Chapter 11 The Chemistry of Solids 2 SOLIDS Solids are either amorphous or crystalline. CRYSTALLINE...

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


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


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



Second Layer

27



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


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…..


35

THE ABA ARRANGEMENT OF LAYERS.A

B

A

OPTION ONE!

LAYERS ONE AND THREE ARE THE SAME!


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.


MORE

41

GREEN SPHERES DO NOT LIE DIRECTLY OVER THE SPHERES OF THE FIRST LAYER.

THE THIRD LAYER IS DIFFERENT FROM THE FIRST…….


OPTION 2

42

A

B

C

NOT THE SAME AS OPTION ONE!

WE CALL THE THIRD LAYER C THIS TIME!


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


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.


A

THIS IS CALLED….

46

A

B

C

THE ABCA ARRANGEMENT………..


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


53

HCP CCP


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?????


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


a DRAWING CUBE FACE

REMEMBER THE SPHERES TOUCH!!Draw a square…..

Now we need to get a in terms of r

80


aCONSTRUCT A TRIANGLE ON THE FACE

Why????

So we can use Pythagoras!

81


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


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


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.


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


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


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??????


117


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


Another one of these….

8 CUBES

SMALL CUBE

125


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!


TIGHT FIT

So we will investigate the size of these holes!

133

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….

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……..


136

2r

Radius of ion = RLook at plane

Radius of hole = r


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


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?


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


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


OCTAHEDRALThe hole filled is octahedral if:

0.414Ranion < rcation < 0.732Ranion

roctahedral < rcation < rcubic

146


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


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


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



A face….



4 Mg’s and 4 O2-

REMEMBER….


R = 86 pm

r = 126 pm

Edge = 424 pm

V = (424)3 = 7.62X107 pm3

174



A face….



4 Mg’s and 4 O2-

REMEMBER….


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


40.61g MgOmole

mole

6.022 X 1023 FU


40.61g MgOmole

4 FUmole

6.022 X 1023 FU unit cell


40.61g MgOmole

4 FUmole

6.022 X 1023 FU unit cell7.62X107 pm3

Unit Cell


40.61g MgOmole

4 FUmole


pm3

7.62X107 pm3

Unit Cell

10-36m3


40.61g MgOmole

4 FUmole


pm3

cm37.62X107 pm3

Unit Cell

10-36m3

10-6m


40.61g MgOmole

4 FUmole


pm3

cm37.62X107 pm3

Unit Cell

10-36m3

10-6m

= 3.54 g/cm3


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


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


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

Chapter 11 The Chemistry of Solids 2 SOLIDS Solids are either amorphous or crystalline. CRYSTALLINE...

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Transcript of Chapter 11 The Chemistry of Solids 2 SOLIDS Solids are either amorphous or crystalline. CRYSTALLINE...