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1 Chapter 5 Structure of Solids 6 Lectures. 2 Solids CrystallineNoncrystalline Gives sharp diffraction patterns Does not give sharp diffraction patterns.

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Presentation on theme: "1 Chapter 5 Structure of Solids 6 Lectures. 2 Solids CrystallineNoncrystalline Gives sharp diffraction patterns Does not give sharp diffraction patterns."— Presentation transcript:

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 7 th. Group (Halogens): single covalent bonds Diatomic molecules Weak van der Waals bond between molecules F 2, Cl 2 : Gas;Br 2 : Liquid; I 2 : orthorhombic xl

7 7 6 th. Group: two covalent bonds: long zig-zag chains Weak van der Waals bonds between chains mostly noncrystalline

8 8 5 th. Group: Three covalent bonds: Puckered sheets Weak van der Waals bond between sheets Mostly noncrystalline

9 9 4 th. Group: Carbon

10 10 Graphite Diamond Buckminster Fullerene 1985 Carbon Nanotubes 1991 Graphene 2004 Allotropes of C

11 11 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

12 12 Graphite x y a = 2.46 Å c = 6.70 Å B A A c Lattice: Simple Hexagonal Motif: 4 carbon atoms

13 13 Graphite Highly Anisotropic: Properties are very different in the a and c directions Uses: Solid lubricant Pencils (clay + graphite, hardness depends on fraction of clay) carbon fibre

14 14 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

15 15 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

16 16 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

17 17 There is no diamond cubic lattice.

18 18 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

19 19 Diamond Cubic Crystal Structures CSiGeGray Sn a (Å)

20 20 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

21 21 USES: Diamond Abrasive in polishing and grinding wire drawing dies Si, Ge, compounds: semiconducting devices SiC abrasives, heating elements of furnaces

22 22 Inorganic Solids Covalent Solids Metals and Alloys Ionic Solids Silica: crystalline and amorphous Polymers Classification Structure Crystallinity Mechanical Behaviour

23 23 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)

24 24 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

25 25 Close packing of equal hard spheres 1-D packing A chain of spheres P.E.=Kissing Number= Close-packed direction of atoms =1 2

26 26 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? L. Fejes Toth: Densest packing of circles in plane

27 27 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)

28 28 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

29 29 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

30 30 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 B A C B A C All atoms are equivalent and their centres form a lattice Motif: single atom 000 ABCABC stacking = CCP crystal = FCC lattice + single atom motif 000  3 a

31 31 Geometrical properties of ABCABC stacking B A C B A C All atoms are equivalent and their centres form a lattice Motif: single atom 000 ABCABC stacking = CCP crystal = FCC lattice + single atom motif 000  3 a

32 32 Geometrical properties of ABCABC stacking B A C B A C All atoms are equivalent and their centres form a lattice Motif: single atom 000 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 ABCABC stacking = CCP crystal = FCC lattice + single atom motif 000  3 a A A A A A A

33 33 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

34 34 Stacking sequence? ABA: HCP

35 35

36 36 pherefig1.gif Find the mistake in the following picture:

37 37 Crystal Structure Coordination number Packing efficiency Table 5.1 Coordination Number and Packing Efficiency Diamond cubic (DC)40.34 Simple cubic (SC)60.52 Body-centred cubic80.68 Face-centred cubic120.74

38 38 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 R R Experiment 2 HW

39 39 Location of Voids in FCC Unit cell

40 40 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)

41 41 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

42 42 TABLE 5.2 SystemCrystal Radius of ValencyElectro- structureatoms, Ǻnegativity Ag-CuAgFCC AuFCC Cu-NiCuFCC NiFCC Ge-SiGeDC SiDC All three systems exhibit complete solid solubility.

43 43 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

44 44 Ordered and Random Substitutional solid solution Random Solid Solution Ordered Solid Solution

45 45 Disordered solid solution of β-Brass: 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 470˚C Above 470˚C Below 470˚C Ordered and random substitutional solid solution β-Brass: (50 at% Zn, 50 at% Cu)

46 46 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.

47 47 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

48 48 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.

49 49 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

50 50 However, when tetrahedral coordination with ligancy 4 becomes stable Recall tetrahedral void in close-packed structure. Thus

51 51 Table 5.3 Ligancy as a Function of Radius Ration LigancyRange of radius ratioConfiguration 20.0 ― 0.155Linear ― 0.225Triangular ― 0.414Tetrahedral ― 0.732Octahedral ― 1.0Cubic 121.0FCC or HCP

52 52 Example 1: NaCl cae2k.com NaCl structure =FCC lattice + 2 atom motif: Cl Na ½ 0 0

53 53 NaCl structure continued CCP of Cl ─ with Na + in ALL octahedral voids

54 54 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

55 55 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+

56 56 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 F ─ ¼ ¼ ¼ F ─ -¼ -¼ -¼

57 57 Example 4: ZnS (Zinc blende or sphalerite) However, actual ligancy is 4 (TETRAHEDRAL COORDINATION) Explanation: nature of bond is more covalent than ionic wikipedia

58 58 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─ Zn 2+ ¼ ¼ ¼

59 59 pixdaus.com

60 60 theoasisxpress.com

61 61

62 62 pixdaus.com What is common to 1, glass of the window 2. sand of the beach, and 3. quartz of the watch?

63 63 Structure of SiO 2 Bond is 50% ionic and 50 % covalent Tetrahedral coordination of O 2─ around Si 4+ Silicate tetrahedron

64 ─2─ 2─2─ 2─2─ 2─2─ 4─ Silicate tetrahedron electrically unbalanced O 2─ need to be shared between two tetrahedra

65 65 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

66 66 2D representation of 3D periodically repeating pattern of tetrahedra in crystalline SiO 2. Note that alternate tetrahedra are inverted

67 67 2 D representation of 3D random network of silicate tetrahedra in the fused silica glass

68 68 Modification leads to breaking of primary bonds between silicate tetrahedra. + Na 2 O = Na Network Modification by addition of Soda

69 69 2 D representation of 3D random network of silicate tetrahedra in the fused silica glass

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

71 71 Freedom of rotation about each bond in space leads to different conformations of C-C backbone 

72 72

73 Crystallinity in long chain polymers Fig. 5.17: semicrystalline polymer

74 74 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

75 75 Branching 2. More is the branching less is the tendency to crystallize

76 76 Tacticity 3. Isotactic and syndiotactic polymers can crystallize but atactic cannot.

77 77 Copoymers: polymeric analog of solid solutions 4. Block and random copolymers promote non crystallinity.

78 78 Plasticizers Low molecular weight additives Impedes chains coming together Reduces crystallization

79 79 Elastomer Polymers with very extensive elastic deformation Stress-strain relationship is non-linear Example: Rubber

80 80 Liquid natural rubber (latex) being collected from the rubber tree

81 81 Isoprene molecule commons.wikimedia.org C=C-C=C HH HH H H3CH3C

82 82 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)

83 83  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

84 84  C  C  C  C  HH HH H CH 3  C  C  C  C  HH HH H CH 3 S Vulcanisation S     Cross- links

85 85 Natural rubber ElastomerEbonite liquidElastic solid Hard & brittle not x-linked lightly x-linked heavily x-linked Effect of cross-linking on polyisoprene

86 86 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.

87 87 Another interesting property of elastomers Thermal behaviour

88 88 Tensile force F Elastomer sample Elastomer sample under tension Coiled chains straight chains heat Higher entropy Lower entropy Still lower entropy Contracts on heating

89 89 Elastomers have  ve thermal expansion coefficient, i.e., they CONTRACT on heating!! EXPERIMENT 4 Section 10.3 of the textbook

90 90 Fapplied tensile force N 0 number of cross-links kBoltzmann constant Tabsolute temperature L 0 initial length (without F) Lfinal length (with F)

91 91 Experimental Theory: Chain uncoiling Bond stretching in straightened out molecules


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