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 specically disclaim any implied warranties ofmerchantability or tness 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 prot 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 engineers Brian S. Mitchell p. cm. Includes bibliographical references and index. ISBN 0-471-43623-2 (cloth) 1. Materials science. I. Title. TA403.M685 2003 620.1 1dc21 2003053451Printed in the United States of America.10 9 8 7 6 5 4 3 2 1

4. To my parents; whose Material was loam; Engineering was labor; Science was lore;And greatest product was love.

5. CONTENTSPreface xiAcknowledgments xv1 The Structure of Materials 1 1.0 Introduction and Objectives 1 1.1 Structure of Metals and Alloys 28 1.2 Structure of Ceramics and Glasses 55 1.3 Structure of Polymers 76 1.4 Structure of Composites 99 1.5 Structure of Biologics 114 References 128 Problems 1302 Thermodynamics of Condensed Phases 136 2.0 Introduction and Objectives 136 2.1 Thermodynamics of Metals and Alloys 140 2.2 Thermodynamics of Ceramics and Glasses 165 2.3 Thermodynamics of Polymers 191 2.4 Thermodynamics of Composites 200 2.5 Thermodynamics of Biologics 204 References 209 Problems 2113 Kinetic Processes in Materials 215 3.0 Introduction and Objectives 215 3.1 Kinetic Processes in Metals and Alloys 219 3.2 Kinetic Processes in Ceramics and Glasses 233 3.3 Kinetic Processes in Polymers 246 3.4 Kinetic Processes in Composites 269 3.5 Kinetic Processes in Biologics 277 References 280 Problems 2824 Transport Properties of Materials 285 4.0 Introduction and Objectives 285 4.1 Momentum Transport Properties of Materials 287 vii

6. viii CONTENTS 4.2 Heat Transport Properties of Materials 316 4.3 Mass Transport Properties of Materials 343 References 374 Problems 3765 Mechanics of Materials 380 5.0 Introduction and Objectives 380 5.1 Mechanics of Metals and Alloys 381 5.2 Mechanics of Ceramics and Glasses 422 5.3 Mechanics of Polymers 448 5.4 Mechanics of Composites 472 5.5 Mechanics of Biologics 515 References 532 Problems 5336 Electrical, Magnetic, and Optical Properties of Materials 537 6.1 Electrical Properties of Materials 538 6.2 Magnetic Properties of Materials 600 6.3 Optical Properties of Materials 644 References 677 Problems 6787 Processing of Materials 681 7.0 Introduction 681 7.1 Processing of Metals and Alloys 681 7.2 Processing of Ceramics and Glasses 704 7.3 Processing of Polymers 754 7.4 Processing of Composites 795 7.5 Processing of Biologics 804 References 811 Problems 8128 Case Studies in Materials Selection 814 8.0 Introduction and Objectives 814 8.1 Selection of Metals for a Compressed Air Tank 821 8.2 Selection of Ceramic Piping for Coal Slurries in a Coal Liquefaction Plant 827 8.3 Selection of Polymers for Packaging 832 8.4 Selection of a Composite for an Automotive Drive Shaft 835 8.5 Selection of Materials as Tooth Coatings 842 References 848 Problems 849

7. CONTENTS ix Appendix 1: Energy Values for Single Bonds 851 Appendix 2: Structure of Some Common Polymers 852 Appendix 3: Composition of Common Alloys 856 Appendix 4: Surface and Interfacial Energies 869 Appendix 5: Thermal Conductivities of Selected Materials 874 Appendix 6: Diffusivities in Selected Systems 880 Appendix 7: Mechanical Properties of Selected Materials 882 Appendix 8: Electrical Conductivity of Selected Materials 893 Appendix 9: Refractive Index of Selected Materials 900Answers to Selected Problems 903Index 907 Sections 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 benet to those students in the biologically relatedengineering disciplines. The key difference is simply the organization of the material,which is intended to benet primarily the chemical and materials engineer. This textbook is organized on two levels: by engineering subject area and by mate-rials class, 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 rst two columns cov-ered in the rst semester and the nal 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-tional course 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-inate topics they deem inappropriate, and students will nd additional topic coverage intheir supplemental reading, which is encouraged through a list of references at the end xi

9. xii PREFACE Metals & Ceramics & Alloys Glasses Polymers Composites BiologicsStructure Crystal Crystal Conguration, Matrices, Biochemistry, structures, structures, Conformation, Reinforce- Tissue Point Defect Molecular ments structure defects, reactions, Weight Dislocations The glassy stateThermo- Phase Ternary Phase separation, Adhesion, Cell dynamics equilibria, systems, Polymer Cohesion, Adhesion, Gibbs Rule Surface solutions, Spreading Cell Lever Rule energy, Polymer spreading Sintering blendsKinetics Trans- Devitrication, Polymerization, Deposition, Receptors, formations, Nucleation, Degradation Inltration Ligand Corrosion Growth bindingTransport Inviscid Newtonian non-Newtonian Porous Flow, Convection, Properties systems, ow, Heat ow, Heat Heat Diffusion Heat capacity, capacity, capacity, capacity, Diffusion Diffusion Diffusion DiffusionMechanical Stress-strain, Fatigue, Viscoelasticity, Laminates Sutures, Properties Elasticity, Fracture, Elastomers Bone, Ductility Creep TeethElectrical, Resistivity, Dielectrics, Ion conductors, Dielectrics, Biosensors, Magnetic & Magnetism, Ferrites, Molecular Storage MRI Optical Reectance Absorbance magnets, LCDs media PropertiesProcessing Casting, Pressing, Extrusion, Pultrusion, Surface Rolling, CVD/CVI, Injection RTM, modication Compaction Sol-Gel molding, Blow CVD/CVI moldingCase Studies Compressed Ceramic Polymeric Composite Tooth air tank piping packaging drive shaft coatingsof 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 specic difcultiesin 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-erative Learning 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-larly difcult 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 signicant 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-standing of 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 difcult 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. MITCHELL Tulane 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. MITCHELL Tulane University xv

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-ical properties 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-odic structure, types of bonding, and potential energy diagrams will be reviewed. Wewill then use this information to look at the specic materials categories in more detail:metals, ceramics, polymers, composites, and biological materials (biologics). There willbe topics that are specic 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 signicantly 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/ionic radii. Identify the type of bonding in a compound. Utilize the concepts of molecular orbital and hybridization theories to explain multiple 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 a plane 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 Wiley & Sons, 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 rst three moments of a polymer molecular weight distribution. Apply principles of glass transition and polymer crystallinity to polymer classi- 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 component properties. Be conversant in terms that relate to the structure of biological materials, such as bronectin 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-minum for 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 laments. 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-cally arranged 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 1000 C, 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 2000 C, 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 to Note 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 3 Table 1.1 Relative Abundance of Elements in the Universe Relative Element Abundance (Si = 1) Hydrogen (H) 12,000 Helium (He) 2,800 Oxygen (O) 16 Nitrogen (N) 8 Carbon (C) 3 Iron (Fe) 2.6 Silicon (Si) 1 Magnesium (Mg) 0.89 Sulfur (S) 0.33 Nickel (Ni) 0.21 Aluminum (Al) 0.09 Calcium (Ca) 0.07 Sodium (Na) 0.045 Chlorine (Cl) 0.025Table 1.2 Relative Abundance of Selected Elements in the Earths Crust Relative 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 1 2 Temperature: 290 K H He 16 C 3 4 62 F 5 6 7 8 9 10 Li Be B C N O F Ne 11 12 13 14 15 16 17 18 Na Mg Al Si P S CI Ar 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd ln Sn Sb Te I Xe 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg TI Pb Bi Po At Rn 87 88 89 Fr Ra Ac Legend Solid Liquid 58 59 60 61 62 63 64 65 66 67 68 69 70 71 Gas Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Not Available 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr (a) 1 2 Temperature: 1280 K H He 1006 C 3 4 1844 F 5 6 7 8 9 10 Li Be B C N O F Ne 11 12 13 14 15 16 17 18 Na Mg Al Si P S CI Ar 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd ln Sn Sb Te I Xe 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg TI Pb Bi Po At Rn 87 88 89 Fr Ra Ac Legend Solid Liquid 58 59 60 61 62 63 64 65 66 67 68 69 70 71 Gas Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Not Available 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr (b)Figure 1.1 The periodic table of the elements at (a) room temperature, (b) 1000 C, (c) 2000 C,and (d) 3500 C.

16. INTRODUCTION AND OBJECTIVES 5 1 2 Temperature: 2280 K H He 2006 C 3 4 3644 F 5 6 7 8 9 10 Li Be B C N O F Ne 11 12 13 14 15 16 17 18 Na Mg Al Si P S CI Ar 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd ln Sn Sb Te I Xe 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg TI Pb Bi Po At Rn 87 88 89 Fr Ra Ac Legend Solid Liquid 58 59 60 61 62 63 64 65 66 67 68 69 70 71 Gas Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Not Available 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr (c) 1 2 Temperature: 3780 K H He 3506 C 3 4 6344 F 5 6 7 8 9 10 Li Be B C N O F Ne 11 12 13 14 15 16 17 18 Na Mg Al Si P S CI Ar 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd ln Sn Sb Te I Xe 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg TI Pb Bi Po At Rn 87 88 89 Fr Ra Ac Legend Solid Liquid 58 59 60 61 62 63 64 65 66 67 68 69 70 71 Gas Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Not Available 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr (d) Figure 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 rst ionization energy, abbreviated IE, issometimes referred to as the ionization potential. It is the energy required to remove

17. 6Table 1.3 Electronic Structure and Melting Points of the Elements 1 es = electronic structure 2 H aw = atomic weight (average including isotopes) He es 1s 1 mp = melting point, C (sublimation temperatures enclosed in parentheses). 1s 2 aw 1.008 4.003 mp 259 3 4 5 6 7 8 9 10 Li Be B C N O F Ne es He2s 1 He2s 2 Be2p 1 Be2p 2 Be2p 3 Be2p 4 Be2p 5 Be2p 6 aw 6.94 9.012 10.81 12.01 14.006 15.999 18.998 20.18 mp 180.5 1289 2103 (3836) 210.0 218.8 219.7 249 THE STRUCTURE OF MATERIALS 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar es Ne3s 1 Ne3s 2 Mg3p 1 Mg3p 2 Mg3p 3 Mg3p 4 Mg3p 5 Mg3p 6 aw 22.99 24.30 26.98 28.09 30.974 32.06 35.45 39.95 mp 97.8 649 660.4 1414 44.1 112.8 101.0 189 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr es Ar4s 1 Ar4s 2 Ca3d 1 Ca3d 2 Ca3d 3 K3d 5 Ca3d 5 Ca3d 6 Ca3d 7 Ca3d 8 K3d 10 Ca3d 10 Ca4p 1 Ca4p 2 Ca4p 3 Ca4p 4 Ca4p 5 Ca4p 6 aw 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 83.80 mp 63.2 840 1541 1672 1929 1863 1246 1538 1494 1455 1084.5 419.6 29.8 938.3 (603) 221 7.2 157 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe es Kr5s 1 Kr5s 2 Sr4d 1 Sr4d 2 Rb4d 4 Rb4d 5 Rb4d 6 Rb4d 7 Rb4d 8 Kr4d 10 Rb4d 10 Sr4d 10 Sr5p 1 Sr5p 2 Sr5p 3 Sr5p 4 Sr5p 5 Sr5p 6 aw 85.47 87.62 88.91 91.22 92.91 95.94 98.91 101.7 102.9 106.4 107.87 112.4 114.8 118.7 121.8 127.6 126.9 131.3 mp 39.5 769 1528 1865 2471 2623 2204 2254 1963 1554 961.9 321.1 156.6 232.0 630.7 449.6 113.6 112 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn es Xe6s 1 Xe6s 2 Ba5d 1 Ba5d 2 Ba5d 3 Ba5d 4 Ba5d 5 Ba5d 6 Xe5d 9 Cs5d 9 Cs5d 10 Ba5d 10 Ba6p 1 Ba6p 2 Ba6p 3 Ba6p 4 Ba6p 5 Ba6p 6 aw 132.9 137.3 138.9 178.5 180.9 183.9 186.2 190.2 192.2 195.1 196.97 200.6 204.4 207.2 208.9 210 210 222 mp 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 atom atom (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 Schr dingers equation (in eV): o 13.6Z 2 IE = (1.2) n2where 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) electronafnity, (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 Afnity (EA). Electron afnity is the reverse process to the ioniza-tion energy; 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 innity.1.0.2.4 Electronegativity. The ionization energy and electron afnity are charac-teristics of 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 afnity: IE + EA = (1.4) 2There 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) = + 0.5 (1.5) rwhere 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 Afnities, and Electronegativities of the Elementsa 1 2 H He IE 1310 2372 EA 67.4 60.2 2.20 3 4 5 6 7 8 9 10 Li Be B C N O F Ne IE 519 900 799 1088 1406 1314 1682 2080 EA 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 18 Na Mg Al Si P S Cl Ar IE 498 736 577 787 1063 1000 1255 2372 EA 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 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr IE 1310 590 632 661 653 653 715 761 757 736 745 904 577 782 966 941 1142 1351 EA 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 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe IE 402 548 636 669 653 695 699 724 745 803 732 866 556 707 833 870 1008 1172 EA 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 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn IE 377 502 540 531 577 770 761 841 887 866 891 1008 590 715 774 1038 EA 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 afnities (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, DE ij : AB = DE AB [(DE AA )(DE BB )]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 of Bond Energy, Melting Point,Bond Substance kJ/mol ( C) CharacteristicsIonic CaCl 651 646 Low electrical conductivity, NaCl 768 801 transparent, brittle, high LiF 1008 870 melting point CuF2 2591 1360 Al2 O3 15,192 3500Covalent Ge 315 958 Low electrical conductivity, GaAs 315 1238 very hard, very high Si 353 1420 melting point SiC 1188 2600 Diamond 714 3550Metallic Na 109 97.5 High electrical and thermal Al 311 660 conductivity, easily Cu 340 1083 deformable, opaque Fe 407 1535 W 844 3370van der Waals Ne 2.5 248.7 Weak binding, low melting Ar 7.6 189.4 and boiling points, very CH4 10 184 compressible Kr 12 157 Cl2 31 103Hydrogen bonding HF 29 92 Higher melting point than van H2 O 50 0 der Waals bonding, tendency to form groups of many molecules

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 satised with this arrangement becausethey oftentimes end up with noble gas electron congurationsthat 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-ativity differences 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 uoride, HF. Though there issignicant 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 1.1 What is the percent ionic character of HF? Answer: According to Table 1.3, the electronegativity of hydrogen is 2.20 and that of uorine 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,rst 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 rst 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 of HISTORICAL HIGHLIGHT Dutch physicist Johannes Diderik van condensation and the critical temperature of der Waals was born on November 23, gases could be predicted. In 1877 J. D. 1837 in Leiden, the Netherlands. He was van der Waals became the rst professor the eldest son of eight children. Initially, of physics at the University Illustre in van der Waals was an elementary school Amsterdam. In 1880 he formulated his teacher during the years 18561861. He law of corresponding states, in 1893 he continued studying to become headmaster devised a theory for capillary phenomena, and attended lectures on mathematics, and in 1891 he introduced his theory for the physics, and astronomy at Leiden University. behavior of two mixtures of two materials. From 1866 onwards he taught physics It was not possible to experimentally show and mathematics at a secondary school in the de-mixing of two gases into two The Hague. After a revision of the law, separate gases under certain circumstances knowledge of Latin and Greek was no longer as predicted by this theory until 1941. a prerequisite for an academic graduation, From 1875 to 1895 J.D. van der Waals and in 1873 J. D. van der Waals graduated was a member of the Dutch Royal Academy on the thesis: Over de continu iteit van of Science. In 1908, at the age of 71, J. D. de gasenvloeistoftoestand (About the van der Waals resigned as a professor. Dur- continuity of gaseous and liquid states). ing his life J. D. van der Waals was honored In this thesis he published the well-known many times. He was one of only 12 foreign law: members of the Academie des Sciences in Paris. In 1910 he received the Nobel prize for a Physics for the incredible work he had done P+ (V b) = RT V2 on the equations of state for gases and u- idsonly the fth Dutch physicist to receive This revision to the ideal gas law accounted this honor. J. D. van der Waals died on March for the specic volume of gas molecules and 8, 1923 at the age of 85. assumed a force between these molecules Source: www.vdwaals.nl which are now known as van der Waals forces. With this law, the existence of

24. INTRODUCTION AND OBJECTIVES 13bonding is of the same order of magnitude in strength as van der Waals bonding, it canhave a profound inuence 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 sh 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 specic 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 rst 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 afnity, 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: a UA = (1.9) rmwhere 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 MATERIALS Table 1.6 Values of the Repulsion Exponent Noble Gas Outer Core Ion Core Conguration n He 1s 2 5 Ne 2s 2 2p 6 7 Ar 3s 2 3p 6 9 Kr 3d 10 4s 2 4p 6 10 Xe 4d 10 5s 2 5p 6 12atoms are, the more they repel each other: b UR = (1.10) rnwhere b and n are constants. The value of n, called the repulsion exponent, dependson the outer core conguration 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: a b U = UA + UR = m + n (1.11) r rThe result is the potential energy well (see Figure 1.3). The well-known Lennard-Jonespotential a b U = 6 + 12 (1.12) r ris a common potential energy function used in a number of models, including col-lision theory for kinetics. It is simply a special case of Eq. (1.11) with n = 12 (Xeconguration) 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 by dU F = (1.13) dr We will see later on that we can use this expression to convert between forceand energy for specic types of atoms and molecules (specic values of n and m).For now, this expression helps us nd 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): dU F = = 0, at r = r0 (1.14) drThis is not the same thing as the maximum attractive force, which we get by maximiz-ing F : dF d 2U Fmax = = 2 =0 (1.15) dr dr

26. INTRODUCTION AND OBJECTIVES 15 + Repulsion Repulsion energy, b rn Potential energy, U s Interatomic or intermolecular distance in 0 r =s rx r a b The net potential energy, U = + rm rn Attraction Attraction energy, a rm r0 Um (a) + Repulsion Repulsion force Force, F r0 Interatomic or intermolecular distance in 0 rx r Attraction Attraction force Fmax (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 JohnWiley & Sons, Inc. This material is used by permission of John Wiley & Sons, 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 gives 1 nb n m r0 = (1.16) ma

27. 16 THE STRUCTURE OF MATERIALS The 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 coefcient. 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 difcult to melt these substances, which have correspondingly large elasticmoduli and low thermal expansion coefcients. Cooperative Learning Exercise 1.1 Work with a neighbor. Consider the Lennard-Jones potential, as given by Eq. (1.12), for which m = 6 and n = 12. You wish to determine the separation distance, r, at which the maximum 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 potential to solve for the constant a in terms of b and the equilibrium bond distance, r0 . Now perform the determination of Fmax as given by Eq. (1.15): substitute this value of a back into Eq. (1.12), differentiate it twice with respect to r (remember that r0 is a constant), and set this equal to zero (determine, then maximize the force function). Solve this equation for 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 potential to solve for the constant b in terms of a and the equilibrium bond distance, r0 . Now perform the determination of Fmax as given by Eq. (1.15); substitute this value of b back into Eq. (1.12), differentiate it twice with respect to r (remember that r0 is a constant), and set this equal to zero (determine, then maximize the force function). Solve this equation for r in terms of r0 . The other constant should drop out. Compare your answers. You should both get the same result. Answer : r = 1.1087r0 U r U r (a) (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 coefcient. Those with a shallow well(b) have a low melting point, low elastic modulus, and high thermal expansion coefcient.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-plete transfer of electrons from one atom to the other. The easiest approach is to rsttransfer 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-tion 1.0.2.1, 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 afnity, EA (see Section 1.0.2.2), which for chlorine is354 kJ/mol. So, the energy required to form ions is given by: Eions = IE Na + 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 = (Z1 e Z2 e)/(40 r 2 ) (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 gives FA = +e2 /(40 r 2 ) (1.21) Energy is released by bringing the ions from innite 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-ing FA : r0 UA,0 = FA dr = e2 /(40 r0 ) (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 denition the attrac-tive energy 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 /(40 nr0 ) (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 : e2 e2 U0 = + + Eions (1.24) 40 r0 40 nr0and simplifying gives: 1 e2 U0 = 1 + Eions (1.25) n 40 r0We 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: 1 1(1.6 1019 C)2 (6.02 1023 mol1 ) U0 = 1 + 142 8 4(8.854 1012 C2 /N m2 )(2.36 1010 m) = 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 constantts directly into the energy expression (Eq. 1.25): 1 e2 UL = M + Eions (1.26) n 40 r0For 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 19 Table 1.7 Typical Values for the Madelung Constant Crystal Lattice Compound (see Section 1.1.1) M NaCl FCC 1.74756 CsCl CsCl 1.76267 CaF2 Cubic 2.51939 CdCl2 Hexagonal 2.244 MgF2 Tetragonal 2.381 ZnS (wurtzite) Hexagonal 1.64132 TiO2 (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-script indicates antibonding. The antibonding orbitals are higher in energy than corre-sponding bonding 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 MATERIALS Atomic orbitals Molecular orbitals (a ) + s A B bonding + + A B + s* 2s 2s antibonding (b) + s A B bonding + + A B + s* 2px 2px antibonding (c ) + A B p bonding + + + A B p* 2py 2py antibonding +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 Wiley & Sons, Inc. This material is used by permissionof John Wiley & Sons, Inc. Antibonding molecular orbital s* Atomic orbital Atomic orbital Energy 1s 1s Atom A s Atom B Bonding molecular orbital MoleculeFigure 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 21 s*2p p*2p 2p p2p 2p s2p s*2s 2s 2s s2s O O2 OFigure 1.7 Molecular orbital diagram for molecular oxygen, O2 . From K. M. Ralls, T. H.Courtney, and J. Wulff, Introduction to Materials Science and Engineeri