Structure, Bonding and Properties

Atomic ArrangementsAtomic 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

Scanning Tunneling Microscope Image of Iron in the (110) plane

LatticesLattices•A grid like pattern

•Composed of unit cells

•Unit cells are stacked together endlessly to form the lattice (with no empty spaces between cells)

• 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 structureStructural elements that may be viewed with the naked eye.

Structure

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 SystemsCrystallographers have shown that only seven different types of unit cells are necessary to create all point latticeCubic a= b = c ; α = β = γ = 90Tetragonal a= b ≠ c ; α = β = γ = 90Rhombohedral a= b = c ; α = β = γ ≠ 90Hexagonal a= b ≠ c ; α = β = 90, γ =120Orthorhombic a≠ b ≠ c ; α = β = γ = 90Monoclinic 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 crespectively, are the lattice parameters of the unit cell. The angles b^c, c^a and a^b, are conventionally labelled α, β and γrespectively.

Bravais LatticesMany 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 in3‐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:

A Bravais lattice is a lattice in which every lattice point has exactly the same environment.

C F

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.

oTranslationoRotationoReflection (Mirror)oGlideoScrew

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 ageometric 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 SymmetryA 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 SymmetryAnother operation that can be performed isinversion through a point.

RotoinversionCombinations 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

The symmetry content of this crystal is thus: i, 1A4, 4A2, 5mLater you will see that this belongs tocrystal class 4/m2/m2/m.

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).

Certain Bravais lattice types are compatible with some symmetry operations:

14 Bravais Lattices + Compatible Symmetry Elements 32 Crystal Symmetry Classes

14 Bravais Lattices + Compatible Symmetry Elements 32 Crystal Symmetry Classes

Inversion Mirror Rotation

Rotations Mirrors Improper rotations

Two Translational Symmetry Elements

32 Crystal Symmetry Classes + Translational Symmetry Operations 230 Space Groups

•A three fold axis can not just have one two fold axis perpendicular to it.•In three dimensions the existence of two perpendicular two foldsimplies the existence of a third perpendicular two fold•The allowed combinations of point symmetry elements are called point groups

All combinations of point  symmetry elements are not possible

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 forcrystallographic pointgroups

Combining symmetry elements

For three dimensions32 point groups14 Bravais latticesbut only 230 space groups

For two dimensions5 lattices10 point groupsbut 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 γ iron δ iron912oC 1400oC 1539oC

liquid iron

BCC FCC BCC

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 DirectionsAtom Positions in Cubic Unit CellsA 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.

y

z

x

1,0,1

1,0,0

1,1,1

½, ½, ½

0,0,0

0,0,1 0,1,1

1,1,0

0,1,0

1,,0 <≤ zyx

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.

y

z

x

1,0,1

1,0,0

1,1,1

½, ½, ½

0,0,0

0,0,1 0,1,1

1,1,0

0,1,0A

Example direction A

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]

y

z

x

1,1,1

0,0,0

0,0,1

B

C

½, 1, 0

Example direction Ba) Two points origin coordinates

1,1,1 and final position coordinates 0,0,0

b) 0,0,0 - 1,1,1 = -1,-1,-1c) No fractions to cleard) Direction ]111[

___

Example direction Ca) Two points origin coordinates

½,1,0 and final position coordinates 0,0,1

b) 0,0,1 - ½,1,0 = -½,-1,1c) There are fractions to clear.

Multiply times 2. 2( -½,-1,1) = -1,-2,2

d) Direction ]221[__

Notes About the Use of Miller Indices for DirectionsA 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 co-ordinates. 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

00]1[],1[000],1[0[001],[010],[100], >100< ≡

Miller Indices for Crystallographic planes in Cubic CellsPlanes in unit cells are also defined by three integer numbers,

called the Miller indices and written (hkl).Miller’s indices can be used as a shorthand notation to identify

crystallographic directions (earlier) AND planes.Procedure for determining Miller Indices

locate the originidentify 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!

take reciprocals of these interceptsclear fractions but do not reduce to lowest integersenclose the resulting numbers in parentheses (h k l). Again, the

negative numbers should be written with a bar over the number.

y

z

x

A

Example: Miller indices for plane Aa) Locate the origin of coordinate.b) Find the intercepts x = 1, y = 1, z = 1c) Find the inverse 1/x=1, 1/y=1, 1/z=1d) No fractions to cleare) (1 1 1)

More Miller Indices - Examples

ab

c

0.50.5

ab

c

2/31/5

ab

c

ab

c

ab

c

ab

c

Notes About the Use of Miller Indices for PlanesA 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 constant.Example: The family of planes {100} consists of three equivalent planes (100), (010) and (001)

222 lkhad

++=

A “family” of crystal planes contains all those planes are crystallo-graphically 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

c

(110) = (1100)- -

(100) = (1010)-

a1

a2

a3

(001) = (0001)

(110) = (1100)- -

Two Types of Indices in the Hexagonal System

a1

a2

a3

c

a1

a2

a3

c

Miller: (hkl) (same as before)

Miller-Bravais: (hkil) → i = - (h+k)

a3 = - (a1 + a2)a1 ,a2 ,and c are independent, a3 is not!

Structures of Metallic Elements

Ru

H

Li

Na

K

Rb

Cs

Fr

Be

Mg

Ca

Sr

Ba

Ra

Sc

Y

La

Ac

Ti

Zr

H f

V

Nb

Ta

Cr

Mo

W

Mn

Tc

Re

Fe

Os

Co

Ir

Rh

Ni

Pd

Pt

Cu

Ag

Au

Zn

Cd

Hg

B

Al

Ga

In

Tl

C

Si

Ge

Sn

Pb

N

P

As

Sb

Bi

O

S

Se

Te

Po

F

Cl

Br

I

At

Ne

Ar

Kr

Xe

Rn

He

Primitive Cubic

Body Centered Cubic

Cubic close packing(Face centered cubic)

Hexagonal close packing

Lattice ConstantStructurea, nm c, nm

Chromium 0.289 0.125Iron 0.287 0.124Molybdenum 0.315 0.136Potassium 0.533 0.231Sodium 0.429 0.186

BCC

Copper 0.361 0.128Gold 0.408 0.144Nickel 0.352 0.125Silver 0.409 0.144Zinc 0.2665 0.5618 0.133Magnesium 0.3209 0.5209 0.160Cobalt 0.2507 0.4069 0.125Titanium 0.2950 0.3584 0.147

HCP

FCC

Tungsten 0.316 0.137Aluminum 0.405 0.143

Atomic Radius, nm

Metal

• Rare due to poor packing (only Po has this structure)• Close-packed directions are cube edges.

• Coordination # = 6 (# nearest neighbors)

SIMPLE CUBIC STRUCTURE (SC)

• Number of atoms per unit cell= 1 atom

6

APF = Volume of atoms in unit cell*

Volume of unit cell

*assume hard spheres

• APF for a simple cubic structure = 0.52

APF = a3

4

3π (0.5a)31

atoms

unit cellatom

volume

unit cellvolume

close-packed directions

a

R=0.5a

contains 8 x 1/8 = 1 atom/unit cell

Atomic Packing Factor (APF)

• Coordination # = 8

• Close packed directions are cube diagonals.--Note: All atoms are identical; the center atom is shaded differently only for ease of viewing.

Body Centered Cubic (BCC)

aR

Close-packed directions: length = 4R

= 3 a

Unit cell contains: 1 + 8 x 1/8 = 2 atoms/unit cell

APF = a3

4

3π ( 3a/4)32

atoms

unit cell atomvolume

unit cell

volume • APF for a BCC = 0.68

• Coordination # = 12

• Close packed directions are face diagonals.--Note: All atoms are identical; the face-centered atoms are shaded differently only for ease of viewing.

Face Centered Cubic (FCC)

a

APF = a3

4

3π ( 2a/4)34

atoms

unit cell atomvolume

unit cell

volume

Unit cell contains: 6 x 1/2 + 8 x 1/8 = 4 atoms/unit cell

Close-packed directions: length = 4R

= 2 a

• APF for a FCC = 0.74

Hexagonal Close-Packed (HCP)The APF and coordination number of the HCP structure is the sameas the FCC structure, that is, 0.74 and 12 respectively.An isolated HCP unit cell has a total of 6 atoms per unit cell.

12 atoms shared by six cells = 2 atoms per cell2 atoms shared by two cells = 1 atom per cell

3 atoms

Close-Packed StructuresBoth the HCP and FCC crystal structures are close-packed structure.Consider the atoms as spheres:

Place one layer of atoms (2 Dimensional solid). Layer APlace the next layer on top of the first. Layer B. Note that there are

two possible positions for a proper stacking of layer B.

A B A : hexagonal close packed A B C : cubic close packed

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 C : cubic close pack

A

B CA

120°

90°A

A

B

A B A : hexagonal close pack

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 inthe 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

Packing of Non-Identical Spheres

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.)

Tetrahedral/Octahedral

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.

FCC Interstitials

Calculate the octahedral void radius as a fraction of the parent atom radius in a FCC structure.

414.0=atom

void

rr

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.

225.0=atom

void

rr

HCP – OctahedralIn 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 – TetragonalIn 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.

HCP 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 OctahedralIn 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 Interstitials

BCC TetrahedralIn 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

155.0=atom

void

rr

Using this diagram calculate the tetrahedral void radius as a fraction of the parent atom radius in a BCC crystal

291.0=atom

void

rr

Interstitial sitesLocations 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 sixother 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

FCC Lattice has:3 [=12(¼)] Oh sites at edge centers+ 1 Oh site at body center

Octahedral, Oh, Sites

FCC Lattice has:8 Th sites at ¼, ¼, ¼ positions

Tetrahedral, Th, Sites

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

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.

Interstitial coordinates

Density CalculationsSince 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/molThe 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

AC NVnA

=ρ

n: number of atoms/unit cellA: 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, 333 216)22( RRaVC ===

32338

/89.8]10023.6)1028.1(216[

)5.63)(4( cmg=×××

=ρ

8.94 g/cm3 in the literature

ExampleRhodium 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

AC NVnA

=ρn: number of atoms/unit cell A: atomic weight

VC: volume of the unit cell NA: Avogadro’s number (6.023x1023 atoms/mole)

structure FCC a has Rhodium

01376.04

)1345.0(627.22

627.22)2

4( and 4 n then FCC is rhodium If

0149.02

)1345.0(316.12

316.12)3

4( and 2 n then BCC is rhodium If

01376.0103768.1.10023.6.41.12

.91.102

333

333

333

333

33231233

13

nmnmxna

rra

nmnmxna

rra

nmcmxmoleatomsxcmg

molgNA

na

nV

A

c

==

===

==

===

===== −−−

−

ρ

Linear And Planar Atomic Densities

Linear atomic density = 2R/Ll

Planar atomic density:

Ll

Crystallographic direction

= 2π R2/(Area A’D’E’B’)

342 RaaLl =⇒=

= 0.612

A’ B’

Structure of CeramicsCeramics

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 elementsAlways 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 insulatorsCan be optically opaque, semi-transparent, or transparentTraditional 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 byThe electric charge: The crystal must remain electrically

neutral. Charge balance dictates chemical formula (Ca2+ and F-

form 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.

Anion

Cation

rr

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 ratiofor a triangle:

1550

2330

21

21

.

cos

)30=( cos

=

==+

°=

+=

=

a

c

ca

a

ca

a

rr

rrrAO

AB

rrAO

rAB

αα

A C

BO

AC

B

O

4140

2245

21

21

.

cos

)45=( cos

=

==+

°=

+=

=

a

c

o

ca

a

ca

a

rr

rrrAO

AB

rrAO

rAB

αα

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.

o/t fcc(ccp) hcpall oct. NaCl NiAsall tetr. CaF2 (ReB2)o/t (all) (Li3Bi) (Na3As)½ t sphalerite (ZnS) wurtzite (ZnS)(½ o CdCl2 CdI2)

Comparison between structures with filled octahedral and tetrahedral holes

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

NaCl structure has Na+ ions at all 4 octahedral sites

Octahedral, Oh, Sites

Na+ ions

Cl- ions

Ionic Crystals prefer the NaClStructure:

• Large interatomic distance• LiH, MgO, MnO, AgBr, PbS, KCl, KBr

Crystals having filled Interstitial Sites

Both the diamond cubic structureAnd the Zinc sulfide structures have 4 tetrahedral sites occupied and 4 tetrahedral sited empty.

Tetrahedral, Th, Sites

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). 

Rock Salt Structure (NaCl)

Cl NaCoordination = 6

NaCl, MgO, LiF, FeO, CoO

NaCl structure: rC = rNa = 0.102 nm,

rA = rCl = 0.181 nm rC/rA = 0.56

AX Type Crystal Structures

Cesium Chloride Structure (CsCl)

CsClCoordination = 8Is this a body centered cubic structure?

CsCl Structure: rC = rCs = 0.170 nm, rA = rCl = 0.181 nm → rC/rA = 0.94

Zinc Blende Structure (ZnS)

ZnSCoordination = 4radius ratio = 0.402

ZnS, ZnTe, SiC have this crystal structure

AmXp-Type Crystal StructuresIf 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).

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). 

CaF2 Fluorite

Fluorite CaF2Fluorite (CaF2): rC = rCa = 0.100 nm, rA= rF = 0.133 nm ⇒ rC/rA = 0.75From the table for stable geometries we see that C.N. = 8Other compounds that have this crystal structure include UO2 , PuO2 , and ThO2.

AmBnXp-Type Crystal StructuresIt 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

• CaTiO3 - dielectric• BaTiO3 - ferroelectric• Pb(Mg1/3Nb2/3)O3 - relaxor

ferroelectric• Pb(Zr1-xTix)O3 - piezoelectric• (Ba1-xLax)TiO3 – semiconductor

• (Y1/3Ba2/3)CuO3-x -superconductor

• NaxWO3 - mixed conductor; electrochromic

• SrCeO3 - H - protonic conductor• RECoO3-x - mixed conductor• (Li0.5-3xLa0.5+x)TiO3 - lithium ion

conductor• LaMnO3-x - Giant magneto-

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 StructureABO3 e.g. KNbO3

SrTiO3LaMnO3

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) A = ( 0, 0, 0)

B = (0, 0, 0) OR B = (½, ½, ½)

X = (½,0,0) (0,½,0) (0,0,½) X = (½,½,0) (½,0,½) (0,½ ½)Draw one of these as a projection.

In SrTiO3, Ti-O ~ 1.95 Åa typical bond length for Ti-O; stable as a cubic structure

In BaTiO3, Ti-O is stretched, > 2.0 ÅToo long for a stable structure.

Ti displaces off its central position towards one oxygen

→ square pyramidal coordination

larger

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)Cubic (Pm3m)T > 393 KT > 393 K

TiTi--O Distances (O Distances (ÅÅ))66××2.002.00

Tetragonal (P4mm)Tetragonal (P4mm)273 K < T < 393 K273 K < T < 393 KTiTi--O Distances (O Distances (ÅÅ))1.83, 41.83, 4××2.00, 2.212.00, 2.21

Toward a cornerToward a cornerOrthorhombic (Amm2)Orthorhombic (Amm2)183 K < T < 273 K183 K < T < 273 KTiTi--O Distances (O Distances (ÅÅ))

22××1.87, 21.87, 2××2.00, 22.00, 2××2.172.17Toward an edgeToward an edge

RhombohedralRhombohedral (R3m)(R3m)183 K < T < 273 K183 K < T < 273 KTiTi--O Distances (O Distances (ÅÅ))

33××1.88, 31.88, 3××2.132.13Toward a faceToward a face

In the cubic structure BaTiOIn the cubic structure BaTiO33 is is paraelectricparaelectric. That is to say that the . That is to say that the

orientations of the ionic orientations of the ionic displacements are not ordered and displacements are not ordered and

dynamic.dynamic.

Below 393 K BaTiOBelow 393 K BaTiO33 becomes becomes ferroelectric and the displacement ferroelectric and the displacement

of the Tiof the Ti4+4+ ions progressively ions progressively displace upon cooling.displace upon cooling.

See Kwei et al. J. Phys. Chem. 97, 2368 (1993),

Density Calculations in Ceramic Structures

Ac

AC

NVAAn

×+×

= ∑ ∑ )('ρ

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 unitΣAA: sum of atomic weights of anions in the formula unitVc: volume of the unit cellNA: Avogadro’s number, 6.023x1023 (formula units)/mol

Example: NaCln’ = 4 in FCC latticeΣAC = ANa = 22.99 g/molΣAA = ACl = 35.45 g/mol

rNa=0.102x10-7 rCl=0.181x10-7 cmVc = a3 = (2rNa+2rCl)3

Vc = (2×0.102×10-7 + 2×0.181×10-7)3 cm3

( )[ ]

3

23377142

10023610181021010202453599224 −

−−=

××××+××

+×= cmg..

.....ρ

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: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:

[ ]

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

ΦΦ−ΦΦΦ+−Φ−−ΦΦ+Φ−

=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−×

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

ΦΦ−ΦΦ×

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=⋅⋅= Φ

coscossinsinsincossincoscoscossinsincossincoscossinsinsinsincoscossincossinsincoscoscos

1000cossin0sincos

cossin0sincos0

001

1000cossin0sincos

11

221122112

221122112

11

11

22

22

12

φφφφφφφφφφφφφφφφφφφφ

φφφφ

φφφφ

φφ

a

aaaa

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.