Chapter #11 Chemistry of Solids

Transcript Chapter #11 Chemistry of Solids

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.

LATTICE EXAMPLE

We can pick out the smallest repeating unit…..

UNIT CELL

We can pick out the smallest repeating unit…..

We call this the UNIT CELL………..

UNIT CELL

The 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…..

SIMPLE CUBIC

Eight equivalent points at the corners of a cube We can imagine an equivalent point at the centre of the spheres

BODY CENTRED CUBIC

Eight equivalent points at the corners of a cube and one at the centre Another possibility……...

FACE CENTRED CUBIC

Eight equivalent points at the corners of a cube and six on the centre of the cube faces Summary……..

THE CUBIC UNIT CELLS

Simple Cubic Unit Cell KNOW THESE!!!!

Body-Centred Cubic Unit Cell

Face-Centred Cubic Unit Cell How do we investigate solids?

Unit Cells in the Cubic Crystal System

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

Electron Sea Model of Metals

10–14

Summary of Crystal Structures

METALS

VIEWED AS CLOSELY PACKED SPHERES HOW CAN WE PACK SPHERES?????

PACKING OF SPHERICAL VEGETABLES

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…….

Packing Spheres into Lattices:

First Layer Lets put in a few more spheres……….

Packing Spheres into Lattices

First Layer

Packing Spheres into Lattices

Next Layer The next spheres fit into

a “dimple” formed by three spheres in the first layer.

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…...

Packing Spheres into Lattices:

Next Layer The next spheres fit into Triangle NOTE: the inverted not triangle inverted The two types of “dimples” formed by three spheres in the first layer.

The second layer…..

Packing Spheres into Lattices:

is formed by choosing one of the sets of dimples Now put on second layer…...

Packing Spheres into Lattices:

Second Layer

Once one is put on the others are forced into half of the dimples of the same type….

Packing Spheres into Lattices

Second Layer

Once one is put on the others are forced into half of the dimples of the same type….

Packing Spheres into Lattices:

Second Layer

Once one is put on the others are forced into half of the dimples of the same type….

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.

Packing Spheres into Lattices

Second Layer Occupied dimple Unoccupied DIMPLE Note that the second layer only occupies half the dimples in the first layer. THE THIRD LAYER…...

PACKING SPHERES INTO LATTICES

SECOND LAYER

HAVE TO CHOOSE A DIMPLE

PACKING SPHERES INTO LATTICES

THIRD LAYER, Choose a dimple 1 1 1 (1) A DIMPLE DIRECTLY ABOVE SPHERE IN THE FIRST LAYER

THIS OR……..

PACKING SPHERES INTO LATTICES

THIRD LAYER NOTE: the inverted triangle 2 1 2 2 1 1 (2) A DIMPLE DIRECTLY ABOVE A DIMPLE IN THE FIRST LAYER……..

CHOOSE OPTION 1…..

PACKING SPHERES INTO LATTICES

THIRD LAYER (option 1) 1 1 1 OPTION ONE!

SPHERE DIRECTLY ABOVE SPHERE IN THE FIRST LAYER ADD SOME MORE……..

PACKING SPHERES INTO LATTICES

THIRD LAYER OPTION ONE!

ADD SOME MORE…..

PACKING SPHERES INTO LATTICES

THIRD LAYER OPTION ONE!

THE ABA ARRANGEMENT OF LAYERS.

A B A

LAYERS ONE AND THREE ARE THE SAME!

PACKING SPHERES INTO LATTICES

THE ABA ARRANGEMENT OF LAYERS.

A B A

CALLED HEXAGONAL CLOSEST PACKIN HCP

PACKING SPHERES INTO LATTICES

THE ABA ARRANGEMENT OF LAYERS, Option 1 .

A HEXAGONAL UNIT CELL.

A B A

HEXAGONAL CLOSEST PACKING

HEXAGONAL UNIT CELL

ABA ARRANGEMENT HAS A HEXAGONAL UNIT CELL.

HCP SUMMARY...

SUMMARY

HEXAGONAL CLOSED PACKED STRUCTURE

EXPANDED VIEW NOW OPTION TWO…..

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

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…….

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!

PACKING SPHERES INTO LATTICES

THIRD LAYER WE CALL THE THIRD LAYER C THIS TIME!

OPTION 2 A B C THE ABC ARRANGEMENT OF LAYERS.

NOW THE FOURTH LAYER…….

PACKING SPHERES INTO LATTICES

FOURTH LAYER PUT SPHERE IN SO THAT A B C FOURTH LAYER THE SAME AS FIRST.

FOURTH LAYER THE SAME AS FIRST.

A C B A

THE ABCA ARRANGEMENT………..

THIS IS CALLED….

PACKING SPHERES INTO LATTICES

THE ABCA ARRANGEMENT………..

A C B A CUBIC CLOSED PACKED….

WHY???

UNIT CELL OF CCP

CUBIC UNIT CELLL

THIS ABCA ARRANGEMENT HAS A FACE- CENTRED CUBIC UNIT CELL (FCC) A COMPARISON…..

HCP

COMPARISON

NOTICE the flip…...

CCP

NEAREST NEIGHBORS…..

COORDINATION NUMBER

The number of nearest neighbors that a lattice point has in a crystalline solid Lets look at hcp and ccp…...

HCP COORDINATION NUMBER

COORDINATION NUMBER

HCP

COORDINATION NUMBER =12

COORDINATION NUMBER

HCP CCP

COORDINATION NUMBER =12

COORDINATION NUMBER

HCP CCP

COORDINATION NUMBER =12

COORDINATION NUMBER

SPHERES IN BOTH HCP AND CCP STRUCTURES EACH HAVE A COORDINATION NUMBER OF 12.

QUESTION ……..

REVIEW QUESTION

Which is the closest packed arrangement?

2 1 Stacking a second close packed layer of spheres directly atop a close-packed layer below Top Row Bottom Row Stacking a second close-packed layer of spheres in the depressions formed by spheres in the close-packed layer below.

Top Row Bottom Row ANSWER……..

REVIEW QUESTION

2 1 Which is the closest packed arrangement?

Stacking a second close packed layer of spheres directly atop a close-packed layer below 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 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

10–62

Interstitial

About Holes in Cubic Arrays

Atomic Size Ratios and the Location of Atoms in Unit Cells Packing hcp or ccp Type of Hole Tetrahedral Radius Ratio pm 0.22 - 0.41

hcp or ccp Simple Cubic Octahedral Cubic 0.41 - 0.73

0.73 - 1.00

COORDINATION NUMBER

SIMPLE CUBIC

How do we count nearest neighbors?

Draw a few more unit cells…...

COORDINATION NUMBER

SIMPLE CUBIC

Highlight the nearest neighbors….

COORDINATION NUMBER

SIMPLE CUBIC How many nearest neighbors???

coordination number of 6 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…….

Body-centered cubic packing (bcc)

COORDINATION NUMBER?????

In bcc lattices, each sphere has a coordination number of 8 What about face centered cubic?

COORDINATION NUMBER

In the three types of cubic unit cells:

Simple cubic CN = 6 Body Centered cubic CN = 8 Face Centered cubic Comes from ccp CN = ?

Just like hcp CN = 12 PACKING EFFICIENCY?

EFFICIENCY OF PACKING

THE FRACTION OF THE VOLUME THAT IS ACTUALLY OCCUPIED BY SPHERES…..

f v



volume occupied by the spheres in the unit cell volume of the unit cell WHAT DOES THIS MEAN???

PACKING EFFICIENCY

FRACTION(F) OF THE VOLUME OCCUPIED BY THE SPHERES IN THE UNIT CELL.

f v



volume occupied by the spheres in the unit cell volume of the unit cell f v



V spheres V unitcell

spheres = number of spheres x volume single sphere V unit cell = a 3 cubic unit cell of edge length Lets get NUMBER OF SPHERES a

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 IN THE LATTICE COUNTS 1 FOR ATOM IN CELL COUNTS AS 1/8 FOR ATOM ON A CORNER.

COUNTS FOR 1/4 ATOM ON A FACE.

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…..

QUESTION

THE NUMBER OF SPHERES IN A BODY CENTRED CUBIC CELL IS?

ANSWER….

QUESTION

THE NUMBER OF SPHERES IN A BODY CENTRED CUBIC CELL IS?

ANSWER….

Atoms = 8(1/8) + 1 = 2

VOLUME OCCUPIED IN FCC….

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 =

4 ( 4 3



3 ) THERE ARE 4 SPHERES IN THE UNIT CELL

FACE-CENTRED CUBIC UNIT CELL

V spheres

4 ( 4 3



3 ) radius of the sphere is r . Now we need the volume of the unit cell.

Why?????

f v



V spheres V unitcell GET DIMENSIONS OF CUBE IN TERMS OF r…..

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!!

a Let side of cube be a DRAWING CUBE FACE REMEMBER THE SPHERES TOUCH!!

Draw a square…..

Now we need to get a in terms of r

a Let side of cube be a CONSTRUCT A TRIANGLE ON THE FACE Why????

So we can use Pythagoras!

a Let side of cube be a GET a in terms of r

r a r 2r

r Let side of cube be a a GET a in terms of r FACE DIAGONAL = r + 2r + r=4r a r PYTHAGORAS !

2 + a 2 = (4r) 2 2a 2 = 16r 2 2r



= 8r

r a Side of cube be in terms of r a



Now we can calculate the volume of the unit cell r 2r

V cell V cell

 

r a

8 3 3 NOW PUT IT ALL TOGETHER

FACE-CENTRED CUBIC UNIT CELL

f v



f v



* 4 3



3 3 V spheres V unitcell

f v



4 * 4 3

 



f v



0 .

740 We conclude…..

FACE-CENTRED CUBIC UNIT CELL f v



V spheres V unitcell

f v



In a cubic closest packed crystal 0 .

740 74% of the volume of a is taken up by spheres and 26% is taken up by empty space.

QUESTION

Body Centered Cubic

d 2 d 2 = e 2 = 2e 2 + e 2 e (4r) 2 16r 2 r 2 = e 2 = e 2 + d 2 + 2e 2 = 3e 2 /16 e = 4r/√3

CUBIC UNIT CELLS

THE EDGE LENGTH IN TERMS OF r SIMPLE CUBIC BODY CENTRED CUBIC FACE CENTRED CUBIC 2r NUMBER OF SPHERES 1

4r 3

2 VOLUME OCCUPIED 4 8

SIMPLE CUBIC CUBIC UNIT CELLS VOLUME OCCUPIED BODY CENTRED CUBIC FACE CENTRED CUBIC 52.4% 2r NUMBER OF SPHERES 1

68.0%

4r 3

QUESTION...

8 4 74.0%

QUESTION

The fraction of space occupied in a hexagonal closest packed arrangement of spheres is the same as that in…..

1 2 3 4 simple cubic unit cell face centered cubic unit cell body centered cubic unit cell none of these ANSWER…..

QUESTION

The fraction of space occupied in a hexagonal closest packed arrangement of spheres is the same as that in…..

1 2 3 4 simple cubic unit cell face centered cubic unit cell body centered cubic unit cell none of these Summary……...

SUMMARY

sc: 52.4% of space occupied by spheres bcc: 68.0% of space occupied by spheres fcc: 74.0% of space occupied by spheres hcp: 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????????

DENSITY

We can calculate the density in a unit cell.

Density



mass volume

Mass is the mass of the number of atoms in the unit cell.

Mass of one atom =atomic mass/6.022x10

23 N 0 = 6.022 x 10 23 atoms per mole Avogadro’s Number!

Volume of a copper unit cell

Cu crystalizes as a fcc r= 128pm = 1.28x10

-10 m = 1.28x10

-8 cm

Volume of unit cell is given by:

V cell

  

V cell



( 8



1 .





) 3

V cell



4 .





COPPER DENSITY CALCULATION

63.54 g Cu mole Cu mole Cu 6.022 X 10 23 atoms 4 atoms Cu unit cell unit cell 4.75X10

-23 cm 3 = 8.89 g/cm 3 Laboratory measured density: 8.92 g/cm 3

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)



4r 3





287



124

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/cm 3 .

55.85 g

Mole Fe

Fe(s) is bcc Two atoms / unit cell

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/cm 3 .

55.85 g cm 3

Mole Fe 7.86 g

Fe(s) is bcc Two atoms / unit cell

AVOGADRO’S NUMBER

The density of Fe(s) is 7.86 g/cm 3 .

length of an edge is 287 pm.

V= e 3 = (287pm) 3 = 2.36x10

-23 cm 3 55.85 g cm 3

Mole Fe 7.86 g pm 3

(10 -12 ) 3 cm 3 Fe(s) is bcc Two atoms / unit cell

AVOGADRO’S NUMBER

The density of Fe(s) is 7.86 g/cm 3 .

length of an edge is 287 pm.

V= e 3 = (287pm) 3 = 2.36x10

-23 cm 3 55.85 g cm 3

Mole Fe 7.86 g pm 3

(10 -12 ) 3 cm 3

unit cell (287 pm) 3

Fe(s) is bcc Two atoms / unit cell

AVOGADRO’S NUMBER

55.85 g cm 3

Mole Fe 7.86 g pm 3

(10 -12 ) 3 cm 3

unit cell (287 pm) 3 2 atoms unit cell 100

AVOGADRO’S NUMBER

55.85 g cm 3

Mole Fe 7.86 g pm 3

(10 -12 ) 3 cm 3

unit cell (287 pm) 3 2 atoms unit cell = 6.022 X 10 -23 atoms/mole 101

IONIC SOLIDS

Binary Ionic Solids: Two types of ions

Examples:

NaCl MgO CaCO 3 MgSO 4 Hard, brittle solids High melting point Electrical insulators except when molten or dissolved in water.

These are lattices of ions…….

102

= Na + = Cl –

IONIC SOLIDS

We notice that this is a cubic array of ions.

Why do ionic solids hold together?????

103

Sodium Chloride

IONIC SOLIDS

The stability of the ionic compound results from the electrostatic attractions between the ions: Li + F – The LiF crystal consists of a Li + lattice of ions.

F – Li + The attractions are stronger than the repulsions , so the Li + F – crystal is stable.

F –

The stability is due to the LATTICE ENERGY

Li + F – Li + F – F – Li + F – Li + How can we describe ionic lattices?

105

NaCl structure

Cl Na +

106

Lets take this apart…...

NaCl structure

Cl Na +

107

Lets look at the black dot lattice….

NaCl structure

What unit cell do the black dots form?

108

The black dots form a fcc lattice!

Now look at the red dots

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…..

110

The red dots form a fcc array!

Now put these back together…..

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

111

ARRAY

NaCl structure

This is the NaCl structure.

Cl Na + 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

114

HOLES IN A LATTICE.

Holes???

What holes?????

Lets look at a fcc lattice!

HOLES IN A FCC LATTICE

115

The black dots form a FCC lattice!

See the holes????

HOLES IN A FCC LATTICE

The black dots form a fcc lattice!

See the holes????

116

HOW MANY HOLES??????

HOLES IN A FCC LATTICE

The holes: How many??

117

THIRTEEN: ONE IN THE CENTRE 12 on the edges.

What shape is the hole ?

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

1 atom

THE OCTAHEDRAL HOLES

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

1 atom

THE 13 OCTAHEDRAL HOLES

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????

The octahedral hole is..

120

THE OCTAHEDRAL HOLES

Between two layers…..

121

Other holes…..

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

123

Take a point at the centre of this cube There are eight of these….

SMALL CUBE

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

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 This hole…….

127

TETRAHEDRAL HOLES

Formed by three spheres in one layer and one sphere in another layer sitting in the dimple they form.

128

There is one more hole……….

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??????

+ or M 2+ 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….

133

LOOK AT HOLE….

134

OCTAHEDRAL HOLES IN FCC

Look at plane Draw a square.

Put in spheres.

These are the anions Fit a small sphere in

135

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 Put in distances……..

136

OCTAHEDRAL HOLES IN FCC

2r Look at plane Radius of ion = R Radius of hole = r R

137

OCTAHEDRAL HOLES IN FCC

R R R R 2r Look at plane Radius of ion = R Radius of hole = r ( 2R) 2 +(2R) 2 = (2R + 2r) 2 2



8R 2 = (2R + 2r) 2 2

( 2 2

 

2 1





R R

2 )

 

R r



0.414R = r

OCTAHEDRAL HOLES IN FCC

138 R R 2r R R

Look at plane Radius of ion = R Radius of hole = r 0.414R = r The size of cation that just fits has a radius that is 0.414 x radius of anion(R)

octahedral hole = 0.414 R What about the tetrahedral hole?

Using similar calculations, we can find the radius of other types of holes as well: DO IT!!!!!!!

fcc

tetrahedral = 0.225 R

octahedral = 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...

cubic = 0.732 R

anion

DO IT!!!!!!!

140

The cubic hole

The coordination number in the cubic hole is ?

cubic = 0.732 R

anion

In contrast for a fcc lattice…...

The coordination number in the fcc tetrahedral hole is ?

141

4 !

The coordination number in the fcc octahedral hole is ?

Face centred cubic:

SUMMARY

: Trigonal hole Too small to be occupied Tetrahedral hole CN = 4 r cation = 0.225R

anion

Octahedral hole CN = 6

cation = 0.414R

anion

Simple cubic: 8 of these 4 of these Cubic hole CN = 8 For a given anion

cation = 0.732R

anion

1 of these r

trigonal

< r

tetrahedral < r octahedral

< r

cubic

142

SUMMARY

Face centred cubic: Trigonal hole Tetrahedral hole Octahedral hole Simple cubic For a given anion Too small to be occupied CN = 4 CN = 6 CN = 8

cation = 0.225R

anion r

cation = 0.414R

anion r

cation = 0.732R

anion

trigonal

< r

tetrahedral < r octahedral

< r

cubic

Which hole will a cation occupy??????

143

8 of these 4 of these 1 of these

INTO WHICH HOLE WILL THE ION GO??

TETRAHEDRAL The hole filled is tetrahedral if: r

tetrahedral < r

cation < r

octahedral

0.225R

anion < r

cation < 0.414R

anion

144

INTO WHICH HOLE WILL THE ION GO??

145

OCTAHEDRAL The hole filled is octahedral if: r

octahedral < r

cation < r

cubic

0.414R

anion < r

cation < 0.732R

anion

INTO WHICH HOLE WILL THE ION GO??

The hole filled is cubic if:

r cubic < r

cation 0.732R

anion < r

cation Lets look at these ideas in action…….

146

CUBIC

NaCl Na + Cl has a radius of 98pm 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

r cation r anion



r Na



r Cl



Na + Cl has a radius of 98pm.

has a radius of 181pm.



181



0 .

tet r cation tet r anion



0 .

225

oct r cation oct r anion



0 .

414

cubic r cation cubic r anion



0 .

732 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???

4 Therefore 4 Cl and 4 Na + So stoichiometry is ok!!

Lets do another example…..

149

Example: Predict the structure of Li 2 S STEP ONE: Li + is 68 pm S 2 is 190pm Examine the cation-anion radius ratios to find which type of holes the smaller ions fill Calculate ratio..

r cation r anion



r r Li



 

190



0 .

36 COMPARE with ratios….

tet r cation tet r anion



0 .

225 Which is the best hole????

150

oct r cation oct r anion



0 .

414 TETRAHEDRAL!!!!

Example: Predict the structure of Li 2 S Li + is 68 pm S 2 is 190pm

r cation r anion



r Li r S

  

190



0 .

36 Calculate ratio..

This requires tetrahedral holes.

Which lattice has tetrahedral holes???

face- centred cubic array Thus the S 2 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

152

So????

FCC unit cell with tetrahedral holes S 2 Li + Four anions in the unit cell.

There are 8 tetrahedral holes.

How many are occupied?

Li 2 S needs two Li + for each S 2 Next Step….

Therefore all the tetrahedral holes are occupied!

153

FCC unit cell with tetrahedral holes filled = S 2– = 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 = S 2– = Li + all the tetrahedral holes have to be occupied.

Li 2 S is a face centered lattice of S 2 with all of the tetrahedral holes filled by Li + ions.

Now do CsCl….

155

CsCl: Cs + is 167 pm Cl is 181pm

r cation r anion



r Cs



r Cl

 

167

181



0 .

92 Calculate ratio Compare…...

cubic r cation cubic r anion



0 .

732 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

ZnS: Zn 2+ is 64 pm S 2 is 190 pm Calculate ratio

r cation r anion



r Zn



r S

 

190



0 .

35 This requires tetrahedral holes.

COMPARE

tet r cation tet r anion



0 .

225

oct r cation oct r anion



0 .

414 The sulfide ions will form a face-centered cubic array because….

that is the only type to possess tetrahedral holes.

What about stoichiometry??????

159

ZnS: Zn 2+ is 64 pm S 2 is 190 pm Calculate ratio

r cation r anion



r Zn



r S

 

190



0 .

35 This requires tetrahedral holes.

COMPARE

tet r cation tet r anion



0 .

225

oct r cation oct r anion



0 .

414 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 S 2 (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 Mg 2+ the edges and K + ions at the corners of a cubic unit cell, F ions at the midpoints of all ions at the body centre. The empirical formula is 1 2 3 KMgF 3 K 3 MgF 2 KMg 2 F 2 4 5 K 2 Mg 2 F K 2 Mg 2 F 3 164

A crystal has Mg 2+

ANSWER

ions at the corners of a cubic unit cell, F at the midpoints of all the edges and K + ions ions at the body centre. The empirical formula is 1 KMgF 3 2 3 K 3 MgF 2 KMg 2 F 2 4 K 2 Mg 2 F 5 K 2 Mg 2 F 3 165

1 4 5 2 3

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...

Na 2 ClO Na 3 ClO NaCl 3 O NaClO 3 NaClO 166

2 3 4 1 5

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... Na 2 ClO Na 3 ClO NaCl 3 O NaClO 3 NaClO 167

This is the flourite structure: MX 2 the anions occupy the tetrahedral holes(8) in a fcc array of the cations(4).

= Ca 2+ = F -

168

Does this fit radius ratios???????

CaF 2 : Ca 2+ is 99 pm F is 136 pm

r cation r anion



r Ca



r F

 

136



0 .

727 FOR TETRAHEDRAL HOLES Calculate ratio

tet r cation tet r anion



0 .

225 OOPS !!!!

The radius ratio is too BIG!!!!

This shows Radius Ratios do not always work properly But CaF 2 can be thought of as a simple cubic of F with Ca 2+ at alternate cubic holes!!!!!!!

169

Ca 2+ SIMPLE CUBIC CaF 2 : Ca 2+

170

Ca 2+ in alternating cubic sites.

Alternative description What is Antiflourite????

This is the flourite structure: MX 2 the anions occupy the tetrahedral holes(8)

= Ca 2+ = F -

in a fcc array of the cations(4).

The antifluorite structure M 2 X (eg K 2 O) the cations occupy the tetrahedral holes

171

and the anions form the fcc array.

Ionic Compound Density

MgO fcc of O 2 Mg 2+ in octahedral holes Calculating the density of an ionic compound A face….

Edge= r oxide ion + 2r Mg ion + r oxide ion Now calculate volume 4 Mg’s and 4 O 2 REMEMBER….

172

Ionic Compound Density

MgO fcc of O 2 Mg 2+ in octahedral holes Calculating the density of an ionic compound R = 86 pm A face….

r = 126 pm Edge = 424 pm V = (424) 3 = 7.62X10

7 pm 3 Edge= r oxide ion + 2r Mg ion + r oxide ion Now calculate volume 4 Mg’s and 4 O 2 REMEMBER….

173

Ionic Compound Density

MgO fcc of O 2 Mg 2+ in octahedral holes Calculating the density of an ionic compound R = 86 pm A face….

r = 126 pm Edge = 424 pm V = (424) 3 = 7.62X10

7 pm 3 Edge= r oxide ion + 2r Mg ion + r oxide ion Now calculate volume 4 Mg’s and 4 O 2 REMEMBER….

174

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 10 23 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 mole 4 FU 6.022 X 10 23 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 mole 4 FU Unit Cell 6.022 X 10 23 FU unit cell 7.62X10

7 pm 3

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 4 FU Unit Cell 6.022 X 10 23 FU unit cell 7.62X10

7 pm 3 pm 3 10 -36 m 3

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 4 FU Unit Cell 6.022 X 10 23 FU unit cell 7.62X10

7 pm 3 pm 3 10 -6 10 -36 m 3 cm 3 m

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 mole 4 FU Unit Cell 6.022 X 10 23 FU unit cell 7.62X10

7 pm 3 pm 3 10 -6 10 -36 m 3 cm 3 m = 3.54 g/cm 3

Diamond and Graphite

Covalently Networked Crystalline Solids

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:



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…….

191

Now imagine reflection of X-rays……….

Bragg Equation Derivation

ө ө ө

d sin ө = x/d x = d sin ө Wave length λ = 2x λ = 2d sin ө

λ = 2d sin ө

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

λ = 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

10–199

Molecular Orbital Energies

Insulator (diamond) Conductor(metal)

10–200

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

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