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 7th. Group (Halogens): single covalent bonds Diatomic molecules Weak van der Waals bond between molecules F2, Cl2: Gas; Br2: Liquid; I2: orthorhombic xl 6 6th. Group: two covalent bonds: long zig-zag chains Weak van der Waals bonds between chains mostly noncrystalline 7 5th. Group: Three covalent bonds: Puckered sheets Weak van der Waals bond between sheets Mostly noncrystalline 8 4th. Group: Carbon 9 Allotropes of C Graphite Buckminster Fullerene 1985 Diamond Carbon Nanotubes 1991 Graphene 2004 10 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 11 Graphite a = 2.46 Å c = 6.70 Å Lattice: Simple Hexagonal Motif: 4 carbon atoms A x B y A www.scifun.ed.ac.uk/ 12 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/ 13 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 14 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 15 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 16 There is no diamond cubic lattice. 17 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 18 3 a 16 Diamond Cubic Crystal Structures C Si Ge Gray Sn a (Å) 3.57 5.43 5.65 6.46 19 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 20 USES: Diamond Abrasive in polishing and grinding wire drawing dies Si, Ge, compounds: semiconducting devices SiC abrasives, heating elements of furnaces 21 Inorganic Solids Covalent Solids Metals and Alloys Ionic Solids Silica: crystalline and amorphous Polymers Classification Structure Crystallinity Mechanical Behaviour 22 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 23 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 24 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 25 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 26 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) 27 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 28 2/3 1/3 1/2 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 29 Geometrical properties of ABCABC stacking C A A C C A B A A A A B B B B C C C 3 a A C C B A B A A A A C C B B B A A A C B A A All atoms are equivalent and their centres form a lattice Motif: single atom 000 ABCABC stacking = CCP crystal = FCC lattice + single atom motif 000 30 Geometrical properties of ABCABC stacking C B A C 3 a B A All atoms are equivalent and their centres form a lattice Motif: single atom 000 ABCABC stacking = CCP crystal = FCC lattice + single atom motif 000 31 Geometrical properties of ABCABC stacking C A A C C A B A A A A B B B B C C C 3 a A C C B A B A A A A C C B B B A A A C B A A All atoms are equivalent and their centres form a lattice Motif: single atom 000 ABCABC stacking = CCP crystal = FCC lattice + single atom motif 000 32 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} 33 Stacking sequence? ABA: HCP 34 35 Find the mistake in the following picture: http://www.tiem.utk.edu/~gross/bioed/webmodules/s pherefig1.gif 36 Table 5.1 Coordination Number and Packing Efficiency Crystal Structure Coordination number Packing efficiency Diamond cubic (DC) 4 0.34 Simple cubic (SC) 6 0.52 Body-centred cubic 8 0.68 Face-centred cubic 12 0.74 37 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 38 Location of Voids in FCC Unit cell 39 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) 40 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 41 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. 42 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 43 Ordered and Random Substitutional solid solution Random Solid Solution Ordered Solid Solution 44 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% probability of being occupied by Cu or Zn Ordered solid solution of β-Brass: Corners are always occupied by Cu, centres always by Zn 45 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. 46 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 47 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. 48 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 49 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 50 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 51 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 52 NaCl structure continued CCP of Cl─ with Na+ in ALL octahedral voids 2RNa" 2RCl a 53 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 54 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+ 55 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 56 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 57 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 58 59 pixdaus.com 60 theoasisxpress 61 What is common to 1, glass of the window 2. sand of the beach, and 3. quartz of the watch? 62 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 63 Silicate tetrahedron electrically unbalanced 4─ 2─ 4+ 2─ 2─ 2─ O2─ need to be shared between two tetrahedra 64 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 65 2D representation of 3D periodically repeating pattern of tetrahedra in crystalline SiO2. Note that alternate tetrahedra 66 are inverted 2 D representation of 3D random network of silicate tetrahedra in the fused silica glass 67 Network Modification by addition of Soda Na + Na2O = Na Modification leads to breaking of primary bonds between silicate tetrahedra. 68 2 D representation of 3D random network of silicate tetrahedra in the fused silica glass 69 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 70 Freedom of rotation about each bond in space leads to different conformations of C-C backbone 109.5 71 72 5.8 Crystallinity in long chain polymers Fig. 5.17: semicrystalline polymer 73 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 74 Branching 2. More is the branching less is the tendency to crystallize 75 Tacticity 3. Isotactic and syndiotactic polymers can crystallize but atactic cannot. 76 Copoymers: polymeric analog of solid solutions 4. Block and random copolymers promote non crystallinity. 77 Plasticizers Low molecular weight additives Impedes chains coming together Reduces crystallization 78 Elastomer Polymers with very extensive elastic deformation Stress-strain relationship is non-linear Example: Rubber 79 Liquid natural rubber (latex) being collected from the rubber tree 80 Isoprene molecule commons.wikimedia.org H H3C H C=C-C=C HH H 81 H H CH3 H H Isoprene molecule CH3 H CCCC H H CCCC H H H Polymerization Polyisoprene mer Liquid (Latex) 82 Vulcanisation H H CH3 H CCCC H H + 2S H H CCCC H H CH H Weak van der Waals bond 3 83 Vulcanisation H H CH3 H CCCC H H S S H H Crosslinks CCCC H H CH H 3 84 Effect of cross-linking on polyisoprene Natural rubber Elastomer Ebonite liquid Elastic solid Hard & brittle not x-linked lightly x-linked heavily x-linked 85 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. 86 Another interesting property of elastomers Thermal behaviour 87 Elastomer sample under tension Elastomer sample heat Tensile force Coiled chains Higher entropy straight chains Lower entropy Contracts on heating Still lower entropy F 88 Elastomers have ve thermal expansion coefficient, i.e., they CONTRACT on heating!! EXPERIMENT 4 Section 10.3 of the textbook 89 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) 90 Bond stretching in straightened out molecules Experimental Theory: Chain uncoiling N kT F 0 L0 L L0 2 L0 L 91