Structuresproperties Cere 114

CLASSICAL MATERIALS SCIENCE

Five categories of materials:

Metals

Polymers

Ceramics

Semiconductors

Composites

Based primarily on

the nature of the

interatomic bonding

on materials conductivity

on materials structure

Definitions:

Metals both pure and alloyed, consist of atoms held together by the delocalized electrons that overcome the mutual repulsion

between the ion cores.

Polymers are macromolecules formed by covalent bonding of many simpler molecular units called mers.

Ceramics are usually associated with mixed bondinga combination of covalent, ionic, and sometimes metallic. They

consist of arrays of interconnected atoms.

Semiconductors are the only class of material based on a property. They are usually defined as having electrical

conductivity between that of a good conductor and an insulator.

Composites are combinations of more than one material or phase. Ceramics are used in many composites, often for

reinforcement.

Classification of solids

-----solid is dimensionally stable and has a volume of its own.

A) Bonding type

Primary bonds Secondary bonds

ionic dipole-dipole

covalent london dispersion

metallic hydrogen

van der waals

B) Atomic arrangement

Ordered Disordered

Atomic arrangement regular random

Order long range short range

Name crystalline amorphous

(crystal) (glass)

The electronegativity values for the elements.

lanthanides

actinides

Metallicity, Size

Metallicity,

Size

mass

Electronegativity

mass Electr

onegat

ivity

Definitions:

a)Metallicity can be defined as the tendency of an atom to donate electrons to metallic or ionic bonds.

-- increases when binding strength between valence electrons

diminishes

-- nuclear charge decreases, binding force decreases with increasing

metallicity

-- valence electron-nuclear separation is greater, binding force

diminishes due to screening of nuclear charge by core electrons

b) Electronegativity the tendency of an atom to attract an electron. According to Pauling, E is scaled between 0 and 4, specifically Fluorine has 4 as the highest

and Cesium the lowest equals 0.7

c) Size and Mass

-- ions shrink with increasing positive charge and expand with increasing

negative charge

-- mass increases with atomic number

Note: Periodic trends are not absolute.

Classification of the elements

-- metallic elements and metal-metal combinations form metallically bonded solids

-- non-metallic elements and nonmetal-nonmetal combinations are covalently

bonded

-- bonds between metals and nonmetals are either ionic or covalent depending on

the electronegativity difference

Paulings expression for the ionicity fraction of a bond:

f > 0.5 (x > 1.7) , the bonds are predominantly ionic f < 0.5 (x 1.7) , the bonds are predominantly covalent

For ternary or more complex compounds, the fractional ionicity can be determined by using stoichiometry weighted averages for the values

of xm and xnm.

After defining the bonding type, the type of atomic structure and properties that the

solid might have can also be inferred.

Knowledge of periodicity allows one to distinguish elements as metals or

nonmetals and to gauge relative electronegativities and sizes.

Based on this, it is possible to assign a bonding type.

From knowledge of the bond type, characteristic structures and properties can be

inferred.

Band gap is the separation in energy, between the highest filled electron energy level in the crystal and the lowest filled empty electron

energy level.

-- radiation energy Eg will be absorbed by the solid and promote electrons to higher energy unfilled states.

-- quantitative parameter that influences the appearance of the solid

-- nondefective solids with Eg > 3 transmit all visible light (1.7 to 3 eV.)

and are thus transparent and colorless

-- solids with Eg < 1.8 are opaque

-- solids with 0 < Eg < 1.7 are black

-- band gap increases with ionicity and the compounds with greater

than 50% ionicity should have large band gap and are colorless

Groups material Ionicity (f%) Band gap

(eV)

Tm oC

IV Ge 0 0.7 1231

III-V GaAs 4 1.4 1510

II -VI ZnSe 15 2.6 1790

I-VIII CuBr 18 5 492

Selected properties of three isoelectronic polar-covalent solids

% ionicity increases with increasing bandgap

Simple bonding models and typical properties

a) Metallic bonding model -- assumes that positively charged

ion cores are arranged periodically in a sea of free electrons

(formed by valence electrons)

Group IA and IIA are metals, the S levels are filled (Alkali or Alkaline earth metals)

B-group or transition metal series d-levels are filled

Lanthanide and actinide series f-levels are filled

Classification of metals

a)Elemental substance (Cu, Ag, Au, Al, Fe, Pb, etc..)

b)Intermetallic compounds (Ni3Al, NiAl, CuZn, cuZn3)

c)Random solid solutions or alloys (AxB1-x, both A and B are

metallic elements)

Typical properties of metallic materials

a)High reflectivity (when polished), high electronic and thermal

conductivity, low to intermediate melting temperatures, high ductility at

temperature less than half of their melting points

b) Intermetallic compounds and refractory metals

-- very high melting points and ductility at room temperature

B) Ionic bond model assumes that charge is transferred from the more metallic atom to the less metallic atom forming oppositely charged species

F12 = kq1q2/r2

12

Electronegativity difference > 1.7

Sample materials: salts (NaCl and CaCl2) and ceramics (MgO, ZrO2, TiO2)

Properties: transparent and colourless, electronically and thermally insulating,

intermediate to high melting temperature, brittle at ambient temperature, and

soluble in polar solvents or acids.

Covalent bonding model assumes that electrons are shared between atoms and that electron charge density accumulates between relatively positively

atomic cores.

Two types of materials

a)3-D covalent networks Si, SiC, GaAs, BN -- there is a covalently bonded path between any two atoms in solid

Covalently bonded networks high melting points and non-

reflective insulating and brittle

Si3N4 and SiO2

b) Molecular solids or polymeric solids

-- atoms within each molecule are linked by covalent bonds, but the molecules

that make up the crystal are held together only by the weak interactions known

collectively by weak interactions (van der Waals, dipolar, hydrogen bonds)

Not all atoms are connected by a path of strong covalent bonds

Example materials: N2, O2, H2O, C60, polyethylene

Low melting temperatures and transparent, insulating, soft, and soluble

Ketalaars triangle illustrates that there is a continuum of bonding types between the three limiting cases

most nearly ideal metallic bond to be Li

CsF and F2 to have the most nearly ideal ionic and covalent bonds, respectively

Most substances exhibit characteristics associated with more than one type of bonding and must be classified by a comparison to the limiting

cases.

The substances listed on the

lateral edges of the triangle are

merely examples chosen based

on periodicity; all materials can

be located on this triangle.

(c) 2003 Brooks/Cole Publishing / Thomson

Learning

Classification of materials based on the type of atomic order.

Amorphous materials - Materials, including glasses, that have no long-range order, or crystal structure.

Glasses - Solid, non-crystalline materials (typically derived from the molten state) that have only short-range atomic order.

Glass-ceramics - A family of materials typically derived from molten inorganic glasses and processed into crystalline materials with very fine grain size and improved mechanical properties.

Amorphous Materials: Principles and Technological Applications

(The dependence of repulsive,

attractive, and net forces on

interatomic separation for two

isolated atoms.

(The dependence of repulsive,

attractive, and net potential

energies on interatomic

separation for two isolated atoms.

Example problem

Ketelaar's Triangle Ketelaar's Triangle finds different structure types in different

regions i.e. according to the electronegativities of A and B,

delta X

Van Arkel

Ketelaar

Orbital hybridization in covalent structures

Covalent structures are formed from atoms that have both

s and p valence electrons.

>>>>in effect, those on the righthand side of the periodic

chart with relatively high electronegativities.

The formation of sp hybrid orbitals results in four

equivalent sp3 orbitals directed towards the vertices of a

tetrahedron.

Examples:

>>compounds formed between group III and V atoms,

such as BN, BP, BAs, AlP, AlAs, AlSb, GaP, GaAs, GaSb,

InP, InAs, and InSb, crystallize in the zinc blende

(sphalerite) structure.

>>C (diamond), Si, Ge, and SiC

>> some of the ligands are

nonbonding lone pairs

>>> s and p orbitals

of group V and VI

atoms

a ligand is an atom, ion, or

molecule (see also:

functional group) that binds

to a central metal-atom to

form a coordination

complex

Long range order requires that the atoms are arrayed in 3-D

pattern that repeats.

The simplest way to describe this pattern is through a unit cell

representation.

Unit cell ---is defined as the smallest region in space that,

when repeated, completely describes the 3-D pattern of the

atoms of a crystal

-Only seven unit cell shapes (7 crystal systems) can be

stacked together to fill 3-D space

They are: cubic, tetragonal, orthorhombic, rhombohedral,

hexagonal, monoclinic, and triclinic

*** they are distinguished by length of cell edges and

interaxial angles (total of six parameters)

(c) 2003 Brooks/Cole Publishing / Thomson Learning

Definition of the lattice parameters and their use in cubic, orthorhombic, and hexagonal crystal systems.

14 Bravais Lattices ( in ball and stick models)


The fourteen types of Bravais lattices grouped in seven crystal systems.

a) Cubic F lattice stretched (c/2) along one axis, resulting into lattice identical

to tetragonal I.

b) New lattice vectors rotated 45o with respect to original vector, defining the

tetragonal I.

An orthorhombic C cell is

formed by stretching the

tetragonal P along the face

diagonal.

Derivations:

Distortions of the

orthorhombic lattices

lead to monoclinic

lattices.

The different

shadings indicate

lattice points at b/2.

To preserve a

consistent coordinate

system, the structure

in (c) is labeled as an

A-type cell.

Complete description of crystal structure:

A)Symmetry of the lattice pattern

Bravais lattices 14 possible lattice arrangements that consider the symmetry within each unit cell.

lattice- can be defined as an indefinitely extending

arrangement of points, each of which is surrounded by an

identical grouping of neighboring points.

B)Location of the atoms on lattice sites

symmetry of the basis

--defined as the atom or grouping of atoms located at

each lattice site.

--- the total number of atomic arrangement increases to

32 point groups

There are three principle crystal structures for metals:

(a) Body-centered cubic (BCC)

(b) Face-centered cubic (FCC)

(c) Hexagonal close-packed (HCP)

Face-centered cubic (FCC)

Determine the number of lattice points per cell in the cubic crystal systems. If there is only one atom located at each lattice point, calculate the number of atoms per unit cell.

Example 1: SOLUTION

In the SC unit cell: lattice point / unit cell = (8 corners)1/8 = 1

In BCC unit cells: lattice point / unit cell = (8 corners)1/8 + (1 center)(1) = 2

In FCC unit cells: lattice point / unit cell = (8 corners)1/8 + (6 faces)(1/2) = 4

The number of atoms per unit cell would be 1, 2, and 4, for the simple cubic, body-centered cubic, and face-centered cubic, unit cells, respectively.

Example 1. Determining the Number of Lattice Points in Cubic Crystal Systems

--volume occupied by atoms/unit cell volume

Determine the relationship between the atomic radius and the lattice parameter in SC, BCC, and FCC structures when one atom is located at each lattice point.

Example 2 Determining the Relationship between Atomic Radius and Lattice Parameters


Learning

The relationships between the atomic radius and the Lattice parameter in cubic systems (for Example 2).

Example 2 SOLUTION Referring to Figure 3.14, we find that atoms touch along the edge of the cube in SC structures.

3

40

ra

In FCC structures, atoms touch along the face diagonal of the cube. There are four atomic radii along this lengthtwo radii from the face-centered atom and one radius from each corner, so:

2

40

ra

ra 20

In BCC structures, atoms touch along the body diagonal. There are two atomic radii from the center atom and one atomic radius from each of the corner atoms on the body diagonal, so


Learning

Illustration of coordinations in (a) SC and (b) BCC unit cells. Six atoms touch each atom in SC, while the eight atoms touch each atom in the BCC unit cell.

Atomic Packing Factor

74.018)2/4(

)3

4(4)(

Factor Packing

24r/ cells,unit FCCfor Since,

)3

4)(atoms/cell (4

Factor Packing

3

3

0

3

0

3

r

r

r

a

a

Calculate the packing factor for the FCC cell.

Example 3: SOLUTION

In a FCC cell, there are four lattice points per cell; if there is one atom per lattice point, there are also four atoms per cell. The volume of one atom is 4r3/3 and the volume of the unit cell is .

3

0a

Example 4 Determining the Density of BCC Iron

Determine the density of BCC iron, which has a lattice parameter of 0.2866 nm.

Example 4 SOLUTION

Atoms/cell = 2, a0 = 0.2866 nm = 2.866 10-8 cm

Atomic mass = 55.847 g/mol

Volume of unit cell = = (2.866 10-8 cm)3 = 23.54 10-24 cm3/cell

Avogadros number NA = 6.02 1023 atoms/mol

3

0a

3

2324/882.7

)1002.6)(1054.23(

)847.55)(2(

number) sadro'cell)(Avogunit of (volume

iron) of mass )(atomicatoms/cell of(number Density

cmg


(a) Illustration showing sharing of face and corner atoms. (b) The models for simple cubic (SC), body centered cubic (BCC), and face-centered cubic (FCC) unit cells, assuming only one atom per lattice point.

Hexagonal close-packed (HCP)


Learning

The hexagonal close-packed (HCP) structure (left) and its unit cell.

--- 6 atoms

per unit cell

POINT COORDINATES

-- the position of any point located within a unit cell may be

specified in terms of its coordinates as fractional multiples

of the unit edge lengths (i.e. in terms of a, b, and c)

Example: Specify point coordinates for all atom

positions for a BCC unit cell.

Problem 1:

The cubic unit cell of perovskite, CaTiO3, is drawn

below. What are the atomic coordinates of the

atoms?

A coppergold alloy has a cubic structure. The atom positions are:

Sketch the unit cell and determine the

formula of the alloy.

CRYSTALLOGRAPHIC DIRECTIONS

Examples illustrating how directions in a crystal are

named.

HCP Crystallographic Directions

1. Vector repositioned (if necessary) to pass

through origin.

2. Read off projections in terms of unit

cell dimensions a1, a2, or a3

3. Adjust to smallest integer values

4. Enclose in square brackets, no commas

[u'v'w']

5. Covert to 4 parameter Miller-Bravais

[u'v'w'] => [ 110 ] ex: 1, 1, 0 =>

Adapted from Fig. 3.8(a), Callister 7e.

- a3

a1

a2

z

Algorithm

Show that the c/a ratio of the ideal

hexagonal close packed structure is

1.633.

HCP Crystallographic Directions Hexagonal Crystals

4 parameter Miller-Bravais lattice coordinates are related to the direction indices (i.e., u'v'w') as follows.

' w w

t

v

u

) v u ( + -

) ' u ' v 2 ( 3

1 -

) ' v ' u 2 ( 3

1 -

] uvtw [ ] ' w ' v ' u [

Fig. 3.8(a), Callister 7e.

- a3

a1

a2

z

=(1/3)(2x1-1)=1/3

=(1/3)(2x1-1)=1/3

= -(1/3+1/3)= -2/3

= 0

Multiplying by 3 to get smallest integers

[ 1120 ]

Crystallographic directions for

HEXAGONAL CRYSTAL

Projections: a(a1 axis), a( a2 axis), and c(z axis) = 1, 1, and 1 respectively

EXAMPLE:

Multiplying the above indices by 3 reduces them to values for u, v,

t, and w of 1, 1, -2 and 3, respectively.

Hence the direction is,

CRYSTALLOGRAPHIC PLANES

65

Crystallographic Planes z

x

y a b

c

4. Miller Indices (110)

example a b c z

x

y a b

c


1. Intercepts 1 1

2. Reciprocals 1/1 1/1 1/

1 1 0 3. Reduction 1 1 0

1. Intercepts 1/2

2. Reciprocals 1/ 1/ 1/

2 0 0 3. Reduction 1 0 0

example a b c

66

Crystallographic Planes z

x

y a b

c


example

1. Intercepts 1/2 1 3/4 a b c

2. Reciprocals 1/ 1/1 1/

2 1 4/3

3. Reduction 6 3 4

(001) (010),

Family of Planes {hkl}

(100), (010), (001), Ex: {100} = (100),

Miller Indices (hkl)

reciprocals

Crystallographic Planes Determine the Miller indices for the plane shown in the

accompanying sketch (a).

Crystallographic Planes

Construct a (011) plane within a cubic unit cell.

Crystallographic Planes

_

(1 1 1) (0 1 1)

Family of Planes All planes that are identical are included in 1 family

denoted by { hkl}

Example {111}

(111) and (111) have equivalent

atoms

74

Crystallographic Planes (HCP) In hexagonal unit cells the same idea is used

example a1 a2 a3 c

4. Miller-Bravais Indices (1011)

1. Intercepts 1 -1 1 2. Reciprocals 1 1/

1 0

-1

-1

1

1

3. Reduction 1 0 -1 1

a2

a3

a1

z

Adapted from Fig. 3.8(a), Callister 7e.

Crystallographic planes for hexagonal crystal

--using four index (hkil) scheme

--where i= -(h+k)

Intersections:

a1=a, a2=-a, z=c, a3=?

In terms of lattice parameters: 1,-1,1

and their reciprocals are equal. It

follows that h=1, k=-1, l=1.

i= -(1-1)=0

Therefore, the (hkil) indices are:

Atomic linear density

a fraction of line length intersected by atoms

LD= Lc/Ll

Atomic planar density

Is simply the fraction of total crystallographic plane area that is occupied by atoms

PD= Ac/Ap

For FCC

For FCC

Calculate the planar density and planar packing fraction for the (010) and (020) planes in simple cubic polonium, which has a lattice parameter of 0.334 nm.

Example 5 Calculating the Planar Density and Packing Fraction

(c) 2003 Brooks/Cole Publishing /

Thomson Learning

The planer densities of the (010) and (020) planes in SC unit cells are not identical (for Example 3.9).

Atomic planar density

Is simply the fraction of total crystallographic plane area that is occupied by atoms

LINEAR & PLANAR

DENSITIES Linear density (LD) = number of

atoms centered on a direction vector / length of direction vector LD (110) = 2 atoms/(4R) = 1/(2R)

Planar density (PD) = number of atoms centered on a plane / area of plane PD (110) = 2 atoms /

[(4R)(2R2)] = 2 atoms / (8R22) = 1/(4R22)

LD and PD are important considerations during deformation and slip; planes tend to slip or slide along planes with high PD along directions with high LD

Example 5 SOLUTION The total atoms on each face is one. The planar density is:

2142

2

atoms/cm 1096.8atoms/nm 96.8

)334.0(

faceper atom 1

face of area

faceper atom (010)density Planar

The planar packing fraction is given by:

79.0)2(

)(

)( atom) 1(

face of area

faceper atoms of area (010)fraction Packing

2

2

20

2

r

r

r

a

However, no atoms are centered on the (020) planes. Therefore, the planar density and the planar packing fraction are both zero. The (010) and (020) planes are not equivalent!

unit cell

volume

a

10

APF for a face-centered cubic structure = 0.74 Metals have relatively large atomic packing factors

Adapted from

Fig. 3.1(a),

Callister 6e.

ATOMIC PACKING FACTOR:

FCC Close-packed directions (diagonal):

Unit cell contains

4 atoms/unit cell

length = 4R a = 2R 2

= 0.74

16 R3

APF =

4

3 4

atoms

unit cell atom

volume R3

2

4R

R

X-ray Diffraction: Determination of Crystal Structures

-- using diffraction of x-ray, atomic interplanar distances and

crystal structures are deduced

The Diffraction Phenomenon

**Diffraction occurs when a wave encounters a series of

regularly spaced obstacles:

(1)are capable of scattering the wave

(2)have spacing that are comparable in magnitude to the

wavelength

---Diffraction is a consequence of specific phase relationships

established between two or more waves that have been

scattered by the obstacles.

Demonstration of how two waves that have the same wavelength and remain in

phase and out of phase after scattering event

-- scattered waves reinforcing one another

-- phase relationship depends upon a

difference in path length

-- difference is integral number of wavelength

-- amplitudes cancel or annul one

another or destructively interfere

Diffraction of x-

rays by a

periodic

arrangement of

atoms

--- Braggs Law

if this is not satisfied, the interference will be nonconstructive and yields very

low-intensity diffracted beam.

-- n is order of reflection defining any integer (1,2,3)

Same (hkl)

Interplanar spacing -- distance between two adjacent and

parallel planes of atoms, which is a function of Miller indices

(hkl) as well as the lattice parameters.

For cubic symmetry:

Diffraction will occur for unit cells having atoms positioned only at cell

corners

Scattering centers that cause of out-of-phase scattering at certain

Bragg angles

--------- face and interior unit cell positions as with FCC and BCC

resulting into absence of beams or peaks------

Condition for diffraction:

For BCC, h+k+l must be even, and FCC, the h, k, and l must all be

either odd or even

Photograph of a XRD diffractometer. (Courtesy of H&M Analytical Services.)

Schematic diagram of an x-ray

diffractometer;

T=x-ray source, S=specimen,

C=detector, and O=the axis around

which the specimen and detector rotate

Diffraction pattern for powdered lead.

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The first 20 positive integers and the corresponding hkl triplets in the primitive (P),

body-centered (I) and face-centered (F) cubic lattices.

Cubic Crystals (Indexing the

Patterns) 222

2

22

4sin lkh

a

:222 lkh Simple cubic: 1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 14, etc. Body-centered cubic: 2, 4, 6, 8, 10, 12, 14, 16, . . .

Face-centered cubic: 3, 4, 8, 11, 12, 16, 19, 20, . . .

Diamond cubic: 3, 8, 11, 16, 19, 24, 27, 32, . . .

Simple cubic bcc fcc

Line sin2 (h2 + k2 + l2)

2

2

4a

(h2 + k2 + l2)

2

2

4a

(h2 + k2 + l2)

2

2

4a

a() hkl

1 0.140 1 0.140 2 0.070 3 0.0467 3.57 1112 0.185 2 0.093 4 0.046 4 0.0463 3.59 2003 0.369 3 0.123 6 0.062 8 0.0461 3.59 2204 0.503 4 0.126 8 0.063 11 0.0457 3.61 3115 0.548 5 0.110 10 0.055 12 0.0457 3.61 2226 0.726 6 0.121 12 0.061 16 0.0454 3.62 4007 0.861 8 0.108 14 0.062 19 0.0453 3.62 3318 0.905 9 0.101 16 0.057 20 0.0453 3.62 420

a lattice parameter

Classroom quiz

The results of a x-ray diffraction experiment using x-

rays with = 0.7107 (a radiation obtained from molybdenum (Mo) target) show that diffracted peaks occur at the following 2 angles:

Example 6 Examining X-ray Diffraction

Determine the crystal structure, the indices of the plane producing each peak, and the lattice parameter of the material.

Example 6 SOLUTION We can first determine the sin2 value for each peak,

then divide through by the lowest denominator, 0.0308.

Example 6 SOLUTION (Continued)

We could then use 2 values for any of the peaks to calculate the interplanar spacing and thus the lattice parameter. Picking peak 8:

2 = 59.42 or = 29.71

868.2)4)(71699.0(

71699.0)71.29sin(2

7107.0

sin2

2224000

400

lkhda

d

This is the lattice parameter for body-centered cubic iron.

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Figure 3.51 Planes in a cubic unit cell for Problem 3.54.

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Figure 3.50 Planes in a cubic unit cell for Problem 3.53.

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Figure 3.49 Directions in a cubic unit cell for Problem 3.52.

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Figure 3.53 Directions in a hexagonal lattice for Problem 3.56.

' w w

t

v

u

) v u ( + -

) ' u ' v 2 ( 3

1 -

) ' v ' u 2 ( 3

1 -

] uvtw [ ] ' w ' v ' u [

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Figure 3.52 Directions in a hexagonal lattice for Problem 3.55.

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Figure 3.54 Planes in a hexagonal lattice for Problem 3.57.

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Figure 3.55 Planes in a hexagonal lattice for Problem 3.58.

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Structuresproperties Cere 114