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Chapter 11 The Chemistry of Solids

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2 SOLIDS Solids are either amorphous or crystalline. CRYSTALLINE SOLIDS: AMORPHOUS SOLIDS: Considerable disorder in structure. Example: rubber, glass 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.

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3 We can pick out the smallest repeating unit….. LATTICE EXAMPLE

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4 We can pick out the smallest repeating unit….. We call this the UNIT CELL……….. UNIT CELL

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5 We call this the UNIT CELL……….. The unit cell drawn here is a simple cubic cell UNIT CELL

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Examples of Unit Cells

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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…..

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

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

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

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11 Simple Cubic Unit Cell Body-Centred Cubic Unit Cell Face-Centred Cubic Unit Cell KNOW THESE!!!! How do we investigate solids? THE CUBIC UNIT CELLS

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Prentice-Hall © 2002General Chemistry: Chapter 13Slide 12 of 35 Unit Cells in the Cubic Crystal System

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13 Good conductors of heat and electricity METALS e.g. copper, gold, steel, sodium, brass. Shiny, ductile and malleable Melting points: soft (Na) or hard (W) low (Hg at -39°C) or high (W at 3370°C) Can be METALS ARE CRYSTALLINE SOLIDS

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Copyright © Houghton Mifflin Company. All rights reserved. 10–14 Electron Sea Model of Metals

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Summary of Crystal Structures

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

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17 PACKING OF SPHERICAL VEGETABLES

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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

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19 Packing Spheres into Lattices: First Layer Lets put in a few more spheres……….

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20 Packing Spheres into Lattices First Layer

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

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22 Packing Spheres into Lattices: Next LayerThe next spheres fit into a “dimple” formed by three spheres in the first layer. There are two sets of dimples…...

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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

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24 Packing Spheres into Lattices: is formed by choosing one of the sets of dimples Now put on second layer…...

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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….

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

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

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28 Packing Spheres into Lattices Second Layer Note that the second layer only occupies half the dimples in the first layer. Inverted triangle dimples are not filled.

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

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30 PACKING SPHERES INTO LATTICES SECOND LAYER HAVE TO CHOOSE A DIMPLE

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

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

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

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34 OPTION ONE! ADD SOME MORE….. PACKING SPHERES INTO LATTICES THIRD LAYER

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35 THE ABA ARRANGEMENT OF LAYERS. A B A OPTION ONE! LAYERS ONE AND THREE ARE THE SAME! PACKING SPHERES INTO LATTICES THIRD LAYER

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36 PACKING SPHERES INTO LATTICES THE ABA ARRANGEMENT OF LAYERS. A B A CALLED HEXAGONAL CLOSEST PACKIN HCP

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

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

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

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

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41 GREEN SPHERES DO NOT LIE DIRECTLY OVER THE SPHERES OF THE FIRST LAYER. THE THIRD LAYER IS DIFFERENT FROM THE FIRST……. PACKING SPHERES INTO LATTICES THIRD LAYER OPTION 2

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42 A B C NOT THE SAME AS OPTION ONE! WE CALL THE THIRD LAYER C THIS TIME! PACKING SPHERES INTO LATTICES THIRD LAYER OPTION 2

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

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44 A B C FOURTH LAYER THE SAME AS FIRST. PACKING SPHERES INTO LATTICES FOURTH LAYER PUT SPHERE IN SO THAT

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45 A B C THE ABCA ARRANGEMENT……….. PHH p 509 FOURTH LAYER THE SAME AS FIRST. PACKING SPHERES INTO LATTICES A THIS IS CALLED….

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46 A B C THE ABCA ARRANGEMENT……….. PACKING SPHERES INTO LATTICES A CUBIC CLOSED PACKED…. WHY???

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

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

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

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50 COORDINATION NUMBER HCP

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51 COORDINATION NUMBER HCP COORDINATION NUMBER =12

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52 COORDINATION NUMBER HCPCCP COORDINATION NUMBER =12

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53 HCP CCP COORDINATION NUMBER =12 COORDINATION NUMBER

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

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55 REVIEW QUESTION 1Stacking a second close- packed layer of spheres directly atop a close-packed layer below Which is the closest packed arrangement? 2Stacking 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

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56 REVIEW QUESTION 1Stacking a second close- packed layer of spheres directly atop a close-packed layer below Which is the closest packed arrangement? 2Stacking 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……..

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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.

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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)

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Bronze

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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

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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.

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10–62 Two Types of Alloys Substitutional Interstitial

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Atomic Size Ratios and the Location of Atoms in Unit Cells PackingType of HoleRadius Ratio hcp or ccpTetrahedral hcp or ccpOctahedral Simple CubicCubic About Holes in Cubic Arrays pm

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64 SIMPLE CUBIC How do we count nearest neighbors? COORDINATION NUMBER Draw a few more unit cells…...

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65 SIMPLE CUBIC Highlight the nearest neighbors…. COORDINATION NUMBER

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66 coordination number of 6 What about body centered cubic????? SIMPLE CUBIC COORDINATION NUMBER How many nearest neighbors???

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67 In the three types of cubic unit cells: Simple cubic COORDINATION NUMBER CN = 6 Body Centered cubic CN = ? Lets look at this…….

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68 In bcc lattices, each sphere has a coordination number of 8 Body-centered cubic packing (bcc) COORDINATION NUMBER????? What about face centered cubic?

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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?

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

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71 FRACTION(F) OF THE VOLUME OCCUPIED BY THE SPHERES IN THE UNIT CELL. V spheres = number of spheres x volume single sphere V unit cell = a 3 cubic unit cell of edge length a Lets get NUMBER OF SPHERES PACKING EFFICIENCY

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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.

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73 What is the number of spheres in the fcc unit cell? Total spheres = 8 (1/8)+ 6 (1/2) = = 4 QUESTION….. FACE-CENTRED CUBIC UNIT CELL Note: 1/8 of a sphere on 8 corners and ½ of a Sphere on 6 faces of the cube

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

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75 QUESTION THE NUMBER OF SPHERES IN A BODY CENTRED CUBIC CELL IS? ANSWER….Atoms = 8(1/8) + 1 = 2 VOLUME OCCUPIED IN FCC….

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76 CUBIC UNIT CELLS WHAT FRACTION OF SPACE IS OCCUPIED INFACE CENTRED CUBIC CELL? NUMBER OF SPHERES IS 4 NOW WE NEED THE VOLUME OF A SPHERE, USING r FOR RADIUS V = total THERE ARE 4 SPHERES IN THE UNIT CELL

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77 Now we need the volume of the unit cell. radius of the sphere is r. GET DIMENSIONS OF CUBE IN TERMS OF r….. Why????? FACE-CENTRED CUBIC UNIT CELL

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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!!

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79 Let side of cube be a a DRAWING CUBE FACE REMEMBER THE SPHERES TOUCH!! Draw a square….. Now we need to get a in terms of r

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80 Let side of cube be a a CONSTRUCT A TRIANGLE ON THE FACE Why???? So we can use Pythagoras!

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81 Let side of cube be a aGET a in terms of r a r 2r r

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

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

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84 FACE-CENTRED CUBIC UNIT CELL We conclude…..

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85 and 26% is taken up by empty space. In a cubic closest packed crystal 74% of the volume of a is taken up by spheres QUESTION FACE-CENTRED CUBIC UNIT CELL

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

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87 CUBIC UNIT CELLS THE EDGE LENGTH IN TERMS OF r SIMPLE CUBIC BODY CENTRED CUBICFACE CENTRED CUBIC NUMBER OF SPHERES 124 2r VOLUME OCCUPIED

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88 CUBIC UNIT CELLS SIMPLE CUBIC BODY CENTRED CUBICFACE CENTRED CUBIC NUMBER OF SPHERES 124 2r VOLUME OCCUPIED 52.4% 68.0% 74.0% QUESTION...

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

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

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91 Make sure you can do the fcc, bcc and sc lattice calculations! 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 SUMMARY What other property of a substance depends on packing efficiency????????

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92 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/6.022x10 23 Avogadro’s Number! DENSITY N 0 = x atoms per mole

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

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COPPER DENSITY CALCULATION g Cu mole Cu X atoms 4 atoms Cu unit cell 4.75X cm 3 unit cell = 8.89 g/cm 3 Laboratory measured density: 8.92 g/cm 3

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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)

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96 AVOGADRO’S NUMBER Fe(s) is bcc Two atoms / unit cell g Mole 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/cm 3.

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97 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. Fe(s) is bcc Two atoms / unit cell g Mole Fe7.86 g cm 3

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98 AVOGADRO’S NUMBER The density of Fe(s) is 7.86 g/cm 3. V= e 3 = (287pm) 3 = 2.36x cm 3 Fe(s) is bcc Two atoms / unit cell length of an edge is 287 pm g Mole Fe7.86 g cm 3 ( ) 3 cm 3 pm 3

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99 AVOGADRO’S NUMBER The density of Fe(s) is 7.86 g/cm 3. V= e 3 = (287pm) 3 = 2.36x cm 3 Fe(s) is bcc Two atoms / unit cell length of an edge is 287 pm g Mole Fe7.86 g cm 3 ( ) 3 cm 3 pm 3 (287 pm) 3 unit cell

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100 AVOGADRO’S NUMBER g Mole Fe7.86 g cm 3 ( ) 3 cm 3 pm 3 (287 pm) 3 unit cell2 atoms unit cell

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101 AVOGADRO’S NUMBER g Mole Fe7.86 g cm 3 ( ) 3 cm 3 pm 3 (287 pm) 3 unit cell2 atoms unit cell = X atoms/mole

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102 IONIC SOLIDS NaCl Binary Ionic Solids: Two types of ions MgOCaCO 3 MgSO 4 Hard, brittle solids High melting point Electrical insulators except when molten or dissolved in water. These are lattices of ions……. Examples:

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

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Prentice-Hall © 2002General Chemistry: Chapter 13Slide 104 of 35 Sodium Chloride

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105 The stability of the ionic compound results from the electrostatic attractions between the ions: Li + F – F – Li + Li + F – 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?

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

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

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

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

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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

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111 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

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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

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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 the two ions (cations and anions) in the solid. NOTE: IONIC SOLIDS HOLES????

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

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115 The black dots form a FCC lattice! See the holes???? HOLES IN A FCC LATTICE

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116 The black dots form a fcc lattice! See the holes???? HOW MANY HOLES?????? HOLES IN A FCC LATTICE

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117 HOLES IN A FCC LATTICE The holes: THIRTEEN:ONE IN THE CENTRE How many?? 12 on the edges. What shape is the hole ?

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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?

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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 = x (1/4) = 4 1/4 atom 1 atom

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120 There are 4 complete octahedral holes per fcc unit cell. THE 13 OCTAHEDRAL HOLES 1/4 atom 1 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..

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121 THE OCTAHEDRAL HOLES Other holes….. Between two layers…..

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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

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

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

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

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126 This is a TETRAHEDRAL HOLE…. 8 CUBES SMALL CUBE

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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…….

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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.

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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????

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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?

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

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132 M + or M 2+ cations always occupy the holes Consequently the radius of the cation must be This causes the X – anions to be pushed apart, greater than the size of the hole! which reduces the X – – X – repulsion. with the largest coordination number without rattling around! Which hole will a cation occupy?????? TIGHT FIT So we will investigate the size of these holes!

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

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

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135 Look at plane Draw a square. Put in spheres. Fit a small sphere in This will be the cation These are the anions Draw diagonal Put in distances…….. OCTAHEDRAL HOLES IN FCC

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136 2r Radius of ion = R Look at plane Radius of hole = r OCTAHEDRAL HOLES IN FCC R

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137 R R R R 2r Radius of ion = R Look at plane Radius of hole = r +(2R) 2 ( 2R) 2 = (2R + 2r) 2 8R 2 = (2R + 2r) R = r OCTAHEDRAL HOLES IN FCC

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138 R R R R 2r Radius of ion = R Look at plane Radius of hole = r 0.414R = r The size of cation that just fits has a radius that is x radius of anion(R) r octahedral hole = R What about the tetrahedral hole? OCTAHEDRAL HOLES IN FCC

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139 Using similar calculations, we can find the radius of other types of holes as well: r tetrahedral = 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 r octahedral = R What about other cubic cell systems?? DO IT!!!!!!!

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140 r cubic = R anion 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!!!!!!!

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

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142 SUMMARY : Face centred cubic: Trigonal holeToo small to be occupied Tetrahedral holeCN = 4r cation = 0.225R anion Octahedral holeCN = 6 8 of these r cation = 0.414R anion 4 of these Simple cubic: Cubic hole CN = 8 r cation = 0.732R anion 1 of these For a given anion r trigonal < r tetrahedral < r octahedral < r cubic

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143 SUMMARY Face centred cubic: Trigonal hole Too small to be occupied Tetrahedral holeCN = 4 r cation = 0.225R anion Octahedral holeCN = 6 8 of these r cation = 0.414R anion 4 of these Simple cubicCN = 8r cation = 0.732R anion 1 of these Which hole will a cation occupy?????? r trigonal < r tetrahedral < r octahedral < r cubic For a given anion

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144 INTO WHICH HOLE WILL THE ION GO?? TETRAHEDRAL The hole filled is tetrahedral if: 0.225R anion < r cation < 0.414R anion r tetrahedral < r cation < r octahedral

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145 INTO WHICH HOLE WILL THE ION GO?? OCTAHEDRAL The hole filled is octahedral if: 0.414R anion < r cation < 0.732R anion r octahedral < r cation < r cubic

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146 INTO WHICH HOLE WILL THE ION GO?? CUBIC The hole filled is cubic if: 0.732R anion < r cation Lets look at these ideas in action……. r cubic < r cation

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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 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

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148 Na + has a radius of 98pm. Cl - has a radius of 181pm lies between and so the sodium cations will occupy octahedral holes in a fcc (ccp) latticeIs the stoichiometry ok??? NaCl

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149 1:1 stoichiometry is required How many complete octahedral holes in face centred cubic array of Cl - ????? So stoichiometry is ok!! 4 How many Cl - needed to form the fcc array??? 4 Therefore 4 Cl - and 4 Na + NaCl Lets do another example…..

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150 Example: Predict the structure of Li 2 S Li + is 68 pmS 2- is 190pm Calculate ratio.. Examine the cation-anion radius ratios to find which type of holes the smaller ions fill STEP ONE: COMPARE with ratios…. Which is the best hole???? TETRAHEDRAL!!!!

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151 This requires tetrahedral holes. Example: Predict the structure of Li 2 S Li + is 68 pmS 2- is 190pm face- centred cubic array Lets look at the structure…... Calculate ratio.. Which lattice has tetrahedral holes??? Thus the S 2- will form a fcc lattice...

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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

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153 FCC unit cell with tetrahedral holes S 2- Li + How many are occupied? Li 2 S needs two Li + for each S 2- Four anions in the unit cell. There are 8 tetrahedral holes. Therefore all the tetrahedral holes are occupied! Next Step….

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154 = S 2– = 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.

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155 = S 2– = Li + all the tetrahedral holes have to be occupied. Now do CsCl…. FCC unit cell with tetrahedral holes filled Li 2 S is a face centered lattice of S 2- with all of the tetrahedral holes filled by Li + ions.

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

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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.

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Cesium Chloride

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159 ZnS: Zn 2+ is 64 pm S 2- is 190 pm Calculate ratio COMPARE This requires tetrahedral holes. The sulfide ions will form a face-centered cubic array because…. that is the only type to possess tetrahedral holes. What about stoichiometry??????

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160 ZnS: Zn 2+ is 64 pm S 2- is 190 pm Calculate ratio COMPARE This requires tetrahedral holes. The sulfide ions will form a face-centered cubic array because…. that is the only type to possess tetrahedral holes. What about stoichiometry??????

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161 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.

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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.

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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.

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164 QUESTION A crystal has Mg 2+ 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 1KMgF 3 2K 3 MgF 2 3KMg 2 F 2 4K 2 Mg 2 F 5K 2 Mg 2 F 3

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165 ANSWER A crystal has Mg 2+ 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 1KMgF 3 2K 3 MgF 2 3KMg 2 F 2 4K 2 Mg 2 F 5K 2 Mg 2 F 3

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166 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... 1Na 2 ClO 2Na 3 ClO 3NaCl 3 O 4NaClO 3 5NaClO

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167 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... 1Na 2 ClO 2Na 3 ClO 3NaCl 3 O 4NaClO 3 5NaClO

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168 This is the flourite structure: MX 2 = Ca 2+ = F - the anions occupy the tetrahedral holes(8) in a fcc array of the cations(4). Does this fit radius ratios???????

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169 CaF 2 : Ca 2+ is 99 pm F - is 136 pm Calculate ratio OOPS !!!! The radius ratio is too BIG!!!! FOR TETRAHEDRAL HOLES 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!!!!!!!

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170 CaF 2 : Alternative description SIMPLE CUBIC Ca 2+ in alternating cubic sites. What is Antiflourite???? Ca 2+

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171 This is the flourite structure: MX 2 The antifluorite structure M 2 X (eg K 2 O) = Ca 2+ = F - the anions occupy the tetrahedral holes(8) in a fcc array of the cations(4). the cations occupy the tetrahedral holes and the anions form the fcc array.

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

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173 Calculating the density of an ionic compound MgO fcc of O 2- Mg 2+ in octahedral holes A face…. Edge= r oxide ion + 2r Mg ion + r oxide ion Now calculate volume 4 Mg’s and 4 O 2- REMEMBER…. Ionic Compound Density R = 86 pm r = 126 pm Edge = 424 pm V = (424) 3 = 7.62X10 7 pm 3

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174 Calculating the density of an ionic compound MgO fcc of O 2- Mg 2+ in octahedral holes A face…. Edge= r oxide ion + 2r Mg ion + r oxide ion Now calculate volume 4 Mg’s and 4 O 2- REMEMBER…. Ionic Compound Density R = 86 pm r = 126 pm Edge = 424 pm V = (424) 3 = 7.62X10 7 pm 3

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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 g MgO mole

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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 g MgO mole X FU

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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 g MgO mole 4 FU mole X FUunit cell

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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 g MgO mole 4 FU mole X FUunit cell7.62X10 7 pm 3 Unit Cell

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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 g MgO mole 4 FU mole X FUunit cell pm X10 7 pm 3 Unit Cell m 3

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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 g MgO mole 4 FU mole X FUunit cell pm 3 cm X10 7 pm 3 Unit Cell m m

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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 g MgO mole 4 FU mole X FUunit cell pm 3 cm X10 7 pm 3 Unit Cell m m = 3.54 g/cm 3

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Copyright © Houghton Mifflin Company. All rights reserved. 10–182 Diamond and Graphite Covalently Networked Crystalline Solids

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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

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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...

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X-Ray Diffraction

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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???

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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 .

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Cell Structure by X-ray Diffraction

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189 BRAGG DIFFRACTION LAW W.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..

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BRAGG’S LAW

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

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Bragg Equation Derivation ө өө x x d 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

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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

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194 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……..

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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

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10–196 Molecular Orbital Energy Levels

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10–197 Molecular Orbital Energy Levels

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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

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Copyright © Houghton Mifflin Company. All rights reserved. 10–199 A Representation of the Energy Levels (Bands) in a Magnesium Crystal

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10–200 Molecular Orbital Energies Insulator (diamond)Conductor(metal)

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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.

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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)

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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

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Copyright © Houghton Mifflin Company. All rights reserved. 10–204 Energy-Level Diagrams for (a) an N-Type Semiconductor and (b) a P-Type Semiconductor As is exampleB is an example

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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

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

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The End

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