chapter 5 structure of solids
DESCRIPTION
Chapter 5 Structure of Solids. 6 Lectures. Solids. Crystalline. Noncrystalline. Long-range periodicity. No long-range periodicity. Gives sharp diffraction patterns. Does not give sharp diffraction patterns. Does not have a sharp meliing point. Has sharp melting point. - PowerPoint PPT PresentationTRANSCRIPT
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
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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
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
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
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