1 Student Name __________________________
Student Number ________________
MATERIALS 4R03 Midterm Examination 2013
DURATION OF EXAMINATION: 2 hours October 2013
McMaster University Midterm Examination Instructor: Tannaz Javadi-Doodran
Answer all multiple choice questions on the examination copy only. Each multiple choice question has 2 marks.
1. The ratio of octahedra and tetrahedra sites to atoms in hcp and fcc crystal structures are as follows:
a) hcp (oct/atoms:2/1, tetra/atoms:4/1), fcc (oct/atoms:2/1, tetra/atoms:4/1)
b) hcp (oct/atoms:1/1, tetra/atoms:2/1), fcc (oct/atoms:1/1, tetra/atoms:2/1)
c) hcp (oct/atoms:2/1, tetra/atoms:1/1), fcc (oct/atoms:1/1, tetra/atoms:2/1)
d) hcp (oct/atoms:1/1, tetra/atoms:2/1), fcc (oct/atoms:2/1, tetra/atoms:1/1)
2. List the most preferable linkage to the least preferable linkage of coordination
a) faces, edges, corners
b) corners, edges, faces
c) edges, corners, faces
d) corners, faces, edges
3. In crystalline silicates, which order of the O/Si ratio is correct?
a) chain of tetrahedron > sheet of tetrahedron > framework of tedrahedron
b) sheet of tedrahedron > chain of tetrahedron > framework of tetrahedron
c) chain of tetrahedron > framework of tetrahedron > sheet of tedrahedron
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4. Which description about glass is not correct?
a) superheating liquid far below the melting temperature can get glass
b) their atomic arrangements lack long-range order
c) some oxide glasses can be used as network modifiers
d) by adding modifiers, the materials’ viscosity will decrease
5. Which of the following procedures cannot produce point defects in ceramics?
a) Thermal excitation
b) Impurity addition
c) Mechanical deformation
d) Oxidation/Reduction process
6. Which description about defects is right?
a) Creating more vacancies can always decrease the systems’ Gibbs free energy
b) Schottky defects are only in ionic compounds
c) Frenkel defects will change the number of lattice sites
d) Extrinsic defects concentration will increase as the temperature goes up
7. Which description is true?
a) ferroelectric are materials that have temporary electric dipoles
b) below the Curie temperature ferroelectric has zero dipole
c) remove the electric field applied on a paraelectric material, the polarisation can’t return
to zero
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8. Under what conditions is the refractive index related to the dielectric constant by k~n2 ?
a) Any dielectric at low frequency, non-polar dielectrics at all frequencies.
b) Any dielectric at optical frequency, non-polar dielectrics at all frequencies
c) Any dielectric at low frequency, polar dielectrics at all frequencies
d) Any dielectric at optical frequency, polar dielectrics at all frequencies
9. A polar liquid is subjected to an alternating current at 50 Hz. The current frequency is then increased to just above the relaxation frequency of the orientational mode of polarisation. Which of these best describes the behaviour of the dielectric constant as the frequency is increased?
a) It remains roughly constant
b) It gradually decreases.
c) It remains roughly constant until the relaxation frequency of the orientational mechanism is reached, at which point it drops sharply.
d) It gradually decreases until the relaxation frequency of the orientational mechanism is reached, at which point it drops sharply.
10. What field has to be applied to give a ferroelectric, with a square hysteresis loop, zero net polarisation?
a) The switching field
b) The polarising field.
c) The coercive field
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11. Which of these will not depolarise a poled ferroelectric piezoelectric?
a) Applying a current perpendicular to the polarisation
b) Applying a reversed electric field to the piezoelectric
c) Applying a very large stress to the piezoelectric
d) Applying a large amount of heat to the piezoelectric
12. Which of these is not a basic requirement for a pyroelectric?
a) A temperature dependent polarisation.
b) A spontaneous polarisation
c) A polar structure
d) A simple cubic structure.
13. How does the resistance vary with temperature in a perfect superconductor?
a) The resistance falls continuously as the temperature approaches absolute zero.
b) The resistance initially falls but passes through a minimum before rising as the temperature approaches absolute zero.
c) The resistance initially falls before discontinuously dropping to zero at a temperature above absolute zero.
d) The resistance falls continuously down to a limiting value as the temperature approaches zero.
14. Polymorphs are materials that have the same ....Compounds... and different
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15. The atomic weight of Si and Al are very similar (28.09 vs. 26.98), yet the densities of SiO2 (cristobalite) and Al2O3 (Corundum) are very different (2.65 vs. 3.95 g/cm3). Explain
this difference in terms of crystal structure and Pauling's rules (12).
One of the major differences in these two systems is the packing of oxygen ions and the result of how the polyhedra are coordinated.
In corundum, the oxygen anions form a hcp structure, with 2/3 of the octahedral interstitial spaces being filled with Al3+ cations. The result is that the coordination octahedra share faces in this system. They are as close packed together as possible.
In SiO2 cristobalite, each Si4+ cation fills a tetrahedral interstitial space and is
coordinated by 4 O2- cations.
This tetrahedral coordination is dictated by the Si4+/O2- radius ratio and the fact that the Si-O bond has a high degree of covalency. The result of Pauling's second rule is that each O2- must be coordinated by two Si4+ cations. The result of this is corner sharing of the SiO44- tetrahedra; they are as least close-packed polyhedra as possible,
and there is a lot of empty space in the cristobalite unit cell.
So there are two main differences that result in the difference in density:
- corundum has close-packed O2- anions, crisobalite has corner-coordinated polyhedra leaving a lot of open space.
- Corundum has more interstitials; 2/3 for every O2-, cristobalite has 1/2 for every O2-.
16. BaTiO3has perovskite structure. what’s the coordination number for Ba2+, Ti4+ and O2-,
please prove that it obeys pauling’s second rule (8).
So the CN for Ba2+, Ti4+ and O2- is 12, 6 and 6, respectively.
and because each oxygen atom is connected with 4 barium atoms and 2 titanium
atoms, so
2 12 4+
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6 2=2
17. Which type of zirconia (ZrO2) is a good oxygen ion conductor? Explain why and how it
becomes a good ion conductor? (8)
Cubic zirconia is a good oxygen ion conductor due to the additives that are added to stabilize the cubic form at room temperature. These additives are (Cao/ MgO/ Y2O3)
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MgO---
ZrO
2---> Mg
Zr''+ O
Ox+ V
O..Based on above equation, it can be seen that the additives produce additional oxygen vacancy into the system and thus the system becomes a good oxygen ion conductor as there are more oxygen vacancy available for oxygen ions to diffuse.
18. Please write down the defect chemical reaction of the dissolution of NaF in YF3. (8)
NaF
YF3Na
Y
''
+
2
V
F.
+
F
F3
NaF
YF3Na
Y''
+
2
Na
i.
+
3
F
F19. The graph below shows how the diffusivity of lightly doped NaCl changes with
temperature, please label the intrinsic and extrinsic part and briefly explain why the two
lines have different slope. (16)
Extrinsic region:
o is dominant where vacancy concentration is constant and independent of temperature as it is
determined by the solution concentration i.e. [VNa'] = [CdNa•] .
o Defect migration
Intrinsic Region:
o the vacancy concentration is governed by thermally intrinsic defect creation mechanism
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(Schottky defect formation)
o high temperature intrinsic diffusivity exhibits a steeper slope with higher activation energy which include not only the energy for defect migration but also for defect creation. (Defect Migration and Creation)
20. Please draw the graph of polarization(P) vs. electric field(E) for a ferroelectric material,
and then point out the remnant and saturation polarization, and the coercive field on the
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21. Use your knowledge about the effect of structure on dielectric constant and match the
following materials into the appropriate boxes to arrange them in order of the dielectric
constant. (12)
Material k Material k
Lead Zirconium Titanate ~2000 Natural Rubber ~ 4
Hydrogen fluoride ~80 Polypropylene ~ 2
Polyvinylidene fluoride ~10 Nitrogen ~ 1
Polyvinylidene fluoride (PVDF) H
F H
F
Hydrogen fluoride
Lead zirconium titanate (PZT)
Polypropylene Natural Rubber