The Chemistry of Solids

The 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 We call this the 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 KNOW THESE!!!! Simple Cubic Unit Cell
Body-Centred Cubic Unit Cell How do we investigate solids? Face-Centred Cubic Unit Cell

Unit Cells in the Cubic Crystal System
Prentice-Hall © 2002 General Chemistry: Chapter 13

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
Copyright © Houghton Mifflin Company. All rights reserved. 10–14

Summary of Crystal Structures
15

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……….

First Layer

Next Layer The next spheres fit into a “dimple” formed by three spheres in the first layer.

Next Layer The next spheres fit into a “dimple” formed by three spheres in the first layer. There are two sets of dimples…...

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

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

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

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

Second Layer Inverted triangle dimples are not filled. Note that the second layer only occupies half the dimples in the first layer.

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

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

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

SPHERE DIRECTLY ABOVE SPHERE IN THE FIRST LAYER
PACKING SPHERES INTO LATTICES THIRD LAYER (option 1) 1 1 1 OPTION ONE! SPHERE DIRECTLY ABOVE SPHERE IN THE FIRST LAYER ADD SOME MORE……..

THIRD LAYER OPTION ONE! ADD SOME MORE…..

THIRD LAYER OPTION ONE! THE ABA ARRANGEMENT OF LAYERS. A B A LAYERS ONE AND THREE ARE THE SAME!

THE ABA ARRANGEMENT OF LAYERS. A B A CALLED HEXAGONAL CLOSEST PACKIN HCP

THE ABA ARRANGEMENT OF LAYERS, Option 1. A HEXAGONAL UNIT CELL. A B A HEXAGONAL CLOSEST PACKING

ABA ARRANGEMENT HAS A HEXAGONAL UNIT CELL.
HCP SUMMARY...

SUMMARY HEXAGONAL CLOSED PACKED STRUCTURE EXPANDED VIEW
NOW OPTION TWO…..

THIRD LAYER OPTION 2! 2 2 2 THIS DIMPLE DOES NOT LIE DIRECTLY OVER THE SPHERES OF THE FIRST LAYER. MORE

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…….

NOT THE SAME AS OPTION ONE! OPTION 2 THIRD LAYER A B C WE CALL THE THIRD LAYER C THIS TIME!

THIRD LAYER WE CALL THE THIRD LAYER C THIS TIME! OPTION 2 A B C THE ABC ARRANGEMENT OF LAYERS. NOW THE FOURTH LAYER…….

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

PHH p 509 FOURTH LAYER THE SAME AS FIRST. A B C A THE ABCA ARRANGEMENT……….. THIS IS CALLED….

THE ABCA ARRANGEMENT……….. A B C 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…..

COMPARISON NOTICE the flip…... CCP HCP NEAREST NEIGHBORS…..

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

COORDINATION NUMBER HCP

COORDINATION NUMBER HCP COORDINATION NUMBER =12

COORDINATION NUMBER HCP CCP COORDINATION NUMBER =12

SPHERES IN BOTH HCP AND CCP STRUCTURES
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?
1 Stacking a second close- packed layer of spheres directly atop a close-packed layer below Top Row Bottom Row 2 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 Which is the closest packed arrangement?
1 Stacking a second close- packed layer of spheres directly atop a close-packed layer below 2 Stacking a second close-packed layer of spheres in the depressions formed by spheres in the close-packed layer below. NEAREST NEIGHBOURS IN OTHER UNIT CELLS……..

Alloys An alloy is a blend of a host metal and one or more other elements which are added to change the properties of the host metal. Ores are naturally occurring compounds or mixtures of compounds from which elements can be extracted. Bronze, first used about 5500 years ago, is an example of a substitutional alloy, where tin atoms replace some of the copper atoms in the cubic array.

Substitutional Alloy Examples
Where a lattice atom is replaced by an atom of similar size Brass, one third of copper atoms are replaced by zinc atoms Sterling silver (93% Silver and 7%Cu) Pewter (85% Sn, 7% Cu, 6% Bi, and 2% Sb) Plumber’s solder (67% Pb and 33% Sn)

Bronze

Alloys Interstitial Alloy
When lattice holes (interstices) are filled with smaller atoms Steel best know interstitial alloy, contains carbon atoms in the holes of an iron crystal Carbon atoms change properties Carbon a very good covalent bonding atom changes the non-directional bonding of the iron, to have some direction Results in increased strength, harder, and less ductile The larger the percent of carbon the harder and stronger the steel Other metals can be used in addition to carbon, thus forming alloy steels

Carbon Steel Unlike bronze the carbon atoms fit into the holes formed by the stacking of the iron atoms. Alloys formed by using the holes are called interstital alloys.

Two Types of Alloys Substitutional Interstitial 10–62

About Holes in Cubic Arrays
Atomic Size Ratios and the Location of Atoms in Unit Cells Packing Type of Hole Radius Ratio hcp or ccp Tetrahedral Octahedral Simple Cubic Cubic pm

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 What about body centered cubic?????
How many nearest neighbors??? coordination number of 6 What about body centered cubic?????

COORDINATION NUMBER Lets look at this…….
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 Simple cubic CN = 6 Body Centered cubic CN = 8
In the three types of cubic unit cells: Simple cubic CN = 6 Body Centered cubic CN = 8 Face Centered cubic CN = ? Comes from ccp Just like hcp CN = 12 PACKING EFFICIENCY?

EFFICIENCY OF PACKING THE FRACTION OF THE VOLUME THAT IS ACTUALLY OCCUPIED BY SPHERES….. WHAT DOES THIS MEAN???

PACKING EFFICIENCY Vspheres= number of spheres x volume single sphere
FRACTION(F) OF THE VOLUME OCCUPIED BY THE SPHERES IN THE UNIT CELL. Vspheres= number of spheres x volume single sphere Vunit cell = a3 cubic unit cell of edge length a Lets get NUMBER OF SPHERES

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 FOR 1/4 ATOM ON A FACE. COUNTS FOR 1/2 ATOM ON A FACE. COUNTS AS 1/8 FOR ATOM ON A CORNER.

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) = = 4 QUESTION…..

QUESTION THE NUMBER OF SPHERES IN A BODY CENTRED CUBIC CELL IS? ANSWER….

QUESTION ANSWER…. Atoms = 8(1/8) + 1 = 2 VOLUME OCCUPIED IN FCC….
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 V = 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 = THERE ARE 4 SPHERES IN THE UNIT CELL total

radius of the sphere is r . Now we need the volume of the unit cell. Why????? 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!

Let side of cube be a a GET a in terms of r r a 2r r

r a 2r r Let side of cube be a a GET a in terms of r
FACE DIAGONAL = r + 2r + r=4r PYTHAGORAS! a2 + a2 = (4r)2 2a2 = 16r 2 a2 = 8r 2 r a 2r r

a a 2r r Side of cube be in terms of r
Now we can calculate the volume of the unit cell r a 2r NOW PUT IT ALL TOGETHER r

We conclude…..

In a cubic closest packed crystal 74% of the volume of a is taken up by spheres and 26% is taken up by empty space. QUESTION

Body Centered Cubic r 2r e (4r)2 = e2 + d2 16r2 = e2 + 2e2 r
d2 = e2 + e2 d2 = 2e2

CUBIC UNIT CELLS THE EDGE LENGTH IN TERMS OF r SIMPLE CUBIC
BODY CENTRED CUBIC FACE CENTRED CUBIC 2r NUMBER OF SPHERES 1 2 4 VOLUME OCCUPIED

CUBIC UNIT CELLS VOLUME OCCUPIED SIMPLE CUBIC BODY CENTRED CUBIC
FACE CENTRED CUBIC 52.4% 74.0% 68.0% 2r QUESTION... NUMBER OF SPHERES 1 2 4

QUESTION The fraction of space occupied in a hexagonal closest packed arrangement of spheres is the same as that in….. 1 simple cubic unit cell 2 face centered cubic unit cell 3 body centered cubic unit cell 4 none of these ANSWER…..

QUESTION The fraction of space occupied in a hexagonal closest packed arrangement of spheres is the same as that in….. 1 simple cubic unit cell 2 face centered cubic unit cell 3 body centered cubic unit cell 4 none of these Summary……...

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.
Mass is the mass of the number of atoms in the unit cell. Mass of one atom =atomic mass/6.022x1023 N0 = x 1023 atoms per mole Avogadro’s Number!

Volume of a copper unit cell
Cu crystalizes as a fcc r= 128pm = 1.28x10-10m = 1.28x10-8cm Volume of unit cell is given by:

COPPER DENSITY CALCULATION
mole Cu 4 atoms Cu unit cell 63.54 g Cu mole Cu unit cell 4.75X10-23 cm3 6.022 X atoms = 8.89 g/cm3 Laboratory measured density: 8.92 g/cm3

DETERMINATION OF ATOMIC RADIUS
At room temperature iron crystallizes with a bcc unit cell. X-ray diffraction shows that the length of an edge is 287 pm. What is the radius of the Fe atom? EDGE LENGTH (e)

AVOGADRO’S NUMBER Sample Problem: Calculate Avogadro’s number of iron if its unit cell length is 287 pm and it has a density of 7.86 g/cm3. 55.85 g Mole Fe Fe(s) is bcc Two atoms / unit cell

AVOGADRO’S NUMBER Sample Problem: Calculate Avogadro’s number of iron if its unit cell length is 287 pm and it has a density of 7.86 g/cm3. 55.85 g cm3 Mole Fe 7.86 g Fe(s) is bcc Two atoms / unit cell 97 97

AVOGADRO’S NUMBER The density of Fe(s) is 7.86 g/cm3.
length of an edge is 287 pm. V= e3 = (287pm)3 = 2.36x10-23cm3 pm3 55.85 g cm3 Mole Fe 7.86 g (10 -12)3 cm3 Fe(s) is bcc Two atoms / unit cell 98 98

AVOGADRO’S NUMBER The density of Fe(s) is 7.86 g/cm3.
length of an edge is 287 pm. V= e3 = (287pm)3 = 2.36x10-23cm3 pm3 55.85 g cm3 unit cell Mole Fe 7.86 g (10 -12)3 cm3 (287 pm)3 Fe(s) is bcc Two atoms / unit cell 99 99

AVOGADRO’S NUMBER 55.85 g cm3 Mole Fe 7.86 g pm3 unit cell 2 atoms
100 100

AVOGADRO’S NUMBER 55.85 g cm3 Mole Fe 7.86 g pm3 unit cell 2 atoms
= X atoms/mole 101 101

IONIC SOLIDS Binary Ionic Solids: Two types of ions Examples: NaCl MgO
CaCO3 MgSO4 Hard, brittle solids High melting point Electrical insulators except when molten or dissolved in water. These are lattices of ions…….

IONIC SOLIDS We notice that this is a cubic array of ions.
= Na+ = Cl– We notice that this is a cubic array of ions. Why do ionic solids hold together?????

General Chemistry: Chapter 13
Sodium Chloride Prentice-Hall © 2002 General Chemistry: Chapter 13

IONIC SOLIDS The stability of the ionic compound results from the electrostatic attractions between the ions: Li+ F– Li+ F– F– Li+ F– Li+ The LiF crystal consists of a lattice of ions. The attractions are stronger than the repulsions, so the crystal is stable. The stability is due to the LATTICE ENERGY How can we describe ionic lattices?

NaCl structure Cl- Na+ Lets take this apart…...

NaCl structure Cl- Na+ Lets look at the black dot lattice….

NaCl structure What unit cell do the black dots form?
The black dots form a fcc lattice! Now look at the red dots

NaCl structure What unit cell is this???? But which one?????
Cubic certainly Lets have another look……….

NaCl structure What unit cell is this???? Bring in a another array…..
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 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???

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????

HOLES IN A LATTICE. Holes??? What holes?????
Lets look at a fcc lattice!

HOLES IN A FCC LATTICE 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???? HOW MANY HOLES??????

HOLES IN A FCC LATTICE The holes: How many?? 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?

THE OCTAHEDRAL HOLES 1 atom 1/4 atom
If each one is occupied by an atom? How many atoms per unit cell? Number of atoms = x (1/4) = 4 There are 4 complete octahedral holes per fcc unit cell.

THE 13 OCTAHEDRAL HOLES 1 atom 1/4 atom The octahedral hole is..
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..

THE OCTAHEDRAL HOLES Between two layers….. 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

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

SMALL CUBE 8 CUBES Take a point at the centre of this cube
Another one of these….

SMALL CUBE 8 CUBES Take a point at the centre of this cube An so on ….

SMALL CUBE 8 CUBES This is a TETRAHEDRAL HOLE….

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…….

one sphere in another layer sitting in the dimple they form.
TETRAHEDRAL HOLES Formed by three spheres in one layer and one sphere in another layer sitting in the dimple they form. There is one more hole……….

TRIGONAL HOLES The smallest hole! Which hole????
Formed by three spheres in one layer. Formed from the space between three ions in a plane. Which hole????

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??????

HOLE OCCUPANTS? Which hole will a cation occupy??????
They occupy the holes that result in maximum attraction and minimum repulsion. To do this…...

Which hole will a cation occupy??????
M+ or M2+ cations always occupy the holes with the largest coordination number without rattling around! TIGHT FIT Consequently the radius of the cation must be greater than the size of the hole! This causes the X– anions to be pushed apart, which reduces the X– – X– repulsion. So we will investigate the size of these holes!

Which hole will a cation occupy??????
Investigate the size of these holes! The size of the hole depends upon the size of the ion (usually anion) that forms the lattice into which the cations are to go……... OCTAHEDRAL HOLE IN FCC…. LOOK AT HOLE….

OCTAHEDRAL HOLES IN FCC
Look at plane Draw a square. Put in spheres. These are the anions Fit a small sphere in

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……..

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

Look at plane Radius of ion = R Radius of hole = r = (2R + 2r)2 (2R)2 +(2R)2 8R2 = (2R + 2r)2 R R 2r R R 0.414R = 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 R 2r 0.414 x radius of anion(R) roctahedral hole = R What about the tetrahedral hole?

fcc RADIUS RATIO: What about other cubic cell systems??
Using similar calculations, we can find the radius of other types of holes as well: fcc rtetrahedral = R DO IT!!!!!!! roctahedral = 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??

SIMPLE CUBIC If the M+ cations (e.g. Cs+) are sufficiently large, they can no longer fit into octahedral holes of a fcc lattice. The next best closest packed X– array adopted by the anions is a simple cubic structure, giving cubic holes which are large enough to hold the cations. YOU can show that... rcubic = Ranion DO IT!!!!!!!

The cubic hole 8 4! 6! The coordination number in the cubic hole is ?
rcubic = Ranion In contrast for a fcc lattice…... The coordination number in the fcc tetrahedral hole is ? The coordination number in the fcc octahedral hole is ? 4! 6!

SUMMARY: Face centred cubic: Simple cubic: Trigonal hole
Too small to be occupied Tetrahedral hole CN = 4 rcation = 0.225Ranion 8 of these Octahedral hole CN = 6 rcation = 0.414Ranion 4 of these Simple cubic: Cubic hole CN = 8 rcation = 0.732Ranion 1 of these For a given anion rtrigonal < rtetrahedral < roctahedral < rcubic

SUMMARY Face centred cubic: Too small to be occupied Trigonal hole
Tetrahedral hole CN = 4 rcation = 0.225Ranion 8 of these Octahedral hole CN = 6 rcation = 0.414Ranion 4 of these Simple cubic CN = 8 rcation = 0.732Ranion 1 of these For a given anion rtrigonal < rtetrahedral < roctahedral < rcubic Which hole will a cation occupy??????

INTO WHICH HOLE WILL THE ION GO??
TETRAHEDRAL The hole filled is tetrahedral if: rtetrahedral < rcation < roctahedral 0.225Ranion < rcation < 0.414Ranion

OCTAHEDRAL The hole filled is octahedral if: roctahedral < rcation < rcubic 0.414Ranion < rcation < 0.732Ranion

CUBIC The hole filled is cubic if: rcubic < rcation 0.732Ranion < rcation Lets look at these ideas in action…….

NaCl Na+ has a radius of 98pm. Cl- has a radius of 181pm.
Consider a fcc array of Cl- then: Radius of the tetrahedral hole is x 181=41pm Radius of the octahedral hole is x 181=75pm Consider a sc array of Cl- then: Radius of the cubic hole is 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 Na+ has a radius of 98pm. Cl- has a radius of 181pm.
0.54 lies between and 0.732 so the sodium cations will occupy octahedral holes in a fcc (ccp) lattice Is the stoichiometry ok???

4 4 So stoichiometry is ok!! NaCl 1:1 stoichiometry is required
How many complete octahedral holes in face centred cubic array of Cl- ????? 4 4 How many Cl- needed to form the fcc array??? Therefore 4 Cl- and 4 Na+ So stoichiometry is ok!! Lets do another example…..

Example: Predict the structure of Li2S
Li+ is 68 pm S2- is 190pm STEP ONE: Examine the cation-anion radius ratios to find which type of holes the smaller ions fill Calculate ratio.. COMPARE with ratios…. Which is the best hole???? TETRAHEDRAL!!!!

Example: Predict the structure of Li2S
Li+ is 68 pm S2- is 190pm Calculate ratio.. This requires tetrahedral holes. Which lattice has tetrahedral holes??? face- centred cubic array Thus the S2- will form a fcc lattice ... Lets look at the structure…...

FCC unit cell with tetrahedral holes
ANION CATION Four anions in the unit cell. There are 8 tetrahedral holes. How many are occupied? STEP TWO: Determine what fraction of those holes must be filled to give the correct chemical formula So????

Four anions in the unit cell.
FCC unit cell with tetrahedral holes S2- Li+ Four anions in the unit cell. There are 8 tetrahedral holes. How many are occupied? Next Step…. Li2S needs two Li+ for each S2- Therefore all the tetrahedral holes are occupied!

FCC unit cell with tetrahedral holes filled
= 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. Which is…..

FCC unit cell with tetrahedral holes filled
= Li+ all the tetrahedral holes have to be occupied. Li2S is a face centered lattice of S2- with all of the tetrahedral holes filled by Li+ ions. Now do CsCl….

the cesium cations will occupy cubic holes of a simple cubic lattice.
Cs+ is 167 pm Cl- is 181pm Calculate ratio CsCl: Compare…... 0.92 is greater than 0.732 the cesium cations will occupy cubic holes of a simple cubic lattice. STOICHIOMETRY?????

Hence stoichiometry OK!
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

ZnS: Zn2+ is 64 pm S2- is 190 pm Calculate ratio
This requires tetrahedral holes. COMPARE The sulfide ions will form a face-centered cubic array because…. that is the only type to possess tetrahedral holes. What about stoichiometry??????

ZnS: Zn2+ is 64 pm S2- is 190 pm Calculate ratio
This requires tetrahedral holes. COMPARE The sulfide ions will form a face-centered cubic array because…. that is the only type to possess tetrahedral holes. What about stoichiometry?????? 160 160

We need an equal number of zinc and sulfide ions.
There are the twice as many tetrahedral holes(8) as S2-(4) that form the fcc lattice. Therefore, half the tetrahedral holes will be filled.

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.

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.

QUESTION A crystal has Mg2+ ions at the corners of a cubic unit cell, F- ions at the midpoints of all the edges and K+ ions at the body centre. The empirical formula is 1 KMgF3 2 K3MgF2 3 KMg2F2 4 K2Mg2F 5 K2Mg2F3

ANSWER A crystal has Mg2+ ions at the corners of a cubic unit cell, F- ions at the midpoints of all the edges and K+ ions at the body centre. The empirical formula is 1 KMgF3 2 K3MgF2 3 KMg2F2 4 K2Mg2F 5 K2Mg2F3

QUESTION A COMPOUND CONTAINS THE ELEMENTS Na,O and Cl. IT CRYSTALLISES IN A CUBIC LATTICE. THE O ATOMS ARE AT THE CORNERS, THE Cl ATOMS ARE AT THE CENTER AND THE Na ARE AT THE CENTRES OF THE FACES OF THE UNIT CELL. THE FORMULA OF THE COMPOUND IS... 1 Na2ClO 2 Na3ClO 3 NaCl3O 4 NaClO3 5 NaClO

QUESTION A COMPOUNDCONTAINS THE ELEMENTS Na,O and Cl. IT CRYSTALLISES IN A CUBIC LATTICE. THEO ATOMS ARE AT THE CORNERS, THE Cl ATOMS ARE AT THE CENTER AND THE Na ARE AT THE CENTRES OF THE FACES OF THE UNIT CELL. THE FORMULA OF THE COMPOUND IS... 1 Na2ClO 2 Na3ClO 3 NaCl3O 4 NaClO3 5 NaClO

This is the flourite structure: MX2
the anions occupy the tetrahedral holes(8) = Ca2+ in a fcc array of the cations(4). = F- Does this fit radius ratios???????

CaF2: Ca2+ is 99 pm F- is 136 pm Calculate ratio FOR TETRAHEDRAL HOLES OOPS!!!! The radius ratio is too BIG!!!! This shows Radius Ratios do not always work properly But CaF2 can be thought of as a simple cubic of F- with Ca2+ at alternate cubic holes!!!!!!!

Ca2+ in alternating cubic sites.
CaF2: SIMPLE CUBIC Ca2+ Ca2+ Ca2+ in alternating cubic sites. Alternative description What is Antiflourite????

This is the flourite structure: MX2
the anions occupy the tetrahedral holes(8) = Ca2+ = F- in a fcc array of the cations(4). The antifluorite structure M2X (eg K2O) the cations occupy the tetrahedral holes and the anions form the fcc array.

Ionic Compound Density
MgO fcc of O2- Mg2+ in octahedral holes Calculating the density of an ionic compound A face…. Edge= roxide ion+ 2rMg ion+ roxide ion Now calculate volume 4 Mg’s and 4 O2- REMEMBER….

MgO fcc of O2- Mg2+ in octahedral holes Calculating the density of an ionic compound R = 86 pm r = 126 pm Edge = 424 pm V = (424)3 = 7.62X107 pm3 A face…. Edge= roxide ion+ 2rMg ion+ roxide ion Now calculate volume 4 Mg’s and 4 O2- REMEMBER…. 173 173

MgO fcc of O2- Mg2+ in octahedral holes Calculating the density of an ionic compound R = 86 pm r = 126 pm Edge = 424 pm V = (424)3 = 7.62X107 pm3 A face…. Edge= roxide ion+ 2rMg ion+ roxide ion Now calculate volume 4 Mg’s and 4 O2- REMEMBER…. 174 174

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

For example calculate the density of MgO a fcc crystal with 762 pm for the length of it’s unit cell. 40.61g MgO mole mole 6.022 X 1023 FU

For example calculate the density of MgO a fcc crystal with 762 pm for the length of it’s unit cell. 40.61g MgO mole 4 FU mole 6.022 X 1023 FU unit cell

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 4 FU Unit Cell mole 6.022 X 1023 FU unit cell 7.62X107 pm3

For example calculate the density of MgO a fcc crystal with 762 pm for the length of it’s unit cell. 40.61g MgO mole 4 FU Unit Cell pm3 mole 6.022 X 1023 FU unit cell 7.62X107 pm3 10-36m3

For example calculate the density of MgO a fcc crystal with 762 pm for the length of it’s unit cell. 40.61g MgO mole 4 FU Unit Cell pm3 10-6m mole 6.022 X 1023 FU unit cell 7.62X107 pm3 10-36m3 cm3

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 4 FU Unit Cell pm3 10-6m mole 6.022 X 1023 FU unit cell 7.62X107 pm3 10-36m3 cm3 = 3.54 g/cm3

Diamond and Graphite Covalently Networked Crystalline Solids

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.

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???

X-ray Diffraction X-ray diffraction (ERD) is a technique for determining the arrangement of atoms or ions in a crystal by analyzing the pattern that results when X-rays are scattered after bombarding the crystal. The Bragg equation relates the angle of diffraction (2) of X-rays to the spacing (d) between the layers of ions or atoms in a crystal: n2dsin.

Cell Structure by X-ray Diffraction

BRAGG DIFFRACTION LAW W.H. Bragg and W.L.Bragg noticed that
q 2 q This is reminiscent of reflection….. So they formulated diffraction in terms of reflection from planes of electron density in the crystal..

BRAGG’S LAW

BRAGG DIFFRACTION LAW A plane of lattice points…….
Now imagine reflection of X-rays……….

Bragg Equation Derivation
ө ө ө d x x 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 = 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……..

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

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

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

No current flows, called reverse bias
Semi Conductors Some electrons flow to create opposite charges No current flows, called reverse bias Current flows, called forward bias

The End

The Chemistry of Solids

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