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Chapter 11The Chemistry of Solids
2
SOLIDSSolids are either amorphous or crystalline.
CRYSTALLINE SOLIDS:
AMORPHOUS SOLIDS: Considerable disorder in structure.Example: rubber, glass
Highly regular structure in the formof 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.
3 We can pick out the smallest repeating unit…..
LATTICE EXAMPLE
4
We can pick out the smallest repeating unit…..
We call this the UNIT CELL………..
UNIT CELL
5
We call this the UNIT CELL………..
The unit cell drawn here is a simple cubic cell
UNIT CELL
Examples of Unit Cells
7
What is a unit cell?The smallest unit that, when stacked together repeatedly without any gaps can reproduce the entire crystal.
UNIT CELL
The three unit cells we deal with are…..
8
Eight equivalent points at the corners of a cube
SIMPLE CUBIC
We can imagine an equivalent point at the centre of the spheres
9
BODY CENTRED CUBIC
Eight equivalent points at the corners of a cube and one at the centre
Another possibility……...
10
FACE CENTRED CUBIC
Eight equivalent points at the corners of a cube and six on the centre of the cube faces
Summary……..
11
Simple Cubic Unit Cell
Body-CentredCubic Unit Cell
Face-CentredCubic Unit Cell
KNOW THESE!!!!
How do we investigate solids?
THE CUBIC UNIT CELLS
Prentice-Hall © 2002 General Chemistry: Chapter 13 Slide 12 of 35
Unit Cells in the Cubic Crystal System
13
Good conductors of heat and electricity
METALSe.g. copper, gold, steel, sodium, brass.
Shiny, ductile and malleableMelting points:
soft (Na) or hard (W)
low (Hg at -39°C)or high (W at 3370°C)
Can be METALS ARE CRYSTALLINE SOLIDS
Copyright © Houghton Mifflin Company. All rights reserved. 10–14
Electron Sea Model of Metals
Summary of Crystal Structures
16
METALS
VIEWED AS CLOSELY PACKED SPHERES
HOW CAN WE PACK SPHERES?????
17
PACKING OF SPHERICAL VEGETABLES
18
Packing Spheres into Lattices
The most efficient way to pack hard spheres is
Spheres are packed in layers in which each sphere is surrounded by six others.
For example…….
CLOSEST PACKING
19
Packing Spheres into Lattices:
First Layer
Lets put in a few more spheres……….
20
Packing Spheres into Lattices
First Layer
21
Packing Spheres into LatticesNext Layer The next spheres fit into
a “dimple” formed by three spheres in the first layer.
22
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…...
23
Packing Spheres into Lattices:Next Layer The next spheres fit into
The two types of “dimples” formed by three spheres in the first layer. The second layer…..
NOTE: the inverted triangle
Triangle not inverted
24
Packing Spheres into Lattices:
is formed by choosing one of the sets of dimples
Now put on second layer…...
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
Once one is put on the others are forced into half of the dimples of the same type….
Second Layer
27
Packing Spheres into Lattices:
Once one is put on the others are forced into half of the dimples of the same type….
And so on….
Second Layer
28
Packing Spheres into LatticesSecond Layer
Note that the second layer only occupies half the dimples in the first layer.
Inverted triangle dimples are not filled.
29
Packing Spheres into LatticesSecond Layer
Note that the second layer only occupies half the dimples in the first layer.
Occupied dimple
Unoccupied DIMPLE
THE THIRD LAYER…...
30
PACKING SPHERES INTO LATTICESSECOND LAYER
HAVE TO CHOOSE A DIMPLE
31
1
11
THIS OR……..
PACKING SPHERES INTO LATTICES
(1) A DIMPLE DIRECTLY ABOVE SPHERE IN THE FIRST LAYER
THIRD LAYER, Choose a dimple
32
1
11
(2) A DIMPLE DIRECTLY ABOVE A DIMPLE IN THE FIRST LAYER……..
2 2
2
NOTE: the inverted triangle
CHOOSE OPTION 1…..
PACKING SPHERES INTO LATTICESTHIRD LAYER
33
PACKING SPHERES INTO LATTICES
OPTION ONE!
1
11
ADD SOME MORE……..
THIRD LAYER (option 1)
SPHERE DIRECTLY ABOVE SPHERE IN THE FIRST LAYER
34
OPTION ONE!
ADD SOME MORE…..
PACKING SPHERES INTO LATTICESTHIRD LAYER
35
THE ABA ARRANGEMENT OF LAYERS.A
B
A
OPTION ONE!
LAYERS ONE AND THREE ARE THE SAME!
PACKING SPHERES INTO LATTICESTHIRD LAYER
36
PACKING SPHERES INTO LATTICESTHE ABA ARRANGEMENT OF LAYERS.
A
B
A
CALLED HEXAGONAL CLOSEST PACKIN HCP
37
PACKING SPHERES INTO LATTICESTHE ABA ARRANGEMENT OF LAYERS, Option 1.
A
B
A
HEXAGONAL CLOSEST PACKING
A HEXAGONAL UNIT CELL.
38
HCP
HEXAGONAL UNIT CELL
ABA ARRANGEMENT HAS A HEXAGONAL UNIT CELL.
SUMMARY...
39
SUMMARY
NOW OPTION TWO…..EXPANDED VIEW
HEXAGONAL CLOSED PACKED STRUCTURE
40
OPTION 2!
THIS DIMPLE DOES NOT
2 2
2
LIE DIRECTLY OVER THE SPHERES OF THE FIRST LAYER.
PACKING SPHERES INTO LATTICESTHIRD LAYER
MORE
41
GREEN SPHERES DO NOT LIE DIRECTLY OVER THE SPHERES OF THE FIRST LAYER.
THE THIRD LAYER IS DIFFERENT FROM THE FIRST…….
PACKING SPHERES INTO LATTICESTHIRD LAYER
OPTION 2
42
A
B
C
NOT THE SAME AS OPTION ONE!
WE CALL THE THIRD LAYER C THIS TIME!
PACKING SPHERES INTO LATTICES
THIRD LAYER OPTION 2
43
A
B
C
THE ABC ARRANGEMENT OF LAYERS.
WE CALL THE THIRD LAYER C THIS TIME!
NOW THE FOURTH LAYER…….
OPTION 2
PACKING SPHERES INTO LATTICESTHIRD LAYER
44
A
B
C
FOURTH LAYER THE SAME AS FIRST.
PACKING SPHERES INTO LATTICESFOURTH LAYERPUT SPHERE IN SO THAT
45
A
B
C
THE ABCA ARRANGEMENT………..
PHH p 509
FOURTH LAYER THE SAME AS FIRST.
PACKING SPHERES INTO LATTICES
A
THIS IS CALLED….
46
A
B
C
THE ABCA ARRANGEMENT………..
PACKING SPHERES INTO LATTICES
A
CUBIC CLOSED PACKED….
WHY???
47
UNIT CELL OF CCP
FACE- CENTRED CUBIC UNIT CELL (FCC)
THIS ABCA ARRANGEMENT HAS A
A COMPARISON…..
CUBIC UNIT CELLL
48
COMPARISON
HCP CCPNOTICE the flip…...
NEAREST NEIGHBORS…..
49
COORDINATION NUMBER
The number of nearest neighbors that a lattice point has in a crystalline solid
Lets look at hcp and ccp…...
50
COORDINATION NUMBER
HCP
51
COORDINATION NUMBER
HCP
COORDINATION NUMBER =12
52
COORDINATION NUMBERHCP CCP
COORDINATION NUMBER =12
53
HCP CCP
COORDINATION NUMBER =12
COORDINATION NUMBER
54
SPHERES IN BOTH HCP AND CCP STRUCTURES
COORDINATION NUMBER
EACH HAVE A COORDINATION NUMBER OF 12.
QUESTION……..
55
REVIEW QUESTION
1 Stacking a second close-packed layer of spheres directly atop a close-packed layer below
Which is the closest packed arrangement?
2 Stacking a second close-packed layer of spheres in the depressions formed by spheres in the close-packed layer below.
ANSWER……..
Top Row
Bottom Row
Top Row
Bottom Row
56
REVIEW QUESTION
1 Stacking a second close-packed layer of spheres directly atop a close-packed layer below
Which is the closest packed arrangement?
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……..
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 AlloyExamplesWhere 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 SteelUnlike 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.
10–62
Two Types of Alloys
Substitutional
Interstitial
Atomic Size Ratios and the Location of Atoms in Unit Cells
Packing Type of Hole Radius Ratio
hcp or ccp Tetrahedral 0.22 - 0.41
hcp or ccp Octahedral 0.41 - 0.73
Simple Cubic Cubic 0.73 - 1.00
About Holes in Cubic Arrays
pm
64
SIMPLE CUBIC
How do we count nearest neighbors?
COORDINATION NUMBER
Draw a few more unit cells…...
65
SIMPLE CUBIC
Highlight the nearest neighbors….
COORDINATION NUMBER
66
coordination number of 6
What about body centered cubic?????
SIMPLE CUBIC
COORDINATION NUMBER
How many nearest neighbors???
67
In the three types of cubic unit cells:
Simple cubic
COORDINATION NUMBER
CN = 6
Body Centered cubic CN = ?
Lets look at this…….
68
In bcc lattices, each sphere has a coordination number of 8
Body-centered cubic packing (bcc)
COORDINATION NUMBER?????
What about face centered cubic?
69
In the three types of cubic unit cells:
Simple cubic
COORDINATION NUMBER
CN = 6
Body Centered cubic CN = 8
Face Centered cubic CN = ?
Just like hcp CN = 12
Comes from ccp
PACKING EFFICIENCY?
70
THE FRACTION OF THE VOLUME THAT IS ACTUALLY OCCUPIED BY SPHERES…..
WHAT DOES THIS MEAN???
cellunit the of volumecellunit the in spheres the by occupied volume
vf
EFFICIENCY OF PACKING
71
FRACTION(F) OF THE VOLUME OCCUPIED BY THE SPHERES IN THE UNIT CELL.
unitcell
spheresv V
Vf
Vspheres= number of spheres x volume single sphere
Vunit cell = a3 cubic unit cell of edge length a
Lets get NUMBER OF SPHERES
cellunit the of volumecellunit the in spheres the by occupied volume
vf
PACKING EFFICIENCY
72
•COUNTING ATOMS IN A UNIT CELL!
•ATOMS CAN BE WHOLLY IN A UNIT CELL OR
PACKING EFFICIENCY
•COUNTING ATOMS IN A UNIT CELL!•ATOMS CAN BE WHOLLY IN A UNIT CELL OR ATOMS SHARED BETWEEN ADJACENT UNIT CELLS IN THE LATTICE COUNTS 1 FOR ATOM IN CELL
COUNTS FOR 1/2 ATOM ON A FACE.
COUNTS FOR 1/4 ATOM ON A FACE.
COUNTS AS 1/8 FOR ATOM ON A CORNER.
73
What is the number of spheres in the fcc unit cell?
Total spheres = 8 (1/8) + 6 (1/2)
= 1 + 3 = 4
QUESTION…..
FACE-CENTRED CUBIC UNIT CELL
Note: 1/8 of a sphere on 8 corners and ½ of aSphere on 6 faces of the cube
74
QUESTIONTHE NUMBER OF SPHERES IN A BODY CENTRED CUBIC CELL IS?
ANSWER….
75
QUESTIONTHE NUMBER OF SPHERES IN A BODY CENTRED CUBIC CELL IS?
ANSWER…. Atoms = 8(1/8) + 1 = 2
VOLUME OCCUPIED IN FCC….
76
CUBIC UNIT CELLSWHAT FRACTION OF SPACE IS OCCUPIED INFACE CENTRED CUBIC CELL?
NUMBER OF SPHERES IS 4
NOW WE NEED THE VOLUME OFA SPHERE, USING r FOR RADIUS
V = total THERE ARE 4 SPHERES IN THE UNIT CELL44
33( )r
77
Now we need the volume of the unit cell.
radius of the sphere is r .
unitcell
spheresv V
Vf
GET DIMENSIONS OF CUBE IN TERMS OF r…..
)33
4(4 rspheresV
Why?????
FACE-CENTRED CUBIC UNIT CELL
78
Let side of cube be a
a
GETTING THE CUBE DIMENSIONS IN TERMS OF r
NOW DRAW A FACE OF THE CUBE
REMEMBER THE SPHERES TOUCH!!
79
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
80
Let side of cube be a
aCONSTRUCT A TRIANGLE ON THE FACE
Why????
So we can use Pythagoras!
81
Let side of cube be a
a GET a in terms of r
a r
2r
r
82
Let side of cube be aa
GET a in terms of r
a r
2r
r
a2 + a2 = (4r)2
2a2 = 16r 2
a2 = 8r 2
FACE DIAGONAL = r + 2r + r=4r
PYTHAGORAS!
8ra
83
Side of cube be in terms of ra
Now we can calculate the volume of the unit cell
ar
2r
r
3aV cell
NOW PUT IT ALL TOGETHER
8ra
3 8rV cell
84
33
34
8
*4
r
rf
v 3
34
8
*4 vf
740.0vf
unitcell
spheresv V
Vf
FACE-CENTRED CUBIC UNIT CELL
We conclude…..
85
and 26% is taken up by empty space.
unitcell
spheresv V
Vf 740.0vf
In a cubic closest packed crystal
74% of the volume of a is taken up by spheres
QUESTION
FACE-CENTRED CUBIC UNIT CELL
Body Centered Cubic
r2r
r
e
ee
d
d2 = e2 + e2
d2 = 2e2
(4r)2 = e2 + d2
16r2 = e2 + 2e2
r2 = 3e2/16 e = 4r/√3
87
CUBIC UNIT CELLSTHE EDGE LENGTH IN TERMS OF r
SIMPLE CUBIC BODY CENTRED CUBIC FACE CENTRED CUBIC
NUMBER OF SPHERES
1 2 4
2r
34r
8r
VOLUME OCCUPIED
88
CUBIC UNIT CELLS
SIMPLE CUBIC BODY CENTRED CUBIC FACE CENTRED CUBIC
NUMBER OF SPHERES
1 2 4
2r
34r 8r
VOLUME OCCUPIED
52.4%68.0% 74.0%
QUESTION...
89
QUESTIONThe 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…..
90
QUESTIONThe 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 cell3 body centered cubic unit cell
4 none of these
Summary……...
91
Make sure you can do the fcc, bcc and sc lattice calculations!
sc: 52.4% of space occupied by spheresbcc: 68.0% of space occupied by spheresfcc: 74.0% of space occupied by sphereshcp: 74.0% of space occupied by spheres
SUMMARY
What other property of a substance depends on packing efficiency????????
92
We can calculate the density in a unit cell.
volumemass
Density
Mass is the mass of the number of atoms in the unit cell.
Mass of one atom =atomic mass/6.022x1023
Avogadro’s Number!
DENSITY
N0 = 6.022 x 1023 atoms per mole
93
Volume of a copper unit cell
r= 128pm = 1.28x10-10m = 1.28x10-8cm Volume of unit cell is given by:
38( )1028.18 cmV
cell
Cu crystalizes as a fcc
3 8rV cell
3231075.4 cmV
cell
COPPER DENSITY CALCULATION
63.54 g Cumole Cu
mole Cu6.022 X 1023 atoms
4 atoms Cuunit cell 4.75X10-23 cm3
unit cell
= 8.89 g/cm3
Laboratory measured density: 8.92 g/cm3
95
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?
EDGE LENGTH (e) 3
4re
43 e
r
pmpm
r 12442873
96
AVOGADRO’S NUMBER
Fe(s) is bcc Two atoms / unit cell
55.85 gMole Fe
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.
97
AVOGADRO’S NUMBERSample 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.
Fe(s) is bcc Two atoms / unit cell
55.85 gMole Fe 7.86 g
cm3
98
AVOGADRO’S NUMBERThe density of Fe(s) is 7.86 g/cm3.
V= e3 = (287pm)3 = 2.36x10-23cm3
Fe(s) is bcc Two atoms / unit cell
length of an edge is 287 pm.
55.85 gMole Fe 7.86 g
cm3
(10 -12)3 cm3
pm3
99
AVOGADRO’S NUMBERThe density of Fe(s) is 7.86 g/cm3.
V= e3 = (287pm)3 = 2.36x10-23cm3
Fe(s) is bcc Two atoms / unit cell
length of an edge is 287 pm.
55.85 gMole Fe 7.86 g
cm3
(10 -12)3 cm3
pm3
(287 pm)3
unit cell
100
AVOGADRO’S NUMBER
55.85 gMole Fe 7.86 g
cm3
(10 -12)3 cm3
pm3
(287 pm)3
unit cell 2 atoms
unit cell
101
AVOGADRO’S NUMBER
55.85 gMole Fe 7.86 g
cm3
(10 -12)3 cm3
pm3
(287 pm)3
unit cell 2 atoms
unit cell
= 6.022 X 10-23 atoms/mole
102
IONIC SOLIDS
NaCl
Binary Ionic Solids: Two types of ions
MgO CaCO3 MgSO4
Hard, brittle solids
High melting pointElectrical insulatorsexcept when molten or dissolved in water.
These are lattices of ions…….
Examples:
103
= Na+
= Cl–
IONIC SOLIDS
We notice that this is a cubic array of ions.
Why do ionic solids hold together?????
Prentice-Hall © 2002 General Chemistry: Chapter 13 Slide 104 of 35
Sodium Chloride
105
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 stability is due to the LATTICE ENERGY
The attractions are stronger than the repulsions, so the crystal is stable.
IONIC SOLIDS
How can we describe ionic lattices?
106
Cl-
Na+
Lets take this apart…...
NaCl structure
107
Cl-
Na+
Lets look at the black dot lattice….
NaCl structure
108
The black dots form a fcc lattice!
What unit cell do the black dots form?
Now look at the red dots
NaCl structure
109
What unit cell is this????
Cubic certainly But which one?????
Lets have another look……….
NaCl structure
110
The red dots form a fcc array!
What unit cell is this????
Now put these back together…..
Bring in a another array…..
NaCl structure
111
FCC OF BLACK DOTSBring 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
NaCl structure
112
This is the NaCl structure.
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???
Cl-
Na+
NaCl structure
113
The anion is usually larger than the cation.
We describe an ionic solid as a lattice of the larger ions with the smaller ions occupying holes in the lattice.
consist of two interpenetrating lattices of thetwo ions (cations and anions) in the solid.
NOTE:
IONIC SOLIDS
HOLES????
114
Holes???
What holes?????
Lets look at a fcc lattice!
HOLES IN A LATTICE.
115
The black dots form a FCC lattice!
See the holes????
HOLES IN A FCC LATTICE
116
The black dots form a fcc lattice!
See the holes????
HOW MANY HOLES??????
HOLES IN A FCC LATTICE
117
HOLES IN A FCC LATTICE
The holes:
THIRTEEN: ONE IN THE CENTRE
How many??
12 on the edges.What shape is the hole ?
118
OCTAHEDRAL HOLES:There is one octahedral hole in the centre of the unit cell.
CENTRAL HOLE
If each one is occupied by an atom?
119
There are 4 complete octahedral holes per fcc unit cell.
THE OCTAHEDRAL HOLES
If each one is occupied by an atom?How many atoms per unit cell?Number of atoms = 1 + 12 x (1/4) = 4
1/4 atom1 atom
120
There are 4 complete octahedral holes per fcc unit cell.
THE 13 OCTAHEDRAL HOLES
1/4 atom1 atom
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????
The octahedral hole is..
121
THE OCTAHEDRAL HOLES
Other holes…..
Between two layers…..
122
There are other holes!
Where are the other holes in the FCC unit cell?
Can you spot them??????
Look at one of the small cubes
OTHER HOLES
123
SMALL CUBE
Take a point at the centre of this cube
There are eight of these….
124
Take a point at the centre of this cube
Another one of these….
8 CUBES
SMALL CUBE
125
Take a point at the centre of this cube
An so on ….
8 CUBES
SMALL CUBE
126
This is a TETRAHEDRAL HOLE….
8 CUBES
SMALL CUBE
127
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
Notice that there are twice as many tetrahedral holes as atoms forming the lattice! That would be 8 holes.
TETRAHEDRAL HOLES
This hole…….
128
TETRAHEDRAL HOLES
Formed by three spheres in one layer and
There is one more hole……….
one sphere in another layer sitting in the dimple they form.
129
Formed from the space between three ions in a plane.
TRIGONAL HOLES
Formed by three spheres in one layer.
The smallest hole!
Which hole????
130
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??????
HOLE OCCUPANTS?
131
They occupy the holes that result in maximum attraction and minimum repulsion.
To do this…...
Which hole will a cation occupy??????
HOLE OCCUPANTS?
132
M+ or M2+ cations always occupy the holes
Consequently the radius of the cation must be
This causes the X– anions to be pushed apart,
greater than the size of the hole!
which reduces the X– – X– repulsion.
with the largest coordination number without rattling around!
Which hole will a cation occupy??????
TIGHT FIT
So we will investigate the size of these holes!
133
Investigate the size of these holes!
Which hole will a cation occupy??????
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….
134
OCTAHEDRAL HOLES IN FCC
Look at planeDraw a square.
Put in spheres.
Fit a small sphere in
These are the anions
135
Look at planeDraw a square.
Put in spheres.
Fit a small sphere in
This will be the cation
These are the anions
Draw diagonal
Put in distances……..
OCTAHEDRAL HOLES IN FCC
136
2r
Radius of ion = RLook at plane
Radius of hole = r
OCTAHEDRAL HOLES IN FCC
R
137
R
R
R
R
2r
Radius of ion = R
Look at plane
Radius of hole = r
+(2R)2
(2R)2 = (2R + 2r)2
8R2 = (2R + 2r)2
rRR 2222 rR 2)222(
rR 12
0.414R = r
OCTAHEDRAL HOLES IN FCC
138
R
R
R
R2r
Radius of ion = R
Look at plane
Radius of hole = r
0.414R = rThe size of cation that just fits has a radius that is
0.414 x radius of anion(R)roctahedral hole = 0.414 R
What about the tetrahedral hole?
OCTAHEDRAL HOLES IN FCC
139
Using similar calculations, we can find the radius of other types of holes as well:
rtetrahedral = 0.225 R
r = radius of ion fitting into hole (usually the cation)
The ratio between the radius of a hole in a cubic lattice
R is the radius of the ion forming the lattice (usually the anion).
fcc
RADIUS RATIO:
and the radius of the ions forming the hole
roctahedral = 0.414 R
What about other cubic cell systems??
DO IT!!!!!!!
140
rcubic = 0.732 Ranion
If the M+ cations (e.g. Cs+) are sufficiently large,
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.
SIMPLE CUBIC
they can no longer fit into octahedral holes of a fcc lattice.
YOU can show that...
DO IT!!!!!!!
141
The cubic hole
The coordination number in the cubic hole is ?
The coordination number in the fcc tetrahedral hole is ?
4!
The coordination number in the fcc octahedral hole is ?
6!
8
In contrast for a fcc lattice…...
rcubic = 0.732 Ranion
142
SUMMARY:
Face centred cubic:Trigonal hole Too small to be occupiedTetrahedral hole CN = 4 rcation = 0.225Ranion
Octahedral hole CN = 68 of these
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
143
SUMMARYFace centred cubic:
Trigonal hole Too small to be occupied
Tetrahedral hole CN = 4 rcation = 0.225Ranion
Octahedral hole CN = 6
8 of these
rcation = 0.414Ranion 4 of these
Simple cubic CN = 8 rcation = 0.732Ranion 1 of these
Which hole will a cation occupy??????
rtrigonal < rtetrahedral < roctahedral < rcubic
For a given anion
144
INTO WHICH HOLE WILL THE ION GO??
TETRAHEDRAL
The hole filled is tetrahedral if:
0.225Ranion < rcation < 0.414Ranion
rtetrahedral < rcation < roctahedral
145
INTO WHICH HOLE WILL THE ION GO??
OCTAHEDRALThe hole filled is octahedral if:
0.414Ranion < rcation < 0.732Ranion
roctahedral < rcation < rcubic
146
INTO WHICH HOLE WILL THE ION GO??
CUBIC
The hole filled is cubic if:
0.732Ranion < rcation
Lets look at these ideas in action…….
rcubic < rcation
147
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…….
NaCl
148
Na+ has a radius of 98pm.Cl- has a radius of 181pm.
54.018198
pmpm
r
r
rr
Cl
Na
anion
cation
225.0tet
anion
tet
cation
rr
414.0oct
anion
oct
cation
rr
732.0cubic
anion
cubic
cation
rr
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???
NaCl
149
1:1 stoichiometry is required
How many complete octahedral holes in face centred cubic array of Cl- ?????
So stoichiometry is ok!!
4How many Cl- needed to form the fcc array??? 4
Therefore 4 Cl- and 4 Na+
NaCl
Lets do another example…..
150
Example: Predict the structure of Li2S
Li+ is 68 pm S2- is 190pm
36.019068
2
pmpm
r
r
rr
S
Li
anion
cationCalculate ratio..
Examine the cation-anion radius ratios to find which type of holes the smaller ions fill
STEP ONE:
225.0tet
anion
tet
cation
rr
414.0oct
anion
oct
cation
rr
COMPARE with ratios….
Which is the best hole???? TETRAHEDRAL!!!!
151
This requires tetrahedral holes.
Example: Predict the structure of Li2S
Li+ is 68 pm S2- is 190pm
36.019068
2
pmpm
r
r
rr
S
Li
anion
cation
face- centred cubic array
Lets look at the structure…...
Calculate ratio..
Which lattice has tetrahedral holes???
Thus the S2- will form a fcc lattice ...
152
So????
FCC unit cell with tetrahedral holes
ANION
CATION
There are 8 tetrahedral holes.
How many are occupied?
Four anions in the unit cell.
STEP TWO: Determine what fraction of those holes must be filled to give the correct chemical formula
153
FCC unit cell with tetrahedral holes
S2-
Li+
How many are occupied?
Li2S needs two Li+ for each S2-
Four anions in the unit cell.
There are 8 tetrahedral holes.
Therefore all the tetrahedral holes are occupied!
Next Step….
154
= S2–
= Li+
all the tetrahedral holes have to be occupied.
Which is…..
FCC unit cell with tetrahedral holes filled
STEP THREE: Describe the solid as an array of the larger ions with the smaller ions occupying the appropriate holes.
155
= S2–
= Li+
all the tetrahedral holes have to be occupied.
Now do CsCl….
FCC unit cell with tetrahedral holes filled
Li2S is a face centered lattice of S2- with all of the
tetrahedral holes filled by Li+ ions.
156
CsCl: Cs+ is 167 pm Cl- is 181pm Calculate ratio
0.92 is greater than 0.732
92.0181167
pmpm
r
r
rr
Cl
Cs
anion
cation
the cesium cations will occupy cubic holes of a simple cubic lattice.
732.0cubic
anion
cubic
cation
rr
Compare…...
STOICHIOMETRY?????
157
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.
Cesium Chloride
159
ZnS: Zn2+ is 64 pm S2- is 190 pm Calculate ratio
35.019064
2
2
pmpm
r
r
rr
S
Zn
anion
cation
225.0tet
anion
tet
cation
rr
414.0oct
anion
oct
cation
rr
COMPARE
This requires tetrahedral holes.
The sulfide ions will form a face-centered cubic array because….
that is the only type to possess tetrahedral holes.
What about stoichiometry??????
160
ZnS: Zn2+ is 64 pm S2- is 190 pm Calculate ratio
35.019064
2
2
pmpm
r
r
rr
S
Zn
anion
cation
225.0tet
anion
tet
cation
rr
414.0oct
anion
oct
cation
rr
COMPARE
This requires tetrahedral holes.
The sulfide ions will form a face-centered cubic array because….
that is the only type to possess tetrahedral holes.
What about stoichiometry??????
161
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.
162
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.
163
There are two forms of ZnS
This is an example of polymorphism.
One is the zinc blende that we have talked about!
The other is wurtzite based on hcp lattice.
164
QUESTIONA 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
ANSWERA 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
166
QUESTIONA 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
167
QUESTIONA 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
168
This is the flourite structure: MX2
= Ca2+
= F-
the anions occupy the tetrahedral holes(8)
in a fcc array of the cations(4).
Does this fit radius ratios???????
169
CaF2: Ca2+ is 99 pm F- is 136 pm Calculate ratio
727.0136992
pmpm
r
r
rr
F
Ca
anion
cation
225.0tet
anion
tet
cation
rr
OOPS!!!!
The radius ratio is too BIG!!!!
FOR TETRAHEDRAL HOLES
This shows Radius Ratios do not always work properlyBut CaF2 can be thought of as a simple cubic of F-
with Ca2+ at alternate cubic holes!!!!!!!
170
CaF2:
Alternative description
SIMPLE CUBIC
Ca2+ in alternating cubic sites.
What is Antiflourite????
Ca2+Ca2+
171
This is the flourite structure: MX2
The antifluorite structure M2X (eg K2O)
= Ca2+
= F-
the anions occupy the tetrahedral holes(8)
in a fcc array of the cations(4).
the cations occupy the tetrahedral holesand the anions form the fcc array.
172
Calculating the density of an ionic compound
MgO fcc of O2- Mg2+ in octahedral holes
A face….
Edge= roxide ion+ 2rMg ion+ roxide ion
Now calculate volume
4 Mg’s and 4 O2-
REMEMBER….
Ionic Compound Density
173
Calculating the density of an ionic compound
MgO fcc of O2- Mg2+ in octahedral holes
A face….
Edge= roxide ion+ 2rMg ion+ roxide ion
Now calculate volume
4 Mg’s and 4 O2-
REMEMBER….
Ionic Compound Density
R = 86 pm
r = 126 pm
Edge = 424 pm
V = (424)3 = 7.62X107 pm3
174
Calculating the density of an ionic compound
MgO fcc of O2- Mg2+ in octahedral holes
A face….
Edge= roxide ion+ 2rMg ion+ roxide ion
Now calculate volume
4 Mg’s and 4 O2-
REMEMBER….
Ionic Compound Density
R = 86 pm
r = 126 pm
Edge = 424 pm
V = (424)3 = 7.62X107 pm3
DENSITY OF IONIC CRYSTALSFor example calculate the density of MgO a fcc crystal with 762 pm for the length of it’s unit cell.
40.61g MgOmole
DENSITY OF IONIC CRYSTALSFor example calculate the density of MgO a fcc crystal with 762 pm for the length of it’s unit cell.
40.61g MgOmole
mole
6.022 X 1023 FU
DENSITY OF IONIC CRYSTALSFor example calculate the density of MgO a fcc crystal with 762 pm for the length of it’s unit cell.
40.61g MgOmole
4 FUmole
6.022 X 1023 FU unit cell
DENSITY OF IONIC CRYSTALSFor example calculate the density of MgO a fcc crystal with 424 pm for the length of it’s unit cell.
40.61g MgOmole
4 FUmole
6.022 X 1023 FU unit cell7.62X107 pm3
Unit Cell
DENSITY OF IONIC CRYSTALSFor example calculate the density of MgO a fcc crystal with 762 pm for the length of it’s unit cell.
40.61g MgOmole
4 FUmole
6.022 X 1023 FU unit cell
pm3
7.62X107 pm3
Unit Cell
10-36m3
DENSITY OF IONIC CRYSTALSFor example calculate the density of MgO a fcc crystal with 762 pm for the length of it’s unit cell.
40.61g MgOmole
4 FUmole
6.022 X 1023 FU unit cell
pm3
cm37.62X107 pm3
Unit Cell
10-36m3
10-6m
DENSITY OF IONIC CRYSTALSFor example calculate the density of MgO a fcc crystal with 424 pm for the length of it’s unit cell.
40.61g MgOmole
4 FUmole
6.022 X 1023 FU unit cell
pm3
cm37.62X107 pm3
Unit Cell
10-36m3
10-6m
= 3.54 g/cm3
Copyright © Houghton Mifflin Company. All rights reserved. 10–182
Diamond and GraphiteCovalently Networked Crystalline Solids
10–183
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
Diamond and Graphite
184
SCATTERING OF X-RAYS BY CRYSTALS
crystal might act as a diffraction grating for the X- rays.
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...
X-Ray Diffraction
186
SCATTERING OF X-RAYS BY CRYSTALS
In 1912 Knipping observed……..
X-RAY DIFFRACTION PATTERN
Von Laue gets Noble Prize…….
How can we understand this???
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: n2dsin.
Cell Structure by X-ray Diffraction
189
BRAGG DIFFRACTION LAWW.H. Bragg and W.L.Bragg noticed that
This is reminiscent of reflection…..
So they formulated diffraction in terms of reflection from planes of electron density in the crystal..
BRAGG’S LAW
191
BRAGG DIFFRACTION LAW
A plane of lattice points…….
Now imagine reflection of X-rays……….
Bragg Equation Derivation
ө
ө ө
x xd
sin ө = x/d x = d sin ө
Wave length λ = 2x 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 = 2dsinөBy adjusting the angle of the x-rays until constructive interference is obtained, distance (d) between atoms is obtained
194
BRAGG DIFFRACTION LAWThe 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……..
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
10–196
Molecular Orbital Energy Levels
10–197
Molecular Orbital Energy Levels
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
Copyright © Houghton Mifflin Company. All rights reserved. 10–199
A Representation of the Energy Levels (Bands) in a Magnesium Crystal
10–200
Molecular Orbital Energies
Insulator (diamond) 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
Copyright © Houghton Mifflin Company. All rights reserved. 10–204
Energy-Level Diagrams for (a) an N-Type Semiconductor and (b) a P-Type Semiconductor
As is example B is an example
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
No current flows, called reverse bias
Current flows, called forward bias
Some electrons flow to create opposite charges
The End