MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2012
Problem Set 2
Due: Tuesday, January 31, 11:00 AM
1. Glass Transition Temperature
a) The glass transition temperature, Tg, is strongly governed by the ability of the polymer
chains to move and rotate, which can be impeded by steric effects or interchain
bonding interactions. If mobility is difficult, the polymer will transition from fluid to
glassy at a higher temperature. Rank the three polymers below from lowest to highest
Tg. Explain your reasoning. (10pts)
Polymer B will have the lowest glass transition temperature. The only interactions
between chains will be Van der Waals, and possibly the steric effect of the CF3
molecule. It also has a short, simple mer that can pack easily.
Polymer A should have the medium Tg. It has a larger sidechain than polymer B, and
may also have hydrogen bonding, due the -COOH group.
Polymer C will likely have the highest Tg. Because of the NH3+ and Cl- ions, the chains
can form ionic bonds with one another. This secondary bond is stronger than the Hbonding of polymer A, with a similarly bulky sidegroup.
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b) Which is more likely to crystallize, atactic polystyrene or syndiotactic polyacrylonitrile
(PAN)? (4pts)
PAN will crystallize more easily. The polystyrene has a bulkier sidegroup, and is also
atactic, both of which act against the formation of a crystalline polymer.
c) You are told the density of a specimen of polymer X (PX) is 1.030 g/cm 3 at 74%
crystallinity. Purely crystalline PX is 1.050 g/cm3. If you need PX to float on water for
your application, what’s the maximum percent crystallinity allowable? (6pts)
The maximum allowable crystallinity is 33%. First, find the density of amorphous PX, ρa,
using the values given for this specimen and the equation given in lecture.
Taking the density of water to be approximately 1 g/cm3,
Therefore, polymer X must be processed to have a crystallinity < 33% for this usage.
2. Crystal Structure - Perovskite
The perovskite crystal structure is illustrated in Figure 12.6 of Callister. SrTiO3 is a material
that exhibits an ideal cubic perovskite crystal structure.
a) For SrTiO3, which elements are cations, and which are anions? (4pts)
In the atomic formula ABO3, A and B are both cations and O is the anion. Thus,
The cations are Sr and Ti, and the anion is O.
b) What are the valence states of each element within this compound? (4pts)
The most probable valence state for Sr and O is 2+ and 2- respectively. Charge
neutrality dictates that Ti must therefor be 4+ and the valence states are Sr2+ Ti4+ O2c) How many Sr, Ti and O atoms are in the unit cell? (4pts)
From figure 12.6 we see that there is a single Ti atom at the center, 8 Sr atoms at the
corners contributing 1/8 for a total of 1 Sr, and 6 O atoms contributing ½ each for a total
of 3. Thus, there are 1 Sr, 1 Ti, and 3 O per unit cell, as the chemical formula implies.
d) Determine the coordination number for Sr, Ti and O in this structure. (4pts)
From figure 12.6 and a bit of imagination, we see that the Sr site has 12 nearest
neighbors, Ti 6 and O 6.
MATERIALS 101
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WINTER 2012
e) Calculate the theoretical density of SrTiO3. Compare this with the experimental
density. (4pts)
The lattice constant of SrTiO3 is 3.905 angstroms. Using equation 12.1 from Callister
and a periodic table gives
(87.62+15.999*3+47.867)(g/mol)/ (3.905*10-8cm)3(6.023*1023/mol)=5.116g cm-3
3. Molecular Weight of Polymers
Molecular Weight
Range (g/mol)
5,000-10,000
10,000-15,000
15,000-20,000
20,000-25,000
25,000-30,000
30,000-35,000
35,000-40,000
Mean molecular
weight Mi
7,500
12,500
17,500
22,500
27,500
32,500
37,500
Fraction of particles
in this range: xi
.04
.11
.16
.27
.22
.14
.06
Fraction of weights in
this weight range: wi
.03
.11
.19
.29
.20
.15
.03
a). Calculate the number-average molecular weight. (5pts, 3 pts if wrong but correct work)
23,400 g/mol
b). Calculate the weight-average molecular weight. (5pts, 3 pts if wrong but correct work)
23,100 g/mol
c). This polymer has a repeat unit molecular weight of 56 g/mol, compute the degree of
polymerization (DP). (5pts, 2 pts if wrong but correct work)
DP = 23,400 (g/mol)/56(g/mol) = 417.86 Or DP=23100 (g/mol)/56(g/mol) = 412
d) Assuming this polymer is a hydrocarbon consisting of only Carbon and Hydrogen
atoms, write down the chemical formula of this hydrocarbon (a mass difference of less
than a gram between your formula and the molecular weight given in part C is OK). Is
this hydrocarbon saturated or unsaturated? (A drawing of the hydrocarbon may help,
but is not required). (3 pts for formula and 2 pts for unsaturated)
C4H8: 2-Butene is the closest match. This is an unsaturated polymer.
MATERIALS 101
Molecular Weight
Range (g/mol)
5,000-10,000
10,000-15,000
15,000-20,000
20,000-25,000
25,000-30,000
30,000-35,000
35,000-40,000
SUM:
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2012
Fraction of
Fraction of
Mean molecular
M_i*
particles in
weights in this
M_i*xi
weight Mi
wi
this range: xi
weight range
.04
7,500
0.03
300
225
.11
12,500
0.1
1375
1250
.16
17,500
0.19
2800
3325
.27
22,500
0.29
6075
6525
.22
27,500
0.22
6050
6050
.14
32,500
0.13
4550
4225
.06
37,500
0.04
2250
1500
157,500
1.00
1.00 23400 23100
4. Polymorphism/Isomorism
a) Sketch the crystallographic planes for the following unit cells. Also, indicate how many
atoms are located in that plane. (Be sure to account for the fact that some atoms may only
have fractional amounts in that plane.) (5pts)
i.the (111) plane of an BCC unit cell
ii.the (012) plane of an BCC unit cell
iii.the (111) plane of an FCC unit cell
iv.the (012) plane of an FCC unit cell
b) At room temperature, pure iron has a BCC crystal structure, but undergoes a polymorphic
transformation at 912°C to an FCC crystal structure. These two forms of iron have
drastically different properties due to their ability to incorporate carbon. Calculate the
planar density along each of the planes above. (5pts)
a) i.(111) plane of a BCC unit cell. Plane with atoms is shown below in the unit cell as
well as a 2D view of the plane. There are three corner atoms on this plane, each of
which is shared with six total similar planes, thus there is ½ atom total on the plane.
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2012
+z
1
a
1
3
2
+y
1
2a
+x
ii. (012) plane of a BCC unit cell. Plane with atoms is shown below in the unit cell as
well as a 2D view of the plane. There are two corner atoms on this plane, each of
which is shared with four total similar planes, thus there is ½ atom total on this
plane.
+z
½
a
1
+y
1
+x
5a
2
iii.(111) plane of a FCC unit cell. Plane with atoms is shown below in the unit cell, a
2D view of the plane is the same as in part i). There are three corner atoms on this
plane, each of which is shared with six total similar planes. Then there are three
face centered atoms on this plane each of which is shared between two planes.
Total there are 2 atoms on this plane.
MATERIALS 101
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WINTER 2012
+z
1
1
+y
1
+x
iv. (0123) plane of an FCC unit cell. Plane with atoms is shown below in the unit cell, a
2D view of the plane is the same as in part ii). There are two corner atoms on this
plane, each of which is share with four total similar planes, thus there is ½ atom total
on this plane. There is also one face centered atom on this plane, which is shared
between two planes. Thus total there is 1 atom on this plane.
+z
½
1
+y
1
+x
b) Planar densities of each of the planes above.
,
,
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2012
i.
ii.
iii.
iv.
c) What is the difference between polymorphism and isomorism? (1pts)
Polymorphism is when a material of given composition has two or more possible crystal
structures. Isomorism is when two or more polymer molecules or repeat units have the
same composition, but different atomic arrangements.
d) Different isomers of a material can also have very different properties due to their atomic
arrangements. Sketch the following isomers. Describe some ways that the different
configurations of these polymers affect their properties. (7 pts)
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
i. isotactic, syndiotactic, and atactic polyvinylchloride.
Isotactic polyvinylchloride
Cl
H
Cl
H
Cl
H
Cl
H
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
n
Syndiotactic polyvinylchloride
Cl
H
H
H
Cl
H
H
H
C
C
C
C
C
C
C
C
H
H
Cl
H
H
H
Cl
H
n
Atactic polyvinylchloride
ii.
Cl
H
Cl
H
H
H
Cl
H
C
C
C
C
C
C
C
C
H
H
H
H
Cl
H
H
H
1-pentene, cis- and trans- pentene.
1-pentene
n
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WINTER 2012
cis-pentene
trans-pentene
The different configurations will affect inter-chain (between chains) interactions. This
can lead to better stacking, which is responsible for higher melting points and
crystallinity.
e) Draw the two molecules which have the same number of carbon atoms as pentene but
are fully saturated and fully unsaturated. (2pts)
Saturated hydrocarbons have only single bonds, and thus new atoms may be added
only by removing others that were already bonded. Unsaturated hydrocarbons have
double or triple covalent bonds.
Fully unsaturated
CH2
C
C
C
CH2
Fully saturated
CH3
CH2
CH2
CH2
CH3
5. Structure - Atomic Ratio
a) Name the expected crystal structure for the following materials using cation to anion
radius ratios: (Hint: tables 12.2 and 12.3 from Callister) (5pts)
i)
MnO
ii) CaF2
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iii)
iv)
v)
NaCl
MgS
CaO
i)
MnO
ii)
CaF2
iii)
NaCl
iv)
MgS
v)
CaO
INTRODUCTION TO STRUCTURE AND PROPERTIES
rMn2
rO2
rCa2
rCl1
rNa1
rCl1
rMg 2
rS 2
rCa2
rO2
WINTER 2012

0.067
 0.479 Coord#=6 =>NaCl (Rock Salt) structure
0.140

0.100
 0.752 Coord#=8 => CsCl (see fig 12.3 p458) structure
0.133

0.102
 0.564 Coord#=6 => NaCl (Rock Salt) structure
0.181

0.072
 0.391 Coord#=4 => ZnS (Zincblende) structure
0.184

0.100
 0.714 Coord#=6 => NaCl (Rock Salt) structure
0.140
b) Both Fluorite (CaF2) and cesium chloride(CsCl) have the same coordination number,
but their structures are slightly different. Ignoring the difference in atom types and
lattice spacing, how are the two structures different and why? (Hint: The difference
may only be seen when comparing more than one cell) (2pts)
Figure 12.3 on pae 458 of Callister shows the CsCl structure. The structure for CaF 2
will be the same type except there will be no Ca2+ cation in every other unit cell (see
figure 12.5). According to the chemical formula for Fluorite there are twice as many
F atoms as there are Ca atoms, which is required in order for the total charge of the
structure to remain at zero.
c) A sample of MgO has Si impurities. For each Si ion incorporated in a Mg site are any
vacancies formed in order to remain charge neutral? (3pts)
Yes, for every Si4+ ion in the place of a Mg2+ another Mg2+ ion must be removed
creating a vacancy in order to remain charge neutral. Charge Neutrality requires the
net charge of the material to remain at zero.
d) Show that the minimum rcation/ranion for a coordination number of 4 is 0.225.(5pts)
Coordination number of 4 implies a tetragonal arrangement of 4 anions around a
cation. The angle formed between the cation and two adjacent anions is 109.5
Degrees. At the minimum rc/ra adjacent anions are touching. Using geometry we can
find the relationship between rc and ra.
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2012
In the illustration to the left we can see that the angle
marked ‘c’ is 0.5*109.5 Degrees = 54.75 Degrees
Using the right triangle constructed we can see that
under the conditions of minimum rc/ra:
The side opposite c has a length of ra and the
hypotenuse has a length of ra+rc.
r
sin  c   a
ra  rc
rc  ra
1

ra
sin  c 
rc
1
1
1

1 
1 
 1  0.225
ra sin  c 
sin  54.75 
0.8166
e) What is the atomic packing factor (APF) of a material with a coordination number of 8
at the minimum cation to anion radius ratio? (5pts)
The minimum rc/ra = 0.732 for a coordination number of 8. At the minimum ration all
atom are in contact with their neighbors. The structure is arranged with anions at all 8
corners of a cube surrounding the cation at the center (as seen in at the bottom of Table
12.2 of Callister).
APF= Va/Vc=(Volume of all atoms in the cell)/(Volume of the cell)
The cubic unit cell has sides of length a = 2ra.
Vcell=a3=8ra3
Va=Ncation x Vcation + Nanion x Vanion
Here: Ncation=1 x 1=1; Nanion=1/8 x 8=1
4
4
Vcation   rc 3 ;Vanion   ra 3
3
3
rc  0.732ra
3
4
4

Va  1    0.732ra   0.729  1   ra 3   5.832ra 3
3

3

3
5.832ra
APF 
 0.729
8ra 3