate

Atomic weight

KeyIA

IIA

IIIB IVB VB VIB VIIBVIII

IB IIB

IIIA IVA VA VIA VIIA

0

Rare earth series

Actinide series

Acti-nideseries

Rareearthseries

2.4 The Periodic Table ● 17

The shape of the sp3 hybrid is what determines the 109� (or tetrahedral) anglefound in polymer chains (Chapter 4).

2.4 THE PERIODIC TABLE

All the elements have been classified according to electron configuration in theperiodic table (Figure 2.6). Here, the elements are situated, with increasing atomicnumber, in seven horizontal rows called periods. The arrangement is such that allelements that are arrayed in a given column or group have similar valence electronstructures, as well as chemical and physical properties. These properties changegradually and systematically, moving horizontally across each period.

The elements positioned in Group 0, the rightmost group, are the inert gases,which have filled electron shells and stable electron configurations. Group VIIAand VIA elements are one and two electrons deficient, respectively, from havingstable structures. The Group VIIA elements (F, Cl, Br, I, and At) are sometimestermed the halogens. The alkali and the alkaline earth metals (Li, Na, K, Be, Mg,Ca, etc.) are labeled as Groups IA and IIA, having, respectively, one and twoelectrons in excess of stable structures. The elements in the three long periods,Groups IIIB through IIB, are termed the transition metals, which have partiallyfilled d electron states and in some cases one or two electrons in the next higherenergy shell. Groups IIIA, IVA, and VA (B, Si, Ge, As, etc.) display characteristicsthat are intermediate between the metals and nonmetals by virtue of their valenceelectron structures.

FIGURE 2.6 The periodic table of the elements. The numbers in parentheses arethe atomic weights of the most stable or common isotopes.

18 ● Chapter 2 / Atomic Structure and Interatomic Bonding

As may be noted from the periodic table, most of the elements really comeunder the metal classification. These are sometimes termed electropositive elements,indicating that they are capable of giving up their few valence electrons to becomepositively charged ions. Furthermore, the elements situated on the right-hand sideof the table are electronegative; that is, they readily accept electrons to formnegatively charged ions, or sometimes they share electrons with other atoms. Figure2.7 displays electronegativity values that have been assigned to the various elementsarranged in the periodic table. As a general rule, electronegativity increases inmoving from left to right and from bottom to top. Atoms are more likely to acceptelectrons if their outer shells are almost full, and if they are less ‘‘shielded’’ from(i.e., closer to) the nucleus.

A T O M I C B O N D I N G I N S O L I D S2.5 BONDING FORCES AND ENERGIES

An understanding of many of the physical properties of materials is predicated ona knowledge of the interatomic forces that bind the atoms together. Perhaps theprinciples of atomic bonding are best illustrated by considering the interactionbetween two isolated atoms as they are brought into close proximity from an infiniteseparation. At large distances, the interactions are negligible; but as the atomsapproach, each exerts forces on the other. These forces are of two types, attractiveand repulsive, and the magnitude of each is a function of the separation or in-teratomic distance. The origin of an attractive force FA depends on the particulartype of bonding that exists between the two atoms. Its magnitude varies with thedistance, as represented schematically in Figure 2.8a. Ultimately, the outer electronshells of the two atoms begin to overlap, and a strong repulsive force FR comesinto play. The net force FN between the two atoms is just the sum of both attractiveand repulsive components; that is,

FN � FA � FR (2.2)

1

H

3

Li4

Be

11

Na12

Mg

19

K20

Ca

37

Rb38

Sr

21

Sc

39

Y

55

Cs56 57 71

Ba La Lu

5

B

13

Al

31

Ga

49

In

81

Tl

6

C

14

Si

32

Ge

50

Sn

82

Pb

7

N

15

P

33

As

51

Sb

83

Bi

8

O

16

S

34

Se

52

Te

84

Po

9

F

17

Cl

35

Br

53

I

85

At

10

Ne

2

He

18

Ar

36

Kr

54

Xe

86

Rn

22

Ti

40

Zr

72

Hf

23

V

41

Nb

73

Ta

24

Cr

42

Mo

74

W

25

Mn

43

Tc

75

Re

26

Fe

44

Ru

76

Os

27

Co

45

Rh

77

Ir

28

Ni

46

Pd

78

Pt

29

Cu

47

Ag

79

Au

30

Zn

48

Cd

80

Hg

87

Fr88

Ra89 102

Ac No

2.1

1.0 1.5

0.9 1.2

0.8 1.0

0.8

1.3

1.0

0.7 0.9 1.1 1.2

2.0

1.5

1.6

1.7

1.8

2.5

1.8

1.8

1.8

1.8

3.0

2.1

2.0

1.9

1.9

3.5

2.5

2.4

2.1

2.0

4.0

3.0

2.8

2.5

2.2

1.5

1.41.2

1.3

1.6

1.6

1.5

1.6

1.8

1.7

1.5

1.9

1.9

1.8

2.2

2.2

1.8

2.2

2.2

1.8

2.2

2.2

1.9

1.9

2.4

1.6

1.7

1.9

0.7 0.9 1.1 1.7

IA

IIA

IIIB IVB VB VIB VIIBVIII

IB IIB

IIIA IVA VA VIA VIIA

0

FIGURE 2.7 The electronegativity values for the elements. (Adapted from LinusPauling, The Nature of the Chemical Bond, 3rd edition. Copyright 1939 and 1940,3rd edition copyright 1960, by Cornell University. Used by permission of thepublisher, Cornell University Press.)

2.5 Bonding Forces and Energies ● 19

which is also a function of the interatomic separation, as also plotted in Figure2.8a. When FA and FR balance, or become equal, there is no net force; that is,

FA � FR � 0 (2.3)

Then a state of equilibrium exists. The centers of the two atoms will remain separatedby the equilibrium spacing r0 , as indicated in Figure 2.8a. For many atoms, r0 isapproximately 0.3 nm (3 A). Once in this position, the two atoms will counteractany attempt to separate them by an attractive force, or to push them together bya repulsive action.

Sometimes it is more convenient to work with the potential energies betweentwo atoms instead of forces. Mathematically, energy (E) and force (F) are related as

E � � F dr (2.4)

Or, for atomic systems,

EN � �r

�FN dr (2.5)

� �r

�FA dr � �r

�FR dr (2.6)

� EA � ER (2.7)

in which EN , EA , and ER are respectively the net, attractive, and repulsive energiesfor two isolated and adjacent atoms.

+

(a)

(b)

Interatomic separation r

Interatomic separation r

Repulsive force FR

Attractive force FA

Net force FN

Att

ract

ion

Rep

ulsi

on

Forc

e F

Repulsive energy ER

Attractive energy EA

Net energy EN

+

0

0

Att

ract

ion

Rep

ulsi

on

Pot

enti

al e

nerg

y E

r0

E0

FIGURE 2.8 (a) Thedependence of repulsive,attractive, and net forces oninteratomic separation fortwo isolated atoms. (b) Thedependence of repulsive,attractive, and net potentialenergies on interatomicseparation for two isolatedatoms.

20 ● Chapter 2 / Atomic Structure and Interatomic Bonding

Figure 2.8b plots attractive, repulsive, and net potential energies as a functionof interatomic separation for two atoms. The net curve, which is again the sum ofthe other two, has a potential energy trough or well around its minimum. Here,the same equilibrium spacing, r0 , corresponds to the separation distance at theminimum of the potential energy curve. The bonding energy for these two atoms,E0 , corresponds to the energy at this minimum point (also shown in Figure 2.8b);it represents the energy that would be required to separate these two atoms to aninfinite separation.

Although the preceding treatment has dealt with an ideal situation involvingonly two atoms, a similar yet more complex condition exists for solid materialsbecause force and energy interactions among many atoms must be considered.Nevertheless, a bonding energy, analogous to E0 above, may be associated witheach atom. The magnitude of this bonding energy and the shape of the energy-versus-interatomic separation curve vary from material to material, and they bothdepend on the type of atomic bonding. Furthermore, a number of material propertiesdepend on E0 , the curve shape, and bonding type. For example, materials havinglarge bonding energies typically also have high melting temperatures; at roomtemperature, solid substances are formed for large bonding energies, whereas forsmall energies the gaseous state is favored; liquids prevail when the energies areof intermediate magnitude. In addition, as discussed in Section 7.3, the mechanicalstiffness (or modulus of elasticity) of a material is dependent on the shape of itsforce-versus-interatomic separation curve (Figure 7.7). The slope for a relativelystiff material at the r � r0 position on the curve will be quite steep; slopes areshallower for more flexible materials. Furthermore, how much a material expandsupon heating or contracts upon cooling (that is, its linear coefficient of thermalexpansion) is related to the shape of its E0-versus-r0 curve �(see Section 17.3).� Adeep and narrow ‘‘trough,’’ which typically occurs for materials having large bondingenergies, normally correlates with a low coefficient of thermal expansion and rela-tively small dimensional alterations for changes in temperature.

Three different types of primary or chemical bond are found in solids—ionic,covalent, and metallic. For each type, the bonding necessarily involves the valenceelectrons; furthermore, the nature of the bond depends on the electron structuresof the constituent atoms. In general, each of these three types of bonding arisesfrom the tendency of the atoms to assume stable electron structures, like those ofthe inert gases, by completely filling the outermost electron shell.

Secondary or physical forces and energies are also found in many solid materials;they are weaker than the primary ones, but nonetheless influence the physicalproperties of some materials. The sections that follow explain the several kinds ofprimary and secondary interatomic bonds.

2.6 PRIMARY INTERATOMIC BONDS

IONIC BONDINGPerhaps ionic bonding is the easiest to describe and visualize. It is always found incompounds that are composed of both metallic and nonmetallic elements, elementsthat are situated at the horizontal extremities of the periodic table. Atoms of ametallic element easily give up their valence electrons to the nonmetallic atoms.In the process all the atoms acquire stable or inert gas configurations and, inaddition, an electrical charge; that is, they become ions. Sodium chloride (NaCl)is the classical ionic material. A sodium atom can assume the electron structure ofneon (and a net single positive charge) by a transfer of its one valence 3s electron

2.6 Primary Interatomic Bonds ● 21

to a chlorine atom. After such a transfer, the chlorine ion has a net negative chargeand an electron configuration identical to that of argon. In sodium chloride, all thesodium and chlorine exist as ions. This type of bonding is illustrated schematicallyin Figure 2.9.

The attractive bonding forces are coulombic; that is, positive and negative ions,by virtue of their net electrical charge, attract one another. For two isolated ions,the attractive energy EA is a function of the interatomic distance according to3

EA � �Ar

(2.8)

An analogous equation for the repulsive energy is

ER �Brn (2.9)

In these expressions, A, B, and n are constants whose values depend on the particularionic system. The value of n is approximately 8.

Ionic bonding is termed nondirectional, that is, the magnitude of the bond isequal in all directions around an ion. It follows that for ionic materials to be stable,all positive ions must have as nearest neighbors negatively charged ions in a three-dimensional scheme, and vice versa. The predominant bonding in ceramic materialsis ionic. Some of the ion arrangements for these materials are discussed in Chapter 3.

Bonding energies, which generally range between 600 and 1500 kJ/mol (3 and8 eV/atom), are relatively large, as reflected in high melting temperatures.4 Table

Na+ Na+

Coulombic bonding force

Na+Cl� Cl�

Na+ Na+ Na+Cl� Cl�

Na+ Na+Cl�Cl� Cl�

Na+ Na+Cl�Cl� Cl�

FIGURE 2.9 Schematic representation of ionicbonding in sodium chloride (NaCl).

3 The constant A in Equation 2.8 is equal to

14��0

(Z1e)(Z2e)

where �0 is the permittivity of a vacuum (8.85 � 10�12 F/m), Z1 and Z2 are the valences ofthe two ion types, and e is the electronic charge (1.602 � 10�19 C).4 Sometimes bonding energies are expressed per atom or per ion. Under these circumstancesthe electron volt (eV) is a conveniently small unit of energy. It is, by definition, the energyimparted to an electron as it falls through an electric potential of one volt. The joule equivalentof the electron volt is as follows: 1.602 � 10�19 J � 1 eV.

22 ● Chapter 2 / Atomic Structure and Interatomic Bonding

2.3 contains bonding energies and melting temperatures for several ionic materials.Ionic materials are characteristically hard and brittle and, furthermore, electricallyand thermally insulative. As discussed in subsequent chapters, these properties area direct consequence of electron configurations and/or the nature of the ionic bond.

COVALENT BONDINGIn covalent bonding stable electron configurations are assumed by the sharing ofelectrons between adjacent atoms. Two atoms that are covalently bonded will eachcontribute at least one electron to the bond, and the shared electrons may beconsidered to belong to both atoms. Covalent bonding is schematically illustratedin Figure 2.10 for a molecule of methane (CH4). The carbon atom has four valenceelectrons, whereas each of the four hydrogen atoms has a single valence electron.Each hydrogen atom can acquire a helium electron configuration (two 1s valenceelectrons) when the carbon atom shares with it one electron. The carbon now hasfour additional shared electrons, one from each hydrogen, for a total of eight valence

Table 2.3 Bonding Energies and Melting Temperatures forVarious Substances

Bonding Energy MeltingkJ/mol eV/Atom, Temperature

Bonding Type Substance (kcal/mol) Ion, Molecule (�C )

IonicNaCl 640 (153) 3.3 801MgO 1000 (239) 5.2 2800

CovalentSi 450 (108) 4.7 1410C (diamond) 713 (170) 7.4 �3550

Hg 68 (16) 0.7 �39

MetallicAl 324 (77) 3.4 660Fe 406 (97) 4.2 1538W 849 (203) 8.8 3410

van der WaalsAr 7.7 (1.8) 0.08 �189Cl2 31 (7.4) 0.32 �101

HydrogenNH3 35 (8.4) 0.36 �78H2O 51 (12.2) 0.52 0

Shared electronfrom hydrogen

Shared electronfrom carbon

HH C

H

H

FIGURE 2.10 Schematic representation ofcovalent bonding in a molecule of methane(CH4).

2.6 Primary Interatomic Bonds ● 23

electrons, and the electron structure of neon. The covalent bond is directional; thatis, it is between specific atoms and may exist only in the direction between oneatom and another that participates in the electron sharing.

Many nonmetallic elemental molecules (H2 , Cl2 , F2 , etc.) as well as moleculescontaining dissimilar atoms, such as CH4 , H2O, HNO3 , and HF, are covalentlybonded. Furthermore, this type of bonding is found in elemental solids such asdiamond (carbon), silicon, and germanium and other solid compounds composedof elements that are located on the right-hand side of the periodic table, such asgallium arsenide (GaAs), indium antimonide (InSb), and silicon carbide (SiC).

The number of covalent bonds that is possible for a particular atom is deter-mined by the number of valence electrons. For N� valence electrons, an atom cancovalently bond with at most 8 � N� other atoms. For example, N� � 7 for chlorine,and 8 � N� � 1, which means that one Cl atom can bond to only one other atom,as in Cl2 . Similarly, for carbon, N� � 4, and each carbon atom has 8 � 4, orfour, electrons to share. Diamond is simply the three-dimensional interconnectingstructure wherein each carbon atom covalently bonds with four other carbon atoms.This arrangement is represented in Figure 3.16.

Covalent bonds may be very strong, as in diamond, which is very hard and hasa very high melting temperature, �3550�C (6400�F), or they may be very weak, aswith bismuth, which melts at about 270�C (518�F). Bonding energies and meltingtemperatures for a few covalently bonded materials are presented in Table 2.3.Polymeric materials typify this bond, the basic molecular structure being a longchain of carbon atoms that are covalently bonded together with two of their availablefour bonds per atom. The remaining two bonds normally are shared with otheratoms, which also covalently bond. Polymeric molecular structures are discussedin detail in Chapter 4.

It is possible to have interatomic bonds that are partially ionic and partiallycovalent, and, in fact, very few compounds exhibit pure ionic or covalent bonding.For a compound, the degree of either bond type depends on the relative positionsof the constituent atoms in the periodic table (Figure 2.6) or the difference in theirelectronegativities (Figure 2.7). The wider the separation (both horizontally—relative to Group IVA—and vertically) from the lower left to the upper-right-handcorner (i.e., the greater the difference in electronegativity), the more ionic the bond.Conversely, the closer the atoms are together (i.e., the smaller the difference inelectronegativity), the greater the degree of covalency. The percent ionic characterof a bond between elements A and B (A being the most electronegative) may beapproximated by the expression

% ionic character � �1 � exp[�(0.25)(XA � XB)2]� � 100 (2.10)

where XA and XB are the electronegativities for the respective elements.

METALLIC BONDINGMetallic bonding, the final primary bonding type, is found in metals and their alloys.A relatively simple model has been proposed that very nearly approximates thebonding scheme. Metallic materials have one, two, or at most, three valence elec-trons. With this model, these valence electrons are not bound to any particularatom in the solid and are more or less free to drift throughout the entire metal.They may be thought of as belonging to the metal as a whole, or forming a ‘‘seaof electrons’’ or an ‘‘electron cloud.’’ The remaining nonvalence electrons andatomic nuclei form what are called ion cores, which possess a net positive chargeequal in magnitude to the total valence electron charge per atom. Figure 2.11 is a

24 ● Chapter 2 / Atomic Structure and Interatomic Bonding

schematic illustration of metallic bonding. The free electrons shield the positivelycharged ion cores from mutually repulsive electrostatic forces, which they wouldotherwise exert upon one another; consequently the metallic bond is nondirectionalin character. In addition, these free electrons act as a ‘‘glue’’ to hold the ion corestogether. Bonding energies and melting temperatures for several metals are listedin Table 2.3. Bonding may be weak or strong; energies range from 68 kJ/mol (0.7eV/atom) for mercury to 850 kJ/mol (8.8 eV/atom) for tungsten. Their respectivemelting temperatures are �39 and 3410�C (�38 and 6170�F).

Metallic bonding is found for Group IA and IIA elements in the periodic table,and, in fact, for all elemental metals.

Some general behaviors of the various material types (i.e., metals, ceramics,polymers) may be explained by bonding type. For example, metals are good conduc-tors of both electricity and heat, as a consequence of their free electrons (seeSections 12.5, 12.6, �and 17.4�). By way of contrast, ionically and covalently bondedmaterials are typically electrical and thermal insulators, due to the absence of largenumbers of free electrons.

Furthermore, in Section 8.5 we note that at room temperature, most metalsand their alloys fail in a ductile manner; that is, fracture occurs after the materialshave experienced significant degrees of permanent deformation. This behavior isexplained in terms of deformation mechanism (Section 8.3), which is implicitlyrelated to the characteristics of the metallic bond. Conversely, at room temperatureionically bonded materials are intrinsically brittle as a consequence of the electricallycharged nature of their component ions (see Section 8.15).

2.7 SECONDARY BONDING OR

VAN DER WAALS BONDING

Secondary, van der Waals, or physical bonds are weak in comparison to the primaryor chemical ones; bonding energies are typically on the order of only 10 kJ/mol(0.1 eV/atom). Secondary bonding exists between virtually all atoms or molecules,but its presence may be obscured if any of the three primary bonding types ispresent. Secondary bonding is evidenced for the inert gases, which have stable

+

�

+

�

+

�

+

+

�

+

�

+

�

+

+

�

+

�

+

�

+

+ + + +

Ion cores

Sea of valence electrons

FIGURE 2.11 Schematic illustration ofmetallic bonding.

2.7 Secondary Bonding or Van der Waals Bonding ● 25

electron structures, and, in addition, between molecules in molecular structuresthat are covalently bonded.

Secondary bonding forces arise from atomic or molecular dipoles. In essence,an electric dipole exists whenever there is some separation of positive and negativeportions of an atom or molecule. The bonding results from the coulombic attractionbetween the positive end of one dipole and the negative region of an adjacent one,as indicated in Figure 2.12. Dipole interactions occur between induced dipoles,between induced dipoles and polar molecules (which have permanent dipoles), andbetween polar molecules. Hydrogen bonding, a special type of secondary bonding,is found to exist between some molecules that have hydrogen as one of the constit-uents. These bonding mechanisms are now discussed briefly.

FLUCTUATING INDUCED DIPOLE BONDSA dipole may be created or induced in an atom or molecule that is normallyelectrically symmetric; that is, the overall spatial distribution of the electrons issymmetric with respect to the positively charged nucleus, as shown in Figure 2.13a.All atoms are experiencing constant vibrational motion that can cause instantaneousand short-lived distortions of this electrical symmetry for some of the atoms ormolecules, and the creation of small electric dipoles, as represented in Figure 2.13b.One of these dipoles can in turn produce a displacement of the electron distributionof an adjacent molecule or atom, which induces the second one also to become adipole that is then weakly attracted or bonded to the first; this is one type of vander Waals bonding. These attractive forces may exist between large numbers ofatoms or molecules, which forces are temporary and fluctuate with time.

The liquefaction and, in some cases, the solidification of the inert gases andother electrically neutral and symmetric molecules such as H2 and Cl2 are realizedbecause of this type of bonding. Melting and boiling temperatures are extremelylow in materials for which induced dipole bonding predominates; of all possibleintermolecular bonds, these are the weakest. Bonding energies and melting tempera-tures for argon and chlorine are also tabulated in Table 2.3.

POLAR MOLECULE-INDUCED DIPOLE BONDSPermanent dipole moments exist in some molecules by virtue of an asymmetricalarrangement of positively and negatively charged regions; such molecules aretermed polar molecules. Figure 2.14 is a schematic representation of a hydrogen

�+�+

Atomic or molecular dipoles

FIGURE 2.12 Schematic illustration of van der Waalsbonding between two dipoles.

�+

Atomic nucleusAtomic nucleus

Electron cloud

(a) (b)

Electron cloud

FIGURE 2.13 Schematicrepresentations of (a) anelectrically symmetricatom and (b) an inducedatomic dipole.

26 ● Chapter 2 / Atomic Structure and Interatomic Bonding

chloride molecule; a permanent dipole moment arises from net positive and negativecharges that are respectively associated with the hydrogen and chlorine ends of theHCl molecule.

Polar molecules can also induce dipoles in adjacent nonpolar molecules, anda bond will form as a result of attractive forces between the two molecules. Further-more, the magnitude of this bond will be greater than for fluctuating induced dipoles.

PERMANENT DIPOLE BONDSVan der Waals forces will also exist between adjacent polar molecules. The associ-ated bonding energies are significantly greater than for bonds involving induced di-poles.

The strongest secondary bonding type, the hydrogen bond, is a special case ofpolar molecule bonding. It occurs between molecules in which hydrogen is cova-lently bonded to fluorine (as in HF), oxygen (as in H2O), and nitrogen (as in NH3).For each HUF, HUO, or HUN bond, the single hydrogen electron is shared withthe other atom. Thus, the hydrogen end of the bond is essentially a positivelycharged bare proton that is unscreened by any electrons. This highly positivelycharged end of the molecule is capable of a strong attractive force with the negativeend of an adjacent molecule, as demonstrated in Figure 2.15 for HF. In essence,this single proton forms a bridge between two negatively charged atoms. Themagnitude of the hydrogen bond is generally greater than that of the other typesof secondary bonds, and may be as high as 51 kJ/mol (0.52 eV/molecule), as shownin Table 2.3. Melting and boiling temperatures for hydrogen fluoride and waterare abnormally high in light of their low molecular weights, as a consequence ofhydrogen bonding.

2.8 MOLECULES

At the conclusion of this chapter, let us take a moment to discuss the concept ofa molecule in terms of solid materials. A molecule may be defined as a group ofatoms that are bonded together by strong primary bonds. Within this context, theentirety of ionic and metallically bonded solid specimens may be considered as asingle molecule. However, this is not the case for many substances in which covalentbonding predominates; these include elemental diatomic molecules (F2 , O2 , H2 ,etc.) as well as a host of compounds (H2O, CO2 , HNO3 , C6H6 , CH4 , etc.). In the

�+

ClH

FIGURE 2.14 Schematic representation of a polar hydrogen chloride(HCl) molecule.

FH

Hydrogenbond

FH

FIGURE 2.15 Schematic representation of hydrogenbonding in hydrogen fluoride (HF).

Important Terms and Concepts ● 27

condensed liquid and solid states, bonds between molecules are weak secondaryones. Consequently, molecular materials have relatively low melting and boilingtemperatures. Most of those that have small molecules composed of a few atomsare gases at ordinary, or ambient, temperatures and pressures. On the other hand,many of the modern polymers, being molecular materials composed of extremelylarge molecules, exist as solids; some of their properties are strongly dependent onthe presence of van der Waals and hydrogen secondary bonds.

S U M M A R YThis chapter began with a survey of the fundamentals of atomic structure, presentingthe Bohr and wave-mechanical models of electrons in atoms. Whereas the Bohrmodel assumes electrons to be particles orbiting the nucleus in discrete paths, inwave mechanics we consider them to be wavelike and treat electron position interms of a probability distribution.

Electron energy states are specified in terms of quantum numbers that giverise to electron shells and subshells. The electron configuration of an atom corre-sponds to the manner in which these shells and subshells are filled with electronsin compliance with the Pauli exclusion principle. The periodic table of the elementsis generated by arrangement of the various elements according to valence electronconfiguration.

Atomic bonding in solids may be considered in terms of attractive and repulsiveforces and energies. The three types of primary bond in solids are ionic, covalent, andmetallic. For ionic bonds, electrically charged ions are formed by the transference ofvalence electrons from one atom type to another; forces are coulombic. There is asharing of valence electrons between adjacent atoms when bonding is covalent.With metallic bonding, the valence electrons form a ‘‘sea of electrons’’ that isuniformly dispersed around the metal ion cores and acts as a form of glue for them.

Both van der Waals and hydrogen bonds are termed secondary, being weak incomparison to the primary ones. They result from attractive forces between electricdipoles, of which there are two types—induced and permanent. For the hydrogenbond, highly polar molecules form when hydrogen covalently bonds to a nonmetallicelement such as fluorine.

I M P O R T A N T T E R M S A N D C O N C E P T S

Atomic mass unit (amu)Atomic numberAtomic weightBohr atomic modelBonding energyCoulombic forceCovalent bondDipole (electric)Electron configurationElectron state

Note: In each chapter, most of the terms listed in the ‘‘Important Terms and Concepts’’section are defined in the Glossary, which follows Appendix E. The others are importantenough to warrant treatment in a full section of the text and can be referenced from thetable of contents or the index.

ElectronegativeElectropositiveGround stateHydrogen bondIonic bondIsotopeMetallic bondMoleMoleculePauli exclusion principle

Periodic tablePolar moleculePrimary bondingQuantum mechanicsQuantum numberSecondary bondingValence electronvan der Waals bondWave-mechanical model

Q U E S T I O N S A N D P R O B L E M S

2.1 (a) What is an isotope?

(b) Why are the atomic weights of the ele-ments not integers? Cite two reasons.

2.2 Cite the difference between atomic mass andatomic weight.

2.3 (a) How many grams are there in 1 amu ofa material?

(b) Mole, in the context of this book, is takenin units of gram-mole. On this basis, howmany atoms are there in a pound-mole ofa substance?

2.4 (a) Cite two important quantum-mechanicalconcepts associated with the Bohr model ofthe atom.

(b) Cite two important additional refine-ments that resulted from the wave-mechanicalatomic model.

2.5 Relative to electrons and electron states, whatdoes each of the four quantum numbersspecify?

2.6 Allowed values for the quantum numbers ofelectrons are as follows:

n � 1, 2, 3, . . .

l � 0, 1, 2, 3, . . . , n � 1

ml � 0, 1, 2, 3, . . . , l

ms � ��

The relationships between n and the shell des-ignations are noted in Table 2.1. Relative tothe 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 numbersfor each of the two electrons in the 1s state, inthe order of nlmlms , are 100( ��) and 100(� ��).

28 ● Chapter 2 / Atomic Structure and Interatomic Bonding

R E F E R E N C E S

Most of the material in this chapter is covered incollege-level chemistry textbooks. Below, two arelisted as references.Kotz, J. C. and P. Treichel, Jr., Chemistry and

Chemical Reactivity, 4th edition, Saunders Col-lege Publishing, Fort Worth, TX, 1999.

Masterton, W. L. and C. N. Hurley, Chemistry,Principles and Reactions, 3rd edition, SaundersCollege Publishing, Philadelphia, 1996.

Write the four quantum numbers for allof the electrons in the L and M shells, and notewhich correspond to the s, p, and d subshells.

2.7 Give the electron configurations for the fol-lowing ions: Fe2�, Fe3�, Cu�, Ba2�, Br�, andS2�.

2.8 Cesium bromide (CsBr) exhibits predomi-nantly ionic bonding. The Cs� and Br� ionshave electron structures that are identical towhich two inert gases?

2.9 With regard to electron configuration, whatdo all the elements in Group VIIA of theperiodic table have in common?

2.10 Without consulting Figure 2.6 or Table 2.2,determine whether each of the electron con-figurations given below is an inert gas, a halo-gen, an alkali metal, an alkaline earth metal,or a transition metal. Justify your choices.

(a) 1s22s22p63s23p63d 74s2.

(b) 1s22s22p63s23p6.

(c) 1s22s22p5.

(d) 1s22s22p63s2.

(e) 1s22s22p63s23p63d 24s2.

(f) 1s22s22p63s23p64s1.

2.11 (a) What electron subshell is being filled forthe rare earth series of elements on the peri-odic table?

(b) What electron subshell is being filled forthe actinide series?

2.12 Calculate the force of attraction between aK� and an O2� ion the centers of which areseparated by a distance of 1.5 nm.

2.13 The net potential energy between two adja-cent ions, EN , may be represented by the sumof Equations 2.8 and 2.9, that is,

EN � �Ar

�Brn (2.11)

Questions and Problems ● 29

Calculate the bonding energy E0 in terms ofthe parameters A, B, and n using the follow-ing procedure:

1. Differentiate EN with respect to r, and thenset the resulting expression equal to zero,since the curve of EN versus r is a minimumat E0 .

2. Solve for r in terms of A, B, and n, whichyields r0 , the equilibrium interionic spacing.

3. Determine the expression for E0 by substi-tution of r0 into Equation 2.11.

2.14 For a K�–Cl� ion pair, attractive and repulsiveenergies EA and ER , respectively, depend onthe distance between the ions r, according to

EA � �1.436

r

ER �5.86 � 10�6

r9

For these expressions, energies are expressedin electron volts per K�–Cl� pair, and r is thedistance in nanometers. The net energy EN isjust the sum of the two expressions above.

(a) Superimpose on a single plot EN , ER , andEA versus r up to 1.0 nm.

(b) On the basis of this plot, determine (i)the equilibrium spacing r0 between the K� andCl� ions, and (ii) the magnitude of the bondingenergy E0 between the two ions.

(c) Mathematically determine the r0 and E0

values using the solutions to Problem 2.13 andcompare these with the graphical results frompart b.

2.15 Consider some hypothetical X� � Y� ion pairfor which the equilibrium interionic spacingand bonding energy values are 0.35 nm and�6.13 eV, respectively. If it is known that nin Equation 2.11 has a value of 10, using theresults of Problem 2.13, determine explicit ex-pressions for attractive and repulsive energ-ies, EA and ER of Equations 2.8 and 2.9.

2.16 The net potential energy EN between two ad-jacent ions is sometimes represented by theexpression

EN � �Cr

� D exp ��r� (2.12)

in which r is the interionic separation and C,D, and are constants whose values dependon the specific material.

(a) Derive an expression for the bonding en-ergy E0 in terms of the equilibrium interionicseparation r0 and the constants D and usingthe following procedure:

1. Differentiate EN with respect to r andset the resulting expression equal to zero.2. Solve for C in terms of D, , and r0 .3. Determine the expression for E0 bysubstitution for C in Equation 2.12.

(b) Derive another expression for E0 in termsof r0 , C, and using a procedure analogousto the one outlined in part a.

2.17 (a) Briefly cite the main differences betweenionic, covalent, and metallic bonding.

(b) State the Pauli exclusion principle.

2.18 Offer an explanation as to why covalentlybonded materials are generally less densethan ionically or metallically bonded ones.

2.19 Compute the percents ionic character of theinteratomic bonds for the following com-pounds: TiO2 , ZnTe, CsCl, InSb, and MgCl2 .

2.20 Make a plot of bonding energy versus meltingtemperature for the metals listed in Table 2.3.Using this plot, approximate the bonding en-ergy for copper, which has a melting tempera-ture of 1084�C.

2.21 Using Table 2.2, determine the number ofcovalent bonds that are possible for atoms ofthe following elements: germanium, phos-phorus, selenium, and chlorine.

2.22 What type(s) of bonding would be expectedfor each of the following materials: brass (acopper-zinc alloy), rubber, barium sulfide(BaS), solid xenon, bronze, nylon, and alumi-num phosphide (AlP)?

2.23 Explain why hydrogen fluoride (HF) has ahigher boiling temperature than hydrogenchloride (HCl) (19.4 vs. �85�C), even thoughHF has a lower molecular weight.

2.24 On the basis of the hydrogen bond, explainthe anomalous behavior of water when itfreezes. That is, why is there volume expan-sion upon solidification?

C h a p t e r 3 / Structures of Metalsand Ceramics

High velocity electron

beams that are produced

when electrons are accelerated

across large voltages become

wavelike in character. Their

wavelengths are shorter than

interatomic spacings, and thus

these beams may be diffracted

by atomic planes in crystalline

materials, in the same manner

as x-rays experience

diffraction.

This photograph shows a

diffraction pattern produced

for a single crystal of gallium

arsenide using a transmission

electron microscope. The

brightest spot near the center

is produced by the incident

electron beam, which is parallel

to a �110� crystallographic

direction. Each of the other

white spots results from an

electron beam that is diffracted

by a specific set of crystallo-

graphic planes. (Photograph

courtesy of Dr. Raghaw S. Rai,

Motorola, Inc., Austin, Texas.)

Why Study Structures of Metals and Ceramics?

The properties of some materials are directly re-lated to their crystal structures. For example, pureand undeformed magnesium and beryllium, havingone crystal structure, are much more brittle (i.e.,fracture at lower degrees of deformation) than arepure and undeformed metals such as gold and sil-ver that have yet another crystal structure (see Sec-tion 8.5). �Also, the permanent magnetic and ferro-electric behaviors of some ceramic materials are

explained by their crystal structures (Sections 18.4and 12.23).�

Furthermore, significant property differences ex-ist between crystalline and noncrystalline materialshaving the same composition. For example, noncrys-talline ceramics and polymers normally are opti-cally transparent; the same materials in crystalline(or semicrystalline) form tend to be opaque or, atbest, translucent.

30

L e a r n i n g O b j e c t i v e sAfter studying this chapter you should be able to do the following:

1. Describe the difference in atomic/molecularstructure between crystalline and noncrystal-line materials.

2. Draw unit cells for face-centered cubic, body-centered cubic, and hexagonal close-packedcrystal structures.

3. Derive the relationships between unit cell edgelength and atomic radius for face-centered cu-bic and body-centered cubic crystal structures.

4. Compute the densities for metals having face-centered cubic and body-centered cubic crystalstructures given their unit cell dimensions.

5. Sketch/describe unit cells for sodium chloride,cesium chloride, zinc blende, diamond cubic,fluorite, and perovskite crystal structures. Dolikewise for the atomic structures of graphiteand a silica glass.

3.1 INTRODUCTION

Chapter 2 was concerned primarily with the various types of atomic bonding, whichare determined by the electron structure of the individual atoms. The presentdiscussion is devoted to the next level of the structure of materials, specifically, tosome of the arrangements that may be assumed by atoms in the solid state. Withinthis framework, concepts of crystallinity and noncrystallinity are introduced. Forcrystalline solids the notion of crystal structure is presented, specified in terms ofa unit cell. Crystal structures found in both metals and ceramics are then detailed,along with the scheme by which crystallographic directions and planes are expressed.Single crystals, polycrystalline, and noncrystalline materials are considered.

C R Y S T A L S T R U C T U R E S3.2 FUNDAMENTAL CONCEPTS

Solid materials may be classified according to the regularity with which atoms orions are arranged with respect to one another. A crystalline material is one in whichthe atoms are situated in a repeating or periodic array over large atomic distances;that is, long-range order exists, such that upon solidification, the atoms will positionthemselves in a repetitive three-dimensional pattern, in which each atom is bondedto its nearest-neighbor atoms. All metals, many ceramic materials, and certainpolymers form crystalline structures under normal solidification conditions. Forthose that do not crystallize, this long-range atomic order is absent; these noncrystal-line or amorphous materials are discussed briefly at the end of this chapter.

Some of the properties of crystalline solids depend on the crystal structure ofthe material, the manner in which atoms, ions, or molecules are spatially arranged.There is an extremely large number of different crystal structures all having long-range atomic order; these vary from relatively simple structures for metals, toexceedingly complex ones, as displayed by some of the ceramic and polymeric

31

6. Given the chemical formula for a ceramic com-pound, the ionic radii of its component ions, de-termine the crystal structure.

7. Given three direction index integers, sketch thedirection corresponding to these indices withina unit cell.

8. Specify the Miller indices for a plane that hasbeen drawn within a unit cell.

9. Describe how face-centered cubic and hexago-nal close-packed crystal structures may be gen-erated by the stacking of close-packed planesof atoms. Do the same for the sodium chloridecrystal structure in terms of close-packedplanes of anions.

10. Distinguish between single crystals and poly-crystalline materials.

11. Define isotropy and anisotropy with respect tomaterial properties.

materials. The present discussion deals with several common metallic and ceramiccrystal structures. The next chapter is devoted to structures for polymers.

When describing crystalline structures, atoms (or ions) are thought of as beingsolid spheres having well-defined diameters. This is termed the atomic hard spheremodel in which spheres representing nearest-neighbor atoms touch one another.An example of the hard sphere model for the atomic arrangement found in someof the common elemental metals is displayed in Figure 3.1c. In this particular caseall the atoms are identical. Sometimes the term lattice is used in the context ofcrystal structures; in this sense ‘‘lattice’’ means a three-dimensional array of pointscoinciding with atom positions (or sphere centers).

3.3 UNIT CELLS

The atomic order in crystalline solids indicates that small groups of atoms form arepetitive pattern. Thus, in describing crystal structures, it is often convenient tosubdivide the structure into small repeat entities called unit cells. Unit cells formost crystal structures are parallelepipeds or prisms having three sets of parallelfaces; one is drawn within the aggregate of spheres (Figure 3.1c), which in this casehappens to be a cube. A unit cell is chosen to represent the symmetry of thecrystal structure, wherein all the atom positions in the crystal may be generated bytranslations of the unit cell integral distances along each of its edges. Thus, the unit

FIGURE 3.1 For the face-centered cubic crystalstructure: (a) a hardsphere unit cellrepresentation, (b) areduced-sphere unit cell,and (c) an aggregate ofmany atoms. (Figure cadapted from W. G.Moffatt, G. W. Pearsall,and J. Wulff, The Structureand Properties ofMaterials, Vol. I, Structure,p. 51. Copyright 1964 byJohn Wiley & Sons, NewYork. Reprinted bypermission of JohnWiley & Sons, Inc.)

32 ● Chapter 3 / Structures of Metals and Ceramics

3.4 Metallic Crystal Structures ● 33

cell is the basic structural unit or building block of the crystal structure and definesthe crystal structure by virtue of its geometry and the atom positions within. Conve-nience usually dictates that parallelepiped corners coincide with centers of the hardsphere atoms. Furthermore, more than a single unit cell may be chosen for aparticular crystal structure; however, we generally use the unit cell having thehighest level of geometrical symmetry.

3.4 METALLIC CRYSTAL STRUCTURES

The atomic bonding in this group of materials is metallic, and thus nondirectionalin nature. Consequently, there are no restrictions as to the number and positionof nearest-neighbor atoms; this leads to relatively large numbers of nearest neigh-bors and dense atomic packings for most metallic crystal structures. Also, for metals,using the hard sphere model for the crystal structure, each sphere represents anion core. Table 3.1 presents the atomic radii for a number of metals. Three relativelysimple crystal structures are found for most of the common metals: face-centeredcubic, body-centered cubic, and hexagonal close-packed.

THE FACE-CENTERED CUBIC CRYSTAL STRUCTUREThe crystal structure found for many metals has a unit cell of cubic geometry, withatoms located at each of the corners and the centers of all the cube faces. It is aptlycalled the face-centered cubic (FCC) crystal structure. Some of the familiar metalshaving this crystal structure are copper, aluminum, silver, and gold (see also Table3.1). Figure 3.1a shows a hard sphere model for the FCC unit cell, whereas in Figure3.1b the atom centers are represented by small circles to provide a better perspectiveof atom positions. The aggregate of atoms in Figure 3.1c represents a section ofcrystal consisting of many FCC unit cells. These spheres or ion cores touch oneanother across a face diagonal; the cube edge length a and the atomic radius R arerelated through

a � 2R �2 (3.1)

This result is obtained as an example problem.For the FCC crystal structure, each corner atom is shared among eight unit

cells, whereas a face-centered atom belongs to only two. Therefore, one eighth of

Table 3.1 Atomic Radii and Crystal Structures for 16 Metals

Atomic AtomicCrystal Radiusb Crystal Radius

Metal Structurea (nm) Metal Structure (nm)Aluminum FCC 0.1431 Molybdenum BCC 0.1363Cadmium HCP 0.1490 Nickel FCC 0.1246Chromium BCC 0.1249 Platinum FCC 0.1387Cobalt HCP 0.1253 Silver FCC 0.1445Copper FCC 0.1278 Tantalum BCC 0.1430Gold FCC 0.1442 Titanium (�) HCP 0.1445Iron (�) BCC 0.1241 Tungsten BCC 0.1371Lead FCC 0.1750 Zinc HCP 0.1332a FCC � face-centered cubic; HCP � hexagonal close-packed; BCC � body-centeredcubic.b A nanometer (nm) equals 10�9 m; to convert from nanometers to angstrom units (A),multiply the nanometer value by 10.

each of the eight corner atoms and one half of each of the six face atoms, or a totalof four whole atoms, may be assigned to a given unit cell. This is depicted in Figure3.1a, where only sphere portions are represented within the confines of the cube.The cell comprises the volume of the cube, which is generated from the centers ofthe corner atoms as shown in the figure.

Corner and face positions are really equivalent; that is, translation of the cubecorner from an original corner atom to the center of a face atom will not alter thecell structure.

Two other important characteristics of a crystal structure are the coordinationnumber and the atomic packing factor (APF). For metals, each atom has the samenumber of nearest-neighbor or touching atoms, which is the coordination number.For face-centered cubics, the coordination number is 12. This may be confirmedby examination of Figure 3.1a; the front face atom has four corner nearest-neighboratoms surrounding it, four face atoms that are in contact from behind, and fourother equivalent face atoms residing in the next unit cell to the front, which isnot shown.

The APF is the fraction of solid sphere volume in a unit cell, assuming theatomic hard sphere model, or

APF �volume of atoms in a unit cell

total unit cell volume(3.2)

For the FCC structure, the atomic packing factor is 0.74, which is the maximumpacking possible for spheres all having the same diameter. Computation of thisAPF is also included as an example problem. Metals typically have relatively largeatomic packing factors to maximize the shielding provided by the free electron cloud.

THE BODY-CENTERED CUBIC CRYSTAL STRUCTUREAnother common metallic crystal structure also has a cubic unit cell with atomslocated at all eight corners and a single atom at the cube center. This is called abody-centered cubic (BCC) crystal structure. A collection of spheres depicting thiscrystal structure is shown in Figure 3.2c, whereas Figures 3.2a and 3.2b are diagrams

34 ● Chapter 3 / Structures of Metals and Ceramics

FIGURE 3.2 For the body-centered cubic crystal structure, (a) a hard sphereunit cell representation, (b) a reduced-sphere unit cell, and (c) an aggregate ofmany atoms. (Figure (c) from W. G. Moffatt, G. W. Pearsall, and J. Wulff, TheStructure and Properties of Materials, Vol. I, Structure, p. 51. Copyright 1964by John Wiley & Sons, New York. Reprinted by permission of John Wiley &Sons, Inc.)

3.4 Metallic Crystal Structures ● 35

of BCC unit cells with the atoms represented by hard sphere and reduced-spheremodels, respectively. Center and corner atoms touch one another along cube diago-nals, and unit cell length a and atomic radius R are related through

a �4R

�3(3.3)

Chromium, iron, tungsten, as well as several other metals listed in Table 3.1 exhibita BCC structure.

Two atoms are associated with each BCC unit cell: the equivalent of one atomfrom the eight corners, each of which is shared among eight unit cells, and thesingle center atom, which is wholly contained within its cell. In addition, cornerand center atom positions are equivalent. The coordination number for the BCCcrystal structure is 8; each center atom has as nearest neighbors its eight corneratoms. Since the coordination number is less for BCC than FCC, so also is theatomic packing factor for BCC lower—0.68 versus 0.74.

THE HEXAGONAL CLOSE-PACKED CRYSTAL STRUCTURENot all metals have unit cells with cubic symmetry; the final common metallic crystalstructure to be discussed has a unit cell that is hexagonal. Figure 3.3a shows areduced-sphere unit cell for this structure, which is termed hexagonal close-packed(HCP); an assemblage of several HCP unit cells is presented in Figure 3.3b. Thetop and bottom faces of the unit cell consist of six atoms that form regular hexagonsand surround a single atom in the center. Another plane that provides three addi-tional atoms to the unit cell is situated between the top and bottom planes. Theatoms in this midplane have as nearest neighbors atoms in both of the adjacenttwo planes. The equivalent of six atoms is contained in each unit cell; one-sixth ofeach of the 12 top and bottom face corner atoms, one-half of each of the 2 centerface atoms, and all the 3 midplane interior atoms. If a and c represent, respectively,

FIGURE 3.3 For the hexagonal close-packed crystal structure, (a) a reduced-sphere unitcell (a and c represent the short and long edge lengths, respectively), and (b) anaggregate of many atoms. (Figure (b) from W. G. Moffatt, G. W. Pearsall, and J. Wulff,The Structure and Properties of Materials, Vol. I, Structure, p. 51. Copyright 1964 byJohn Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.)

the short and long unit cell dimensions of Figure 3.3a, the c/a ratio should be 1.633;however, for some HCP metals this ratio deviates from the ideal value.

The coordination number and the atomic packing factor for the HCP crystalstructure are the same as for FCC: 12 and 0.74, respectively. The HCP metals includecadmium, magnesium, titanium, and zinc; some of these are listed in Table 3.1.

EXAMPLE PROBLEM 3.1

Calculate the volume of an FCC unit cell in terms of the atomic radius R.

S O L U T I O N

In the FCC unit cell illustrated,

the atoms touch one another across a face-diagonal the length of which is 4R.Since the unit cell is a cube, its volume is a 3, where a is the cell edge length.From the right triangle on the face,

a2 � a2 � (4R)2

or, solving for a,

a � 2R �2 (3.1)

The FCC unit cell volume VC may be computed from

VC � a 3 � (2R �2) 3 � 16R 3 �2 (3.4)

EXAMPLE PROBLEM 3.2

Show that the atomic packing factor for the FCC crystal structure is 0.74.

S O L U T I O N

The APF is defined as the fraction of solid sphere volume in a unit cell, or

APF �total sphere volume

total unit cell volume�

VS

VC

Both the total sphere and unit cell volumes may be calculated in terms of theatomic radius R. The volume for a sphere is ���R 3, and since there are four

36 ● Chapter 3 / Structures of Metals and Ceramics

a

a

4R

R

3.5 Density Computations—Metals ● 37

atoms per FCC unit cell, the total FCC sphere volume is

VS � (4)43

�R 3 �163

�R 3

From Example Problem 3.1, the total unit cell volume is

VC � 16R 3 �2

Therefore, the atomic packing factor is

APF �VS

VC�

(��� )�R 3

16R 3 �2� 0.74

3.5 DENSITY COMPUTATIONS—METALS

A knowledge of the crystal structure of a metallic solid permits computation of itstheoretical density � through the relationship

� �nA

VC NA(3.5)

where

n � number of atoms associated with each unit cell

A � atomic weight

VC � volume of the unit cell

NA � Avogadro’s number (6.023 � 1023 atoms/mol)

EXAMPLE PROBLEM 3.3

Copper has an atomic radius of 0.128 nm (1.28 A), an FCC crystal structure, andan atomic weight of 63.5 g/mol. Compute its theoretical density and compare theanswer with its measured density.

S O L U T I O N

Equation 3.5 is employed in the solution of this problem. Since the crystalstructure is FCC, n, the number of atoms per unit cell, is 4. Furthermore, theatomic weight ACu is given as 63.5 g/mol. The unit cell volume VC for FCCwas determined in Example Problem 3.1 as 16R3 �2, where R, the atomicradius, is 0.128 nm.

Substitution for the various parameters into Equation 3.5 yields

� �nACu

VC NA�

nACu

(16R 3 �2)NA

�(4 atoms/unit cell)(63.5 g/mol)

[16 �2(1.28 � 10�8 cm)3/unit cell](6.023 � 1023 atoms/mol)

� 8.89 g/cm3

The literature value for the density of copper is 8.94 g/cm3, which is in veryclose agreement with the foregoing result.

3.6 CERAMIC CRYSTAL STRUCTURES

Because ceramics are composed of at least two elements, and often more, theircrystal structures are generally more complex than those for metals. The atomicbonding in these materials ranges from purely ionic to totally covalent; many ceram-ics exhibit a combination of these two bonding types, the degree of ionic characterbeing dependent on the electronegativities of the atoms. Table 3.2 presents thepercent ionic character for several common ceramic materials; these values weredetermined using Equation 2.10 and the electronegativities in Figure 2.7.

For those ceramic materials for which the atomic bonding is predominantlyionic, the crystal structures may be thought of as being composed of electricallycharged ions instead of atoms. The metallic ions, or cations, are positively charged,because they have given up their valence electrons to the nonmetallic ions, oranions, which are negatively charged. Two characteristics of the component ionsin crystalline ceramic materials influence the crystal structure: the magnitude ofthe electrical charge on each of the component ions, and the relative sizes of thecations and anions. With regard to the first characteristic, the crystal must beelectrically neutral; that is, all the cation positive charges must be balanced by anequal number of anion negative charges. The chemical formula of a compoundindicates the ratio of cations to anions, or the composition that achieves this chargebalance. For example, in calcium fluoride, each calcium ion has a �2 charge (Ca2�),and associated with each fluorine ion is a single negative charge (F�). Thus, theremust be twice as many F� as Ca2� ions, which is reflected in the chemical for-mula CaF2 .

The second criterion involves the sizes or ionic radii of the cations and anions,rC and rA , respectively. Because the metallic elements give up electrons whenionized, cations are ordinarily smaller than anions, and, consequently, the ratiorC/rA is less than unity. Each cation prefers to have as many nearest-neighbor anionsas possible. The anions also desire a maximum number of cation nearest neighbors.

Stable ceramic crystal structures form when those anions surrounding a cationare all in contact with that cation, as illustrated in Figure 3.4. The coordinationnumber (i.e., number of anion nearest neighbors for a cation) is related to thecation–anion radius ratio. For a specific coordination number, there is a critical orminimum rC/rA ratio for which this cation–anion contact is established (Figure 3.4),which ratio may be determined from pure geometrical considerations (see ExampleProblem 3.4).

38 ● Chapter 3 / Structures of Metals and Ceramics

Table 3.2 For SeveralCeramic Materials, PercentIonic Character of theInteratomic Bonds

Percent IonicMaterial Character

CaF2 89MgO 73NaCl 67Al2O3 63SiO2 51Si3N4 30ZnS 18SiC 12

3.6 Ceramic Crystal Structures ● 39

The coordination numbers and nearest-neighbor geometries for various rC/rA

ratios are presented in Table 3.3. For rC/rA ratios less than 0.155, the very smallcation is bonded to two anions in a linear manner. If rC/rA has a value between0.155 and 0.225, the coordination number for the cation is 3. This means each cationis surrounded by three anions in the form of a planar equilateral triangle, with thecation located in the center. The coordination number is 4 for rC/rA between 0.225and 0.414; the cation is located at the center of a tetrahedron, with anions at eachof the four corners. For rC/rA between 0.414 and 0.732, the cation may be thoughtof as being situated at the center of an octahedron surrounded by six anions, oneat each corner, as also shown in the table. The coordination number is 8 for rC/rA

between 0.732 and 1.0, with anions at all corners of a cube and a cation positionedat the center. For a radius ratio greater than unity, the coordination number is 12.The most common coordination numbers for ceramic materials are 4, 6, and 8.Table 3.4 gives the ionic radii for several anions and cations that are common inceramic materials.

EXAMPLE PROBLEM 3.4

Show that the minimum cation-to-anion radius ratio for the coordination num-ber 3 is 0.155.

S O L U T I O N

For this coordination, the small cation is surrounded by three anions to forman equilateral triangle as shown below—triangle ABC; the centers of all fourions are coplanar.

This boils down to a relatively simple plane trigonometry problem. Consid-eration of the right triangle APO makes it clear that the side lengths are relatedto the anion and cation radii rA and rC as

AP � rA

Cation

Anion

rA

rC

�

BOC

AP

Stable Stable Unstable

FIGURE 3.4 Stable and unstableanion–cation coordinationconfigurations. Open circles representanions; colored circles denote cations.

and

AO � rA � rC

Furthermore, the side length ratio AP/AO is a function of the angle � as

APAO

� cos �

40 ● Chapter 3 / Structures of Metals and Ceramics

Table 3.3 Coordination Numbers andGeometries for Various Cation–AnionRadius Ratios (rC/rA)

Coordination Cation–Anion CoordinationNumber Radius Ratio Geometry

2 �0.155

3 0.155–0.225

4 0.225–0.414

6 0.414–0.732

8 0.732–1.0

Source: W. D. Kingery, H. K. Bowen, and D. R. Uhlmann,Introduction to Ceramics, 2nd edition. Copyright 1976 byJohn Wiley & Sons, New York. Reprinted by permission ofJohn Wiley & Sons, Inc.

3.6 Ceramic Crystal Structures ● 41

The magnitude of � is 30�, since line AO bisects the 60� angle BAC. Thus,

APAO

�rA

rA � rC� cos 30� �

�32

Or, solving for the cation–anion radius ratio,

rC

rA�

1 � �3/2

�3/2� 0.155

AX-TYPE CRYSTAL STRUCTURESSome of the common ceramic materials are those in which there are equal numbersof cations and anions. These are often referred to as AX compounds, where Adenotes the cation and X the anion. There are several different crystal structuresfor AX compounds; each is normally named after a common material that assumesthe particular structure.

Rock Salt StructurePerhaps the most common AX crystal structure is the sodium chloride (NaCl), orrock salt, type. The coordination number for both cations and anions is 6, andtherefore the cation–anion radius ratio is between approximately 0.414 and 0.732.A unit cell for this crystal structure (Figure 3.5) is generated from an FCC arrange-ment of anions with one cation situated at the cube center and one at the centerof each of the 12 cube edges. An equivalent crystal structure results from a face-centered arrangement of cations. Thus, the rock salt crystal structure may be thoughtof as two interpenetrating FCC lattices, one composed of the cations, the other ofanions. Some of the common ceramic materials that form with this crystal structureare NaCl, MgO, MnS, LiF, and FeO.

Cesium Chloride StructureFigure 3.6 shows a unit cell for the cesium chloride (CsCl) crystal structure; thecoordination number is 8 for both ion types. The anions are located at each of thecorners of a cube, whereas the cube center is a single cation. Interchange of anions

Table 3.4 Ionic Radii for Several Cations and Anions(for a Coordination Number of 6)

Ionic Radius Ionic RadiusCation (nm) Anion (nm)Al3� 0.053 Br� 0.196Ba2� 0.136 Cl� 0.181Ca2� 0.100 F� 0.133Cs� 0.170 I� 0.220Fe2� 0.077 O2� 0.140Fe3� 0.069 S2� 0.184K� 0.138Mg2� 0.072Mn2� 0.067Na� 0.102Ni2� 0.069Si4� 0.040Ti4� 0.061

with cations, and vice versa, produces the same crystal structure. This is not a BCCcrystal structure because ions of two different kinds are involved.

Zinc Blende StructureA third AX structure is one in which the coordination number is 4; that is, all ionsare tetrahedrally coordinated. This is called the zinc blende, or sphalerite, structure,after the mineralogical term for zinc sulfide (ZnS). A unit cell is presented in Figure3.7; all corner and face positions of the cubic cell are occupied by S atoms, whilethe Zn atoms fill interior tetrahedral positions. An equivalent structure results ifZn and S atom positions are reversed. Thus, each Zn atom is bonded to four Satoms, and vice versa. Most often the atomic bonding is highly covalent in com-pounds exhibiting this crystal structure (Table 3.2), which include ZnS, ZnTe,and SiC.

AmXp-TYPE CRYSTAL STRUCTURESIf the charges on the cations and anions are not the same, a compound can existwith the chemical formula AmXp , where m and/or p � 1. An example would beAX2 , for which a common crystal structure is found in fluorite (CaF2). The ionicradii ratio rC/rA for CaF2 is about 0.8 which, according to Table 3.3, gives a coordina-tion number of 8. Calcium ions are positioned at the centers of cubes, with fluorine

42 ● Chapter 3 / Structures of Metals and Ceramics

Na+ Cl�

FIGURE 3.5 A unit cell for the rock salt, or sodiumchloride (NaCl), crystal structure.

FIGURE 3.6 A unit cell for the cesium chloride (CsCl)crystal structure.

Cl�Cs+

3.6 Ceramic Crystal Structures ● 43

ions at the corners. The chemical formula shows that there are only half as manyCa2� ions as F� ions, and therefore the crystal structure would be similar to CsCl(Figure 3.6), except that only half the center cube positions are occupied by Ca2�

ions. One unit cell consists of eight cubes, as indicated in Figure 3.8. Other com-pounds that have this crystal structure include UO2 , PuO2 , and ThO2 .

AmBnXp-TYPE CRYSTAL STRUCTURESIt is also possible for ceramic compounds to have more than one type of cation;for two types of cations (represented by A and B), their chemical formula may bedesignated as AmBnXp . Barium titanate (BaTiO3), having both Ba2� and Ti4� cations,falls into this classification. This material has a perovskite crystal structure and ratherinteresting electromechanical properties to be discussed later. At temperaturesabove 120�C (248�F), the crystal structure is cubic. A unit cell of this structure isshown in Figure 3.9; Ba2� ions are situated at all eight corners of the cube and asingle Ti4� is at the cube center, with O2� ions located at the center of each of thesix faces.

Table 3.5 summarizes the rock salt, cesium chloride, zinc blende, fluorite, andperovskite crystal structures in terms of cation–anion ratios and coordination num-bers, and gives examples for each. Of course, many other ceramic crystal structuresare possible.

FIGURE 3.7 A unit cell for the zinc blende (ZnS)crystal structure.

FIGURE 3.8 A unit cell for the fluorite (CaF2)crystal structure.

SZn

Ca2+ F�

EXAMPLE PROBLEM 3.5

On the basis of ionic radii, what crystal structure would you predict for FeO?

S O L U T I O N

First, note that FeO is an AX-type compound. Next, determine the cation–anionradius ratio, which from Table 3.4 is

rFe2�

rO2��

0.077 nm0.140 nm

� 0.550

This value lies between 0.414 and 0.732, and, therefore, from Table 3.3 thecoordination number for the Fe2� ion is 6; this is also the coordination numberof O2�, since there are equal numbers of cations and anions. The predictedcrystal structure will be rock salt, which is the AX crystal structure having acoordination number of 6, as given in Table 3.5.

44 ● Chapter 3 / Structures of Metals and Ceramics

FIGURE 3.9 A unit cell for the perovskite crystalstructure.

Table 3.5 Summary of Some Common Ceramic Crystal Structures

CoordinationNumbersStructure

Anion PackingTypeStructure Name Cation Anion ExamplesRock salt (sodium AX FCC 6 6 NaCl, MgO, FeO

chloride)Cesium chloride AX Simple cubic 8 8 CsClZinc blende AX FCC 4 4 ZnS, SiC

(sphalerite)Fluorite AX2 Simple cubic 8 4 CaF2 , UO2 , ThO2

Perovskite ABX3 FCC 12(A) 6 BaTiO3 , SrZrO3 ,6(B) SrSnO3

Spinel AB2X4 FCC 4(A) 4 MgAl2O4 , FeAl2O4

6(B)

Source: W. D. Kingery, H. K. Bowen, and D. R. Uhlmann, Introduction to Ceramics, 2nd edition. Copyright 1976 by John Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons, Inc.

Ti4+ Ba2+ O2�

3.7 Density Computations—Ceramics ● 45

3.7 DENSITY COMPUTATIONS—CERAMICS

It is possible to compute the theoretical density of a crystalline ceramic materialfrom unit cell data in a manner similar to that described in Section 3.5 for metals.In this case the density � may be determined using a modified form of Equation3.5, as follows:

� �n�(�AC � �AA)

VC NA(3.6)

where

n� � the number of formula units1 within the unit cell

�AC � the sum of the atomic weights of all cations in the formula unit

�AA � the sum of the atomic weights of all anions in the formula unit

VC � the unit cell volume

NA � Avogadro’s number, 6.023 � 1023 formula units/mol

EXAMPLE PROBLEM 3.6

On the basis of crystal structure, compute the theoretical density for sodiumchloride. How does this compare with its measured density?

S O L U T I O N

The density may be determined using Equation 3.6, where n�, the number ofNaCl units per unit cell, is 4 because both sodium and chloride ions form FCClattices. Furthermore,

�AC � ANa � 22.99 g/mol

�AA � ACl � 35.45 g/mol

Since the unit cell is cubic, VC � a3, a being the unit cell edge length. For theface of the cubic unit cell shown below,

a � 2rNa� � 2rCl�

rNa� and rCl� being the sodium and chlorine ionic radii, given in Table 3.4 as0.102 and 0.181 nm, respectively.

Thus,

VC � a3 � (2rNa� � 2rCl�)3

1 By ‘‘formula unit’’ we mean all the ions that are included in the chemical formula unit.For example, for BaTiO3 , a formula unit consists of one barium ion, a titanium ion, andthree oxygen ions.

And finally,

� �n�(ANa � ACl)

(2rNa� � 2rCl�)3NA

�4(22.99 � 35.45)

[2(0.102 � 10�7) � 2(0.181 � 10�7)]3(6.023 � 1023)

� 2.14 g/cm3

This compares very favorably with the experimental value of 2.16 g/cm3.

3.8 SILICATE CERAMICS

Silicates are materials composed primarily of silicon and oxygen, the two mostabundant elements in the earth’s crust; consequently, the bulk of soils, rocks, clays,and sand come under the silicate classification. Rather than characterizing the crystalstructures of these materials in terms of unit cells, it is more convenient to usevarious arrangements of an SiO4

4� tetrahedron (Figure 3.10). Each atom of silicon

is bonded to four oxygen atoms, which are situated at the corners of the tetrahedron;the silicon atom is positioned at the center. Since this is the basic unit of the silicates,it is often treated as a negatively charged entity.

Often the silicates are not considered to be ionic because there is a significantcovalent character to the interatomic Si–O bonds (Table 3.2), which bonds aredirectional and relatively strong. Regardless of the character of the Si–O bond,there is a �4 charge associated with every SiO4

4� tetrahedron, since each of the

four oxygen atoms requires an extra electron to achieve a stable electronic structure.Various silicate structures arise from the different ways in which the SiO4

4� units

can be combined into one-, two-, and three-dimensional arrangements.

46 ● Chapter 3 / Structures of Metals and Ceramics

a

2(rNa+ + rCl�)

rNa+

rCl�

Na+ Cl�

3.9 Carbon ● 47

SILICAChemically, the most simple silicate material is silicon dioxide, or silica (SiO2).Structurally, it is a three-dimensional network that is generated when every corneroxygen atom in each tetrahedron is shared by adjacent tetrahedra. Thus, the materialis electrically neutral and all atoms have stable electronic structures. Under thesecircumstances the ratio of Si to O atoms is 1 : 2, as indicated by the chemical formula.

If these tetrahedra are arrayed in a regular and ordered manner, a crystallinestructure is formed. There are three primary polymorphic crystalline forms ofsilica: quartz, cristobalite (Figure 3.11), and tridymite. Their structures are relativelycomplicated, and comparatively open; that is, the atoms are not closely packedtogether. As a consequence, these crystalline silicas have relatively low densities;for example, at room temperature quartz has a density of only 2.65 g/cm3. Thestrength of the Si–O interatomic bonds is reflected in a relatively high meltingtemperature, 1710�C (3110�F).

Silica can also be made to exist as a noncrystalline solid or glass; its structureis discussed in Section 3.20.

THE SILICATES (CD-ROM)

3.9 CARBON

Carbon is an element that exists in various polymorphic forms, as well as in theamorphous state. This group of materials does not really fall within any one of thetraditional metal, ceramic, polymer classification schemes. However, it has beendecided to discuss these materials in this chapter since graphite, one of the polymor-phic forms, is sometimes classified as a ceramic. This treatment focuses on the

�

�

�

� Si4+ O2�

FIGURE 3.10 Asilicon–oxygen (SiO4

4�)

tetrahedron.

Si4+ O2�

FIGURE 3.11 The arrangement of silicon and oxygenatoms in a unit cell of cristobalite, a polymorph ofSiO2 .

structures of graphite and diamond �and the new fullerenes.� The characteristicsand current and potential uses of these materials are discussed in Section 13.11.

DIAMONDDiamond is a metastable carbon polymorph at room temperature and atmosphericpressure. Its crystal structure is a variant of the zinc blende, in which carbon atomsoccupy all positions (both Zn and S), as indicated in the unit cell shown in Figure3.16. Thus, each carbon bonds to four other carbons, and these bonds are totallycovalent. This is appropriately called the diamond cubic crystal structure, which isalso found for other Group IVA elements in the periodic table [e.g., germanium,silicon, and gray tin, below 13�C (55�F)].

GRAPHITEGraphite has a crystal structure (Figure 3.17) distinctly different from that of dia-mond and is also more stable than diamond at ambient temperature and pressure.The graphite structure is composed of layers of hexagonally arranged carbon atoms;within the layers, each carbon atom is bonded to three coplanar neighbor atomsby strong covalent bonds. The fourth bonding electron participates in a weak vander Waals type of bond between the layers.

FULLERENES (CD-ROM)

48 ● Chapter 3 / Structures of Metals and Ceramics

C

FIGURE 3.16 A unit cell for the diamond cubic crystalstructure.

Carbonatom

FIGURE 3.17 The structure of graphite.

3.11 Crystal Systems ● 49

3.10 POLYMORPHISM AND ALLOTROPY

Some metals, as well as nonmetals, may have more than one crystal structure, aphenomenon known as polymorphism. When found in elemental solids, the condi-tion is often termed allotropy. The prevailing crystal structure depends on both thetemperature and the external pressure. One familiar example is found in carbonas discussed in the previous section: graphite is the stable polymorph at ambientconditions, whereas diamond is formed at extremely high pressures. Also, pure ironhas a BCC crystal structure at room temperature, which changes to FCC iron at912�C (1674�F). Most often a modification of the density and other physical proper-ties accompanies a polymorphic transformation.

3.11 CRYSTAL SYSTEMS

Since there are many different possible crystal structures, it is sometimes convenientto divide them into groups according to unit cell configurations and/or atomicarrangements. One such scheme is based on the unit cell geometry, that is, theshape of the appropriate unit cell parallelepiped without regard to the atomicpositions in the cell. Within this framework, an x, y, z coordinate system is establishedwith its origin at one of the unit cell corners; each of the x, y, and z axes coincideswith one of the three parallelepiped edges that extend from this corner, as illustratedin Figure 3.19. The unit cell geometry is completely defined in terms of six parame-ters: the three edge lengths a, b, and c, and the three interaxial angles �, �, and �.These are indicated in Figure 3.19, and are sometimes termed the lattice parametersof a crystal structure.

On this basis there are found crystals having seven different possible combina-tions of a, b, and c, and �, �, and �, each of which represents a distinctcrystal system. These seven crystal systems are cubic, tetragonal, hexagonal,orthorhombic, rhombohedral, monoclinic, and triclinic. The lattice parameterrelationships and unit cell sketches for each are represented in Table 3.6. Thecubic system, for which a � b � c and � � � � � � 90�, has the greatestdegree of symmetry. Least symmetry is displayed by the triclinic system, sincea � b � c and � � � � �.

From the discussion of metallic crystal structures, it should be apparent thatboth FCC and BCC structures belong to the cubic crystal system, whereas HCPfalls within hexagonal. The conventional hexagonal unit cell really consists of threeparallelepipeds situated as shown in Table 3.6.

z

y

x

a

b

c

�

FIGURE 3.19 A unit cell with x, y, and z coordinate axes,showing axial lengths (a, b, and c) and interaxial angles (�,�, and �).

50 ● Chapter 3 / Structures of Metals and Ceramics

Table 3.6 Lattice Parameter Relationships and Figures ShowingUnit Cell Geometries for the Seven Crystal Systems

AxialCrystal System Relationships Interaxial Angles Unit Cell Geometry

Cubic a � b � c � � � � � � 90�

Hexagonal a � b � c � � � � 90�, � � 120�

Tetragonal a � b � c � � � � � � 90�

Rhombohedral a � b � c � � � � � � 90�

Orthorhombic a � b � c � � � � � � 90�

Monoclinic a � b � c � � � � 90� � �

Triclinic a � b � c � � � � � � 90�

aa

a

aaa

c

aa

c

aaa

�

ab

c

a

b

c�

ab

c�

�

�

C R Y S T A L L O G R A P H I C D I R E C T I O N SA N D P L A N E S

When dealing with crystalline materials, it often becomes necessary to specify someparticular crystallographic plane of atoms or a crystallographic direction. Labelingconventions have been established in which three integers or indices are used todesignate directions and planes. The basis for determining index values is the unitcell, with a coordinate system consisting of three (x, y, and z) axes situated at oneof the corners and coinciding with the unit cell edges, as shown in Figure 3.19. Forsome crystal systems—namely, hexagonal, rhombohedral, monoclinic, and tri-clinic—the three axes are not mutually perpendicular, as in the familiar Cartesiancoordinate scheme.

3.12 CRYSTALLOGRAPHIC DIRECTIONS

A crystallographic direction is defined as a line between two points, or a vec-tor. The following steps are utilized in the determination of the three directionalindices:

1. A vector of convenient length is positioned such that it passes throughthe origin of the coordinate system. Any vector may be translatedthroughout the crystal lattice without alteration, if parallelism is main-tained.

2. The length of the vector projection on each of the three axes is de-termined; these are measured in terms of the unit cell dimensions a, b,and c.

3. These three numbers are multiplied or divided by a common factor toreduce them to the smallest integer values.

4. The three indices, not separated by commas, are enclosed in squarebrackets, thus: [uvw]. The u, v, and w integers correspond to the reducedprojections along the x, y, and z axes, respectively.

For each of the three axes, there will exist both positive and negative coordi-nates. Thus negative indices are also possible, which are represented by a bar overthe appropriate index. For example, the [111] direction would have a componentin the �y direction. Also, changing the signs of all indices produces an antiparallel

3.12 Crystallographic Directions ● 51

FIGURE 3.20 The [100], [110], and [111] directions within aunit cell.

z

y

x

[111]

[110]

[100]

EXAMPLE PROBLEM 3.8

Draw a [110] direction within a cubic unit cell.

S O L U T I O N

First construct an appropriate unit cell and coordinate axes system. In theaccompanying figure the unit cell is cubic, and the origin of the coordinatesystem, point O, is located at one of the cube corners.

S O L U T I O N

The vector, as drawn, passes through the origin of the coordinate system, andtherefore no translation is necessary. Projections of this vector onto the x, y,and z axes are, respectively, a/2, b, and 0c, which become ��, 1, and 0 in termsof the unit cell parameters (i.e., when the a, b, and c are dropped). Reductionof these numbers to the lowest set of integers is accompanied by multiplicationof each by the factor 2. This yields the integers 1, 2, and 0, which are thenenclosed in brackets as [120].

This procedure may be summarized as follows:

x y zProjections a/2 b 0cProjections (in terms of a, b, and c) �� 1 0Reduction 1 2 0Enclosure [120]

52 ● Chapter 3 / Structures of Metals and Ceramics

direction; that is, [111] is directly opposite to [111]. If more than one direction orplane is to be specified for a particular crystal structure, it is imperative for themaintaining of consistency that a positive–negative convention, once established,not be changed.

The [100], [110], and [111] directions are common ones; they are drawn in theunit cell shown in Figure 3.20.

EXAMPLE PROBLEM 3.7

Determine the indices for the direction shown in the accompanying figure.

z

y

x

a

b

Projection ony axis (b)

Projection onx axis (a/2)

c

3.12 Crystallographic Directions ● 53

This problem is solved by reversing the procedure of the preceding example.For this [110] direction, the projections along the x, y, z axes are a, �a, and0a, respectively. This direction is defined by a vector passing from the originto point P, which is located by first moving along the x axis a units, and fromthis position, parallel to the y axis �a units, as indicated in the figure. Thereis no z component to the vector, since the z projection is zero.

For some crystal structures, several nonparallel directions with different indicesare actually equivalent; this means that the spacing of atoms along each directionis the same. For example, in cubic crystals, all the directions represented by thefollowing indices are equivalent: [100], [100], [010], [010], [001], and [001]. As aconvenience, equivalent directions are grouped together into a family, which areenclosed in angle brackets, thus: �100�. Furthermore, directions in cubic crystalshaving the same indices without regard to order or sign, for example, [123] and[213], are equivalent. This is, in general, not true for other crystal systems. Forexample, for crystals of tetragonal symmetry, [100] and [010] directions are equiva-lent, whereas [100] and [001] are not.

HEXAGONAL CRYSTALSA problem arises for crystals having hexagonal symmetry in that some crystallo-graphic equivalent directions will not have the same set of indices. This is circum-vented by utilizing a four-axis, or Miller–Bravais, coordinate system as shown inFigure 3.21. The three a1 , a2 , and a3 axes are all contained within a single plane

z

+y yO

a

aa

[110] Direction

P

x

a

a

a1

a2

a3

z

120°

FIGURE 3.21 Coordinate axis system for a hexagonal unit cell(Miller–Bravais scheme).

(called the basal plane), and at 120� angles to one another. The z axis is perpendicularto this basal plane. Directional indices, which are obtained as described above, willbe denoted by four indices, as [uvtw]; by convention, the first three indices pertainto projections along the respective a1 , a2 , and a3 axes in the basal plane.

Conversion from the three-index system to the four-index system,

[u �v �w �] —� [uvtw]

is accomplished by the following formulas:

u �n3

(2u� � v�) (3.7a)

v �n3

(2v� � u�) (3.7b)

t � �(u � v) (3.7c)

w � nw � (3.7d)

where primed indices are associated with the three-index scheme and unprimed,with the new Miller–Bravais four-index system; n is a factor that may be requiredto reduce u, v, t, and w to the smallest integers. For example, using this conversion,the [010] direction becomes [1210]. Several different directions are indicated in thehexagonal unit cell (Figure 3.22a).

3.13 CRYSTALLOGRAPHIC PLANES

The orientations of planes for a crystal structure are represented in a similar manner.Again, the unit cell is the basis, with the three-axis coordinate system as representedin Figure 3.19. In all but the hexagonal crystal system, crystallographic planes arespecified by three Miller indices as (hkl). Any two planes parallel to each otherare equivalent and have identical indices. The procedure employed in determinationof the h, k, and l index numbers is as follows:

1. If the plane passes through the selected origin, either another parallelplane must be constructed within the unit cell by an appropriate transla-tion, or a new origin must be established at the corner of another unit cell.

2. At this point the crystallographic plane either intersects or parallels eachof the three axes; the length of the planar intercept for each axis isdetermined in terms of the lattice parameters a, b, and c.

54 ● Chapter 3 / Structures of Metals and Ceramics

[0001]

(0001)

[1120]`

[1100]`

(1010)

(1011)`

a1 a1

a2

a3a3

zz

(a) (b)

`

FIGURE 3.22 For thehexagonal crystal system,(a) [0001], [1100], and [1120]directions, and (b) the (0001),(1011), and (1010) planes.

3.13 Crystallographic Planes ● 55

z

x

y

z

x

y

z

x

y

(b)

(c)

(a)

O

(001) Plane referenced tothe origin at point O

(111) Plane referenced tothe origin at point O

(110) Plane referenced to theorigin at point O

Other equivalent(001) planes

Other equivalent(111) planes

Other equivalent(110) planes

O

O

FIGURE 3.23 Representations of aseries each of (a) (001), (b) (110),and (c) (111) crystallographic planes.

3. The reciprocals of these numbers are taken. A plane that parallels anaxis may be considered to have an infinite intercept, and, therefore, azero index.

4. If necessary, these three numbers are changed to the set of smallestintegers by multiplication or division by a common factor.2

5. Finally, the integer indices, not separated by commas, are enclosed withinparentheses, thus: (hkl).

An intercept on the negative side of the origin is indicated by a bar or minussign positioned over the appropriate index. Furthermore, reversing the directionsof all indices specifies another plane parallel to, on the opposite side of and equidis-tant from, the origin. Several low-index planes are represented in Figure 3.23.

One interesting and unique characteristic of cubic crystals is that planes anddirections having the same indices are perpendicular to one another; however, forother crystal systems there are no simple geometrical relationships between planesand directions having the same indices.

2 On occasion, index reduction is not carried out �(e.g., for x-ray diffraction studies that aredescribed in Section 3.19);� for example, (002) is not reduced to (001). In addition, for ceramicmaterials, the ionic arrangement for a reduced-index plane may be different from that fora nonreduced one.

56 ● Chapter 3 / Structures of Metals and Ceramics

z

y

x

a

b

Oc

z

y

x(a) (b)

xb

c/2

(012) Plane

z

OO

�

�

�

EXAMPLE PROBLEM 3.9

Determine the Miller indices for the plane shown in the accompanyingsketch (a).

S O L U T I O N

Since the plane passes through the selected origin O, a new origin must bechosen at the corner of an adjacent unit cell, taken as O� and shown in sketch(b). This plane is parallel to the x axis, and the intercept may be taken as �a.The y and z axes intersections, referenced to the new origin O�, are �b andc/2, respectively. Thus, in terms of the lattice parameters a, b, and c, theseintersections are �, �1, and ��. The reciprocals of these numbers are 0, �1, and2; and since all are integers, no further reduction is necessary. Finally, enclosurein parentheses yields (012).

These steps are briefly summarized below:

x y zIntercepts �a �b c/2Intercepts (in terms of lattice parameters) � �1 ��

Reciprocals 0 �1 2Reductions (unnecessary)Enclosure (012)

EXAMPLE PROBLEM 3.10

Construct a (011) plane within a cubic unit cell.

S O L U T I O N

To solve this problem, carry out the procedure used in the preceding examplein reverse order. To begin, the indices are removed from the parentheses, andreciprocals are taken, which yields �, �1, and 1. This means that the particularplane parallels the x axis while intersecting the y and z axes at �b and c,respectively, as indicated in the accompanying sketch (a). This plane has beendrawn in sketch (b). A plane is indicated by lines representing its intersectionswith the planes that constitute the faces of the unit cell or their extensions. Forexample, in this figure, line ef is the intersection between the (011) plane and

3.13 Crystallographic Planes ● 57

the top face of the unit cell; also, line gh represents the intersection betweenthis same (011) plane and the plane of the bottom unit cell face extended.Similarly, lines eg and fh are the intersections between (011) and back andfront cell faces, respectively.

ATOMIC ARRANGEMENTSThe atomic arrangement for a crystallographic plane, which is often of interest,depends on the crystal structure. The (110) atomic planes for FCC and BCC crystalstructures are represented in Figures 3.24 and 3.25; reduced-sphere unit cells arealso included. Note that the atomic packing is different for each case. The circlesrepresent atoms lying in the crystallographic planes as would be obtained from aslice taken through the centers of the full-sized hard spheres.

A ‘‘family’’ of planes contains all those planes that are crystallographicallyequivalent—that is, having the same atomic packing; and a family is designated byindices that are enclosed in braces—e.g., �100�. For example, in cubic crystals the(111), (111), (111), (111), (111), (111), (111), and (111) planes all belong to the�111� family. On the other hand, for tetragonal crystal structures, the �100� familywould contain only the (100), (100), (010), and (010) since the (001) and (001)planes are not crystallographically equivalent. Also, in the cubic system only, planeshaving the same indices, irrespective of order and sign, are equivalent. For example,both (123) and (312) belong to the �123� family.

HEXAGONAL CRYSTALSFor crystals having hexagonal symmetry, it is desirable that equivalent planes havethe same indices; as with directions, this is accomplished by the Miller–Bravais

z

y

x

a

b

y

bO

c

y

f

e

(a) (b)x

Point of intersectionalong y axis

z

g

h

(011) Plane

A A B C

D E F

B

C

F

D

E

(a) (b)

FIGURE 3.24 (a) Reduced-sphere FCC unit cell with(110) plane. (b) Atomicpacking of an FCC (110)plane. Correspondingatom positions from (a)are indicated.

58 ● Chapter 3 / Structures of Metals and Ceramics

system shown in Figure 3.21. This convention leads to the four-index (hkil) scheme,which is favored in most instances, since it more clearly identifies the orientationof a plane in a hexagonal crystal. There is some redundancy in that i is determinedby the sum of h and k through

i � �(h � k) (3.8)

Otherwise the three h, k, and l indices are identical for both indexing systems.Figure 3.22b presents several of the common planes that are found for crystalshaving hexagonal symmetry.

3.14 LINEAR AND PLANAR ATOMIC DENSITIES

(CD-ROM)

3.15 CLOSE-PACKED CRYSTAL STRUCTURES

METALSIt may be remembered from the discussion on metallic crystal structures (Section3.4) that both face-centered cubic and hexagonal close-packed crystal structureshave atomic packing factors of 0.74, which is the most efficient packing of equal-sized spheres or atoms. In addition to unit cell representations, these two crystalstructures may be described in terms of close-packed planes of atoms (i.e., planeshaving a maximum atom or sphere-packing density); a portion of one such planeis illustrated in Figure 3.27a. Both crystal structures may be generated by thestacking of these close-packed planes on top of one another; the difference betweenthe two structures lies in the stacking sequence.

Let the centers of all the atoms in one close-packed plane be labeled A. Associ-ated with this plane are two sets of equivalent triangular depressions formed bythree adjacent atoms, into which the next close-packed plane of atoms may rest.Those having the triangle vertex pointing up are arbitrarily designated as B positions,while the remaining depressions are those with the down vertices, which are markedC in Figure 3.27a.

A second close-packed plane may be positioned with the centers of its atomsover either B or C sites; at this point both are equivalent. Suppose that the Bpositions are arbitrarily chosen; the stacking sequence is termed AB, which isillustrated in Figure 3.27b. The real distinction between FCC and HCP lies in where

A

B

C

E

D(a) (b)

A B

D E

C

�

�

�

�

�

� �

�

� �

FIGURE 3.25 (a)Reduced-sphereBCC unit cell with(110) plane. (b)Atomic packing of aBCC (110) plane.Corresponding atompositions from (a)are indicated.

3.15 Close-Packed Crystal Structures ● 59

FIGURE 3.27 (a) A portionof a close-packed plane ofatoms; A, B, and Cpositions are indicated.(b) The AB stackingsequence for close-packedatomic planes. (Adaptedfrom W. G. Moffatt, G. W.Pearsall, and J. Wulff, TheStructure and Properties ofMaterials, Vol. I, Structure,p. 50. Copyright 1964 byJohn Wiley & Sons, NewYork. Reprinted bypermission of John Wiley &Sons, Inc.)

the third close-packed layer is positioned. For HCP, the centers of this layer arealigned directly above the original A positions. This stacking sequence, ABABAB. . . , is repeated over and over. Of course, the ACACAC . . . arrangement wouldbe equivalent. These close-packed planes for HCP are (0001)-type planes, and thecorrespondence between this and the unit cell representation is shown in Figure 3.28.

For the face-centered crystal structure, the centers of the third plane are situatedover the C sites of the first plane (Figure 3.29a). This yields an ABCABCABC . . .stacking sequence; that is, the atomic alignment repeats every third plane. It ismore difficult to correlate the stacking of close-packed planes to the FCC unit cell.However, this relationship is demonstrated in Figure 3.29b; these planes are of the(111) type. The significance of these FCC and HCP close-packed planes will becomeapparent in Chapter 8.

FIGURE 3.28 Close-packed plane stackingsequence for hexagonal close-packed.(Adapted from W. G. Moffatt, G. W. Pearsall,and J. Wulff, The Structure and Properties ofMaterials, Vol. I, Structure, p. 51. Copyright 1964 by John Wiley & Sons, New York.Reprinted by permission of John Wiley &Sons, Inc.)

60 ● Chapter 3 / Structures of Metals and Ceramics

CERAMICSA number of ceramic crystal structures may also be considered in terms of close-packed planes of ions (as opposed to atoms for metals), as well as unit cells.Ordinarily, the close-packed planes are composed of the large anions. As theseplanes are stacked atop each other, small interstitial sites are created between themin which the cations may reside.

These interstitial positions exist in two different types, as illustrated in Figure3.30. Four atoms (three in one plane, and a single one in the adjacent plane)surround one type, labeled T in the figure; this is termed a tetrahedral position,since straight lines drawn from the centers of the surrounding spheres form a four-sided tetrahedron. The other site type, denoted as O in Figure 3.30, involves sixion spheres, three in each of the two planes. Because an octahedron is producedby joining these six sphere centers, this site is called an octahedral position. Thus,the coordination numbers for cations filling tetrahedral and octahedral positionsare 4 and 6, respectively. Furthermore, for each of these anion spheres, one octahe-dral and two tetrahedral positions will exist.

Ceramic crystal structures of this type depend on two factors: (1) the stackingof the close-packed anion layers (both FCC and HCP arrangements are possible,which correspond to ABCABC . . . and ABABAB . . . sequences, respectively),and (2) the manner in which the interstitial sites are filled with cations. For example,consider the rock salt crystal structure discussed above. The unit cell has cubicsymmetry, and each cation (Na� ion) has six Cl� ion nearest neighbors, as may beverified from Figure 3.5. That is, the Na� ion at the center has as nearest neighbors

FIGURE 3.29 (a) Close-packed stacking sequence for face-centered cubic. (b) Acorner has been removed to show the relation between the stacking of close-packed planes of atoms and the FCC crystal structure; the heavy triangle outlinesa (111) plane. (Figure (b) from W. G. Moffatt, G. W. Pearsall, and J. Wulff, TheStructure and Properties of Materials, Vol. I, Structure, p. 51. Copyright 1964 byJohn Wiley & Sons, New York. Reprinted by permission of John Wiley & Sons,Inc.)

3.15 Close-Packed Crystal Structures ● 61

the six Cl� ions that reside at the centers of each of the cube faces. The crystalstructure, having cubic symmetry, may be considered in terms of an FCC array ofclose-packed planes of anions, and all planes are of the {111} type. The cationsreside in octahedral positions because they have as nearest neighbors six anions.Furthermore, all octahedral positions are filled, since there is a single octahedralsite per anion, and the ratio of anions to cations is 1 : 1. For this crystal structure,the relationship between the unit cell and close-packed anion plane stacking schemesis illustrated in Figure 3.31.

Other, but not all, ceramic crystal structures may be treated in a similar manner;included are the zinc blende and perovskite structures. The spinel structure is oneof the AmBnXp types, which is found for magnesium aluminate or spinel (MgAl2O4).With this structure, the O2� ions form an FCC lattice, whereas Mg2� ions fill tetrahe-dral sites and Al3� reside in octahedral positions. Magnetic ceramics, or ferrites,have a crystal structure that is a slight variant of this spinel structure; and themagnetic characteristics are affected by the occupancy of tetrahedral and octahedralpositions �(see Section 18.5).�

FIGURE 3.30 The stacking of one plane of close-packed spheres (anions) on topof another; tetrahedral and octahedral positions between the planes aredesignated by T and O, respectively. (From W. G. Moffatt, G. W. Pearsall, andJ. Wulff, The Structure and Properties of Materials, Vol. 1, Structure. Copyright 1964 by John Wiley & Sons, New York. Reprinted by permission of JohnWiley & Sons, Inc.)

FIGURE 3.31 A section of the rock saltcrystal structure from which a corner hasbeen removed. The exposed plane of anions(light spheres inside the triangle) is a {111}-type plane; the cations (dark spheres)occupy the interstitial octahedral positions.

62 ● Chapter 3 / Structures of Metals and Ceramics

C R Y S T A L L I N E A N D N O N C R Y S T A L L I N EM A T E R I A L S3.16 SINGLE CRYSTALS

For a crystalline solid, when the periodic and repeated arrangement of atoms isperfect or extends throughout the entirety of the specimen without interruption,the result is a single crystal. All unit cells interlock in the same way and have thesame orientation. Single crystals exist in nature, but they may also be producedartificially. They are ordinarily difficult to grow, because the environment must becarefully controlled.

If the extremities of a single crystal are permitted to grow without any externalconstraint, the crystal will assume a regular geometric shape having flat faces, aswith some of the gem stones; the shape is indicative of the crystal structure. Aphotograph of several single crystals is shown in Figure 3.32. Within the past fewyears, single crystals have become extremely important in many of our moderntechnologies, in particular electronic microcircuits, which employ single crystals ofsilicon and other semiconductors.

3.17 POLYCRYSTALLINE MATERIALS

Most crystalline solids are composed of a collection of many small crystals or grains;such materials are termed polycrystalline. Various stages in the solidification of apolycrystalline specimen are represented schematically in Figure 3.33. Initially, smallcrystals or nuclei form at various positions. These have random crystallographicorientations, as indicated by the square grids. The small grains grow by the successiveaddition from the surrounding liquid of atoms to the structure of each. The extremi-ties of adjacent grains impinge on one another as the solidification process ap-proaches completion. As indicated in Figure 3.33, the crystallographic orientationvaries from grain to grain. Also, there exists some atomic mismatch within theregion where two grains meet; this area, called a grain boundary, is discussed inmore detail in Section 5.8.

FIGURE 3.32 Photograph showing several singlecrystals of fluorite, CaF2 . (Smithsonian Institutionphotograph number 38181P.)

3.18 Antisotropy ● 63

(a) (b)

(c) (d)

FIGURE 3.33 Schematic diagrams of the various stages in the solidification of apolycrystalline material; the square grids depict unit cells. (a) Small crystallitenuclei. (b) Growth of the crystallites; the obstruction of some grains that areadjacent to one another is also shown. (c) Upon completion of solidification, grainshaving irregular shapes have formed. (d) The grain structure as it would appearunder the microscope; dark lines are the grain boundaries. (Adapted from W.Rosenhain, An Introduction to the Study of Physical Metallurgy, 2nd edition,Constable & Company Ltd., London, 1915.)

3.18 ANISOTROPY

The physical properties of single crystals of some substances depend on the crystallo-graphic direction in which measurements are taken. For example, the elastic modu-lus, the electrical conductivity, and the index of refraction may have differentvalues in the [100] and [111] directions. This directionality of properties is termedanisotropy, and it is associated with the variance of atomic or ionic spacing withcrystallographic direction. Substances in which measured properties are indepen-dent of the direction of measurement are isotropic. The extent and magnitude ofanisotropic effects in crystalline materials are functions of the symmetry of thecrystal structure; the degree of anisotropy increases with decreasing structural sym-metry—triclinic structures normally are highly anisotropic. The modulus of elasticityvalues at [100], [110], and [111] orientations for several materials are presented inTable 3.7.

For many polycrystalline materials, the crystallographic orientations of theindividual grains are totally random. Under these circumstances, even though each

64 ● Chapter 3 / Structures of Metals and Ceramics

Table 3.7 Modulus of Elasticity Values forSeveral Metals at Various CrystallographicOrientations

Modulus of Elasticity (GPa)Metal [100] [110] [111]Aluminum 63.7 72.6 76.1Copper 66.7 130.3 191.1Iron 125.0 210.5 272.7Tungsten 384.6 384.6 384.6

Source: R. W. Hertzberg, Deformation and FractureMechanics of Engineering Materials, 3rd edition. Copy-right 1989 by John Wiley & Sons, New York. Re-printed by permission of John Wiley & Sons, Inc.

grain may be anisotropic, a specimen composed of the grain aggregate behavesisotropically. Also, the magnitude of a measured property represents some averageof the directional values. Sometimes the grains in polycrystalline materials have apreferential crystallographic orientation, in which case the material is said to havea ‘‘texture.’’

3.19 X-RAY DIFFRACTION: DETERMINATION OF

CRYSTAL STRUCTURES (CD-ROM)

3.20 NONCRYSTALLINE SOLIDS

It has been mentioned that noncrystalline solids lack a systematic and regulararrangement of atoms over relatively large atomic distances. Sometimes such materi-als are also called amorphous (meaning literally without form), or supercooledliquids, inasmuch as their atomic structure resembles that of a liquid.

An amorphous condition may be illustrated by comparison of the crystallineand noncrystalline structures of the ceramic compound silicon dioxide (SiO2), whichmay exist in both states. Figures 3.38a and 3.38b present two-dimensional schematicdiagrams for both structures of SiO2 , in which the SiO4�

4 tetrahedron is the basicunit (Figure 3.10). Even though each silicon ion bonds to four oxygen ions for bothstates, beyond this, the structure is much more disordered and irregular for thenoncrystalline structure.

Whether a crystalline or amorphous solid forms depends on the ease with whicha random atomic structure in the liquid can transform to an ordered state duringsolidification. Amorphous materials, therefore, are characterized by atomic or mo-lecular structures that are relatively complex and become ordered only with somedifficulty. Furthermore, rapidly cooling through the freezing temperature favorsthe formation of a noncrystalline solid, since little time is allowed for the order-ing process.

Metals normally form crystalline solids; but some ceramic materials are crystal-line, whereas others (i.e., the silica glasses) are amorphous. Polymers may be com-pletely noncrystalline and semicrystalline consisting of varying degrees of crystallin-

3.20 Noncrystalline Solids ● 65

(a) (b)

Silicon atomOxygen atom

FIGURE 3.38 Two-dimensional schemes of the structure of (a) crystalline silicondioxide and (b) noncrystalline silicon dioxide.

ity. More about the structure and properties of these amorphous materials isdiscussed below and in subsequent chapters.

SILICA GLASSESSilicon dioxide (or silica, SiO2) in the noncrystalline state is called fused silica, orvitreous silica; again, a schematic representation of its structure is shown in Figure3.38b. Other oxides (e.g., B2O3 and GeO2) may also form glassy structures (andpolyhedral oxide structures �similar to those shown in Figure 3.12�); these materials,as well as SiO2 , are network formers.

Si4+ O2� Na+

FIGURE 3.39 Schematic representation ofion positions in a sodium–silicate glass.

66 ● Chapter 3 / Structures of Metals and Ceramics

The common inorganic glasses that are used for containers, windows, and soon are silica glasses to which have been added other oxides such as CaO andNa2O. These oxides do not form polyhedral networks. Rather, their cations areincorporated within and modify the SiO4

4� network; for this reason, these oxide

additives are termed network modifiers. For example, Figure 3.39 is a schematicrepresentation of the structure of a sodium–silicate glass. Still other oxides, suchas TiO2 and Al2O3 , while not network formers, substitute for silicon and becomepart of and stabilize the network; these are called intermediates. From a practicalperspective, the addition of these modifiers and intermediates lowers the meltingpoint and viscosity of a glass, and makes it easier to form at lower temperatures�(Section 14.7).�

S U M M A R YAtoms in crystalline solids are positioned in an orderly and repeated pattern thatis in contrast to the random and disordered atomic distribution found in noncrystal-line or amorphous materials. Atoms may be represented as solid spheres, and, forcrystalline solids, crystal structure is just the spatial arrangement of these spheres.The various crystal structures are specified in terms of parallelepiped unit cells,which are characterized by geometry and atom positions within.

Most common metals exist in at least one of three relatively simple crystalstructures: face-centered cubic (FCC), body-centered cubic (BCC), and hexagonalclose-packed (HCP). Two features of a crystal structure are coordination number(or number of nearest-neighbor atoms) and atomic packing factor (the fraction ofsolid sphere volume in the unit cell). Coordination number and atomic packingfactor are the same for both FCC and HCP crystal structures.

For ceramics both crystalline and noncrystalline states are possible. The crystalstructures of those materials for which the atomic bonding is predominantly ionicare determined by the charge magnitude and the radius of each kind of ion. Someof the simpler crystal structures are described in terms of unit cells; several of thesewere discussed (rock salt, cesium chloride, zinc blende, diamond cubic, graphite,fluorite, perovskite, and spinel structures).

Theoretical densities of metallic and crystalline ceramic materials may be com-puted from unit cell and atomic weight data.

Generation of face-centered cubic and hexagonal close-packed crystal structuresis possible by the stacking of close-packed planes of atoms. For some ceramic crystalstructures, cations fit into interstitial positions that exist between two adjacent close-packed planes of anions.

For the silicates, structure is more conveniently represented by means of inter-connecting SiO4

4� tetrahedra. Relatively complex structures may result when other

cations (e.g., Ca2�, Mg2�, Al3�) and anions (e.g., OH�) are added. The structuresof silica (SiO2), silica glass, �and several of the simple and layered silicates� were pre-sented.

Structures for the various forms of carbon—diamond, graphite, �and the fuller-enes�—were also discussed.

Crystallographic planes and directions are specified in terms of an indexingscheme. The basis for the determination of each index is a coordinate axis systemdefined by the unit cell for the particular crystal structure. Directional indices arecomputed in terms of vector projections on each of the coordinate axes, whereasplanar indices are determined from the reciprocals of axial intercepts. For hexagonalunit cells, a four-index scheme for both directions and planes is found to bemore convenient.

References ● 67

�Crystallographic directional and planar equivalencies are related to atomiclinear and planar densities, respectively.� The atomic packing (i.e., planar density)of spheres in a crystallographic plane depends on the indices of the plane as wellas the crystal structure. For a given crystal structure, planes having identical atomicpacking yet different Miller indices belong to the same family.

Single crystals are materials in which the atomic order extends uninterruptedover the entirety of the specimen; under some circumstances, they may have flatfaces and regular geometric shapes. The vast majority of crystalline solids, however,are polycrystalline, being composed of many small crystals or grains having differentcrystallographic orientations.

Other concepts introduced in this chapter were: crystal system (a classificationscheme for crystal structures on the basis of unit cell geometry); polymorphism (orallotropy) (when a specific material can have more than one crystal structure); andanisotropy (the directionality dependence of properties).

�X-ray diffractometry is used for crystal structure and interplanar spacing deter-minations. A beam of x-rays directed on a crystalline material may experiencediffraction (constructive interference) as a result of its interaction with a series ofparallel atomic planes according to Bragg’s law. Interplanar spacing is a functionof the Miller indices and lattice parameter(s) as well as the crystal structure.�

I M P O R T A N T T E R M S A N D C O N C E P T S

AllotropyAmorphousAnionAnisotropyAtomic packing factor (APF)Body-centered cubic (BCC)Bragg’s lawCationCoordination numberCrystal structure

Crystal systemCrystallineDiffractionFace-centered cubic (FCC)GrainGrain boundaryHexagonal close-packed (HCP)IsotropicLattice

Lattice parametersMiller indicesNoncrystallineOctahedral positionPolycrystallinePolymorphismSingle crystalTetrahedral positionUnit cell

R E F E R E N C E S

Azaroff, L. F., Elements of X-Ray Crystallography,McGraw-Hill Book Company, New York,1968. Reprinted by TechBooks, Marietta,OH, 1990.

Barrett, C. S. and T. B. Massalski, Structure ofMetals, 3rd edition, Pergamon Press, Oxford,1980.

Barsoum, M. W., Fundamentals of Ceramics, TheMcGraw-Hill Companies, Inc., New York,1997.

Budworth, D. W., An Introduction to Ceramic Sci-ence, Pergamon Press, Oxford, 1970.

Buerger, M. J., Elementary Crystallography, JohnWiley & Sons, New York, 1956.

Charles, R. J., ‘‘The Nature of Glasses,’’ Scientific

American, Vol. 217, No. 3, September 1967,pp. 126–136.

Chiang, Y. M., D. P. Birnie, III, and W. D. Kingery,Physical Ceramics: Principles for Ceramic Sci-ence and Engineering, John Wiley & Sons, Inc.,New York, 1997.

Cullity, B. D., Elements of X-Ray Diffraction, 3rdedition, Addison-Wesley Publishing Co.,Reading, MA, 1998.

Curl, R. F. and R. E. Smalley, ‘‘Fullerenes,’’ Scien-tific American, Vol. 265, No. 4, October 1991,pp. 54–63.

Gilman, J. J., ‘‘The Nature of Ceramics,’’ ScientificAmerican, Vol. 217, No. 3, September 1967,pp. 112–124.

68 ● Chapter 3 / Structures of Metals and Ceramics

Hauth, W. E., ‘‘Crystal Chemistry in Ceramics,’’American Ceramic Society Bulletin, Vol. 30,1951: No. 1, pp. 5–7; No. 2, pp. 47–49; No. 3,pp. 76–77; No. 4, pp. 137–142; No. 5, pp. 165–167; No. 6, pp. 203–205. A good overview ofsilicate structures.

Kingery, W. D., H. K. Bowen, and D. R. Uhlmann,Introduction to Ceramics, 2nd edition, JohnWiley & Sons, New York, 1976. Chapters 1–4.

Richerson, D. W., Modern Ceramic Engineering,2nd edition, Marcel Dekker, New York, 1992.

Schwartz, L. H. and J. B. Cohen, Diffraction fromMaterials, 2nd edition, Springer-Verlag, NewYork, 1987.

Van Vlack, L. H., Physical Ceramics for Engineers,Addison-Wesley Publishing Company, Read-ing, MA, 1964. Chapters 1–4 and 6–8.

Wyckoff, R. W. G., Crystal Structures, 2nd edition,Interscience Publishers, 1963. Reprinted byKrieger Publishing Company, Melbourne,FL, 1986.

Q U E S T I O N S A N D P R O B L E M S

Note: To solve those problems having an asterisk (*) by their numbers, consultation of supplementarytopics [appearing only on the CD-ROM (and not in print)] will probably be necessary.

3.1 What is the difference between atomic struc-ture and crystal structure?

3.2 What is the difference between a crystalstructure and a crystal system?

3.3 If the atomic radius of aluminum is 0.143nm, calculate the volume of its unit cell incubic meters.

3.4 Show for the body-centered cubic crystalstructure that the unit cell edge length a andthe atomic radius R are related through a �4R/�3.

3.5 For the HCP crystal structure, show that theideal c/a ratio is 1.633.

3.6 Show that the atomic packing factor for BCCis 0.68.

3.7 Show that the atomic packing factor for HCPis 0.74.

3.8 Iron has a BCC crystal structure, an atomicradius of 0.124 nm, and an atomic weightof 55.85 g/mol. Compute and compare itsdensity with the experimental value foundinside the front cover.

3.9 Calculate the radius of an iridium atom giventhat Ir has an FCC crystal structure, a densityof 22.4 g/cm3, and an atomic weight of 192.2g/mol.

3.10 Calculate the radius of a vanadium atom,given that V has a BCC crystal structure, adensity of 5.96 g/cm3, and an atomic weightof 50.9 g/mol.

3.11 Some hypothetical metal has the simple cu-bic crystal structure shown in Figure 3.40. Ifits atomic weight is 70.4 g/mol and the atomicradius is 0.126 nm, compute its density.

3.12 Zirconium has an HCP crystal structure anda density of 6.51 g/cm3.

(a) What is the volume of its unit cell incubic meters?

(b) If the c/a ratio is 1.593, compute thevalues of c and a.

3.13 Using atomic weight, crystal structure, andatomic radius data tabulated inside the frontcover, compute the theoretical densities oflead, chromium, copper, and cobalt, and then

FIGURE 3.40 Hard-sphere unit cell representation of thesimple cubic crystal structure.

Questions and Problems ● 69

compare these values with the measureddensities listed in this same table. The c/aratio for cobalt is 1.623.

3.14 Rhodium has an atomic radius of 0.1345 nm(1.345 A) and a density of 12.41 g/cm3. De-termine whether it has an FCC or BCC crys-tal structure.

3.15 Below are listed the atomic weight, density,and atomic radius for three hypotheticalalloys. For each determine whether its crystalstructure is FCC, BCC, or simple cubic andthen justify your determination. A simplecubic unit cell is shown in Figure 3.40.

Atomic AtomicWeight Density Radius

Alloy (g/mol) (g/cm3) (nm)

A 77.4 8.22 0.125B 107.6 13.42 0.133C 127.3 9.23 0.142

3.16 The unit cell for tin has tetragonal symmetry,with a and b lattice parameters of 0.583 and0.318 nm, respectively. If its density, atomicweight, and atomic radius are 7.30 g/cm3,118.69 g/mol, and 0.151 nm, respectively,compute the atomic packing factor.

3.17 Iodine has an orthorhombic unit cell forwhich the a, b, and c lattice parameters are0.479, 0.725, and 0.978 nm, respectively.

(a) If the atomic packing factor and atomicradius are 0.547 and 0.177 nm, respectively,determine the number of atoms in eachunit cell.

(b) The atomic weight of iodine is 126.91 g/mol; compute its density.

3.18 Titanium has an HCP unit cell for which theratio of the lattice parameters c/a is 1.58. Ifthe radius of the Ti atom is 0.1445 nm, (a)determine the unit cell volume, and (b) cal-culate the density of Ti and compare it withthe literature value.

3.19 Zinc has an HCP crystal structure, a c/a ratioof 1.856, and a density of 7.13 g/cm3. Calcu-late the atomic radius for Zn.

3.20 Rhenium has an HCP crystal structure, anatomic radius of 0.137 nm, and a c/a ratio of1.615. Compute the volume of the unit cellfor Re.

3.21 This is a unit cell for a hypothetical metal:

+z

+yO

+x

0.40 nm

0.30 nm

0.30 nm

90°

90°

90°

(a) To which crystal system does this unitcell belong?

(b) What would this crystal structure becalled?

(c) Calculate the density of the material,given that its atomic weight is 141 g/mol.

3.22 Using the Molecule Definition File (MDF)on the CD-ROM that accompanies thisbook, generate a three-dimensional unit cellfor the intermetallic compound AuCu3 giventhe following: 1) the unit cell is cubic withan edge length of 0.374 nm, 2) gold atomsare situated at all cube corners, and 3) copperatoms are positioned at the centers of all unitcell faces.

3.23 Using the Molecule Definition File (MDF)on the CD-ROM that accompanies thisbook, generate a three-dimensional unit cellfor the intermetallic compound AuCu giventhe following: 1) the unit cell is tetragonalwith a � 0.289 nm and c � 0.367 nm (seeTable 3.6), 2) gold atoms are situated at allunit cell corners, and 3) a copper atom ispositioned at the center of the unit cell.

3.24 Sketch a unit cell for the body-centered or-thorhombic crystal structure.

3.25 For a ceramic compound, what are the twocharacteristics of the component ions thatdetermine the crystal structure?

3.26 Show that the minimum cation-to-anion ra-dius ratio for a coordination number of 4is 0.225.

3.27 Show that the minimum cation-to-anion ra-dius ratio for a coordination number of 6 is0.414. Hint: Use the NaCl crystal structure(Figure 3.5), and assume that anions and cat-ions are just touching along cube edges andacross face diagonals.

3.28 Demonstrate that the minimum cation-to-anion radius ratio for a coordination numberof 8 is 0.732.

3.29 On the basis of ionic charge and ionic radii,predict the crystal structures for the follow-ing materials: (a) CsI, (b) NiO, (c) KI, and(d) NiS. Justify your selections.

3.30 Which of the cations in Table 3.4 would youpredict to form iodides having the cesiumchloride crystal structure? Justify yourchoices.

3.31 Compute the atomic packing factor for thecesium chloride crystal structure in whichrC/rA � 0.732.

3.32 Table 3.4 gives the ionic radii for K� and O2�

as 0.138 and 0.140 nm, respectively. Whatwould be the coordination number for eachO2� ion? Briefly describe the resulting crystalstructure for K2O. Explain why this is calledthe antifluorite structure.

3.33 Using the Molecule Definition File (MDF)on the CD-ROM that accompanies thisbook, generate a three-dimensional unit cellfor lead oxide, PbO, given the following: (1)the unit cell is tetragonal with a � 0.397nm and c � 0.502 nm, (2) oxygen ions aresituated at all cube corners, and, in addition,at the centers of the two square faces, (3)one oxygen ion is positioned on each of twoof the other opposing faces (rectangular) atthe 0.5a-0.237c coordinate, and (4) for theother two rectangular and opposing faces,oxygen ions are located at the 0.5a-0.763c co-ordinate.

3.34 Calculate the density of FeO, given that ithas the rock salt crystal structure.

3.35 Magnesium oxide has the rock salt crystalstructure and a density of 3.58 g/cm3.

(a) Determine the unit cell edge length.

(b) How does this result compare with theedge length as determined from the radii inTable 3.4, assuming that the Mg2� and O2�

ions just touch each other along the edges?

3.36 Compute the theoretical density of diamondgiven that the CUC distance and bond angleare 0.154 nm and 109.5�, respectively. Howdoes this value compare with the mea-sured density?

3.37 Compute the theoretical density of ZnSgiven that the ZnUS distance and bondangle are 0.234 nm and 109.5�, respectively.How does this value compare with the mea-sured density?

3.38 Cadmium sulfide (CdS) has a cubic unit cell,and from x-ray diffraction data it is knownthat the cell edge length is 0.582 nm. If themeasured density is 4.82 g/cm3, how manyCd2� and S2� ions are there per unit cell?

3.39 (a) Using the ionic radii in Table 3.4, com-pute the density of CsCl. Hint: Use a modifi-cation of the result of Problem 3.4.

(b) The measured density is 3.99 g/cm3. Howdo you explain the slight discrepancy be-tween your calculated value and the mea-sured one?

3.40 From the data in Table 3.4, compute thedensity of CaF2 , which has the fluoritestructure.

3.41 A hypothetical AX type of ceramic materialis known to have a density of 2.65 g/cm3 anda unit cell of cubic symmetry with a cell edgelength of 0.43 nm. The atomic weights of theA and X elements are 86.6 and 40.3 g/mol,respectively. On the basis of this informa-tion, which of the following crystal structuresis (are) possible for this material: rock salt,cesium chloride, or zinc blende? Justifyyour choice(s).

3.42 The unit cell for MgFe2O4 (MgO-Fe2O3) hascubic symmetry with a unit cell edge lengthof 0.836 nm. If the density of this materialis 4.52 g/cm3, compute its atomic packingfactor. For this computation, you will needto use ionic radii listed in Table 3.4.

3.43 The unit cell for Al2O3 has hexagonal sym-metry with lattice parameters a � 0.4759 nmand c � 1.2989 nm. If the density of thismaterial is 3.99 g/cm3, calculate its atomicpacking factor. For this computation useionic radii listed in Table 3.4.

3.44 Compute the atomic packing factor for thediamond cubic crystal structure (Figure 3.16).

70 ● Chapter 3 / Structures of Metals and Ceramics

Questions and Problems ● 71

Assume that bonding atoms touch one an-other, that the angle between adjacent bondsis 109.5�, and that each atom internal to theunit cell is positioned a/4 of the distance awayfrom the two nearest cell faces (a is the unitcell edge length).

3.45 Compute the atomic packing factor for ce-sium chloride using the ionic radii in Table3.4 and assuming that the ions touch alongthe cube diagonals.

3.46 In terms of bonding, explain why silicate ma-terials have relatively low densities.

3.47 Determine the angle between covalentbonds in an SiO4

4� tetrahedron.

3.48 Draw an orthorhombic unit cell, and withinthat cell a [121] direction and a (210) plane.

3.49 Sketch a monoclinic unit cell, and within thatcell a [011] direction and a (002) plane.

3.50 Here are unit cells for two hypotheticalmetals:

(a) What are the indices for the directionsindicated by the two vectors in sketch (a)?

Direction 2

0.5 nm

+y

0.4 nm

+z

+x

Direction 1

0.3 nm

(a)

0.4 nm

0.4 nm

Plane 1Plane 2

0.2 nm

+x

+y

+z

(b)

(b) What are the indices for the two planesdrawn in sketch (b)?

3.51 Within a cubic unit cell, sketch the follow-ing directions:

(a) [110];

(b) [121];

(c) [012];

(d) [133];

(e) [111];

(f) [122];

(g) [123];

(h) [103].

3.52 Determine the indices for the directionsshown in the following cubic unit cell:

A

+z

+x

+y

12

12

12

B

CD

12

12

,

3.53 Determine the indices for the directionsshown in the following cubic unit cell:

A23

23

13

13

13

12

B

C

D

12

12

,

23

+x

+y

+z

3.54 For tetragonal crystals, cite the indices ofdirections that are equivalent to each of thefollowing directions:

(a) [101];

(b) [110];

(c) [010].

72 ● Chapter 3 / Structures of Metals and Ceramics

3.55 (a) Convert the [100] and [111] directionsinto the four-index Miller–Bravais schemefor hexagonal unit cells.

(b) Make the same conversion for the (010)and (101) planes.

3.56 Determine the Miller indices for the planesshown in the following unit cell:

A

B

+z

+x

+y

12

23

3.57 Determine the Miller indices for the planesshown in the following unit cell:

A

B

+z

+x

+y

12

12

12

12

13

(a)

z

a3

a2

a1

(b)

z

a3

a2

a1

3.60 Determine the indices for the planes shownin the hexagonal unit cells shown below.

3.59 Sketch the (1101) and (1120) planes in a hex-agonal unit cell.

A

B

+z

+x

+y

23

12

3.58 Determine the Miller indices for the planesshown in the following unit cell:

Questions and Problems ● 73

3.61 Sketch within a cubic unit cell the follow-ing planes:

(a) (011);

(b) (112);

(c) (102);

(d) (131);

(e) (111);

(f) (122);

(g) (123);

(h) (013).

3.62 Sketch the atomic packing of (a) the (100)plane for the FCC crystal structure, and (b)the (111) plane for the BCC crystal structure(similar to Figures 3.24b and 3.25b).

3.63 For each of the following crystal structures,represent the indicated plane in the mannerof Figures 3.24 and 3.25, showing both anionsand cations: (a) (100) plane for the rocksalt crystal structure, (b) (110) plane forthe cesium chloride crystal structure,(c) (111) plane for the zinc blende crystalstructure, and (d) (110) plane for the perov-skite crystal structure.

3.64 Consider the reduced-sphere unit cell shownin Problem 3.21, having an origin of the coor-dinate system positioned at the atom labeledwith an O. For the following sets of planes,determine which are equivalent:

(a) (100), (010), and (001).

(b) (110), (101), (011), and (110).

(c) (111), (111), (111), and (111).

3.65 Cite the indices of the direction that resultsfrom the intersection of each of the followingpair of planes within a cubic crystal: (a)(110) and (111) planes; (b) (110) and(110) planes; and (c) (101) and (001) planes.

3.66 The zinc blende crystal structure is one thatmay be generated from close-packed planesof anions.

(a) Will the stacking sequence for this struc-ture be FCC or HCP? Why?

(b) Will cations fill tetrahedral or octahedralpositions? Why?

(c) What fraction of the positions will be oc-cupied?

3.67 The corundum crystal structure, found forAl2O3 , consists of an HCP arrangement of

O2� ions; the Al3� ions occupy octahedral po-sitions.

(a) What fraction of the available octahe-dral positions are filled with Al3� ions?

(b) Sketch two close-packed O2� planesstacked in an AB sequence, and note octahe-dral positions that will be filled with theAl3� ions.

3.68 Iron sulfide (FeS) may form a crystal struc-ture that consists of an HCP arrangement ofS2� ions.

(a) Which type of interstitial site will theFe2� ions occupy?

(b) What fraction of these available intersti-tial sites will be occupied by Fe2� ions?

3.69 Magnesium silicate, Mg2SiO4 , forms in theolivine crystal structure which consists of anHCP arrangement of O2� ions.

(a) Which type of interstitial site will theMg2� ions occupy? Why?

(b) Which type of interstitial site will theSi4� ions occupy? Why?

(c) What fraction of the total tetrahedralsites will be occupied?

(d) What fraction of the total octahedralsites will be occupied?

3.70* Compute and compare the linear densities ofthe [100], [110], and [111] directions for FCC.

3.71* Compute and compare the linear densitiesof the [110] and [111] directions for BCC.

3.72* Calculate and compare the planar densitiesof the (100) and (111) planes for FCC.

3.73* Calculate and compare the planar densitiesof the (100) and (110) planes for BCC.

3.74* Calculate the planar density of the (0001)plane for HCP.

3.75 Here are shown the atomic packing schemesfor several different crystallographic direc-tions for some hypothetical metal. For eachdirection the circles represent only thoseatoms contained within a unit cell, which cir-cles are reduced from their actual size.

74 ● Chapter 3 / Structures of Metals and Ceramics

0.40 nm

[100], [010]

0.50 nm

[001]

0.64 nm

[011], [101]

0.566 nm

[110]

(a) To what crystal system does the unitcell belong?

(b) What would this crystal structure becalled?

3.76 Below are shown three different crystallo-graphic planes for a unit cell of some hypo-thetical metal; the circles represent atoms:

(a) To what crystal system does the unitcell belong?

(b) What would this crystal structure becalled?

(c) If the density of this metal is 8.95 g/cm3,determine its atomic weight.

3.77 Explain why the properties of polycrystallinematerials are most often isotropic.

(001) (110)

0.4

0 n

m

0.5

0 n

m

0.4

6 n

m

0.30 nm

0.35 nm

(101)

0.40 nm

3.78* Using the data for molybdenum in Table3.1, compute the interplanar spacing for the(111) set of planes.

3.79* Determine the expected diffraction angle forthe first-order reflection from the (113) setof planes for FCC platinum when monochro-matic radiation of wavelength 0.1542 nm isused.

3.80* Using the data for aluminum in Table 3.1,compute the interplanar spacings for the(110) and (221) sets of planes.

3.81* The metal iridium has an FCC crystal struc-ture. If the angle of diffraction for the (220)set of planes occurs at 69.22� (first-order re-flection) when monochromatic x-radiationhaving a wavelength of 0.1542 nm is used,compute (a) the interplanar spacing for thisset of planes, and (b) the atomic radius foran iridium atom.

3.82* The metal rubidium has a BCC crystal struc-ture. If the angle of diffraction for the (321)set of planes occurs at 27.00� (first-order re-flection) when monochromatic x-radiationhaving a wavelength of 0.0711 nm is used,compute (a) the interplanar spacing for thisset of planes, and (b) the atomic radius forthe rubidium atom.

3.83* For which set of crystallographic planes willa first-order diffraction peak occur at a dif-fraction angle of 46.21� for BCC iron whenmonochromatic radiation having a wave-length of 0.0711 nm is used?

3.84* Figure 3.37 shows an x-ray diffraction pat-tern for �-iron taken using a diffractometerand monochromatic x-radiation having awavelength of 0.1542 nm; each diffractionpeak on the pattern has been indexed. Com-pute the interplanar spacing for each set ofplanes indexed; also determine the latticeparameter of Fe for each of the peaks.

3.85* The diffraction peaks shown in Figure 3.37are indexed according to the reflection rulesfor BCC (i.e., the sum h � k � l must beeven). Cite the h, k, and l indices for thefirst four diffraction peaks for FCC crystalsconsistent with h, k, and l all being eitherodd or even.

Questions and Problems ● 75

40 50 60 70 80 90

Diffraction angle 2q

Inte

nsit

y (r

elat

ive)

FIGURE 3.41 Diffraction patternfor polycrystalline copper.

3.86* Figure 3.41 shows the first four peaks of thex-ray diffraction pattern for copper, whichhas an FCC crystal structure; monochro-matic x-radiation having a wavelength of0.1542 nm was used.

(a) Index (i.e., give h, k, and l indices for)each of these peaks.

(b) Determine the interplanar spacing foreach of the peaks.

(c) For each peak, determine the atomic ra-dius for Cu and compare these with the valuepresented in Table 3.1.

3.87 Would you expect a material in which theatomic bonding is predominantly ionic in na-

ture to be more or less likely to form a non-crystalline solid upon solidification than acovalent material? Why? (See Section 2.6.)

Design Problem

3.D1* Gallium arsenide (GaAs) and gallium phos-phide (GaP) both have the zinc blende crys-tal structure and are soluble in one anotherat all concentrations. Determine the con-centration in weight percent of GaP thatmust be added to GaAs to yield a unit celledge length of 0.5570 nm. The densities ofGaAs and GaP are 5.307 and 4.130 g/cm3,respectively.

C h a p t e r 3 / Structures of Metalsand Ceramics

3.8 SILICATE CERAMICS

THE SILICATESFor the various silicate minerals, one, two, or three of the corner oxygen atoms ofthe SiO4

4� tetrahedra are shared by other tetrahedra to form some rather complex

structures. Some of these, represented in Figure 3.12, have formulas SiO44

�,Si2O6

7�, Si3O6

9�, and so on; single-chain structures are also possible, as in Figure

3.12e. Positively charged cations such as Ca2�, Mg2�, and Al3� serve two roles.First, they compensate the negative charges from the SiO4

4� units so that charge

neutrality is achieved; and second, these cations ionically bond the SiO44

� tetra-hedra together.

Simple SilicatesOf these silicates, the most structurally simple ones involve isolated tetrahedra(Figure 3.12a). For example, forsterite (Mg2SiO4) has the equivalent of two Mg2�

ions associated with each tetrahedron in such a way that every Mg2� ion has sixoxygen nearest neighbors.

The Si2O67

� ion is formed when two tetrahedra share a common oxygen atom(Figure 3.12b). Akermanite (Ca2MgSi2O7) is a mineral having the equivalent oftwo Ca2� ions and one Mg2� ion bonded to each Si2O6

7� unit.

Layered SilicatesA two-dimensional sheet or layered structure can also be produced by the sharingof three oxygen ions in each of the tetrahedra (Figure 3.13); for this structure therepeating unit formula may be represented by (Si2O5)2�. The net negative chargeis associated with the unbonded oxygen atoms projecting out of the plane of thepage. Electroneutrality is ordinarily established by a second planar sheet structurehaving an excess of cations, which bond to these unbonded oxygen atoms from theSi2O5 sheet. Such materials are called the sheet or layered silicates, and their basicstructure is characteristic of the clays and other minerals.

One of the most common clay minerals, kaolinite, has a relatively simple two-layer silicate sheet structure. Kaolinite clay has the formula Al2(Si2O5)(OH)4 inwhich the silica tetrahedral layer, represented by (Si2O5)2�, is made electricallyneutral by an adjacent Al2(OH)2

4� layer. A single sheet of this structure is shown

in Figure 3.14, which is exploded in the vertical direction to provide a betterperspective of the ion positions; the two distinct layers are indicated in the figure.The midplane of anions consists of O2� ions from the (Si2O5)2� layer, as well asOH� ions that are a part of the Al2(OH)2�

4 layer. Whereas the bonding within thistwo-layered sheet is strong and intermediate ionic-covalent, adjacent sheets areonly loosely bound to one another by weak van der Waals forces.

S-1

A crystal of kaolinite is made of a series of these double layers or sheets stackedparallel to each other, which form small flat plates typically less than 1 �m indiameter and nearly hexagonal. Figure 3.15 is an electron micrograph of kaolinitecrystals at a high magnification, showing the hexagonal crystal plates some of whichare piled one on top of the other.

These silicate sheet structures are not confined to the clays; other minerals alsoin this group are talc [Mg3(Si2O5)2(OH)2] and the micas [e.g., muscovite, KAl3Si3

O10(OH)2], which are important ceramic raw materials. As might be deduced fromthe chemical formulas, the structures for some silicates are among the most complexof all the inorganic materials.

S-2 ● Chapter 3 / Structures of Metals and Ceramics

SiO44–

(a)Si2O7

6–

(b)Si3O9

6–

(c)

Si6O1812–

Si4+ O2–

(d)

(SiO3)n(e)

2n–

FIGURE 3.12 Fivesilicate ion structuresformed from SiO4

4�

tetrahedra.

Si4+ O2�

FIGURE 3.13 Schematic representation of thetwo-dimensional silicate sheet structure having arepeat unit formula of (Si2O5)2�.

3.9 Carbon ● S-3

3.9 CARBON

FULLERENESAnother polymorphic form of carbon was discovered in 1985. It exists in discretemolecular form, and consists of a hollow spherical cluster of sixty carbon atoms; asingle molecule is denoted by C60 . Each molecule is composed of groups of carbonatoms that are bonded to one another to form both hexagon (six-carbon atom) andpentagon (five-carbon atom) geometrical configurations. One such molecule, shownin Figure 3.18, is found to consist of 20 hexagons and 12 pentagons, which arearrayed such that no two pentagons share a common side; the molecular surfacethus exhibits the symmetry of a soccer ball. The material composed of C60 moleculesis known as buckminsterfullerene, named in honor of R. Buckminster Fuller, who

OH�

O2�

Al3+

Si4+

Anion midplaneAl2(OH)4

2+ Layer

(Si2O5)2� Layer

FIGURE 3.14 The structure ofkaolinite clay. (Adapted from W. E.Hauth, ‘‘Crystal Chemistry ofCeramics,’’ American Ceramic SocietyBulletin, Vol. 30, No. 4, 1951, p. 140.)

FIGURE 3.15 Electron micrograph ofkaolinite crystals. They are in theform of hexagonal plates, some ofwhich are stacked on top of oneanother. 15,000�. (Photographcourtesy of Georgia Kaolin Co., Inc.)

invented the geodesic dome; each C60 is simply a molecular replica of such a dome,which is often referred to as ‘‘buckyball’’ for short. The term fullerene is used todenote the class of materials that are composed of this type of molecule.

Diamond and graphite are what may be termed network solids, in that all ofthe carbon atoms form primary bonds with adjacent atoms throughout the entiretyof the solid. By way of contrast, the carbon atoms in buckminsterfullerene bondtogether so as to form these spherical molecules. In the solid state, the C60 unitsform a crystalline structure and pack together in a face-centered cubic array.

As a pure crystalline solid, this material is electrically insulating. However, withproper impurity additions, it can be made highly conductive and semiconductive.As a final note, molecular shapes other than the ball clusters recently have beendiscovered; these include nanoscale tubular and polyhedral structures. It is antici-pated that, with further developments, the fullerenes will become technologicallyimportant materials.

3.14 LINEAR AND PLANAR ATOMIC DENSITIES

The two previous sections discussed the equivalency of nonparallel crystallographicdirections and planes. Directional equivalency is related to the atomic linear densityin the sense that equivalent directions have identical linear densities. The directionvector is positioned so as to pass through atom centers, and the fraction of linelength intersected by these atoms is equal to the linear density.

Correspondingly, crystallographic planes that are equivalent have the sameatomic planar density. The plane of interest is positioned so as to pass through atomcenters. And planar density is simply the fraction of total crystallographic planearea that is occupied by atoms (represented as circles). It should be noted that theconcepts of linear and planar densities are one- and two-dimensional analogs ofthe atomic packing factor; their determinations are illustrated in the following twoexample problems.

EXAMPLE PROBLEM 3.11

Calculate the linear density of the [100] direction for BCC.

S O L U T I O N

A BCC unit cell (reduced sphere) and the [100] direction therein are shownin Figure 3.26a; represented in Figure 3.26b is the linear packing in this direction.

S-4 ● Chapter 3 / Structures of Metals and Ceramics

FIGURE 3.18 The structure of a C60 molecule.

3.14 Linear and Planar Atomic Densities ● S-5

As a basis for our computation let us use the line length within the unit cell,Ll , which in this case is the lattice parameter a—the distance between thecenters of atoms M and N. In terms of the atomic radius R,

Ll � a �4R

�3(see Equation 3.3)

Now, the total line length intersecting circles (atoms M and N), Lc , is equalto 2R. And the linear density LD is just the following ratio:

LD �Lc

Ll�

2R

4R/�3� 0.866

EXAMPLE PROBLEM 3.12

Calculate the planar density of the (110) plane for FCC.

S O L U T I O N

The atomic packing of this plane is represented in Figure 3.24b. Consider thatportion of the plane that intersects a unit cell (Figure 3.24b), and then computeboth this planar area and total circle area in terms of the atomic radius R.Planar density, then, is just the ratio of these two areas.

The unit cell plane area, Ap , is simply that of the rectangle circumscribedby the centers of the atoms A, C, D, and F (Figure 3.24b). The rectangle length(AC) and width (AD) are, respectively,

AC � 4R

AD � 2R �2 (see Equation 3.1)

Therefore,

Ap � (AC)(AD)

� (4R)(2R �2) � 8R2�2

Now, for the total circle area, one fourth of each of atoms A, C, D, and Fand one half of atoms B and E reside within this rectangle, which gives a total

(a)

[100]

(b)

N

N

M

M

aa

FIGURE 3.26 (a) Reduced-sphere BCC unit cell with the [100] direction indicated.(b) Atomic spacing in the [100] direction for the BCC crystal structure—betweenatoms M and N in (a).

of 2 equivalent circles. Thus the total circle area Ac is just

Ac � (2)�R 2

Finally, the planar density PD is just

PD �Ac

Ap�

2�R 2

8R 2 �2� 0.555

Linear and planar densities are important considerations relative to the processof slip—that is, the mechanism by which metals plastically deform (Section 8.5).Slip occurs on the most densely packed crystallographic planes and, in those planes,along directions having the greatest atomic packing.

3.19 X-RAY DIFFRACTION: DETERMINATION OF

CRYSTAL STRUCTURES

Historically much of our understanding regarding the atomic and molecular arrange-ments in solids has resulted from x-ray diffraction investigations; furthermore, x-raysare still very important in developing new materials. A brief overview of the diffrac-tion phenomenon and how, using x-rays, atomic interplanar distances and crystalstructures are deduced will now be given.

THE DIFFRACTION PHENOMENONDiffraction occurs when a wave encounters a series of regularly spaced obstaclesthat (1) are capable of scattering the wave, and (2) have spacings that are comparablein magnitude to the wavelength. Furthermore, diffraction is a consequence of specificphase relationships that are established between two or more waves that have beenscattered by the obstacles.

Consider waves 1 and 2 in Figure 3.34a, which have the same wavelength (�)and are in phase at point O–O�. Now let us suppose that both waves are scatteredin such a way that they traverse different paths. The phase relationship betweenthe scattered waves, which will depend upon the difference in path length, is impor-tant. One possibility results when this path length difference is an integral numberof wavelengths. As noted in Figure 3.34a, these scattered waves (now labeled 1�and 2�) are still in phase. They are said to mutually reinforce (or constructivelyinterfere with) one another; and, when amplitudes are added, the wave shown onthe right side of the figure results. This is a manifestation of diffraction, and werefer to a diffracted beam as one composed of a large number of scattered wavesthat mutually reinforce one another.

Other phase relationships are possible between scattered waves that will notlead to this mutual reinforcement. The other extreme is that demonstrated in Figure3.34b, wherein the path length difference after scattering is some integral numberof half wavelengths. The scattered waves are out of phase—that is, correspondingamplitudes cancel or annul one another, or destructively interfere (i.e., the resultantwave has zero amplitude), as indicated on the extreme right side of the figure. Ofcourse, phase relationships intermediate between these two extremes exist, resultingin only partial reinforcement.

X-RAY DIFFRACTION AND BRAGG’S LAWX-rays are a form of electromagnetic radiation that have high energies and shortwavelengths—wavelengths on the order of the atomic spacings for solids. When a

S-6 ● Chapter 3 / Structures of Metals and Ceramics

3.19 X-Ray Diffraction: Determination of Crystal Structures ● S-7

beam of x-rays impinges on a solid material, a portion of this beam will be scatteredin all directions by the electrons associated with each atom or ion that lies withinthe beam’s path. Let us now examine the necessary conditions for diffraction ofx-rays by a periodic arrangement of atoms.

Consider the two parallel planes of atoms A–A� and B–B� in Figure 3.35, whichhave the same h, k, and l Miller indices and are separated by the interplanar spacingdhkl . Now assume that a parallel, monochromatic, and coherent (in-phase) beamof x-rays of wavelength � is incident on these two planes at an angle �. Tworays in this beam, labeled 1 and 2, are scattered by atoms P and Q. Constructiveinterference of the scattered rays 1� and 2� occurs also at an angle � to the planes,if the path length difference between 1–P–1� and 2–Q–2� (i.e., SQ � QT) is equalto a whole number, n, of wavelengths. That is, the condition for diffraction is

n� � SQ � QT (3.9)

or

n� � dhkl sin � � dhkl sin � � 2dhkl sin � (3.10)

Equation 3.10 is known as Bragg’s law; also, n is the order of reflection, whichmay be any integer (1, 2, 3, . . .) consistent with sin � not exceeding unity. Thus,we have a simple expression relating the x-ray wavelength and interatomic spacingto the angle of the diffracted beam. If Bragg’s law is not satisfied, then the interfer-

Wave 1 Wave 1

Wave 2

Position

Wave 2

+

Scatteringevent

Am

plit

ude

O

(a)

(b)

O'

A

λ λ

A

A

A

λ λ

�

2A

Wave 3 Wave 3

Wave 4

Position

Wave 4

+

Scatteringevent

Am

plit

ude

P

P'

A

λ λ

A

A

A

λ

λ

�

�

�

�

FIGURE 3.34 (a)Demonstration of

how two waves(labeled 1 and 2)

that have the samewavelength � and

remain in phaseafter a scattering

event (waves 1� and2�) constructivelyinterfere with one

another. Theamplitudes of the

scattered waves addtogether in the

resultant wave. (b)Demonstration of

how two waves(labeled 3 and 4)

that have the samewavelength and

become out of phaseafter a scattering

event (waves 3� and4�) destructively

interfere with oneanother. The

amplitudes of thetwo scattered wavescancel one another.

ence will be nonconstructive in nature so as to yield a very low-intensity dif-fracted beam.

The magnitude of the distance between two adjacent and parallel planes ofatoms (i.e., the interplanar spacing dhkl) is a function of the Miller indices (h, k,and l) as well as the lattice parameter(s). For example, for crystal structures havingcubic symmetry,

dhkl �a

�h2 � k2 � l2(3.11)

in which a is the lattice parameter (unit cell edge length). Relationships similar toEquation 3.11, but more complex, exist for the other six crystal systems noted inTable 3.6.

Bragg’s law, Equation 3.10, is a necessary but not sufficient condition for diffrac-tion by real crystals. It specifies when diffraction will occur for unit cells havingatoms positioned only at cell corners. However, atoms situated at other sites (e.g.,face and interior unit cell positions as with FCC and BCC) act as extra scatteringcenters, which can produce out-of-phase scattering at certain Bragg angles. The netresult is the absence of some diffracted beams that, according to Equation 3.10,should be present. For example, for the BCC crystal structure, h � k � l must beeven if diffraction is to occur, whereas for FCC, h, k, and l must all be either oddor even.

DIFFRACTION TECHNIQUESOne common diffraction technique employs a powdered or polycrystalline specimenconsisting of many fine and randomly oriented particles that are exposed to mono-chromatic x-radiation. Each powder particle (or grain) is a crystal, and having alarge number of them with random orientations ensures that some particles areproperly oriented such that every possible set of crystallographic planes will beavailable for diffraction.

The diffractometer is an apparatus used to determine the angles at which diffrac-tion occurs for powdered specimens; its features are represented schematically inFigure 3.36. A specimen S in the form of a flat plate is supported so that rotationsabout the axis labeled O are possible; this axis is perpendicular to the plane of the

S-8 ● Chapter 3 / Structures of Metals and Ceramics

FIGURE 3.35Diffraction of x-rays

by planes of atoms(A–A� and B–B�).

� �

��

λ

λIncident

beamDiffracted

beam

P

S T

Q

A

B

1

2

A�

B�

1�

2�

dhkl

3 It should be pointed out that the symbol � has been used in two different contexts for thisdiscussion. Here, � represents the angular locations of both x-ray source and counter relativeto the specimen surface. Previously (e.g., Equation 3.10), it denoted the angle at which theBragg criterion for diffraction is satisfied.

3.19 X-Ray Diffraction: Determination of Crystal Structures ● S-9

page. The monochromatic x-ray beam is generated at point T, and the intensitiesof diffracted beams are detected with a counter labeled C in the figure. The specimen,x-ray source, and counter are all coplanar.

The counter is mounted on a movable carriage that may also be rotated aboutthe O axis; its angular position in terms of 2� is marked on a graduated scale.3

Carriage and specimen are mechanically coupled such that a rotation of the specimenthrough � is accompanied by a 2� rotation of the counter; this assures that theincident and reflection angles are maintained equal to one another (Figure 3.36).Collimators are incorporated within the beam path to produce a well-defined andfocused beam. Utilization of a filter provides a near-monochromatic beam.

As the counter moves at constant angular velocity, a recorder automaticallyplots the diffracted beam intensity (monitored by the counter) as a function of 2�;2� is termed the diffraction angle, which is measured experimentally. Figure 3.37shows a diffraction pattern for a polycrystalline specimen of iron. The high-intensitypeaks result when the Bragg diffraction condition is satisfied by some set of crystallo-graphic planes. These peaks are plane-indexed in the figure.

Other powder techniques have been devised wherein diffracted beam intensityand position are recorded on a photographic film instead of being measured bya counter.

One of the primary uses of x-ray diffractometry is for the determination ofcrystal structure. The unit cell size and geometry may be resolved from the angularpositions of the diffraction peaks, whereas arrangement of atoms within the unitcell is associated with the relative intensities of these peaks.

X-rays, as well as electron and neutron beams, are also used in other types ofmaterial investigations. For example, crystallographic orientations of single crystalsare possible using x-ray diffraction (or Laue) photographs. The chapter-openingphotograph for this chapter was generated using an incident electron beam thatwas directed on a gallium arsenide crystal; each spot (with the exception of thebrightest one near the center) resulted from an electron beam that was diffracted

O

θ

2θ

S

T

C

160°

140°

120°

100°80°

60°

40°

20°0°

FIGURE 3.36 Schematic diagram of anx-ray diffractometer; T � x-ray source,S � specimen, C � detector, and O �the axis around which the specimenand detector rotate.

by a specific set of crystallographic planes. Other uses of x-rays include qualitativeand quantitative chemical identifications, and the determination of residual stressesand crystal size.

EXAMPLE PROBLEM 3.13

For BCC iron, compute (a) the interplanar spacing, and (b) the diffractionangle for the (220) set of planes. The lattice parameter for Fe is 0.2866 nm(2.866 A). Also, assume that monochromatic radiation having a wavelength of0.1790 nm (1.790 A) is used, and the order of reflection is 1.

S O L U T I O N

(a) The value of the interplanar spacing dhkl is determined using Equation 3.11,with a � 0.2866 nm, and h � 2, k � 2, and l � 0, since we are considering the(220) planes. Therefore,

dhkl �a

�h2 � k2 � l2

�0.2866 nm

�(2)2 � (2)2 � (0)2� 0.1013 nm (1.013 A)

(b) The value of � may now be computed using Equation 3.10, with n � 1,since this is a first-order reflection:

sin � �n�

2dhkl�

(1)(0.1790 nm)(2)(0.1013 nm)

� 0.884

� � sin�1(0.884) � 62.13�

The diffraction angle is 2�, or

2� � (2)(62.13�) � 124.26�

S-10 ● Chapter 3 / Structures of Metals and Ceramics

(110)

Inte

nsit

y (r

elat

ive)

20 30 40 50 60 70 80 90 100

Diffraction angle 2θ

(200)

(211)

FIGURE 3.37 Diffraction pattern for polycrystalline �-iron.