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Chapter 5 Structure of Solids 6 Lectures 1 Solids Crystalline Long-range periodicity Gives sharp diffraction patterns Has sharp melting point Has higher density Noncrystalline No long-range periodicity Does not give sharp diffraction patterns Does not have a sharp meliing point Has a lower density 2 3 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: 106 K s-1 4 Inorganic Solids Covalent Solids Metals and Alloys Ionic Solids Silica: crystalline and amorphous Polymers Classification Structure Crystallinity Mechanical Behaviour 5 Inorganic Solids Covalent Solids Metals and Alloys Ionic Solids Silica: crystalline and amorphous Polymers Classification Structure Crystallinity Mechanical Behaviour 6 Metals and Alloys 1. Metallic bond: Nondrectional (Fact) As many bonds as geometrically possible (to lower the energy) Close packing 2. Atoms as hard sphere (Assumption) 3. Elements (identical atoms) 1, 2 & 3 Elemental metal crystals: close packing of equal hard spheres 7 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? Kepler’s conjecture, 1611 P.E 3 2 0.74 Kissing Number Problem What is the maximum number of spheres that can touch a given sphere? Coding Theory Internet data transmission 8 Lecture 9 13.08.2013 9 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 10 Close packing of equal hard spheres 1-D packing A chain of spheres occupiedlength P.E.= =1 total length Kissing Number= 2 Close-packed direction of atoms 11 Close packing of equal hard spheres 2-D packing A hexagonal layer of atoms Close-packed directions? Close-packed plane of atoms 3 occupiedarea .907 P.E.= total area 2 3 Kissing Number=6 1940 L. Fejes Toth : Densest packing of circles in plane 12 Close packing of equal hard spheres 3-D packing First layer A A A C C A B A A A B B A B C C Third layer A or C A C C B Second layer B B A A A C C B B A A A C B A A Close packed crystals: …ABABAB… Hexagonal close packed (HCP) …ABCABC… Cubic close packed (CCP) 13 Geometrical properties of ABAB stacking A C B a b=a A =120 C A B B C B A A A c B A C B B A B C A C C A C A A A B A B A A C B A A A and B do not have identical neighbours Either A or B as lattice points, not both Unit cell: a rhombus based prism with a=bc; ==90, =120 The unit cell contains only one lattice point (simple) but two atoms (motif) ABAB stacking = HCP crystal = Hexagonal P lattice + 2 atom motif 000 14 2/3 1/3 1/2 z Hexagonal close-packed (HCP) crystal ½ y ½ ½ ½ Corner and inside atoms do not have the same neighbourhood x Lattice: Simple hexagonal hcp lattice Motif: Two atoms: 000; 2/3 1/3 1/2 hcp crystal 16 c/a ratio of an ideal HCP crystal A A C C B B C B A B A C B A c C A B A B C A B C A C A A A B A A A C B A B A A 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 c = 2 height of a tetrahedron of edge length a 2 2 c a 3 17 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 18 Geometrical properties of ABCABC stacking A A C C A B A A B A B C C B A C C B A B A A A C C B B A A A C B A A 19 Geometrical properties of ABCABC stacking C B All atoms are equivalent A and their centres form a C lattice 3 a B A Motif: single atom 000 What is the Bravais lattice? ABCABC stacking = CCP crystal = FCC lattice + single atom motif 000 20 Close packed planes in the FCC unit cell of cubic close packed crystal Body diagonal A B C A Close packed planes: {1 1 1} 21 Stacking sequence? 22 Stacking sequence? 23 Find the mistake in the following figure from a website: http://www.tiem.utk.edu/~gross/bioed/webmodules/s pherefig1.gif 24 Table 5.1 Coordination Number and Packing Efficiency CW Crystal Structure Diamond cubic (DC) Coordination number 4 HW Packing efficiency 0.32 0.52 Simple cubic (SC) 6 Body-centred cubic 8 0.68 Face-centred cubic 12 0.74 Empty spaces are distributed in various voids 25 End of lecture 10 (16.08.2013) Beginning of lecture 11 26 Voids in Close-Packed Crystals TETRAHEDRAL VOID OCTAHEDRAL VOID B A A A B C A A B A No. of atoms defining the void 4 No. of voids per atom 2 Edge length of void 2R Size of the void 0.225 R B A A 6 A Experiment 2 1 2R HW 0.414 R 27 Locate of Voids in CCP Unit cell 28 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) 29 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 30 TABLE 5.2 System Crystal structure Radius of atoms, Ǻ Valency Electronegativity Ag-Cu Ag Au FCC FCC 1.44 1.44 1 1 1.9 1.9 Cu-Ni Cu Ni FCC FCC 1.28 1.25 1 2 1.9 1.8 Ge-Si Ge Si DC DC 1.22 1.18 4 4 1.8 1.8 All three systems exhibit complete solid solubility. 31 BRASS Cu FCC + Zn HCP Unfavourable structure factor Limited Solubility: Max solubility of Cu in Zn: 1 wt% Cu Max Solubility of Zn in Cu: 35 wt% Zn 32 Ordered and Random Substitutional solid solution Random Solid Solution Ordered Solid Solution 33 Ordered and random substitutional solid solution β-Brass: (50 at% Zn, 50 at% Cu) Disordered solid solution of β-Brass: Above 470˚C 470˚C Below 470˚C 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 34 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. Fe3C in steels. 35 End of lecture 10 (16.08.2013) Beginning of lecture 11 36 4th. Group: Carbon 37 Allotropes of C Graphite Buckminster Fullerene 1985 Diamond Carbon Nanotubes 1991 Graphene 2004 38 Graphite Sp2 hybridization 3 covalent bonds Hexagonal sheets a = 2 d cos 30° = √3 d y x a =120 b=a d = 1.42 Å a = 2.46 Å 39 Graphite a = 2.46 Å c = 6.70 Å 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 A x B y A www.scifun.ed.ac.uk/ 40 c Graphite Highly Anisotropic: Properties are very different in the a and c directio Uses: Solid lubricant Pencils (clay + graphite, hardness depends on fraction of clay) carbon fibre www.sciencemuseum.org.uk/ 41 Diamond Sp3 hybridization 4 covalent bonds Tetrahedral bonding Location of atoms: 8 Corners 6 face centres 4 one on each of the 4 body diagonals 42 Diamond Cubic Crystal: yLattice1 & motif? 0,1 M S P N D A T B 3 4 Q L K D y R C 1 2 R M S A 0,1 C 1 4 L Q T 0,1 1 4 x 0,1 2 K 3 4 1 2 N B P 1 2 0,1 x Projection of the unit cell on the bottom face of the cube Diamond Cubic Crystal = FCC lattice + motif: 000; ¼¼¼ 43 Crystal Structure = Lattice + Motif Diamond Cubic Crystal Structure = FCC Lattice + 2 atom Motif 000 1 1 1 4 4 4 There are only three Bravais Lattices: SC, BCC, FCC. Diamond Cubic Lattice 44 There is no diamond cubic lattice. 45 Diamond Cubic Structure Coordination 4 number Corners Face Inside 1 1 8 6 1 4 8 Effective number of atoms in the unit cell = 8 2 Relaton between lattice parameter and atomic radius 3a 2r 4 8r a 3 Packing efficiency 4 3 8 r 3 3 0.34 46 3 a 16 Diamond Cubic Crystal Structures C Si Ge Gray Sn a (Å) 3.57 5.43 5.65 6.46 47 Equiatomic binary AB compounds having diamond cubic y 1 2 0,1 0,1 like structure IV-IV compound: SiC III-V compound: AlP, AlAs, AlSb, GaP, GaAs, GaSb, InP, InAs, InSb 1 4 3 4 1 2 S 1 2 0,1 1 4 1 2 3 4 II-VI compound: ZnO, ZnS, CdS, CdSe, CdTe I-VII compound: CuCl, AgI 0,1 0,1 48 USES: Diamond Abrasive in polishing and grinding wire drawing dies Si, Ge, compounds: semiconducting devices SiC abrasives, heating elements of furnaces 49 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 50 Allotropes of C Graphite Buckminster Fullerene 1985 Diamond Carbon Nanotubes 1991 Graphene 2004 C60 Buckminsterfullerene H.W. Kroto, J.R. Heath, S.C. O’Brien, R.F. Curl and R.E. Smalley Nature 318 (1985) 162-163 Long-chain carbon molecules in interstellar space 1996 Nobel Prize A carbon atom at each vertex American architect, author, designer, futurist, inventor, and 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. Montreal Biosphere in Montreal, Canada Truncated Icosahedron Icosahedron: A Platonic solid (a regular solid) Truncated Icosahedron: An Archimedean solid 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 Regular Polygons: All sides equal all angles equal Triangle square 3 4 pentagon hexagon… 5 6… A regular n-gon with any n >= 3 is possible There are infinitely many regular polygons 3D: Regular Polyhedra or Platonic Solids All faces regular congruent polygons, all corners identical. Cube How many regular solids? Tetrahedron There are 5 and only 5 Platonic or regular solids ! 59 There are 5 and only 5 Platonic or regular solids ! V E F 1. Tetrahedron 4 6 4 2. Octahedron 6 12 8 3. Cube 8 12 6 4. Icosahedron 12 30 20 5. Dodecahedron 20 30 12 Duals Duals Euler’s Polyhedr on Formula V-E+F=2 Duality Tetrahedron Self-Dual Octahedron-Cube Icosahedron-Dodecahedron 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) 3 Tetrahedron 4 Octahedron 5 Icosahedron 6 or more: not possible Square (90) 3 Cube 4 or more: not possible Pentagon (108) 3 Dodecahedron Truncated Icosahedron: V=60, E=90, F=32 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 Structural features of carbon nanotubes a a 2 1 =chiral angle zigzig (n,0) (n,m)=(6, 3) Material Young's Modulus (TPa) Tensile Strength (GPa) SWNT ~1 (from 1 to 5) 13-53 Armchair SWNT 0.94 Zigzag SWNT 0.94 Chiral SWNT 0.92 MWNT 0.8-0.9 150 Stainless Steel ~0.2 ~0.65-1 15-50 Kevlar ~0.15 ~3.5 ~2 0.25 29.6 Kevlar T T 126.2 T 94.5 E E T T Elongation at Break (%) 16 23.1 15.6-17.5 Source: wiki 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. 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]. 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," The Space Engineering and Science Institute presents The 2009 Space Elevator Conference Pioneer the next frontier this summer with a four-day conference on the Space Elevator in Redmond, Washington at the Microsoft Conference Center. Thursday, August 13 through Sunday, August 16, 2009 Register Today! End of lecture 12 (20.08.2013) Beginning of lecture 13 (21.08.2013) 73 IONIC SOLIDS Cation radius: R+ Anion radius: RUsually R R 1. Cation and anion attract each other. 2. Cation and anion spheres touch each other 3. Ionic bonds are non-directional 1, 2, 3 => Close packing of unequal spheres 74 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. 75 Effect of radius ratio Rc 0.155 Ra unstable Anions not touching the central cation, Anions touching each other Rc 0.155 Ra Rc 0.155 Ra Critically stable stable Anions touching the central cation Anions touching Rc 0.155 Ligancy 2 Ra Anions touching central cation Anions not touching each other Rc 0.155 Ligancy 3 Ra 76 Rc 0.155 Ligancy 3 Ra However, when Rc 0.225 Ra tetrahedral coordination with ligancy 4 becomes stable Recall tetrahedral void in close-packed structure. Thus Rc 0.155 0.225 Ligancy 3 Ra 77 Table 5.3 Ligancy as a Function of Radius Ration Ligancy Range of radius ratio 2 0.0 ― 0.155 3 0.155 ― 0.225 Triangular 4 0.225 ― 0.414 Tetrahedral 6 0.414 ― 0.732 Octahedral 8 0.732 ― 1.0 Cubic 12 1.0 Configuration Linear FCC or HCP 78 Example 1: NaCl RNa RCl 0.54 0.414 0.54 0.732 Ligancy6 OctahedralCoordination cae2k.com NaCl structure =FCC lattice + 2 atom motif: Cl- 0 0 0 Na ½ 0 0 79 NaCl structure continued CCP of Cl─ with Na+ in ALL octahedral voids 2RNa" 2RCl a 80 Example 2 : CsCl Structure RCs RCl 0.91 0.732 0.91 1 Ligancy 8 Cubic coordination of Claround Cs+ seas.upenn.edu CsCl structure = SC lattice + 2 atom motif: Cl 000 Cs ½ ½ ½ BCC 2RCs 2RCl 3a 81 Example 3: CaF2 (Fluorite or fluorospar) RCa 2 RF 0.73 0.73 0.732 Octahedral or cubic coordination Actually cubic coordination of F─ around Ca2+ But the ratio of number of F─ to Ca2+ is 2:1 So only alternate cubes of F─ are filled with Ca2+ 82 Simple cubic crystal of F─ with Ca2+ in alternate cube centres Alternately, Ca2+ at FCC sites with F─ in ALL tetrahedral voids CaF structure= FCC lattice + 3 atom motif Ca2+ F─ F─ 000 ¼¼¼ -¼ -¼ -¼ 83 Example 4: ZnS (Zinc blende or sphalerite) RZn 2 RS 2 0.48 0.414 0.48 0.732 Ligancy6 OctahedralCoordination However, actual ligancy is 4 (TETRAHEDRAL COORDINATION) wikipedia Explanation: nature of bond is more covalent than ionic 84 ZnS structure CCP of S2─ with Zn2+ in alternate tetrahedral voids seas.upenn.edu ZnS structure = FCC lattice + 2 atom motif S2─ 0 0 0 Zn2+ ¼ ¼ ¼ 85 86 pixdaus.com 87 theoasisxpress 88 What is common to 1, glass of the window 2. sand of the beach, and 3. quartz of the watch? 89 pixdaus.com Structure of SiO2 Bond is 50% ionic and 50 % covalent RSi 4 RO 2 0.29 0.225 0.29 .414 Tetrahedral coordination of O2─ around Si4+ Silicate tetrahedron 90 Silicate tetrahedron electrically unbalanced 4─ 2─ 4+ 2─ 2─ 2─ O2─ need to be shared between two tetrahedra 91 1. O2─ need to be shared between two tetrahedra. 2. Si need to be as far apart as possible Face sharing Edge sharing Corner sharing Silicate tetrahedra share corners 92 2D representation of 3D periodically repeating pattern of tetrahedra in crystalline SiO2. Note that alternate tetrahedra 93 are inverted 2 D representation of 3D random network of silicate tetrahedra in the fused silica glass 94 Network Modification by addition of Soda Na + Na2O = Na Modification leads to breaking of primary bonds between silicat tetrahedra. 95 2 D representation of 3D random network of silicate tetrahedra in the fused silica glass 96 End of lecture 13 (21.08.2013) Syllabus for minor I (upto lecture 13) Beginning of lecture 14 (23.08 2013) 97 5.7 Structure of Long Chain Polymers 109.5 H Degree of Polymerization: No. of repeating monomers in a chain H C A C C 98 Freedom of rotation about each bond in space leads to different conformations of C-C backbone 109.5 99 100 5.8 Crystallinity in long chain polymers Fig. 5.17: semicrystalline polymer 101 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 102 Branching 2. More is the branching less is the tendency to crystallize 103 Tacticity 3. Isotactic and syndiotactic polymers can crystallize but atactic cannot. 104 Copoymers: polymeric analog of solid solutions 4. Block and random copolymers promote non crystallinity. 105 Plasticizers Low molecular weight additives Impedes chains coming together Reduces crystallization 106 Elastomer Polymers with very extensive elastic deformation Stress-strain relationship is non-linear Example: Rubber 107 Liquid natural rubber (latex) being collected from the rubber tree 108 Isoprene molecule commons.wikimedia.org H H3C H C=C-C=C HH H 109 H H CH3 H H Isoprene molecule CH3 H CCCC H H CCCC H H H Polymerization Polyisoprene mer Liquid (Latex) 110 Vulcanisation H H CH3 H CCCC H H + 2S H H CCCC H H CH H Weak van der Waals bond 3 111 Vulcanisation H H CH3 H CCCC H H S S H H Crosslinks CCCC H H CH H 3 112 Effect of cross-linking on polyisoprene Natural rubber Elastomer Ebonite liquid Elastic solid Hard & brittle not x-linked lightly x-linked heavily x-linked 113 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. 114 End of lecture 14 (23.08.2013) Beginning of Lecture 15 (27.08.2013) (rubber elasticity + pre minor I discussion) 115 Another interesting property of elastomers Thermal behaviour 116 Elastomer sample under tension Elastomer sample heat Tensile force Coiled chains Higher entropy straight chains Lower entropy Contracts on heating Still lower entropy F 117 Elastomers have ve thermal expansion coefficient, i.e., they CONTRACT on heating!! EXPERIMENT 4 Section 10.3 of the textbook 118 N 0 kT F L0 F N0 k T L0 L L L0 L0 L 2 applied tensile force number of cross-links Boltzmann constant absolute temperature initial length (without F) final length (with F) 119 Bond stretching in straightened out molecules Experimental Theory: Chain uncoiling N kT F 0 L0 L L0 2 L0 L 120