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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. 2 LATTICE EXAMPLE We can pick out the smallest repeating unit….. 3 UNIT CELL We can pick out the smallest repeating unit….. We call this the UNIT CELL……….. 4 UNIT CELL We call this the UNIT CELL……….. The 5 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….. 7 SIMPLE CUBIC Eight equivalent points at the corners of a cube We can imagine an equivalent point at the centre of the spheres 8 BODY CENTRED CUBIC Eight equivalent points at the corners of a cube and one at the centre Another possibility……... 9 FACE CENTRED CUBIC Eight equivalent points at the corners of a cube and six on the centre of the cube faces Summary…….. 10 THE CUBIC UNIT CELLS Simple Cubic Unit Cell KNOW THESE!!!! Body-Centred Cubic Unit Cell Face-Centred Cubic Unit Cell 11 How do we investigate solids? Unit Cells in the Cubic Crystal System Prentice-Hall © 2002 General Chemistry: Chapter 13 Slide 12 of 35 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 13 Electron Sea Model of Metals Copyright © Houghton Mifflin Company. All rights reserved. 10–14 Summary of Crystal Structures METALS VIEWED AS CLOSELY PACKED SPHERES HOW CAN WE PACK SPHERES????? 16 PACKING OF SPHERICAL VEGETABLES 17 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……. 18 Packing Spheres into Lattices: First Layer Lets put in a few more spheres………. 19 Packing Spheres into Lattices First Layer 20 Packing Spheres into Lattices Next Layer The next spheres fit into a “dimple” formed by three spheres in the first layer. 21 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…... 22 Packing Spheres into Lattices: Next Layer Triangle not inverted The next spheres fit into NOTE: the inverted triangle The two types of “dimples” formed by three spheres in the first layer. The second layer….. 23 Packing Spheres into Lattices: is formed by choosing one of the sets of dimples Now put on second layer…... 24 Packing Spheres into Lattices: Second Layer Once one is put on the others are forced into half of the dimples of the same type…. 25 Packing Spheres into Lattices Second Layer Once one is put on the others are forced into half of the dimples of the same type…. 26 Packing Spheres into Lattices: Second Layer Once one is put on the others are forced into half of the dimples of the same type…. 27 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. 28 Packing Spheres into Lattices Second Layer Occupied dimple Unoccupied DIMPLE Note that the second layer only occupies half the dimples in the first layer. 29 THE THIRD LAYER…... PACKING SPHERES INTO LATTICES SECOND LAYER HAVE TO CHOOSE A DIMPLE 30 PACKING SPHERES INTO LATTICES THIRD LAYER, Choose a dimple 1 1 1 (1) A DIMPLE DIRECTLY ABOVE SPHERE IN THE FIRST LAYER 31 THIS OR…….. PACKING SPHERES INTO LATTICES THIRD LAYER NOTE: the inverted triangle 2 1 1 1 2 2 (2) A DIMPLE DIRECTLY ABOVE A DIMPLE IN THE FIRST CHOOSE OPTION 1….. LAYER…….. 32 PACKING SPHERES INTO LATTICES THIRD LAYER (option 1) 1 1 1 OPTION ONE! SPHERE DIRECTLY ABOVE SPHERE IN THE FIRST LAYER ADD SOME MORE…….. 33 PACKING SPHERES INTO LATTICES THIRD LAYER OPTION ONE! ADD SOME MORE….. 34 PACKING SPHERES INTO LATTICES THIRD LAYER OPTION ONE! THE ABA ARRANGEMENT OF LAYERS. A B A LAYERS ONE AND THREE ARE THE SAME! 35 PACKING SPHERES INTO LATTICES THE ABA ARRANGEMENT OF LAYERS. A B A CALLED HEXAGONAL CLOSEST PACKIN HCP 36 PACKING SPHERES INTO LATTICES THE ABA ARRANGEMENT OF LAYERS, Option 1. A HEXAGONAL UNIT CELL. A B A HEXAGONAL CLOSEST PACKING 37 HEXAGONAL UNIT CELL ABA ARRANGEMENT HAS A HEXAGONAL UNIT CELL. HCP 38 SUMMARY... SUMMARY HEXAGONAL CLOSED PACKED STRUCTURE EXPANDED VIEW NOW OPTION TWO….. 39 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 40 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……. 41 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! 42 PACKING SPHERES INTO LATTICES THIRD LAYER WE CALL THE THIRD LAYER C THIS TIME! OPTION 2 A B C THE ABC ARRANGEMENT OF LAYERS. 43 NOW THE FOURTH LAYER……. PACKING SPHERES INTO LATTICES FOURTH LAYER PUT SPHERE IN SO THAT A B C FOURTH LAYER THE SAME AS FIRST. 44 PACKING SPHERES INTO LATTICES PHH p 509 FOURTH LAYER THE SAME AS FIRST. A B C A THE ABCA ARRANGEMENT……….. THIS IS CALLED…. 45 PACKING SPHERES INTO LATTICES THE ABCA ARRANGEMENT……….. A B C A CUBIC CLOSED PACKED…. 46 WHY??? UNIT CELL OF CCP CUBIC UNIT CELLL THIS ABCA ARRANGEMENT HAS A FACE- CENTRED CUBIC UNIT CELL (FCC) A COMPARISON….. 47 COMPARISON NOTICE the flip…... HCP NEAREST NEIGHBORS….. 48 CCP COORDINATION NUMBER The number of nearest neighbors that a lattice point has in a crystalline solid Lets look at hcp and ccp…... 49 COORDINATION NUMBER HCP 50 COORDINATION NUMBER HCP COORDINATION NUMBER =12 51 COORDINATION NUMBER HCP COORDINATION NUMBER =12 52 CCP COORDINATION NUMBER HCP COORDINATION NUMBER =12 53 CCP COORDINATION NUMBER SPHERES IN BOTH HCP AND CCP STRUCTURES EACH HAVE A COORDINATION NUMBER OF 12. QUESTION…….. 54 REVIEW QUESTION Which is the closest packed arrangement? 1 Stacking a second closeTop Row packed layer of spheres directly atop a close-packed layer below Bottom Row 2 Stacking a second close-packed Top Row layer of spheres in the depressions formed by spheres in Bottom Row the close-packed layer below. ANSWER…….. 55 REVIEW QUESTION Which is the closest packed arrangement? 1 Stacking a second closepacked 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…….. 56 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 pm hcp or ccp Tetrahedral 0.22 - 0.41 hcp or ccp Octahedral 0.41 - 0.73 Simple Cubic Cubic 0.73 - 1.00 COORDINATION NUMBER SIMPLE CUBIC How do we count nearest neighbors? Draw a few more unit cells…... 64 COORDINATION NUMBER SIMPLE CUBIC Highlight the nearest neighbors…. 65 COORDINATION NUMBER SIMPLE CUBIC How many nearest neighbors??? coordination number of 6 66 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……. 67 Body-centered cubic packing (bcc) COORDINATION NUMBER????? In bcc lattices, each sphere has a coordination number of 8 What about face centered cubic? 68 COORDINATION NUMBER 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 69 CN = 12 PACKING EFFICIENCY? EFFICIENCY OF PACKING THE FRACTION OF THE VOLUME THAT IS ACTUALLY OCCUPIED BY SPHERES….. volume occupied by the spheres in the unit cell fv volume of the unit cell WHAT DOES THIS MEAN??? 70 PACKING EFFICIENCY FRACTION(F) OF THE VOLUME OCCUPIED BY THE SPHERES IN THE UNIT CELL. volume occupied by the spheres in the unit cell fv volume of the unit cell fv Vspheres Vunitcell Vspheres= number of spheres x volume single sphere Vunit cell = a3 cubic unit cell of edge length a Lets get NUMBER OF SPHERES 71 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 COUNTS 1 FOR ATOM IN CELL IN THE LATTICE COUNTS FOR 1/4 ATOM ON A FACE. COUNTS AS 1/8 FOR ATOM ON A CORNER. 72 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….. 73 QUESTION THE NUMBER OF SPHERES IN A BODY CENTRED CUBIC CELL IS? ANSWER…. 74 QUESTION THE NUMBER OF SPHERES IN A BODY CENTRED CUBIC CELL IS? ANSWER…. Atoms = 8(1/8) + 1 = 2 VOLUME OCCUPIED IN FCC…. 75 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 76 4 3 = 4 ( r ) 3 THERE ARE 4 SPHERES IN THE UNIT CELL FACE-CENTRED CUBIC UNIT CELL 4 3 Vspheres 4( r ) 3 radius of the sphere is r . Now we need the volume of the unit cell. Why????? fv Vspheres Vunitcell GET DIMENSIONS OF CUBE IN TERMS OF r….. 77 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!! 78 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 79 Let side of cube be a a CONSTRUCT A TRIANGLE ON THE FACE Why???? So we can use Pythagoras! 80 Let side of cube be a a a GET a in terms of r r 2r r 81 Let side of cube be a a GET a in terms of r FACE DIAGONAL = r + 2r + r=4r PYTHAGORAS! a2 + a2 = (4r)2 a r 2a2 = 16r 2 2r 82 r a2 = 8r 2 ar 8 Side of cube be in terms of r a ar 8 Now we can calculate the volume of the unit cell Vcell a3 a r Vcell r 8 2r 83 r NOW PUT IT ALL TOGETHER 3 FACE-CENTRED CUBIC UNIT CELL fv fv Vspheres Vunitcell 4 4 * 3 r 3 r 83 fv 0.740 84 fv 4 4* 3 83 We conclude….. FACE-CENTRED CUBIC UNIT CELL fv Vspheres Vunitcell fv 0.740 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 85 Body Centered Cubic e d2 = e2 + e2 d2 = 2e2 (4r)2 = e2 + d2 16r2 = e2 + 2e2 r2 = 3e2/16 e = 4r/√3 CUBIC UNIT CELLS THE EDGE LENGTH IN TERMS OF r SIMPLE CUBIC 2r NUMBER OF SPHERES 1 87 BODY CENTRED CUBIC 4r 3 FACE CENTRED CUBIC r 8 2 4 VOLUME OCCUPIED CUBIC UNIT CELLS VOLUME OCCUPIED SIMPLE CUBIC BODY CENTRED CUBIC FACE CENTRED CUBIC 52.4% 74.0% 68.0% 2r NUMBER OF SPHERES 1 88 4r 3 2 r 8 QUESTION... 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….. 89 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……... 90 SUMMARY sc: bcc: fcc: hcp: 52.4% of space occupied by spheres 68.0% of space occupied by spheres 74.0% of space occupied by spheres 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???????? 91 DENSITY We can calculate the density in a unit cell. mass Density volume Mass is the mass of the number of atoms in the unit cell. Mass of one atom =atomic mass/6.022x1023 N0 = 6.022 x 1023 atoms per mole Avogadro’s Number! 92 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: Vcell r 8 8 3 Vcell ( 8 1.28 10 cm ) 3 Vcell 4.75 10 93 23 cm 3 COPPER DENSITY CALCULATION 63.54 g Cu mole Cu 4 atoms Cu unit cell 4.75X10-23 cm3 mole Cu 6.022 X 1023 atoms unit cell = 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? 4r e 3 3e r 4 EDGE LENGTH (e) 3 287 pm r 124 pm 4 95 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 96 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 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 cm3 pm3 55.85 g Mole Fe 7.86 g (10 -12)3 cm3 Fe(s) is bcc Two atoms / unit cell 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 cm3 pm3 55.85 g Mole Fe 7.86 g (10 -12)3 cm3 unit cell (287 pm)3 Fe(s) is bcc Two atoms / unit cell 99 AVOGADRO’S NUMBER cm3 pm3 55.85 g Mole Fe 7.86 g (10 -12)3 cm3 100 unit cell 2 atoms (287 pm)3 unit cell AVOGADRO’S NUMBER cm3 pm3 55.85 g Mole Fe 7.86 g (10 -12)3 cm3 unit cell 2 atoms (287 pm)3 unit cell = 6.022 X 10-23 atoms/mole 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……. 102 IONIC SOLIDS = Na+ = Cl– We notice that this is a cubic array of ions. Why do ionic solids hold together????? 103 Sodium Chloride Prentice-Hall © 2002 General Chemistry: Chapter 13 Slide 104 of 35 IONIC SOLIDS The stability of the ionic compound results from the electrostatic attractions between the ions: Li+ F– Li+ F– F– Li+ F– Li+ Li+ F– Li+ F– F– Li+ F– Li+ The LiF crystal consists of a lattice of ions. The 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? 105 NaCl structure ClNa+ Lets take this apart…... 106 NaCl structure ClNa+ Lets look at the black dot lattice…. 107 NaCl structure What unit cell do the black dots form? The black dots form a fcc lattice! Now look at the red dots 108 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….. The red dots form a fcc array! Now put these back together….. 110 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 111 NaCl structure This is the NaCl structure. ClNa+ 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 HOLES IN A LATTICE. Holes??? What holes????? Lets look at a fcc lattice! 114 HOLES IN A FCC LATTICE The black dots form a FCC lattice! See the holes???? 115 HOLES IN A FCC LATTICE The black dots form a fcc lattice! See the holes???? HOW MANY HOLES?????? 116 HOLES IN A FCC LATTICE The holes: How many?? THIRTEEN: ONE IN THE CENTRE 12 on the edges. What shape is the hole ? 117 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 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 = 1 + 12 x (1/4) = 4 There are 4 complete octahedral holes per fcc unit cell. 119 THE 13 OCTAHEDRAL HOLES 1 atom 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???? 120 The octahedral hole is.. THE OCTAHEDRAL HOLES Between two layers….. Other holes….. 121 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 Take a point at the centre of this cube There are eight of these…. 123 SMALL CUBE 8 CUBES Take a point at the centre of this cube Another one of these…. 124 SMALL CUBE 8 CUBES Take a point at the centre of this cube An so on …. 125 SMALL CUBE 8 CUBES This is a TETRAHEDRAL HOLE…. 126 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 127 This hole……. 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………. 128 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?????? 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! 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…. LOOK AT HOLE…. 133 OCTAHEDRAL HOLES IN FCC Look at plane Draw a square. Put in spheres. These are the anions Fit a small sphere in 134 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 135 Put in distances…….. OCTAHEDRAL HOLES IN FCC Look at plane Radius of ion = R Radius of hole = r 2r 136 R OCTAHEDRAL HOLES IN FCC Look at plane Radius of ion = R Radius of hole = r 2 (2R)2 +(2R)2 = (2R + 2r) 8R2 = (2R + 2r)2 R R 2r R R 137 2 2R 2 R 2r ( 2 2 2) R 2r 2 1R r 0.414R = r OCTAHEDRAL HOLES IN FCC Look at plane Radius of ion = R Radius of hole = r R R R 0.414R = r The size of cation that just fits has a radius that is 0.414 x radius of anion(R) 2r R roctahedral hole = 0.414 R What about the tetrahedral hole? 138 Using similar calculations, we can find the radius of other types of holes as well: fcc DO IT!!!!!!! rtetrahedral = 0.225 R roctahedral = 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... rcubic = 0.732 Ranion DO IT!!!!!!! 140 The cubic hole The coordination number in the cubic hole is ? 8 rcubic = 0.732 Ranion In contrast for a fcc lattice…... The coordination number in the fcc tetrahedral hole is ? 4! 141 The coordination number in the fcc octahedral hole is ? 6! SUMMARY: Face centred 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 142 SUMMARY Face centred cubic: Trigonal hole Too small to be occupied Tetrahedral hole CN = 4 rcation = 0.225Ranion 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?????? 143 8 of these INTO WHICH HOLE WILL THE ION GO?? TETRAHEDRAL The hole filled is tetrahedral if: rtetrahedral < rcation < roctahedral 0.225Ranion < rcation < 0.414Ranion 144 INTO WHICH HOLE WILL THE ION GO?? OCTAHEDRAL The hole filled is octahedral if: roctahedral < rcation < rcubic 0.414Ranion < rcation < 0.732Ranion 145 INTO WHICH HOLE WILL THE ION GO?? CUBIC The hole filled is cubic if: rcubic < rcation 0.732Ranion < rcation Lets look at these ideas in action……. 146 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 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 Na+ has a radius of 98pm. Cl- has a radius of 181pm. rcation rNa 98 pm 0.54 ranion rCl 181 pm tet rcation 0.225 tet ranion oct rcation 0.414 oct ranion cubic rcation 0.732 cubic ranion 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??? Therefore 4 Cl- and 4 Na+ So stoichiometry is ok!! Lets do another example….. 149 4 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.. rcation rLi 68 pm 0.36 ranion rS 2 190 pm COMPARE with ratios…. tet rcation 0.225 tet ranion Which is the best hole???? 150 oct rcation 0.414 oct ranion TETRAHEDRAL!!!! Example: Predict the structure of Li2S Li+ is 68 pm S2- is 190pm Calculate ratio.. rcation rLi 68 pm 0.36 ranion rS 2 190 pm 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…... 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 So???? 152 FCC unit cell with tetrahedral holes S2- Li+ Four anions in the unit cell. There are 8 tetrahedral holes. How many are occupied? Li2S needs two Li+ for each S2- Next Step…. Therefore all the tetrahedral holes are occupied! 153 FCC unit cell with tetrahedral holes filled = S2– = 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 = S2– = 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. 155 Now do CsCl…. + is 167 pm Cs CsCl: Cl- is 181pm Calculate ratio rcation rCs 167 pm 0.92 ranion rCl 181 pm cubic Compare…... rcation 0.732 cubic ranion 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 Zn2+ is 64 pm S2- is 190 pm ZnS: rcation rZn2 64 pm 0.35 ranion rS 2 190 pm tet COMPARE rcation 0.225 tet ranion Calculate ratio This requires tetrahedral holes. oct rcation 0.414 oct ranion The sulfide ions will form a face-centered cubic array because…. that is the only type to possess tetrahedral holes. What about stoichiometry?????? 159 Zn2+ is 64 pm S2- is 190 pm ZnS: rcation rZn2 64 pm 0.35 ranion rS 2 190 pm tet COMPARE rcation 0.225 tet ranion Calculate ratio This requires tetrahedral holes. oct rcation 0.414 oct ranion 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 S2(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 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 164 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 165 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 166 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 167 This is the flourite structure: MX2 the anions occupy the tetrahedral holes(8) = Ca2+ in a fcc array of the cations(4). Does this fit radius ratios??????? 168 = F- CaF2: Ca2+ is 99 pm F- is 136 pm Calculate ratio rcation rCa 2 99 pm 0.727 ranion rF 136 pm FOR TETRAHEDRAL HOLES tet rcation 0.225 tet ranion 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 Fwith Ca2+ at alternate cubic holes!!!!!!! 169 SIMPLE CUBIC CaF2: Ca2+ Ca2+ Ca2+ in alternating cubic sites. Alternative description What is Antiflourite???? 170 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. 171 Ionic Compound Density fcc of O2- Mg2+ in octahedral holes Calculating the density of an ionic compound A face…. MgO Edge= roxide ion+ 2rMg ion+ roxide ion Now calculate volume 4 Mg’s and 4 O2172 REMEMBER…. Ionic Compound Density fcc of O2- Mg2+ in octahedral holes Calculating the density of an ionic compound R = 86 pm A face…. r = 126 pm MgO Edge = 424 pm V = (424)3 = 7.62X107 pm3 Edge= roxide ion+ 2rMg ion+ roxide ion Now calculate volume 4 Mg’s and 4 O2173 REMEMBER…. Ionic Compound Density fcc of O2- Mg2+ in octahedral holes Calculating the density of an ionic compound R = 86 pm A face…. r = 126 pm MgO Edge = 424 pm V = (424)3 = 7.62X107 pm3 Edge= roxide ion+ 2rMg ion+ roxide ion Now calculate volume 4 Mg’s and 4 O2174 REMEMBER…. 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 1023 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 4 FU mole 6.022 X 1023 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 Unit Cell 4 FU mole 6.022 X 1023 FU unit cell 7.62X107 pm3 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 Unit Cell pm3 4 FU mole 6.022 X 1023 FU unit cell 7.62X107 pm3 10-36m3 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 Unit Cell pm3 10-6m 4 FU mole 6.022 X 1023 FU unit cell 7.62X107 pm3 10-36m3 cm3 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 Unit Cell pm3 10-6m 4 FU 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 Copyright © Houghton Mifflin Company. All rights reserved. 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: n2dsin. 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……. Now imagine reflection of X-rays………. 191 Bragg Equation Derivation ө өө d sin ө = x/d Wave length λ = 2x x = d sin ө 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…….. 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 Copyright © Houghton Mifflin Company. All rights reserved. 10–199 Molecular Orbital Energies Insulator (diamond) 10–200 Conductor(metal) Semi Conductors • For metalloids the distance between the conducting band and the nonconduction band are lower, in between that for metals and nonmentals, thus called semiconductors. • For example silicon is a semiconductor, with the same structure as diamond, since it is in the same group. • Diamond has a large gap in its band model, but silicon, being a semi conductor, has a smaller gap, thus promoting conduction. Semi Conductors • At higher temperatures more electrons are promoted into the conduction band and conductivity increases for semiconductors. • adding impurities, such as phosphorus or gallium in metalloids (usually silicon) changes the conduction characteristics of the metalloid (silicon). • When a small fraction of silicon atoms are replaced with phosphous atoms, each with one more electron than silicon, then extra electrons are available for conduction (Called an n-type semi conductor) Semi Conductors • N-type semi conductors, using a phosphorus impurity, provide more electrons than the original semi conductor, usually Silicon. – These electrons lie closer to the conduction band and less energy is required for conduction – This is called an n-type due to extra negative charge • Conductivity can be enhanced by an element such as boron that has one less valence electron than silicon – These are called P-semiconductors – Since we are missing an electron then there is a hole, which an electron fills thus creating another hole – Holes flow in a direction opposite to the flow of electrons, since lower lying electrons are promoted to fill the hole – Called p for positive charge, due to one less electron Energy-Level Diagrams for (a) an N-Type Semiconductor and (b) a P-Type Semiconductor As is example Copyright © Houghton Mifflin Company. All rights reserved. B is an example 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 The End