1

Chapter 5

Structure of Solids

6 Lectures

2

Solids

Crystalline Noncrystalline

Gives sharp diffraction patterns

Does not give sharp diffraction patterns

Long-range periodicity No long-range periodicity

Has sharp melting point

Does not have a sharp meliing point

Has higher density Has a lower density

3

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: 106 K s-1

5

Inorganic SolidsCovalent SolidsMetals and AlloysIonic SolidsSilica: crystalline and amorphous

PolymersClassification StructureCrystallinityMechanical Behaviour

6

Inorganic SolidsCovalent SolidsMetals and AlloysIonic SolidsSilica: crystalline and amorphous

PolymersClassification StructureCrystallinityMechanical Behaviour

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

Close packing of equal hard spheresArrangement of equal nonoverlapping

spheres to fill space as densely as possibleSphere packing problem: What is the densest packing of spheres in 3D?

Kissing Number Problem

Kepler’s conjecture, 1611 74.023

. EP

What is the maximum number of spheres that can touch a given sphere?

Coding TheoryInternet data transmission

9

Lecture 913.08.2013

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

Close packing of equal hard spheres

1-D packing

A chain of spheres

P.E.=

Kissing Number=

Close-packed direction of atoms

=1 2lengthtotallengthoccupied

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

areatotalareaoccupied 907.

32

1940 L. Fejes Toth : Densest packing of circles in plane

13

Close packing of equal hard spheres

3-D packing

A A A

AA AA

AA

A

A

A AA

A

A

B BB

B B B

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

Geometrical properties of ABAB stacking

A A A

AA AA

AA

A

A

A AA

A

A

BB

B B B

B B B

C C C

C

C

C C

C C

B

A and B do not have identical neighboursEither A or B as lattice points, not both

a

b = a=120

Unit cell: a rhombus based prism with a=bc; ==90, =120

A

AB

Bc

The unit cell contains only one lattice point (simple) but two atoms (motif)ABAB stacking = HCP crystal = Hexagonal P lattice + 2 atom motif000

2/3 1/3 1/2

½

½

½

½

Lattice: Simple hexagonal

hcp lattice hcp 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

17

c/a ratio of an ideal HCP crystal

A A A

AA AA

AA

A

A

A AA

A

A

BB

B B B

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 = 2RThe same B atom also forms an inverted tetrahedron with three A atoms sitting above it

A

AB

Bc

c = 2 height of a tetrahedron of edge length a

ac322

18

Lec 10: Structure of metals and alloys (Close Packing) continued16.08.2013

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

Geometrical properties of ABCABC stacking

A A A

AA AA

AA

A

A

A AA

A

A

B BB

B B B

B B B

C C C

C

C

C C

C C

A

A

A

A AA

20

Geometrical properties of ABCABC stacking B

A

CB

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

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

Stacking sequence?

23

Stacking sequence?

24

http://www.tiem.utk.edu/~gross/bioed/webmodules/spherefig1.gif

Find the mistake in the following figure from a website:

25

Crystal Coordination PackingStructure number efficiency

Diamond cubic (DC) 4

Simple cubic (SC) 6

Body-centred cubic 8

Face-centred cubic 12

Table 5.1Coordination Number and Packing Efficiency

Empty spaces are distributed in various voids

HWCW

0.320.52

0.68

0.74

26

End of lecture 10 (16.08.2013)

Beginning of lecture 11

27

Voids in Close-Packed Crystals

A

AAB

A

AAA

A

B

B B

C

TETRAHEDRAL VOID OCTAHEDRAL VOID

A

No. of atoms defining 4 6the void

No. of voids per atom 2 1

Edge length of void 2 R 2 R

Size of the void 0.225 R 0.414 R

Experiment 2

HW

28

Locate of Voids in CCP Unit cell

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

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

TABLE 5.2

System Crystal Radius of Valency Electro-structure atoms, Ǻ negativity

Ag-Cu Ag FCC 1.44 1 1.9Au FCC 1.44 1 1.9

Cu-Ni Cu FCC 1.28 1 1.9Ni FCC 1.25 2 1.8

Ge-Si Ge DC 1.22 4 1.8Si DC 1.18 4 1.8

All three systems exhibit complete solid solubility.

32

BRASSCu + 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

Ordered and RandomSubstitutional solid solution

Random Solid Solution

Ordered Solid Solution

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

Intermediate Structures

Crystal structure of Cu:

FCCCrystal 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 STRUCURESIf an intermediate structure occurs only at a fixed composition it is called an INTERMETALLIC COMPOUND, e.g. Fe3C in steels.

36

End of lecture 10 (16.08.2013)

Beginning of lecture 11

37

4th. Group: Carbon

38

Graphite Diamond

Buckminster Fullerene1985

Carbon Nanotubes1991

Graphene2004

Allotropes of C

39

GraphiteSp2 hybridization 3 covalent bonds

Hexagonal sheets

x ya b=a=120

a = 2 d cos 30°

= √3 dd = 1.42 Åa = 2.46 Å

40

Graphite

x y

a = 2.46 Å c = 6.70 Å

B

A

A

www.scifun.ed.ac.uk/

c

Lattice: Simple HexagonalMotif: 4 carbon atoms

000; 2/3 1/3 0; 2/3 1/3 1/2; 1/3 2/3 1/2

41

Graphite Highly Anisotropic:

Properties are very different in the a and c directions

www.sciencemuseum.org.uk/

Uses:Solid lubricantPencils (clay + graphite, hardness

depends on fraction of clay)carbon fibre

42

DiamondSp3 hybridization 4 covalent bonds

Location of atoms:8 Corners6 face centres4 one on each of the 4 body diagonals

Tetrahedral bonding

43

Diamond Cubic Crystal: Lattice & motif?

AA BB

C

CD

D

x

y

P

P

QQ

RR

S

S

T

T

KK

L

L

MM

N

N

0,1

0,1

0,1

0,1

0,1

41

41

43

43

Diamond Cubic Crystal= FCC lattice + motif:

x

y

21

21

21

21

Projection of the unit cell on the bottom face of the cube

000; ¼¼¼

44

Crystal Structure = Lattice + Motif

Diamond Cubic Crystal Structure

FCCLattice

2 atomMotif

41

41

41

000= +

There are only three Bravais Lattices: SC, BCC, FCC.

Diamond Cubic Lattice

45

There is no diamond cubic lattice.

46

Diamond Cubic

Structure

Effective number of atoms in the unit cell = 881

Corners

Relaton between lattice parameter and atomic radius

ra 243

38ra

Packing efficiency

34.01633

483

3

a

r

Coordination number 4

8621

41

InsideFace

47

Diamond Cubic Crystal StructuresC Si Ge Gray Sn

a (Å) 3.57 5.43 5.65 6.46

480,1 0,1

21

IV-IV compound: SiCIII-V compound:

AlP, AlAs, AlSb, GaP, GaAs, GaSb,

InP, InAs, InSbII-VI compound:

ZnO, ZnS,CdS, CdSe, CdTe

I-VII compound:CuCl, AgI

y

S

0,1 0,1

0,1

41

41

43

43

21

21

21

Equiatomic binary AB compounds having diamond cubic like structure

49

USES:

DiamondAbrasive in polishing and grindingwire drawing dies

Si, Ge, compounds: semiconducting devices

SiCabrasives, heating elements of furnaces

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

Graphite Diamond

Buckminster Fullerene1985

Carbon Nanotubes1991

Graphene2004

Allotropes of C

C60 BuckminsterfullereneH.W. Kroto, J.R. Heath, S.C. O’Brien, R.F. Curl and R.E. SmalleyNature 318 (1985) 162-163

1996 Nobel Prize

Long-chain carbon molecules in interstellar space

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

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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

Rhombusunequal angles not regular

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…

3D: Regular Polyhedra or Platonic SolidsAll faces regular congruent polygons, all corners identical.

Cube

How many regular solids?

Tetrahedron

59

There are 5 and only 5 Platonic or

regular solids !

There are 5 and only 5 Platonic or regular solids !

1. Tetrahedron4 6 42. Octahedron6 12 83. Cube 8 12 64. Icosahedron 12 30

205. Dodecahedron 20 30

12

Duals

Duals

Euler’s Polyhedron FormulaV-E+F=2

V E F

Duality

Tetrahedron Self-DualOctahedron-CubeIcosahedron-Dodecahedron

Proof of Five Platonic Solids

At any vertex at least three faces should meetThe sum of polygonal angles at any vertex should be less the 360

Triangles (60) 3 Tetrahedron4 Octahedron5 Icosahedron6 or more: not possible

Square (90) 3 Cube4 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 nanotubesJeroen W. G. Wilder, Liesbeth C. Venema, Andrew G. Rinzler, Richard E. Smalley & Cees Dekker

zigzig (n,0)

armchair (n,n)

(n,m)=(6,3)

a1a

2

wrapping vector

Structural features of carbon nanotubes

=chiral angle

Material Young's Modulus

(TPa)

Tensile Strength

(GPa)

Elongation at Break (%)

SWNT ~1 (from 1 to 5) 13-53E 16Armchair SWNT

0.94T 126.2T 23.1

Zigzag SWNT

0.94T 94.5T 15.6-17.5

Chiral SWNT

0.92

MWNT 0.8-0.9E 150

Stainless Steel

~0.2 ~0.65-1 15-50Kevlar ~0.15 ~3.5 ~2KevlarT 0.25 29.6

Source: wiki

ElectricalFor 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].

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73

End of lecture 12 (20.08.2013)

Beginning of lecture 13 (21.08.2013)

74

IONIC SOLIDS

Cation radius: R+ Anion radius: R-

1. Cation and anion attract each other.

Usually

RR

2. Cation and anion spheres touch each other

1, 2, 3 => Close packing of unequal spheres

3. Ionic bonds are non-directional

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

Anions not touching the central cation, Anions touching each other

Anions touching the central cationAnions touching

Anions touching central cationAnions not touching each other

155.0a

c

RR 155.0

a

c

RR 155.0

a

c

RR

unstable Critically stable stable

Effect of radius ratio

2155.0 LigancyRR

a

c 3155.0 LigancyRR

a

c

77

3155.0 LigancyRR

a

c

However, when tetrahedral coordinationwith ligancy 4 becomes stable

225.0a

c

RR

Recall tetrahedral void in close-packed structure.

Thus

3225.0155.0 LigancyRR

a

c

78

Table 5.3Ligancy as a Function of Radius Ration

Ligancy Range of radius ratioConfiguration2 0.0 ― 0.155 Linear

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 FCC or HCP

79

Example 1: NaCl

cae2k.com

onCoordinatiOctahedralLigancy

RR

Cl

Na

6732.054.0414.0

54.0

NaCl structure =FCC lattice + 2 atom motif: Cl- 0 0 0

Na ½ 0 0

80

aRRClNa

22 "

NaCl structure continued

CCP of Cl─ with Na+ in ALL octahedral voids

81

seas.upenn.edu

Example 2 : CsCl Structure

191.0732.0

91.0Cl

Cs

RR

Ligancy 8Cubic coordination of Cl- around Cs+

CsCl structure = SC lattice + 2 atom motif: Cl 000

Cs ½ ½ ½ aRR

ClCs322 BCC

82

Example 3: CaF2 (Fluorite or fluorospar)

732.073.0

73.02

F

Ca

RR

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+

83

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+ 000F─ ¼ ¼ ¼F─ -¼ -¼ -¼

84

Example 4: ZnS (Zinc blende or sphalerite)

onCoordinatiOctahedralLigancy

RR

S

Zn

6732.048.0414.0

48.02

2

However, actual ligancy is 4 (TETRAHEDRAL COORDINATION)

Explanation: nature of bond is more covalent than ionic

wikipedia

85

seas.upenn.edu

ZnS structure

CCP of S2─ with Zn2+ in alternate tetrahedral voids

ZnS structure = FCC lattice + 2 atom motif S2─ 0 0 0 Zn2+ ¼ ¼ ¼

86

pixdaus.com

87

theoasisxpress.com

88

89

pixdaus.com

What is common to 1, glass of the window2. sand of the beach, and 3. quartz of the watch?

90

Structure of SiO2

414.29.0225.0

29.02

4

O

Si

RR

Bond is 50% ionic and 50 % covalent

Tetrahedral coordination of O2─ around Si4+

Silicate tetrahedron

91

4+

2─

2─

2─

2─

4─

Silicate tetrahedron electrically unbalanced

O2─ need to be shared between two tetrahedra

92

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

93

2D representation of 3D periodically repeating pattern of tetrahedra in crystalline SiO2. Note that alternate tetrahedra are inverted

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

95

Modification leads to breaking of primary bonds between silicat tetrahedra.

+ Na2O =Na

Na

Network Modification by addition of Soda

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

97

End of lecture 13 (21.08.2013)

Syllabus for minor I (upto lecture 13)

Beginning of lecture 14 (23.08 2013)

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

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

109.5

100

101

5.8 Crystallinity in long chain polymers

Fig. 5.17: semicrystalline polymer

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

Branching

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

104

Tacticity

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

105

Copoymers: polymeric analog of solid solutions

4. Block and random copolymers promote non crystallinity.

106

Plasticizers

Low molecular weight additives

Impedes chains coming together

Reduces crystallization

107

ElastomerPolymers with very extensive elastic deformation

Stress-strain relationship is non-linearExample: Rubber

108

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

109

Isoprene molecule

commons.wikimedia.org

C=C-C=CH H

HH

H

H3C

110

C C C CH H

HH

H

CH3

Isoprene molecule

Polyisoprene mer

C C C C H H

HH H CH3

Polymerization

Liquid(Latex)

111

C C C C H H

HH H CH3

C C C C H H

HH

H CH3

+ 2S

Vulcanisation

Weak van der

Waals bond

112

C C C C H H

HH H CH3

C C C C H H

HH

H CH3

S

Vulcanisation

S

Cross-links

113

Natural rubber Elastomer Ebonite

liquid Elastic solid

Hard & brittle

not x-linked

lightly x-linked

heavilyx-linked

Effect of cross-linking on polyisoprene

114

Charles GoodyearDecember 29, 1800-July 1,

1860Debt at the time of

death $200,000Life 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

End of lecture 14 (23.08.2013)

Beginning of Lecture 15 (27.08.2013) (rubber elasticity + pre minor I

discussion)

116

Another interesting property of elastomers

Thermal behaviour

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

Elastomers have ve thermal expansion coefficient, i.e., they

CONTRACT on heating!!

EXPERIMENT 4

Section 10.3 of the textbook

119

20

00

0

LL

LL

LkTNF

F applied tensile forceN0 number of cross-linksk Boltzmann constantT absolute temperatureL0 initial length (without F)L final length (with F)

120

Experimental

Theory: Chain uncoiling

20

00

0

LL

LL

LkTNF

Bond stretching in straightened out molecules