Chapter 3

Chapter 3
What is the difference between a crystal and a quasicrystal?
Crystals are solids in which all of the atoms occupy well-defined locations, ordered
across the whole of the material. They are defined in terms of a unit cell, a brick-like shape
which contains all of the atomic species in the crystal, and which, when stacked in threedimensions, reproduces the macroscopic crystal. Because of space-filling requirements, a
unit cell with overall five-fold, or greater than six-fold, rotation symmetry cannot exist,
because none can be packed to form a crystal.
In 1984, a metallic alloy Al88Mn12 was reported which possessed a ten-fold rotational
symmetry axis and long-range translational order and since then, many other alloys that
show five-fold, eight-fold, ten-fold and twelve-fold rotation symmetry have been discovered.
As these structures are incompatible with classical crystallography they are known as
quasiperiodic crystals, or more compactly, as quasicrystals. A quasicrystal is made up
certain structural units, all oriented in the same way, and separated by variable amounts of
disordered material.
The essential difference between the two is that crystals show complete long-range
order whereas quasicrystals show long-range orientational order but not translational order.
What are point defects?
Point defects are mistakes in a crystal at a single atom site. In a pure crystal, these
are atom vacancies or atoms in normally unoccupied sites, called interstitials.
occur in all pure solids and are called intrinsic defects.
They may
Foreign atoms, impurities or
dopants constitute extrinsic defects. If these atoms occupy sites normally occupied by the
parent atom they form substitutional defects and if they occupy normally unoccupied sites
they constitute interstitial defects.
In ionic compounds the charges must remain balanced when point defects are
introduced into the crystal.
The defects arising from balanced populations of cation and
anion vacancies are known as Schottky defects. Defects that arise when one ionic species
is displaced to an interstitial site leaving a vacancy are known as Frenkel defects.
Non-stoichiometric compounds often exhibit a composition range due to unbalanced
populations of point defects.
For example, calcia-stabilized zirconia, exists over a wide
composition range and for practical purposes calcia-stabilised zirconia Ca2+xZr4+1-xO2-x has
a variable composition due to the presence of vacancies on oxygen ion sites.
What is the microstructure of a solid?
The microstructure of a solid is a general name for the structures that exist at a submillimetre scale and which have a dominant effect on the properties of the material. As the
name implies, the microstructure of an object is observed using optical microscopy.
The
scale range of the microstructure of a solid can be thought of as approximately 10-4 – 10-6
m.
For example, good quality ceramics have a microstructure that is a mixture of
crystallites in glass.
At the lower end of the microstructure scale defects such as
precipitates play an important role in controlling properties.
Quick quiz
1. c; 2. a; 3. c; 4. b; 5. c; 6. a; 7. b; 8. a; 9. b; 10. a; 11. b; 12. b; 13. c; 14. c;
15. a; 16. c; 17. b; 18. b; 19. c.
Calculations and questions
1. (b) 1.64 x 10-21 J; (c) 3.86 x 10-10 m.
2. 119 K.
3. 5.93 kJ mol-1.
4. (a) 4.97 x 10-22 J; (b) 3.13 x 10-10 m.
5. A = -6.72 x 10-134; B = -2.47 x 10-77.
7. 1.99 x 10-4.
8. 4.52 x 10-4.
9. 5.05 x 1023 m-3.
10. 2.80 x 1024 m-3.
11. 500 K, 1.80 x 10-13; 1000 K, 4.24 x 10-7.
12. 1.74 x 1020 m-3.
13. 3.26 x 10-10.
14. 1.30 x 10-9 m-3.
15. 194 kJ mol-1
16. 2.98 x 1016 m-3; 6.12 x 1021 m-3.
17. 2.09 x 1027 m-3.
18. 2.57 x 1014 m-3.
19. 5.62 x 1025 m-3.
20. 6.9 x 1024 m-3.
21. 225.2 kJ mol-1.
23. Y0.165Zr0.835O1.917; Y3+ substitutes for Zr4+; every two Y3+ gives one oxygen vacancy.
24. (a) Ca0.105Bi1.895O2.947; (b) Ca2+ substitutes for Bi3+; 1 oxygen vacancy for every 2 Ca2+
added.
25.
a. Li+ substitutes for Ca2+, Br- vacancies; b. Ca2+ substitutes for Li+, Li+ vacancies; c.
Mg2+ substitutes for Fe3+, O2- vacancies; Mg2+ substitutes for Ni2+.
26.
a. Cd2+ substitutes for Na+, Na+ vacancies; b. Na+ substitutes for Cd2+, Cl- vacancies;
c. Sc3+ substitutes for Zr4+, O2- vacancies; Zr4+ substitutes for Hf4+.
28. 2.24 x 1028 m-3; 2.23 x 1028 m-3.
Solutions
1 The Lennard-Jones constants for argon are A = 1.78 x 10-134 J m12 and B = 1.08 x 10-77
J m6.
(a) Plot the attractive and repulsive potential energies; (b) estimate the minimum
potential energy of the pair, which can be considered the bonding energy of a molecule of
argon; (c) the equilibrium interatomic separation of this molecule.
The Lennard-Jones potential is:
VLJ = Ar--12 - Br--6
(a)
(b, c) From the plots:
Vmin ~ -1.6 x 10-21 J; rmin ~ 0.386 nm;
Note: These values can be calculated analytically:
differentiate VLJ to give:
dV / dr = -12 Ar-13 + 6 Br-7
At a mimimum dV / dr = 0 = -12 Ar-13 + 6 Br-7 at rmin so
12 A rmin-13 = 6 B rmin-7
rmin
= [2A / B]1/6
Substituting for A and B gives rmin = 3.858 x 10-10 m
Vmin occurs at rmin so
Vmin = A rmin --12 - B rmin --6
rmin
= [2A / B]1/6
B1/6 = 2 A1/6 / rmin
B = 2 A / rmin6
Vmin = A rmin --12 - 2 A rmin --12 = - A rmin --12
= -1.78 x 10-134 / (3.858 x 10-10)-12 = 1.64 x 10-21 J
2
By equating the thermal energy (kBT, where kB is the Boltzmann constant), with the
bonding energy, estimate the temperature at which argon atoms are likely to start to form
pairs, and so form a liquid, using the data in question 1. Compare this to the boiling point
of argon (Q. 22).
Vmin is equal to ~ -1.6 x 10-21 J
If the thermal energy is equated to the bonding energy
Vmin ~ kBTb
where Tb is the boiling point of argon. Hence:
Tb = 1.6 x 10-21 / 1.38 x 10-23 = 116 K
or
Tb = 1.64 x 10-21 / 1.38 x 10-23 = 119 K
3
If the atoms in liquid argon are surrounded by twelve nearest neighbours on average,
estimate the energy of evaporation of the liquid.
To dissociate a cluster of 12 atoms we need Vsub where Vsub is an approximation to the
sublimation energy. If the energy between a pair of atoms is Vmin, then approximately:
Vsub ~ ½ (12 Vmin) = 6 Vmin
Vsub ~ 6 x 1.64 x 10-21 per atom
~ 6 x 1.64 x 10-21 x NA J mol-1
~ 5.93 kJ mol-1
4 The Lennard-Jones constants for neon are A = 4.39 x 10-136 J m12 and B = 9.30 x 10-79 J
m6.
Calculate (a) the bonding energy; (b) the equilibrium separation of a pair of neon
atoms.
(b) Using the methods set out above:
rmin
(a)
= [2A / B]1/6 = [2 x 4.39 x 10-136 / 9.30 x 10-79]1/6 = 3.13 x 10-10 m
Vmin = - A rmin --12
= -4.39 x 10-136 / (3.13 x 10-10)12 = -4.97 x 10-22 J
5 The Lennard-Jones constants for helium are A = 4.91 x 10-137 J m12 and B = 4.16 x 10-80
J m6 and for xenon are A = 2.54 x 10-133 J m12 and B = 5.66 x 10-77 J m6.
Using these
values and those in questions 1 and 4, estimate the Lennard-Jones constants for krypton.
The simplest way to do this is to interpolate between the values for Ar and Xe arithmetically
or graphically. To draw a graph, take log10 of A and B and plot these.
A
log A
B
log B
He
4.91 x 10-137
-136.3089
4.16 x 10-80
-79.3809
Ne
4.39 x 10-136
-135.3575
9.30 x 10-79
-78.0315
Ar
1.78 x 10-134
-133.7496
1.08 x 10-77
-76.9666
2.54 x 10-133
-132.5952
5.66 x 10-77
-76.2472
Kr
Xe
Arithmetically or from a graph: log A = 133.17, A = 6.72 x 10-134
log B = 76.61, B = 2.47 x 10-77
6 Derive the relationship
VLJ = 4 Vmin [(r(0) / r)6 - (r(0) / r)12]
from
VLJ = Ar--12 - Br--6
The following steps lead to the required equation, although the
derivation can be shortened.
When V = 0:
0 = Ar(0)--12 - Br(0)--6
Ar(0)--6 = B
(1)
When V = Vmin : dV / dr = 0, hence: -12 Armin-13 -(-6) Brmin-7 = 0
-12 Armin-13 = -6 Brmin-7
-2 Armin-6 = -B
Equating (1) and (2):
r(0)--6 = 2 rmin-6
rmin = 21/6 r(0)
(2)
Ar(0)--6 = 2 Armin-6
Vmin
=
Armin-12 -(2 Armin-6) rmin-6
Vmin
=
-Armin-12
A =
From (1)
-Vmin rmin12 = -Vmin (21/6 r(0))12 = -4Vmin r(0)12
B = Ar(0)--6
= -4Vmin r(0)6
Substituting: V = -4Vmin r(0)12 r-12 - (-4Vmin r(0)6) r-6
= 4Vmin [ -(r(0) / r-)12 + (r(0) / r-)12]
= 4Vmin [ (r(0) / r-)6 - (r(0) / r-)12]
Note : This equation is often found in the form :
V = 4  [ ( / r-)612 - ( / r-)12]
or
V = 4  [ (r(0) / r-)612 - (r(0) / r-)12]
This equation works if , the energy minimum, is taken as positive.
However,
conventionally a binding energy is negative, as here.
7 The enthalpy of formation of vacancies in pure nickel is H = 97.3 kJ mol-1. What is the
fraction of sites vacant at 1100C?
The fraction of vacant sites is nd / N, where:
nd / N = exp [-97.3 x 103 / (8.31451 x 1373)] = 1.99 x 10-4
8
The enthalpy of formation of vacancies in pure copper is H = 86.9 kJ mol-1.
the fraction of sites vacant at 1084C?
The fraction of vacant sites is nd / N, where:
nd / N = exp [-86.9 x 103 / (8.31451 x 1357)] = 4.52 x 10-4
What is
9 The enthalpy of formation of vacancies in pure gold is H = 123.5 kJ mol-1. The density
of gold is 19281 kg m-3. What number of atom positions is vacant at 1000C?
nd ≈ N exp (-H / kBT)
The relative molar mass of an element of density  contains NA atoms and the value of
N is given by:
N =  x NA / molar mass
where NA is Avogadro’s constant.
N = (19281 x 6.022 x 1023) / 0.197
Hence, the number of vacancies, nd, is:
nd = [(19281 x 6.022 x 1023) / 0.197] exp [-123.5 x 103 / (8.31451 x 1273)]
5.05 x 1023 m-3
10 The enthalpy of formation of vacancies in pure aluminium is H = 72.4 kJ mol-1. The
density of aluminium is 2698 kg m-3. What number of atom positions is vacant at 600C?
nd ≈ N exp (-H / kBT)
The relative molar mass of an element of density  contains NA atoms and the value of
N is given by:
N =  x NA / molar mass
where NA is Avogadro’s constant.
N = (2698 x 6.022 x 1023) / 0.02698
Hence, the number of vacancies, nd, is:
nd = [(2698 x 6.022 x 1023) / 0.02698] exp [-72.4 x 103 / (8.31451 x 873)]
2.80 x 1024 m-3
11
Calculate how the fraction of Schottky defects in a crystal of KCl varies with
temperature if HS is 244 kJ mol-1.
nS / N
≈ exp (-H / RT)
= exp [244000 / (2 x 8.31451 x T)
T/K
nS / N
T/K
nS / N
300
5.73 x 10-22
700
7.88 x 10-10
400
1.17 x 10-16
800
1.08 x 10-8
500
1.80 x 10-13
900
8.31 x 10-8
600
2.39x 10-11
1000
4.24 x 10-7
Note: melting point is 1043 K
12 Calculate the number of Schottky defects in a crystal of KCl at 800 K. The cubic unit
cell of this material has a cell edge of 0.629 nm.
Each unit cell contains 4 K + and 4 Cl-
ions.
nS
≈ N exp (-H / 2RT)
= N exp [244000 / (2 x 8.31451 x 800)
The number of sites, N, is given by [4 / (0.629 x 10-9)3] m-3 = 1.61 x 1028 m-3
Hence,
nS
= 1.74 x 1020 m-3
13 The enthalpy of formation of a Frenkel defect in AgBr, is 1.81 x 10-19 J. Estimate the
fraction of interstitial silver atoms due to Frenkel defect formation in a crystal of AgBr at 300
K.
Take the fraction of interstitial Ag atoms as nF / (N N*)½
nF / (N N*)½
= exp [-1.81 x 10-19 / (2 x kB x 300)]
= exp [-1.81 x 10-19 / (2 x 1.38066 x 10-23 x 300)]
= 3.26 x 10-10
14 AgBr has cubic unit cell with an edge of 0.576 nm. There are four Ag atoms in the unit
cell, and assume that there are four interstitial positions available for Ag atoms.
Calculate
the absolute number of interstitial defects present per cubic metre at 300 K.
Using the result of Q13:
nF =
(N N*)½ 3.26 x 10-10
N = N* = 4, hence:
15
nF =
4 x 3.26 x 10-10 m-3 = 1.30 x 10-9 m-3
Calculate the enthalpy of formation of Frenkel defects in NaBr, using the data on the
number of defects present given in the Table.
Temperature / K
nF / m-3
200
1.428 x 102
300
7.257 x 1010
400
1.636 x 1015
500
6.693 x 1017
600
3.687 x 1019
700
6.468 x 1020
800
5.538 x 1021
900
2.943 x 1022
nF
ln nF
≈ (N N*)½ exp (-HF / 2RT)
= ln (N N*)½ - (HF / 2RT)
To obtain HF plot ln nF versus 1 / T. the slope is equal to (-HF / 2R)
nF / m-3
ln nF
T / K 1/T / (1/K)
1.428 x 102
4.961
200
0.00500
7.257 x 1010 25.008 300
0.00333
1.636 x 1015 35.031 400
0.00250
6.693 x 1017 41.045 500
0.00200
3.687 x 1019 45.054 600
0.00167
6.468 x 1020 47.919 700
0.00143
5.538 x 1021 50.066 800
0.00125
2.943 x 1022 51.736 900
0.00111
The slope of the graph  -11847
Thus -11847 = -HF / 2 x 8.31451 = 194 kJ mol-1
16
The energy of formation of Schottky defects in a crystal of CaO, is given as 6.1 eV.
Calculate the number of Schottky defects present in CaO at 1000C and 2000C.
How
many vacancies are present at these temperatures? CaO has a density of 3300 kg m -3.
nS
≈ N exp (-H / 2RT)
6.1 eV = 6.1 x 96.485 kJ mol-1 = 588.6 kJ mol-1
The number of atom sites, N, = density x NA / molar mass
= (3300 x 6.02214 x 1023) / (0.056077)
At 1000C
nS
= (3300 x 6.02214 x 1023) /(0.056077) exp [-588.6 x 103 / (2 x 8.31451 x 1273)]
= 2.98 x 1016 m-3
At 2000C
nS
= (3300 x 6.02214 x 1023) /(40.078 + 15.999) x 103 exp [-588.6 x 103 / (2 x .31451 x
2273)]
= 6.12 x 1021 m-3
17 Calculate the number of Schottky defects in a crystal of MgO, at 1500C if HS is 96.5
kJ mol-1 and the density of MgO is 3580 kg m-3.
nS
≈ N exp (-H / 2RT)
The number of atom sites, N, = density x NA / molar mass
= (3580 x 6.02214 x 1023) / (0.040304)
= (3580 x 6.02214 x 1023) /(0.040304) exp [-95.6 x 103 / (2 x 8.31451 x 1773)]
nS
= 2.09 x 1027 m-3
18
Calculate the number of Frenkel defects present in a crystal of AgCl, at 300 K, given
that the material has a cubic unit cell of edge 0.555 nm that contains four Ag atoms.
Assume that the interstitial atoms occupy any of eight tetrahedral sites in the unit cell. HF
is 2.69 x 10-19 J.
nF
≈ (N N*) exp (-HF / 2kBT)
To find N, in one unit cell there are 4 Ag atoms, so:
N = 4 / (0.555 x 10-9)3 m-3
To find N*, in one unit cell there are 8 Ag interstitial sites, so:
N* = 8 / (0.555 x 10-9)3 m-3
Hence: nF
23
= [4 / (0.555 x 10-9)3] [8 / (0.555 x 10-9)3] exp [-2.69 x 10-19 / 2 x 1.38066 x 10-
x 300]
nF
19
= 2.57 x 1014 m-3
Calculate the number of vacancies in a crystal of NiO containing Schottky defects, at
1000C, given that HS is 160 kJ mol-1, and the density is 6670 kg m-3.
The number of vacancies in each case is twice the number of Schottky defects, as a
Schottky defect consists of an anion plus a cation vacancy. Thus:
nS
≈ N exp (-H / 2RT)
The number of atom sites, N, = density x NA / molar mass
= (6670 x 6.02214 x 1023) / (0.074692)
nS
=
(6670 x 6.02214 x 1023) /(0.074692) exp [160 x 103 / (2 x 8.31451 x 1273)]
= 2.81 x 1025 m-3
Number of vacancies = 5.62 x 1025 m-3
20
The fraction of Schottky defects in NiO, at 1000C is 1.25 x 10-4.
contains four Ni atoms, and has a cell edge of 0.417 nm.
The cubic unit cell
Calculate the number of nickel
vacancies present.
nS / N = 1.25 x 10-4
The number of nickel vacancies present as a fraction of the cation sites will be the same.
The number of cation sites is N(cation sites) = 4 / (0.417 x 10-9)3 m-3
n(Ni vacancies) = 1.25 x 10-4 x 4 / (0.417 x 10-9)3 m-3
= 6.9 x 1024 m-3
21
The number of Schottky defects in LiF, which has a cubic unit cell containing four Li
and four Cl atoms, with a cell-edge of 0.4026 nm, is 1.12 x 1022 m-3 at 600C.
the energy of formation of these defects.
nS
= 1.12 x 1022 m-3 ≈ N exp (-HS / 2RT)
N = 4 / (0.4026 x 10-9)3 m-3
nS / N = 1.12 x 1022 x (0.4026 x 10-9)3 / 4 = 1.827 x 10-7
exp [-HS / 2x 8.31451 x 873] =
-HS / 2x 8.31451 x 873 =
-HS = 225.2 kJ mol-1
1.827 x 10-7
-15.515
Calculate
22
The melting points and boiling points of the noble gases are given in the table.
Explain these trends, and why the melting and boiling points are so close.
Element Melting point / C Boiling point / C
Helium
-
-268.9
Neon
-248.6
-246.1
Argon
-189.4
-185.9
Krypton
-157.4
-157.4
Xenon
-111.8
-108.0
Radon
-71
-61.7
The interaction responsible for condensation and solidification are weak London dispersion
forces.
These increase as the electron shell volumes (or size of the atoms) grows.
An
intensification in the force means that the gas will condense more readily, i.e. at a higher
temperature and the solid will melt at a higher temperature.
The boiling points and melting points are similar because n additional force between the
atoms arises on condensation.
23 9 mole % of Y2O3 is mixed with 91 mole % ZrO2 and heated until a uniform product with
high oxygen ion conductivity is obtained. The resulting crystal is a stabilised zirconia with
the formula YxZryOz. Determine x, y and z.
We have 0.09 Y2O3 + 0.91 ZrO2
The overall composition is Y0.18Zr0.91O2.09 = M1.09O2.09
However the Y3+ substitutes for Zr4+ so that the metal total remains at 1.0.
For each Y3+
there is 1 O2- vacancy, (Y2O3 instead of Zr2O4), so that the real composition is MO1.917:
Y(0.18 / 1.09)Zr(0.91 / 1.09)O(2.09 / 1.09)
= Y0.165Zr0.835O1.917
24 CaO forms a solid solution with Bi2O3 to give a material with a high anionic conductivity.
If 10 mole % CaO is reacted with 90 mol % Bi2O3; (a) what is the formula of the final solid;
(b) what numbers and types of vacancies have been created?
(a) We have 0.10 CaO + 0.90 Bi2O3
The overall composition is Ca0.1Bi1.8O2.80 = M1.9O2.80
However the Ca2+ substitutes for Bi3+ so that the metal total remains at 2.0. For each Ca 2+
there is 0.5 O2- vacancy, (CaO instead of BiO1.5), so that the real composition is M2O2.947:
Ca(0.1 / 1.9)x2Bi(1.8 / 1.9)x2O(2.8 / 1.9)x2
= Ca1.05Bi1.895O2.947
(b) There is 1 oxygen vacancy per 2 Ca2+ substitutional defect.
25 What defects will form in the crystals made by adding small amounts of compound a to
compound b.
a. LiBr, b. CaBr2.
b. CaBr2, b. LiBr.
a. MgO, b. Fe2O3.
a. MgO, b. NiO.
(a) CaBr2 + LiBr (small quantity): Li+ substitutes on Ca2+ sites; vacancy on Br- sites.
(b) LiBr + CaBr2 (small quantity): Ca2+ substitutes on Li+ sites; Br- interstitials (unlikely due
to size) or Li+ vacancies.
(c) Fe2O3 + MgO (small quantity): Mg2+ substitutes on Fe3+ sites; vacancies on O2- sites.
(d) NiO + MgO (small quantity): Mg2+ substitutes on Ni2+ sites.
26 What defects will form in the crystals made by adding small amounts of compound a to
compound b.
a. CdCl2, b. NaCl.
b. NaCl, b. CdCl2.
c. Sc2O3, b. ZrO2.
a. ZrO2, b. HfO2.
(a) NaCl + CdCl2 (small quantity): Cd2+ substitutes on Na+ sites; Cl- interstitials (unlikely
due to size) or Na+ vacancies.
(b) CdCl2 + NaCl (small quantity): Na+ substitutes on Cd2+sites; vacancies on Cl- sites.
(c) ZrO2 + Sc2O3 (small quantity): Sc3+ substitutes on Zr4+ sites; vacancies on O2- sites.
(d) HfO2 + ZrO2 (small quantity): Zr4+ substitutes on Hf4+ sites.
27 Show that the number of metal atom sites N in a crystal of composition MX is given by
N =  NA / relative molar mass of MX
where  is the density of MX and NA is the Avogadro constant.
The number of moles in  kg of compound is  / molar mass (kg mol-1 not g mol-1).
Each mole contains NA atoms of M and NA atoms of X, so:
N (m-3) =  (kg m-3) NA (mol-1)/ relative molar mass of MX (kg mol-1)
28 Sodium chloride has a density of 2165 kg m -3. The unit cell, which is cubic, with a cell
of edge 0.563 nm, contains four Na and four Cl atoms. Calculate the number of atoms of
Na per cubic metre using (a) density; (b) unit cell data.
(a) From the unit cell: N = 4 / (0.563 x 10-9)3 = 2.24 x 1028 m-3
(b) From the density: N = (2165 x 6.02214 x 1023) / 0.0585 = 2.23 x 1028 m-3