materials discovery camp

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Crystallography World ofWonders

Claudia J. [email protected]

University of Tennessee/Oak Ridge National Laboratory

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Al vs Al2O3AlMelts at 660 oCFCC a = 4.0495 ÅDensity = 2.71 gm/cm3

Al2O3Melts at 2000 oCBased on HCP a = 4.7589 and c = 12.991 ÅDensity = 3.98 gm/cm3

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• Bond length, r

• Bond energy, Eo

F F

r

• Melting Temperature, Tm

Tm is larger if Eo is larger.

PROPERTIES FROMBONDING: TM

From W.D. Callister: Materials Science and Engineering: An Introduction

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!metals! !ceramics! !polymers

Why? Metals have... • close-packing (metallic bonding) • large atomic mass Ceramics have... • less dense packing (covalent bonding) • often lighter elements Polymers have... • poor packing (often amorphous) • lighter elements (C,H,O) Composites have... • intermediate values Data from Table B1, Callister 6e.

DENSITIES OF MATERIAL CLASSES


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• Occurs between + and - ions.• Requires electron transfer.• Large difference in electronegativity required.• Example: NaCl

IONIC BONDING


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• Predominant bonding in Ceramics

Give up electrons Acquire electrons

He-

Ne-

Ar-

Kr-

Xe-

Rn-

F4.0

Cl3.0

Br2.8

I2.5

At2.2

Li1.0

Na0.9

K0.8

Rb0.8

Cs0.7

Fr0.7

H2.1

Be1.5

Mg1.2

Ca1.0

Sr1.0

Ba0.9

Ra0.9

Ti1.5

Cr1.6

Fe1.8

Ni1.8

Zn1.8

As2.0

CsCl

MgO

CaF2

NaCl

O3.5

Adapted from Fig. 2.7, Callister 6e. (Fig. 2.7 is adapted from Linus Pauling, The Nature of theChemical Bond, 3rd edition, Copyright 1939 and 1940, 3rd edition. Copyright 1960 by CornellUniversity.

EXAMPLES: IONIC BONDING


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• Requires shared electrons• Example: CH4

C: has 4 valence e, needs 4 more

H: has 1 valence e, needs 1 more

Electronegativities are comparable.

Adapted from Fig. 2.10, Callister 6e.

COVALENT BONDING

H, atomic number 1, electronic configuration 1s1

C, atomic number 6, electronic configuration 1s22s22p2


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• Molecules with nonmetals

• Molecules with metals and nonmetals• Elemental solids (RHS of Periodic Table)

• Compound solids (about column IVA)

He-

Ne-

Ar-

Kr-

Xe-

Rn-

F4.0

Cl3.0

Br2.8

I2.5

At2.2

Li1.0

Na0.9

K0.8

Rb0.8

Cs0.7

Fr0.7

H2.1

Be1.5

Mg1.2

Ca1.0

Sr1.0

Ba0.9

Ra0.9

Ti1.5

Cr1.6

Fe1.8

Ni1.8

Zn1.8

As2.0

SiC

C(diamond)

H2O

C2.5

H2

Cl2

F2

Si1.8

Ga1.6

GaAs

Ge1.8

O2.0

co

lum

n I

VA

Sn1.8

Pb1.8

Adapted from Fig. 2.7, Callister 6e. (Fig. 2.7 isadapted from Linus Pauling, The Nature of the Chemical Bond, 3rd edition, Copyright1939 and 1940, 3rd edition. Copyright 1960 by Cornell University.

EXAMPLES: COVALENT BONDING


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Metallic bonding


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Ceramics(Ionic & covalent bonding):

Metals(Metallic bonding):

Polymers(Covalent & Secondary):

Large bond energylarge Tm

large Esmall α

Variable bond energymoderate Tm

moderate Emoderate α

Directional PropertiesSecondary bonding dominates

small Tsmall Elarge α

SUMMARY: PRIMARY BONDS


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• Non dense, random packing

• Dense, regular packing

Dense, regular-packed structures tend to have lower energy.

ENERGY AND PACKING


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FCC

a

a

r

4r


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• Coordination # = 12

• APF = 0.74

• 3D Projection • 2D Projection

A sites

B sites

A sites

Adapted from Fig. 3.3, Callister 6e.

HEXAGONAL CLOSE-PACKED STRUCTURE (HCP)


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FCC and HCP close-packed lattices

• Both lattices are formed by a sequentialstacking of planar layers of close packedatoms.

• Within each layer each atom has six nearestneighbors.

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FCC and HCP close-packed latticesThe “A” layerall positions that are directly abovethe centers of the A atoms arereferred to as “A” positions,whether they are occupied or not

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FCC and HCP close-packed lattices• Both FCC and HCP lattices are formed by

stacking like layers on top of this first layer in aspecific order to make a three dimensional lattice.

• These become close-packed in three dimensionsas well as within each planar hexagonal layer.

• Close packing is achieved by positioning theatoms of the next layer in the troughs between theatoms in the “A” layer

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• Each one of these low positions occursbetween a triangle of atoms. Some pointtowards the top of the page and some pointtowards the bottom of the page.

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• Any two of these immediately adjacenttriangles are too close to be both occupiedby the next layer of atoms.

• Instead the next close-packed “B” layer willfill every other triangle, which will all pointin the same direction.

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• The “B” layer is identical to the A-layerexcept for its slight off translation.

• Continued stacking of close-packed layerson top of the B-layers generates both theFCC and HCP lattices.

“A” layer

“B” layer

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The FCC close-packed lattice

• The FCC lattice is formed when the thirdlayer is stacked so that its atoms arepositioned in downward-pointing trianglesof oxygen atoms in the “B” layer.

• These positions do not lie directly over theatoms in either the A or B layers, so it isdenoted as the “C” layer

“A” layer

“B” layer

“C” layer

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The FCC close-packed lattice

• The stacking sequence finally repeats itself when afourth layer is added over the C atoms with itsatoms directly over the A layer (the occupiedtriangles in the C layer again point downward) soit is another A layer.

• The FCC stacking sequence (ABCA) is repeatedindefinitely to form the lattice:

• …ABCABCABCABC...

“A” layer

“B” layer

“C” layer

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The FCC close-packed latticeEven though this lattice is made by stacking hexagonal planarlayers, in three dimensions its unit cell is cubic. A perspectiveshowing the cubic FCC unit cell is shown below, where the body-diagonal planes of the atoms are the original A, B, C, and layersof oxygen atoms

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The HCP close-packed lattice

• The HCP lattice is formed when the third layer isstacked so that its atoms are positioned directlyabove the “A” layer (in the upward facingtriangles of the “B” layer).

• The HCP stacking sequence (ABAB) is repeatedindefinitely to form the lattice:

• …ABABABAB...

“A” layer

“B” layer

Repeat “A” layer

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• ABCABC... Stacking Sequence• 2D Projection

• FCC Unit Cell

FCC STACKING SEQUENCE


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Perspective of FCC and HCP structuresviewed parallel to close-packed planes

FCC (...ABCABCABC…) HCP (...ABABABAB…)

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Ceramics• Characteristics

– Hard– Brittle– Heat- and corrosion-resistant

• Made by firing clay or other mineralstogether and consisting of one of moremetals in combination with one or morenonmetals (usually oxygen)

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Nomenclature

• The letter a is added to the end of anelement name implies that the oxide of thatelement is being referred to:

• SiO2 - silica Si4+ + 2(O2-)• Al2O3 - alumina 2(Al3+) + 3(O2-)• MgO - magnesia Mg2+ + O2-

Positively charged ions cations example: Si4+, Al3+, Mg2+

Negatively charged ions - anions example: O2-

Charge balanced

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Closed Packed LatticesThe Basis for Many Ceramic Crystal Structures

• Ionic crystal structures are primarily formedas derivatives of the two simple closepacked lattices: face center cubic (FCC)and hexagonal close packed (HCP).

• Most ionic crystals are easily derived fromthese by substituting atoms into theinterstitial sites in these structures.

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Closed Packed Lattices• The larger of the ions, generally the anion, forms the

closed-packed structure, and the cations occupy theinterstices.– We will often consider the anion to be oxygen (O2-) for

convenience since so many important ceramics are oxides.However, the anion could be a halogen or sulfur.

– In the case of particularly heavy cations, such as zirconium anduranium, the cations are larger than the oxygen and the structurecan be more easily represented as a closed packed arrangement ofcations with oxygen inserted in the interstices.

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Location and Density ofInterstitial Sites

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Interstitial Sites

• The interstitial sites exist between the layersin the close-packed structures

• There are two types of interstitial sites– tetrahedral– octahedral

• These are the common locations for cationsin ceramic structures

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Interstitial Sites• Each site is defined by the local coordination shell formed

between any two adjoining close-packed layers– the configuration of the third layer does not matter– the nearest neighbor configuration of oxygen atoms around the

octahedral and tetrahedral cations is independent of whether thebasic structure is derived from FCC or HCP

• FCC and HCP have the same density of these sites

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Interstitial sites

• Octahedral: 3-6-7-b-c-f– 3 from the A layer and 3 from the B layer– an octahedron has eight sides and six vertices– the octahedron centered between these six

atoms, equidistant from each - exactly half waybetween the two layers

31 2 4

5 6 78 9 10

a b c

d e f g

h i j

Numbers = A siteslower case letters = B sites

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Interstitial sites

The octahedral site neither directly above nor directlybelow any of the atoms of the A and B layers thatsurround the site

– The octahedral site will be directly above or below a C-layered atom (if it is FCC)

– These octahedral sites form a hexagonal array, centeredexactly half-way between the close-packed layers

31 2 4

5 6 78 9 10

a b c

d e f g

h i j


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Interstitial sites

• Tetrahedral: 1-2-5-a and e-h-i-9– 1 negative tetrahedron– 1 positive tetrahedron– three of one layer and one of the second layer– 3A and 1B – one apex pointing out of the plane of the

board– 3B and 1 A – one apex pointing into the plane of the

board

31 2 4

5 6 78 9 10

a b c

d e f g

h i j


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Interstitial sites

• Tetrahedral: 1-2-5-a and e-h-i-9– For both tetrahedral sites the center of the

tetrahedron is either directly above or below anatom in either the A or B layers

– The geometric centers are not halfway betweenthe adjacent oxygen planes but slightly closer tothe plane that forms the base of the tetrahedron

31 2 4

5 6 78 9 10

a b c

d e f g

h i j


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Interstitial sites31 2 4

5 6 78 9 10

a b c

d e f g

h i j


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Octahedral sites in the FCC Unit Cell

oo

oo

oo

o

o

o

o

oo

The FCC cell contains fouratoms

six faces that eachcontribute one half and atom

eight corners thateach contribute one-eighth anatomFCC cell contains fouroctahedral sites

12 edges each withone quarter of a site

one site in the center

o

One octahedral site halfwayalong each edge and one atthe cube center

The ratio of octahedral sites to atomsIs 1:1

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Tetrahedral sites in the FCC Unit CellOne tetrahedral site insideeach corner

Eight tetrahedral sites

The ratio of tetrahedral sites to atomsIs 2:1

ttt

tt

t

t

t

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General Structural formula forclose-packed structures

• T2nOnXn

T – Tetrahedral sitesO – Octahedral sitesX – Anions

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Example:

A2nBnXnA = tetrahedral sitesB = octahedral sitesX = anions

MgAl2O4If fully occupied A8B4X4

Mg in tetrahedral sites - 1/8 of the sites occupiedAl in octahedral sites - 1/2 of the sites occupied

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Linus Pauling

• Nobel Prize in Chemistry 1954– “for his research into the nature of the chemical bond

and its application to the elucidation of the structure ofcomplex substances”

• Nobel Peace Prize 1962• Born in 1901 and died in 1994• We may use Pauling’s rules to predict the

tendency for a specific compound to form aspecific crystal structure

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Pauling’s Rules

• Pauling’s rules are based on the geometricstability of packing for ions of differentsizes and simple electrostatic stabilityarguments.– These geometric arguments treat the ions as

hard spheres which is an over implication

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Ionic crystal radii

• Ionic radii (as defined by interatomicspacings) do vary from compound tocompound– they tend to vary most strongly with the

valance state of the ion and the number ofnearest neighbor ions of the opposite charge

• We may consider an ionic radius to be constant for agiven valance state and nearest-neighborcoordination number

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Pauling’s Rule 1

stable stable unstable

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CN Disposition of ions rc/raabout

central atom

8 corners of ≥ 0.732 a cube

6 corners of ≥ 0.414 an octahedron

4 corners of ≥ 0.225 a tetrahedron

2 corners of ≥ 0.155 a triangle

1 linear ≥ 0

When the radiusratio is less than thisgeometricallydetermined criticalvalue the next lowercoordination ispreferred

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Cubic a = b = c α = β = γ = 90ο

Hexagonal a = b ≠ c α = β = 90ο, γ = 120ο

Tetragonal a = b ≠ c α = β = γ = 90ο

Rhombohedral a = b = c α = β = γ ≠ 90ο

Orthorhombic a ≠ b ≠ c α = β = γ = 90ο

Monoclinic a ≠ b ≠ c α = γ = 90ο, ≠ β

Triclinic a ≠ b ≠ c α ≠ β ≠ γ

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Rocksalt

• NaCl, KCl, LiF, MgO, CaO, SrO, NiO,CoO, MnO, PbO– for all of these the anion is larger than cation

and forms the basic FCC lattice

The lattice parameterof the cubic unit cell is“ao” and each unitcontains 4 formulaunits

ao

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Calculating density• NiO -rocksalt structurea = 4.1771 Å space group Fm3mAtom Ox Wy x y zNi +2 4a 0 0 0O -2 4b 0.5 0.5 0.5

ρ =n’(∑MNi +∑� MO)

Vunit cellNAV

4(58.69 + 15.999) g/mol((4.1771 x 10-8 cm)3)(6.022 X 1023 atom/mol) =

g/molcm3 x molecules/mol

= 6.81 g/cm3

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Anti-fluoriteLi2Oa = 4.619 Åspace group Fm3m (225)

atom x y z WyO 0 0 0 4aLi 1/4 1/4 1/4 8c

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