Transcript 第7章固态(E).ppt

Physical Chemistry I
Chapter VII Solid State
Physical Chemistry
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Physical Chemistry I
Chapter VII Solid State
Chapter VII Solid State
we couldn't live without their
mechanical properties…
oxide
Ca10(PO4)6OH2
polymer
metal
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polymer
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Chapter VII Solid State
we couldn't live without their
electronic properties…
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Chapter VII Solid State
Chemical classification:
molecular
ionic
covalent
metallic
© Mark S.
2016/8/6 Golden 2002
C M S M a s t e r s : Electronic Structure and Chemistry of Solids.
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Solids
Amorphous solids: no repeating pattern, only short
range order, extensively disordered - non crystalline
Crystals: highly regular arrangement of atoms, ions,
molecules - periodic (repeating)
correlation with physical properties:
Regular shapes;
Anisotropy: a substance whose physical properties
are difference in different directions.
A sharp melting point
Homogeneous
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STM下的Si晶体
Chapter VII Solid State
来自于巴基斯坦的黄玉晶体
来自于加拿大魁北克的钙柱石（黄）和透辉石（蓝）
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Early ideas
 A crystal has flat faces that are at chatacteristic angles to
each other
 Crystals have symmetry (Kepler) and long range order
 Spheres and small shapes can be packed to produces
regular shapes (Hooke, Hauy)
?
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Group discussion
Kepler wondered why snowflakes have 6 corners,
never 5 or 7. By considering the packing of polygons
in 2 dimensions, demonstrate why pentagons and
heptagons shouldn’t occur.
Empty space
not allowed
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All M.C. Escher works (c) Cordon Art-Baarn-the Netherlands.
All rights reserved.
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UNIT CELL = The
smallest component of
the crystal, which when
stacked together with
pure
translational
repetition
reproduces
the
whole crystal
Primitive Cell: simplest cell, contain one lattice point
Not necessary have the crystal symmetry
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How do we describe crystalline solids ？
— Lattice
A lattice is an infinite 1,2, or 3-D regular arrangement
of points, each of which has identical surroundings.
Any periodic pattern can be described by placing
lattice points at equivalent positions within each unit
of the pattern.
To recover original pattern we add the motif to each
lattice point.
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 Lattice points must link up - cannot have gaps
between adjacent points
 All lattice points must be identical (with identical
environments)
 Lattice points must show the full symmetry of
the structure  next section
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2D example – Square lattice
(sodium chloride, NaCl)
We define lattice points ; these are points
with identical environments
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Choice of origin is arbitrary - lattice points need not be
atoms - but unit cell size should always be the same.
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This is also a unit cell it doesn’t matter if you start from Na or Cl
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- or if you don’t start from an atom
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This is NOT a unit cell even though they are all the
same - empty space is not allowed!
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In 2D, this IS a unit cell
In 3D, it is NOT
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


Translation Tmnp  ma  nb  pc
Unit Cell choice
By convention the unit cell is chosen so that it is
as small as possible while reflecting the full
symmetry of the lattice
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Definitions
1. The unit cell
“The smallest repeat unit of a crystal structure, in 3D,
which shows the full symmetry of the structure”
The unit cell is a
box with:
• 3 sides - a, b, c
• 3 angles - , , 
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Translational
vector
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P
Chapter VII Solid State
P
NP
5 Bravais Lattice in 2D
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Square
a=b
 =90
Rectangular
a b
=90
Centered
Rectangular
Hexagonal
a b
=90
a=b
 =120
Oblique
a b
 90
5 Bravais Lattice in 2D
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Lattice Planes and Miller Indices
Imagine representing a crystal structure on a grid (lattice)
which is a 3D array of points (lattice points). Can imagine
dividing the grid into sets of “planes” in different orientations
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 All planes in a set are identical
 The planes are “imaginary”
 The perpendicular distance between pairs of adjacent
planes is the d-spacing
 Need to label planes to be able to identify them
Find intercepts on
a,b,c: 1/4, 2/3, 1/2
Take reciprocals 4,
3/2, 2
Multiply up to
integers: (8 3 4) [if
necessary]
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General label is (h k l) which intersects at a/h, b/k, c/l
(hkl) is the MILLER INDEX of that plane.
Plane perpendicular to y
cuts at , 1, 
 (0 1 0) plane
This diagonal cuts at 1, 1, 
 (1 1 0) plane
NB an index 0 means that the
plane is parallel to that axis
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Using the same set of axes draw the
planes with the following Miller
indices:
(0 0 1)
(1 1 1)
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Using the same set of axes draw the planes with the
following Miller indices:
(0 0 1)
(1 1 1)
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(100)
(200)
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(111)
(110)
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Lattice spacing
1
h2  k 2  l 2

2
d hkl
a2
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For cubic system
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d-spacing formula
2
2
1 h
k
l
 2 2 2
2
d
a
b
c
For cubic crystals (special case
of orthogonal) a=b=c :-
1 h2  k 2  l 2

2
d
a2
e.g. for
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For orthogonal crystal systems
(i.e. ===90) :-
(1 0 0)
(2 0 0)
(1 1 0)
d=a
d = a/2
d = a/2
etc.
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A cubic crystal has a=5.2 Å (=0.52nm). Calculate the dspacing of the (1 1 0) plane
2
2
2
1 h k l
11


d2
a2
5 .2 2
O
5 .2 2
d
 3. 7 A
2
A tetragonal crystal has a=4.7 Å, c=3.4 Å. Calculate the
separation of the:
(1 0 0)
4.7 Å
(0 0 1)
3.4 Å
(1 1 1) planes
2.4 Å
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1 h 2  k 2 l2

 2
2
2
d
a
c
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[a  b ]
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The difference between real crystal and lattice
model
Point Defects
Dislocations
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Line Defects
partial dislocation
Edge dislocations
Screw dislocations
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晶体的对称性和分类
宏观对称性
晶体的对称性
微观对称性
晶体的宏观对称性和32点群
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有限图形
点群
宏观对称性
点阵结构
空间群
微观对称性
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晶体中宏观对称元素符号
分子
晶体
对称元素
对称操作
对称元素
对称操作
旋转轴 C n
旋转 C n
旋转轴 n
旋转 L2 / n
反映面 
反映 
反映面 m
反映 M
对称中心 i
反演 i
对称中心 i
倒反 I
象转轴 S n
象转 S n
反轴 n
旋转倒反 IL2 / n 
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m
m
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Element
Operation
Rotation axis, Cn
n-fold rotation
Improper rotation axis, Sn
n-fold improper rotation
Plane of symmetry, 
Reflection
Center of symmetry, i
Inversion
Identity, E
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Types of symmetry:
1. Crystallographers make use of all the symmetry in a crystal
to minimize the number of independent coordinates
2. Lattice symmetry (translation)
3. Point symmetry (rotation axes, mirror plane, rotationinversion axes, inversion center)
4. Other translational symmetry elements (screw axes and
glide planes)
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Allowed rotation axis:
1, 2, 3, 4, 6
NOT 5, > 6
A
2 / n
2
CD  2 cos
OC
n
CD  m AO
C
O
2 / n
Quasicrystal: AlFeCu
B
2 / n
D
2
2 cos
m
n
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理想晶体只有4次反轴。
 2
2  IL
 2
 2 
1  IL
I
 1 

  2
3   IL
  3
2
6
2
  2
  IL
  6
2


   I

2


   M


   3


   3

 2
4  IL
 4


M


 3 

L


M

 2 
3

2



 
  2
3   IL
  3
3
  2
6   IL
  6
3

  2
4   IL
  4
2

4   IL
   L  M
 2
  4 
3
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3
3
2

   2

不
能
独
立
存
在
独
立
存
在
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理想晶体独立存在的对称元素:
m
i
2
1
4
3
6
4
将上述八个对称元素，取一个或几个组合，可以得到32个
点群（即自然生长的宏观晶体不会超出32点群的范围）。
C1
C2
C3
C4
C2 h C3 h C4 h
C6 h
D2
D6
D3
D4
C6
Ci
S4
C2v C3v C4v
Ci 
C6v
S 6 C4 h  C6 h 
D2 h D3 h D4 h D6 h D2 d D3 d
T Td Th
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O Oh
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3D: 14 Bravais Lattice, 7 Crystal System
Definition:
Bravais Lattice: an infinite array of discrete points with an arrangement and
orientation that appears exactly the same from whichever of the points the array
is viewed.
Name
Triclinic
Monoclinic
Orthorhombic
Tetragonal
Cubic
Number of Bravais lattices
1 (P)
2 (P, C)
4 (P, F, I, A)
2 (P, I)
3 (P, F, I)
Trigonal
1 (P)
Hexagonal
1 (P)
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Conditions
a1  a2  a3

a1  a2  a3
 =  = 90°  
a1  a2  a3
 =  =  = 90°
a1 = a2  a3
 =  =  = 90°
a1 = a2 = a3
 =  =  = 90°
a1 = a2 = a3
 =  =  < 120°  90°
a1 = a2  a3
 =  = 90°
 = 120°
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Unit cell symmetries - cubic

4 fold rotation axes
(passing through pairs of
opposite face centers,
parallel to cell axes)
TOTAL = 3
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Unit cell symmetries - cubic


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4 fold rotation axes
TOTAL = 3
3-fold rotation axes
(passing through cube
body diagonals)
TOTAL = 4
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Chapter VII Solid State
One 4-fold axes
Why not F tetragonal?
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Orthorhombic: P, I, F, C
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Trigonal: P : 3-fold rotation
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Chapter VII Solid State
Hexagonal
Triclinic
Monoclinic
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Chapter VII Solid State
The choice of unit cell: reflect the crystal symmetry
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晶体的微观对称性和230空间群
微观对称元素和操作
Translation
Screw: a combination of a rotation
and a translation
 2   m 
L
T  a   nm
 n  n 
Glide: a combination of a reflection
and a translation
点群+微观对称元素
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空间群
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Chapter VII Solid State
Close packed structures - metals
Most efficient way of packing equal sized spheres.
In 2D, have close packed layers
Coordination number
(CN) = 6. This is the
maximum possible for
2D packing.
Can stack close packed (c.p.) to give 3D structures.
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Chapter VII Solid State
Two main stacking
sequences:
If we start with one cp layer, two possible ways of adding a
second layer (can have one or other, but not a mixture) :
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Two main stacking
sequences:
If we start with one cp layer, two possible ways of adding a
second layer (can have one or other, but not a mixture) :
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Let’s assume the second layer is B (red). What
about the third layer?
Two possibilities:
(1) Can have A position again (blue). This leads
to the regular sequence …ABABABA…..
Hexagonal close packing (hcp)
(2) Can have layer in C position, followed by the
same repeat, to give …ABCABCABC…
Cubic close packing (ccp)
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Hexagonal close packed
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Cubic close packed
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No matter what type of packing, the coordination number of
each equal size sphere is always 12
We will see that other coordination numbers are
possible for non-equal size spheres
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ccp = fcc ?
Build up ccp layers
(ABC… packing)
Add construction lines
- can see fcc unit cell
c.p layers are oriented perpendicular to the body
diagonal of the cube
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4 1.33r 3
%
 74.05%
3
( 2 2r )
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So the face of the unit
cell looks like:
Unit cell side
2a2 = (4r)2
a = 2r 2
Volume = (162) r3
Face centred cubic - so number of atoms per unit cell
=corners + face centres = (8  1/8) + (6  1/2) = 4
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Packing fraction
The fraction of space which is occupied by atoms is
called the “packing fraction”, , for the structure
space occupied by atoms
 =
available space
For cubic close packing:
4 3
r

3
 4

 074
.
3
16 2r
3 2
The spheres have been packed together as closely as
possible, resulting in a packing fraction of 0.74
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CN=12
(0,0,0)
(1/3,2/3,1/2)
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Hexagonal close packed structures
(hcp)
hcp
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
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bcc
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A1
A3
重复方式
ABCABC
ABABAB
空间点阵型式
立方面心
六方
球：四面题：八面体
4：8：4
2：4：2
空间占有率
74.06%
74.06%
配位数
12
12
分数坐标
 1 1
0,0,0  0, , 
 2 2
1 1 1 1 
 ,0,   , ,0 
 2 2  2 2 
0,0,0  2 , 1 , 1 
密置层方向
3
6
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 3 3 2
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Primitive
a = 2r
a3 = 8r3
No. of atoms = (8 x 1/8) = 1
4 r 3
3

3
8r
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

6
= 0.52
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Metallic solids
FCC
Chapter VII Solid State
Close-packed structures
HCP
…..or nearly
close packed
BCC
Detail factors for determining the structure involve the
band structure (see later lectures)
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Band Theory
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Bands in Metals
3p
3p
Energy
3s
3s
Na
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Mg
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Electrical Conductivity
Conductivity of metals decreases with temperature as
atomic vibrations scatter free electrons.
Conductivity of semiconductors increases with temperature
as the number of carriers increases.
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Electrical Conductivity
insulators
semiconductors metals
diamond
fused
silica
germanium
silicon
glass
10 -24 10 -20 10 -16
10 -12
10 -8
10 -4
100
copper
iron
10 4
10 8
Conductivity ( W –1 -cm –1 )
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Optical Properties of Metals
Some possible electronic
transitions in a half-filled
band of a metal
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Color that
corresponds to
band gap energy
Bandgap energy (eV)
4
3 violet
green
2
colorless
yellow
blue
1
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ultraviolet
Apparent color
of material
(unabsorbed light)
orange
yellow
red
infrared
red
.
black
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CsCl
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•Motif: Cl at (0,0,0); Cs at (1/2,1/2,1/2)
•1CsCl in unit cell
•Coordination: 8:8 (cubic)
CsCl
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•CCP Ca2+ with F- in all Tetrahedral holes
•Lattice: fcc
•Motif: Ca2+ at (0,0,0); 2F- at (1/4,1/4,1/4) & (3/4,3/4,3/4)
•4CaF2 in unit cell
•Coordination: Ca2+ 8 (cubic) : F- 4 (tetrahedral)
•In the related Anti-Fluorite structure Cation and Anion
positions are reversed
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CaF2
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Zinc Blende
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Chapter VII Solid State
ZnS
Wurtzite
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Chapter VII Solid State
ZnS
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Chapter VII Solid State
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Zinc Blende: ZnS
•CCP S2- with Zn2+ in half Tetrahedral holes (only T+ {or T-}
filled)
•Lattice: fcc
•4ZnS in unit cell
•Motif: S at (0,0,0); Zn at (1/4,1/4,1/4)
•Coordination: 4:4 (tetrahedral)
•Cation and anion sites are topologically identical
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NaCl
•Very common (inc. 'ionics', 'covalents' & 'intermetallics' )
•Most alkali halides (CsCl, CsBr, CsI excepted)
•Most oxides / chalcogenides of alkaline earths
•Many nitrides, carbides, hydrides (e.g. ZrN, TiC, NaH)
CaF2 (Fluorite)
•Fluorides of large divalent cations, chlorides of Sr, Ba
•Oxides of large quadrivalent cations (Zr, Hf, Ce, Th, U)
Na2O (Anti-Fluorite)
•Oxides /chalcogenides of alkali metals
ZnS (Zinc Blende/Sphalerite)
•Formed from Polarizing Cations (Cu+, Ag+, Cd2+, Ga3+...)
and Polarizable Anions (I-, S2-, P3-, ...);
•e.g. Cu(F,Cl,Br,I), AgI, Zn(S,Se,Te), Ga(P,As), Hg(S,Se,Te)
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晶体
Chapter VII Solid State
NaCl
CaCl
CaF2
立方 ZnS
六方 ZnS
TiO2
空隙
正八面体
立方体
立方体
正四面体
正四面体
八面体
比例
1
1
1/2
1/2
1/2
1/2
配位数
6
8
8
4
4
6
空隙
正八面体
立方体
正四面体
正四面体
正四面体
三角形
比例
1
1
1
1/2
1/2
1
配位数
6
8
4
4
4
3
点阵型式
立方 F
立方 P
立方 F
立方 F
六方
四方
正粒子数
4
1
4
4
2
2
负离子数
4
1
8
4
2
4
正
离
子
负
离
子
分数坐标+
(0,0,0)
(0,1 / 2,1 / 2)
(1 / 2,1 / 2,0)
(0,1 / 2,1 / 2)
(0,0,0)
(1 / 2,1 / 2,1 / 2 )
(0,1 / 2,0)
分数坐标-
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(1 / 2,0,0)
(0,0,1 / 2)
(0,0,0)
)
(0,1 / 2,1 / 2) ((10,/ 03,,0
2 / 3,1 / 2 )
(1 / 2,1 / 2,0)
(0,1 / 2,1 / 2)
(1 / 2,1 / 2,1 / 2 )
(1 / 2,1 / 2,1 / 2)
(0,1 / 2,0)
(1 / 2,0,0)
(0,0,1 / 2)
Chemistry Department of Fudan University
(1 / 3, 2 / 3,7 / 8 )
( 0,0, 3 / 8 )
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Lattice Energy
Born-Habor Cycle/Thermochemical Calculation
H f
1
Na( s )  Cl2 ( g )        NaCl ( s )
2
1
U
H  D
H  S sublimation
2
dissociation
Cl ( g )     EA
      Cl  ( g )
+
IP
Na( g )                      Na  ( g )
H f
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1
 S  Ddissociation  IP  EA  U
2
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Lattice Energy= net potential energy of the arrangement of charges
that form the structure.
= energy required to sublime the crystal and convert
it into a collection of gaseous ions.
NaCl ( s)    Na  ( g )  Cl  ( g )
H  U
Electrostatic force
(Attraction & repulsion)
Z  Z e2
V 
r
Short range repulsion:
B
V n
r
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B: Born exponent; n =5-12
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Ion Conf. n
He
5
Ne
7
Ar, Cu+ 9
Kr, Ag+ 10
Xe, Au+ 12
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Chapter VII Solid State
Z  Z e2
12
8
6
V 
(6 


 ....)
r
2
3
4
Madelung Constant
Structure Type
Madelung Constant
CsCl
1.763
NaCl
1.748
ZnS (Wurtzite)
1.641
ZnS (Zinc Blende)
1.638
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Z  Z  e 2 NA BN
U 
 n
r
r
dU Z  Z  e 2 NA nBN

 n 1  0
2
dr
r
r
Z  Z  e 2 Ar n 1
B
n
2
Z  Z  e NA
1
U 
(1  )
n
re
N: Avogadro No.
re
-U
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Some Lattice Energies:
MgO
3938
kJ/mol
LiF
1024
NaF
911
CaO
3566
LiCl
861
KF
815
SrO
3369
LiBr
803
RbF
777
BaO
3202
LiI
744
CsF
748
Melting point: MgO: 2800 C CaO: 2572 C
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BaO: 1923 C
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Chapter VII Solid State
For NaCl:
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S
109kJ/mol
IP
493.7kJ/mol
1/2D
121kJ/mol
EA
-356kJ/mol
U
-764.4kJ/mol
H f
=-396.7kJ/mol
Experimentally:
-410.9kJ/mol
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U
Ucalc.
UBorn-Haber
AgF
920
953
33
AgCl
832
903
71
AgBr
815
895
80
AgI
777
882
105
Increasing covalent!
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离子的极化
理论值  /实验值
R Ag  X
rAg  rX 
晶体
实验值
AgF
954
921
3%
2.46
2.46
0
AgCl
904
833
7.5%
2.77
2.94
0.17
AgBr
895
816
8.8%
2.88
3.09
0.21
AgI
883
778
11.9%
2.99
3.33
0.34
诱导偶极矩  
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
E
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离子极化的规律
离子半径越大，离子的极化率越大
负离子的极化率大于正离子
正离子的价数越高极化率越小，负离子价数越
高极化率越大
正离子容易使负离子极化、负离子容易被极化
键型变异
一种晶体采用什么键合形式，不仅与组成晶体的原子有关，而
且与晶体的结构有关。同一元素的原子可以出现多种键型，同
一晶体中也存在多种类型的化学键。
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Radius ratio rules
Rationalisation for
octahedral coordination:
R= radius of large ion,
r=radius of small ion
R
1

 cos 45 
Rr
2
 2R  R  r
 ( 2  1) R  r
r

 0.414
R
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If r/R < 0.414, the cation is too small and can
“rattle” inside the octahedral site
If r/R > 0.414, the anions are pushed apart
If r/R << or >> 0.414, coordination changes:
Coordination
Minimum r/R
Linear, 2
Trigonal, 3
0.155
Tetrahedral, 4
0.225
Octahedral, 6
0.414
Cubic, 8
0.732
Close packed, 12
1.000
A simple prediction tool, but beware
- it doesn’t always work!
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Radius ratio rules
Rationalisation for 8-fold coordination:
Unit cell edge a = 2R
Atoms touch along diagonal (if
small ion fits perfectly into
space) so
a3 = 2(R+r)
Divide: 3 = (R+r)/R
Multiply out 3R = R+r
R(3 -1) = r
r/R = 3 -1 = 0.732
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Pauling Rule 1: Coordination Polyhedra
"A coordination polyhedron of anions is formed around every cation (and viceversa) - it will only be stable if the cation is in contact with each of its neighbours.
•Ionic crystals may thus be considered as sets of linked polyhedra.
•The cation-anion distance is regarded as the sum of the ionic radii."
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The Coordination Number of the Cation will be Maximized subject
to the criterion of Maintaining Cation-Anion Contact
Determined by comparison of the ratio of the ionic radii, r+/r- with values
derived from the geometric contact criterion
Radius Ratio Rule
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Limiting Radius Ratios - anions in the coordination polyhedron of
cation are in contact with the cation and with each other
If cation were to shrink further (i.e. r+/r- decrease) cation-anion
contact would be lost in contravention of Pauling's 1st Rule.
Radius Ratio
Coordination no.
Binary (AB) Structure-type
r+/r- = 1
12
none known
1 > r+/r- > 0.732
8
CsCl
0.732 > r+/r- > 0.414
6
NaCl
0.414 > r+/r- > 0.225
4
ZnS
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离子半径
这些经验性资料使我们明白原子间距的数据中可能存在着某种系统性，晶
体结构的其他方面也如此。我想我们已经处在这样一个时期：在没有x射
线衍射图的条件下能够预言晶体的结构。
—————L.鲍林
哥希密特半径
鲍林半径
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正负离子不等径圆球密堆积的几
何关系为基础
从原子核对外层电子的静电作用
力原理出发
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哥希密特半径
Chapter VII Solid State
-
-
-
-
a0  2rS 2  2rCa 2
rCa 2  100 pm
晶体
MgO
MnO
CaO
MgS
MnS
CaS
a 0 (pm)
421
444
480
519
521
568
a0  2rO 2  2rCa 2
rO 2  140 pm
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2a0  4rS 2
rS 2  184 pm
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鲍林半径
对NaCl型晶体
Cn
Cn
r

Z  Z *
以KCl为例， rKCl  314 pm
  7  0.35  8  0.85  2 1  11.25
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rCl   rK   314
Cn  1030 pm
rCl   Cn / 17  11.25 
rCl   133 pm
rK   Cn / 19  11.25 
rK   181 pm
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• Allow Structure Prediction/Rationalization
e.g. In Perovskite, CaTiO3
•Ca2+ is 12-coordinated by O2Ca-O bond has e.b.s. =
2/ = 1/
12
6
•Ti4+ is 6-coordinated by O2Ti-O bond has e.b.s. = 4/6 =
2/
3
•O2- has a total valency of 2
Ca2+(1/6)} +{2 x Ti4+(2/3)}
satisfied by {4 x
• Each Oxygen must be common to 4 CaO12
cuboctahedra & 2 TiO6 octahedra
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Pauling Rule 3: Polyhedral Linking
"The stability of structures with different types of polyhedral linking is vertexsharing > edge-sharing > face-sharing"
•effect is largest for cations with high charge and low coordination number
•especially large when r+/r- approaches the lower limit of the polyhedral
stability
Sharing edges/faces brings ions at the center of each polyhedron closer together,
hence increasing electrostatic repulsions
i.e. disposition of ions of similar charge will be such as to minimize the
Electrostatic Energy between them
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Ni-Ni interaction
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Pauling Rule 4: Cation Evasion in >Binaries
"In a crystal containing different cations those of high valency and small
coordination number tend not to share polyhedron elements with each other"
CaII 12-coordinate CaO12 cuboctahedra share FACES
TiIV 6-coordinate TiO6 octahedra share only VERTICES
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Pauling Rule 2: Electrostatic Valence Principle ("Bond Strength")
"In a stable ionic structure the charge on an ion is balanced by the sum of
electrostatic bond strengths to the ions in its coordination polyhedron“
i.e. A stable ionic structure must be arranged to preserve Local Electroneutrality
Electrostatic Bond Strength (e.b.s.)
•For a cation Mm+ surrounded by n anions Xx- the electrostatic bond strength of the
cation is defined as:-
•For each anion (cation) the sum of the electrostatic bond strengths of the
surrounding cations (anions) must balance the negative (positive) charge on the
anion (cation)
For a binary compound A B the coordination numbers of A and B are in the
ratio y:x
x
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y
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Pauling Rule 5: Environmental Homogeneity
"The number of essentially different kinds of constituent in a crystal tend to be
small"
i.e. as far as possible, similar environments for chemically similar atoms
Treating the mineral Garnet Ca3Al2Si3O12 as an ionic crystal
Ca2+
Al3+
Si4+
coordination
8
6
4
e.b.s.
1/
4
1/
2
1
O2- bond strength of 2 is satisfied by a number of alternative combination of bonds
•Pauling Rule 5: Each O2- would prefer the same environment
• Only one possible arrangement:
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When Pauling's Rules are NOT Obeyed
Special structural influences in the bonding
Distorted coordination polyhedra
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Face-sharing
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Indirect evidence
that the structure is NOT
ionic
Increasing Polarization in
bonding
low-dimensionality layers/chains
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P=4, Tectosilicate, SiO2
e.g. Mn+x/n[(AlO2)x(SiO2)y].mH2O Zeolite ZSM-5
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CARBON with P = 4 the Diamond FRAMEWORK
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Chapter VII Solid State
Graphite
Types of Solids: Example Crystalline
Planar sp2
Structures
Diamond
bonding
3 bonding
Tetrahedral
sp
(good lubricant)
(very hard!)
Vertical bonds
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Bond angle = 109.5º
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