1/24/2013
LECTURE #05
Chapter 3: Packing Densities and Coordination
Learning Objecti es
Learning Objectives
• How does atomic packing factor change with different atom types?
• How do you calculate the density of a material?
l?
Relevant Reading for this Lecture...
• Pages 46-58.
2
Re‐‐cap: Atomic Packing Factor (APF)
Re
Volume of atoms in unit cell*
APF = Volume of unit cell
*
*assume hard spheres
h d h
Describes How efficiently atoms fill space within a given unit cell
unit cell
Unit cell contains:
6 x 1/2 + 8 x 1/8 = 4 atoms/unit cell
4
2a
2 a
APF = 4
3
( 2a/4)
(
)3
a3
a
Adapted from
Fig. 3.1(a),
Callister 7e. 3
volume
atom
volume
unit cell
Close‐packed directions: • APF for a face‐centered cubic structure = 0.74
length = 4R = 2 a
1
1/24/2013
If the unit cell has different
atoms – watch out!
Ceramic Crystal
y
Structures!
4
Watch Out! Different Atoms. Must Modify APF Equations.
3a
Rblue
a
a
2a
Close‐packed directions:
Adapted from Fig. 3.2(a), Callister 7e.
a
diagonal = 2Rblue + 2Rred = 3 a
Must put in correct value for each atom type
for each
atoms
unit cell
2
APF = 5
4
3
 ( 3a/4) 3
a3
volume
atom
volume
unit cell
2
1/24/2013
What if ions from the crystal structure?
Cs+
Clԟ
Same rules as metals
w/respect to crystal
symmetry.
Structure: CsCl type
Bravais lattice: simple cubic
Ions/unit cell: 1 Cs+ + 1 Clԟ
Ionic packing factor
(IPF), similar definition
at APF.
Cesium chloride (CsCl) unit cell showing (a) ion positions and the two ions per lattice point and (b) full‐size ions. Note that the Cs+−Cl− pair associated with a given lattice point is not a molecule because the ionic bonding is non‐directional and because a given Cs+ is equally bonded to eight adjacent Cl−, and vice versa.
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 r’s for Cs+ and Cl-?
r Cs+ = 0.170 nm
r Cl- = 0.181 nm
These numbers came from inside cover of text book!
r +/r – = 0.939
7
3
1/24/2013
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]
IPF =
volume of unit cell
One atom of Cs + One atom of Cl
4/3π(0.170nm)3 + 4/3π(0.181nm)3
VCl
VCs
3a
= 0.0454
0 0454 nm3
a
r Cs+ = 0.170 nm
r Cl- = 0.181 nm
√3 a = 2rCs+ + 2rCla = 0.405 nm
V = a3 = 0.0664nm3
2a
IPF = 0.683
8
For kicks, what if the ions touched along the edge length (not the body diagonal)?
[# of atoms/unit cell] [volume of a sphere]
IPF =
volume of unit cell
O
t
t
One atom
off C
Cs + O
One atom
off Cl
4/3π(0.170nm)3 + 4/3π(0.181nm)3
r Cs+ = 0.170 nm
r Cl- = 0.181 nm
√3 a = 2rCs+ + 2rCla = 0.405 nm
V = a3 = 0.0664nm3
VCl
VCs
= 0.0454 nm3
which r?
average them?
a = 2r
a = 0.405 nm 0.352nm
V = a3 = 0.0436nm3
ravg = 0.176
0 176
IPF = 0.683 1.04
Not possible! >100% packing!!!
9
It is critical that you identify the correct ‘hard sphere’ touching directions!
4
1/24/2013
Let’s do another example: Consider NaCl
What is the CN?
What is the IPF?
How do we determine CN?
Correct, CN → r +/r –
What are the r’s for Na+ and Cl-?
r
Na+
= 0.102 nm
r Cl- = 0.181 nm
Again, got these
numbers from
inside cover of
text book!
r +/r – = 0.564
10
Not a very clear representation
– let’s expand it!
Motif
2 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.
11
5
1/24/2013
Calculate the IPF of NaCl
IPF =
[# of atoms/unit cell] [volume of a sphere]
volume of unit cell
What type of unit cell do we have? 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 4/3π(0.102 nm)3 + 4 x 4/3π(0.181 nm)3
= 0.117nm3
12
What is the volume of the unit cell?
a = 2r Na + 2rCl
a = 0.566nm
V = a3 = 0.181nm3
IPF = 0.117nm3 (previous slide)
0.181nm3
= 0.65
W t h O t! Diff
Watch Out! Different Ions. tI
2 a
a
CORRECTION!
Though it looks like FCC symmetry, the face diagonal atoms don’t touch; but the edge atoms do touch!
Why? Cations and anions do not have the same size! Before we considered atoms of the same size
13
6
1/24/2013
Atoms can occupy ‘interstitial’ sites
Unoccupied space in center can accommodate small atoms, e.g. He in UO2 fuel rods
Fl it (C F2) Fluorite (CaF
)
unit cell showing (a) ion positions (b) full‐size ions. FCC interstitial sites
FCC interstitial sites
Note: useful info on atom
placement
Structure: fluorite (CaF2) type
Bravais lattice: FCC
Ions/unit cell: 4Ca2+ + 8FTypical ceramics: UO2, ThO2, TeO2
14
CLASS ROOM EXAMPLE:
Calculate the ionic packing factor for UO2, which has the CaF2 structure U and O ions touch along a
portion of the body diagonal
15
7
1/24/2013
CLASS ROOM EXAMPLE: SOLUTION
Calculate the ionic packing factor for UO2, which has the CaF2 structure This problem is tricky!
The face diagonal has a length of 2a.
The body diagonal has a length of 3a.
Along with the cell edge, they form a right
triangle within the unit cell.
U2+
a
3a
O2a
a
a
The Ca2+ and F- ions touch a short distance
along the body diagonal.
By the principle of similitude, this smaller
triangle, measures as ¼ the size of the large
triangle
one.
Because of this, the length of the bond becomes:
1
 3a  RU 4  RO 2  0.105  0.132 nm  0.548 nm
4
Now you can calculate a and the corresponding unit cell volume.
16
CLASS ROOM EXAMPLE: SOLUTION ‐ continued
3a
 RU 4  RO2  0.105  0.132 nm
4
IPF 
Vions in unit cell
Vunit cell
Solving for a we get: a  0.548 nm
Vunit cell  a 3   0.548 nm   0.164 nm3
3
4
Vsingle ion   R 3
3
4
4
16
32
3
3
Vions  4   RU3 4  8   RO3 2    0.105     0.132   0.0965 nm3
3
3
3
3
IPF 
Vions
0.0965 nm3

 0.588
Vunit cell
0.164 nm3
17
8
1/24/2013
THEORETICAL DENSITY, 
Density = mass/volume
mass = number of atoms per unit cell  mass of each atom
mass = number of atoms per unit cell 
mass of each atom
mass of each atom = atomic weight / Avogadro’s number
# atoms/unit cell
 nA
VcNA
Volume/unit cell
(cm3/unit cell)
Atomic weight (g/mol)
Avogadro's number
(6.023 x 10 23 atoms/mol)
18
THEORETICAL DENSITY, 
# atoms/unit cell
 nA
VcNA
Volume/unit cell
(cm3/unit cell)
Atomic weight (g/mol)
Avogadro's number
(6.023 x 10 23 atoms/mol)
Example: Copper
• crystal structure = FCC: 4 atoms/unit cell
• atomic weight = 63.55 g/mol (1 amu = 1 g/mol)
• atomic radius R 0 128 nm (1 nm 10‐7 cm)
• atomic radius R = 0.128 nm (1 nm = 10
Vc = a3 ; For FCC, a = 4R/ 2 ; Vc = 4.75 x 10-23cm3
Result: theoretical Cu = 8.89 g/cm3
Compare to actual: Cu = 8.94 g/cm3
Why the difference?
19
9
1/24/2013
Theoretical Density, 
• Ex: Cr (BCC) A = 52.00 g/mol
R = 0.125 nm
n = 2
R
a
a = 4R/ 3 = 0.2887 nm
atoms
g
mol
2 52.00
unit cell
 = a 3 6.023 x 1023
volume
theoretical = 7.18 g/cm3
actual = 7.19 g/cm3
atoms
mol
unit cell
20
20
Densities of Material Classes
In general
metals > ceramics > polymers
Why?
Metals/ Alloys
30
Metals have...
have
Ceramics have...
• less dense packing
• often lighter elements
Polymers
y
have...
• low packing density
(often amorphous)
• lighter elements (C,H,O)
Composites have...
• intermediate values
3
 (g/cm )
• close‐packing
(metallic bonding)
• often large atomic masses
20
Platinum
Gold, W
Tantalum
10
Silver, Mo
Cu,Ni
Steels
Tin, Zinc
5
4
3
2
Titanium
Aluminum
Magnesium
Graphite/ Ceramics/ Semicond
Polymers
Composites/ fibers
B ased on data in Table B1, Callister *GFRE, CFRE, & AFRE are Glass,
Carbon & Aramid Fiber‐Reinforced
Carbon, & Aramid Fiber
Reinforced
Epoxy composites (values based on
60% volume fraction of aligned fibers
in an epoxy matrix).
Zirconia
Al oxide
Diamond
Si nitride
Glass ‐soda
Concrete
Silicon
G raphite
1
0.5
0.4
0.3
PTFE
Silicone
PVC
PET
PC
H DPE, PS
PP, LDPE
Glass fibers
GFRE*
Carbon fibers
CFRE *
A ramid fibers
AFRE *
Wood
Data from Table B1, Callister 7e. 21
10
1/24/2013
Summary
• If there are different ions in the unit cell, APF/IPF equations must be modified! Find out
APF/IPF equations must be modified! Find out which direction the ‘hard spheres’ touch • The ionic packing factor can be calculated by
IPF 
•
Vions in unit cell
Vunit cell
Theoretical Density can be calculated by
Theoretical Density can be calculated by
# atoms/unit cell
 nA
VcNA
Volume/unit cell
(cm3/unit cell)
Atomic weight (g/mol)
Avogadro's number
(6.023 x 10 23 atoms/mol)
22
11