Presentation on theme: "The Chemistry of Solids"— Presentation transcript:
1 The Chemistry of Solids Chapter 11The Chemistry of Solids
2 SOLIDS Solids are either amorphous or crystalline. AMORPHOUS SOLIDS:Considerable disorder in structure.Example: rubber, glassCRYSTALLINE SOLIDS:Highly regular structure in the formof a repeating lattice of atoms or moleculesCrystalline solids are classified as:atomic, metallic, ionic, or covalent network, depending on the type of force holding the particles together, and most often involve a metal.
3 LATTICE EXAMPLEWe can pick out the smallest repeating unit…..
4 UNIT CELL We can pick out the smallest repeating unit….. We call this the UNIT CELL………..
5 UNIT CELL We call this the UNIT CELL……….. The unit cell drawn here is a simple cubic cell
13 METALS e.g. copper, gold, steel, sodium, brass. Good conductors of heat and electricityShiny, ductile and malleableMelting points:low (Hg at -39°C)or high (W at 3370°C)Can besoft (Na) or hard (W)METALS ARE CRYSTALLINE SOLIDS
21 Packing Spheres into Lattices Next LayerThe next spheres fit intoa “dimple” formed by three spheres in the first layer.
22 Packing Spheres into Lattices: Next LayerThe next spheres fit intoa “dimple” formed by three spheres in the first layer.There are two sets of dimples…...
23 Packing Spheres into Lattices: Next LayerThe next spheres fit intoNOTE: the inverted triangleTriangle not invertedThe two types of “dimples” formed by three spheres in the first layer.The second layer…..
24 Packing Spheres into Lattices: is formed by choosing one of the sets of dimplesNow put on second layer…...
25 Packing Spheres into Lattices: Second LayerOnce one is put on the others are forced into half of the dimples of the same type….
26 Packing Spheres into Lattices Second LayerOnce one is put on the others are forced into half of the dimples of the same type….
27 Packing Spheres into Lattices: Second LayerOnce one is put on the others are forced into half of the dimples of the same type….And so on….
28 Packing Spheres into Lattices Second LayerInverted triangle dimples are not filled.Note that the second layer only occupies half the dimples in the first layer.
29 Packing Spheres into Lattices Second LayerOccupied dimpleUnoccupied DIMPLENote that the second layer only occupies half the dimples in the first layer.THE THIRD LAYER…...
30 PACKING SPHERES INTO LATTICES SECOND LAYERHAVE TO CHOOSE A DIMPLE
31 PACKING SPHERES INTO LATTICES THIRD LAYER, Choose a dimple1(1) A DIMPLE DIRECTLY ABOVE SPHERE IN THE FIRST LAYERTHIS OR……..
32 PACKING SPHERES INTO LATTICES THIRD LAYERNOTE: the inverted triangle1222(2) A DIMPLE DIRECTLY ABOVE A DIMPLE IN THE FIRST LAYER……..CHOOSE OPTION 1…..
33 SPHERE DIRECTLY ABOVE SPHERE IN THE FIRST LAYER PACKING SPHERES INTO LATTICESTHIRD LAYER (option 1)111OPTION ONE!SPHERE DIRECTLY ABOVE SPHERE IN THE FIRST LAYERADD SOME MORE……..
34 PACKING SPHERES INTO LATTICES THIRD LAYEROPTION ONE!ADD SOME MORE…..
35 PACKING SPHERES INTO LATTICES THIRD LAYEROPTION ONE!THE ABA ARRANGEMENT OF LAYERS.ABALAYERS ONE AND THREE ARE THE SAME!
36 PACKING SPHERES INTO LATTICES THE ABA ARRANGEMENT OF LAYERS.ABACALLEDHEXAGONAL CLOSEST PACKIN HCP
37 PACKING SPHERES INTO LATTICES THE ABA ARRANGEMENT OF LAYERS, Option 1.A HEXAGONAL UNIT CELL.ABAHEXAGONAL CLOSEST PACKING
38 ABA ARRANGEMENT HAS A HEXAGONAL UNIT CELL. HCPSUMMARY...
52 COORDINATION NUMBERHCPCCPCOORDINATION NUMBER =12
53 COORDINATION NUMBERHCPCCPCOORDINATION NUMBER =12
54 SPHERES IN BOTH HCP AND CCP STRUCTURES COORDINATION NUMBERSPHERES IN BOTH HCP AND CCP STRUCTURESEACH HAVE A COORDINATION NUMBER OF 12.QUESTION……..
55 REVIEW QUESTION Which is the closest packed arrangement? 1 Stacking a second close- packed layer of spheres directly atop a close-packed layer belowTop RowBottom Row2 Stacking a second close-packed layer of spheres in the depressions formed by spheres in the close-packed layer below.Top RowBottom RowANSWER……..
56 REVIEW QUESTION Which is the closest packed arrangement? 1 Stacking a second close- packed layer of spheres directly atop a close-packed layer below2 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……..
57 AlloysAn 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.
58 Substitutional Alloy Examples Where a lattice atom is replaced by an atom of similar sizeBrass, one third of copper atoms are replaced by zinc atomsSterling silver (93% Silver and 7%Cu)Pewter (85% Sn, 7% Cu, 6% Bi, and 2% Sb)Plumber’s solder (67% Pb and 33% Sn)
60 Alloys Interstitial Alloy When lattice holes (interstices) are filled with smaller atomsSteel best know interstitial alloy, contains carbon atoms in the holes of an iron crystalCarbon atoms change propertiesCarbon a very good covalent bonding atom changes the non-directional bonding of the iron, to have some directionResults in increased strength, harder, and less ductileThe larger the percent of carbon the harder and stronger the steelOther metals can be used in addition to carbon, thus forming alloy steels
61 Carbon SteelUnlike bronze the carbon atoms fit into the holes formed by the stacking of the iron atoms. Alloys formed by using the holes are called interstital alloys.
62 Two Types of AlloysSubstitutionalInterstitial10–62
63 About Holes in Cubic Arrays Atomic Size Ratios and the Location of Atoms in Unit CellsPackingType of HoleRadius Ratiohcp or ccpTetrahedralOctahedralSimple CubicCubicpm
64 COORDINATION NUMBER SIMPLE CUBIC How do we count nearest neighbors? Draw a few more unit cells…...
65 COORDINATION NUMBERSIMPLE CUBICHighlight the nearest neighbors….
66 COORDINATION NUMBER SIMPLE CUBIC What about body centered cubic????? How many nearest neighbors???coordination number of 6What about body centered cubic?????
67 COORDINATION NUMBER Lets look at this……. In the three types of cubic unit cells:Simple cubicCN = 6Body Centered cubicCN = ?Lets look at this…….
68 Body-centered cubic packing (bcc) COORDINATION NUMBER?????In bcc lattices, each sphere has a coordination number of 8What about face centered cubic?
69 COORDINATION NUMBER Simple cubic CN = 6 Body Centered cubic CN = 8 In the three types of cubic unit cells:Simple cubicCN = 6Body Centered cubicCN = 8Face Centered cubicCN = ?Comes from ccpJust like hcpCN = 12PACKING EFFICIENCY?
70 EFFICIENCY OF PACKINGTHE FRACTION OF THE VOLUME THAT IS ACTUALLY OCCUPIED BY SPHERES…..WHAT DOES THIS MEAN???
71 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 sphereVunit cell = a3cubic unit cell of edge length aLets get NUMBER OF SPHERES
72 PACKING EFFICIENCY COUNTING ATOMS IN A UNIT CELL! ATOMS CAN BE WHOLLY IN A UNIT CELL ORCOUNTING ATOMS IN A UNIT CELL!ATOMS CAN BE WHOLLY IN A UNITCELL OR ATOMS SHAREDBETWEEN ADJACENT UNIT CELLSIN THE LATTICECOUNTS 1 FOR ATOM IN CELLCOUNTS FOR 1/4 ATOM ON A FACE.COUNTS FOR 1/2 ATOM ON A FACE.COUNTS AS 1/8 FOR ATOM ON A CORNER.
73 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 aSphere on 6 faces of the cubeTotal spheres =8 (1/8)+ 6 (1/2)= = 4QUESTION…..
74 QUESTIONTHE NUMBER OF SPHERES IN A BODY CENTRED CUBIC CELL IS?ANSWER….
75 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 = 2VOLUME OCCUPIED IN FCC….
76 CUBIC UNIT CELLS V = WHAT FRACTION OF SPACE IS OCCUPIED INFACE CENTRED CUBIC CELL?NUMBER OF SPHERES IS 4NOW WE NEED THE VOLUME OFA SPHERE, USING r FOR RADIUSV =THERE ARE 4 SPHERES IN THE UNIT CELLtotal
77 FACE-CENTRED CUBIC UNIT CELL radius of the sphere is r .Now we need the volume of the unit cell.Why?????GET DIMENSIONS OF CUBE IN TERMS OF r…..
78 GETTING THE CUBE DIMENSIONS IN TERMS OF r Let side of cube be aaNOW DRAW A FACE OF THE CUBEREMEMBER THE SPHERES TOUCH!!
79 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
80 a Let side of cube be a CONSTRUCT A TRIANGLE ON THE FACE Why???? So we can use Pythagoras!
85 FACE-CENTRED CUBIC UNIT CELL In a cubic closest packed crystal74% of the volume of a is taken up by spheresand 26% is taken up by empty space.QUESTION
86 Body Centered Cubic r 2r e (4r)2 = e2 + d2 16r2 = e2 + 2e2 r d2 = e2 + e2d2 = 2e2
87 CUBIC UNIT CELLS THE EDGE LENGTH IN TERMS OF r SIMPLE CUBIC BODY CENTRED CUBICFACE CENTRED CUBIC2rNUMBER OF SPHERES124VOLUME OCCUPIED
88 CUBIC UNIT CELLS VOLUME OCCUPIED SIMPLE CUBIC BODY CENTRED CUBIC FACE CENTRED CUBIC52.4%74.0%68.0%2rQUESTION...NUMBER OF SPHERES124
89 QUESTIONThe fraction of space occupied in a hexagonal closest packed arrangement of spheres is the same as that in…..1 simple cubic unit cell2 face centered cubic unit cell3 body centered cubic unit cell4 none of theseANSWER…..
90 QUESTIONThe fraction of space occupied in a hexagonal closest packed arrangement of spheres is the same as that in…..1 simple cubic unit cell2 face centered cubic unit cell3 body centered cubic unit cell4 none of theseSummary……...
91 SUMMARY sc: 52.4% of space occupied by spheres bcc: 68.0% of space occupied by spheresfcc: 74.0% of space occupied by sphereshcp: 74.0% of space occupied by spheresMake sure you can do the fcc, bcc and sc lattice calculations!What other property of a substance depends on packing efficiency????????
92 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/6.022x1023N0 = x 1023 atoms per moleAvogadro’s Number!
93 Volume of a copper unit cell Cu crystalizes as a fccr= 128pm = 1.28x10-10m = 1.28x10-8cmVolume of unit cell is given by:
94 COPPER DENSITY CALCULATION mole Cu4 atoms Cuunit cell63.54 g Cumole Cuunit cell4.75X10-23 cm36.022 X atoms= 8.89 g/cm3Laboratory measured density: 8.92 g/cm3
95 DETERMINATION OF ATOMIC RADIUS At room temperature iron crystallizes with a bcc unit cell.X-ray diffraction shows that the length of an edge is 287 pm.What is the radius of the Fe atom?EDGE LENGTH (e)
96 AVOGADRO’S NUMBERSample Problem: Calculate Avogadro’s number of iron if its unit cell length is 287 pm and it has a density of 7.86 g/cm3.55.85 gMole FeFe(s) is bcc Two atoms / unit cell
97 AVOGADRO’S NUMBERSample Problem: Calculate Avogadro’s number of iron if its unit cell length is 287 pm and it has a density of 7.86 g/cm3.55.85 gcm3Mole Fe7.86 gFe(s) is bcc Two atoms / unit cell9797
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 = 2.36x10-23cm3pm355.85 gcm3Mole Fe7.86 g(10 -12)3 cm3Fe(s) is bcc Two atoms / unit cell9898
99 AVOGADRO’S NUMBER The density of Fe(s) is 7.86 g/cm3. length of an edge is 287 pm.V= e3 = (287pm)3 = 2.36x10-23cm3pm355.85 gcm3unit cellMole Fe7.86 g(10 -12)3 cm3(287 pm)3Fe(s) is bcc Two atoms / unit cell9999
100 AVOGADRO’S NUMBER 55.85 g cm3 Mole Fe 7.86 g pm3 unit cell 2 atoms 100100
101 AVOGADRO’S NUMBER 55.85 g cm3 Mole Fe 7.86 g pm3 unit cell 2 atoms = X atoms/mole101101
102 IONIC SOLIDS Binary Ionic Solids: Two types of ions Examples: NaCl MgO CaCO3MgSO4Hard, brittle solidsHigh melting pointElectrical insulatorsexcept when molten or dissolved in water.These are lattices of ions…….
103 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?????
105 IONIC SOLIDSThe 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 ENERGYHow can we describe ionic lattices?
107 NaCl structureCl-Na+Lets look at the black dot lattice….
108 NaCl structure What unit cell do the black dots form? The black dots form a fcc lattice!Now look at the red dots
109 NaCl structure What unit cell is this???? But which one????? Cubic certainlyLets have another look……….
110 NaCl structure What unit cell is this???? Bring in a another array….. The red dots form a fcc array!Now put these back together…..
111 NaCl structure FCC OF BLACK DOTS Bring in red dots NOTICE THE RED DOTS FIT NICELY IN BETWEEN BLACK DOTSTHE RED DOTS SIT IN THE HOLES OF THE BLACK DOT FCC ARRAY
112 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-) latticeSO HOW WE DESCRIBE IONIC SOLIDS???
113 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????
114 HOLES IN A LATTICE. Holes??? What holes????? Lets look at a fcc lattice!
115 HOLES IN A FCC LATTICE The black dots form a FCC lattice! See the holes????
116 HOLES IN A FCC LATTICE The black dots form a fcc lattice! See the holes????HOW MANY HOLES??????
117 HOLES IN A FCC LATTICE The holes: How many?? THIRTEEN: ONE IN THE CENTRE12 on the edges.What shape is the hole ?
118 CENTRAL HOLE OCTAHEDRAL HOLES: There is one octahedral hole in the centre of the unit cell.If each one is occupied by an atom?
119 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) = 4There are 4 complete octahedral holes per fcc unit cell.
120 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..
121 THE OCTAHEDRAL HOLESBetween two layers…..Other holes…..
122 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
123 SMALL CUBE Take a point at the centre of this cube There are eight of these….
124 SMALL CUBE 8 CUBES Take a point at the centre of this cube Another one of these….
125 SMALL CUBE8 CUBESTake a point at the centre of this cubeAn so on ….
127 TETRAHEDRAL HOLESNotice 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 cellThis hole…….
128 one sphere in another layer sitting in the dimple they form. TETRAHEDRAL HOLESFormed by three spheres in one layer andone sphere in another layer sitting in the dimple they form.There is one more hole……….
129 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????
130 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??????
131 HOLE OCCUPANTS? Which hole will a cation occupy?????? They occupy the holes that result in maximum attraction and minimum repulsion.To do this…...
132 Which hole will a cation occupy?????? M+ or M2+ cations always occupy the holeswith the largest coordination number without rattling around!TIGHT FITConsequently the radius of the cation must begreater 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!
133 Which hole will a cation occupy?????? Investigate the size of these holes!The size of the hole depends upon thesize of the ion (usually anion) that forms the lattice into which the cations are to go……...OCTAHEDRAL HOLE IN FCC….LOOK AT HOLE….
134 OCTAHEDRAL HOLES IN FCC Look at planeDraw a square.Put in spheres.These are the anionsFit a small sphere in
135 OCTAHEDRAL HOLES IN FCC Look at planeDraw a square.Put in spheres.These are the anionsFit a small sphere inThis will be the cationDraw diagonalPut in distances……..
136 OCTAHEDRAL HOLES IN FCC Look at planeRadius of ion = RRadius of hole = rR2r
137 OCTAHEDRAL HOLES IN FCC Look at planeRadius of ion = RRadius of hole = r= (2R + 2r)2(2R)2+(2R)28R2 = (2R + 2r)2RR2rRR0.414R = r
138 OCTAHEDRAL HOLES IN FCC Look at planeRadius of ion = RRadius of hole = r0.414R = rThe size of cation that just fits has a radius that isR2r0.414 x radius of anion(R)roctahedral hole = RWhat about the tetrahedral hole?
139 fcc RADIUS RATIO: What about other cubic cell systems?? Using similar calculations, we can find the radius of other types of holes as well:fccrtetrahedral = RDO IT!!!!!!!roctahedral = Rr = 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 latticeand the radius of the ions forming the holeWhat about other cubic cell systems??
140 SIMPLE CUBICIf 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 = RanionDO IT!!!!!!!
141 The cubic hole 8 4! 6! The coordination number in the cubic hole is ? rcubic = RanionIn 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!
142 SUMMARY: Face centred cubic: Simple cubic: Trigonal hole Too small to be occupiedTetrahedral holeCN = 4rcation = 0.225Ranion8 of theseOctahedral holeCN = 6rcation = 0.414Ranion4 of theseSimple cubic:Cubic holeCN = 8rcation = 0.732Ranion1 of theseFor a given anionrtrigonal < rtetrahedral < roctahedral < rcubic
143 SUMMARY Face centred cubic: Too small to be occupied Trigonal hole Tetrahedral holeCN = 4rcation = 0.225Ranion8 of theseOctahedral holeCN = 6rcation = 0.414Ranion4 of theseSimple cubicCN = 8rcation = 0.732Ranion1 of theseFor a given anionrtrigonal < rtetrahedral < roctahedral < rcubicWhich hole will a cation occupy??????
144 INTO WHICH HOLE WILL THE ION GO?? TETRAHEDRALThe hole filled is tetrahedral if:rtetrahedral < rcation < roctahedral0.225Ranion < rcation < 0.414Ranion
145 INTO WHICH HOLE WILL THE ION GO?? OCTAHEDRALThe hole filled is octahedral if:roctahedral < rcation < rcubic0.414Ranion < rcation < 0.732Ranion
146 INTO WHICH HOLE WILL THE ION GO?? CUBICThe hole filled is cubic if:rcubic < rcation0.732Ranion < rcationLets look at these ideas in action…….
147 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=41pmRadius of the octahedral hole is x 181=75pmConsider a sc array of Cl- then:Radius of the cubic hole is x 181=132pmSo the best fit is the octahedral hole in the fcc array!The 98pm is bigger than 75pm but less than 132!OR USING RATIOS…….
148 NaCl Na+ has a radius of 98pm. Cl- has a radius of 181pm. 0.54 lies between and 0.732so the sodium cations will occupy octahedral holesin a fcc (ccp) latticeIs the stoichiometry ok???
149 4 4 So stoichiometry is ok!! NaCl 1:1 stoichiometry is required How many complete octahedral holes in face centred cubic array of Cl- ?????44How many Cl- needed to form the fcc array???Therefore 4 Cl- and 4 Na+So stoichiometry is ok!!Lets do another example…..
150 Example: Predict the structure of Li2S Li+ is 68 pmS2- is 190pmSTEP ONE:Examine the cation-anion radius ratios to find which type of holes the smaller ions fillCalculate ratio..COMPARE with ratios….Which is the best hole????TETRAHEDRAL!!!!
151 Example: Predict the structure of Li2S Li+ is 68 pmS2- is 190pmCalculate ratio..This requires tetrahedral holes.Which lattice has tetrahedral holes???face- centred cubic arrayThus the S2- will form a fcc lattice ...Lets look at the structure…...
152 FCC unit cell with tetrahedral holes ANIONCATIONFour 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 formulaSo????
153 Four anions in the unit cell. FCC unit cell with tetrahedral holesS2-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!
154 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…..
155 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….
156 the cesium cations will occupy cubic holes of a simple cubic lattice. Cs+ is 167 pmCl- is 181pmCalculate ratioCsCl:Compare…...0.92 is greater than 0.732the cesium cations will occupy cubic holes of a simple cubic lattice.STOICHIOMETRY?????
157 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.
159 ZnS: Zn2+ is 64 pm S2- is 190 pm Calculate ratio This requires tetrahedral holes.COMPAREThe sulfide ions will form a face-centered cubic array because….that is the only type to possess tetrahedral holes.What about stoichiometry??????
160 ZnS: Zn2+ is 64 pm S2- is 190 pm Calculate ratio This requires tetrahedral holes.COMPAREThe sulfide ions will form a face-centered cubic array because….that is the only type to possess tetrahedral holes.What about stoichiometry??????160160
161 We need an equal number of zinc and sulfide ions. There are the twice as many tetrahedral holes(8) as S2-(4) that form the fcc lattice.Therefore, half the tetrahedral holes will be filled.
162 We need an equal number of zinc and sulfide ions. Half the tetrahedral holes will be filled.ZnS is composed of a fcc lattice of sulfide anions with zinc cations in half the tetrahedral holes.
163 There are two forms of ZnS One is the zinc blende that we have talked about!This is an example of polymorphism.The other is wurtzite based on hcp lattice.
164 QUESTIONA crystal has Mg2+ ions at the corners of a cubic unit cell, F- ions at the midpoints of all the edges and K+ ions at the body centre. The empirical formula is1 KMgF32 K3MgF23 KMg2F24 K2Mg2F5 K2Mg2F3
165 ANSWERA crystal has Mg2+ ions at the corners of a cubic unit cell, F- ions at the midpoints of all the edges and K+ ions at the body centre. The empirical formula is1 KMgF32 K3MgF23 KMg2F24 K2Mg2F5 K2Mg2F3
166 QUESTIONA COMPOUND CONTAINS THE ELEMENTS Na,O and Cl. IT CRYSTALLISES IN A CUBIC LATTICE. THE O ATOMS ARE AT THE CORNERS, THE Cl ATOMS ARE AT THE CENTER AND THE Na ARE AT THE CENTRES OF THE FACES OF THE UNIT CELL. THE FORMULA OF THE COMPOUND IS...1 Na2ClO2 Na3ClO3 NaCl3O4 NaClO35 NaClO
167 QUESTIONA COMPOUNDCONTAINS THE ELEMENTS Na,O and Cl. IT CRYSTALLISES IN A CUBIC LATTICE. THEO ATOMS ARE AT THE CORNERS, THE Cl ATOMS ARE AT THE CENTER AND THE Na ARE AT THE CENTRES OF THE FACES OF THE UNIT CELL. THE FORMULA OF THE COMPOUND IS...1 Na2ClO2 Na3ClO3 NaCl3O4 NaClO35 NaClO
168 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???????
169 CaF2:Ca2+ is 99 pm F- is 136 pmCalculate ratioFOR TETRAHEDRAL HOLESOOPS!!!!The radius ratio is too BIG!!!!This shows Radius Ratios do not always work properlyBut CaF2 can be thought of as a simple cubic of F-with Ca2+ at alternate cubic holes!!!!!!!
170 Ca2+ in alternating cubic sites. CaF2:SIMPLE CUBICCa2+Ca2+Ca2+ in alternating cubic sites.Alternative descriptionWhat is Antiflourite????
171 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 holesand the anions form the fcc array.
172 Ionic Compound Density MgOfcc of O2-Mg2+ in octahedral holesCalculating the density of an ioniccompoundA face….Edge= roxide ion+ 2rMg ion+ roxide ionNow calculate volume4 Mg’s and 4 O2-REMEMBER….
173 Ionic Compound Density MgOfcc of O2-Mg2+ in octahedral holesCalculating the density of an ioniccompoundR = 86 pmr = 126 pmEdge = 424 pmV = (424)3 = 7.62X107 pm3A face….Edge= roxide ion+ 2rMg ion+ roxide ionNow calculate volume4 Mg’s and 4 O2-REMEMBER….173173
174 Ionic Compound Density MgOfcc of O2-Mg2+ in octahedral holesCalculating the density of an ioniccompoundR = 86 pmr = 126 pmEdge = 424 pmV = (424)3 = 7.62X107 pm3A face….Edge= roxide ion+ 2rMg ion+ roxide ionNow calculate volume4 Mg’s and 4 O2-REMEMBER….174174
175 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 MgOmole
176 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 MgOmolemole6.022 X 1023 FU
177 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 MgOmole4 FUmole6.022 X 1023 FUunit cell
178 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 MgOmole4 FUUnit Cellmole6.022 X 1023 FUunit cell7.62X107 pm3
179 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 MgOmole4 FUUnit Cellpm3mole6.022 X 1023 FUunit cell7.62X107 pm310-36m3
180 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 MgOmole4 FUUnit Cellpm310-6mmole6.022 X 1023 FUunit cell7.62X107 pm310-36m3cm3
181 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 MgOmole4 FUUnit Cellpm310-6mmole6.022 X 1023 FUunit cell7.62X107 pm310-36m3cm3= 3.54 g/cm3
183 Diamond and GraphiteThe p Orbitals (a) Perpendicular to the Plane of the Carbon Ring System in Graphite can Combine to Form (b) an Extensive pi Bonding Network10–183
184 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.
186 SCATTERING OF X-RAYS BY CRYSTALS In 1912 Knipping observed……..X-RAY DIFFRACTION PATTERNVon Laue gets Noble Prize…….How can we understand this???
187 X-ray DiffractionX-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.
189 BRAGG DIFFRACTION LAW W.H. Bragg and W.L.Bragg noticed that q2 qThis is reminiscent of reflection…..So they formulated diffraction in terms of reflection from planes of electron density in the crystal..
191 BRAGG DIFFRACTION LAW A plane of lattice points……. Now imagine reflection of X-rays……….
192 Bragg Equation Derivation өөөdxxsin ө = x/d x = d sin өWave length λ = 2x λ = 2d sin өnλ = 2d sin өn due to multiple layers of particles
193 BRAGG’S LAWOnly at certain angles of ө will the waves from different planes be in phase, thus nλ = 2dsinөBy adjusting the angle of the x-rays until constructive interference is obtained, distance (d) between atoms is obtained
194 BRAGG DIFFRACTION LAW The Braggs also demonstrated diffraction…. And formulated a diffraction law…...When electromagnetic radiation passes through matter…….It interacts with the electrons andIs scattered in all directionsthe waves interfere……..
195 The Band Theory (MO theory) Review the Li MO diagramMany vacant MO’sIn fact only sigma is filledThis is for two atomsNow how about four atoms more MO’sHow about a mole of atoms, tons of MO’sFor magnesium, which is HCC, look at the bandsThe lower band holds electrons, but the next highest vacant MO is just a small energy jump awayElectrons do not flow in the lower band since they bump into each otherBut a slight amount of energy promotes them to the conduction band where they flow freely
198 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 filledWhy then do nonmetals not conductThere is a large energy difference between conduction and non conduction bandThere are more valence electrons
200 Molecular Orbital Energies Insulator (diamond)Conductor(metal)10–200
201 Semi ConductorsFor 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.
202 Semi ConductorsAt 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)
203 Semi ConductorsN-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 conductionThis is called an n-type due to extra negative chargeConductivity can be enhanced by an element such as boron that has one less valence electron than siliconThese are called P-semiconductorsSince we are missing an electron then there is a hole, which an electron fills thus creating another holeHoles flow in a direction opposite to the flow of electrons, since lower lying electrons are promoted to fill the holeCalled p for positive charge, due to one less electron
205 Semi ConductorsImportant application is to combine an n-type and a p-type together, called a p-n junctionWhen they are connected some of the electrons from the n-type flow into the open holes of the p-type, thus creating a charge differenceOnce the charge difference is achieved then electron flow ceases, this is called contact potential, or junction potentialIf an external voltage is applied then electrons will only flow in one wayFrom the n-type to the p-typeThe holes flow in the opposite directionP-n-junctions makes an excellent rectifier, a device that produces a pulsating direct current from an alternating currentThe overall effect is to convert alternating current into direct currentOld rectifiers were vacuum tubes, which were not very reliable
206 No current flows, called reverse bias Semi ConductorsSome electrons flow to create opposite chargesNo current flows, called reverse biasCurrent flows, called forward bias