1. 1 Chapter 5 Structure of Solids 6 Lectures

2. 2 Solids CrystallineNoncrystalline Gives sharp diffraction patterns Does not give sharp diffraction patterns Long-range periodicityNo long-range periodicity Has sharp melting point Does not have a sharp meliing point Has higher density Has a lower density

3. 3

4. 4 Factors promoting the formation of noncrystalline structures 1.Primary bonds do not extend in all three directions and the secondary bonds are not strong enough. 2.The difference in the free energy of the crystalline and non crystalline phases is small. 3.The rate of cooling from the liquid state is higher than a critical cooling rate. Metallic Glass: 10 6 K s -1

5. 5 Inorganic Solids Covalent Solids Metals and Alloys Ionic Solids Silica: crystalline and amorphous Polymers Classification Structure Crystallinity Mechanical Behaviour

6. 6 Inorganic Solids Covalent Solids Metals and Alloys Ionic Solids Silica: crystalline and amorphous Polymers Classification Structure Crystallinity Mechanical Behaviour

7. 7 Metals and Alloys  As many bonds as geometrically possible (to lower the energy) 2. Atoms as hard sphere (Assumption) 1, 2 & 3  Elemental metal crystals: close packing of equal hard spheres 1. Metallic bond: Nondrectional (Fact)  Close packing 3. Elements (identical atoms)

8. 8 Close packing of equal hard spheres Arrangement of equal nonoverlapping spheres to fill space as densely as possible Sphere packing problem: What is the densest packing of spheres in 3D? Kissing Number Problem Kepler’s conjecture, 1611 What is the maximum number of spheres that can touch a given sphere? Coding Theory Internet data transmission

9. 9 Lecture 9 13.08.2013

10. 10 We are currently preparing students for jobs that do not yet exist, using technologies that haven’t been invented in order to solve problems that we don’t even know are problems yet. http://www.youtube.com/watch?v=XVQ1ULfQawk

11. 11 Close packing of equal hard spheres 1-D packing A chain of spheres P.E.=Kissing Number= Close-packed direction of atoms =1 2

12. 12 Close packing of equal hard spheres 2-D packing A hexagonal layer of atoms P.E.= Kissing Number=6 Close-packed plane of atoms Close-packed directions?3 1940 L. Fejes Toth: Densest packing of circles in plane

13. 13 Close packing of equal hard spheres 3-D packing A A A A A A A A A A A A A A A A B B B B BB B B B C C C C C C C C C First layer A Second layer B Third layer A or C Close packed crystals: …ABABAB… Hexagonal close packed (HCP) …ABCABC… Cubic close packed (CCP)

14. 14 Geometrical properties of ABAB stacking A A A A A A A A A A A A A A A A B B B BB B B B C C C C C C C C C B A and B do not have identical neighbours Either A or B as lattice points, not both a b = a  =120  Unit cell: a rhombus based prism with a=b  c;  =  =90 ,  =120  A A B B c The unit cell contains only one lattice point (simple) but two atoms (motif) ABAB stacking = HCP crystal = Hexagonal P lattice + 2 atom motif 000 2/3 1/3 1/2

15. ½ ½ ½ ½ Lattice: Simple hexagonal hcp latticehcp crystal Hexagonal close-packed (HCP) crystal x y z Corner and inside atoms do not have the same neighbourhood Motif: Two atoms: 000; 2/3 1/3 1/2

16. 16

17. 17 c/a ratio of an ideal HCP crystal A A A A A A A A A A A A A A A A B B B BB B B B C C C C C C C C C B A single B atom sitting on a base of three A atoms forms a regular tetrahedron with edge length a = 2R The same B atom also forms an inverted tetrahedron with three A atoms sitting above it A A B B c  c = 2  height of a tetrahedron of edge length a

18. 18 Lec 10: Structure of metals and alloys (Close Packing) continued 16.08.2013 Rescheduled Lecture for the missed lecture on Wednesday: Today 7-8 pm VLT1 Office hours for discussions: 5-6 pm on tuesdays, wednesdays and thursdays Doubt clearning class on request

19. 19 Geometrical properties of ABCABC stacking A A A A A A A A A A A A A A A A B B B B BB B B B C C C C C C C C C A A A A A A

20. 20 Geometrical properties of ABCABC stacking B A C B A C ABCABC stacking = CCP crystal = FCC lattice + single atom motif 000  3 a All atoms are equivalent and their centres form a lattice Motif: single atom 000 What is the Bravais lattice?

21. 21 A C A Body diagonal B Close packed planes in the FCC unit cell of cubic close packed crystal Close packed planes: {1 1 1} B

22. 22 Stacking sequence?

23. 23 Stacking sequence?

24. 24 http://www.tiem.utk.edu/~gross/bioed/webmodules/s pherefig1.gif Find the mistake in the following figure from a website:

25. 25 Crystal Coordination Packing Structure number efficiency Diamond cubic (DC)4 Simple cubic (SC)6 Body-centred cubic8 Face-centred cubic12 Table 5.1 Coordination Number and Packing Efficiency Empty spaces are distributed in various voids HW CW 0.32 0.52 0.68 0.74

26. 26 End of lecture 10 (16.08.2013) Beginning of lecture 11

27. 27 Voids in Close-Packed Crystals A A A B A A AA A B B B C TETRAHEDRAL VOIDOCTAHEDRAL VOID A No. of atoms defining 46 the void No. of voids per atom 21 Edge length of void 2 R 2 R Size of the void 0.225 R 0.414 R Experiment 2 HW

28. 28 Locate of Voids in CCP Unit cell

29. 29 Solid Solution A single crystalline phase consisting of two or more elements is called a solid solution. Substitutional Solid solution of Cu and Zn (FCC) Interstitial solid solution of C in Fe (BCC)

30. 30 Hume-Rothery Rules for Extensive Solid Solution (Unlimited solubility) Interstitial solid solution Substitutional solid solution 1.Structure factor Crystal structure of the two elements should be the same 2.Size factor: Size of the two elements should not differ by more than 15% 3. Electronegativity factor: Electronegativity difference between the elements should be small 4. Valency factor: Valency of the two elements should be the same

31. 31 TABLE 5.2 SystemCrystal Radius ofValencyElectro- structureatoms, Ǻnegativity Ag-CuAgFCC1.4411.9 AuFCC1.4411.9 Cu-NiCuFCC1.2811.9 NiFCC1.2521.8 Ge-SiGeDC1.2241.8 SiDC1.1841.8 All three systems exhibit complete solid solubility.

32. 32 BRASS Cu + Zn FCC HCP Limited Solubility: Max solubility of Cu in Zn: 1 wt% Cu Max Solubility of Zn in Cu: 35 wt% Zn Unfavourable structure factor

33. 33 Ordered and Random Substitutional solid solution Random Solid Solution Ordered Solid Solution

34. 34 Disordered solid solution of β-Brass: Corner and centre both have 50% proibability of being occupied by Cu or Zn34 Ordered solid solution of β-Brass: Corners are always occupied by Cu, centres always by Zn 470˚C Above 470˚C Below 470˚C Ordered and random substitutional solid solution β-Brass: (50 at% Zn, 50 at% Cu)

35. 35 Intermediate Structures Crystal structure of Cu: FCC Crystal structure of Zn: HCP Crystal structure of random β-brass: BCC Such phases that have a crystal structure different from either of the two components are called INTERMEDIATE STRUCURES If an intermediate structure occurs only at a fixed composition it is called an INTERMETALLIC COMPOUND, e.g. Fe 3 C in steels.

36. 36 End of lecture 10 (16.08.2013) Beginning of lecture 11

37. 37 4 th. Group: Carbon

38. 38 Graphite Diamond Buckminster Fullerene 1985 Carbon Nanotubes 1991 Graphene 2004 Allotropes of C

39. 39 Graphite Sp 2 hybridization  3 covalent bonds  Hexagonal sheets x y a b=a  =120  a = 2 d cos 30° = √3 d d = 1.42 Å a = 2.46 Å

40. 40 Graphite x y a = 2.46 Å c = 6.70 Å B A A www.scifun.ed.ac.uk/ c Lattice: Simple Hexagonal Motif: 4 carbon atoms 000; 2/3 1/3 0; 2/3 1/3 1/2; 1/3 2/3 1/2

41. 41 Graphite Highly Anisotropic: Properties are very different in the a and c directions www.sciencemuseum.org.uk/ Uses: Solid lubricant Pencils (clay + graphite, hardness depends on fraction of clay) carbon fibre

42. 42 Diamond Sp 3 hybridization  4 covalent bonds Location of atoms: 8 Corners 6 face centres 4 one on each of the 4 body diagonals  Tetrahedral bonding

43. 43 Diamond Cubic Crystal: Lattice & motif? A A B B C C D D x y P P Q Q R R S S T T K K L L M M N N 0,1 Diamond Cubic Crystal = FCC lattice + motif: x y Projection of the unit cell on the bottom face of the cube 000; ¼¼¼

44. 44 Crystal Structure = Lattice + Motif Diamond Cubic Crystal Structure FCC Lattice 2 atom Motif =+ There are only three Bravais Lattices: SC, BCC, FCC. Diamond Cubic Lattice

45. 45 There is no diamond cubic lattice.

46. 46 Diamond Cubic Structure Effective number of atoms in the unit cell = Corners Relaton between lattice parameter and atomic radius Packing efficiency Coordination number 4 InsideFace

47. 47 Diamond Cubic Crystal Structures CSiGeGray Sn a (Å)3.575.435.656.46

48. 48 0,1 IV-IV compound: SiC III-V compound: AlP, AlAs, AlSb, GaP, GaAs, GaSb, InP, InAs, InSb II-VI compound: ZnO, ZnS, CdS, CdSe, CdTe I-VII compound: CuCl, AgI y S 0,1 Equiatomic binary AB compounds having diamond cubic like structure

49. 49 USES: Diamond Abrasive in polishing and grinding wire drawing dies Si, Ge, compounds: semiconducting devices SiC abrasives, heating elements of furnaces

50. 50 End of lecture 11 (16.08.2013) (Evening class, a postponed class for the missed class on Wednesday 14.08.2013) Beginning of lecture 12

51. Graphite Diamond Buckminster Fullerene 1985 Carbon Nanotubes 1991 Graphene 2004 Allotropes of C

52. C 60 Buckminsterfullerene H.W. Kroto, J.R. Heath, S.C. O’Brien, R.F. Curl and R.E. Smalley Nature 318 (1985) 162-163 1996 Nobel Prize Long-chain carbon molecules in interstellar space A carbon atom at each vertex

53. AmericanAmerican architect, author, designer, futurist, inventor, and visionary.architect authordesigner futuristinventor visionary He was expelled from Harvard twice: 1. first for spending all his money partying with a Vaudeville troupe, 2. for his "irresponsibility and lack of interest". what he, as an individual, could do to improve humanity's condition, which large organizations, governments, and private enterprises inherently could not do.

54. Montreal Biosphere in Montreal, Canada

55. Truncated Icosahedron Icosahedron: A Platonic solid (a regular solid) Truncated Icosahedron: An Archimedean solid

56. A regular polygon A polygon with all sides equal and all angles equal Square regular Rectangle unequal sides not regular Rhombus unequal angles not regular

57. Regular Polygons: All sides equal all angles equal A regular n-gon with any n >= 3 is possible 3 4 5 6… There are infinitely many regular polygons Triangle square pentagon hexagon…

58. 3D: Regular Polyhedra or Platonic Solids All faces regular congruent polygons, all corners identical. Cube How many regular solids? Tetrahedron

59. 59 There are 5 and only 5 Platonic or regular solids !

61. 1.Tetrahedron464 2.Octahedron6128 3.Cube8126 4.Icosahedron123020 5.Dodecahedron203012 Duals Euler’s Polyhedr on Formula V-E+F=2 V E F

62. Duality TetrahedronSelf-Dual Octahedron-Cube Icosahedron-Dodecahedron

63. Proof of Five Platonic Solids At any vertex at least three faces should meet The sum of polygonal angles at any vertex should be less the 360  Triangles (60  )3Tetrahedron 4Octahedron 5Icosahedron 6 or more: not possible Square (90  )3Cube 4 or more: not possible Pentagon (108  )3Dodecahedron

64. Truncated Icosahedron: V=60, E=90, F=32

67. Nature 391, 59-62 (1 January 1998) Electronic structure of atomically resolved carbon nanotubes Jeroen W. G. Wilder, Liesbeth C. Venema, Andrew G. Rinzler, Richard E. Smalley & Cees Dekker

68. zigzig (n,0) armchair (n,n) (n,m)=(6, 3) a1a1 a2a2 wrapping vector Structural features of carbon nanotubes  =chiral angle

69. MaterialYoung's ModulusYoung's Modulus (TPa) Tensile Strength Tensile Strength (GPa) Elongation at Break (%) SWNT~1 (from 1 to 5)13-53 E 16 Armchair SWNT 0.94 T 126.2 T 23.1 Zigzag SWNT 0.94 T 94.5 T 15.6-17.5 Chiral SWNT 0.92 MWNT0.8-0.9 E 150 Stainless Steel ~0.2~0.65-115-50 Kevlar~0.15~3.5~2 Kevlar T 0.2529.6 Source: wiki

70. Electrical For a given (n,m) nanotube, if n = m, the nanotube is metallic; if n − m is a multiple of 3, then the nanotube is semiconducting with a very small band gap, otherwise the nanotube is a moderate semiconductor.semiconductor Thus all armchair (n=m) nanotubes are metallic, and nanotubes (5,0), (6,4), (9,1), etc. are semiconducting. In theory, metallic nanotubes can carry an electrical current density of 4×109 A/cm2 which is more than 1,000 times greater than metals such as copper[23].copper[23]

71. While the fantasy of a space elevator has been around for about 100 years, the idea became slightly more realistic by the 1991 discovery of "carbon nanotubes,"

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73. 73 End of lecture 12 (20.08.2013) Beginning of lecture 13 (21.08.2013)

74. 74 IONIC SOLIDS Cation radius: R + Anion radius: R - 1. Cation and anion attract each other. Usually 2. Cation and anion spheres touch each other 1, 2, 3 => Close packing of unequal spheres 3. Ionic bonds are non-directional

75. 75 IONIC SOLIDS Local packing geometry 1. Anions and cations considered as hard spheres always touch each other. 2.Anions generally will not touch, but may be close enough to be in contact with each other in a limiting situation. 3. As many anions as possible surround a central cation for the maximum reduction in electrostatic energy.

76. 76 Anions not touching the central cation, Anions touching each other Anions touching the central cation Anions touching Anions touching central cation Anions not touching each other unstableCritically stablestable Effect of radius ratio

77. 77 However, when tetrahedral coordination with ligancy 4 becomes stable Recall tetrahedral void in close-packed structure. Thus

78. 78 Table 5.3 Ligancy as a Function of Radius Ration LigancyRange of radius ratioConfiguration 20.0 ― 0.155Linear 30.155 ― 0.225Triangular 40.225 ― 0.414Tetrahedral 60.414 ― 0.732Octahedral 80.732 ― 1.0Cubic 121.0FCC or HCP

79. 79 Example 1: NaCl cae2k.com NaCl structure =FCC lattice + 2 atom motif: Cl - 0 0 0 Na ½ 0 0

80. 80 NaCl structure continued CCP of Cl ─ with Na + in ALL octahedral voids

81. 81 seas.upenn.edu Example 2 : CsCl Structure Ligancy 8 Cubic coordination of Cl - around Cs + CsCl structure = SC lattice + 2 atom motif:Cl 000 Cs ½ ½ ½ BCC

82. 82 Example 3: CaF 2 (Fluorite or fluorospar) Octahedral or cubic coordination Actually cubic coordination of F ─ around Ca 2+ But the ratio of number of F ─ to Ca 2+ is 2:1 So only alternate cubes of F ─ are filled with Ca 2+

83. 83 Simple cubic crystal of F ─ with Ca 2+ in alternate cube centres Alternately, Ca 2+ at FCC sites with F ─ in ALL tetrahedral voids CaF structure= FCC lattice + 3 atom motif Ca 2+ 000 F ─ ¼ ¼ ¼ F ─ -¼ -¼ -¼

84. 84 Example 4: ZnS (Zinc blende or sphalerite) However, actual ligancy is 4 (TETRAHEDRAL COORDINATION) Explanation: nature of bond is more covalent than ionic wikipedia

85. 85 seas.upenn.edu ZnS structure CCP of S 2─ with Zn 2+ in alternate tetrahedral voids ZnS structure = FCC lattice + 2 atom motif S 2─ 0 0 0 Zn 2+ ¼ ¼ ¼

86. 86 pixdaus.com

87. 87 theoasisxpress.com

88. 88

89. 89 pixdaus.com What is common to 1, glass of the window 2. sand of the beach, and 3. quartz of the watch?

90. 90 Structure of SiO 2 Bond is 50% ionic and 50 % covalent Tetrahedral coordination of O 2─ around Si 4+ Silicate tetrahedron

91. 91 4+4+ 2─2─ 2─2─ 2─2─ 2─2─ 4─ Silicate tetrahedron electrically unbalanced O 2─ need to be shared between two tetrahedra

92. 92 1. O 2─ need to be shared between two tetrahedra. 2. Si need to be as far apart as possible Face sharingEdge sharingCorner sharing Silicate tetrahedra share corners

93. 93 2D representation of 3D periodically repeating pattern of tetrahedra in crystalline SiO 2. Note that alternate tetrahedra are inverted

94. 94 2 D representation of 3D random network of silicate tetrahedra in the fused silica glass

95. 95 Modification leads to breaking of primary bonds between silicat tetrahedra. + Na 2 O = Na Network Modification by addition of Soda

96. 96 2 D representation of 3D random network of silicate tetrahedra in the fused silica glass

97. 97 End of lecture 13 (21.08.2013) Syllabus for minor I (upto lecture 13) Beginning of lecture 14 (23.08 2013)

98. 98 5.7 Structure of Long Chain Polymers Degree of Polymerization: No. of repeating monomers in a chain 109.5  A C C C H H

99. 99 Freedom of rotation about each bond in space leads to different conformations of C-C backbone 109.5 

100. 100

101. 101 5.8 Crystallinity in long chain polymers Fig. 5.17: semicrystalline polymer

102. 102 Factors affecting crystallinity of a long chain polymer 1.Higher the degree of polymerization lower is the degree of crystallization. Longer chains get easily entagled

103. 103 Branching 2. More is the branching less is the tendency to crystallize

104. 104 Tacticity 3. Isotactic and syndiotactic polymers can crystallize but atactic cannot.

105. 105 Copoymers: polymeric analog of solid solutions 4. Block and random copolymers promote non crystallinity.

106. 106 Plasticizers Low molecular weight additives Impedes chains coming together Reduces crystallization

107. 107 Elastomer Polymers with very extensive elastic deformation Stress-strain relationship is non-linear Example: Rubber

108. 108 Liquid natural rubber (latex) being collected from the rubber tree

109. 109 Isoprene molecule commons.wikimedia.org C=C-C=C HH HH H H3CH3C

110. 110 C  C  C  C HH HH H CH 3 Isoprene molecule Polyisoprene mer  C  C  C  C  HH HH H CH 3 Polymerization Liquid (Latex)

111. 111  C  C  C  C  HH HH H CH 3  C  C  C  C  HH HH H CH 3 + 2S Vulcanisation Weak van der Waals bond

112. 112  C  C  C  C  HH HH H CH 3  C  C  C  C  HH HH H CH 3 S Vulcanisation S     Cross- links

113. 113 Natural rubber ElastomerEbonite liquidElastic solid Hard & brittle not x-linked lightly x-linked heavily x-linked Effect of cross-linking on polyisoprene

114. 114 Charles Goodyear December 29, 1800-July 1, 1860 Debt at the time of death $200,000 Life should not be estimated exclusively by the standard of dollars and cents. I am not disposed to complain that I have planted and others have gathered the fruits. One has cause to regret only when he sows and no one reaps.

115. 115 End of lecture 14 (23.08.2013) Beginning of Lecture 15 (27.08.2013) (rubber elasticity + pre minor I discussion)

116. 116 Another interesting property of elastomers Thermal behaviour

117. 117 Tensile force F Elastomer sample Elastomer sample under tension Coiled chains straight chains heat Higher entropy Lower entropy Still lower entropy Contracts on heating

118. 118 Elastomers have  ve thermal expansion coefficient, i.e., they CONTRACT on heating!! EXPERIMENT 4 Section 10.3 of the textbook

119. 119 Fapplied tensile force N 0 number of cross-links kBoltzmann constant Tabsolute temperature L 0 initial length (without F) Lfinal length (with F)

120. 120 Experimental Theory: Chain uncoiling Bond stretching in straightened out molecules