Transcript
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
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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
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Inorganic Solids
Covalent Solids
Metals and Alloys
Ionic Solids
Silica: crystalline and amorphous
PolymersClassification
Structure
Crystallinity
Mechanical Behaviour
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7th. Group (Halogens): single covalent bondsDiatomic molecules
Weak van der Waals bond between molecules
F2, Cl2: Gas; Br2: Liquid; I2: orthorhombic xl
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6th. Group: two covalent bonds: long zig-zag chains
Weak van der Waals bonds between chains
mostly noncrystalline
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5th. Group: Three covalent bonds: Puckered sheets
Weak van der Waals bond between sheets
Mostly noncrystalline
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4th. Group: Carbon
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Graphite Diamond
Buckminster Fullerene1985
Carbon Nanotubes1991
Graphene2004
Allotropes of C
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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
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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
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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
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DiamondSp3 hybridization 4 covalent bonds
Location of atoms:
8 Corners
6 face centres
4 one on each of the 4 body diagonals
Tetrahedral bonding
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Diamond Cubic Crystal: Lattice & motif?
AA BB
C
CD
D
x
y
P
P
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
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Crystal Structure = Lattice + Motif
Diamond Cubic Crystal Structure
FCCLattice
2 atomMotif
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41
41
000= +
There are only three Bravais Lattices: SC, BCC, FCC.
Diamond Cubic Lattice
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There is no diamond cubic lattice.
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Diamond Cubic
Structure
Effective number of atoms in the unit cell = 88
1
Corners
Relaton between lattice parameter and atomic radius
ra
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3
3
8ra
Packing efficiency
34.016
334
8
3
3
a
r
Coordination number
4
862
1 41
InsideFace
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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
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1
2
1
2
1
Equiatomic binary AB compounds having diamond cubic like structure
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USES:
DiamondAbrasive in polishing and grindingwire drawing dies
Si, Ge, compounds: semiconducting devices
SiCabrasives, heating elements of furnaces
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Inorganic Solids
Covalent Solids
Metals and Alloys
Ionic Solids
Silica: crystalline and amorphous
PolymersClassification
Structure
Crystallinity
Mechanical Behaviour
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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)
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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
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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
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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
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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
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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
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
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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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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
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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)
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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
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Ordered and RandomSubstitutional solid solution
Random Solid Solution
Ordered Solid Solution
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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
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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
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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
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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
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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
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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
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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
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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
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pixdaus.com
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theoasisxpress.com
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pixdaus.com
What is common to 1, glass of the window2. sand of the beach, and 3. quartz of the watch?
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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
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4+
2─
2─
2─
2─
4─
Silicate tetrahedron electrically unbalanced
O2─ need to be shared between two tetrahedra
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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
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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
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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
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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
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Freedom of rotation about each bond in space leads to different conformations of C-C backbone
109.5
72
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5.8 Crystallinity in long chain polymers
Fig. 5.17: semicrystalline polymer
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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
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Branching
2. More is the branching less is the tendency to crystallize
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Tacticity
3. Isotactic and syndiotactic polymers can crystallize but atactic cannot.
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Copoymers: polymeric analog of solid solutions
4. Block and random copolymers promote non crystallinity.
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Plasticizers
Low molecular weight additives
Impedes chains coming together
Reduces crystallization
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Elastomer
Polymers with very extensive elastic deformation
Stress-strain relationship is non-linear
Example: Rubber
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Liquid natural rubber (latex) being collected from the rubber tree
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Isoprene molecule
commons.wikimedia.org
C=C-C=CH H
HH
H
H3C
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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)
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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
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C C C C
H H
HH H CH3
C C C C
H H
HH
H CH3
S
Vulcanisation
S
Cross-links
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Natural rubber
Elastomer Ebonite
liquid Elastic solid
Hard & brittle
not x-linked
lightly x-linked
heavilyx-linked
Effect of cross-linking on polyisoprene
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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.
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Another interesting property of elastomers
Thermal behaviour
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Tensile force
F
Elastomer sample
Elastomer sample
under tension
Coiled chains
straight
chains
heat
Higher entropy
Lower entropy Still
lower entropy
Contracts on heating
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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)
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Experimental
Theory: Chain uncoiling
2
0
00
0
L
L
L
L
L
kTNF
Bond stretching in straightened out molecules
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