Structure, Bonding and Properties
Atomic Arrangements
•In gases there is no order
•In liquids there is short range order
•In solids there is long range range
•The order is determined by the type of atomic bonds
Lattices
•A grid like pattern
•Composed of unit cells
•Unit cells are stacked together endlessly to form the lattice (with no empty spaces between cells)
Scanning Tunneling Microscope Image of Iron in the (110) plane
Structure
• Subatomic level Electronic structure of individual atoms that defines interaction among atoms (interatomic bonding).
• Atomic level Arrangement of atoms in materials (for the same atoms can have different properties, e.g. two forms of carbon: graphite and diamond) • Microscopic structure Arrangement of small grains of material that can be identified by microscopy.
• Macroscopic structure
Structural elements that may be viewed with the naked eye.
Monarch butterfly
~ 0.1 m
Amorphous Solids:
The atoms are not orderly arranged in 3‐D. Some can have order only in two dimensions such as the layered materials (clays, graphite, MoS2 ).
While there is no long range order in the amorphous materials, certain bond distances are maintained and some short range order can be achieved. Crystalline Materials:
Atoms are orderly arranged in 3‐D for long distances. Crystalline solids can be classified as single crystals or monocrystals and polycrystals. Polycrystals exhibit a 2‐D defect known as grain boundaries.
2-D lattice
Lattice: A 3‐dimensional system of points that designate the positions of the components (atoms, ions, or molecules) that make up the substance. Unit Cell: The smallest repeating unit of the lattice.
The lattice is generated by repeating the unit cell in all three dimensions
Crystal Systems
Crystallographers have shown that only seven different types of unit cells are necessary to create all point lattice
Cubic
a= b = c ; α = β = γ = 90
Tetragonal
a= b ≠ c ; α = β = γ = 90
Rhombohedral
a= b = c ; α = β = γ ≠ 90
Hexagonal
a= b ≠ c ; α = β = 90, γ =120
Orthorhombic
a≠ b ≠ c ; α = β = γ = 90
Monoclinic
a≠ b ≠ c ; α = γ = 90 ≠ β
Triclinic
a≠ b ≠ c ; α ≠ γ ≠ β ≠ 90
The basis vectors a, b and c define the unit cell; their magnitudes a, b and c
respectively, are the lattice parameters of the unit cell. The angles b^c, c^a and a^b, are conventionally labelled α, β and γ
respectively.
Bravais Lattices
Many of the seven crystal systems have variations of the basic unit cell. August Bravais (1811‐1863) showed that 14 standards unit cells could describe all possible lattice networks.
The number of ways in which points can be arranged regularly in
3‐D, such that the stacking of unit cells fills space, is not limitless.
Certain unit cells are compatible with body‐centering, face centering or side‐centering. For example the orthorhombic unit cells can be:
C
F
A Bravais lattice is a lattice in which every lattice point has exactly the same environment.
Symmetry
Although the properties of a crystal can be anisotropic, there may be different directions along which they are identical. These directions are said to be equivalent and the crystal is said to possess symmetry.
For example, that a particular edge of a cube cannot be distinguished from any other is a measure of its symmetry; an orthorhombic parallelepiped has lower symmetry, since its edges can be distinguished by length.
oTranslation
oRotation
oReflection (Mirror)
oGlide
oScrew
Translation: Operation required as definition of unit cell.
Rotation:
1‐2‐3‐4 and 6 Fold Rotation Axis corresponding to angles of rotation of 360, 180, 120, 90 and 60 degrees.
Although objects themselves may appear to have 5‐fold, 7‐fold, 8‐
fold, or higher‐fold rotation axes, these are not possible in crystals. The reason is that the external shape of a crystal is based on a
geometric arrangement of atoms. Note that if we try to combine objects with 5‐foldand 8‐fold apparent symmetry, that we cannot combine them in such a way that they completely fill space, as illustrated below.
Mirror Symmetry
A mirror symmetry operation is an imaginary operation that can be performed to reproduce an object. The operation is done by imagining that you cut the object in half, then place a mirror next to one of the halves of the object along the cut. If the reflection in the mirror reproduces the other half of the object, then the object is said to have mirror symmetry. The plane of the mirror is an element of symmetry referred to as a mirror plane, and is symbolized with the letter m.
Center of Symmetry
Another operation that can be performed is
inversion through a point.
Rotoinversion
Combinations of rotation with a center of symmetry perform the symmetry operation of rotoinversion.
2‐fold Rotoinversion ‐ The operation of 2‐
fold rotoinversion involves first rotating the object by 180o then inverting it through an inversion center. This operation is equivalent to having a mirror plane perpendicular to the 2‐fold rotoinversion axis.
3‐fold Rotoinversion ‐ This involves rotating the object by 120o (360/3 = 120), and inverting through a center. A cube is good example of an object that possesses 3‐
fold rotoinversion axes. A 3‐fold rotoinversion axis is denoted as
Thus, this crystal has the following symmetry elements:
1 ‐ 4‐fold rotation axis (A4)
4 ‐ 2‐fold rotation axes (A2), 2 cutting the faces & 2 cutting the edges.
5 mirror planes (m), 2 cutting across the faces, 2 cutting through the edges, and one cutting horizontally through the center.
Note also that there is a center of symmetry (i).
The symmetry content of this crystal is thus: i, 1A4, 4A2, 5m
Later you will see that this belongs to
crystal class 4/m2/m2/m.
Certain Bravais lattice types are compatible with some symmetry operations:
14 Bravais Lattices + Compatible Symmetry Elements
Î32 Crystal Symmetry Classes
Inversion
Mirror
Rotation
Rotations
Mirrors
Improper rotations
14 Bravais Lattices + Compatible Symmetry Elements
Î32 Crystal Symmetry Classes
Two Translational Symmetry Elements
32 Crystal Symmetry Classes + Translational Symmetry Operations Æ
230 Space Groups
All combinations of point symmetry elements are not possible
•A three fold axis can not just have one two fold axis perpendicular to it.
•In three dimensions the existence of two perpendicular two folds
implies the existence of a third perpendicular two fold
•The allowed combinations of point symmetry elements are called point groups
Point symmetry elements compatible with 3D translations
Only 32 point groups are consistent with periodicity in 3D.
The 32 point groups
Schönflies and Hermann‐
Maugin symbols for
crystallographic point
groups
Combining symmetry elements
For three dimensions
32 point groups
14 Bravais lattices
but only 230 space groups
For two dimensions
5 lattices
10 point groups
but only 17 plane groups
Examples: Two‐fold Rotation ………….. Mirror Plane
Space groups are numbered (1-230) from the lowest to the highest
symmetry.
First letter of space group notation indicates the type of unit cell:
P = primitive
I = Body-centered
F = Face-centered
C = side-centered
Other symbols indicate symmetry operations:
m = mirror plane
3 = three-fold rotation axis
21= two-fold screw axis
c = glide axis
Polymorphism or Allotropy
Many elements or compounds exist in more than one crystalline
form under different conditions of temperature and pressure.
This phenomenon is termed polymorphism and if the material is
an elemental solid is called allotropy.
Example:
Iron (Fe – Z = 26)
liquid above 1539 C.
δ-iron (BCC) between 1394 and 1539 C.
γ-iron (FCC) between 912 and 1394 C.
α-iron (BCC) between -273 and 912 C.
α iron
BCC
912oC
γ iron
FCC
1400oC
δ iron
BCC
1539oC
liquid iron
Another example of allotropy is carbon.
Pure, solid carbon occurs in three crystalline forms – diamond,
graphite; and large, hollow fullerenes. Two kinds of fullerenes are
shown here: buckminsterfullerene (buckyball) and carbon nanotube.
Crystallographic Planes and Directions
Atom Positions in Cubic Unit Cells
A cube of lattice parameter a is considered to have a side equal to
unity. Only the atoms with coordinates x, y and z greater than or
equal to zero and less than unity belong to that specific cell.
z
0,0,1
1,0,1
0,1,1
1,1,1
½, ½, ½
1,0,0
0,1,0
0,0,0
1,1,0
x
y
0 ≤ x, y , z < 1
Directions in The Unit Cell
For cubic crystals the crystallographic directions indices are the
vector components of the direction resolved along each of the
coordinate axes and reduced to the smallest integer.
z
Example direction A
0,0,1
1,0,1
0,1,1
1,1,1
½, ½, ½
1,0,0
0,1,0
0,0,0 A
1,1,0
x
y
a) Two points origin coordinates
0,0,0 and final position
coordinates 1,1,0
b) 1,1,0 - 0,0,0 = 1,1,0
c) No fractions to clear
d) Direction [110]
z
0,0,1
C
B
0,0,0
x
Example direction B
a) Two points origin coordinates
1,1,1 and final position
coordinates 0,0,0
b) 0,0,0 - 1,1,1 = -1,-1,-1
c) No fractions _to_ clear
_
d) Direction [111]
Example direction C
1,1,1
a) Two points origin coordinates
½,1,0 and final position
coordinates 0,0,1
y
b) 0,0,1 - ½,1,0 = -½,-1,1
½, 1, 0
c) There are fractions to clear.
Multiply times 2. 2( -½,-1,1) =
_ _
-1,-2,2
d) Direction [1 2 2]
Notes About the Use of Miller Indices for Directions
‰ A direction and its negative are not identical; [100] is not equal to
[bar100]. They represent the same line but opposite directions. .
‰ direction and its multiple are identical: [100] is the same direction
as [200]. We just forgot to reduce to lowest integers.
‰ Certain groups of directions are equivalent; they have their
particular indices primarily because of the way we construct the coordinates. For example, a [100] direction is equivalent to the [010]
direction if we re-define the co-ordinates system. We may refer to
groups of equivalent directions as directions of the same family. The
special brackets < > are used to indicate this collection of directions.
Example:
The family of directions <100> consists of six equivalent directions
< 100 > ≡ [100], [010], [001], [010], [001 ], [ 100]
Miller Indices for Crystallographic planes in Cubic Cells
¾ Planes in unit cells are also defined by three integer numbers,
called the Miller indices and written (hkl).
¾ Miller’s indices can be used as a shorthand notation to identify
crystallographic directions (earlier) AND planes.
Procedure for determining Miller Indices
ƒ locate the origin
ƒ identify the points at which the plane intercepts the x, y and z
coordinates as fractions of unit cell length. If the plane passes
through the origin, the origin of the co-ordinate system must be
moved!
ƒ take reciprocals of these intercepts
ƒ clear fractions but do not reduce to lowest integers
ƒ enclose the resulting numbers in parentheses (h k l).
Again, the
negative numbers should be written with a bar over the number.
z
A
x
Example: Miller indices for plane A
a) Locate the origin of coordinate.
b) Find the intercepts x = 1, y = 1, z = 1
c) Find the inverse 1/x=1, 1/y=1, 1/z=1
d) No fractions to clear
y e) (1 1 1)
More Miller Indices - Examples
c
c
c
1/5
0.5
a
2/3
b
b
b
a
0.5
a
c
c
c
b
a
b
a
b
a
Notes About the Use of Miller Indices for Planes
‰ A plane and its negative are parallel and identical.
‰ Planes and its multiple are parallel planes: (100) is parallel to the
plane (200) and the distance between (200) planes is half of the
distance between (100) planes.
Certain groups of planes are equivalent (same atom distribution);
they have their particular indices primarily because of the way we
construct the co-ordinates. For example, a (100) planes is
equivalent to the (010) planes. We may refer to groups of
equivalent planes as planes of the same family. The special
brackets { } are used to indicate this collection of planes.
‰ In cubic systems the direction of miller indices [h k l] is normal
o perpendicular to the (h k l) plane.
‰ in cubic systems, the distance d between planes (h k l ) is given
by the formula
where a is the lattice
a
d=
constant.
h2 + k 2 + l 2
Example:
The family of planes {100} consists of three equivalent planes (100)
, (010) and (001)
A “family” of crystal planes contains all those planes are crystallographically equivalent.
• Planes have the same atomic packing density
• a family is designated by indices that are enclosed by braces.
- {111}:
(111 ), (111 ), (111), (111), (111), (111), (111), (111)
• Single
Crystal
• Polycrystalline materials
• Anisotropy and isotropy
Two Types of Indices in the Hexagonal System
a1 ,a2 ,and c are independent, a3 is not!
c
a3 =
a3
Miller:
a2
(hkl)
(a1 + a2)
-
(same as before)
Miller-Bravais: (hkil) → i = - (h+k)
a1
(001) = (0001)
c
-
-
c
(110) = (1100)
a3
a3
a2
a2
a1
a1
-
(110) = (1100)
(100) = (1010)
Structures of Metallic Elements
H
He
Li
Be
B
C
N
O
F
Ne
Na
Mg
Al
Si
P
S
Cl
Ar
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
Cs
Ba
La
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
Fr
Ra
Ac
Prim itive Cubic
Cubic close packing
(Face centered cubic)
Body Centered Cubic
Hexagonal close packing
Structure
Metal
Lattice Constant
a, nm
BCC
FCC
HCP
c, nm
Atomic
Radius, nm
Chromium
0.289
0.125
Iron
0.287
0.124
Molybdenum
0.315
0.136
Potassium
0.533
0.231
Sodium
0.429
0.186
Tungsten
0.316
0.137
Aluminum
0.405
0.143
Copper
0.361
0.128
Gold
0.408
0.144
Nickel
0.352
0.125
Silver
0.409
0.144
Zinc
0.2665
0.5618
0.133
Magnesium
0.3209
0.5209
0.160
Cobalt
0.2507
0.4069
0.125
Titanium
0.2950
0.3584
0.147
SIMPLE CUBIC STRUCTURE (SC)
• Rare due to poor packing (only Po has this structure)
• Close-packed directions are cube edges.
• Coordination # = 6 (# nearest neighbors)
• Number of atoms per unit cell= 1 atom
Atomic Packing Factor (APF)
Volume of atoms in unit cell*
APF =
Volume of unit cell
*assume hard spheres
• APF for a simple cubic structure = 0.52
atoms
unit cell
a
R=0.5a
close-packed directions
contains 8 x 1/8 =
1 atom/unit cell
APF =
volume
atom
4
π (0.5a)3
1
3
a3
volume
unit cell
6
Body Centered Cubic (BCC)
• Close packed directions are cube diagonals.
--Note: All atoms are identical; the center atom is shaded differently only for ease of
viewing.
• Coordination # = 8
Unit cell contains:
1 + 8 x 1/8
= 2 atoms/unit cell
Close-packed directions:
length = 4R
= 3a
R
a
atoms
volume
4
3
π ( 3a/4)
2
unit cell
atom
3
APF =
• APF for a BCC = 0.68
volume
a3
unit cell
Face Centered Cubic (FCC)
• Close packed directions are face diagonals.
--Note: All atoms are identical; the face-centered atoms are shaded differently only for
ease of viewing.
• Coordination # = 12
Close-packed directions:
length = 4R
= 2a
Unit cell contains:
6 x 1/2 + 8 x 1/8
= 4 atoms/unit cell
a
atoms
volume
4
3
π ( 2a/4)
4
unit cell
atom
3
APF =
• APF for a FCC = 0.74
volume
3
a
unit cell
Hexagonal Close-Packed (HCP)
The APF and coordination number of the HCP structure is the same
as the FCC structure, that is, 0.74 and 12 respectively.
An isolated HCP unit cell has a total of 6 atoms per unit cell.
2 atoms shared by two cells = 1
atom per cell
3 atoms
12 atoms shared by six cells = 2 atoms per cell
Close-Packed Structures
Both the HCP and FCC crystal structures are close-packed structure.
Consider the atoms as spheres:
‰Place one layer of atoms (2 Dimensional solid). Layer A
‰Place the next layer on top of the first. Layer B. Note that there are
two possible positions for a proper stacking of layer B.
The third layer (Layer C) can be placed in also teo different
positions to obtain a proper stack.
(1)exactly above of atoms of Layer A (HCP) or
(2)displaced
A B A : hexagonal close packed
A B C : cubic close packed
A
B
C
A
A B C : cubic close pack
A B A : hexagonal close pack
90°
A
B
A
120°
Packing of Non-Identical Spheres
The holes between the atoms of a crystal, called interstices,
can house smaller atoms without appreciable distortion of the
host.
Many compounds of two or more elements have a structure,
which can be described by the smaller atoms/ions filling the
interstices between the larger atoms/ions. Different structures
arise from the different numbers and sizes of the interstices in
the fcc, hcp, bcc and simple cubic structures. The way the
interstices are distributed is also important.
Two important interstices are the tetragonal and the
octahedral interstices in close packed structures
Tetrahedral/Octahedral
The tetragonal interstice, surrounded by four
atoms.
The octahedral interstice, surrounded by six
atoms. The six atoms surround (or coordinate)
the interstice in the shape of an octahedron.
(Also consider as three atoms below and three
atoms above.)
FCC Interstitials
FCC – Octahedral
In the fcc structure, consider the interstitial site
shown. Six host atoms surround it. These six
atoms surround (or coordinate) the interstitial site
in the shape of an octahedron. There is one
octahedral site at the centre of the FCC cell
(½,½,½) and one on each of the twelve cell edges
(½,0,0). A total of 13 octahedral sites.
Calculate the octahedral void radius as a fraction of the parent atom radius in a FCC structure.
rvoid
= 0.414
ratom
FCC ‐ Tetrahedron In the fcc structure, consider the interstitial site shown. Four atoms surround it. These four atoms surround the interstitial site in the shape of a tetrahedron. There are eight tetrahedral sites in the FCC unit cell located at (¼,¼,¼). Calculate the tetrahedral void radius as a fraction of the parent atom radius in a FCC structure.
rvoid
= 0.225
ratom
HCP Interstitials
HCP – Octahedral
In the hcp structure, consider the interstitial
site shown. Six host atoms surround it.
These six atoms surround the interstitial
site in the shape of an octahedron. There
are six octahedral sites.
HCP – Tetragonal
In the hcp structure, consider the interstitial
site shown. Four atoms surround it. These
four atoms surround the interstitial site in
the shape of a tetrahedron. Total of 8
tetrahedral sites.
BCC Interstitials
Note: the bcc structure is not close
packed. In the bcc structure the
octahedron and tetrahedron are not
regular, they do not have edges of equal
lengths.
BCC Octahedral
In the bcc structure, consider the
interstitial site shown. Six host atoms
surround it. These six atoms surround
the interstitial site in the shape of an
octahedron. There is one octahedral site
on each of the six BCC cell faces (½,½,0)
and one on each of the twelve cell edges
(½,0,0). Total of 18 sites.
BCC Tetrahedral
In the bcc structure, consider the interstitial site shown. Four atoms surround it. These four atoms surround the interstitial site in the shape of a tetrahedron. There are four tetrahedral sites on each of the six BCC cell faces (½,¼,0).
Total of 24 sites.
Using this diagram calculate the octahedral void radius as a fraction of the parent atom radius in a BCC crystal
rvoid
= 0.155
ratom
Using this diagram calculate the tetrahedral void radius as a fraction of the parent atom radius in a BCC crystal
rvoid
= 0.291
ratom
Interstitial sites
Locations between the ‘‘normal’’ atoms or ions in a crystal into
which another - usually different - atom or ion is placed.
o Cubic site - An interstitial position that has a coordination number
of eight. An atom or ion in the cubic site touches eight other atoms
or ions.
o Octahedral site - An interstitial position that has a coordination
number of six. An atom or ion in the octahedral site touches six
other atoms or ions.
o Tetrahedral site - An interstitial position that has a coordination
number of four. An atom or ion in the tetrahedral site touches four
other atoms or ions.
Crystals having filled Interstitial Sites
Octahedral, Oh, Sites
FCC Lattice has:
3 [=12(¼)] Oh sites at edge centers
+ 1 Oh site at body center
Tetrahedral, Th, Sites
FCC Lattice has:
8 Th sites at ¼, ¼, ¼ positions
Interstitial sites are important because we can derive more structures from these basic
FCC, BCC, HCP structures by partially or completely different sets of these sites
Interstitial coordinates
It is easy to identify the atomic coordinates of the
interstitials in the fcc and bcc structures. Use the
unit cell diagrams below to help identify the
interstitial positions in the hcp structure.
Density Calculations
Since the entire crystal can be generated by the repetition of the
unit cell, the density of a crystalline material, ρ = the density of the
unit cell = (atoms in the unit cell, n ) × (mass of an atom, M) / (the
volume of the cell, Vc)
Atoms in the unit cell, n = 2 (BCC); 4 (FCC); 6 (HCP)
Mass of an atom, M = Atomic weight, A, in amu (or g/mol) is
given in the periodic table. To translate mass from amu to grams
we have to divide the atomic weight in amu by the Avogadro
number NA = 6.023 × 1023 atoms/mol
The volume of the cell, Vc = a3 (FCC and BCC)
a = 2R√2 (FCC);
a = 4R/√3 (BCC)
where R is the atomic radius.
Density Calculation
n: number of atoms/unit cell
nA
ρ=
VC N A
A: atomic weight
VC: volume of the unit cell
NA: Avogadro’s number
(6.023x1023 atoms/mole)
Example Calculate the density of copper.
RCu =0.128nm, Crystal structure: FCC, ACu= 63.5 g/mole
n = 4 atoms/cell,
VC = a 3 = (2 R 2 ) 3 = 16 2 R 3
(4)(63.5)
3
ρ=
=
8
.
89
g
/
cm
[16 2 (1.28 × 108 ) 3 × 6.023 × 10 23 ]
8.94 g/cm3 in the literature
Example
Rhodium has an atomic radius of 0.1345nm (1.345A) and a density of 12.41g.cm-3.
Determine whether it has a BCC or FCC crystal structure. Rh (A = 102.91g/mol)
Solution
nA
ρ=
VC N A
n: number of atoms/unit cell
VC: volume of the unit cell
A: atomic weight
NA: Avogadro’s number
(6.023x1023 atoms/mole)
Vc a 3
A
102.91g .mol −1
=
=
=
= 1.3768 x10 − 23 cm 3 = 0.01376nm 3
−3
−1
23
n
n
ρN A 12.41g .cm 6.023x10 atoms.mole
If rhodium is BCC then n = 2 and a 3 = (4r ) 3 = 12.316r 3
3
a 3 12.316 x(0.1345nm) 3
=
= 0.0149nm 3
2
n
If rhodium is FCC then n = 4 and a 3 = (4r ) 3 = 22.627 r 3
2
a 3 22.627 x(0.1345nm) 3
=
= 0.01376nm 3
4
n
Rhodium has a FCC structure
Linear And Planar Atomic Densities
4R
Ll = 2a ⇒ a =
3
Linear atomic density = 2R/Ll
= 0.612
Planar atomic density:
Crystallographic direction
A’
B’
Ll
= 2π R2/(Area A’D’E’B’)
Structure of Ceramics
Ceramics
keramikos - burnt stuff in Greek - desirable properties of
ceramics are normally achieved through a high temperature heat
treatment process (firing).
Usually a compound between metallic and nonmetallic elements
Always composed of more than one element (e.g., Al2O3, NaCl,
SiC, SiO2)
Bonds are partially or totally ionic, can have combination of ionic
and covalent bonding (electronegativity)
Generally hard, brittle and electrical and thermal insulators
Can be optically opaque, semi-transparent, or transparent
Traditional ceramics – based on clay (china, bricks, tiles,
porcelain), glasses.
“New ceramics” for electronic, computer, aerospace industries.
Crystal Structures in Ceramics with
predominantly ionic bonding
Crystal structure is defined by
The electric charge: The crystal must remain electrically
neutral. Charge balance dictates chemical formula (Ca2+ and Fform CaF2).
Relative size of the cation and anion. The ratio of the atomic
radii (rcation/ranion) dictates the atomic arrangement. Stable
structures have cation/anion contact.
rCation
rAnion
Coordination Number: the number of anions nearest neighbors for a
cation.
As the ratio gets larger (that is as rcation/ranion ~ 1) the coordination
number gets larger and larger.
Holes in sphere packing
Triangular
Tetrahedral
Octahedral
Calculating minimum radius ratio
for a triangle:
O
B
B
O
A
C
C
A
1
AB = ra
2
AO = ra + rc
1
AB = ra
2
AO = ra + rc
1 AB
2
= cosα (α = 30°)
AO
ra
3
= cos 30 =
ra + rc
2
1 AB
2
= cosα (α = 45°)
AO
ra
2
o
= cos 45 =
ra + rc
2
rc
= 0.155
ra
rc
= 0.414
ra
for an octahedral hole
C.N. = 2
rC/rA < 0.155
C.N. = 3
0.155 < rC/rA < 0.225
C.N. = 4
0.225 < rC/rA < 0.414
C.N. = 6
0.414 < rC/rA < 0.732
C.N. = 8
0.732 < rC/rA < 1.0
Ionic (and other) structures may be derived from the occupation of
interstitial sites in close-packed arrangements.
Comparison between structures with filled octahedral and tetrahedral holes
o/t
all oct.
all tetr.
o/t (all)
½t
(½ o
fcc(ccp)
NaCl
CaF2
(Li3Bi)
sphalerite (ZnS)
CdCl2
hcp
NiAs
(ReB2)
(Na3As)
wurtzite (ZnS)
CdI2)
Location and number of tetrahedral
holes in a fcc (ccp) unit cell
- Z = 4 (number of atoms in the unit cell)
- N = 8 (number of tetrahedral holes in the
unit cell)
Crystals having filled Interstitial Sites
Octahedral, Oh, Sites
Ionic Crystals prefer the NaCl
Structure:
NaCl structure has Na+ ions
at all 4 octahedral sites
Na+ ions
Cl- ions
• Large interatomic distance
• LiH, MgO, MnO, AgBr, PbS,
KCl, KBr
Crystals having filled Interstitial Sites
Tetrahedral, Th, Sites
Both the diamond cubic structure
And the Zinc sulfide structures
have 4 tetrahedral sites occupied
and 4 tetrahedral sited empty.
Zn atoms
S atoms
Covalently Bonded Crystals Prefer
this Structure
• Shorter Interatomic Distances
than ionic
• Group IV Crystals (C, Si, Ge, Sn)
• Group III--Group V Crystals
(AlP, GaP, GaAs, AlAs, InSb)
• Zn, Cd – Group VI Crystals
(ZnS, ZnSe, CdS)
• Cu, Ag – Group VII Crystals
(AgI, CuCl, CuF)
The "zinc blende" lattice is face centered cubic (fcc) with two atoms in the base at (0,0,0) and (¼, ¼, ¼). AX Type Crystal Structures
Rock Salt Structure (NaCl)
NaCl structure: rC = rNa = 0.102 nm,
rA = rCl = 0.181 nm rC/rA = 0.56
Coordination = 6
Cl
Na
NaCl, MgO, LiF, FeO, CoO
Cesium Chloride Structure (CsCl)
CsCl Structure: rC = rCs = 0.170 nm,
rA = rCl = 0.181 nm → rC/rA = 0.94
Coordination = 8
Cl
Cs
Is this a body centered cubic structure?
Zinc Blende Structure (ZnS)
radius ratio = 0.402
Coordination = 4
S
Zn
ZnS, ZnTe, SiC have this crystal structure
AmXp-Type Crystal Structures
If the charges on the cations and anions are not the same, a compound
can exist with the chemical formula AmXp , where m and/or p ≠ 1. An
example would be AX2 , for which a common crystal structure is
found in fluorite (CaF2).
CaF2 Fluorite
The lattice is face centered cubic (fcc) with three atoms in the base, one kind (the cations) at (0,0,0), and the other two (anions of the same kind) at (¼, ¼, ¼), and (¼, ¾, ¼). Fluorite CaF2
Fluorite (CaF2): rC = rCa = 0.100 nm, rA
= rF = 0.133 nm ⇒ rC/rA = 0.75
From the table for stable geometries we
see that C.N. = 8
Other compounds that have this crystal
structure include UO2 , PuO2 , and ThO2.
AmBnXp-Type Crystal Structures
It is also possible for ceramic compounds to have more than one
type of cation; for two types of cations (represented by A and B),
their chemical formula may be designated as AmBnXp . Barium
titanate (BaTiO3), having both Ba2+ and Ti4+ cations, falls into this
classification. This material has a perovskite crystal structure and
rather interesting electromechanical properties
Perovskite an Inorganic Chameleon
ABX3 - three compositional variables, A,
B and X
•
•
•
•
•
• (Y1/3Ba2/3)CuO3-x superconductor
• NaxWO3 - mixed conductor;
electrochromic
• SrCeO3 - H - protonic conductor
CaTiO3 - dielectric
• RECoO3-x - mixed conductor
BaTiO3 - ferroelectric
• (Li0.5-3xLa0.5+x)TiO3 - lithium ion
Pb(Mg1/3Nb2/3)O3 - relaxor
conductor
ferroelectric
Pb(Zr1-xTix)O3 - piezoelectric • LaMnO3-x - Giant magneto(Ba1-xLax)TiO3 – semiconductor resistance
The lattice is essentially cubic primitive, but may be distorted to some extent and then becomes orthorhombic or worse. It is also known as the BaTiO3 or CaTiO3 lattice and has three different atoms in the base. In the example it would be Ba at (0,0,0), O at (½, ½, ,0)
and Ti at (½, ½, ½). A particular interesting perovskite (at high pressures) is MgSiO3. It is assumed to form the bulk of the mantle of the earth, so it is the most abundant stuff on this planet, neglecting its Fe/Ni core. The mechanical properties (including the movement of dislocations) of this (and related) minerals are essential for geotectonics ‐ forming the continents, making and quenching volcanoes, earthquakes The perovskite structure CaTiO3
- TiO6 – octahedra
- CaO12 – cuboctahedra
(Ca2+ and O2- form a cubic close
packing)
→ preferred basis structure of
piezoelectric, ferroelectric and
superconducting materials
Perovskite Structure
ABO3
e.g. KNbO3
SrTiO3
LaMnO3
SrTiO3 cubic, a = 3.91 Å
In SrTiO3, Ti-O = a/2 = 1.955 Å
Sr-O = a√2/2 = 2.765 Å
CN of A=12, CN of B=6
OR
The fractional coordinates for cubic perovskite are:
A = (½, ½, ½)
B = (0, 0, 0)
A = ( 0, 0, 0)
OR
X = (½,0,0) (0,½,0) (0,0,½)
B = (½, ½, ½)
X = (½,½,0) (½,0,½) (0,½ ½)
Draw one of these as a projection.
In SrTiO3, Ti-O ~ 1.95 Å
a typical bond length for Ti-O;
stable as a cubic structure
larger
In BaTiO3, Ti-O is stretched, > 2.0 Å
Too long for a stable structure.
Ti displaces off its central position
towards one oxygen
→ square pyramidal coordination
This creates a net dipole moment :
Displacement by 5-10% Ti-O bond length
Random dipole orientations
paraelectric
Aligned dipole orientations
ferroelectric
Under an applied electric field, dipole orientations can be
reversed, i.e. the structure is polarisable
Dipoles tend to be ‘frozen in’ at room temperature; as increase
temperature, thermal vibrations increase the polarisability
BaTiO3 Phase
Transitions
Cubic (Pm3m)
T > 393 K
Ti-O Distances (Å)
In the cubic structure BaTiO3 is
paraelectric. That is to say that the
orientations of the ionic
displacements are not ordered and
dynamic.
6×2.00
Tetragonal (P4mm)
273 K < T < 393 K
Ti-O Distances (Å)
1.83, 4×2.00, 2.21
Below 393 K BaTiO3 becomes
Toward a corner
ferroelectric and the displacement
of the Ti4+ ions progressively
Orthorhombic (Amm2)
displace upon cooling.
183 K < T < 273 K
Ti-O Distances (Å)
2×1.87, 2×2.00, 2×2.17
Toward an edge
Rhombohedral (R3m)
183 K < T < 273 K
Ti-O Distances (Å)
3×1.88, 3×2.13
Toward a face
See Kwei et al. J. Phys. Chem. 97, 2368 (1993),
Density Calculations in Ceramic Structures
ρ=
n'×( ∑ AC + ∑ AA )
Vc × N A
n’: number of formula units in unit cell (all ions that are included
in the chemical formula of the compound = formula unit)
ΣAC: sum of atomic weights of cations in the formula unit
ΣAA: sum of atomic weights of anions in the formula unit
Vc: volume of the unit cell
NA: Avogadro’s number, 6.023x1023 (formula units)/mol
Example: NaCl
n’ = 4 in FCC lattice
ΣAC = ANa = 22.99 g/mol
ΣAA = ACl = 35.45 g/mol
rCl=0.181x10-7 cm
rNa=0.102x10-7
Vc = a3 = (2rNa+2rCl)3
Vc = (2×0.102×10-7 + 2×0.181×10-7)3 cm3
ρ=
[2 × 0.102 ×10
4 × (22.99 + 35.45)
−7
+ 2 × 0.181× 10
] × 6.023 ×10
−7 3
23
= 2.14g.cm −3
Structure, Bonding and Properties
•BaTiO3 : Ferroelectric (TC ~ 130°C, εr > 1000)
– Ba2+ ion stretches the octahedra (Ti-O dist. ~ 2.00Å), this lowers
energy of CB (LUMO) and stabilizes SOJT dist.
•SrTiO3 : Insulator, Normal dielectric (εr ~ x)
– Sr2+ ion is a good fit (Ti-O dist. ~ 1.95Å), this compound is close to a
ferroelectric instability and is called a quantum paraelectric.
•PbTiO3 : Ferroelectric (TC ~ 490°C)
– Displacements of both Ti4+ and Pb2+ (6s26p0 cation) stabilize
ferroelectricity
•BaSnO3 : Insulator, Normal dielectric (εr ~ x)
– Main group Sn4+ has no low lying t2g orbitals and no tendency toward
SOJT dist.
•KNbO3 : Ferroelectric (TC ~ x)
– Behavior is very similar to BaTiO3
•KTaO3 : Insulator, Normal dielectric (εr ~ x)
– Ta 5d orbitals are more electropositive and have a larger spatial extent
than Nb 4d orbitals (greater spatial overlap with O 2p), both effects raise
the energy of the t2g LUMO, diminishing the driving force for a SOJT
dist.
Transformations
Many physical properties depend on the crystallographic directions and the anisotropy of a material is best described by tensors. (a)Tensors can be used to describe physical properties
(b) Symmetry effects on physical properties can be described by how the tensor transforms under a symmetry operation (c) the magnitude of a property in any arbitrary direction can be evaluated by transforming the tensor.
(d) It can be used to draw a geometric representation of the property.
(e) It provides a way of averaging the properties over a certain direction.
(f) It relates properties of single crystals with polycrystals
Tensor: A specific type of matrix representation that can relate the directionality of either a material property (property tensors –
conductivity, elasticity) or a condition/state (condition tensors – stress, strain).
Tensor of zero‐rank: scalar quantity (density, temperature).
Tensor of first‐rank: vector quantity (force, electric field, flux of atoms).
Tensor of second‐rank: relates two vector quantities (flux of atoms with concentration gradient).
Tensor third‐rank: relates vector with a second rank tensor (electric field with strain in a piezoelectric material)
Tensor Fourth‐rank: relates two second rank tensors (relates strain and stress – Elasticity) The key to understanding property or condition tensors is to recognize that tensors can be specified with reference to some coordinate system which is usually defined in 3‐D space by orthogonal axes that obey a right‐hand rule.
Rotation Matrix and Euler Angles: Several schemes can be used to produce a rotation matrix. The three Euler angles are given as three counterclockwise rotations: (a)A rotation about a z‐axis, defined as φ1 (b)A rotation about the new x‐axis, defined as Φ
(c)A rotation about the second z‐position, defined as φ2
The rotation matrix a is given by the matrix multiplication of the rotation matrices of each individual rotations:
⎡ cos φ2
[a ] = aφ 2 ⋅ aΦ ⋅ aφ1 = ⎢⎢− sin φ2
⎢⎣ 0
sin φ2
cos φ2
0
0 ⎤ ⎡ cos φ1 sin φ1 0⎤ ⎞
0⎤ ⎛ ⎡1
⎜
⎟
⎥
⎢
⎥
⎢
⎥
0⎥ × ⎜ ⎢0 cos Φ sin Φ ⎥ × ⎢− sin φ1 cos φ1 0⎥ ⎟
0
1⎥⎦ ⎟⎠
0
1⎥⎦ ⎜⎝ ⎢⎣0 − sin Φ cos Φ ⎥⎦ ⎢⎣ 0
cos φ2 sin φ1 + cos Φ cos φ1 sin φ2 sin Φ sin φ2 ⎤
⎡ cos φ2 cos φ1 − cos Φ sin φ1 sin φ2
[a ] = ⎢⎢− sin φ2 cosφ1 − cos Φ sin φ1 cosφ2 − sin φ2 sin φ1 + cos Φ cosφ1 cosφ2 sin Φ cosφ2 ⎥⎥
⎢⎣
− sin Φ cos φ1
sin Φ sin φ1
cos Φ ⎥⎦
Mathematically, the transformation converts a set of orthogonal axes (X1, Y1, Z1) into another (X2, Y2, Z2). The two set of axes are related to one another by nine direction cosines (a11, a12, a13, a21, a22, a23, a31, a32, a33) . The first subscript refers to the new axis.