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CHAPTER 4: STRUCTURE ANDPROPERTIES OF CERAMICS

Chapter 12-


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ISSUES TO ADDRESS...ISSUES TO ADDRESS...ISSUES TO ADDRESS...ISSUES TO ADDRESS...

Structures of ceramic

CHAPTER 4: STRUCTURE AND

PROPERTIES OF CERAMICS

Chapter 12-

How do they differ fromthat of metals?

1


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A ceramic has traditionally been defined asAn inorganic solid that is prepared from

powdered materials, is fabricated intoproducts through the application of heat, anddisplays such characteristic properties ashardness, strength, low electrical

conductivity, and brittleness.The word ceramic comes from the Greekword "keramikos", which means "pottery.

Chapter 12-

ey are yp ca y crys a ne n na ure anare compounds formed between metallic andnonmetallic elements such as aluminum andoxygen (alumina-Al2O3), calcium and oxygen

(calcia - CaO), and silicon and nitrogen(silicon nitride-Si3N4).


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SINGLE VS POLYCRYSTALS

SINGLE CRYSTAL: Properties vary with direction: anisotropic

Example: the modulus of elasticity (E)in BCC iron:

E (diagonal) = 273 GPa

Chapter 12-

POLYCRYSTAL: Properties may/may not vary with direction.

If grains are randomly oriented: isotropic.(E poly iron = 210 GPa)

E (edge) = 125 GPa


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The atoms in ceramic materials are heldtogether by a chemical bond

Briefly though, the two most common chemicalbonds for ceramic materials are covalent andionic.

Chapter 12-

Covalent and ionic bonds are much strongerthan in metallic bonds and, generally speaking,this is why ceramics are brittle and metals areductile.

The main compositional classes of engineeringceramics are the oxides, nitrides and carbides.


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Ceramics have ionic or covalent bonding

Crystal structure is strongly influencedby bonding

This influences mechanical and physical

Chapter 12-

. . .

! most ceramics are not electricalconductors;

! little ductility at room temperature


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Bonding:--Mostly ionic, some covalent.--% ionic character increases with difference in

electronegativity.

He-

H2.1

CaF2: large

Large vs small ionic bond character:

CERAMIC BONDING

Chapter 12-

Ne-

Ar-

Kr

-

Xe

-Rn

-

Cl3.0

Br2.8

I

2.5At

2.2

Li1.0

Na0.9

K0.8

Rb

0.8Cs0.7

Fr0.7

Be1.5

Mg1.2

Sr

1.0Ba

0.9

Ra0.9

Ti

1.5

Cr

1.6

Fe

1.8

Ni

1.8

Zn

1.8

As

2.0

C2.5

Si1.8

F4.0

Ca1.0

Table of Electronegativities

SiC: small


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Different types ofceramic structures

Oxide structures:cubic closed packed structure

Rock Salt (NaCl), Wurtzite (BeO),

Zinc Blende (ZnS),Spinel (MgAl2O4),

Chapter 12-

Cesium Chloride, Fluorite (CaF2),

Perovskite (BaTiO3),Ilmenite (MgTiO3)

Silicate structures:

Silica (SiO2), Orthosilicates(Fe2SiO4),Pyrosilicates(Si2O7)


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Conditions for a stable configuration

Anions and cations should always toucheach other

Anions generally will not touch eachother,but they should be in close contact

Maximum possible number of anions

Chapter 12-

max. reduction in the electrostaticenergy,

rc


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GROUPING OF IONS AND PAULINGS FIRST RULE

STABLE AND UNSTABLE CONFIGURATIONS

Coordination number is the no. ofanions surrounding the central cation

Chapter 12-

oor na on po y e ron o an ons s orme a ou eac ca on n e s ruc ure* Anion is also surrounded by a coordination polyhedron of cations

Cation coordination polyhedron building block of an ionic structure

Coordination number/ Ligancy(L) = RADIUS OF CATION / RADIUS OF ANION* A central cation of given size cannot remain in contact with all surrounding anions,

if the radius of anion is larger than a certain critical value.* A given cordination number is thus stable only when the ratio of cation to anion radiusis greater than a critical value.

* Anions are larger than cations, so the critical radius ratio for a structure is alwaysdetermined by the coordination of anions about the cations.

* Magnitude of electrical charge on each of the componentions must be the same for charge neutrality


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Coordination # increases withIssue: How many anions can you

arrange around a cation?

rcationranion

rcationranion

Coord #

0.414 in monoatomic FCC*edge centered Na atom force to move the corner atom

a little more than in FCC monoatomic cell


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Cesium Chloride Structure

Simple cubic anion packing ,two interpenetratingmonoatomic SC corresponding to anions and cations

EN of Cl=1/8x8=1, EN of Cs=1 , 1:1,satisfies neutrality in

the unit cell Anions located at each corners of the cube,cube center

is a simple cation r / r = 0.170/0.181=0.939 (not exactly 0.732 as for

Chapter 12-

monoatomic SC lattice) , therefore L=8,ie 8 foldcoordination number Coordination no. 8 (eightfold coordination),Cl is in eight

fold coordination Cl- ions are in a simple cubic array, with all the

interstices filled with Cs+ ions Interchange of cations with anions and vice versa

produce the same structure Not a BCC structure, since two different ions are

involved


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Cesium Chloride (CsCl)

Cs-Cl:(3/2)a

Chapter 12-

Cl-

Cs+

Cs-Cs:aCs:

(1/2 1/2 1/2)

Cl: (0 0 0)


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SRUCTURE OF CsCl

Chapter 12-


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Zinc Blende Structure

Cubic closed pack structure

All ions are tetrahedrally

coordinated (coordination no. 4) rC / rA = 0.074/0.184 = 0.4021

Chapter 12-

cell occupied by S and Zn atoms fillinterior tetrahedral positions

Each Zn atom bonded to four Satoms

Bonding is more covalent than ionic


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Zinc Blende (ZnS)

Zn-S :(3/4)aZn-Zn

Chapter 12-

S

Zn

EN of Zn : 4(at 4 tetrahedral positions)EN of S: 1/8 x8 + x 6 = 4

a/2

S: (000)

(1/2 1/2 0)Zn:

(3/4 1/4 1/4)


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Am Xp CRYSTAL STRUCTURES

If the charges on cation and anion are notthe same, compound exists with chemicalformula Am Xp ,where m and/p is not equal

to 1 Coordination no. is 8 Eg. CaF2 , half as many Ca

2+ ions as F- ions

Chapter 12-

crystal structure similar to CsCl, but onlyhalf of the cation sites are filled Unit cell is based on the FCC packing of

the cations

Eg: ThO2, TeO2, UO2


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Charge Neutrality:--Net charge in the

structure should

be zero.

--

CaF2:Ca2+

cationF-

F-

anions+

IONIC BONDING & STRUCTURE

Chapter 12-

m, p determined by charge neutrality Stable structures:

--maximize the # of nearest oppositely charged neighbors.

- -

- -+

unstable

- -

- -+

stable

- -

- -+

stable


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Consider CaF2 :rcation

ranion

=

0.100

0.133 0.8

Based on this ratio, coord # = 8 and structure = CsCl. Result: CsCl structure with only half the cation sitesoccupied.

Am

Xp

STRUCTURES

,(One unit cell - 8cubes)

Chapter 12-6

are occupied since#Ca2+ ions = 1/2 # F- ions.


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FLUORITE (CaF2)

Ca-F :(3/4)a

F-F : 0.5 a

Chapter 12-

FFFF ----CaCaCaCa2+2+2+2+

oneoneoneone unit cell consists of 8 cubes, Funit cell consists of 8 cubes, Funit cell consists of 8 cubes, Funit cell consists of 8 cubes, F---- at corners of 8 cubesat corners of 8 cubesat corners of 8 cubesat corners of 8 cubesFFFF---- ions: (1/8 x 8) x 8 cubes=8ions: (1/8 x 8) x 8 cubes=8ions: (1/8 x 8) x 8 cubes=8ions: (1/8 x 8) x 8 cubes=8

To satisfy electroneutrality, there is a requirement of of no. of FTo satisfy electroneutrality, there is a requirement of of no. of FTo satisfy electroneutrality, there is a requirement of of no. of FTo satisfy electroneutrality, there is a requirement of of no. of F---- ions,ions,ions,ions,so Caso Caso Caso Ca2+2+2+2+ ions = 4, only 4 Body centered positions are filledions = 4, only 4 Body centered positions are filledions = 4, only 4 Body centered positions are filledions = 4, only 4 Body centered positions are filled

Ca-Ca: a/2


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Perovskite structure

Chemical Formula: AmBnXP, A and Bare cations and X are anions

Based on closed packing of anions Ceramic compounds with more than one

t e of cation

Chapter 12-

In CaTiO3, Ca2+ and O2- ions combine toform a close packed structure with thesmaller,more highly charged Ti4+ ions inoctahedral interstices

rTi / rO = 0.48 and rBa / rO = 0.97 (cubic)

CN = 6 w.r.t Ti ion


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PEROVSKITE STRUCTURE (BaTiO3)

Ti4+

O2-

Chapter 12-

Ba2+

1. Each oxygen ion is surrounded by four Ba ions , Ba at the corner, 1/8 x 8 = 12. Each Ba ion is surrounded by twelve oxygen ions.

3. In the center Ti ion is octahedrally coordinated to six oxygen ions

Ti at the center, so BCC and thus number = 1

4. Oxygen ion ic FCC ,so x 6 = 3

5. In BaTiO3, ratio is Ba:Ti:O = 1:1:3


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CERAMIC DENSIY COMPUTATION

Theoretical density , of a crystalline ceramicmaterial,

= n(Ac + AA) / VC NA

n=the number of formula units (all the ions that areincluded in the chemical formula unit) within theunit cell, in BaTiO3, there is one Barium ion, a

Chapter 12-

Ac=the sum of the atomic weights of all cations inthe formula unit

AA=the sum of the atomic weights of all anions inthe formula unit

V c=unit cell volume

NA= Avogadros number


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Compute theoretical density of NaCl ?

n, the no. of NaCl units per unit cell is 4 because both sodiumand chloride ions form FCC lattices.

Ac=ANa=22.99gm/mol,

AA=ACl = 35.45gm/mol, Since unit cell is cubictherefore Vc=a

3

rNa+

Chapter 12-

rCl-

2(rNa+ + r Cl-) = a,

Vc = a3 = [2(rNa+ + r Cl-)]3 , from = n(Ac + AA) / VC NA = 2.14gm/cc

a


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Compute APF for NaCl?

rC / rA = 0.414 , APF=VS/VC,

there are four cations and four anions per unit cell,

therefore, Vs= 4(4 /3 rA3 + 4 /3 rC

3 )

Chapter 12-

C A ,

and Vs= 17.94rA3, Vc = 22.62 rA3

Therefore, APF = VS/VC= 17.94rA3 / 22.62 rA3

= 0.79

c A C


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Interpenetrating FCC LatticeInterpenetrating FCC LatticeInterpenetrating FCC LatticeInterpenetrating FCC Lattice

Chapter 12-


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CaFCaFCaFCaF2222 StructureStructureStructureStructure

Chapter 12-

www.uwgb.edu/dutchs/PETROLGY/Fluorite%20Structure.HTM


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BaBaBaBa----Coordination No. in Perovskite: 12Coordination No. in Perovskite: 12Coordination No. in Perovskite: 12Coordination No. in Perovskite: 12

Chapter 12-

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