Chapter 4:
Structures and energetics of metallic and
ionic solids
4.1. Introduction
Atoms (and later ions) will be viewed as hard spheres. In the case of pure
metals, the packing pattern often provides the greatest spatial efficiency
(closest packing).
Ionic crystals can often be viewed as a close-packed arrangement of the larger
ion, with the smaller ion placed in the “holes” of the structure.
Unit Cells
Crystals consist of repeating asymmetric units which may be atoms, ions or
molecules. The space lattice is the pattern formed by the points that
represent these repeating structural units.
A unit cell of the crystal is an imaginary parallel-sided region from which the
entire crystal can be built up.
Usually the smallest unit cell which exhibits the greatest symmetry is chosen. If
repeated (translated) in 3 dimensions, the entire crystal is recreated.
2
1
# of Atoms/Unit Cell
For atoms in a cubic unit cell:
Atoms in corners are ⅛ within the cell
Atoms on edges are ¼ within the cell
Atoms on faces are ½ within the cell
3
4.2. Packing of Spheres
Since metal atoms and ions lack directional bonding, they will
often pack with greatest efficiency
How can we stack metal atoms to minimize empty space?
2-dimensions
close packing
(high space filling)
• Coordination # = 6
(# nearest neighbors)
4
versus
primitive packing
(low space filling)
• Coordination # = 4
(# nearest neighbors)
Now stack these 2-D layers to make 3-D structures
2
close packing
One layer (layer A) of close-packed spheres contains hollows
(dimples) that exhibit a regular pattern :
Two dimple types:
Type 1 point UP
Type 2 point DOWN
Note:
dimple types are equivalent since you could rotate
the whole structure 180° and exchange them.
A second layer (layer B) of close-packed spheres can be
formed by occupying every other hollow in layer A.
Two hollow types:
Type 1 : lies over a sphere in layer A
Type 2 : lies over hollows in layer A
5
By stacking spheres over these different types of hollow, two different
third layers of spheres can be produced :
The blue spheres form a new layer C;
this gives an ABC sequence of layers
The third layer replicates layer A;
this gives an ABA sequence
6
3
In an ABC pattern, the unit cell that results is a face-centered cube.
This structure is called cubic close-packed structure (CCP)
In an ABA pattern, the arrangement has a hexagonal unit cell.
This structure is said to be hexagonal close-packed structure (HCP)
7
No matter what type of packing:
the coordination number of each equal size sphere is always 12
The Inter-layer distance
Atomic Packing Factor (APF) or space filling :
d AB = 2R
2
3
APF =
π
= 0.74
3 2
Small holes make up the other 26% of space :
There are two types of holes created by a close-packed arrangement:
For n atoms in a close-packed
structure, there are:
n octahedral holes
2n tetrahedral holes
8
Two layers of close-packed atoms:
4
Coordination number
ABA close-packed arrangement
ABC close-packed arrangement
A
3
A
B
6
B
A
3
12
C
6 coplanar
3 above the plane
3 below the plane
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HCP
Inter-layer distance
d AB = 2R
2
3
Calculation of the distance between two successive layers: dAB
D
e = a = 2R
Layer (B)
D
Layer (A)
C
A
C
A
In tetrahedron ABCD
B
AD 2 = e 2 = AO 2 + DO 2
O
M
Calculation of AO shown in ABC triangle below
e
B
2
e2 =
e
+ DO 2
3
DO = e
2
=h
3
In triangle ABC
e2 =
e2
+ AM 2
4
AM =
AO =
3
e
2
d AB = h = 2R
2
3
2
2 3
e
AM =
e=
3
3 2
3
10
5
Atomic Packing Factor (APF) or space filling
Volume of atoms (assumed to hard spheres) in unit cell
Space filling =
Volume of unit cell
atoms
unit cell
volume
atom
4
APF =
a 2
a
4 3
πr
3
volume
a3
unit cell
4 3
4
πr
4 π r3
3
APF =
= 3 3
3
(2 2 r ) 16 2 r
4
a 2 = 4r
a=2 2r
# atoms /unit cell
8 x 1/8 + 6 x ½ = 4
APF =
π
= 0.74
3 2
The close-packed arrangement utilizes 74% of the available space, producing a
dense arrangement of atoms. Small holes make up the other 26% of the unit cell.
11
Holes in Close Packed Crystals
Octahedral hole
Octahedral holes lie within two staggered triangular
planes of atoms.
The coordination number of an atom occupying an
octahedral hole is 6.
The size of the octahedral hole = rO = 0.414 R ;
where R is the radius of the close-packed atom or ion.
For
n atoms in a close-packed structure, there are
n octahedral holes.
Example: cubic close-packed
The green atoms are in a cubic closepacked arrangement
The small orange spheres show the
position of octahedral holes in the unit cell
Each
hole has a coordination number of 6
12
# atoms
8x⅛ +6x½ = 4
# Oct. holes
1 + 12x¼ = 4
6
tetrahedral hole
Tetrahedral holes are formed by a planar
triangle of atoms, with a 4th atom covering
the indentation in the center.
The resulting hole has a coordination
number of 4
The size of the tetrahedral holes = rT =
of the close-packed atom or ion.
.225 R ; where R is the radius
For n atoms in a close-packed structure, there are
2n tetrahedral holes.
Example: cubic close-packed
The orange spheres show atoms in a cubic
close-packed arrangement
The small white spheres behind each corner
indicate the location of the tetrahedral holes
Each
hole has a coordination number of 4
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# atoms
8x⅛ +6x½ = 4
# T. holes
8
primitive packing
A A arrangement
A
A
ABA arrangement
A
B
A
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Simple
cubic (or primitive cubic) unit cell
Body-centered cubic unit cell
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primitive packing
Simple cubic (or primitive cubic) unit cells
coordination number is 6,
packing efficiency is only 52.4%.
Body-centered cubic unit cells
packing efficiency is 68%
coordination number = 8
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The packing-of-spheres model
applied to the structures of elements
4.3. Metallic Crystal Structures
Tend to be densely packed.
Reasons for dense packing:
₋ Typically, only one element is present, so all atomic
radii are the same.
₋
Metallic bonding is not directional.
₋
Nearest neighbor distances tend to be small in order
to lower bond energy.
₋
Electron cloud shields cores from each other
Have the simplest crystal structures.
We will examine three such structures...
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8
Hexagonal Close-Packed Structure (HCP)
• ABAB... Stacking Sequence
ex: Be, Mg, Ti, Zr, Zn, Cd, Ru, . . .
• 3D Projection
c
• 2D Projection
A sites
Top layer
B sites
Middle layer
A sites
Bottom layer
a
• Coordination # = 12
• APF = 0.74
• c/a = 1.632
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IDEAL c/a
c 2h
2
=
=2
= 1.632
a a
3
HCP
Nearly Ideal c/a for Mg
Important Note
Metal
c/a
Cd
Zn
Mg
Zr
Ti
Be
1.886
1.856
1.624
1.590
1.588
1.586
As most (so called) ‘HCP’ metals do not have an ideal c/a ratio there are not
close packed in the true sense.
Often we discuss all such HCP metals (with non-ideal c/a ratios) as being ‘hard
sphere packing’ with ideal c/a ratio.
We
have to remember that this is just an approximation.
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9
Face Centered Cubic Structure (FCC)
ABCABC... Stacking Sequence (cubic close-packed structure : CCP)
ex: Al, Cu, Au, Pb, Ni, Pt, Ag
Atoms touch each other along face diagonals.
Note: All atoms are identical; the face-centered atoms are shaded
differently only for ease of viewing.
Coordination # = 12
4 atoms/unit cell: 6 face x 1/2 + 8 corners x 1/8
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Body Centered Cubic Structure (BCC)
ex: Cr, W, Fe (α), Tantalum, Molybdenum
Atoms touch each other along cube diagonals.
Note: All atoms are identical; the center atom is shaded
differently only for ease of viewing.
Coordination # = 8
2 atoms/unit cell: 1 center + 8 corners x 1/8
Note:
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bcc
≠
hcp
10
Simple Cubic Structure (SC)
ex: Rare due to low packing density (only Po has this structure)
• Close-packed directions are cube edges.
• Coordination # = 6
(# nearest neighbors)
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Polymorphism
If a substance exists in more than one crystalline
form, it is polymorphic.
Many metals exhibit different crystal structures with
changes in pressure and temperature. Typically, denser
forms occur at higher pressures.
Higher temperatures often cause close-packed
structures to become body-center cubic structures due
to atomic vibrations (example : Fe ).
“P-T “phase diagram for iron.
Atomic Radii of Metals
Metallic radii are defined as half the internuclear distance as determined by X-ray
crystallography. However, this distance varies with coordination number of the
atom; increasing with increasing coordination number.
Goldschmidt radii correct all metallic radii for a coordination number of 12:
Coord #
Relative radius
22
12
1.000
8
0.97
6
0.96
4
0.88
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4.4. Ionic Crystal Structures
Periodic trends in ionic radii
1) The radii of ions increase as you go down within a group: More electrons are
present and the outer electrons are further away from the nucleus.
5) As you move from left to right across a row of the periodic table, Radii of ions
of similar charge decrease.
3) In a series of isoelectronic anions (e.g. F-, O2-) the radius increases with
increasing negative charge.
4) For elements with more than one oxidation state (e.g. Ru4+, Ru5+), the radii
decreases as the oxidation state increases.
2) In a series of isoelectronic cations (e.g. Na+, Mg2+, Al3+) the radius decreases
with increasing positive charge.
6) The spin state (high or low spin) affects the ionic radius of transition metals.
7) Like metallic radii, ionic radii seem to vary with coordination number. As the
coordination number increases, the apparent ionic radius increases.
Note: These trends in the sizes of ions may be explained by consideration
of the shielding and effective nuclear charge. Zeff = Z – S
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Ionic Radii
Ionic radii are difficult to determine, as x-ray data only shows the position of
the nuclei, and not the electrons.
Experimental data only give the internuclear distance:
Internuclear distance
between a cation and the = rcation + ranion
closest anion in a lattice
Most systems assign a radius to the oxide ion (often 1.26Å), and the radius
of the cation is determined relative to this assigned value.
Radius ratio rules
Since anions are often larger than cations, ionic structures are often viewed as a
close-packed array of anions with cations added, and sometimes distorting the
close-packed arrangement.
r+
can be used to make a first guess at the likely coordination
r−
number and geometry around the cation using a set of simple rules:
The radius ratio
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12
Predicted
coordination
number of
cation
Predicted
coordination
geometry of
cation
<0.15
2
Linear
0.15–0.22
3
Trigonal planar
0.22–0.41
4
Tetrahedral
0.41–0.73
6
Octahedral
>0.73
8
Cubic
r+
Value of
r−
For a given compound stoichiometry, predictions about the
coordination type of the cation necessarily make predictions
about the coordination type of the anion. Use of radius ratios
meets with some success, but there are many limitations.
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Common Crystal Types
The Rock Salt (NaCl) structure
Can be viewed as a face-centered cubic
array of the anions, with the cations in all
of the octahedral holes
or
A face-centered cubic array of
the cations with anions in all of
the octahedral holes
The coordination number is 6 for both ions
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13
The CsCl structure
Chloride ions occupy the corners of a cube, with a
cesium ion in the center (called a cubic hole)
or vice versa:
Both ions have a coordination number of 8,
with the two ions fairly similar in size
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The Zinc-blende or Sphalerite structure
Anions (S2-) ions are in a face-centered cubic arrangement,
with cations (Zn2+) in half of the tetrahedral holes.
The coordination number is 4 for both ions
Note: Both sites are equivalent, and the unit cell could be
drawn with the S2− ions in the red (Zn2+) sites .
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14
Wurtzite (a second polymorph of ZnS)
wurtzite (a second polymorph
of ZnS) define a hexagonal prism;
the Zn2+ ions are shown in grey
and the S2− ions in yellow. Both
ions are tetrahedrally sited
A close hexagonal arrangement
of S2- ions with Zn2+ ions in
half of
the tetrahedral holes
The coordination number
is 4 for both ions
Note:
an alternative unit cell
could be drawn by interchanging
the ion positions.
Three unit cells of wurtzite
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The Fluorite (CaF2) and Antifluorite structures
A face-centered cubic arrangement
of Ca2+ ions with F- ions in all of the
tetrahedral holes.
Ca2+
F-
The coordination number is
• 4 for F- ions (4 Ca2+ )
• 8 for Ca2+ ions (8 F- )
The antifluorite structure reverses the
positions of the cations and anions. An
example is K2O.
The unit cell of CaF2
Zn2+
S2-
Note:
CaF2 and zinc blende (ZnS) have
the same structure except that anions (S2-)
in half of the tetrahedral holes.
The unit cell of zinc blende (ZnS)
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15
TiO2 rutile - type
The rutile lattice is adopted by:
SnO2 (cassiterite, the most important tin-bearing mineral),
MnO2 (pyrolusite)
and PbO2.
The unit cell of rutile (one polymorph of TiO2). Colour code: Ti, silver; O, red.
The coordination number is
• 6 for Ti (octahedral)
•3 for O (trigonal planar)
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consistent with the 1:2 stoichiometry of rutile
CdI2 – type (layer structures)
The structure which can be described in terms of
‘stacked sandwiches’, each ‘sandwich’ consisting
of a layer of I- ions, a parallel layer of Cd2+ ions,
and another parallel layer of I- ions; each
‘sandwich’ is electrically neutral.
Only weak van der Waals forces operate
between the ‘sandwiches’ (the central gap
between the layers in the Figure) and this leads
to CdI2 crystals exhibiting pronounced cleavage
planes parallel to the layers.
If a crystal breaks along a plane related to the lattice
Parts of two layers of the CdI2 lattice;
structure, the plane is called a cleavage plane.
Cd2+ ions are shown in pale grey and
−
−
Other compounds crystallizing with a CdI2 lattice I ions in gold. The I ions are
arranged in an hcp array.
include MgBr2, MgI2, CaI2, iodides of many d-block
metals, and many metal hydroxides including Mg(OH)2
(the mineral brucite) in which the [OH-] ions are treated
as spheres for the purposes of structural description.
The CdCl2 lattice is related to the CdI2 layer-structure but with the Cl ions in a cubic close32
packed
arrangement. Examples of compounds adopting this structure are FeCl2 and CoCl2.
16
4.5. Lattice energy
The lattice energy is a measure of the strength of ionic
bonds within a specific crystal structure.
It is usually defined as the energy change when a mole
of a crystalline solid is formed from its gaseous ions.
M+(g) + X-(g)
∆E = Lattice Energy
MX(s)
Lattice energies cannot be measured directly, so they
are obtained using Hess’ Law.
They will vary greatly with ionic charge, and, to a lesser
degree, with ionic size.
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Lattice energy: estimates from an electrostatic model
The lattice energy, ΔU(0 K), of an ionic compound is the change in internal energy
that accompanies the formation of one mole of the solid from its constituent gasphase ions at 0 K.
the Coulombic attraction between the ions
For an isolated ion-pair:
where:
 z z e2 
∆U = − + − 
 4πε 0 r 
ΔU =change in internal energy (unit = joules)
|z+| = modulus of the positive charge (for K+, |z+| = 1; for Mg2+ |z+| = 2)
|z-| = modulus of the negative charge (for F-, |z-| = 1; for O2-, |z-| = 2)
e = charge on the electron = 1.602x10-19 C
ε0= permittivity of a vacuum = 8.854 x 10-12Fm-1;
r = internuclear distance between the ions (units = m).
by considering the ions that surround Mz+ (or Xz- ) ion ; the lattice energy is estimated in joules
per mole of compound.
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L A z+ z − e 2
∆U = −
4πε 0 r
where: L = Avogadro number = 6.022 x 1023mol-1
A = Madelung constant (no units).
17
Madelung constants, A, for selected lattice types. Values of A are numerical and have no units.
Born forces
Coulombic interactions are not the only forces operating in a real ionic lattice. The
ions have finite size, and electron–electron and nucleus–nucleus repulsions also
arise; these are Born forces. The following equation gives the simplest expression
for the increase in repulsive energy upon assembling the lattice from gaseous ions.
∆U =
35
LB
rn
Where:
B = repulsion coefficient
n = Born exponent.
Values of the Born exponent, n, given for an ionic compound MX in terms of the
electronic configuration of the ions [M+][X−]. The value of n for an ionic
compound is determined by averaging the component values
e.g. for MgO, n = 7;
5+9
for LiCl, n = 2 = 7
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18
The Born–Lande´ equation
For the lattice energy that takes into account both the Coulombic and Born
interactions in an ionic lattice, we combine the two equations to give:
∆U = −
L A z+ z− e 2 L B
+ n
4πε 0 r
r
We evaluate B in terms of the other components of the equation by making use
of the fact that at the equilibrium separation where r = r0, the differential dU = 0
dr
Differentiating with respect to r gives:
0=
L A z+ z− e 2 n L B
− n+1
4πε 0 r02
r0
and rearrangement gives:
B=
A z + z − e 2 r0n−1
4πε 0 n
Combining equations gives the Born–Lande´ equation:
L = Avogadro number = 6.022 x 1023mol-1
L A z + z − e  1  A = Madelung constant (no units).
∆U (0 K ) = −
1 −  |z | and |z | = modulus of the ions’ charges
+
4πε 0 r0
 n
e = charge on the electron = 1.602x10-19 C
Where: Δ U(OK) in J/mol and: ε0= permittivity of a vacuum = 8.854 x 10-12Fm-1
r0 = internuclear distance between the ions (units = m).
n = Born exponent.
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2
Example: Use of the Born–Lande´ equation
Sodium fluoride adopts the NaCl type lattice. Estimate the lattice energy
of NaF using an electrostatic model.
Data required:
L = 6:022x1023 mol-1
A = 1.7476
e = 1.602x10-19 C
ε0 = 8.854x10-12 Fm-1
Born exponent for NaF = 7
Internuclear NaF distance = 231 pm
Solution:
The change in internal energy (the lattice energy) is given by the Born–Lande´
equation:
L A z+ z− e 2  1 
∆U (0 K ) = −
4πε 0 r0
1 − 
 n
r0 must be in m: 231pm = 2.31x10-10 m
∆U0 = -
6:022x1023 x 1.7476 x 1 x 1 x (1.602x10-19)2
4 x 3.142 x
8.854x10-12
x
2.31x10-10
(1 -
1
)
7
= - 900 624 J mol-1
≈ - 901 kJ mol-1
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19
Lattice energy: the Born–Haber cycle
By considering the definition of lattice energy, it is easy to see why these
quantities are not measured directly.
Lattice Energy
Mn+(g) +n X-(g)
MXn(s)
However, an associated lattice enthalpy of a salt can be related to several other
quantities by a thermochemical cycle called the Born–Haber cycle
A Born–Haber thermochemical cycle for the formation of a salt MXn. This gives an
enthalpy change associated with the formation of the ionic lattice MXn.
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Example: Application of the Born–Haber cycle
Given that the standard enthalpy of formation at 298K of CaF2 is 1228 kJ mol-1,
determine the lattice energy for CaF2 using appropriate data from the Appendices.
Solution:
First, construct an appropriate thermochemical cycle:
Values that need to be found in the Appendices are:
Appendix 10:
∆aHo(Ca,s) = 178 kJ mol-1
D(F2;g )=2∆aHo(F;g) = 158 kJ mol-1
Appendix 8:
IE1(Ca;g) = 590 ; IE2(Ca;g) =1145 kJ mol-1
Appendix 9:
∆EAH(F;g) = 328 kJ mol-1
Use of Hess’s Law gives:
∆U(0K) ≈ ∆fHo(CaF2,s) - ∆aHo(Ca,s) - D(F2;g ) - ƩIEi(Ca;g) - 2∆EAH(F;g)
≈ 1228 – 178 – 158 – (590 +1145) – 2 x 328
40
∆U(0K) ≈ - 2643 kJ mol-1
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4.6. Defects in solid state lattices: an introduction
ideal lattices in which every site is occupied by the correct type of atom
or ion appertain only at 0K, and above this temperature, lattice defects
are always present.
There are various types of lattice defects, but we shall introduce only the
Schottky and Frenkel defects.
Schottky defect
A Schottky defect consists of an atom or ion vacancy in a crystal lattice, but the
stoichiometry of a compound (and thus electrical neutrality) must be retained.
compare this with
41
A Schottky defect involves vacant cation and
anion sites; equal numbers of cations and anions
must be absent to maintain electrical neutrality
Part of one face of an
ideal NaCl structure
Examples of Schottky defects
In a metal lattice, a vacant atom site may be present.
in ionic lattices are: a vacant cation and a vacant anion site in an MX salt,
or
a vacant cation and two vacant anion sites in an MX2 salt.
Frenkel defect
In a Frenkel defect, an atom or ion occupies a normally vacant site, leaving its
‘own’ lattice site vacant.
Silver bromide adopts an NaCl structure
(a)
An ideal lattice can be described in terms of Ag+ ions occupying
octahedral holes in a cubic close-packed array of bromide ions.
(b)
A Frenkel defect in AgBr involves the migration of Ag+ ions
into tetrahedral holes; in the diagram, one Ag+ ion occupies a
tetrahedral hole which was originally vacant in (a), leaving the
central octahedral hole empty.
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21