1. AN INTRODUCTION TOMATERIALS ENGINEERINGAND SCIENCE

2. AN INTRODUCTION TOMATERIALS ENGINEERINGAND SCIENCEFOR CHEMICAL ANDMATERIALS ENGINEERSBrian S. MitchellDepartment of Chemical Engineering,Tulane UniversityA JOHN WILEY & SONS, INC., PUBLICATION 3. This book is printed on acid-free paper.Copyright 2004 by John Wiley & Sons, Inc., Hoboken, New Jersey. All rights reserved.Published simultaneously in Canada.No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form orby any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except aspermitted under Section 107 or 108 of the 1976 United States Copyright Act, without either the priorwritten permission of the Publisher, or authorization through payment of the appropriate per-copy fee tothe Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400, fax978-750-4470, or on the web at www.copyright.com. Requests to the Publisher for permission should beaddressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030,(201) 748-6011, fax (201) 748-6008, e-mail: permreq@wiley.com.Limit of Liability/Disclaimer of Warranty: While the publisher and author have used their best efforts inpreparing this book, they make no representations or warranties with respect to the accuracy orcompleteness of the contents of this book and specifically disclaim any implied warranties ofmerchantability or fitness for a particular purpose. No warranty may be created or extended by salesrepresentatives or written sales materials. The advice and strategies contained herein may not be suitablefor your situation. You should consult with a professional where appropriate. Neither the publisher norauthor shall be liable for any loss of profit or any other commercial damages, including but not limited tospecial, incidental, consequential, or other damages.For general information on our other products and services please contact our Customer Care Departmentwithin the U.S. at 877-762-2974, outside the U.S. at 317-572-3993 or fax 317-572-4002.Wiley also publishes its books in a variety of electronic formats. Some content that appears in print,however, may not be available in electronic format.Library of Congress Cataloging-in-Publication Data:Mitchell, Brian S., 1962-An introduction to materials engineering and science: for chemical and materials engineersBrian S. Mitchellp. cm.Includes bibliographical references and index.ISBN 0-471-43623-2 (cloth)1. Materials science. I. Title.TA403.M685 2003620.11dc21 2003053451Printed in the United States of America.10 9 8 7 6 5 4 3 2 1 4. To my parents; whoseMaterial was loam;Engineering was labor;Science was lore;And greatest product was love. 5. CONTENTSPreface xiAcknowledgments xv1 The Structure of Materials 11.0 Introduction and Objectives 11.1 Structure of Metals and Alloys 281.2 Structure of Ceramics and Glasses 551.3 Structure of Polymers 761.4 Structure of Composites 991.5 Structure of Biologics 114References 128Problems 1302 Thermodynamics of Condensed Phases 1362.0 Introduction and Objectives 1362.1 Thermodynamics of Metals and Alloys 1402.2 Thermodynamics of Ceramics and Glasses 1652.3 Thermodynamics of Polymers 1912.4 Thermodynamics of Composites 2002.5 Thermodynamics of Biologics 204References 209Problems 2113 Kinetic Processes in Materials 2153.0 Introduction and Objectives 2153.1 Kinetic Processes in Metals and Alloys 2193.2 Kinetic Processes in Ceramics and Glasses 2333.3 Kinetic Processes in Polymers 2463.4 Kinetic Processes in Composites 2693.5 Kinetic Processes in Biologics 277References 280Problems 2824 Transport Properties of Materials 2854.0 Introduction and Objectives 2854.1 Momentum Transport Properties of Materials 287vii 6. viii CONTENTS4.2 Heat Transport Properties of Materials 3164.3 Mass Transport Properties of Materials 343References 374Problems 3765 Mechanics of Materials 3805.0 Introduction and Objectives 3805.1 Mechanics of Metals and Alloys 3815.2 Mechanics of Ceramics and Glasses 4225.3 Mechanics of Polymers 4485.4 Mechanics of Composites 4725.5 Mechanics of Biologics 515References 532Problems 5336 Electrical, Magnetic, and Optical Properties of Materials 5376.1 Electrical Properties of Materials 5386.2 Magnetic Properties of Materials 6006.3 Optical Properties of Materials 644References 677Problems 6787 Processing of Materials 6817.0 Introduction 6817.1 Processing of Metals and Alloys 6817.2 Processing of Ceramics and Glasses 7047.3 Processing of Polymers 7547.4 Processing of Composites 7957.5 Processing of Biologics 804References 811Problems 8128 Case Studies in Materials Selection 8148.0 Introduction and Objectives 8148.1 Selection of Metals for a Compressed Air Tank 8218.2 Selection of Ceramic Piping for Coal Slurries in a CoalLiquefaction Plant 8278.3 Selection of Polymers for Packaging 8328.4 Selection of a Composite for an Automotive Drive Shaft 8358.5 Selection of Materials as Tooth Coatings 842References 848Problems 849 7. CONTENTS ixAppendix 1: Energy Values for Single Bonds 851Appendix 2: Structure of Some Common Polymers 852Appendix 3: Composition of Common Alloys 856Appendix 4: Surface and Interfacial Energies 869Appendix 5: Thermal Conductivities of Selected Materials 874Appendix 6: Diffusivities in Selected Systems 880Appendix 7: Mechanical Properties of Selected Materials 882Appendix 8: Electrical Conductivity of Selected Materials 893Appendix 9: Refractive Index of Selected Materials 900Answers to Selected Problems 903Index 907Sections marked with an asterisk can be omitted in an introductory course. 8. PREFACEThis textbook is intended for use in a one- or two-semester undergraduate course inmaterials science that is primarily populated by chemical and materials engineeringstudents. This is not to say that biomedical, mechanical, electrical, or civil engineeringstudents will not be able to utilize this text, nor that the material or its presentation isunsuitable for these students. On the contrary, the breadth and depth of the materialcovered here is equivalent to most traditional metallurgy-based approaches to thesubject that students in these disciplines may be more accustomed to. In fact, thetreatment of biological materials on the same level as metals, ceramics, polymers, andcomposites may be of particular benefit to those students in the biologically relatedengineering disciplines. The key difference is simply the organization of the material,which is intended to benefit primarily the chemical and materials engineer.This textbook is organized on two levels: by engineering subject area and by mate-rialsclass, as illustrated in the accompanying table. In terms of topic coverage, thisorganization is transparent: By the end of the course, the student will have coveredmany of the same things that would be covered utilizing a different materials sciencetextbook. To the student, however, the organization is intended to facilitate a deeperunderstanding of the subject material, since it is presented in the context of coursesthey have already had or are currently takingfor example, thermodynamics, kinetics,transport phenomena, and unit operations. To the instructor, this organization meansthat, in principle, the material can be presented either in the traditional subject-orientedsequence (i.e., in rows) or in a materials-oriented sequence (i.e., in columns). The latterapproach is recommended for a two-semester course, with the first two columns cov-eredin the first semester and the final three columns covered in the second semester.The instructor should immediately recognize that the vast majority of traditionalmaterials science concepts are covered in the columns on metals and ceramics, andthat if the course were limited to concepts on these two materials classes only, thestudent would receive instruction in many of the important topics covered in a tradi-tionalcourse on materials. Similarly, many of the more advanced topics are found inthe sections on polymers, composites, and biological materials and are appropriate fora senior-level, or even introductory graduate-level, course in materials with appropriatesupplementation and augmentation.This textbook is further intended to provide a unique educational experience forthe student. This is accomplished through the incorporation of instructional objectives,active-learning principles, design-oriented problems, and web-based information andvisualization utilization. Instructional objectives are included at the beginning of eachchapter to assist both the student and the instructor in determining the extent of topicsand the depth of understanding required from each topic. This list should be used as aguide only: Instructors will require additional information they deem important or elim-inatetopics they deem inappropriate, and students will find additional topic coverage intheir supplemental reading, which is encouraged through a list of references at the endxi 9. xii PREFACEMetals AlloysCeramics Glasses Polymers Composites BiologicsStructure Crystalstructures,Pointdefects,DislocationsCrystalstructures,Defectreactions,The glassystateConfiguration,Conformation,MolecularWeightMatrices,Reinforce-mentsBiochemistry,TissuestructureThermo-dynamicsPhaseequilibria,Gibbs RuleLever RuleTernarysystems,Surfaceenergy,SinteringPhase separation,Polymersolutions,PolymerblendsAdhesion,Cohesion,SpreadingCellAdhesion,CellspreadingKinetics Trans-formations,CorrosionDevitrification,Nucleation,GrowthPolymerization,DegradationDeposition,InfiltrationReceptors,LigandbindingTransportPropertiesInviscidsystems,Heatcapacity,DiffusionNewtonianflow, Heatcapacity,Diffusionnon-Newtonianflow, Heatcapacity,DiffusionPorous Flow,Heatcapacity,DiffusionConvection,DiffusionMechanicalPropertiesStress-strain,Elasticity,DuctilityFatigue,Fracture,CreepViscoelasticity,ElastomersLaminates Sutures,Bone,TeethElectrical,Magnetic OpticalPropertiesResistivity,Magnetism,ReflectanceDielectrics,Ferrites,AbsorbanceIon conductors,Molecularmagnets, LCDsDielectrics,StoragemediaBiosensors,MRIProcessing Casting,Rolling,CompactionPressing,CVD/CVI,Sol-GelExtrusion,Injectionmolding, BlowmoldingPultrusion,RTM,CVD/CVISurfacemodificationCase Studies Compressedair tankCeramicpipingPolymericpackagingCompositedrive shaftToothcoatingsof each chapter. Active-learning principles are exercised through the presentation ofexample problems in the form of Cooperative Learning Exercises. To the student, thismeans that they can solve problems in class and can work through specific difficultiesin the presence of the instructor. Cooperative learning has been shown to increase thelevel of subject understanding when properly utilized. No class is too large to allowstudents to take 510 minutes to solve these problems. To the instructor, the Coop-erativeLearning Exercises are to be used only as a starting point, and the instructoris encouraged to supplement his or her lecture with more of these problems. Particu-larlydifficult concepts or derivations are presented in the form of Example Problemsthat the instructor can solve in class for the students, but the student is encouraged tosolve these problems during their own group or individual study time. Design-orientedproblems are offered, primarily in the Level III problems at the end of each chapter,Smith, K. Cooperative Learning and College Teaching, 3(2), 1012 (1993). 10. PREFACE xiiithat incorporate concepts from several chapters, that involve significant informationretrieval or outside reading, or that require group activities. These problems may ormay not have one best answer and are intended to promote a deeper level of under-standingof the subject. Finally, there is much information on the properties of materialsavailable on the Internet. This fact is utilized through the inclusion of appropriate weblinks. There are also many excellent visualization tools available on the Internet forconcepts that are too difficult to comprehend in a static, two-dimensional environment,and links are provided to assist the student in their further study on these topics.Finally, the ultimate test of the success of any textbook is whether or not it stays onyour bookshelf. It is hoped that the extent of physical and mechanical property data,along with the depth with which the subjects are presented, will serve the student wellas they transition to the world of the practicing engineer or continue with their studiesin pursuit of an advanced degree.BRIAN S. MITCHELLTulane University 11. ACKNOWLEDGMENTSThe author wishes to thank the many people who have provided thoughtful input tothe content and presentation of this book. In particular, the insightful criticisms andcomments of Brian Grady and the anonymous reviewers are very much appreciated.Thanks also go to my students who have reviewed various iterations of this textbook,including Claudio De Castro, Shawn Haynes, Ryan Shexsnaydre, and Amanda Moster,as well as Dennis Johnson, Eric Hampsey, and Tom Fan. The support of my colleaguesduring the writing of this book, along with the support of the departmental staff, aregratefully acknowledged. Finally, the moral support of Bonnie, Britt, Rory, and Chelsieis what ultimately has led to the completion of this textbookthank you.BRIAN S. MITCHELLTulane Universityxv 12. CHAPTER 1The Structure of Materials1.0 INTRODUCTION AND OBJECTIVESA wealth of information can be obtained by looking at the structure of a material.Though there are many levels of structure (e.g., atomic vs. macroscopic), many phys-icalproperties of a material can be related directly to the arrangement and types ofbonds that make up that material. We will begin by reviewing some general chemicalprinciples that will aid us in our description of material structure. Such topics as peri-odicstructure, types of bonding, and potential energy diagrams will be reviewed. Wewill then use this information to look at the specific materials categories in more detail:metals, ceramics, polymers, composites, and biological materials (biologics). There willbe topics that are specific to each material class, and there will also be some that arecommon to all types of materials. In subsequent chapters, we will explore not onlyhow the building blocks of a material can significantly impact the properties a materialpossesses, but also how the material interacts with its environment and other materialssurrounding it.By the end of this chapter you should be able to: Identify trends in the periodic table for IE, EA, electronegativity, and atomic/ionicradii. Identify the type of bonding in a compound. Utilize the concepts of molecular orbital and hybridization theories to explainmultiple bonds, bond angle, diamagnetism, and paramagnetism. Identify the seven crystal systems and 14 Bravais lattices. Calculate the volume of a unit cell from the lattice translation vectors. Calculate atomic density along directions, planes, and volumes in a unit cell. Calculate the density of a compound from its crystal structure and atomic mass. Locate and identify the interstitial sites in a crystal structure. Assign coordinates to a location, indices to a direction, and Miller indices to aplane in a unit cell. Use Braggs Law to convert between diffraction angle and interplanar spacing. Read and interpret a simple X-ray diffraction pattern. Identify types of point and line defects in solids.An Introduction to Materials Engineering and Science: For Chemical and Materials Engineers,by Brian S. MitchellISBN 0-471-43623-2 Copyright 2004 John WileySons, Inc.1 13. 2 THE STRUCTURE OF MATERIALS Calculate the concentration of point defects in solids. Draw a Burgers circuit and identify the direction of dislocation propagation. Use Paulings rules to determine the stability of a compound. Predict the structure of a silicate from the Si/O ratio. Apply Zachariasens rules to determine the glass forming ability of an oxide. Write balanced defect reaction equations using KrogerVink notation. Classify polymers according to structure or formability. Calculate the first three moments of a polymer molecular weight distribution. Apply principles of glass transition and polymer crystallinity to polymer classifi-cation. Identify nematic, smectic, and cholesteric structures in liquid crystalline polymers. Identify the components in a composite material. Approximate physical properties of a composite material based on componentproperties. Be conversant in terms that relate to the structure of biological materials, such asfibronectin and integrins.1.0.1 The ElementsElements are materials, too. Oftentimes, this fact is overlooked. Think about all thematerials from our daily lives that are elements: gold and silver for our jewelry; alu-minumfor our soda cans; copper for our plumbing; carbon, both as a luminescentdiamond and as a mundane pencil lead; mercury for our thermometers; and tungstenfor our light bulb filaments. Most of these elements, however, are relatively scarce inthe grand scheme of things. A table of the relative abundance of elements (Table 1.1)shows that most of our universe is made up of hydrogen and helium. A little closerto home, things are much different. A similar table of relative abundance (Table 1.2)shows that helium on earth is relatively scarce, while oxygen dominates the crust ofour planet. Just think of how much molecular oxygen, water, and aluminosilicate rocksare contained in the earths crust. But those are moleculeswe are concentrating onatoms for the moment. Still, elements are of vital importance on earth, and the oneswe use most often are primarily in the solid form.Recall from your introductory chemistry course that the elements can be systemati-callyarranged in a periodic table according to their electronic structure (see Table 1.3).An overall look at the periodic table (Figure 1.1) shows that many elements are solids(white boxes) at room temperature. The fact that many of these elements remain solidwell above ambient temperatures is important. As we heat to 1000C, note that manyof the IIIAVA elements have melted (light shaded); also note how many of the alkalimetals (IA) have vaporized (dark shaded), but how most of the transition elements arestill in solid form. At 2000C, the alkali earths are molten, and many of the transitionelements have begun to melt, too. Note that the highest melting point element is carbon(Figure 1.1d). Keep in mind that this is in an inert atmosphere. What should happen toNote that the Lanthanide (atomic numbers 5871) and Actinide (90103) series elements, as well as thesynthetic elements of atomic number greater than 87, are omitted from all the periodic tables in this text.With the possible exception of nuclear fuels such as uranium and plutonium, these elements are of littlegeneral engineering interest. 14. INTRODUCTION AND OBJECTIVES 3Table 1.1 Relative Abundance of Elements in theUniverseElementRelativeAbundance (Si = 1)Hydrogen (H) 12,000Helium (He) 2,800Oxygen (O) 16Nitrogen (N) 8Carbon (C) 3Iron (Fe) 2.6Silicon (Si) 1Magnesium (Mg) 0.89Sulfur (S) 0.33Nickel (Ni) 0.21Aluminum (Al) 0.09Calcium (Ca) 0.07Sodium (Na) 0.045Chlorine (Cl) 0.025Table 1.2 Relative Abundance of Selected Elements in the Earths CrustRelative RelativeElement Abundance (ppm) Element Abundance (ppm)Oxygen (O) 466,000 Fluorine (F) 300Silicon (Si) 277,200 Strontium (Sr) 300Aluminum (Al) 81,300 Barium (Ba) 250Iron (Fe) 50,000 Zirconium (Zr) 220Calcium (Ca) 36,300 Chromium (Cr) 200Sodium (Na) 28,300 Vanadium (V) 150Potassium (K) 25,900 Zinc (Zn) 132Magnesium (Mg) 20,900 Nickel (Ni) 80Titanium (Ti) 4,400 Molybdenum (Mo) 15Hydrogen (H) 1,400 Uranium (U) 4Phosphorus (P) 1,180 Mercury (Hg) 0.5Manganese (Mn) 1,000 Silver (Ag) 0.1Sulfur (S) 520 Platinum (Pt) 0.005Carbon (C) 320 Gold (Au) 0.005Chlorine (Cl) 314 Helium (He) 0.003this element in the presence of oxygen? Such elements as tungsten, platinum, molyb-denum,and tantalum have exceptional high-temperature properties. Later on we willinvestigate why this is so.In addition, many elements are, in and of themselves, materials of construction.Aluminum and copper are just a few examples of elements that are used extensivelyfor fabricating mechanical parts. Elements have special electrical characteristics, too.Silver and gold are used not just for jewelry, but also for a wide variety of electricalcomponents. We will visit all of these topics in the course of this textbook. 15. 4 THE STRUCTURE OF MATERIALS1.0.2 Trends in the Periodic TableA closer look at the periodic table points out some interesting trends. These trendsnot only help us predict how one element might perform relative to another, but alsogive us some insight into the important properties of atoms and ions that determinetheir performance. For example, examination of the melting points of the elements inTable 1.3 shows that there is a general trend to decrease melting point as we go down(a)4Be12Mg20Ca38Sr56Ba88Ra1H3Li11Na19K37Rb55Cs87Fr5B13Al31Ga49ln81TI6C14Si32Ge50Sn82Pb22Ti40Zr72Hf21Sc39Y57La89Ac24Cr42Mo74W23V41Nb73Ta26Fe44Ru76Os25Mn43Tc75Re28Ni46Pd78Pt27Co45Rh77Ir30Zn48Cd29Cu47Ag80Hg59Pr58Ce61Pm60Nd63Eu62Sm65Tb64Gd66Dy67Ho68Er69Tm70Yb91Pa90Th93Np92U95Am94Pu97Bk96Cm98Cf99Es100Fm101Md102NoLegendSolidLiquidGasNot AvailableTemperature: 290 K16 C62 F15P33As51Sb83Bi71Lu103Lr16S17CI2He18Ar34Se35Br36Kr52Te84Po53I54Xe85At86Rn7N8O9F10Ne79Au(b)4Be12Mg20Ca38Sr56Ba88Ra1H3Li11Na19K37Rb55Cs87Fr5B13Al31Ga49ln81TI6C14Si32Ge50Sn82Pb22Ti40Zr72Hf21Sc39Y57La89Ac24Cr42Mo74W23V41Nb73Ta26Fe44Ru76Os25Mn43Tc75Re28Ni46Pd78Pt27Co45Rh77Ir30Zn48Cd29Cu47Ag79Au80Hg59Pr58Ce61Pm60Nd63Eu62Sm65Tb64Gd66Dy67Ho68Er69Tm70Yb91Pa90Th93Np92U95Am94Pu97Bk96Cm98Cf99Es100Fm101Md102NoLegendSolidLiquidGasNot AvailableTemperature: 1280 K1006 C1844 F15P33As51Sb83Bi71Lu103Lr16S17CI2He18Ar34Se35Br36Kr52Te84Po53I54Xe85At86Rn7N8O9F10NeFigure 1.1 The periodic table of the elements at (a) room temperature, (b) 1000C, (c) 2000C,and (d) 3500C. 16. INTRODUCTION AND OBJECTIVES 5(c)4Be12Mg20Ca38Sr56Ba88Ra1H3Li11Na19K37Rb55Cs87Fr5B13Al31Ga49ln81TI6C14Si32Ge50Sn82Pb22Ti40Zr72Hf21Sc39Y57La89Ac24Cr42Mo74W23V41Nb73Ta26Fe44Ru76Os25Mn43Tc75Re28Ni46Pd78Pt27Co45Rh77Ir30Zn48Cd29Cu47Ag80Hg59Pr58Ce61Pm60Nd63Eu62Sm65Tb64Gd66Dy67Ho68Er69Tm70Yb91Pa90Th93Np92U95Am94Pu97Bk96Cm98Cf99Es100Fm101Md102NoLegendSolidLiquidGasNot AvailableTemperature: 2280 K2006 C3644 F15P33As51Sb83Bi71Lu103Lr16S17CI2He18Ar34Se35Br36Kr52Te84Po53I54Xe85At86Rn7N8O9F10Ne79Au(d)4Be12Mg20Ca38Sr56Ba88Ra1H3Li11Na19K37Rb55Cs87Fr5B13Al31Ga49ln81TI6C14Si32Ge50Sn82Pb22Ti40Zr72Hf21Sc39Y57La89Ac24Cr42Mo74W23V41Nb73Ta26Fe44Ru76Os25Mn43Tc75Re28Ni46Pd78Pt27Co45Rh77Ir30Zn48Cd29Cu47Ag80Hg59Pr58Ce61Pm60Nd63Eu62Sm65Tb64Gd66Dy67Ho68Er69Tm70Yb91Pa90Th93Np92U95Am94Pu97Bk96Cm98Cf99Es100Fm101Md102NoLegendSolidLiquidGasNot AvailableTemperature: 3780 K3506 C6344 F15P33As51Sb83Bi71Lu103Lr16S17CI2He18Ar34Se35Br36Kr52Te84Po53I54Xe85At86Rn7N8O9F10Ne79AuFigure 1.1 (continued).a column for the alkali metals and alkali earth elements (columns IA and IIA), butthat the column trend for the transition metals appears to be different. There are sometrends that are more uniform, however, and are related to the electronic structure ofthe element.1.0.2.1 First Ionization Energy (IE). The first ionization energy, abbreviated IE, issometimes referred to as the ionization potential. It is the energy required to remove 17. 6 THE STRUCTURE OF MATERIALSTable 1.3 Electronic Structure and Melting Points of the Elements1 es = electronic structure 2H aw = atomic weight (average including isotopes) Hees 1s1 mp = melting point, C (sublimation temperatures enclosed in parentheses). 1s2aw 1.008 4.003mp 259 3 4 5 6 7 8 9 10Li Be B C N O F Nees He2s1 He2s2 Be2p1 Be2p2 Be2p3 Be2p4 Be2p5 Be2p6aw 6.94 9.012 10.81 12.01 14.006 15.999 18.998 20.18mp 180.5 1289 2103 (3836) 210.0 218.8 219.7 24911 12 13 14 15 16 17 18Na Mg Al Si P S Cl Ares Ne3s1 Ne3s2 Mg3p1 Mg3p2 Mg3p3 Mg3p4 Mg3p5 Mg3p6aw 22.99 24.30 26.98 28.09 30.974 32.06 35.45 39.95mp 97.8 649 660.4 1414 44.1 112.8 101.0 18919 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kres Ar4s1 Ar4s2 Ca3d1 Ca3d2 Ca3d3 K3d5 Ca3d5 Ca3d6 Ca3d7 Ca3d8 K3d10 Ca3d10 Ca4p1 Ca4p2 Ca4p3 Ca4p4 Ca4p5 Ca4p6aw 39.10 40.08 44.96 47.9 50.94 51.9 54.93 55.85 58.93 58.71 63.55 65.37 69.72 72.59 74.92 78.96 79.90 83.80mp 63.2 840 1541 1672 1929 1863 1246 1538 1494 1455 1084.5 419.6 29.8 938.3 (603) 221 7.2 15737 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xees Kr5s1 Kr5s2 Sr4d1 Sr4d2 Rb4d4 Rb4d5 Rb4d6 Rb4d7 Rb4d8 Kr4d10 Rb4d10 Sr4d10 Sr5p1 Sr5p2 Sr5p3 Sr5p4 Sr5p5 Sr5p6aw 85.47 87.62 88.91 91.22 92.91 95.94 98.91 101.7 102.9 106.4 107.87 112.4 114.8 118.7 121.8 127.6 126.9 131.3mp 39.5 769 1528 1865 2471 2623 2204 2254 1963 1554 961.9 321.1 156.6 232.0 630.7 449.6 113.6 11255 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rnes Xe6s1 Xe6s2 Ba5d1 Ba5d2 Ba5d3 Ba5d4 Ba5d5 Ba5d6 Xe5d9 Cs5d9 Cs5d10 Ba5d10 Ba6p1 Ba6p2 Ba6p3 Ba6p4 Ba6p5 Ba6p6aw 132.9 137.3 138.9 178.5 180.9 183.9 186.2 190.2 192.2 195.1 196.97 200.6 204.4 207.2 208.9 210 210 222mp 28.4 729 921 2231 3020 3387 3186 3033 2447 1772 1064.4 38.9 304 327.5 271.4 254 71Source: Ralls, Courtney, Wulff, Introduction to Materials Science and Engineering, Wiley, 1976. 18. INTRODUCTION AND OBJECTIVES 7the most weakly bound (usually outermost) electron from an isolated gaseous atomatom (g) + IE positive ion (g) + e(1.1)and can be calculated using the energy of the outermost electron as given by the Bohrmodel and Schrodingers equation (in eV):IE = 13.6Z2n2 (1.2)where Z is the effective nuclear charge and n is the principal quantum number.As shown in Figure 1.2a, the general trend in the periodic table is for the ionizationenergy to increase from bottom to top and from left to right (why?). A quantity relatedto the IE is the work function. The work function is the energy necessary to removean electron from the metal surface in thermoelectric or photoelectric emission. We willdescribe this in more detail in conjunction with electronic properties of materials inChapter 6.(a) (b)(c) (d)Figure 1.2 Some important trends in the periodic table for (a) ionization energy, (b) electronaffinity, (c) atomic and ionic radii, and (d) electronegativity. Increasing values are in the directionof the arrow. 19. 8 THE STRUCTURE OF MATERIALS1.0.2.2 Electron Affinity (EA). Electron affinity is the reverse process to the ioniza-tionenergy; it is the energy change (often expressed in eV) associated with an isolatedgaseous atom accepting one electron:atom (g) + e negative ion (g) (1.3)Unlike the ionization energy, however, EA can have either a negative or positivevalue, so it is not included in Eq. (1.3). The EA is positive if energy is released uponformation of the negative ion. If energy is required, EA is negative. The general trendin the periodic table is again toward an increase in EA as we go from the bottom to top,and left to right (Figure 1.2b), though this trend is much less uniform than for the IE.1.0.2.3 Atomic and Ionic Radii. In general, positive ions are smaller than neutralatoms, while negative ions are larger (why?). The trend in ionic and atomic radii isopposite to that of IE and EA (Figure 1.2c). In general, there is an increase in radiusfrom top to bottom, right to left. In this case, the effective nuclear charge increases fromleft to right, the inner electrons cannot shield as effectively, and the outer electronsare drawn close to the nucleus, reducing the atomic radius. Note that the radii are onlyapproximations because the orbitals, in theory, extend to infinity.1.0.2.4 Electronegativity. The ionization energy and electron affinity are charac-teristicsof isolated atoms; they say very little about how two atoms will interact witheach other. It would be nice to have an independent measure of the attraction an atomhas for electrons in a bond formed with another atom. Electronegativity is such a quan-tity.It is represented by the lowercase Greek letter chi, . Values can be calculatedusing one of several methods discussed below. Values of are always relative to oneanother for a given method of calculation, and values from one method should not beused with values from another method.Based upon a scale developed by Mulliken, electronegativity is the average of theionization energy and the electron affinity: = IE + EA2(1.4)There are other types of electronegativity scales as well, the most widely utilized ofwhich is the one from the developer of the electronegativity concept, Linus Pauling: = 0.31(n + 1 c)r+ 0.5 (1.5)where n is the number of valence electrons, c is any formal valence charge on the atomand the sign corresponding to it, and r is the covalent radius. Typical electronegativityvalues, along with values of IE and EA, are listed in Table 1.4. We will use the conceptof electronegativity to discuss chemical bonding.1.0.3 Types of BondsElectronegativity is a very useful quantity to help categorize bonds, because it providesa measure of the excess binding energy between atoms A and B, AB (in kJ/mol):AB = 96.5(A B)2 (1.6) 20. Table 1.4 Ionization Energies, Electron Affinities, and Electronegativities of the Elementsa1 2H HeIE 1310 2372EA 67.4 60.2 2.20 3 4 5 6 7 8 9 10Li Be B C N O F NeIE 519 900 799 1088 1406 1314 1682 2080EA 77.0 18.4 31.8 119.7 4.6 141.8 349.4 54.8 0.98 1.57 2.04 2.55 3.04 3.44 3.98 11 12 13 14 15 16 17 18Na Mg Al Si P S Cl ArIE 498 736 577 787 1063 1000 1255 2372EA 117.2 0 50.2 138.1 75.3 199.6 356.1 60.2 0.93 1.31 1.61 1.90 2.19 2.58 3.16 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br KrIE 1310 590 632 661 653 653 715 761 757 736 745 904 577 782 966 941 1142 1351EA 67.4 333.0 2.20 1.00 1.36 1.54 1.63 1.66 1.55 1.8 1.88 1.91 1.90 1.65 1.81 2.01 2.18 2.55 2.96 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I XeIE 402 548 636 669 653 695 699 724 745 803 732 866 556 707 833 870 1008 1172EA 304.2 0.82 0.95 1.22 1.33 1.6 2.16 1.9 2.28 2.2 2.20 1.93 1.69 1.78 1.96 2.05 2.1 2.66 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At RnIE 377 502 540 531 577 770 761 841 887 866 891 1008 590 715 774 1038EA 0.79 0.89 1.10 1.3 1.5 2.36 1.9 2.2 2.2 2.28 2.54 2.00 2.04 2.33 2.02 2.0 2.2 a Ionization energy (IE) and Electron affinities (EA) are expressed as kilojoules per mole. 1 eV = 96,490 J/mol.Source: R. E. Dickerson, H. B. Gray, and G. P. Haight, Chemical Principles, 3rd ed., Pearson Education, Inc., 1979.9 21. 10 THE STRUCTURE OF MATERIALSThe excess binding energy, in turn, is related to a measurable quantity, namely thebond dissociation energy between two atoms, DEij :AB = DEAB [(DEAA)(DEBB)]1/2 (1.7)The bond dissociation energy is the energy required to separate two bonded atoms(see Appendix 1 for typical values). The greater the electronegativity difference, thegreater the excess binding energy. These quantities give us a method of characterizingbond types. More importantly, they relate to important physical properties, such asmelting point (see Table 1.5). First, let us review the bond types and characteristics,then describe each in more detail.1.0.3.1 Primary Bonds. Primary bonds, also known as strong bonds, are createdwhen there is direct interaction of electrons between two or more atoms, either throughtransfer or as a result of sharing. The more electrons per atom that take place in this pro-cess,the higher the bond order (e.g., single, double, or triple bond) and the strongerthe connection between atoms. There are four general categories of primary bonds:ionic, covalent, polar covalent, and metallic. An ionic bond, also called a heteropolarTable 1.5 Examples of Substances with Different Types of Interatomic BondingType ofBond SubstanceBond Energy,kJ/molMelting Point,C) Characteristics(Ionic CaCl 651 646 Low electrical conductivity,NaCl 768 801 transparent, brittle, highLiF 1008 870 melting pointCuF2 2591 1360Al2O3 15,192 3500Covalent Ge 315 958 Low electrical conductivity,GaAs 315 1238 very hard, very highSi 353 1420 melting pointSiC 1188 2600Diamond 714 3550Metallic Na 109 97.5 High electrical and thermalAl 311 660 conductivity, easilyCu 340 1083 deformable, opaqueFe 407 1535W 844 3370van der Waals Ne 2.5 248.7 Weak binding, low meltingAr 7.6 189.4 and boiling points, veryCH4 10 184 compressibleKr 12 157Cl2 31 103Hydrogen bonding HF 29 92 Higher melting point than vanH2O 50 0 der Waals bonding, tendencyto form groups of manymolecules 22. INTRODUCTION AND OBJECTIVES 11bond, results when electrons are transferred from the more electropositive atom to themore electronegative atom, as in sodium chloride, NaCl. Ionic bonds usually resultwhen the electronegativity difference between two atoms in a diatomic molecule isgreater than about 2.0. Because of the large discrepancy in electronegativities, oneatom will generally gain an electron, while the other atom in a diatomic moleculewill lose an electron. Both atoms tend to be satisfied with this arrangement becausethey oftentimes end up with noble gas electron configurationsthat is, full electronicorbitals. The classic example of an ionic bond is NaCl, but CaF2 and MgO are alsoexamples of molecules in which ionic bonding dominates.A covalent bond, or homopolar bond, arises when electrons are shared betweentwo atoms (e.g., HH). This means that a binding electron in a covalent diatomicmolecule such as H2 has equal likelihood of being found around either hydrogen atom.Covalent bonds are typically found in homonuclear diatomics such as O2 and N2,though the atoms need not be the same to have similar electronegativities. Electroneg-ativitydifferences of less than about 0.4 characterize covalent bonds. For two atomswith an electronegativity difference of between 0.4 and 2.0, a polar covalent bond isformedone that is neither truly ionic nor totally covalent. An example of a polarcovalent bond can be found in the molecule hydrogen fluoride, HF. Though there issignificant sharing of the electrons, some charge distribution exists that results in apolar or partial ionic character to the bond. The percent ionic character of the bondcan again be related to the electronegativities of the individual atoms:% ionic character = 100{1 exp[0.25(A B)2]} (1.8)Example Problem 1.1What is the percent ionic character of HF?Answer: According to Table 1.3, the electronegativity of hydrogen is 2.20 and that offluorine 3.98. Putting these values into Eq. (1.8) gives% ionic character of HF = 100[1 exp{0.25(2.20 3.98)2}] = 55%The larger the electronegativity difference, the more ionic character the bond has. Ofcourse, if the electronegativity difference is greater than about 2.0, we know that anionic bond should result.Finally, a special type of primary bond known as a metallic bond is found inan assembly of homonuclear atoms, such as copper or sodium. Here the bondingelectrons become decentralized and are shared by the core of positive nuclei. Metallicbonds occur when elements of low electronegativity (usually found in the lower leftregion of the periodic table) bond with each other to form a class of materials we callmetals. Metals tend to have common characteristics such as ductility, luster, and highthermal and electrical conductivity. All of these characteristics can to some degreebe accounted for by the nature of the metallic bond. The model of a metallic bond,first proposed by Lorentz, consists of an assembly of positively charged ion coressurrounded by free electrons or an electron gas. We will see later on, when we 23. 12 THE STRUCTURE OF MATERIALSdescribe intermolecular forces and bonding, that the electron cloud does indeed havestructure in the quantum mechanical sense, which accounts nicely for the observedelectrical properties of these materials.1.0.3.2 Secondary Bonds. Secondary bonds, or weak bonds, occur due to indirectinteraction of electrons in adjacent atoms or molecules. There are three main types ofsecondary bonding: hydrogen bonding, dipoledipole interactions, and van der Waalsforces. The latter, named after the famous Dutch physicist who first described them,arise due to momentary electric dipoles (regions of positive and negative charge) thatcan occur in all atoms and molecules due to statistical variations in the charge density.These intermolecular forces are common, but very weak, and are found in inert gaseswhere other types of bonding do not exist.Hydrogen bonding is the attraction between hydrogen in a highly polar moleculeand the electronegative atom in another polar molecule. In the water molecule, oxygendraws much of the electron density around it, creating positively charged centers at thetwo hydrogen atoms. These positively charged hydrogen atoms can interact with thenegative center around the oxygen in adjacent water molecules. Although this type ofHISTORICAL HIGHLIGHTDutch physicist Johannes Diderik vander Waals was born on November 23,1837 in Leiden, the Netherlands. He wasthe eldest son of eight children. Initially,van der Waals was an elementary schoolteacher during the years 18561861. Hecontinued studying to become headmasterand attended lectures on mathematics,physics, and astronomy at Leiden University.From 1866 onwards he taught physicsand mathematics at a secondary school inThe Hague. After a revision of the law,knowledge of Latin and Greek was no longera prerequisite for an academic graduation,and in 1873 J. D. van der Waals graduatedon the thesis: Over de continuiteit vande gasenvloeistoftoestand (About thecontinuity of gaseous and liquid states).In this thesis he published the well-knownlaw:P + aV 2(V b) = RTThis revision to the ideal gas law accountedfor the specific volume of gas molecules andassumed a force between these moleculeswhich are now known as van der Waalsforces. With this law, the existence ofcondensation and the critical temperature ofgases could be predicted. In 1877 J. D.van der Waals became the first professorof physics at the University Illustre inAmsterdam. In 1880 he formulated hislaw of corresponding states, in 1893 hedevised a theory for capillary phenomena,and in 1891 he introduced his theory for thebehavior of two mixtures of two materials.It was not possible to experimentally showthe de-mixing of two gases into twoseparate gases under certain circumstancesas predicted by this theory until 1941.From 1875 to 1895 J.D. van der Waalswas a member of the Dutch Royal Academyof Science. In 1908, at the age of 71, J. D.van der Waals resigned as a professor. Dur-inghis life J. D. van der Waals was honoredmany times. He was one of only 12 foreignmembers of the Academie des Sciences inParis. In 1910 he received the Nobel prize forPhysics for the incredible work he had doneon the equations of state for gases and flu-idsonly the fifth Dutch physicist to receivethis honor. J. D. van derWaals died on March8, 1923 at the age of 85.Source: www.vdwaals.nl 24. INTRODUCTION AND OBJECTIVES 13bonding is of the same order of magnitude in strength as van der Waals bonding, it canhave a profound influence on the properties of a material, such as boiling and meltingpoints. In addition to having important chemical and physical implications, hydrogenbonding plays an important role in many biological and environmental phenomena. It isresponsible for causing ice to be less dense than water (how many other substances doyou know that are less dense in the solid state than in the liquid state?), an occurrencethat allows fish to survive at the bottom of frozen lakes.Finally, some molecules possess permanent charge separations, or dipoles, such asare found in water. The general case for the interaction of any positive dipole with anegative dipole is called dipoledipole interaction. Hydrogen bonding can be thought ofas a specific type of dipoledipole interaction. A dipolar molecule like ammonia, NH3,is able to dissolve other polar molecules, like water, due to dipoledipole interactions.In the case of NaCl in water, the dipoledipole interactions are so strong as to breakthe intermolecular forces within the molecular solid.Now that the types of bonds have been reviewed, we will concentrate on the primarybond because it correlates more directly with physical properties in solids than dosecondary bonds. Be aware that the secondary forces exist, though, and that they playa larger role in liquids and gases than in solids.1.0.4 Intermolecular Forces and BondingWe have described the different types of primary bonds, but how do these bonds formin the first place? What is it that causes a sodium ion and a chloride ion to form acompound, and what is it that prevents the nuclei from fusing together to form oneelement? These questions all lead us to the topics of intermolecular forces and bondformation. We know that atoms approach each other only to a certain distance, andthen, if they form a compound, they will maintain some equilibrium separation distanceknown as the bond length. Hence, we expect that there is some attractive energy thatbrings them together, as well as some repulsive energy that keeps the atoms a certaindistance apart.Also known as chemical affinity, the attractive energy between atoms is what causesthem to approach each other. This attraction is due to the electrostatic force betweenthe nucleus and electron clouds of the separate atoms. It should make sense to youthat the attractive energy (UA) is inversely proportional to the separation distance, r;that is, the further the atoms are apart, the weaker the attraction:UA = arm(1.9)where a is a constant that we will describe in more detail in a moment, and m is aconstant with a value of 1 for ions and 6 for molecules. Notice that there is a negativesign in Eq. (1.9). By convention, we will refer to the attractive energy as a negativeenergy.Once the atoms begin to approach each other, they can only come so close togetherdue to the impenetrability of matter. The result is a repulsive energy, which we assign apositive value, again, by convention. The primary constituents of this repulsive energyare nucleusnucleus and electronelectron repulsions. As with the attractive energy,the repulsive energy is inversely proportional to the separation distance; the closer the 25. 14 THE STRUCTURE OF MATERIALSTable 1.6 Values of the Repulsion ExponentNoble GasIon CoreOuter CoreConfiguration nHe 1s2 5Ne 2s22p6 7Ar 3s23p6 9Kr 3d104s24p6 10Xe 4d105s25p6 12atoms are, the more they repel each other:UR = brn(1.10)where b and n are constants. The value of n, called the repulsion exponent, dependson the outer core configuration of the atom. Values of the repulsion exponent are givenin Table 1.6.The total, or potential energy of the system is then the sum of the attractive andrepulsive components:U = UA + UR =arm+ brn(1.11)The result is the potential energy well (see Figure 1.3). The well-known Lennard-JonespotentialU = ar6+ br12 (1.12)is a common potential energy function used in a number of models, including col-lisiontheory for kinetics. It is simply a special case of Eq. (1.11) with n = 12 (Xeconfiguration) and m = 6 (molecules).It is oftentimes useful to know the forces involved in bonding, as well as the energy.Recall that energy and force, F, are related byF = dUdr(1.13)We will see later on that we can use this expression to convert between forceand energy for specific types of atoms and molecules (specific values of n and m).For now, this expression helps us find the equilibrium bond distance, r0, which occurswhen forces are equal (the sum of attractive and repulsive forces is zero) or at minimumpotential energy (take the derivative and set it equal to zero):F = dUdr= 0, at r = r0 (1.14)This is not the same thing as the maximum attractive force, which we get by maximiz-ingF:Fmax = dFdr= d2Udr2= 0 (1.15) 26. INTRODUCTION AND OBJECTIVES 15Repulsion energy,br nInteratomic or intermolecular distance in The net potential energy, U = aAttraction energy, armrmbr n +rxsr = sr0UmAttraction RepulsionPotential energy, U+0 (a)Repulsion forceInteratomic or intermolecular distance in Attraction forcerxrrr0Attraction RepulsionForce, F+0Fmax (b)Figure 1.3 The interatomic (a) potential energy and (b) force diagrams. From Z. Jastrzebski,The Nature and Properties of Engineering Materials, 2nd ed. by Copyright 1976 by JohnWileySons, Inc. This material is used by permission of John WileySons, Inc.The forces are equal when the potential energy is a minimum and the separationdistance is at the bond length, r0. Differentiation of Eq. (1.11) and solving for r0 interms of a, b, n, and m givesr0 =nbma 1n m(1.16) 27. 16 THE STRUCTURE OF MATERIALSThe potential energy well concept is an important one, not only for the calculationof binding energies, as we will see in a moment, but for a number of importantphysical properties of materials. How tightly atoms are bound together in a compoundhas a direct impact on such properties as melting point, elastic modulus and thermalexpansion coefficient. Figure 1.4 shows a qualitative comparison of a material that hasa deep and narrow potential energy wells versus one in which the potential energy wellis wide and shallow. The deeper well represents stronger interatomic attraction; henceit is more difficult to melt these substances, which have correspondingly large elasticmoduli and low thermal expansion coefficients.Cooperative Learning Exercise 1.1Work with a neighbor. Consider the Lennard-Jones potential, as given by Eq. (1.12), forwhich m = 6 and n = 12. You wish to determine the separation distance, r, at which themaximum force occurs, Fmax, in terms of the equilibrium bond distance, r0.Person 1: Use Eq. (1.16) with the values of m and n of the Lennard-Jones potentialto solve for the constant a in terms of b and the equilibrium bond distance, r0. Nowperform the determination of Fmax as given by Eq. (1.15): substitute this value of a backinto Eq. (1.12), differentiate it twice with respect to r (remember that r0 is a constant), andset this equal to zero (determine, then maximize the force function). Solve this equationfor r in terms of r0. The other constant should drop out.Person 2: Use Eq. (1.16) with the values of m and n of the Lennard-Jones potentialto solve for the constant b in terms of a and the equilibrium bond distance, r0. Nowperform the determination of Fmax as given by Eq. (1.15); substitute this value of b backinto Eq. (1.12), differentiate it twice with respect to r (remember that r0 is a constant), andset this equal to zero (determine, then maximize the force function). Solve this equationfor r in terms of r0. The other constant should drop out.Compare your answers. You should both get the same result.Answer: r = 1.1087r0U r(a)U r(b)Figure 1.4 Schematic representation of the relationship between the shape of the potentialenergy well and selected physical properties. Materials with a deep well (a) have a high meltingpoint, high elastic modulus, and low thermal expansion coefficient. Those with a shallow well(b) have a low melting point, low elastic modulus, and high thermal expansion coefficient.Adapted from C. R. Barrett, W. D. Nix, and A. S. Tetelman, The Principles of EngineeringMaterials. Copyright 1973 by Prentice-Hall, Inc. 28. INTRODUCTION AND OBJECTIVES 171.0.4.1 The Ionic Bond. To form an ionic bond, we must account for the com-pletetransfer of electrons from one atom to the other. The easiest approach is to firsttransfer the electrons to form ions, then bring the two ions together to form a bond.Sodium chloride is a simple example that allows us to obtain both the bond energyand equilibrium bond distance using the potential energy approach. For this case, thepotential energy, U, not only is the sum of the attractive and repulsive energies, UAand UR, respectively, but must also take into account the energy required to form ionsfrom sodium and chlorine atoms, Eions . So, our energy expression looks like:U = UA + UR + Eions (1.17)which at the equilibrium bond distance gives the equilibrium potential energy, U0:U0 = UA,0 + UR,0 + Eions (1.18)Let us examine each of the three energies in Eq. (1.18) individually.Energy is required to form an ion of Na+ from elemental sodium. From Sec-tion1.0.2.1, we already know that this process of removing an electron from an isolatedatom is the ionization energy, IE, which for sodium is 498 kJ/mol. Similarly, energyis given off when Cl is formed by adding an electron to an isolated gaseous atom ofchlorine. This is the electron affinity, EA (see Section 1.0.2.2), which for chlorine is354 kJ/mol. So, the energy required to form ions is given by:Eions = IENa + EACl = 498 354 = 144 kJ/mol (1.19)For a diatomic molecule, the attraction between the two ions is strictly due toopposite charges, so the attractive force is given by Coulombs Law:FA = (Z1e Z2e)/(40r2) (1.20)where o is a constant called the electric permittivity (8.854 1012 C2/N m2), e is thecharge of an electron (1.6 1019 C), Z is respective numbers of charge of positiveand negative ions (Z1 = Z2 = 1), and r is the separation distance between ions inmeters. (We will learn more of the electric permittivity in Chapter 6.) Substituting thevalues of Z in for sodium and chloride ions givesFA = +e2/(40r2) (1.21)Energy is released by bringing the ions from infinite separation to their equilibriumseparation distance, r0. Recall that energy and force are related to one another byEq. (1.13), so that the equilibrium attractive energy, UA,0, can be found by integrat-ingFA:UA,0 = r0FA dr = e2/(40r0) (1.22)Note the similarity in form of this expression for the attractive energy with that ofEq. (1.9). The exponent on r is 1, as it should be for ions (m = 1), and the other 29. 18 THE STRUCTURE OF MATERIALSparameters can be grouped to form the constant, a. Recall that by definition the attrac-tiveenergy is a negative energy, so we will end up inserting a minus sign in front ofthe expression in Eq. (1.22) to match the form shown in Eq. (1.9).The repulsive energy can be derived simply by using the general expression given inEq. (1.10) and solving for the constant b by minimizing the potential energy function[Eq. (1.11)] with a knowledge of the constant a from Eq. (1.22) and (1.9) (you shouldtry this) to obtain:UR,0 = e2/(40nr0) (1.23)where e and 0 are the same as for the attractive force, r0 is again the equilibriumseparation distance, and n is the repulsion exponent.Inserting UA,0 and UR,0 into the main energy expression, Eq. (1.18), (recall that theattractive energy must be negative) gives us the equilibrium potential energy, U0:U0 =e240r0+ e240nr0+ Eions (1.24)and simplifying gives:U0 =1 1n e240r0+ Eions (1.25)We solved for the equilibrium bond distance, r0, in Eq. (1.16), and the constants a andb have, in effect, just been evaluated. Inserting these values into Eq. (1.25), along withEq. (1.19) and using n = 8 (why?), gives:U0 =1 181(1.6 1019 C)2(6.02 1023 mol1)4(8.854 1012 C2/N m2)(2.36 1010 m)+ 142= 371 kJ/molSimilarly, we can calculate bond energies for any type of bond we wish to create.Refer to Appendix 1 for bond energy values.When we have an ordered assembly of atoms called a lattice, there is more thanone bond per atom, and we must take into account interactions with adjacent atomsthat result in an increased interionic spacing compared to an isolated atom. We do thiswith the Madelung constant, M. This parameter depends on the structure of the ioniccrystal, the charge on the ions, and the relative size of the ions. The Madelung constantfits directly into the energy expression (Eq. 1.25):UL =M 1n e240r0+ Eions (1.26)For NaCl, M = 1.75 so the lattice energy, UL, is 811 kJ/mol. Typical values ofthe Madelung constant are given in Table 1.7 for different crystal structures (seeSection 1.1.1). In general, the lattice energy increases (becomes more negative) withdecreasing interionic distance for ions with the same charge. This increase in latticeenergy translates directly into an increased melting point. For example, if we replace 30. INTRODUCTION AND OBJECTIVES 19Table 1.7 Typical Values for the Madelung ConstantCompoundCrystal Lattice(see Section 1.1.1) MNaCl FCC 1.74756CsCl CsCl 1.76267CaF2 Cubic 2.51939CdCl2 Hexagonal 2.244MgF2 Tetragonal 2.381ZnS (wurtzite) Hexagonal 1.64132TiO2 (rutile) Tetragonal 2.408-SiO2 Hexagonal 2.2197the chlorine in sodium chloride with other halogens, while retaining the cubic struc-ture,the interionic spacing should change, as well as the melting point. We could alsoaccount for additional van der Waals interactions, but this effect is relatively smallin lattices.1.0.4.2 The Covalent Bond. Recall that covalent bonding results when electronsare shared by similar atoms. The simplest example is that of a hydrogen molecule,H2. We begin by using molecular orbital theory to represent the bonding. Two atomicorbitals (1s) overlap to form two molecular orbitals (MOs), represented by : onebonding orbital (1s), and one antibonding orbital, (1s), where the asterisk super-scriptindicates antibonding. The antibonding orbitals are higher in energy than corre-spondingbonding orbitals.The shapes of the electron cloud densities for various MOs are shown in Figure 1.5.The overlap of two s orbitals results in one -bonding orbital and one -antibondingorbital. When two p orbitals overlap in an end-to-end fashion, as in Figure 1.5b, theyare interacting in a manner similar to s s overlap, so one -bonding orbital and one-antibonding orbital once again are the result. Note that all orbitals are symmetricabout a plane between the two atoms. Side-to-side overlap of p orbitals results in one-bonding orbital and one -antibonding orbital. There are a total of four orbitals:two for px and two for py . Note that there is one more node (region of zero electrondensity) in an antibonding orbital than in the corresponding bonding orbital. This iswhat makes them higher in energy.As in the case of ionic bonding, we use a potential energy diagram to show howorbitals form as atoms approach each other, as shown in Figure 1.6. The electronsfrom the isolated atoms are then placed in the MOs from bottom to top. As long as thenumber of bonding electrons is greater than the number of antibonding electrons, themolecule is stable. For atoms with p and d orbitals, diagrams become more complexbut the principles are the same. In all cases, there are the same number of molecularorbitals as atomic orbitals. Be aware that there is some change in the relative energiesof the and orbitals as we go down the periodic chart, particularly around O2.As a result, you might see diagrams that have the 2p orbitals lower in energy thanthe 2p. Do not let this confuse you if you see some variation in the order of theseorbitals in other texts or references. For our purposes, it will not affect whether themolecule is stable or not. 31. 20 THE STRUCTURE OF MATERIALSAtomic orbitals Molecular orbitals+2s+2s+ + 2px 2px2py 2pyA B AB +++ + sbondings*antibonding+ sA Bbondingpbonding AB ++A Bs*antibondingp*antibonding++A B(a )(b)(c )Figure 1.5 The shape of selected molecular orbitals formed from the overlap of two atomicorbitals. From K. M. Ralls, T. H. Courtney, and J. Wulff, Introduction to Materials Science andEngineering. Copyright 1976 by John WileySons, Inc. This material is used by permissionof John WileySons, Inc.Antibonding molecular orbitals*Atomic orbital Atomic orbital1s 1sAtom A Atom BsBonding molecular orbitalMoleculeEnergyFigure 1.6 Molecular orbital diagram for the hydrogen molecule, H2. Reprinted, by permission,from R. E. Dickerson, H. B. Gray, and G. P. Haight, Jr., Chemical Principles, 3rd ed., p. 446.Copyright 1979 by Pearson Education, Inc. 32. INTRODUCTION AND OBJECTIVES 212p 2p2ss2pO O2 O2sp*2ps*2ps*2ss2sp2pFigure 1.7 Molecular orbital diagram for molecular oxygen, O2. From K.M. Ralls, T. H.Courtney, and J. Wulff, Introduction to Materials Science and Engineering. Copyright 1976by John WileySons, Inc. This material is used by permission of John WileySons, Inc.Let us use molecular oxygen, O2, as an example. As shown in Figure 1.7, eachoxygen atom brings six outer-core electrons to the molecular orbitals. (Note that the1s orbitals are not involved in bonding, and are thus not shown. They could be shownon the diagram, but would be at a very low relative energy at the bottom of thediagram.) The 12 total electrons in the molecule are placed in the MOs from bottom totop; according to Hunds rule, the last two electrons must be placed in separate 2porbitals before they can be paired.The pairing of electrons in the MOs can manifest itself in certain physical prop-ertiesof the molecule. Paramagnetism results when there are unpaired electrons inthe molecular orbitals. Paramagnetic molecules magnetize in magnetic fields due tothe alignment of unpaired electrons. Diamagnetism occurs when there are all pairedelectrons in the MOs. We will revisit these properties in Chapter 6.We can use molecular orbital theory to explain simple heteronuclear diatomicmolecules, as well. A molecule such as hydrogen fluoride, HF, has molecular orbitals,but we must remember that the atomic orbitals of the isolated atoms have much dif-ferentenergies from each other to begin with. How do we know where these energiesare relative to one another? Look back at the ionization energies in Table 1.4, and yousee that the first ionization energy for hydrogen is 1310 kJ/mol, whereas for fluorine itis 1682 kJ/mol. This means that the outer-shell electrons have energies of 1310 (1selectron) and 1682 kJ/mol (2p electron), respectively. So, the electrons in fluorineare more stable (as we would expect for an atom with a much larger nucleus rela-tiveto hydrogen), and we can construct a relative molecular energy diagram for HF(see Figure 1.8) This is a case where the electronegativity of the atoms is useful. Itqualitatively describes the relative energies of the atomic orbitals and the shape of theresulting MOs. The molecular energy level diagram for the general case of moleculeAB where B is more electronegative than A is shown in Figure 1.9, and the corre-spondingmolecular orbitals are shown in Figure 1.10. In Figure 1.9, note how the Batomic orbitals are lower in energy than those of atom A. In Figure 1.10, note how thenumber of nodes increases from bonding to antibonding orbitals, and also note howthe electron probability is greatest around the more electronegative atom. 33. 22 THE STRUCTURE OF MATERIALSExample Problem 1.2Construct the molecular orbital diagram for diatomic nitrogen, N2. Is this molecule param-agneticor diamagnetic?Answer: Diatomic nitrogen has all paired electrons, so it is diamagnetic.*2p2pN2*2p2p*2s2s2p2s 2s2pH orbital1sEnergyHF orbitals F orbitalss*pxpy 2ps2s 2sFigure 1.8 Molecular orbital diagram for HF. Reprinted, by permission, from R. E. Dickerson,H. B. Gray, and G. P. Haight, Jr., Chemical Principles, 3rd ed., p. 461. Copyright 1979 byPearson Education, Inc. 34. INTRODUCTION AND OBJECTIVES 23A orbitalsnpnsAB orbitals B orbitalsnpnss*zszpx pys*sssp*xp*yEnergyFigure 1.9 Molecular orbital diagram for the general case of a diatomic molecule AB, whereB is more electronegative than A. Reprinted, by permission, from R. E. Dickerson, H. B. Gray,and G. P. Haight, Jr., Chemical Principles, 3rd ed., p. 464. Copyright 1979 by PearsonEducation, Inc.Molecular orbitals dont explain everything and become increasingly more difficultto draw with more than two atoms. We use a model called hybridization to explainother effects, particularly in carbon compounds. Hybridization is a mixing of atomicorbitals to create new orbitals that have a geometry better suited to a particular type ofbonding environment. For example, in the formation of the compound BeH2, we wouldlike to be able to explain why this molecule is linear; that is, the HBeH bond is 180.A B zz(a)A B(b)0+Figure 1.10 The shape of selected molecular orbitals for the diatomic molecule AB, where Bis more electronegative than A: (a) , (b) , (c) and (d) . 35. 24 THE STRUCTURE OF MATERIALSx x+A Bz0(c)x xA Bz0+ +(d)0Figure 1.10 (continued).The hydrogen atoms only have one electron each to donate, both in their respective1s orbitals, but beryllium has two electrons in the 2s orbital, and because its principalquantum number is two, it also has 2p orbitals, even though they are empty.The trick is to make two equivalent orbitals in Be out of the atomic orbitals so thateach hydrogen will see essentially the same electronic environment. We can accomplishthis by mixing the 2s orbital and one of the empty 2p orbitals (say, the 2pz) toform two equivalent orbitals we call sp hybrids, since they have both s and pcharacteristics. As with molecular orbital theory, we have to end up with the samenumber of orbitals we started with. The bonding lobes on the new spa and spb orbitalson Be are 180 apart, just as we need to form BeH2. In this manner, we can mixany type of orbitals we wish to come up with specific bond angles and numbers ofequivalent orbitals. The most common combinations are sp, sp2, and sp3 hybrids. Insp hybrids, one s and one p orbital are mixed to get two sp orbitals, both of which 36. INTRODUCTION AND OBJECTIVES 25are 180 apart. A linear molecule results e.g., BeH2, as shown in Figure 1.11. In sp2hybrids, one s and two p orbitals are mixed to obtain three sp2 orbitals. Each orbitalhas 1/3 s and 2/3 p characteristic. A trigonal planar orbital arrangement results, with120 bond angles. An example of a trigonal planar molecule is BF3, as in Figure 1.12.Finally, in sp3 hybrids, when one s and three p orbitals are mixed, four sp3 orbitalsresult, each having 1/4 s and 3/4 p characteristic. The tetrahedral arrangement oforbitals creates a 109.5 bond angle, as is found in methane, CH4 (Figure 1.13).The concept of hybridization not only gives us a simple model for determiningthe correct geometry in simple molecules, but also provides us with a rationalizationfor multiple bonds. A double bond can result from sp2 hybridization: one sp2sp2bond and one bond that forms between the p orbitals not involved in hybridization.An example is in C2H4 (ethylene, Figure 1.14a), where each carbon undergoes sp2hybridization so that it can form an sp21s bond with two hydrogens and an sp2sp2bond with the other carbon. The remaining p orbitals on each carbon (say, pz) shareelectrons, which form the CC double bond. A triple bond can be explained in termsof sp hybridization. It is formed from one spsp bond and two bonds which formH Be HFigure 1.11 The linear structure of BeH2.FBFFFigure 1.12 The trigonal planar structure of BF3.HHHHCFigure 1.13 The tetrahedral structure of CH4. 37. 26 THE STRUCTURE OF MATERIALS sp 2sp 2sp 2 sp 2sp 2 sp 2HH 1s1sC CH H(a)1s1sH 1s sp C sp sp C sp 1s H (b)Figure 1.14 Hybridization resulting in (a) double bond and (b) triple bond.between the two remaining p orbitals after hybridization. Acetylene (Figure 1.14b),C2H2, is such a compound in which both carbons undergo sp hybridization so thatthey can accommodate one bond with each other and one with hydrogen. Bonds canform between the remaining p orbitals, which in this case could be the pz and pyorbitals on each carbon, for a total of three bonds between the carbon atoms.Example Problem 1.3Consider the molecule NF3. How can we explain the observation that the FNF bondangles in NF3 are 107.3 and not 120, as we might predict?Answer: Nitrogen undergoes sp3 hybridization, not sp2, so it is tetrahedral. The additionalsp3 orbital is occupied by a lone pair of electrons from the nitrogen. This lone pair results inelectronelectron repulsion that causes the other sp3 orbitals bonded to fluorines to be closertogether than the normal 109 tetrahedral bond angle, hence the 107.3 FNF bond angle.1.0.4.3 The Metallic Bond. Some elements, namely those in the first two columnsof the periodic table (IA and IIA) and the transition metals (e.g., Co, Mn, Ni), notonly have a propensity for two atoms to share electrons such as in a covalent bond,but also have a tendency for large groups of atoms to come together and share valence 38. INTRODUCTION AND OBJECTIVES 27electrons in what is called a metallic bond. In this case, there are now N atoms in thelattice, where N is a large number. There are N atomic orbitals, so there must be NMOs, many of which are degenerate, or of the same energy. This leads to bands ofelectrons, as illustrated in Figure 1.15 for sodium. The characteristics of the metallicbond are that the valence electrons are not associated with any particular atom in thelattice, so they form what is figuratively referred to as an electron gas around thesolid core of metallic nuclei. As a result, the bonds in metals are nondirectional, unlikecovalent or ionic bonds in which the electrons have a finite probability of being arounda particular atomThe electrons not involved in bonding remain in what is called the core band,whereas the valence electrons that form the electron gas enter into the valence band.10203040500102030600 0.5 1.0 1.503d4s1019 E (J)E (eV)Interatomic spacing (nm)3p2p3sObserved equilibriumvalue, r0 = 0.366 nmFigure 1.15 Energy band diagram for a sodium lattice. From K. M. Ralls, T. H. Courtney, andJ. Wulff, Introduction to Materials Science and Engineering. Copyright 1976 by John Wiley Sons, Inc. This material is used by permission of John WileySons, Inc. After J. C. Slater,Phys. Rev., 45, 794 (1934). 39. 28 THE STRUCTURE OF MATERIALSThe remaining unfilled orbitals form higher-energy bands, called the conduction band.Keep in mind that even though the d and f orbitals may not be filled with electrons,they still exist for many of the heavier elements, so they must be included in themolecular orbital diagram. We will see later in Chapter 6 that the conduction bandplays a very important role in the electrical, thermal, and optical properties of metals.1.1 STRUCTURE OF METALS AND ALLOYSSince the electrons in a metallic lattice are in a gas, we must use the core electronsand nuclei to determine the structure in metals. This will be true of most solids we willdescribe, regardless of the type of bonding, since the electrons occupy such a smallvolume compared to the nucleus. For ease of visualization, we consider the atomic coresto be hard spheres. Because the electrons are delocalized, there is little in the way ofelectronic hindrance to restrict the number of neighbors a metallic atom may have. Asa result, the atoms tend to pack in a close-packed arrangement, or one in which themaximum number of nearest neighbors (atoms directly in contact) is satisfied.Refer to Figure 1.16. The most hard spheres one can place in the plane arounda central sphere is six, regardless of the size of the spheres (remember that all ofthe spheres are the same size). You can then place three spheres in contact with thecentral sphere both above and below the plane containing the central sphere. Thisresults in a total of 12 nearest-neighbor spheres in contact with the central sphere inthe close-packed structure.Closer inspection of Figure 1.16a shows that there are two different ways to placethe three nearest neighbors above the original plane of hard spheres. They can bedirectly aligned with the layer below in an ABA type of structure, or they can berotated so that the top layer does not align core centers with the bottom layer, resultingin an ABC structure. This leads to two different types of close-packed structures.The ABAB. . . structure (Figure 1.16b) is called hexagonal close-packed (HCP) andthe ABCABC. . . structure is called face-centered cubic (FCC). Remember that bothAB(a) (b) (c)ACBA AFigure 1.16 Close-packing of spheres. (a) Top view, (b) side view of ABA structure, (c) sideview of ABC structure. From Z. Jastrzebski, The Nature and Properties of EngineeringMaterials,2nd ed. Copyright 1976 by John WileySons, Inc. This material is used by permission ofJohn WileySons, Inc. 40. STRUCTURE OF METALS AND ALLOYS 29Figure 1.17 The extended unit cell of the hexagonal close-packed (HCP) structure.of these close-packed arrangements have a coordination number (number of nearestneighbors surrounding an atom) of 12: 6 in plane, 3 above, and 3 below.Keep in mind that for close-packed structures, the atoms touch each other in alldirections, and all nearest neighbors are equivalent. Let us first examine the HCPstructure. Figure 1.17 is a section of the HCP lattice, from which you should be ableto see both hexagons formed at the top and bottom of what is called the unit cell.You should also be able to identify the ABA layered structure in the HCP unit cell ofFigure 1.17 through comparison with Figure 1.16. Let us count the number of atomsin the HCP unit cell. The three atoms in the center of the cell are completely enclosed.The atoms on the faces, however, are shared with adjacent cells in the lattice, whichextends to infinity. The center atoms on each face are shared with one other HCP unitcell, either above (for the top face) or below (for the bottom face), so they contributeonly half of an atom each to the HCP unit cell under consideration. This leaves the sixcorner atoms on each face (12 total) unaccounted for. These corner atoms are at theintersection of a total of six HCP unit cells (you should convince yourself of this!), soeach corner atom contributes only one-sixth of an atom to our isolated HCP unit cell.So, the total number of whole atoms in the HCP unit cell is3 1 = 3 center atoms2 (1/2) = 1 face atom12 (1/6) = 2 corner atoms6 total atomsAt this point, you may find it useful to get some styrofoam spheres or hard balls to help you visualizesome of the structures we will describe. We will use a number of perspectives, views, and diagrams to buildthese structures. Some will treat the atoms as solids spheres that can touch each other, some will use dots atthe center of the atoms to help you visualize the larger structure. Not all descriptions will help youfindthose perspectives that work best for you. 41. 30 THE STRUCTURE OF MATERIALS(a) (b)Figure 1.18 The face-centered cubic (FCC) structure showing (a) atoms touching and(b) atoms as small spheres. Reprinted, by permission, from W. Callister, Materials Science andEngineering: An Introduction, 5th ed., p. 32. Copyright 2000 by John WileySons, Inc.Counting the atoms in the FCC structure is performed in a similar manner, exceptthat visualizing the FCC structure takes a little bit of imagination and is virtuallyimpossible to show on a two-dimensional page. Take the ABC close-packed structureshown in Figure 1.16c, and pick three atoms along a diagonal. These three atoms formthe diagonal on the face of the FCC unit cell, which is shown in Figure 1.18. There isa trade-off in doing this: It is now difficult to see the close-packed layers in the FCCstructure, but it is much easier to see the cubic structure (note that all the edges of thefaces have the same length), and it is easier to count the total number of atoms in theFCC cell. In a manner similar to counting atoms in the HCP cell, we see that thereare zero atoms completely enclosed by the FCC unit cell, six face atoms that are eachshared with an adjacent unit cell, and eight corner atoms at the intersection of eightunit cells to give6 (1/2) = 3 face atoms8 (1/8) = 1 corner atom4 total atomsRemember that both HCP and FCC are close-packed structures and that each has acoordination number of 12, but that their respective unit cells contain 6 and 4 totalatoms. We will now see how these two special close-packed structures fit into a largerassembly of crystal systems.1.1.1 Crystal StructuresOur description of atomic packing leads naturally into crystal structures. While someof the simpler structures are used by metals, these structures can be employed byheteronuclear structures, as well. We have already discussed FCC and HCP, but thereare 12 other types of crystal structures, for a total of 14 space lattices or Bravaislattices. These 14 space lattices belong to more general classifications called crystalsystems, of which there are seven. 42. STRUCTURE OF METALS AND ALLOYS 31abcyzxFigure 1.19 Definition of a coordinate system for crystal structures.Before describing each of the space lattices, we need to define a coordinate system.The easiest coordinate system to use depends upon which crystal system we are lookingat. In other words, the coordinate axes are not necessarily orthogonal and are definedby the unit cell. This may seem a bit confusing, but it simplifies the description ofcell parameters for those systems that do not have crystal faces at right angles toone another. Refer to Figure 1.19. For each crystal system, we will define the spacelattice in terms of three axes, x, y, and z, with interaxial angles , , . Note that theinteraxial angle is defined by the angle formed between axes z and y, and also notethat angles and are defined similarly. Only in special circumstances are , , equal to 90. The distance along the y axis from the origin to the edge of the unit cellis called the lattice translation vector, b. Lattice translation vectors a and c are definedsimilarly along the axes x and z, respectively. The magnitudes (lengths) of the latticetranslation vectors are called the lattice parameters, a, b, and c. We will now examineeach of the seven crystal systems in detail.1.1.1.1 Crystal Systems. The cubic crystal system is composed of three spacelattices, or unit cells, one of which we have already studied: simple cubic (SC), body-centeredcubic (BCC), and face-centered cubic (FCC). The conditions for a crystal tobe considered part of the cubic system are that the lattice parameters be the same (sothere is really only one lattice parameter, a) and that the interaxial angles all be 90.The simple cubic structure, sometimes called the rock salt structure because it isthe structure of rock salt (NaCl), is not a close-packed structure (see Figure 1.20). Infact, it contains about 48% void space; and as a result, it is not a very dense structure.The large space in the center of the SC structure is called an interstitial site, whichis a vacant position between atoms that can be occupied by a small impurity atom oralloying element. In this case, the interstitial site is surrounded by eight atoms. Alleight atoms in SC are equivalent and are located at the intersection of eight adjacentunit cells, so that there are 8 (1/8) = 1 total atoms in the SC unit cell. Notice that 43. 32 THE STRUCTURE OF MATERIALSCrystalStructureLatticeParametersInteraxialAnglesSimple Body-centered Face-centeredSimple Base-centered Body-centered Face-centeredSimple Body-centeredSimple Base-centeredCubic a = b = g =90a = b = cOrthorhombic a = b = g =90a b cRhombohedral a = b = c a = b = g 90,120 Tetragonal a = b c a = b = g = 90Monoclinic a b c a = g = 90, b 90Triclinic a b c a = b = g 90Hexagonal a = b, a c a = b = 90, g = 120Figure 1.20 Summary of the 14 Bravais space lattices. 44. STRUCTURE OF METALS AND ALLOYS 33Figure 1.20 shows the atoms as points, but the atoms actually occupy a larger spacethan that. In fact, for SC, the atoms touch along the edge of the crystal.Body-centered cubic (BCC) is the unit cell of many metals and, like SC, is nota close-packed structure. The number of atoms in the BCC unit cell are calculatedas follows:1 1 = 1 center atom8 (1/8) = 1 corner atom2 total atomsFinally, face-centered cubic (FCC) has already been described (Figure 1.18). Eventhough FCC is a close-packed structure, there are interstitial sites, just as in SC. Thereare actually two different types of interstitial sites in FCC, depending on how manyatoms surround the interstitial site. A group of four atoms forms a tetrahedral interstice,as shown in Figure 1.21. A group of six atoms arranged in an octahedron (an eight-sidedgeometric figure), creates an octahedral interstice (Figure 1.22). Figure 1.23 shows thelocations of these interstitial sites within the FCC lattice. Note that there are eight totaltetrahedral interstitial sites in FCC and there are four total octahedral interstitial sitesin FCC (prove it!), which are counted in much the same way as we previously countedthe total number of atoms in a unit cell. We will see later on that these interstitialsites play an important role in determining solubility of impurities and phase stabilityof alloys.Interstitial sites are the result of packing of the spheres. Recall from Figure 1.18 thatthe spheres touch along the face diagonal in FCC. Similarly, the spheres touch along thebody diagonal in BCC and along an edge in SC. We should, then, be able to calculatethe lattice parameter, a, or the length of a face edge, from a knowledge of the sphereradius. In SC, it should be evident that the side of a unit cell is simply 2r. Application ofa little geometry should prove to you that in FCC, a = 4r/2. The relationship betweena and r for BCC is derived in Example Problem 1.4; other geometric relationships,including cell volume for cubic structures, are listed in Table 1.8. Finally, atomic radiifor the elements can be found in Table 1.9. The radius of an atom is not an exactlydefined quantity, and it can vary depending upon the bonding environment in whichFigure 1.21 A tetrahedral interstice. From W. D. Kingery, H. K. Bowen, and D. R. Uhlmann,Introduction to Ceramics. Copyright 1976 by John WileySons, Inc. This material is usedby permission of John WileySons, Inc. 45. 34 THE STRUCTURE OF MATERIALSFigure 1.22 An octahedral interstice. From W. D. Kingery, H. K. Bowen, and D. R. Uhlmann,Introduction to Ceramics. Copyright 1976 by John WileySons, Inc. This material is usedby permission of John WileySons, Inc.Figure 1.23 Location of interstitial sites in FCC. From W. D. Kingery, H. K. Bowen, andD. R. Uhlmann, Introduction to Ceramics. Copyright 1976 by John WileySons, Inc. Thismaterial is used by permission of John WileySons, Inc. 46. STRUCTURE OF METALS AND ALLOYS 35Example Problem 1.4Molybdenum has a BCC structure with an atomic radius of 1.36 A . Calculate the latticeparameter for BCC Mo.HGaaabCEABEAGaC2r r ra 2Reprinted, by permission, from Z. Jastrzebski, The Nature andProperties of Engineering Materials, p. 47, 2nd ed. Copyright 1976,John WileySons, Inc.Answer: We knowthat the molybdenumatoms touch along thebody diagonal in BCC,as shown in the projec-tionat right. The lengthof the body diagonal,then, is 4r, and isrelated to the latticeparameter, a (which isthe length of the cubeedge, not the lengthof the face diagonal,which is a2) byapplication of the Pythagorean theorem:(4r)2 = a2 + (a2)2 = 3a23 = 4(1.36)/a = 4r/3 = 3.14The lattice parameter for BCC Mo is 3.14 A , which is consistent with the value in Table 1.11.Table 1.8 Summary of Important Parameters in the Cubic Space LatticesSimple Cubic Face-Centered Cubic Body-Centered Cubic2 4r/Unit cell side, a 2r 4r/3Face diagonal2(2r) 4r2/3(4r)Body diagonal3(2r)3/2(4r) 4rNumber of atoms 1 4 2Cell volume 8r3 32r3264r333r = atomic radius.it finds itself. As a result, three types of radii are listed for each element in Table 1.9:an atomic radius of an isolated atom, an ionic radius, and a metallic radius. Just asin Figure 1.2 for electronic structure, there are some important trends in the atomicradii. The atomic radius tends to increase as one goes down the column in a series.This is due to the addition of energy levels and more electron density. Radii tend todecrease as we move across a row, because there is less shielding from inner electrons 47. 36 THE STRUCTURE OF MATERIALSTable 1.9 Atomic, Ionic, and Metallic Radii of the Elements1Pb4 2.15 Mn2+ 0.91 Mn3+ 0.70 Fe2+ 0.87Pb2+ 1.32 Pt4+ 0.55 Co2+ 0.82 W6+ 0.65S6+ 0.34 Rh4+ 0.65 Mo4+ 0.68 Cr3+ 0.64Te4+ 0.89 V3+ 0.65 V4+ 0.61 Sn4 2.15Si4 1.98 Nb4+ 0.74 Tl+ 1.49Ti3+ 0.69 Ti2+ 0.76 Se6+ 0.300.042H Many elements have multiple valence states. Additional ionic radii are listedbelow.HeAtomic 0.46 2+ 3+ 4+ 5+ 2 1 Ionic 1.54 ions ions ions ions ions ions Metallic 3 4 5 6 7 8 9 10Li Be B C N O F NeAtomic 1.52 1.14 0.97 0.77 0.71 0.6 1.60Ionic 0.78 0.54 0.2 0.2 0.1 1.32 1.33 Metallic 1.23 0.89 0.81 11 12 13 14 15 16 17 18Na Mg Al Si P S Cl ArAtomic 1.86 1.60 3+ 4+ 5+ 6+ 4+ 4+ 3+ 2+ 1+ 2+ 1.43 1.17 1.09 1.06 1.07 1.92Ionic 0.98 0.78 ions ions ions ions ions ions ions ions ions ions 0.57 0.39 0.3 1.74 1.81 Metallic 1.57 1.36 1.25 1.17 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br KrAtomic 2.31 1.97 1.60 1.47 1.32 1.25 1.12 1.24 1.25 1.25 1.28 1.33 1.35 1.22 1.25 1.16 1.19 1.97Ionic 1.33 1.06 0.82+ 0.64 0.45+ 0.3 0.52 0.673+ 0.65 0.78 0.96 0.83 0.62 0.44 0.69 1.91 1.96 Metallic 2.03 1.74 1.44 1.32 1.22 1.18 1.17 1.17 1.16 1.15 1.17 1.25 1.25 1.22 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I XeAtomic 2.51 2.15 1.81 1.58 1.43 1.36 1.34 1.34 1.37 1.44 1.50 1.57 1.58 1.61 1.43 1.36 2.18Ionic 1.49 1.27 1.06 0.89 0.74 0.65 0.65 0.68 0.50 1.13 1.03 0.91 0.74 0.93+ 0.89 2.20 Metallic 2.16 1.91 1.62 1.45 1.34 1.30 1.27 1.25 1.25 1.28 1.34 1.41 1.50 1.40 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86Cs Ba La Hf Ta W Re Os Ir Pt Au Hg TI Pb Bi Po At RnAtomic 2.65 2.17 1.87 1.59 1.47 1.37 1.38 1.35 1.35 1.38 1.44 1.50 1.71 1.75 1.82 1.40 Ionic 1.65 1.3 1.22 0.84 0.68 0.68 0.72 0.67 0.664+ 0.52 1.37 1.12 1.49 0.84 1.23+ 0.76+ 0.67+ Metallic 2.35 1.98 1.69 1.44 1.34 1.30 1.28 1.26 1.27 1.30 1.34 1.44 1.55 1.54 Source: Materials Science and Engineering Handbook; Pauling, Nature of the Chemical Bond. All values in angstroms, A(1A= 0.1 nm). 48. STRUCTURE OF METALS AND ALLOYS 37and the outer-core electrons are drawn more tightly toward the nucleus. There aresome notable exceptions, or jumps, in the row trend (why?). In general, the ionicradius is much smaller for positive ions and much larger for negative ions than thecorresponding isolated atom (why?), and it follows the same general trend as for theisolated atoms. For the discussion of elemental, crystalline solids, the metallic radiusis most appropriate. We will find later that the ionic values will be equally importantfor heteronuclear structures. There are other types of radii, such as covalent radii andvan der Waals radii. The former is highly dependent upon the type of covalent bond.For example, a carbon atom in a carboncarbon single bond has a covalent radius of1.54 A , whereas the same atom in a carboncarbon triple bond is only 1.35 A .Continuing with our survey of the seven crystal systems, we see that the tetragonalcrystal system is similar to the cubic system in that all the interaxial angles are 90.However, the cell height, characterized by the lattice parameter, c, is not equal tothe base, which is square (a = b). There are two types of tetragonal space lattices:simple tetragonal, with atoms only at the corners of the unit cell, and body-centeredtetragonal, with an additional atom at the center of the unit cell.Orthorhombic crystals are similar to both tetragonal and cubic crystals becausetheir coordinate axes are still orthogonal, but now all the lattice parameters are unequal.There are four types of orthorhombic space lattices: simple orthorhombic, face-centeredorthorhombic, body-centered orthorhombic, and a type we have not yet encountered,base-centered orthorhombic. The first three types are similar to those we have seenfor the cubic and tetragonal systems. The base-centered orthorhombic space lattice hasa lattice point (atom) at each corner, as well as a lattice point only on the top andbottom faces (called basal faces). All four orthorhombic space lattices are shown inFigure 1.20.There is only one space lattice in the rhombohedral crystal system. This crystalis sometimes called hexagonal R or trigonal R, so dont confuse it with the othertwo similarly-named crystal systems. The rhombohedral crystal has uniform latticeparameters in all directions and has equivalent interaxial angles, but the angles arenonorthogonal and are less than 120.The crystal descriptions become increasingly more complex as we move to the mon-oclinicsystem. Here all lattice parameters are different, and only two of the interaxialangles are orthogonal. The third angle is not 90. There are two types of monoclinicspace lattices: simple monoclinic and base-centered monoclinic. The triclinic crystal,of which there is only one type, has three different lattice parameters, and none of itsinteraxial angles are orthogonal, though they are all equal.Finally, we revisit the hexagonal system in order to provide some additional details.The lattice parameter and interaxial angle conditions shown in Figure 1.20 for thehexagonal cell refer to what is called the primitive cell for the hexagonal crystal,which can be seen in the front quadrant of the extended cell in Figure 1.17. Theprimitive hexagonal cell has lattice points only at its corners and has one atom in thecenter of the primitive cell, for a basis of two atoms. A basis is a unit assembly ofatoms identical in composition, arrangement, and orientation that is placed in a regularmanner on the lattice to form a space lattice. You should be able to recognize that thereare three equivalent primitive cells in the extended HCP structure. The HCP extendedcell, which is more often used to represent the hexagonal structure, contains a total ofsix atoms, as we calculated earlier. In the extended structure, the ratio of the height of 49. 38 THE STRUCTURE OF MATERIALSTable 1.10 Axial Ratios for Some HCP MetalsMetal c/aBe, Y 1.57Hf, Os, Ru, Ti 1.58Sc, Zr 1.59Tc, Tl 1.60La 1.61Co, Re 1.62Mg 1.63Zn 1.85Cd 1.89Ideal (sphere packing) 1.633the cell to its base, c/a, is called the axial ratio. Table 1.10 lists typical values of theaxial ratio for some common HCP crystals.A table of crystal structures for the elements can be found in Table 1.11 (excludingthe Lanthanide and Actinide series). Some elements can have multiple crystal struc-tures,depending on temperature and pressure. This phenomenon is called allotropyand is very common in elemental metals (see Table 1.12). It is not unusual for close-packedcrystals to transform from one stacking sequence to the other, simply through ashift in one of the layers of atoms. Other common allotropes include carbon (graphiteat ambient conditions, diamond at high pressures and temperature), pure iron (BCC atroom temperature, FCC at 912C and back to BCC at 1394C), and titanium (HCP toBCC at 882C).1.1.1.2 Crystal Locations, Planes, and Directions. In order to calculate suchimportant quantities as cell volumes and densities, we need to be able to specify loca-tionsand directions within the crystal. Cell coordinates specify a position in the latticeand are indicated by the variables u, v, w, separated by commas with no brackets:u distance along the lattice translation vector av distance along the lattice translation vector bw distance along the lattice translation vector cHISTORICAL HIGHLIGHTOn warming, gray (or ) tin, with a cubicstructure changes at 13.2C into white (or) tin, the ordinary form of the metal, whichhas a tetragonal structure. When tin is cooledbelow 13.2C, it changes slowly from whiteto gray. This change is affected by impuri-tiessuch as aluminum and zinc and can beprevented by small additions of antimony orbismuth. The conversion was first noted asgrowths on organ pipes in European cathe-drals,where it was thought to be the devilswork. This conversion was also speculated tobe caused by microorganisms and was calledtin plague or tin disease.Source: www.webelements.com/webelements/elements/text/key/Sn.html 50. Table 1.11 Common Crystal Structures, Densities, and Lattice Parameters of the Elements1 2H HeStruct. hcp fcc, g/cc n/a n/aa, A4.70 4.243 4 5 6 7 8 9 10Li Be B C N O F NeStruct. bcc hcp rhom diam hcp mon mon fcc, g/cc 0.533 1.85 2.47 3.51 n/a n/a n/a n/aa, A3.51 2.29 5.06 3.56 3.86 5.40 5.50 4.4211 12 13 14 15 16 17 18Na Mg Al Si P S Cl ArStruct. bcc hcp fcc diam tricl orth orth fcc, g/cc 0.966 1.74 2.70 2.33 1.82 2.09 n/a n/aa, A4.29 3.21 4.05 5.43 11.5 10.4 6.22 5.2619 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br KrStruct. bcc fcc hcp hcp bcc bcc cubic bcc hcp fcc fcc hcp orth diam rhom mon orth fcc, g/cc 0.862 1.53 2.99 4.51 6.09 7.19 7.47 7.87 8.8 8.91 8.93 7.13 5.91 5.32 5.78 4.81 n/a n/aa, A5.33 5.56 3.31 2.95 3.03 2.88 8.91 2.86 2.51 3.52 3.61 2.66 4.52 5.65 3.76 9.05 6.72 5.7137 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I XeStruct. bcc fcc hcp hcp bcc bcc hcp hcp fcc fcc fcc hcp tetrag tetrag trig trig orth fcc, g/cc 1.53 2.58 4.48 6.51 8.58 10.22 11.5 12.36 12.42 12.0 10.5 8.65 7.29 7.29 6.69 6.25 4.95 n/aa, A5.59 6.08 3.65 3.61 3.30 3.14 2.74 2.71 3.80 3.88 4.08 2.98 3.25 5.83 4.31 4.46 7.18 6.2055 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At RnStruct. bcc bcc hcp hcp bcc bcc hcp hcp fcc fcc fcc rhom hcp fcc rhom cub n/a n/a, g/cc 1.91 3.59 6.17 13.3 16.7 19.3 21.0 22.58 22.55 21.4 19.28 n/a 11.87 11.34 9.80 9.2 n/a n/aa, A6.14 5.01 3.77 3.20 3.30 3.17 2.76 2.73 3.84 3.92 4.07 3.01 3.46 4.95 4.74 3.36 n/a n/aNote that many elements have allotrones. Cited structures are generally the most stable.Source: Materials Science and Engineering Handbook; Kittel, Solid State Physics; http://www.webelements.com/39 51. 40 THE STRUCTURE OF MATERIALSTable 1.12 Some Metal AllotropesMetalR.T. CrystalStructureStructure atOther TemperaturesCa FCC BCC (447C)Co HCP FCC (427C)Hf HCP BCC (1742C)Fe BCC FCC (912C)BCC (1394C)Li BCC BCC ( 193C)Na BCC BCC ( 233C)Sn BCT Cubic (13C)Tl HCP BCC (234C)Ti HCP BCC (883C)Y HCP BCC (1481C)Zr HCP BCC (872C)For example, the center atom in the BCC space lattice (see Figure 1.20) has