Chapter – 3
Structure of Nonmetallic Materials
In Chapter 2 we discussed the structure of metallic materials. In this chapter we shall discuss the
structure of nonmetallic material. The atomic bonding in nonmetallic materials ranges from purely
ionic to totally covalent; many materials exhibit a combination of these two types of bonding.
IONIC CRYSTALS
Typical ionic solids contain discrete charged ions separated by regions of negligible electron
density. These ions are held together by strong electrostatic forces, which account for the large
binding energies, high melting points, hardness and incompressibility of ionic solids. The simplest
examples of ionic solids are the compounds formed by electropositive metals, such as sodium or
potassium with electronegative nonmetals such as chlorine and oxygen. Such compounds mostly
contain simple monatomic ions which pack together as if they were spheres of different sizes.
Hence the crystal structures of these substances are fairly simple, as they are determined by two
factors: the maintenance of electrical neutrality and the relative sizes of the ions.
Electrical Neutrality
If the charges on the anion and the cation are identical, the compound has the formula MX, and the
coordination number for each ion is identical to assure a proper balance of charge. Examples of
compounds having the formula MX include CsCl and NaCl. Although CsCl is a useful example of
Introduction to Crystallography
a compound structure, it does not represent any commercially important ceramics. By contrast the
NaCl structure is shared by many important ceramic materials.
If the valence of the cation is +2 and that of the anion is –1, then twice as may anions must be
present, and the formula is of the form MX2. The structure of the MX2 compound must assure that
the coordination number of the cation is twice the coordination number of the anion. For example,
each cation may have 8 anion nearest neighbors, while only 4 cations will touch each anion.
Included in the MX2 category is the perhaps the most important ceramic compound SiO2.
Ionic Radii
Ionic solids consist of cations and anions. In ionic bonding some atoms lose their outer electrons to
become cations and others gain electrons to become anions. Thus the cations are normally smaller
than the anions they bond with. The number of anions that surround a central cation in an ionic
solid is called the coordination number and corresponds to the number of nearest neighbours
surrounding a central cation. For stability as many anions as possible surround a central cation.
However the anions must make contact with the central cation.
Table 3.1 : Ionic Radii for Several Cations and Anions
Cation
Ionic Radius (nm)
Anion
Ionic Radius (nm)
Ba2+
0.136
Br -
0.196
Ca2+
0.100
Cl -
0.181
Cs+
0.170
F-
0.133
Fe2+
0.077
I-
0.220
Fe3+
0.069
O2-
0.140
K+
0.138
S2-
0.184
Mg2+
0.072
Mn2+
0.067
Na+
0.102
Ni2+
0.069
Si 4+
0.040
Ti 4+
0.061
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Structure of Nonmetallic Materials
The ratio of the radius of the central cation to that of the surrounding anion is called the radius
ratio. The radius ratio when the anions just touch each other and contact the central cation is called
the critical (minimum) radius ratio. The radius ratio influences both the manner of packing and the
coordination number. Interstitial atoms whose radii are slightly larger than the radius of the
interstitial site may enter that site, pushing the surrounding atoms slightly apart. However, atoms
whose radii are smaller than the radii of the hole are not allowed to fit into the interstitial site
because the central cation can rattle around in the cage of anions.
The packing geometry of one type of cation and one type of anion with the cation as the smaller
ion is a function of the ion sizes. This can be worked out from space filling geometry, when the
following conditions corresponding to a stable configuration are satisfied simultaneously.
(i) anions and cations considered as hard spheres always touch each other
(ii) anions generally will not touch but may be close enough to be in contact with one another and
(iii) as many anions as possible surround a central cation for the maximum reduction of
electrostatic energy.
When the cation is very small compared to the anion, only two anions can be neighbours to the
cation in order to satisfy the above three conditions.
From the simple geometry it can be shown that the ratio of the cation to anion radius r c/ra for the
configuration in which the three surrounding anions are touching one another and also the central
cation (Fig. 3.1) is 0.155. This triangular configuration is one of the critical configurations and the
radius ratio is said to be a critical value, because for smaller values of r c/ra not all the three anions
would touch the central cation [Fig. 3.1(b)].
(a)
(b)
(c)
Fig. 3.1: Triangular Coordination of Anions around a Central Cation
(a) The Critical Condition (b) Unstable Configuration (c) Stable but not Critical
This violates condition (i) above and leads to instability. Here, the only way to satisfy all three
conditions is to reduce the number of anions to two. For values of r c/ra greater than 0.155, all the
anions do not touch one another. But all three conditions are still satisfied. This situation will
prevail till the radius ratio is increased to 0.225, the next higher critical value corresponding to a
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Introduction to Crystallography
tetrahedral (four) coordination. The possible values of coordination number and radius ratio ranges
in which they are stable are listed in Table 3.2.
Table: 3.2: Coordination Number as a Function of Radius Ratio
Coordination Number
Range of Radius Ratio
Configuration
2
0 – 0.155
3
0.155 – 0.225
Triangular
4
0.225 – 0.414
Tetrahedral
6
0.414 – 0.732
Octahedral
8
0.732 – 1.00
Cubic
12
1.0
Linear
FCC or HCP
The three conditions for stability is not always valid. If the directional characteristics of bonding
persist to any significant degree (as in partial covalency) the considerations based on close packing
geometry alone will fail. The nearest neighbour interactions can also affect the local packing
geometry. However, in a number of cases the coordination rules outlined above are obeyed.
Example 3.1: Find the critical radius ratio for a triangular coordination.
Solution:
In Fig. 3.2 triangle ABC is an equilateral triangle and line AD bisects angle CAB. Thus angle
DAE = 30o. Thus
AD = R + r
cos 30o =
AE
R
=
= 0.866
AD
Rr
R = 0.866(R + r)
= 0.866R + 0.866r
0.866r = R – 0.866R = R(0.134)
r
= 0.155
R
Fig. 3.2 : Determination of Critical Radius for
Triangular Coordination
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Structure of Nonmetallic Materials
Example 3.2: Find the critical radius ratio for tetrahedral coordination
P
Q
R
T
U
S
(a)
(b)
P
Fig. 3.3: Determination of Critical Radius Ratio for Tetrahedral Coordination
Solution
At the critical condition in this case, four anions at the four corners of a tetrahedron touch one
another and also the cation at the centre of the hole between. Referring to Fig. 3.3, the normal from
the apex to the base of the tetrahedron PT is
2 a/ 3 , where a is the side of the tetrahedron. Here
a = 2 ra
The centre of the tetrahedral hole lies at three-fourths of the distance from the apex. Therefore,
rc + ra
= ¾ PT = ¾ x
2 / 3 x 2 ra
= 1.225 ra
rc/ra
= 0.225
If the interstitial atom becomes too large, it prefers to enter a site having a larger coordination
number. Therefore, an atom whose radius ratio is between 0.225 and 0.414 will enter a tetrahedral
site; if the radius is somewhat larger than 0.414, it will enter an octahedral site. When the atoms
have the same size, as in pure metals, the radius ratio is one and the coordination number is 12,
which is the case for metals with FCC and HCP structures.
Ionic solids cannot form close packed structures like FCC or HCP because the atoms are not of the
same size. The charge on the ions requires an alternating arrangement of anions and cations. The
type of structure is mainly determined by the packing of the larger anions which are normally the
negative ions. The structure frame work is made by anions within which the small cations (positive
ions) fit interstitially. Three common ionic crystal structures are the sodium chloride and caesium
chloride and zinc blende structures.
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Introduction to Crystallography
Sodium Chloride Structure
The radius ratio for sodium and chloride ions is rNa /rCl = 0.97/1.81 = 0.536; the sodium ion has a
charge of + 1, while the chloride ion has a charge of – 1. Therefore, based on the charge balance
and radius ratio, each anion and cation must have a coordination number of six. The FCC
structure, with Na cations at FCC positions and Cl anions at the four octahedral sites, will satisfy
these requirements [Fig. 3.4(a)]. We can also consider this structure to be FCC with two ions – one
Na and one Cl – associated with each lattice point. Many ceramics, including MgO, CaO, and FeO
have this structure.
Cl- (b)
Cs+
Cl- (a)
Na+
Fig. 3.4: Ionic Crystal Structures
(a) Unit Cell of NaCl (b) Unit Cell of CsCl
Cesium Chloride Structure
Cesium chloride (CsCl) is an ionic structure showing two interpenetrating simple cubic lattices,
one for each kind of ion [Fig. 3.4(b)]. The radius ratio, rCs /rCl = 1.67/1.81 = 0.92, dictates that a
cation is surrounded by eight anions. We can characterise the structure as a simple cubic structure
with two ions – one Cs and one Cl – associated with each lattice point. This structure is possible
when the anion and the cation have the same valence.
Example 3.3: Predict the coordination number for the ionic solids CsCl and NaCl. Use the
following ionic radii for the prediction:
Cs+ = 0.170 nm Na+ = 0.102 nm Cl- = 0.181 nm
Solution
The radius ratio for CsCl is
r(Cs  )
R (Cl )
=
0.170 nm
= 0.94
0.181nm
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Structure of Nonmetallic Materials
This ratio is greater than 0.732, CsCl should show cubic coordination (CN = 8), which it does.
The radius ratio for NaCl is
r(Na  )
R (Cl )
=
0.102nm
= 0.56
0.181nm
Since this ratio is greater than 0.414 but less than 0.732, NaCl should show octahedral
coordination (CN = 6), which it does.
Example 3.4: Calculate the ionic packing factor for CsCl. Ionic radii are Cs+ = 0.170 nm and Cl- =
0.181 nm.
Solution:
The ions touch each other across the cube diagonal of the CsCl unit cell, as shown in Fig. 3.5.
Fig. 3.5: Cube Diagonal of CsCl Unit Cell
Let r = Cs+ ion and R = Cl- ion. Thus
√3a
= 2r + 2R
= 2(0.170 nm + 0.181 nm)
a
= 0.405 nm
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Introduction to Crystallography
4 3
4
π r (l Cs  ion)  π R 3 (l Cl  ion)
3
3
CsCl ionic packing factor =
3
a
4
4
π (0.170 nm)3  π (0.181nm)3
3
= 3
(0.405 nm)3
= 0.68
Example 3.5: Show that MgO can have the sodium chloride crystal structure and calculate the
density of MgO. [Given: rMg = 0.66 and rO = 1.32 A]
Solution
rMg
rO
=
0.66
= 0.50
1.32
Since 0.414 < 0.50 < 0.732, the coordination number is six and the NaCl structure is possible.
The atomic weights are 24.3 and 16 g/g. mole for magnesium and oxygen, respectively. The ions
touch along the edge of the cube, so
a0
= 2rMg + 2ro = 2(0.66) + (1.32) = 3.96 A
ρ
=
(4 Mg ions) (24.3)  (4 0 ions) (16)
(3.96 x 10 8 cm) 3 (6.02 x 10 23 )
= 4.31 Mgm-3
Example3.6: For KCl, (a) verify that the compound may have the cesium chloride structure and
(b) calculate the packing factor for the compound. [Given: rK = 1.33 A and rCl = 1.81 A]
Answer
(a)
rk
1.33
=
= 0.735
rCl
1.81
Since 0.732 < 0.735 < 1.000, the coordination number is eight and the CsCl structure is likely.
76
Structure of Nonmetallic Materials
(b) The ions touch along the body diagonal of the unit cell, so
3 a 0 = 2 rK + 2 rCl
= 2 (1.33) + 2 (1.81) = 6.28A
a 0 = 3.63A
4π 3
4π 3
rK (1 K ion) 
rCl (1 Cl ion)
3
3
Packing factor =
a 30
4π
4π
(1.33) 3 
(1.81)3
3
3
=
= 0.725
(3.63) 3
Example 3.7: Calculate the density of NaCl from its crystal structure, the ionic radii of Na+ and
Cl- ions, and the atomic masses of Na and Cl. The ionic radius of Na+ = 0.102 nm and that of Cl =
0.181 nm. The atomic mass of Na = 22.99 g/mol and that of Cl = 35.45 g/mol.
Solution
The Cl- ions in the NaCl unit cell form an FCC-type atom latice, and the Na+ ions occupy the
interstitial spaces between the Cl- ions [Fig. 3.6(a)].
(a)
(b)
Fig. 3.6: (a) NaCl Unit Cell lattice Points Showing the Positions of the Ions.
(b) Octahedral Coordination of Cl-anions around a central Sodium Cation.
There is equivalent of one Cl- ion at the corners of the NaCl unit cell since 8 corners × 1/8 ion, and
there is the equivalent of three Cl- ions at the faces of the NaCl unit cell since 6 faces ½ ion = 3 Clions, making a total of four Cl- ions per NaCl unit cell. To maintain charge neutrality in the NaCl
77
Introduction to Crystallography
unit cell, there must also be the equivalent of four Na+ ions per unit cell. Thus there are four Na+
Cl- ion pairs in the NaCl unit cell.
Fig. 3.7: Cube face of NaCl Unit Cell
(Ions in contact with the cube edges)
To calculate the density of the NaCl unit cell, we shall first determine the mass of one NaCl unit
cell and then its volume. Knowing these two quantities, we can calculate the density m/V.
The mass of a NaCl unit cell is
m=
(4Na   22.99 g/mol)  (4Cl   35.45 g/mol)
6.02  10 23 atoms (ions)/mol
= 3.88 × 10-22 g
the volume of the NaCl unit cell is equal to α3, where α is the lattice constant of the NaCl unit cell.
The Cl- and Na+ ions contact each other along the cube edges of the unit cell, as shown in Fig. 3.7,
and thus
α = 2 ( rNa  R Cl  )
= 2 (0.102 nm + 0.181 nm) = 0.566 nm
= 0.566 nm × 10-7 cm/nm = 5.66 × 10-8 cm
V =a3 = 1.81 × 10-22 cm3
The density of NaCl is
3.88  10 22 g
g
m
p=
=
= 2.14
 22
3
V
1.81 10 cm
cm 3
The handbook value for the density of NaCl is 2.16 g/cm3.
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Structure of Nonmetallic Materials
Example 3.8: Calculate the linear density of Ca2+ and O2- ions in ions per nanometer in the [110]
direction of CaO which has the NaCl structure. (Ionic radii: Ca2+ = 0.106 nm and O2- = 0.132 nm.)
Solution:
The length of the [110] distance across the base face of a unit cube is √2α, where a is the length of
a side of the cube or lattice constant [Fig. 3.8(b)].
Fig. 3.8: (a) NaCl Lattice-points Unit Cell
(b) The [110] Direction
From the cube face of the NaCl unit cell (Fig. 3.7), we see that
a = 2r + 2R.
Thus, for CaO,
a = 2( rCa 2  + R O 2  ) = 2 (0.106 nm + 0.132 nm) = 0.476 nm
The linear density of the O2- ions in the [110] direction is
ρL =
2O
2
2a
=
2O
2
= 2.97 O2-/nm
2(0.47nm)
The linear density of Ca2+ ions in the [110] direction is also 2.97Ca2+/nm if we shift the origin of
the [110] direction from (0, 0, 0) to (0, ½, 0),
Thus, there are 2.97(Ca2+ or O2-)/nm in the [110] direction.
Example 3.9: Calculate the planar density of Ca2+ and O2- ions in ions per square nanometer on
the (111) plane of CaO (NaCl structure, Ionic radii; Ca2+ = 0.106 nm and O2- = 0.132 nm.)
Solution:
If we consider, the anions (O2- ions) to be located at the DCC positions (Fig. 3.9) for the Cl- ions of
then the (111) plane contains the equivalent of two anions.
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Introduction to Crystallography
[ 3 × 60o = 180o = ½ anion + (3 ×½) anions at each midpoint of the sides of the (111) planar
triangle of Fig. 3.9(b) = a total of 2 anions within the (111) triangle].
Anions or
cations
(a)
(b)
Fig. 3.9: (a) NaCl Unit Cell lattice Points Showing the Positions of the Ions.
(b) The (111) Plane in the Unit Cell
The lattice constant for the unit cell a = 2(r + R) = 2(0.106 nm + 0.132 nm) = 0.476 nm. The
planar area A = ½ bh, where h = 3 /2 a2. Thus,
A = (½ √2α)(√ 3 /2 a) =
3 /2 a2 = √ 3 /2 (0.476 nm)2 = 0.196 nm2
The planar density for the O2- anions is
2(O 2  ions)
0.196nm 2
= 10.2 O2- ions/nm2
The planar density for the Ca2+ cations is the same if we consider the Ca2+ to be located at the
FCC lattice points of the unit cell, and thus
ρ planar (CaO) = 10.2(Ca2+ or O2-)/nm2.
Zinc Blende Structure
The zinc blende structure has the chemical formula ZnS. Although the Zn ions have a charge of
+2 and S has a charge of –2, zinc blende (ZnS) cannot have the sodium chloride structure because
rZn /rS = 0.74/1.84 = 0.402.
This radius ratio demands a coordination number of four, which in turn means that the sulfide ions
will enter tetrahedral sites in a unit cell. The FCC structure, with Zn cations at the normal lattice
points and S anions at one-half of the tetrahedral sites, can accommodate the restrictions of both
charge balance and coordination number.
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Structure of Nonmetallic Materials
(a)
Zn
S
(b)
Ti4+
Ba2+
O2-
Fig. 3.10: (a) Crystal Structure of Cubic Zinc Sulphide
(b) Perovskite Crystal Structure
The zinc blende structure [Fig. 3.10(a)] may be regarded as two interpenetrating(a)face centred cubic
lattices of the elements, with the corner of one located at the position of ¼, ¼, ¼ of the other.
There are four molecules of ZnS per conventional cell. About each atom there are four equally
distant atoms of the opposite kind arranged at the corners of a regular tetrahedron.
It is also possible for ceramic compounds to have more than one type of cation, BaTiO3 having
both Ba2+ and Ti4+ cations falls into this classification. This material has a perovskite crystal
structure. The unit cell of this structure is shown in Fig. 3.10(b); the Ba2+ ions are situated at all
eight corners of the cube and a single Ti4+ is at the cube centre, with O2- ions located at the centre
of each of the six faces.
Example 3.10 : Based on charge balance and radius ratio, which of the cubic structures we have
discussed would BeO most likely have? [Given: rBe = 0.35, rO = 1.32]
Solution
The radius ratio is
rBe
0.35
=
= 0.265
1.32
ro
This radius ratio requires a coordination number of four. Of the cubic structures we have
examined, only the zinc blende structure, with half of the tetrahedral sites filled, satisfies both the
coordination number and charge balance requirements.
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Introduction to Crystallography
Example 3.11: Calculate the density of zinc blends (ZnS). Assume the structure to consist of ions
and that the ionic radius of Zn2+ = 0.060 nm and that of S2- = 0.174 nm.
Solution:
Fig. 3.11: ZnS Structure showing Relationship between Lattice
Constant and radii of Sulphur and Zinc Atoms/Ions
Density =
massofunitcell
volumeofunitcell
There are four zinc ions and four sulfur ions per unit cell. Thus Mass of unit cell
=
(4Zn 2   65.37g/mol)  (4S 2   32.06g/mol)
6.02 10 23 atoms/mol
= 6.47 × 10-22 g
Volume of unit cell = α3
From Fig. 1.54,
3
α = rzn2+ + Rs2- = 0.060 nm + 0.174 nm = 0.234 nm
4
α = 5.40 × 10-8 cm
α = 1.57 × 10-22+ cm3
mass
6.47  10 22 g
=
= 4.12 g/cm3
 22
3
volume
1.57 10 cm
The handbook value for the density of ZnS (cubic) is 4.10 g/cm3.
Thus,
Density =
82
Structure of Nonmetallic Materials
Representing Crystals in projections: Crystal Plans
The more complicated the crystal structure and the larger the unit cell, the more difficult it is to
visualize the atom or ion positions from diagrams or photographs of models, atoms or ions may be
hidden behind others and therefore not seen. Another form of representation, the crystal plane or
crystal projection, is needed, which precisely shows the atomic or ionic positions in the unit cell.
The first step is to specify the axes x, y and z from a common origin and along the sides of the cell.
The atomic or ionic positions or coordinates in the unit cell are specified as fractions of the cell
edge length in the order x, y, z. Thus in the bcc structure the coordinates of the atom/ion at the
origin is (000) and that at the centre of the cube is (½ ½ ½). As all eight corners of the cube are
equivalent positions i.e., any of the eight corners can be chosen as the origin, (000) specifies all the
corner atoms, and the two coordinates (000) and (½ ½ ½) are equal to the two atoms/ions in the
bcc unit cell. In the fcc, with four atoms/ions per unit cell, the coordinates are: (000), (½ ½ 0),
(½ 0 ½), (0 ½ ½).
(a)
(b)
Fig. 3.12: Plan of (a) BCC Structure (b) CCP or HCP Structure
Fig. 3.13: Alternate Unit Cells of the Perovskite Strucutre
Crystal projections or plans are usually drawn perpendicular to the z-axis, and Fig. 3.12 are plans
of the bcc and fcc structures respectively. Note that only the z-coordinates are indicated in these
diagrams; the x and y coordinates need not be written down because they are clear from the plan.
Similarly, no z coordinates are indicated for all the corner atoms because all eight corners are
equivalent positions in the structure, as mentioned above.
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Introduction to Crystallography
Sketching crystal plans helps one to understand the similarities and differences between structures,
in fact, it is very difficult to understand them otherwise. For example, Fig. 3.13 (a) and (b) show
how the same crystal structure (perovskite, CaTiO3). They look different because the origins of the
cell have been chosen to coincide with different atoms/ions.
COVALENT CRYSTALS
Elements from the central groups of the Periodic table are not readily ionised. The energy required
to remove all the valence electrons is too large for ionic bonding to be possible. However, it is still
possible for each atom to complete its outer shell by sharing electrons with its neighbours. Thus in
covalent bonding arrangement each atom has 8 – N neighbours.
The constituent atoms are linked by covalent bonds in one, two or three dimensions. The covalent
bonds themselves are strong, but they only confer strength and resistance to compression in the
directions in which they occur. Hence, the mechanical properties of covalent solids depend
markedly on whether the covalent bonds extend in one, two or three dimensions. Covalent solids
do not form close packed structures because the directional nature of the covalent bonds must be
maintained. The simplest covalent structure is that of diamond (and of silicon and germanium).
(a)
(b)
Fig. 3.14: (a) Covalent Bonding: Atoms of Group IV each has Four Neighbours.
(b) Crystal Structure of Diamond Showing Tetrahedral Bond Arrangement:
Covalently bonded materials frequently must have complex structures in order to satisfy the
directional restraints imposed by the bonding.
Diamond Cubic Structure
Elements such as silicon, germanium, and carbon in its diamond form are bonded by four covalent
bonds and produce a tetrahedron (Fig. 3.14). The coordination number for each silicon atom is
only four, due to the nature of the covalent bonding.
84
Structure of Nonmetallic Materials
As these tetrahedral groups are combined, a large cube can be constructed [Fig. 3.14(b)]. This
large cube contains eight smaller cubes that are the size of the tetrahedral cube; however, only four
of the cubes contain tetrahedra. The large cube is the diamond cubic, or DC, unit cell. The lattice is
a special FCC structure. The atoms on the corners of the tetrahedral cubes provide atoms at each of
the regular FCC lattice points. However, four additional atoms are present within the DC unit cell
from the atoms in the center of the tetrahedral cubes.
We can describe the DC lattice as an FCC lattice with two atoms associated with each lattice point.
Therefore, there must be eight atoms per unit cell.
Example 3.11: Determine the packing factor for DC silicon.
Solution
The atoms touch along the body diagonal of the cell (Fig. 3.15). Although atoms are not present at
all locations along the body diagonal, there are voids that have the same diameter as atoms.
a0
r
2r
2r
r
2r
8r =
3 a0
Fig. 3.15: Relation between Lattice parameter and
Atomic Radius in Diamond Cubic
Consequently,
3 a0 = 8r
Packing Factor
4
(8 atoms/cell) ( π r 3 )
3
=
a 30
4
(8) ( π r 3 )
3
=
(8 r/ 3 ) 3
= 0.34
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Introduction to Crystallography
The low pressure crystal form of carbon is graphite which differs in many respects from diamond.
It is a layer structure in which the atoms are covalently bonded to only three nearest neighbours in
the layer, and the layers themselves are only weakly bonded by van der Waals forces (Fig. 3.14).
Fig. 3.14: The Layer Structure of Graphite
Since there is one electron per atom that is not employed in the covalent bonding there is an
abundant supply of free electrons, and so graphite is a good electrical conductor whereas diamond
is an almost perfect insulator. However, the free electrons cannot cross easily from layer to layer
and so the resulting conductivity is highly anisotropic. It was believed that graphite owed its
lubricating properties to the weak van der Waals bonds between adjacent layers. It has now been
shown that lubrication is associated with adsorbed gas layers such as oxygen or organic molecules.
CRYSTALLINE POLYMERS
Polymers involve molecules rather than just atoms or ions, as with metals and ceramics. The
atomic arrangement is, therefore, more complex for polymers. We think of polymer crystallinity as
the packing of molecular chains so as to produce an ordered atomic array.
As a polymer crystallizes from the molten state, a certain amount of amorphous polymer will also
be present within the structure (to accommodate mismatch and other defects within the crystal).
For his reason most crystalline polymers are actually semicrystalline, in which the percentage of
crystalline phase can range from 40 to 90 percent.
Polymer properties are significantly affected by the degree of crystallinity. Crystalllinity tends to
increase the mechanical properties such as tensile strength and hardness while diminishing
ductility, toughness and elongation.
86
Structure of Nonmetallic Materials
The density of a crystalline polymer will be greater than an amorphous one of the same material
and molecular weight, since the chains are more densely packed together for the crystalline
structure.
The degree of crystallinity by weight may be determined from accurate density measurements
according to
% Crystallinity =
ρ c (ρ s  ρ a }
x 100
ρ s (ρ c  ρ a)
Where, ρ s = the density of specimen for which the percent crystallinity is to be determined
ρ c = density of the perfectly crystalline polymer
ρ a = density of the totally amorphous polymer.
The values of ρ c and ρ a must be measured by other experimental means.
Mer Unit
Fig. 3.15: The Unit Cell of Crystalline Polyethylene
The dashed lines in Fig. 3.15 outline the unit cell for the lattice of polyethylene. Polyethylene is
obtained by joining C2H4 molecules together to produce long polymer chains that form an
orthorhombic unit cell. Some polymers, including nylon, can have several polymorphic forms.
The degree of crystallinity may range from completely amorphous to almost entirely (up to 90
percent crystalline); by way of contrast metal specimens are almost always entirely crystalline.
The degree of crystallinity of a polymer depends on the rate of cooling during solidification as
well as on the chain configuration. During crystallization upon cooling through the melting
temperature, the chains which are highly random and entangles in viscous liquid, must assume an
ordered configuration. For this to occur, sufficient time must be allowed for the chains to move
and align themselves.
On the other hand, crystallization is not easily prevented in chemically simple polymers such as
polyethylene, even for very rapid cooling rates. For linear polymers, crystallization is easily
accomplished because there are virtually no restrictions to prevent chain alignment. Any side
branches interfere with crystallization, such that branched polymers never are highly crystalline; in
fact excessive branching may prevent any crystallization whatsoever.
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Introduction to Crystallography
For copolymers as a general rule, the more irregular and random the mer arrangement, the greater
is the tendency for the development of noncrystallinity. For alternating and block copolymers there
is some likelihood of crystallization. On the other hand, random and graft copolymers are normally
amorphous.
Example 3.12: How many carbon and hydrogen atoms are in each unit cell of crystalline
polyethylene? There are twice as many hydrogen atoms as carbon atoms in the chain. .ρ of
polyethylene= 0.95 Mgm-3.
Solution
If we let x be the number of carbon atoms, then 2x is the number of hydrogen atoms.
p=
(x) (12 g/g. mole)  (2x) (1g/g. mole)
(7.41 x 10 -8 ) (4.94 x 10 8 ) (2.55 x 10 8 ) (6.02 x 10 23 )
0.95 =
14 x
56.2
x = 3.8 ≃ 4 carbon atoms per cell
2 x = 8 hydrogen atoms per cell
To some extent, the physical properties of polymeric materials are influenced by the degree of
crystallinity. Crystalline polymers are usually stronger and more resistant to dissolution and
softening by heat.
Polymer single crystals as thin platelets and having chain folded structures may be grown from
dilute solutions. Many semicrystalline polymers form spherulites. Spherulites are considered to be
polymer analog of grains in polycrystalline metals and ceramics. Each spherulite is really
composed of many different lamellar crystals and in addition some amorphous material.
Polyethylene, polypropylene, polyvinyl chloride and nylon form a spherulite structure when they
crystallize from a melt.
PROBLEMS AND QUESTIONS
3.1
Show that the minimum cation to anion ratio for a coordination number of 4 is 0.225.
3.2
Show that the minimum cation to anion ratio for a coordination number of 6 is 0.414.
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Structure of Nonmetallic Materials
[Hint: Use the NaCl structure and assume that anions and cations are just touching along the
cube edges and across face diagonals.]
3.3
Show that the minimum cation to anion ratio for a coordination number of 8 is 0.732.
3.4.
Using the ionic radii given in Table 3. , determine the coordination number expect6ed for
the following compounds.
(a) FeO,
(b) CaO
(c) SiC
(d) PbS
(e) B2O3
[ Ans. (a) NaCl (d) CsCl ]
3.5.
Using the ionic radii given in Table 3. , determine the coordination number expect6ed for
the following compounds.
(a) Al2O3
(b) TiO2
(c) MgO
(d) SiO2
(e)
3.6
Based on the ionic radius ratio and the necessity for charge balance, which of the cubic
structures would you expect CdS to possess?
3.7
Based on the ionic radius ratio and the necessity for charge balance, which of the cubic
structures would you expect CoO to possess?
3.8
The compound NiO has the sodium chloride crystal structure. Calculate (a) the lattice
parameter (b) density and (c) packing factor for NiO.
3.9
One of the forms of BeO has the ZnS structure at high temperature. Determine (a) the lattice
parameter, (b) the density and (c) the packing factor for the compound.
3.10 (a) Germanium has diamond cubic structure with a lattice parameter of 5.6575A. Calculate
the size of he germanium atom in the unit cell.
(b) Does this best match up with germanium’s atomic radius or ionic radius?
3.11 Calculate the fraction of [111] direction and the fraction of the (111) plane actually covered
by atoms in the diamond cubic unit cell.
3.12 Calculate the fraction of [111] direction and the fraction of the (111) plane actually covered
by sodium ions in the NaCl cubic unit cell.
3.13 Compute the atomic packing factor for the CsCl crystal structure in which the radius ratio is
0.732.
3.14 The ZnS crystal structure is one that may be generated from close packed planes of anions.
(a) Will the packing sequence for this structure be FCC or HCP? Why?
(b) Will cations fit tetrahedral or octahedral positions? Why?
(c) What fraction of positions will be occupied?
[ Ans. (a) FCC
(b) Tetrahedral
(c) one=half]
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Introduction to Crystallography
3.15 Iron oxide (FeO) has the rock salt crystal structure and a density of 5.70 g/cm3.
(a) Determine the unit cell edge length.
(b) How does the result compare with the edge length as determined from radii in Table 3.
assuming that Fe 2+ and O 2- ions just touch each other along the edges.
[Ans. (a) a = 0.437 nm
(b) a = 434 nm]
3.16 A hypothetical AX type of ceramic material is known to have a density of 2.65 g/cm3 and a
unit cell of cubic symmetry with a cell edge length of 0.43 nm. The atomic weights of the A
and X elements are 86.6 and 40.3 g/mol respectively. On the basis of the information, which
of the following crystal structures ia (are) possible for this material: rock salt, cesium
chloride or ZnS? Justify your choice(s).
[ Ans. Cesium Chloride]
3.17 Compute the atomic packing factor for the diamond cubic crystal structure. Assume that the
bonding atoms touch one another, that the angle between adjacent bonds is 109.5o, and that
each atom internal to the unit cell is positioned a/4 of the distance away from the two
nearest cell faces (a is the unit cell edge length).
[ Ans. APF = 0.68]
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