Chapter 5 July 2011

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 Solids

Covalent Solids

Metals and Alloys

Ionic Solids

Silica: crystalline and amorphous

PolymersClassification

Structure

Crystallinity

Mechanical Behaviour

6

7th. Group (Halogens): single covalent bondsDiatomic molecules

Weak van der Waals bond between molecules

F2, Cl2: Gas; Br2: Liquid; I2: orthorhombic xl

7

6th. Group: two covalent bonds: long zig-zag chains

Weak van der Waals bonds between chains

mostly noncrystalline

8

5th. Group: Three covalent bonds: Puckered sheets

Weak van der Waals bond between sheets

Mostly noncrystalline

9

4th. Group: Carbon

10

Graphite Diamond

Buckminster Fullerene1985

Carbon Nanotubes1991

Graphene2004

Allotropes of C

11

Graphite

Sp2 hybridization 3 covalent bonds

Hexagonal sheets

x y

a b=a=120

a = 2 d cos 30°

= √3 dd = 1.42 Å

a = 2.46 Å

12

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

13

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

14

DiamondSp3 hybridization 4 covalent bonds

Location of atoms:

8 Corners

6 face centres

4 one on each of the 4 body diagonals

Tetrahedral bonding

15

Diamond Cubic Crystal: Lattice & motif?

AA BB

C

CD

D

x

y

P

P

QQ

R

R

S

S

T

T

KK

L

L

MM

N

N

0,1

0,1

0,1

0,1

0,1

4

1

4

1

4

3

4

3

Diamond Cubic Crystal

= FCC lattice + motif:

x

y

2

1

2

1

2

1

2

1

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

000; ¼¼¼

16

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

17

There is no diamond cubic lattice.

18

Diamond Cubic

Structure

Effective number of atoms in the unit cell = 88

1

Corners

Relaton between lattice parameter and atomic radius

ra

24

3

3

8ra

Packing efficiency

34.016

334

8

3

3

a

r

Coordination number

4

862

1 41

InsideFace

19

Diamond Cubic Crystal StructuresC Si Ge Gray Sn

a (Å) 3.57 5.43 5.65 6.46

200,1 0,1

2

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 0,1

0,1

4

1

4

1

4

3

4

32

1

2

1

2

1

Equiatomic binary AB compounds having diamond cubic like structure

21

USES:

DiamondAbrasive in polishing and grindingwire drawing dies

Si, Ge, compounds: semiconducting devices

SiCabrasives, heating elements of furnaces

22

Inorganic Solids

Covalent Solids

Metals and Alloys

Ionic Solids

Silica: crystalline and amorphous

PolymersClassification

Structure

Crystallinity

Mechanical Behaviour

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

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 74.023

.

EP

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

Coding TheoryInternet data transmission

25

Close packing of equal hard spheres

1-D packing

A chain of spheres

P.E.=

Kissing Number=

Close-packed direction of atoms

=1 2lengthtotal

lengthoccupied

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?

3

areatotal

areaoccupied 907.32

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

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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)

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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 neighbours

Either A or B as lattice points, not both

a

b = a=120

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

A

A

B

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

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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 = 2R

The same B atom also forms an inverted tetrahedron with three A atoms sitting above it

A

A

B

Bc

c = 2 height of a tetrahedron of edge length a

ac3

22

30

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

B

A

C

B

A

C

All atoms are equivalent and their centres form a latticeMotif: single atom 000

ABCABC stacking = CCP crystal

= FCC lattice + single atom motif 000

3 a

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Geometrical properties of ABCABC stacking B

A

C

B

A

C

All atoms are equivalent and their centres form a latticeMotif: single atom 000

ABCABC stacking = CCP crystal

= FCC lattice + single atom motif 000

3 a

32

Geometrical properties of ABCABC stacking B

A

C

B

A

C

All atoms are equivalent and their centres form a latticeMotif: single atom 000

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

ABCABC stacking = CCP crystal

= FCC lattice + single atom motif 000

3 a

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

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Stacking sequence?

ABA: HCP

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36

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

Find the mistake in the following picture:

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Crystal Coordination PackingStructure number efficiency

Diamond cubic (DC) 4 0.32

Simple cubic (SC) 6 0.52

Body centred cubic 8 0.68

Face-centred cubic 12 0.74

Table 5.1

Coordination Number and Packing Efficiency

Empty spaces are distributed in various voids

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Voids in Close-Packed Crystals

A

AA

B

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

39

Location of Voids in FCC Unit cell

C C

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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

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

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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.

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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

Ordered and RandomSubstitutional solid solution

Random Solid Solution

Ordered Solid Solution

45

Disordered solid solution of β-Brass:

Corner and centre both have 50% proibability of being

occupied by Cu or Zn45

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)

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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.

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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

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.

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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

155.0a

c

R

R155.0

a

c

R

R 155.0a

c

R

R

unstable Critically stable stable

Effect of radius ratio

2155.0 LigancyR

R

a

c3155.0 Ligancy

R

R

a

c

50

3155.0 LigancyR

R

a

c

However, when tetrahedral coordination

with ligancy 4 becomes stable

225.0a

c

R

R

Recall tetrahedral void in close-packed structure.

Thus

3225.0155.0 LigancyR

R

a

c

51

Table 5.3

Ligancy 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

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Example 1: NaCl

cae2k.com

onCoordinatiOctahedral

Ligancy

R

R

Cl

Na

6

732.054.0414.0

54.0

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

Na ½ 0 0

53

aRRClNa

22 "

NaCl structure continued

CCP of Cl─ with Na+ in ALL octahedral voids

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seas.upenn.edu

Example 2 : CsCl Structure

191.0732.0

91.0Cl

Cs

R

R

Ligancy 8Cubic coordination of Cl- around Cs+

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

Cs ½ ½ ½

aRRClCs

322 BCC

55

Example 3: CaF2 (Fluorite or fluorospar)

732.073.0

73.02

F

Ca

R

R

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+

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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

57

Example 4: ZnS (Zinc blende or sphalerite)

onCoordinatiOctahedral

Ligancy

R

R

S

Zn

6

732.048.0414.0

48.02

2

However, actual ligancy is 4 (TETRAHEDRAL COORDINATION)

Explanation: nature of bond is more covalent than ionic

wikipedia

58

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

59

pixdaus.com

60

theoasisxpress.com

61

62

pixdaus.com

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

63

Structure of SiO2

414.29.0225.0

29.02

4

O

Si

R

R

Bond is 50% ionic and 50 % covalent

Tetrahedral coordination of O2─ around Si4+

Silicate tetrahedron

64

4+

2─

2─

2─

2─

4─

Silicate tetrahedron electrically unbalanced

O2─ need to be shared between two tetrahedra

65

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

66

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

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

68

Modification leads to breaking of primary bonds between silicat tetrahedra.

+ Na2O =Na

Na

Network Modification by addition of Soda

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

70

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

71

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

109.5

72

73

5.8 Crystallinity in long chain polymers

Fig. 5.17: semicrystalline polymer

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

Branching

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

76

Tacticity

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

77

Copoymers: polymeric analog of solid solutions

4. Block and random copolymers promote non crystallinity.

78

Plasticizers

Low molecular weight additives

Impedes chains coming together

Reduces crystallization

79

Elastomer

Polymers with very extensive elastic deformation

Stress-strain relationship is non-linear

Example: Rubber

80

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

81

Isoprene molecule

commons.wikimedia.org

C=C-C=CH H

HH

H

H3C

82

C C C C

H H

HH

H

CH3

Isoprene molecule

Polyisoprene mer

C C C C

H H

HH H CH3

Polymerization

Liquid

(Latex)

83

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

84

C C C C

H H

HH H CH3

C C C C

H H

HH

H CH3

S

Vulcanisation

S

Cross-links

85

Natural rubber

Elastomer Ebonite

liquid Elastic solid

Hard & brittle

not x-linked

lightly x-linked

heavilyx-linked

Effect of cross-linking on polyisoprene

86

Charles GoodyearDecember 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

Another interesting property of elastomers

Thermal behaviour

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

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

CONTRACT on heating!!

EXPERIMENT 4

Section 10.3 of the textbook

90

2

0

00

0

L

L

L

L

L

kTNF

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

91

Experimental

Theory: Chain uncoiling

2

0

00

0

L

L

L

L

L

kTNF

Bond stretching in straightened out molecules