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Chapter3:PackingDensitiesandChapter3:PackingDensitiesand
CoordinationCoordination
LECTURE #05
earn ng ec ves
Howdoesatomicpackingfactorchangewith
differentatomtypes?
Howdoyoucalculatethedensityofa
material?
2
Pages 46-58.
Relevant Reading for this Lecture...
ReRecap:AtomicPackingFactor(APF)cap:AtomicPackingFactor(APF)
APF=Volumeofatomsinunitcell*
Volumeofunitcell
Describes
Howefficiently
atomsfillspace
withinagiven
unit cellassume ar sp eres
4(2a/4)34
volume
Unitcellcontains:
6x1/2+8x1/8
= 4atoms/unitcell
a
Adaptedfrom
Fig.3.1(a),
Callister7e.
APF= 3
a3
unitcell
volume
Closepackeddirections:
length=4R= 2a
APFforafacecentered
cubicstructure=0.743
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atoms watch out!
Ceramic Crystal Structures!
4
WatchOut!DifferentAtoms. MustModifyAPFEquations.WatchOut!DifferentAtoms. MustModifyAPFEquations.
a
Rblue
a3
aAdaptedfromFig.3.2(a),Callister7e.
a2a
Closepackeddirections:
Mustputincorrectvalue
3adiagonal =2Rblue +2Rred =
APF=
4
3 ( 3a/4) 32
atomsunit cell
a3
unit cell
volume
oreac a om ype
atom
volume
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Whatifionsfromthecrystalstructure?Whatifionsfromthecrystalstructure?
Cl
Cs+
Cesiumchloride(CsCl)unitcellshowing(a)ionpositionsandthetwoionsperlatticepoint
and(b)fullsizeions.NotethattheCs+Clpairassociatedwithagivenlatticepointisnot
amoleculebecausetheionicbondingisnondirectionalandbecauseagivenCs+ isequally
bondedtoeightadjacentCl,andviceversa.
(IPF), similar definition
at APF.
w/respect to crystal
symmetry.
Bravaislattice: simple cubicIons/unit cell: 1 Cs++1 Cl
6
Calculate the following: (1) The CN of CsCl (2) Ionic Packing Factor of CsCl
Recall from the last lecture CN r/R =r+/r
What are the rs for Cs+ and Cl-?
rCs+ = 0.170 nm
rCl- = 0.181 nmThese numbers came from inside cover of text book!
r+/r = 0.939
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Now calculate the Ionic Packing Factor of CsCl
For these ions, they touch along the body diagonal
(make sure you know which direction the hard
spheres are touching)
[# of atoms/unit cell] [volume of a sphere]
volume of unit cell
a3
One atom of Cs + One atom of Cl
4/3(0.170nm)3 + 4/3(0.181nm)3
VCs VCl
=
rCs+ = 0.170 nm
rCl- = 0.181 nm
a2
a
3 a = 2rCs+ + 2rCl-a = 0.405 nm
V = a3 = 0.0664nm3
IPF = 0.683
= . nm
8
For kicks, what if the ions touched along the edge length (not the body diagonal)?
[# of atoms/unit cell] [volume of a sphere]
volume of unit cellIPF =
ne a om o s + ne a om o
4/3(0.170nm)3 + 4/3(0.181nm)3
VCs VCl
= 0.0454 nm3
rCs+ = 0.170 nm
rCl- = 0.181 nmwhich r?
average them?
3 a = 2rCs+ + 2rCl-a = 0.405 nm a = 2r
ravg = .V = a3 = 0.0664nm3 a = 0.405 nm 0.352nmV = a3 = 0.0436nm3
Not possible! >100% packing!!!
It is critical that you identify the correct hard sphere touching directions!
IPF = 0.683 1.04
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Lets do another example: Consider NaCl
What is the CN?
What is the IPF?
How do we determine CN?
Correct, CN r+/r
rNa+ = 0.102 nm
rCl- = 0.181 nm
What are the rs for Na+ and Cl-?
Again, got these
numbers from
inside cover of
text book!
r+/r = 0.564
10
Not a very clear representation
lets expand it!
Motif2 ions per lattice point
Do you see how the green spheres form a FCC structure?!
The blue spheres do the same thing, form
a FCC structure
We have two interpenetrating FCC lattices
Commonly called a Rock Salt structure
structure seen in other materials including
TiC, TaC, MgO, CaO, etc.
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Calculate the IPF of NaCl
[# of atoms/unit cell] [volume of a sphere]
volume of unit cellIPF =
Two FCC lattices one for Na and one for Cl
4 atoms/Na FCC + 4 atoms/Cl FCC = 8 atoms/unit cell total
Four atoms of Na + Four atoms of Cl
4 x 4/3(0.102 nm)3 + 4 x 4/3(0.181 nm)3= 0.117nm3
What type of unit cell do we have?
12
a = 2rNa + 2rCl
a = 0.566nm0.181nm3
IPF = 0.117nm3 (previous slide)
= 0.65V = a3 = 0.181nm3
What is the volume of the unit cell?
2a
Watc Out D erentIons.Watc Out D erentIons.
CORRECTION!a
Thoughit
looks
like
FCC
symmetry,
theface diagonalatomsdonttouch;
buttheedgeatomsdotouch!
Why? Cationsandanionsdonot
havethesamesize!Beforewe
consideredatomsofthesamesize13
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Unoccupiedspace
incentercan
accommodatesmall
atoms,e.g.Hein
UO2 fuelrods
AtomscanoccupyinterstitialsitesAtomscanoccupyinterstitialsites
uor e a 2
unitcellshowing
(a)ionpositions
(b)fullsizeions.
Not e: useful info on atom
placementFCC interstitial sites
Structure: fluorite (CaF2) type
Bravaislattice: FCCIons/unit cell: 4Ca2++8F-
Typical ceramics: UO2, ThO2, TeO2
14
Calculate
the
ionic
packing
factor
for
UO2,
which
has
the
CaF2 structure
CLASSROOMEXAMPLE:
U and O ions touch along a
portion of the body diagonal
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CLASSROOMEXAMPLE: SOLUTION
This problem is tricky!
The face diagonal has a length of 2a.
CalculatetheionicpackingfactorforUO2,whichhastheCaF2 structure
The body diagonal has a length of 3a.
Along with the cell edge, they form a right
triangle within the unit cell.
The Ca2+ and F- ions touch a short distance
along the body diagonal.
By the principle of similitude, this smaller
3aU2+
O-
a
,
one.2 a
4 2
13 0.105 0.132 0.548
4 U OR R nm nm a
a
Because of this, the length of the bond becomes:
Now you can calculate aand the corresponding unit cell volume.
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CLASSROOMEXAMPLE: SOLUTION continued
ions in unit cell
unit cell
V
IPF V
Solving for we get: 0.548nma a
4 23 0.105 0.1324 U OR R nm a
33 30.548 0.164
unit cellV a nm nm
3
singleion
4
3V R
4 2 3 33 3 34 8 0.105 0.132 0.09653 3 3 3
ions U OV R R nm
3
3
0.09650.588
0.164ions
unit cell
V nmIPF
V nm
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THEORETICALDENSITY,THEORETICALDENSITY,Density=mass/volume
massofeachatom=atomicweight/Avogadrosnumber
# atoms/unit cell Atomic weight (g/mol)
VcNAVolume/unit cell(cm3/unit cell)Avogadro's number(6.023 x 1023 atoms/mol)
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n A# atoms/unit cell Atomic weight (g/mol)
THEORETICALDENSITY,
THEORETICAL
DENSITY,
c AVolume/unit cell(cm3/unit cell)
Avogadro's number
(6.023 x 1023 atoms/mol)
Example: Copper
crystalstructure=FCC: 4atoms/unitcell
atomicweight=63.55g/mol (1amu =1g/mol)
a om c
ra us
=
.
nm
nm
=
cm
Vc = a3 ; For FCC, a = 4R/ 2 ; Vc = 4.75 x 10
-23cm3
Compare to actual: Cu = 8.94 g/cm3
Result: theoretical Cu = 8.89 g/cm3
Why the difference?
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Ex:Cr(BCC)
A = 52.00g/mol
TheoreticalDensity,TheoreticalDensity,
R =0.125nm
n =2
a=4R/3=0.2887nmaR
atoms
= a 3
52.002unitcellmol
unitcell
volume atoms
mol
6.023x1023
theoretical =7.18g/cm3
actual =7.19
g/cm
3
2020
DensitiesofMaterialClassesDensitiesofMaterialClasses
metals >ceramics >polymers
Why?
Graphite/
Ceramics/
Semicond
Metals/
Alloys
Composites/
fibersPolymers
20
30B asedondatainTableB1,Callister
*GFRE,CFRE,&AFREareGlass,
Gold,WPlatinum
Ingeneral
(g/cm
)32
,
Epoxycomposites(valuesbasedon
60%volumefractionofalignedfibers
inanepoxymatrix).10
3
4
5
Magnesium
Aluminum
Steels
Titanium
Cu,Ni
Tin,Zinc
Silver,Mo
an a um
G raphite
Silicon
Glass sodaConcrete
SinitrideDiamondAloxide
Zirconia
PTFE
CFRE *
GFRE*
Glassfibers
Carbon fibers
...
closepacking
(metallicbonding)
oftenlargeatomicmasses
Ceramics have...lessdensepacking
oftenlighterelements
Pol mers have...
DatafromTableB1,Callister7e.
1
0.3
0.4
0.5
HDPE,PSPP,LDPE
PCPETPVC
Wood
AFRE *A ramidfibers
low
packing
density
(oftenamorphous)
lighterelements(C,H,O)
Composites have...intermediatevalues
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Iftherearedifferentionsintheunitcell,
SummarySummary
whichdirectionthehardspherestouch
Theionicpackingfactorcanbecalculatedby
ionsin unit cell
unit cell
VIPF
V
n A
VcNA
# atoms/unit cell Atomic weight (g/mol)
Volume/unit cell
(cm3/unit cell)
Avogadro's number
(6.023 x 1023 atoms/mol)
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