Chapter #11 Chemistry of Solids

Report
Chapter 11
The Chemistry of Solids
SOLIDS
Solids are either amorphous or crystalline.
AMORPHOUS SOLIDS: Considerable disorder in structure.
Example: rubber, glass
CRYSTALLINE SOLIDS: Highly regular structure in the form
of a repeating lattice of atoms or molecules
Crystalline solids are classified as:
atomic, metallic, ionic, or covalent network,
depending on the type of force holding the
particles together, and most often involve a
metal.
2
LATTICE EXAMPLE
We
can pick out the smallest repeating unit…..
3
UNIT CELL
We can pick out the smallest repeating unit…..
We call this the UNIT CELL………..
4
UNIT CELL
We call this the UNIT CELL………..
The
5 unit cell drawn here is a simple cubic cell
Examples of Unit Cells
UNIT CELL
What is a unit cell?
The smallest unit that, when stacked together
repeatedly without any gaps can reproduce the
entire crystal.
The three unit cells we deal with are…..
7
SIMPLE CUBIC
Eight equivalent points at
the corners of a cube
We can imagine an equivalent point at the centre of the
spheres
8
BODY CENTRED CUBIC
Eight equivalent points at
the corners of a cube and
one at the centre
Another possibility……...
9
FACE CENTRED CUBIC
Eight equivalent points at
the corners of a cube and six
on the centre of the cube
faces
Summary……..
10
THE CUBIC UNIT CELLS
Simple Cubic Unit Cell
KNOW
THESE!!!!
Body-Centred
Cubic Unit Cell
Face-Centred
Cubic Unit Cell
11
How do we investigate
solids?
Unit Cells in the Cubic Crystal System
Prentice-Hall © 2002
General Chemistry: Chapter 13
Slide 12 of 35
METALS
e.g. copper, gold, steel, sodium, brass.
Good conductors of heat and electricity
Shiny, ductile and malleable
Melting points: low (Hg at -39°C)
or high (W at 3370°C)
Can be soft (Na) or hard (W)
METALS ARE CRYSTALLINE SOLIDS
13
Electron Sea Model of Metals
Copyright © Houghton Mifflin
Company. All rights reserved.
10–14
Summary of Crystal Structures
METALS
VIEWED AS CLOSELY PACKED SPHERES
HOW CAN WE PACK SPHERES?????
16
PACKING OF SPHERICAL VEGETABLES
17
Packing Spheres into Lattices
The most efficient way to pack hard spheres is
CLOSEST PACKING
Spheres are packed in layers in which each
sphere is surrounded by six others.
For example…….
18
Packing Spheres into Lattices:
First Layer
Lets put in a few more spheres……….
19
Packing Spheres into Lattices
First Layer
20
Packing Spheres into Lattices
Next Layer
The next spheres fit into
a “dimple” formed by three spheres in the first layer.
21
Packing Spheres into Lattices:
Next Layer
The next spheres fit into
a “dimple” formed by three spheres in the first layer.
There are two sets of dimples…...
22
Packing Spheres into Lattices:
Next Layer
Triangle
not
inverted
The next spheres fit into
NOTE: the
inverted
triangle
The two types of “dimples” formed by three spheres
in the first layer.
The second layer…..
23
Packing Spheres into Lattices:
is formed by choosing one of the sets of dimples
Now put on second layer…...
24
Packing Spheres into Lattices:
Second Layer
Once one is put on the others are forced into half
of the dimples of the same type….
25
Packing Spheres into Lattices
Second Layer
Once one is put on the others are forced into half
of the dimples of the same type….
26
Packing Spheres into Lattices:
Second Layer
Once one is put on the others are forced into half
of the dimples of the same type….
27
And so on….
Packing Spheres into Lattices
Second Layer
Inverted triangle
dimples are not filled.
Note that the second layer only occupies half the
dimples in the first layer.
28
Packing Spheres into Lattices
Second Layer
Occupied dimple
Unoccupied DIMPLE
Note that the second layer only occupies half the dimples in the first layer.
29
THE THIRD LAYER…...
PACKING SPHERES INTO LATTICES
SECOND LAYER
HAVE TO CHOOSE A DIMPLE
30
PACKING SPHERES INTO LATTICES
THIRD LAYER, Choose a dimple
1
1
1
(1) A DIMPLE DIRECTLY ABOVE SPHERE IN THE FIRST LAYER
31
THIS OR……..
PACKING SPHERES INTO LATTICES
THIRD LAYER
NOTE: the
inverted
triangle
2
1
1
1
2
2
(2) A DIMPLE DIRECTLY ABOVE A DIMPLE IN THE FIRST
CHOOSE OPTION 1…..
LAYER……..
32
PACKING SPHERES INTO LATTICES
THIRD LAYER (option 1)
1
1
1
OPTION ONE!
SPHERE DIRECTLY ABOVE SPHERE IN THE FIRST LAYER
ADD SOME MORE……..
33
PACKING SPHERES INTO LATTICES
THIRD LAYER
OPTION ONE!
ADD SOME MORE…..
34
PACKING SPHERES INTO LATTICES
THIRD LAYER OPTION ONE!
THE ABA ARRANGEMENT OF LAYERS.
A
B
A
LAYERS ONE AND THREE ARE THE SAME!
35
PACKING SPHERES INTO LATTICES
THE ABA ARRANGEMENT OF LAYERS.
A
B
A
CALLED HEXAGONAL CLOSEST PACKIN HCP
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PACKING SPHERES INTO LATTICES
THE ABA ARRANGEMENT OF LAYERS, Option 1.
A HEXAGONAL UNIT CELL.
A
B
A
HEXAGONAL CLOSEST PACKING
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HEXAGONAL UNIT CELL
ABA ARRANGEMENT HAS A HEXAGONAL UNIT CELL.
HCP
38
SUMMARY...
SUMMARY
HEXAGONAL CLOSED PACKED STRUCTURE
EXPANDED VIEW
NOW OPTION TWO…..
39
PACKING SPHERES INTO LATTICES
THIRD LAYER
OPTION 2!
2
2
2
THIS DIMPLE DOES NOT
LIE DIRECTLY OVER THE SPHERES OF THE FIRST LAYER.
MORE
40
PACKING SPHERES INTO LATTICES
THIRD LAYER
GREEN SPHERES DO NOT LIE DIRECTLY OVER THE SPHERES
OF THE FIRST LAYER.
OPTION 2
THE THIRD LAYER IS DIFFERENT FROM THE FIRST…….
41
PACKING SPHERES INTO LATTICES
NOT THE SAME AS OPTION ONE!
THIRD LAYER
OPTION 2
A
B
C
WE CALL THE THIRD LAYER C THIS TIME!
42
PACKING SPHERES INTO LATTICES
THIRD LAYER
WE CALL THE THIRD LAYER C THIS TIME!
OPTION 2
A
B
C
THE ABC ARRANGEMENT OF LAYERS.
43
NOW THE FOURTH LAYER…….
PACKING SPHERES INTO LATTICES
FOURTH LAYER
PUT SPHERE IN SO THAT
A
B
C
FOURTH LAYER THE SAME AS FIRST.
44
PACKING SPHERES INTO LATTICES
PHH p 509
FOURTH LAYER THE SAME AS FIRST.
A
B
C
A
THE ABCA ARRANGEMENT………..
THIS IS CALLED….
45
PACKING SPHERES INTO LATTICES
THE ABCA ARRANGEMENT………..
A
B
C
A
CUBIC CLOSED PACKED….
46
WHY???
UNIT CELL OF CCP
CUBIC UNIT CELLL
THIS ABCA ARRANGEMENT HAS A
FACE- CENTRED CUBIC UNIT CELL (FCC)
A COMPARISON…..
47
COMPARISON
NOTICE the flip…...
HCP
NEAREST NEIGHBORS…..
48
CCP
COORDINATION NUMBER
The number of nearest neighbors that a
lattice point has in a crystalline solid
Lets look at hcp and ccp…...
49
COORDINATION NUMBER
HCP
50
COORDINATION NUMBER
HCP
COORDINATION NUMBER =12
51
COORDINATION NUMBER
HCP
COORDINATION NUMBER =12
52
CCP
COORDINATION NUMBER
HCP
COORDINATION NUMBER =12
53
CCP
COORDINATION NUMBER
SPHERES IN BOTH HCP AND CCP STRUCTURES
EACH HAVE A COORDINATION NUMBER OF 12.
QUESTION……..
54
REVIEW QUESTION
Which is the closest packed arrangement?
1
Stacking a second closeTop Row
packed layer of spheres directly
atop a close-packed layer below Bottom Row
2
Stacking a second close-packed
Top Row
layer of spheres in the
depressions formed by spheres in Bottom Row
the close-packed layer below.
ANSWER……..
55
REVIEW QUESTION
Which is the closest packed arrangement?
1
Stacking a second closepacked layer of spheres directly
atop a close-packed layer below
2
Stacking a second close-packed
layer of spheres in the
depressions formed by spheres in
the close-packed layer below.
NEAREST NEIGHBOURS IN OTHER UNIT CELLS……..
56
Alloys
• An alloy is a blend of a host metal and one or more
other elements which are added to change the
properties of the host metal.
• Ores are naturally occurring compounds or mixtures
of compounds from which elements can be
extracted.
• Bronze, first used about 5500 years ago, is an
example of a substitutional alloy, where tin atoms
replace some of the copper atoms in the cubic array.
Substitutional Alloy
Examples
Where a lattice atom is replaced by an atom of
similar size
• Brass, one third of copper atoms are replaced by
zinc atoms
• Sterling silver (93% Silver and 7%Cu)
• Pewter (85% Sn, 7% Cu, 6% Bi, and 2% Sb)
• Plumber’s solder (67% Pb and 33% Sn)
Bronze
Alloys
– Interstitial Alloy
• When lattice holes (interstices) are filled with smaller
atoms
• Steel best know interstitial alloy, contains carbon
atoms in the holes of an iron crystal
–Carbon atoms change properties
»Carbon a very good covalent bonding atom
changes the non-directional bonding of the
iron, to have some direction
»Results in increased strength, harder, and less
ductile
»The larger the percent of carbon the harder and
stronger the steel
• Other metals can be used in addition to carbon, thus
forming alloy steels
Carbon Steel
Unlike bronze the carbon atoms fit into the holes
formed by the stacking of the iron atoms. Alloys
formed by using the holes are called interstital
alloys.
Two Types of
Alloys
Substitutional
Interstitial
10–62
About Holes in Cubic Arrays
Atomic Size Ratios and the Location of Atoms in Unit
Cells
Packing
Type of Hole
Radius Ratio
pm
hcp or ccp
Tetrahedral
0.22 - 0.41
hcp or ccp
Octahedral
0.41 - 0.73
Simple Cubic
Cubic
0.73 - 1.00
COORDINATION NUMBER
SIMPLE CUBIC
How do we count nearest neighbors?
Draw a few more unit cells…...
64
COORDINATION NUMBER
SIMPLE CUBIC
Highlight the nearest neighbors….
65
COORDINATION NUMBER
SIMPLE CUBIC How many nearest neighbors???
coordination number of 6
66
What about body centered cubic?????
COORDINATION NUMBER
In the three types of cubic unit
cells:
Simple cubic CN = 6
Body Centered cubic CN = ?
Lets look at this…….
67
Body-centered cubic packing (bcc)
COORDINATION NUMBER?????
In bcc lattices, each sphere has a coordination
number of 8
What about face centered cubic?
68
COORDINATION NUMBER
In the three types of cubic unit cells:
Simple cubic
CN = 6
Body Centered cubic
CN = 8
Face Centered cubic CN = ?
Comes from ccp
Just like hcp
69
CN = 12
PACKING EFFICIENCY?
EFFICIENCY OF PACKING
THE FRACTION OF THE VOLUME THAT IS ACTUALLY
OCCUPIED BY SPHERES…..
 volume occupied by the spheres in the unit cell 
fv  

volume of the unit cell


WHAT DOES THIS MEAN???
70
PACKING EFFICIENCY
FRACTION(F) OF THE VOLUME OCCUPIED BY THE SPHERES IN THE UNIT
CELL.
 volume occupied by the spheres in the unit cell 
fv  

volume of the unit cell


fv 
Vspheres
Vunitcell
Vspheres= number of spheres x volume single sphere
Vunit cell = a3
cubic unit cell of edge length a
Lets get NUMBER OF SPHERES
71
PACKING EFFICIENCY
•COUNTING ATOMS IN A UNIT CELL!
•ATOMS CAN BE WHOLLY IN A UNIT CELL OR
•COUNTING ATOMS IN A UNIT CELL!
•ATOMS CAN BE WHOLLY IN A UNIT
CELL OR ATOMS SHARED
BETWEEN ADJACENT UNIT CELLS
COUNTS 1 FOR ATOM IN CELL
IN THE LATTICE
COUNTS FOR 1/4 ATOM ON A FACE.
COUNTS AS 1/8 FOR ATOM ON A CORNER.
72
COUNTS FOR 1/2 ATOM ON A FACE.
FACE-CENTRED CUBIC UNIT CELL
What is the number of spheres in the fcc unit cell?
Note: 1/8 of a sphere on 8 corners and ½ of a
Sphere on 6 faces of the cube
Total spheres = 8 (1/8) + 6 (1/2)
=1+3=4
QUESTION…..
73
QUESTION
THE NUMBER OF SPHERES IN A BODY CENTRED CUBIC CELL
IS?
ANSWER….
74
QUESTION
THE NUMBER OF SPHERES IN A BODY CENTRED CUBIC CELL
IS?
ANSWER…. Atoms = 8(1/8) + 1 = 2
VOLUME OCCUPIED IN FCC….
75
CUBIC UNIT CELLS
WHAT FRACTION OF SPACE IS OCCUPIED INFACE
CENTRED CUBIC CELL?
NUMBER OF SPHERES IS 4
NOW WE NEED THE VOLUME OF
A SPHERE, USING r FOR RADIUS
V total
76
4 3
= 4 ( r )
3
THERE ARE 4 SPHERES IN THE UNIT CELL
FACE-CENTRED CUBIC UNIT CELL
4 3
Vspheres
 4( r )
3
radius of the sphere is r .
Now we need the volume of the unit cell.
Why?????
fv 
Vspheres
Vunitcell
GET DIMENSIONS OF CUBE IN TERMS OF r…..
77
GETTING THE CUBE DIMENSIONS IN TERMS OF r
Let side of cube be a
a
NOW DRAW A FACE OF THE CUBE
REMEMBER THE SPHERES TOUCH!!
78
Let side of cube be a
a
DRAWING CUBE FACE
REMEMBER THE SPHERES TOUCH!!
Draw a square…..
Now we need to get a in terms of r
79
Let side of cube be a
a
CONSTRUCT A TRIANGLE
ON THE FACE
Why????
So we can use Pythagoras!
80
Let side of cube be a
a
a
GET a in terms of r
r
2r
r
81
Let side of cube be a
a
GET a in terms of r
FACE DIAGONAL = r + 2r + r=4r
PYTHAGORAS!
a2 + a2 = (4r)2
a
r
2a2 = 16r 2
2r
82
r
a2 = 8r 2
ar 8
Side of cube be in terms of r
a
ar 8
Now we can calculate the volume of
the unit cell
Vcell  a3
a
r
Vcell  r 8 
2r
83
r
NOW PUT IT ALL TOGETHER
3
FACE-CENTRED CUBIC UNIT CELL
fv 
fv 
Vspheres
Vunitcell
4
4 * 3 r 3
r 83


fv  0.740
84
fv 
4
4* 3
83
 
We conclude…..
FACE-CENTRED CUBIC UNIT CELL
fv 
Vspheres
Vunitcell
fv  0.740
In a cubic closest packed crystal
74% of the volume of a is taken up by spheres
and 26% is taken up by empty space.
QUESTION
85
Body Centered Cubic
e
d2 = e2 + e2
d2 = 2e2
(4r)2 = e2 + d2
16r2 = e2 + 2e2
r2 = 3e2/16
e = 4r/√3
CUBIC UNIT CELLS
THE EDGE LENGTH IN TERMS OF r
SIMPLE CUBIC
2r
NUMBER OF SPHERES
1
87
BODY CENTRED CUBIC
4r
3
FACE CENTRED CUBIC
r 8
2
4
VOLUME OCCUPIED
CUBIC UNIT CELLS
VOLUME OCCUPIED
SIMPLE CUBIC
BODY CENTRED CUBIC
FACE CENTRED CUBIC
52.4%
74.0%
68.0%
2r
NUMBER OF SPHERES
1
88
4r
3
2
r 8
QUESTION...
4
QUESTION
The fraction of space occupied in a hexagonal closest packed
arrangement of spheres is the same as that in…..
1
simple cubic unit cell
2
face centered cubic unit cell
3
body centered cubic unit cell
4
none of these
ANSWER…..
89
QUESTION
The fraction of space occupied in a hexagonal closest packed
arrangement of spheres is the same as that in…..
1
simple cubic unit cell
2
face centered cubic unit cell
3
body centered cubic unit cell
4
none of these
Summary……...
90
SUMMARY
sc:
bcc:
fcc:
hcp:
52.4% of space occupied by spheres
68.0% of space occupied by spheres
74.0% of space occupied by spheres
74.0% of space occupied by spheres
Make sure you can do the fcc, bcc and sc lattice
calculations!
What other property of a substance depends on
packing efficiency????????
91
DENSITY
We can calculate the density in a unit cell.
mass
Density 
volume
Mass is the mass of the number of atoms in the
unit cell.
Mass of one atom =atomic mass/6.022x1023
N0 = 6.022 x 1023 atoms per mole
Avogadro’s Number!
92
Volume of a copper unit cell
Cu crystalizes as a fcc
r= 128pm = 1.28x10-10m = 1.28x10-8cm
Volume of unit cell is given by:
Vcell  r 8 
8
3
Vcell  ( 8  1.28  10 cm )
3
Vcell  4.75  10
93
 23
cm
3
COPPER DENSITY CALCULATION
63.54 g Cu mole Cu
4 atoms Cu unit cell
4.75X10-23 cm3
mole Cu 6.022 X 1023 atoms unit cell
= 8.89 g/cm3
Laboratory measured density: 8.92 g/cm3
DETERMINATION OF ATOMIC RADIUS
At room temperature iron crystallizes with a bcc unit cell.
X-ray diffraction shows that the length of an edge is 287 pm.
What is the radius of the Fe atom?
4r
e
3
3e
r
4
EDGE LENGTH (e)
3  287 pm
r
 124 pm
4
95
AVOGADRO’S NUMBER
Sample Problem: Calculate Avogadro’s
number of iron if its unit cell length is 287 pm
and it has a density of 7.86 g/cm3.
55.85 g
Mole Fe
Fe(s) is bcc Two atoms / unit cell
96
AVOGADRO’S NUMBER
Sample Problem: Calculate Avogadro’s
number of iron if its unit cell length is 287
pm and it has a density of 7.86 g/cm3.
55.85 g cm3
Mole Fe 7.86 g
Fe(s) is bcc Two atoms / unit cell
97
AVOGADRO’S NUMBER
The density of Fe(s) is 7.86 g/cm3.
length of an edge is 287 pm.
V= e3 = (287pm)3 = 2.36x10-23cm3
cm3
pm3
55.85 g
Mole Fe 7.86 g (10 -12)3 cm3
Fe(s) is bcc Two atoms / unit cell
98
AVOGADRO’S NUMBER
The density of Fe(s) is 7.86 g/cm3.
length of an edge is 287 pm.
V= e3 = (287pm)3 = 2.36x10-23cm3
cm3
pm3
55.85 g
Mole Fe 7.86 g (10 -12)3 cm3
unit cell
(287 pm)3
Fe(s) is bcc Two atoms / unit cell
99
AVOGADRO’S NUMBER
cm3
pm3
55.85 g
Mole Fe 7.86 g (10 -12)3 cm3
100
unit cell 2 atoms
(287 pm)3 unit cell
AVOGADRO’S NUMBER
cm3
pm3
55.85 g
Mole Fe 7.86 g (10 -12)3 cm3
unit cell 2 atoms
(287 pm)3 unit cell
= 6.022 X 10-23 atoms/mole
101
IONIC SOLIDS
Binary Ionic Solids: Two types of ions
Examples: NaCl MgO CaCO3 MgSO4
Hard, brittle solids
High melting point
Electrical insulators
except when molten or dissolved in water.
These are lattices of ions…….
102
IONIC SOLIDS
= Na+
= Cl–
We notice that this is a cubic array of ions.
Why do ionic solids hold together?????
103
Sodium Chloride
Prentice-Hall © 2002
General Chemistry: Chapter 13
Slide 104 of 35
IONIC SOLIDS
The stability of the ionic compound results from the
electrostatic attractions between the ions:
Li+
F–
Li+
F–
F–
Li+
F–
Li+
Li+
F–
Li+
F–
F–
Li+
F–
Li+
The LiF crystal consists of a
lattice of ions.
The attractions are stronger
than the repulsions, so the
crystal is stable.
The stability is due to the LATTICE ENERGY
How can we describe ionic lattices?
105
NaCl structure
ClNa+
Lets take this apart…...
106
NaCl structure
ClNa+
Lets look at the black dot lattice….
107
NaCl structure
What unit cell do the black dots form?
The black dots form a fcc lattice!
Now look at the red dots
108
NaCl structure
What unit cell is this????
Cubic certainly
But which one?????
Lets have another look……….
109
NaCl structure
What unit cell is this????
Bring in a another array…..
The red dots form a fcc array!
Now put these back together…..
110
NaCl structure
FCC OF BLACK DOTS
Bring in red dots
NOTICE THE RED DOTS FIT NICELY IN BETWEEN BLACK
DOTS
THE RED DOTS SIT IN THE HOLES OF THE BLACK DOT FCC
ARRAY
111
NaCl structure
This is the NaCl structure.
ClNa+
Two interpenetrating fcc arrays, one of Na+ ions and one of Cl- ions.
The Na+ sit in the holes of the black (Cl-) lattice
SO HOW WE DESCRIBE IONIC SOLIDS???
112
IONIC SOLIDS
consist of two interpenetrating lattices of the
two ions (cations and anions) in the solid.
We describe an ionic solid as a lattice of the larger ions
with the smaller ions occupying holes in the lattice.
NOTE:
The anion is usually larger than the cation.
HOLES????
113
HOLES IN A LATTICE.
Holes???
What holes?????
Lets look at a fcc lattice!
114
HOLES IN A FCC LATTICE
The black dots form a FCC lattice!
See the holes????
115
HOLES IN A FCC LATTICE
The black dots form a fcc lattice!
See the holes????
HOW MANY HOLES??????
116
HOLES IN A FCC LATTICE
The holes:
How many??
THIRTEEN: ONE IN THE CENTRE
12 on the edges.
What shape is the hole ?
117
CENTRAL HOLE
OCTAHEDRAL HOLES:
There is one octahedral hole in the centre of the
unit cell.
If each one is occupied by an atom?
118
THE OCTAHEDRAL HOLES
1 atom
1/4 atom
If each one is occupied by an atom?
How many atoms per unit cell?
Number of atoms = 1 + 12 x (1/4) = 4
There are 4 complete octahedral holes per fcc unit cell.
119
THE 13 OCTAHEDRAL HOLES
1 atom
1/4 atom
There are 4 complete octahedral holes per fcc unit cell.
Notice that the number of octahedral holes is the same
as the number of atoms forming the unit cell!!
( 8x(1/8) + 6x(1/2) = 4) remember????
120
The octahedral hole is..
THE OCTAHEDRAL HOLES
Between two layers…..
Other holes…..
121
OTHER HOLES
There are other holes!
Can you spot them??????
Where are the other holes in the FCC unit cell?
Look at one of the small cubes
122
SMALL CUBE
Take a point at the centre of this cube
There are eight of these….
123
SMALL CUBE
8 CUBES
Take a point at the centre of this cube
Another one of these….
124
SMALL CUBE
8 CUBES
Take a point at the centre of this cube
An so on ….
125
SMALL CUBE
8 CUBES
This is a TETRAHEDRAL HOLE….
126
TETRAHEDRAL HOLES
Notice that there
are twice as
many tetrahedral
holes as atoms
forming the
lattice! That
would be 8 holes.
There is one tetrahedral hole in each of the eight
smaller cubes in the unit cell.
All the holes are completely within the cell, so there are
8 tetrahedral holes per fcc unit cell
127
This hole…….
TETRAHEDRAL HOLES
Formed by three spheres in one layer and
one sphere in another layer sitting in the dimple they form.
There is one more hole……….
128
TRIGONAL HOLES
The smallest hole!
Formed by three spheres in one layer.
Formed from the space between three ions in a plane.
Which hole????
129
HOLE OCCUPANTS?
Which holes are used by the cation??
Which of the holes is used depends upon the size of the
cation and…..
The size of the hole in the anion lattice…..
Why??????
130
HOLE OCCUPANTS?
Which hole will a cation occupy??????
They occupy the holes that result in maximum attraction
and minimum repulsion.
To do this…...
131
Which hole will a cation occupy??????
M+ or M2+ cations always occupy the holes
with the largest coordination number without
rattling around!
TIGHT FIT
Consequently the radius of the cation must be
greater than the size of the hole!
This causes the X– anions to be pushed apart,
which reduces the X– – X– repulsion.
So we will investigate the size of these holes!
132
Which hole will a cation occupy??????
Investigate the size of these holes!
The size of the hole depends upon the
size of the ion (usually anion) that forms the
lattice into which the cations are to go……...
OCTAHEDRAL HOLE IN FCC….
LOOK AT HOLE….
133
OCTAHEDRAL HOLES IN FCC
Look at plane
Draw a square.
Put in spheres.
These are the anions
Fit a small sphere in
134
OCTAHEDRAL HOLES IN FCC
Look at plane
Draw a square.
Put in spheres.
These are the anions
Fit a small sphere in
This will be the cation
Draw diagonal
135
Put in distances……..
OCTAHEDRAL HOLES IN FCC
Look at plane
Radius of ion = R
Radius of hole = r
2r
136
R
OCTAHEDRAL HOLES IN FCC
Look at plane
Radius of ion = R
Radius of hole = r
2
(2R)2 +(2R)2 = (2R + 2r)
8R2 = (2R + 2r)2
R
R
2r
R
R
137
2 2R  2 R  2r

( 2 2  2) R  2r
2  1R  r
0.414R = r
OCTAHEDRAL HOLES IN FCC
Look at plane
Radius of ion = R
Radius of hole = r
R
R
R
0.414R = r
The size of cation that just
fits has a radius that is
0.414 x radius of anion(R)
2r
R
roctahedral hole = 0.414 R
What about the tetrahedral hole?
138
Using similar calculations, we can find the radius of
other types of holes as well:
fcc
DO IT!!!!!!!
rtetrahedral = 0.225 R
roctahedral = 0.414 R
r = radius of ion fitting into hole (usually the cation)
R is the radius of the ion forming the lattice (usually the anion).
RADIUS RATIO:
The ratio between the radius of a hole in a cubic lattice
and the radius of the ions forming the hole
What about other cubic cell systems??
139
SIMPLE CUBIC
If the M+ cations (e.g. Cs+) are sufficiently large,
they can no longer fit into octahedral holes of a fcc lattice.
The next best closest packed X– array adopted by the
anions is a simple cubic structure, giving cubic holes
which are large enough to hold the cations.
YOU can show that...
rcubic = 0.732 Ranion
DO IT!!!!!!!
140
The cubic hole
The coordination number in the cubic hole is ?
8
rcubic = 0.732 Ranion
In contrast for a fcc lattice…...
The coordination number in
the fcc tetrahedral hole is ?
4!
141
The coordination number in
the fcc octahedral hole is ?
6!
SUMMARY:
Face centred cubic:
Trigonal hole Too small to be occupied
Tetrahedral hole CN = 4 rcation = 0.225Ranion
8 of these
Octahedral hole CN = 6 rcation = 0.414Ranion
4 of these
Simple cubic:
Cubic hole CN = 8 rcation = 0.732Ranion 1 of these
For a given anion
rtrigonal < rtetrahedral < roctahedral < rcubic
142
SUMMARY
Face centred cubic:
Trigonal hole
Too small to be occupied
Tetrahedral hole
CN = 4
rcation = 0.225Ranion
Octahedral hole
CN = 6
rcation = 0.414Ranion
4 of these
Simple cubic
CN = 8
rcation = 0.732Ranion
1 of these
For a given anion
rtrigonal < rtetrahedral < roctahedral < rcubic
Which hole will a cation occupy??????
143
8 of these
INTO WHICH HOLE WILL THE ION GO??
TETRAHEDRAL
The hole filled is tetrahedral if:
rtetrahedral < rcation < roctahedral
0.225Ranion < rcation < 0.414Ranion
144
INTO WHICH HOLE WILL THE ION GO??
OCTAHEDRAL
The hole filled is octahedral if:
roctahedral < rcation < rcubic
0.414Ranion < rcation < 0.732Ranion
145
INTO WHICH HOLE WILL THE ION GO??
CUBIC
The hole filled is cubic if:
rcubic < rcation
0.732Ranion < rcation
Lets look at these ideas in action…….
146
NaCl
Na+ has a radius of 98pm.
Cl- has a radius of 181pm.
Consider a fcc array of Cl- then:
Radius of the tetrahedral hole is 0.225 x 181=41pm
Radius of the octahedral hole is 0.414 x 181=75pm
Consider a sc array of Cl- then:
Radius of the cubic hole is 0.732 x 181=132pm
So the best fit is the octahedral hole in the fcc array!
The 98pm is bigger than 75pm but less than 132!
OR USING RATIOS…….
147
NaCl
Na+ has a radius of 98pm.
Cl- has a radius of 181pm.
rcation rNa 98 pm


 0.54
ranion rCl  181 pm
tet
rcation
 0.225
tet
ranion
oct
rcation
 0.414
oct
ranion
cubic
rcation
 0.732
cubic
ranion
0.54 lies between 0.414 and 0.732
so the sodium cations will occupy octahedral holes
in a fcc (ccp) lattice Is the stoichiometry ok???
148
NaCl
1:1 stoichiometry is required
How many complete octahedral holes in face centred
cubic array of Cl- ????? 4
How many Cl- needed to form the fcc array???
Therefore 4 Cl- and 4 Na+
So stoichiometry is ok!!
Lets do another example…..
149
4
Example: Predict the structure of Li2S
Li+ is 68 pm
S2- is 190pm
STEP ONE:
Examine the cation-anion radius ratios to find which
type of holes the smaller ions fill
Calculate ratio..
rcation rLi
68 pm


 0.36
ranion rS 2 190 pm
COMPARE with ratios….
tet
rcation
 0.225
tet
ranion
Which is the best hole????
150
oct
rcation
 0.414
oct
ranion
TETRAHEDRAL!!!!
Example: Predict the structure of Li2S
Li+ is 68 pm
S2- is 190pm
Calculate ratio..
rcation rLi
68 pm


 0.36
ranion rS 2 190 pm
This requires tetrahedral holes.
Which lattice has tetrahedral holes???
face- centred cubic array
Thus the S2- will form a fcc lattice ...
Lets look at the structure…...
151
FCC unit cell with tetrahedral holes
ANION
CATION
Four anions in the
unit cell.
There are 8
tetrahedral holes.
How many are occupied?
STEP TWO: Determine what fraction of those holes
must be filled to give the correct chemical formula So????
152
FCC unit cell with tetrahedral holes
S2-
Li+
Four anions in
the unit cell.
There are 8
tetrahedral holes.
How many are occupied?
Li2S needs two
Li+
for each
S2-
Next Step….
Therefore all the tetrahedral holes are occupied!
153
FCC unit cell with tetrahedral holes filled
= S2–
= Li+
all the tetrahedral
holes have to be
occupied.
STEP THREE: Describe the solid as an array of the larger
ions with the smaller ions occupying the appropriate holes.
154
Which is…..
FCC unit cell with tetrahedral holes filled
= S2–
= Li+
all the tetrahedral
holes have to be
occupied.
Li2S is a face centered lattice of S2- with all of the
tetrahedral holes filled by Li+ ions.
155
Now do CsCl….
+ is 167 pm
Cs
CsCl:
Cl- is 181pm
Calculate ratio
rcation rCs  167 pm


 0.92
ranion rCl  181 pm
cubic
Compare…...
rcation
 0.732
cubic
ranion
0.92 is greater than 0.732
the cesium cations will occupy cubic holes of a
simple cubic lattice.
STOICHIOMETRY?????
156
There are the same number of cubic holes and lattice
points in the cubic lattice.
Hence stoichiometry OK!
CsCl is composed of a simple cubic lattice of chloride
anions with cesium cations in all the cubic holes.
157
Cesium Chloride
Zn2+ is 64 pm S2- is 190 pm
ZnS:
rcation rZn2
64 pm


 0.35
ranion rS 2 190 pm
tet
COMPARE
rcation
 0.225
tet
ranion
Calculate ratio
This requires
tetrahedral holes.
oct
rcation
 0.414
oct
ranion
The sulfide ions will form a face-centered cubic
array because….
that is the only type to possess tetrahedral holes.
What about stoichiometry??????
159
Zn2+ is 64 pm S2- is 190 pm
ZnS:
rcation rZn2
64 pm


 0.35
ranion rS 2 190 pm
tet
COMPARE
rcation
 0.225
tet
ranion
Calculate ratio
This requires
tetrahedral holes.
oct
rcation
 0.414
oct
ranion
The sulfide ions will form a face-centered cubic
array because….
that is the only type to possess tetrahedral holes.
What about stoichiometry??????
160
We need an equal number of zinc and sulfide ions.
There are the twice as many tetrahedral holes(8) as S2(4) that form the fcc lattice.
Therefore, half the tetrahedral holes will be filled.
161
We need an equal number of zinc and sulfide ions.
Half the tetrahedral holes will be filled.
ZnS is composed of a fcc lattice of sulfide anions with zinc
cations in half the tetrahedral holes.
162
There are two forms of ZnS
One is the zinc blende that we have talked about!
This is an example
of polymorphism.
The other is wurtzite based on hcp lattice.
163
QUESTION
A crystal has Mg2+ ions at the corners of a cubic unit cell, F- ions at the midpoints of all
the edges and K+ ions at the body centre. The empirical formula is
1
KMgF3
2
K3MgF2
3
KMg2F2
4
K2Mg2F
5
K2Mg2F3
164
ANSWER
A crystal has Mg2+ ions at the corners of a cubic unit cell, F- ions
at the midpoints of all the edges and K+ ions at the body centre.
The empirical formula is
1
KMgF3
2
K3MgF2
3
KMg2F2
4
K2Mg2F
5
K2Mg2F3
165
QUESTION
A COMPOUND CONTAINS THE ELEMENTS Na,O and Cl. IT CRYSTALLISES IN A CUBIC LATTICE.
THE O ATOMS ARE AT THE CORNERS, THE Cl ATOMS ARE AT THE CENTER AND THE Na ARE
AT THE CENTRES OF THE FACES OF THE UNIT CELL. THE FORMULA OF THE COMPOUND IS...
1
Na2ClO
2
Na3ClO
3
NaCl3O
4
NaClO3
5
NaClO
166
QUESTION
A COMPOUNDCONTAINS THE ELEMENTS Na,O and Cl. IT CRYSTALLISES IN A CUBIC LATTICE.
THEO ATOMS ARE AT THE CORNERS, THE Cl ATOMS ARE AT THE CENTER AND THE Na ARE AT
THE CENTRES OF THE FACES OF THE UNIT CELL. THE FORMULA OF THE COMPOUND IS...
1
Na2ClO
2
Na3ClO
3
NaCl3O
4
NaClO3
5
NaClO
167
This is the flourite structure: MX2
the anions occupy the tetrahedral holes(8)
= Ca2+
in a fcc array of
the cations(4).
Does this fit radius ratios???????
168
= F-
CaF2:
Ca2+ is 99 pm F- is 136 pm
Calculate ratio
rcation rCa 2 99 pm


 0.727
ranion rF  136 pm
FOR TETRAHEDRAL HOLES
tet
rcation
 0.225
tet
ranion
OOPS!!!!
The radius ratio is too BIG!!!!
This shows Radius Ratios do not always work properly
But CaF2 can be thought of as a simple cubic of Fwith Ca2+ at alternate cubic holes!!!!!!!
169
SIMPLE CUBIC
CaF2:
Ca2+
Ca2+
Ca2+ in alternating cubic sites.
Alternative description
What is Antiflourite????
170
This is the flourite structure: MX2
the anions occupy the tetrahedral holes(8)
= Ca2+
= F-
in a fcc array of
the cations(4).
The antifluorite structure M2X (eg K2O)
the cations occupy the tetrahedral holes
and the anions form the fcc array.
171
Ionic Compound Density
fcc of O2- Mg2+ in octahedral holes
Calculating the density of an ionic
compound
A face….
MgO
Edge= roxide ion+ 2rMg ion+ roxide ion
Now calculate volume
4 Mg’s and 4 O2172
REMEMBER….
Ionic Compound Density
fcc of O2- Mg2+ in octahedral holes
Calculating the density of an ionic
compound
R = 86 pm
A face….
r = 126 pm
MgO
Edge = 424 pm
V = (424)3 =
7.62X107 pm3
Edge= roxide ion+ 2rMg ion+ roxide ion
Now calculate volume
4 Mg’s and 4 O2173
REMEMBER….
Ionic Compound Density
fcc of O2- Mg2+ in octahedral holes
Calculating the density of an ionic
compound
R = 86 pm
A face….
r = 126 pm
MgO
Edge = 424 pm
V = (424)3 =
7.62X107 pm3
Edge= roxide ion+ 2rMg ion+ roxide ion
Now calculate volume
4 Mg’s and 4 O2174
REMEMBER….
DENSITY OF IONIC CRYSTALS
For example calculate the density of MgO a fcc crystal with
762 pm for the length of it’s unit cell.
40.61g MgO
mole
DENSITY OF IONIC CRYSTALS
For example calculate the density of MgO a fcc crystal with
762 pm for the length of it’s unit cell.
40.61g MgO mole
mole
6.022 X 1023 FU
DENSITY OF IONIC CRYSTALS
For example calculate the density of MgO a fcc crystal with
762 pm for the length of it’s unit cell.
40.61g MgO mole
4 FU
mole
6.022 X 1023 FU unit cell
DENSITY OF IONIC CRYSTALS
For example calculate the density of MgO a fcc crystal with
424 pm for the length of it’s unit cell.
40.61g MgO mole
Unit Cell
4 FU
mole
6.022 X 1023 FU unit cell 7.62X107 pm3
DENSITY OF IONIC CRYSTALS
For example calculate the density of MgO a fcc crystal with
762 pm for the length of it’s unit cell.
40.61g MgO mole
Unit Cell
pm3
4 FU
mole
6.022 X 1023 FU unit cell 7.62X107 pm3 10-36m3
DENSITY OF IONIC CRYSTALS
For example calculate the density of MgO a fcc crystal with
762 pm for the length of it’s unit cell.
40.61g MgO mole
Unit Cell
pm3 10-6m
4 FU
mole
6.022 X 1023 FU unit cell 7.62X107 pm3 10-36m3 cm3
DENSITY OF IONIC CRYSTALS
For example calculate the density of MgO a fcc crystal with
424 pm for the length of it’s unit cell.
40.61g MgO mole
Unit Cell
pm3 10-6m
4 FU
mole
6.022 X 1023 FU unit cell 7.62X107 pm3 10-36m3 cm3
= 3.54 g/cm3
Diamond and Graphite
Covalently Networked Crystalline Solids
Copyright © Houghton Mifflin
Company. All rights reserved.
10–182
Diamond and Graphite
The p Orbitals (a) Perpendicular to the Plane of
the Carbon Ring System in Graphite can
Combine to Form (b) an Extensive pi Bonding
Network
10–183
SCATTERING OF X-RAYS BY CRYSTALS
In 19th century crystals were identified by their
shape…..
Crystallographers did not know atomic positions within
the crystal…….
In 1895 Roentgen discovered X-rays…...
And Max von Laue suggested that...
crystal might act as a diffraction grating for the X-
rays.
184
X-Ray Diffraction
SCATTERING OF X-RAYS BY CRYSTALS
In 1912 Knipping
observed……..
X-RAY DIFFRACTION PATTERN
Von Laue gets Noble Prize…….
How can we understand
this???
186
X-ray Diffraction
• X-ray diffraction (ERD) is a technique for determining
the arrangement of atoms or ions in a crystal by
analyzing the pattern that results when X-rays are
scattered after bombarding the crystal.
• The Bragg equation relates the angle of diffraction
(2) of X-rays to the spacing (d) between the layers
of ions or atoms in a crystal: n2dsin.
Cell Structure by X-ray Diffraction
BRAGG DIFFRACTION LAW
W.H. Bragg and W.L.Bragg noticed that

2
This is reminiscent of reflection…..
So they formulated diffraction in terms of
reflection from planes of electron density in the
crystal..
189
BRAGG’S LAW
BRAGG DIFFRACTION LAW
A plane of lattice points…….
Now imagine reflection of X-rays……….
191
Bragg Equation Derivation
ө
өө
d
sin ө = x/d
Wave length λ = 2x
x = d sin ө
λ = 2d sin ө
nλ = 2d sin ө
n due to multiple layers of particles
BRAGG’S LAW
Only at certain angles of ө will the waves from different
planes be in phase, thus nλ = 2dsinө
By adjusting the angle of the x-rays until constructive
interference is obtained, distance (d) between atoms is
obtained
BRAGG DIFFRACTION LAW
The Braggs also demonstrated diffraction….
And formulated a diffraction law…...
When electromagnetic radiation passes through matter…….
It interacts with the electrons and
Is scattered in all directions
the waves interfere……..
194
The Band Theory (MO theory)
• Review the Li MO diagram
– Many vacant MO’s
•
•
•
•
•
•
In fact only sigma is filled
This is for two atoms
Now how about four atoms more MO’s
How about a mole of atoms, tons of MO’s
For magnesium, which is HCC, look at the bands
The lower band holds electrons, but the next highest
vacant MO is just a small energy jump away
• Electrons do not flow in the lower band since they
bump into each other
• But a slight amount of energy promotes them to the
conduction band where they flow freely
Molecular Orbital
Energy Levels
10–196
Molecular Orbital
Energy Levels
10–197
Magnesium Band Model
• Looking at the band diagram for Mg
– The 1S, 2s, 2P electrons are in the well(localized
electrons)
– The valance electrons occupy closely spaced
orbitals that are partially filled
• Why then do nonmetals not conduct
– There is a large energy difference between
conduction and non conduction band
– There are more valence electrons
A Representation of the Energy Levels
(Bands) in a Magnesium Crystal
Copyright © Houghton Mifflin
Company. All rights reserved.
10–199
Molecular Orbital Energies
Insulator (diamond)
10–200
Conductor(metal)
Semi Conductors
• For metalloids the distance between the
conducting band and the nonconduction band
are lower, in between that for metals and
nonmentals, thus called semiconductors.
• For example silicon is a semiconductor, with
the same structure as diamond, since it is in
the same group.
• Diamond has a large gap in its band model,
but silicon, being a semi conductor, has a
smaller gap, thus promoting conduction.
Semi Conductors
• At higher temperatures more electrons are promoted
into the conduction band and conductivity increases
for semiconductors.
• adding impurities, such as phosphorus or gallium in
metalloids (usually silicon) changes the conduction
characteristics of the metalloid (silicon).
• When a small fraction of silicon atoms are replaced
with phosphous atoms, each with one more electron
than silicon, then extra electrons are available for
conduction (Called an n-type semi conductor)
Semi Conductors
• N-type semi conductors, using a phosphorus impurity,
provide more electrons than the original semi conductor,
usually Silicon.
– These electrons lie closer to the conduction band and
less energy is required for conduction
– This is called an n-type due to extra negative charge
• Conductivity can be enhanced by an element such as boron
that has one less valence electron than silicon
– These are called P-semiconductors
– Since we are missing an electron then there is a hole,
which an electron fills thus creating another hole
– Holes flow in a direction opposite to the flow of
electrons, since lower lying electrons are promoted to fill
the hole
– Called p for positive charge, due to one less electron
Energy-Level Diagrams for (a) an N-Type
Semiconductor and (b) a P-Type Semiconductor
As is example
Copyright © Houghton Mifflin
Company. All rights reserved.
B is an example
10–204
Semi Conductors
• Important application is to combine an n-type and a p-type
together, called a p-n junction
– When they are connected some of the electrons from the n-type
flow into the open holes of the p-type, thus creating a charge
difference
– Once the charge difference is achieved then electron flow ceases,
this is called contact potential, or junction potential
– If an external voltage is applied then electrons will only flow in
one way
• From the n-type to the p-type
• The holes flow in the opposite direction
– P-n-junctions makes an excellent rectifier, a device that produces
a pulsating direct current from an alternating current
– The overall effect is to convert alternating current into direct
current
– Old rectifiers were vacuum tubes, which were not very reliable
Semi Conductors
Some electrons flow to create opposite charges
No current flows, called reverse bias
Current flows, called forward bias
The End

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