1
Structure of Crystalline Compounds
Periodical repeat of the
same building units
2
Lattice and Structure
Structural Motiff
Lattice point
3
Unit Cell
Periodical repeat of unit cells = crystal
4
Five Planar Lattices
5
STM Nb/Se
6
Lattice and Unit Cell
Unit CellLattice point
Unit Cell Parameters
a, b, c – edge lengths
α, β, γ – angles
7
Seven Crystal Systems
8
14 Bravais Lattices
9
Z
Y
X
( 1 1 1)
Miller Indices
(h k l)
x
h
∗
=
intercept
1
z
l
∗
=
intercept
1
y
k
∗
=
intercept
1
a
b
c
10
STM Picture of Fe in (110) Plane
11
Miller Indices
h = 1 / ∞ = 0
k = 1 / 1 = 1
l = 1 / ∞ = 0
( 0 1 0)
x
h
∗
=
intercept
1
z
l
∗
=
intercept
1
y
k
∗
=
intercept
1
12
Miller Indices
13
Three Cubic Cells
Primitive (P) Body centered (I) Face centered (F)
14
Primitive (P)
Body centered (I)
Face centered (F)
15
a
a
a
d
D
a = edge
d = face diagonal
(d2 = a2 + a2 = 2a2)
D = body diagonal
(D2 = d2 + a2 = 2a2 + a2 = 3a2)
a2 ⋅=d a3 ⋅=D
Cube
16
Space filling
52%
Coord. No. 6
Primitive Cubic Cell, Po - Litviněnko
17
Primitive Cubic Cell
atoms touch along edge (a)
a = 2r then r =
Cell volume V = a3 = 8r3
Volume of atoms in the cell
VA = 4/3 π r3
Space filling = Va/V 100 = 52%
a
2
a
r
x 8 vertices =
1/8 atom
vertes
1 atom
cell
Number of lattice points in the cell
Space filling
18
Space filling 68%
Coord. no 8
Body Centered Cell, W
19
x 8 vertices = 1 atom
+ center = 1 atom
2 atoms/cell
1/8 atom
vertex
D = 4r =
a = then r =
V = a3 =
Atoms touch along body diagonal (D)
a3 ⋅
3
r4
4
a3 ⋅
3
3
r4
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
Body Centered Cell, W
a
d
D
r
Number of lattice points in the cell
20
21
Space filling 74%
Coord. no 12
Face Centered Cell, Cu
(= Close Cubic Packing)
22
x 8 vertices = 1 atom
x 6 faces = 3 atoms
4 atoms/cell
1/8 atom
vertex
d = 4r =
a = or r =
V = a3 =
Atoms touch along face diagonal (d)
a2 ⋅
2
r4
4
a2 ⋅
1/2 atom
face
3
2
r4
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
Face Centered Cell
a
d
r
Number of lattice points in the cell
23
Space Filling
74%4√2a/4Face
centered
34%8√3a/8Diamond
68%2√3a/4Body
centered
52%1a/2Primitive
cubic
Space
Filling
No. of
atoms
Radius
24
Close Packing on a Plane
Square packing
Lots of free space
4 neighboring atoms
Hexagonal packing
The best use of space
6 neighboring atoms
25
Holes B and C cannot be
filled at the same time by
atoms in the second layer
26
hexagonal
cubic
Two layers of close
packing
Johannes Kepler 1611
27
Close Packing in Space
cubichexagonal
Johannes Kepler 1611
28
cubichexagonal Close Packing
29
cubichexagonal
Close Packing
30
cubic
hexagonal
Mg, Be, Zn, Ni, Li, Be, Os, He
Cu, Ca, Sr, Ag, Au, Ar, F2, C60,
opal (300 nm)
Close Packing
31
Close Packing of Large Structures
32
Structure of Dry Ice
33
Coordination Polyhedra
34
Cubic Close Packing = Face Centered Cell
Layers (ABC)
Close packed layers are oriented perpendicular to body
diagonal of cubic cell
35
Tetrahedral T+ Tetrahedral T-Octahedral O
For N close packed
atoms, there are N
octahedral and 2N
tetrahedral holes per one
cell
36
Two Types of Holes
Cubic Close Packing = Face Centered Cell
Number of atoms in a cell N = 4
Tetrahedral (2N = 8)Octahedral (N = 4)
37
Cation/Anion Radius Ratio
0.225 – 0.4144 – Tetrahedral
0.414 – 0.7326 – Octahedral
0.732 – 1.008 – Cubic
1.00 (substitution)12 – Cub. and Hex.
r/RCoord. No.
Void
radius
decreases
38
39
Structures Derived from Cubic Close
Packing
Li2O
BiF3
40
Sodium Chloride, NaCl
Cubic Close Packing of Cl,
Na occupies octahedral holes
Z = ?
Coord. No.
Na = 6
Cl = 6
41Two close packed lattices of cations and anions
42
Structure of Pyrite - FeS2
Na+ ClFe2+
S2
2Derive
more complex structures from simple structural types
43K2[PtCl6], Cs2[SiF6], [Fe(NH3)6][TaF6]2
Fluorite, CaF2 (Inverse Type Li2O)
F / Li
Ca / O
44
Sfalerite, ZnS
Cubic Close Packing of S
Zn occupies ½ of tetrahedral holes
Cubic Close Packing of Zn
S occupies ½ of tetrahedral holes
45
Sfalerite, ZnS
46
Diamond, C
47
6,16Å
2,50 Å
4,10Å
cubic hexagonal
SiO2
cristobalite
SiO2 tridymite
Ice
Diamond, C
lonsdaleite
48
Structure of Group 14 Elements
Diamond structure – cell size increases down the group
49
Wurzite, ZnS
Hexagonal Close
Packing of S
Zn occupies
½ of tetrahedral holes
Polymorphs of ZnS
50
13-15 and 12-16 Semiconductors
Sfalerite Wurzite
InP, GaAs
HgTe, CdTe
ZnO, CdSe
AlN, GaN
51
[Cr(NH3)6]Cl3, K3[Fe(CN)6]
BiF3/Li3Bi
Cubic Close Packing of Bi (4)
F occupies tetrahedral holes (8) and
octahedral holes (4)
Cubic Close Packing of Bi (4)
Li occupies tetrahedral holes (8) and
octahedral holes (4)
52
CsCl
53
CsCl is not a body centered
cubic cell
54
Primitive cubic
ReO3
55
Perovskite, CaTiO3
Two equivalent views at perovskite unit cell
Ti
CaO
Ti
O
Ca
56
CsCl
Perovskite, CaTiO3
57
Rutile, TiO2
Coordination Number Rule
AxBy
c.n.(A) / c.n.(B) = y / x
Coordination Numbers are in an inverse ratio of
stoichiometric coefficients
58
Phase Transitions at High Pressure
Incr. coordination number
Incr. density
Elongation of bond lengths
Nonmetal-Metal transition
Sfalerite Sodium Chloride
High Pressure Effects
59
Lattice Energy
L = Ecoul + Erep
Ion pair
n = Born’s exponent
(experimental data from
compressibility measurements)
Energy released upon the formation of 1 mol of ionic solid
from its ions in the gas phase
d
eZZ
E BA
coul
2
04
1
πε
=
nrep
d
B
E =
60
Madelung’s Constant
Ecoul = (e2 / 4 π e0) × (zA zB / d) × [+2(1/1) - 2(1/2) + 2(1/3) - 2(1/4) + ....]
Ecoul = (e2 / 4 π e0) × (zA zB / d) × (2 ln 2)
Counts all interactions in the crystal lattice
Madelung’s constant M
(for linear arrangement of ions)
= sum of a convergent series
61
Madelung’s Constant for NaCl
Ecoul = (e2 / 4 π e0) * (zA zB / d) × [6(1/1) - 12(1/√2) + 8(1/√3) - 6(1/√4) + 24(1/√5) ....]
Ecoul = (e2 / 4 π e0) × (zA zB / d) × M
Convergent series
62
Madelung’s Constants for Different
Structures
1.64132ZnS Wurtzite
1.63805ZnS Sfalerite
2.519CaF2
1.76267CsCl
1.74756NaCl
MStructural type
63
Lattice Energy
1 mol of ions
Attractive
Repulsive
L = Ecoul + Erep
Find minimum dL/d(d) = 0
nA
BA
A
d
B
N
d
eZZ
MNL +=
0
2
4πε
d
eZZ
MNE BA
ACoul
0
2
4πε
=
nArep
d
B
NE =
64
Lattice Energy
⎟
⎠
⎞
⎜
⎝
⎛
+=
nd
eZZ
MNL BA
A
1
1
4 0
2
πε
nEl. config.
10Kr
12Xe
9Ar
7Ne
5He
Born – Mayer equation
d* = 0.345 Å
Born – Lande equation
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−=
d
d
d
eZZ
MNL BA
A
*
0
2
1
4πε
65
Lattice Energy
Kapustinski
M/v is approx. constant for all types of structures
v = number of ions in formula unit
M replaced by 0.87 v, no need to know the structure
⎟
⎠
⎞
⎜
⎝
⎛
−=
dd
ZZ
vL BA 345,0
11210
66
structure M CN stoichio M / v
CsCl 1.763 (8,8) AB 0.882
NaCl 1.748 (6,6) AB 0.874
ZnS sfalerite 1.638 (4,4) AB 0.819
ZnS wurtzite 1.641 (4,4) AB 0.821
CaF2 fluorite 2.519 (8,4) AB2 0.840
TiO2 rutile 2.408 (6,3) AB2 0.803
CdI2 2.355 (6,3) AB2 0.785
Al2O3 4.172 (6,4) A2B3 0.834
v = number of ions in formula unit
Kapustinski
67
∆Hf
o = - 411 kJ mol−1
∆Hsubl
o = 108 kJ mol−1
½ D= 121 kJ mol−1
EA = - 354 kJ mol−1
IE = 502 kJ mol−1
L=?Na(s) + 1/2 Cl2 (g)
Na(g) + 1/2 Cl2 (g)
Na(g) + Cl (g)
Na+
(g) + Cl (g)
Na+
(g) + Cl-
(g)
NaCl (s)
0 = −∆Hf
o + ∆Hsubl
o + 1/2 D + IE + EA + L
0 = 411 + 108 +121 + 502 + (-354) + L
L = − 788 kJ mol−1
Born-Haber Cycle
68
Lattice Energy of NaCl
Calculated from Born – Lande eq. L = − 765 kJ mol−1
Considers only ionic contribution
Measurement from Born–Haber cycle L = − 788 kJ mol−1
Ionic and covalent contribution