APS104S – Introduction to Materials & Chemistry Lecture 4 January 12, 2016 Instructor: Gisele Azimi 1) Bonding energy 2) Ionic bonding (Madelung constant) Bonding energy, attractive, repulsive - A + B EN = EA + ER = n r r A3rac5ve Repulsive energy energy A, B, n = constants At r = r0 ! dE/dr = 0 & E = E0 r0 = equilibrium separaon distance (bond distance) E0 = Bond energy r0 = rcaon + ranion Example 2 • The potential energy between Mg2+ and O2- ions in a MgO crystal can be estimated as • where A = 1.61 x 10-27 J m, B = 3.87×10-96 J m8, r is the separation (between ion centers) in m, and E is the potential energy in J. • Calculate the equilibrium separation between the ion centers. • Calculate the bonding energy between the Mg and O ions at this equilibrium distance. • If the ionic radius of Mg2+ is 0.072 nm, what is the ionic radius of O2- ? Example 2-solution Ionic. Bonding (example NaCl) Bonding is described as electrostatic attraction between point charges. Na+ Cl- Ionic. Bonding (example NaCl) 6 nearest neighbours @ r = a (+ -) Let’s call this “a” a = r0 Ionic. Bonding (example NaCl) 12 nearest neighbours @ r = √2 a (+ +) √2 a Let’s call this “a” a = r0 Ionic. Bonding (example NaCl) 8 nearest neighbours @ r = √3 a (+ -) √3 a !! 8 ! = − ( ) 4!!! 3! Let’s call this “a” a = r0 Ionic. Bonding (example NaCl) e2 " 6 12 8 % E = − $ − + −!' 4πεo # a a 2 a 3 & e2 1 E = −M 4πεo a 12 8 Let’s call this “a” M = 6 − + −... M: Madelung constant a = r0 2 3 M depends on the geometry; for rock salt = 1.748 1 eV = 1.60 x 10-19 J e = 1.602 × 10-19 C a = r0 = 0.28 nm -12 2 2 ε0 = 8.854 x 10 C /(Nm ) E = -8.97 eV/atom (only 10% of the real value) Madelung. constant for CsCl 8 nearest neighbours @ r = √3/2 a (+ -) √3/2 a Let’s call this “a” Madelung. constant for CsCl 6 nearest neighbours @ r = a (+ +) Madelung. constant for CsCl e2 1 E = −M 4πεo a Let’s call this “a” the crystalline solid state CHAPTER 3_C topics: unit cell metallic crystal structures density computaons metallic/ceramic ceramic crystal structures direc/ons planes & densies 5.2 & 5.3 Light & the Electromagne/c Spectrum 5.4 The Plank Equa/on 5.5 The de Broglie Equaon 5.6 Heisenberg Uncertainty Principle 5.7 & 5.10 Quantum Numbers 5.8 Shapes of Orbitals 5.9 Quantum Mechanics & Atomic Spectra 5.11-14 Mul/electron Atoms 5.15 Periodic Properes Subjects we will study • What are crystals • Crystal systems • Crystal structure in metals (with a focus on cubic systems) Why study crystal structure? Many properties of materials are directly related to their crystallinty and crystal structure Crystallographic - points Crystalline Crystal structure Unit cell - Direcons - planes Non-crystalline Classification of solids by atomic arrangement " Has a well-defined ordered structure in 3D (over large range). Atoms self-organize in a periodic Dense, array. ORDERED packing " Well defined shaped due to well- ordered par5cles. • metals • many ceramics • some polymers " Has a disordered structure, lacking a systemac atomic arrangement. " Poorly defined shapes because they don’t have extensive ordering of Non dense, par5cles. RANDOM packing • complex structures • rapid cooling Some crystals (well-defined ordered structure; well-defined shape) Au Cu Cr PbS FeS2 NiSO4 NaCl Some amorphous solids (disordered structure, poorly defined shapes) Gummite amorphous solid Obsidian, a volcanic glass Gummite: a yellow, amorphous mineral, a mixture of oxides and silicates of uranium, formed by weathering of pitchblende (a mineral used as a source of uranium). Stacking objects in 2-D with no voids square rectangle NO trapezoid Unit cell: the repeating group – 7 ways to stack crystals in 3D with no voids Unit cell: the repeating group A unit cell is the most basic and least volume consuming repeang structure of any solid. Unit cell: where atoms sit: Bravais Lattices Let us look at “cubic” crystal structure Note: for metals, all atoms are iden5cal; the FCC and BCC atoms are shaded 1 differently only for ease of viewing. ½ ¼ An ion in the body of a cell belongs en5rely to that cell and counts as 1. An ion on a face is shared by two cells and contributes ½. An ion on an edge is shared by four cells and hence contributes ¼. ⅛ An ion at a vertex is shared by eight cells, and so contributes ⅛. Crystal structure in metals Metallic bonds (such as Au, Ag, Cu) Can be pictured as an array of positive ions in a “sea of electrons". Cu (FCC) Tend to be densely packed. REASONS FOR DENSE PACKING - Typically, only one element is present, so all atomic radii are the same. - Nearest neighbor distances tend to be small in order to lower bond energy. - Electron cloud shields cores from each other. - Metallic bonding is not directional • Have the simplest crystal structures. Lattice points per cell: how many atoms we have in each unit cell 2 4 1 The relation between unit cell length (a) and radius of atom (r) BCC SC FCC r a 2a = 4r 3a= 4r a = 2r Coordination #: the number of nearest neighbors SC 6 BCC 8 FCC 12 Atomic packing factor (APF) volume of atoms in unit cell APF = volume of unit cell Example 1: for FCC APF=(4x(4/3)Πr3/a3 4r = √2 x a APF=0.74 a= unit cell length r= atoms radii Example: Atomic packing factor (APF) for BCC volume of atoms in unit cell APF = volume of unit cell 3 a 4r = 3 a r = a/4 3 a 2 a R Volume of each atom a atoms / unit cell 2 (4/3)π(r)3 APF = a3 volume / unit cell APF = 0.68 Summary Face centered cubic (fcc) Body centered cubic (bcc) Simple Cubic FCC BCC Al, Cu Cr, Fe, Au, Pt Mo r = radius of atom a = length of unit cell FCC BCC SC # of atoms in cell 4 2 1 Coordina>on # 12 8 6 Filling Fracon 74 % 68% 52% Rela>on 2a = 4r 3a= 4r a = 2r Calculating theoretical density of metals Example: Calculate the density of iron, which has a body-centered cubic unit cell with an edge length of 286.6 pm. The molar mass of iron is 55.85 g/ mol. Measured (literature) value for iron density: 7.874 g/cm3 Calculating theoretical density of metals Theoretical density formula atomic mass (g / mol) #atoms / unit cell n A ρ = VC NA volume / unit cell Avogadro’s number 23 (cm3 / unit cell) 6.02 × 10 atoms / mol Characteristics of selected elements at 20°C At. Weight Density Atomic radius Element Symbol (amu) (g/cm3) (nm) Aluminum Al 26.98 2.71 0.143 Argon Ar 39.95 ------ ------ Barium Ba 137.33 3.5 0.217 Beryllium Be 9.012 1.85 0.114 Boron B 10.81 2.34 ------ Bromine Br 79.90 ------ ------ Cadmium Cd 112.41 8.65 0.149 Calcium Ca 40.08 1.55 0.197 Carbon C 12.011 2.25 0.071 Cesium Cs 132.91 1.87 0.265 Chlorine Cl 35.45 ------ ------ Chromium Cr 52.00 7.19 0.125 Cobalt Co 58.93 8.9 0.125 Copper Cu 63.55 8.94 0.128 Flourine F 19.00 ------ ------ Gallium Ga 69.72 5.90 0.122 Germanium Ge 72.59 5.32 0.122 Gold Au 196.97 19.32 0.144 Helium He 4.003 ------ ------ Hydrogen H 1.008 ------ ------ Example Nickel crystallizes in a fcc lattice. Its density is 8.90 g/cm3. What is the atomic radius of nickel in pm? 1 m = 1012 pm 1 cm = 1010 pm Example - Solution .