CH3. Intro to Solids Lattice geometries Common structures Lattice energies Born-Haber model Thermodynamic effects Electronic structure . C. A. B. Stacked 2D hexagonal arrays. Packing efficiency.

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CH3. Intro to Solids Lattice geometries Common structures Lattice energies Born-Haber model Thermodynamic effects

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Packing efficiency • It can be easily shown that all close-packed arrays have a packing efficiency (Vocc/Vtot) of 0.74 • This is the highest possible value for same-sized spheres, though this is hard to prove “…And suppose…that there were one form, which we will call ice-nine - a crystal as hard as this desk - with a melting point of, let us say, one-hundred degrees Fahrenheit, or, better still, … one-hundred-and-thirty degrees.” Kurt Vonnegut, Jr. Cat’s Cradle

Some alloys AlloyComposition Cu, Ni any Cu and Ni are ccp, r(Cu) = 1.28, r(Ni) = 1.25 Å Cast iron Fe, C (2+ %), Mn, Si r(Fe) = 1.26, r(C) = 0.77 Stainless Steels Fe, Cr, Ni, C … Brass CuZn (b) = bcc r(Zn) = 1.37, hcp substitutional interstitial intermetallic

Chemical Composition % (Max unless noted) Stainless C Mn P S Si Cr Ni Mo N 410 0.15 1.00 0.040 0.030 0.500 11.50-13.00 430 0.12 1.00 0.040 0.030 1.000 16.00-18.00 0.75 304 0.08 2.00 0.045 0.030 1.000 18.00-20.00 8.00-10.50 316 0.08 2.00 0.045 0.030 1.000 16.00-18.00 10.00-14.00 2.00-3.00 2205 0.02 2.00 0.045 0.030 1.000 22.00-23.00 5.50-6.00 3.00-3.50 0.17 A few stainless steels

Cl Na A c B a C b NaCl (rocksalt) • fcc anion array with all Oh sites filled by cations • the stoichiometry is 1:1 (AB compound) • CN = 6,6 • Look down the body diagonal to see 2D hex arrays in the sequence (AcBaCb)n • The sequence shows coordination, for example the c layer in AcB Oh coordination

CaC2 Tetragonal distortion of rocksalt structure (a = b ≠ c) Complex anion also decreases (lowers) symmetry

Antifluorite / Fluorite • Antifluorite is an fcc anion array with cations filling all Td sites • 8 Td sites / unit cell and 4 spheres, so this must be an A2B-type salt. • Stacking sequence is (AabBbcCca)n • CN = 4,8. Anion coordination is cubic. • Fluorite structure reverses cation and anion positions. An example is the mineral fluorite CaF2

c C b B a A Sphalerite (ZnS) • fcc anion array with cations filling ½ Td sites • Td sites are filled as shown • Look down body diagonal of the cube to see the sequence (AaBbCc)n… • If all atoms were C, this is diamond structure.

NiAs • hcp anion array with cations filling all Oh sites • cation layers all eclipsing one another • stacking sequence is (AcBc)n • CN = 6,6 • AcB and BcA gives Oh cation coordination, but cBc and cAc gives trigonal prismatic (D3h) anion coordination

CdI2 • hcp anion array with cations filling ½ Oh sites in alternating layers • Similar to NiAs, but leave out every other cation layer • stacking sequence is (AcB)n • CN = (6, 3) • anisotropic structure, strong bonding within AcB layers, weak bonding between layers • the layers are made from edge-sharing CdI6 octahedra

Rutile (TiO2) • hcp anion array with cations filling ½ Oh sites in alternating rows • the filled cation rows are staggered • CN = 6, 3 • the filled rows form chains of edge-sharing octahedra. These chains are not connected within one layer, but are connected by the row of octahedra in the layers above and below. • Lattice symmetry is tetragonal due to the arrangement of cations.

Wurtzite (ZnS) • hcp anion array with cations filling ½ Td sites • Stacking sequence = (AaBb)n • CN = 4, 4 • wurtzite and sphalerite are closely related structures, except that the basic arrays are hcp and ccp, respectively. • Many compounds can be formed in either structure type: ZnS, has two common allotropes, sphalerite and wurtzite

ReO3 • Re is Oh, each O is shared between 2 Re, so there are ½ * 6 = 3 O per Re, overall stoichiometry is thus ReO3 • Neither ion forms a close-packed array. The oxygens fill 3/4 of the positions for fcc (compare with NaCl structure). • The structure has ReO6 octahedra sharing all vertices.

An ordered AA’BX3 perovskite Perovskite (CaTiO3) • Similar to ReO3, with a cation (CN = 12) at the unit cell center. • Simple perovskites have an ABX3 stoichiometry. A cations and X anions, combined, form a close-packed array, with B cations filling 1/4 of the Oh sites.

Superconducting copper oxides • Many superconducting copper oxides have structures based on the perovskite lattice. An example is: • YBa2Cu3O7. In this structure, the perovskite lattice has ordered layers of Y and Ba cations. The idealized stoichiometry has 9 oxygens, the anion vacancies are located mainly in the Y plane, leading to a tetragonal distortion and anisotropic (layered) character.

Charged spheres Assumes a uniform charge distribution (unpolarizable ions). With softer ions, higher order terms (d-2, d-3, ...) can be included. For 2 spherical ions in contact, the electrostatic interaction energy is: Eel = (e2 / 4 pe0) (ZA ZB / d) e = e- charge = 1.602 x 10-19 C e0 = vac. permittivity = 8.854 x 10-12 C2J-1m-1 ZA = charge on ion A ZB = charge on ion B d = separation of ion centers

Infinite linear chains • Consider an infinite linear chain of alternating cations and anions with charges +e or –e • The electrostatic terms are: Eel = (e2/4pe0)(ZAZB/ d) [2(1) - 2(1/2) + 2(1/3) - 2(1/4) +…] = (e2/4pe0)(ZAZB/d) (2 ln2)

Madelung constants Generalizing the equation for 3D ionic solids, we have: Eel = (e2 / 4pe0) * (ZA ZB / d) * A where A is called the Madelung constant and is determined by the lattice geometry

Some values for A and A / n: lattice A CN stoich A / n CsCl 1.763 (8,8) AB 0.882 NaCl 1.748 (6,6) AB 0.874 sphalerite 1.638 (4,4) AB 0.819 wurtzite 1.641 (4,4) AB 0.821 fluorite 2.519 (8,4) AB2 0.840 rutile 2.408 (6,3) AB2 0.803 Madelung constants

Born-Meyer model • Electrostatic forces are net attractive, so d → 0 (the lattice collapse to a point) without a repulsive term • Add a pseudo hard-shell repulsion: C‘ e-d/d* where C' and d* are scaling factors (d* has been empirically fit as 0.345 Å) • Vrep mimics a step function for hard sphere compression (0 where d > hard sphere radius, very large where d < radius)

Note sign conventions !!! E = -DHL = (e2/4pe0) (NAZAZB/d0) (1 - d*/d0) (Born-Meyer equation) Born-Meyer eqn • The total interaction energy, E: E = Eel + Erepulsive = (e2 / 4pe0)(NAZAZB /d) + NC'e-d/d* • Since E has a single minimum d, set dE/dd = 0 and solve for C‘:

Further refinements • Eel’ include higher order terms • Evdw NC’’r-6 instantaneous polarization • EZPE Nhnolattice vibrations For NaCl: Etotal = Eel’+ Erep + Evdw + EZPE -859 + 99 - 12 + 7 kJ/mol

Kapustinskii approximation: • The ratio A/n is approximately constant, where n is the number of ions per formula unit (n is 2 for an AB - type salt, 3 for an AB2 or A2B - type salt, ...) • Substitute the average value into the B-M eqn, combine constants, to get the Kapustinskii equation: DHL = -1210 kJÅ/mol (nZAZB / d0) (1 - d*/d0) with d0 in Å

Kapustinskii eqn • Using theaverage A / n value decreases the accuracy of calculated E’s. Use only when lattice structure is unknown. • DHL (ZA,ZB,n,d0). The first 3 of these parameters are given from in the formula unit, the only other required info is d0. • d0 can be estimated for unknown structures by summing tabulated cation and anion radii. The ionic radii depend on both charge and CN.

Example: Use the Kapustiskii eqn to estimate DHL for MgCl2 • ZA = +2, ZB = -1, n = 3 • r(Mg2+) CN 8 = 1.03 Å r(Cl-) CN 6 = 1.67 Å • d0 ≈ r+ + r- ≈ 2.7 Å • DHL(Kap calc) = 2350 kJ/mol • DHL(best calc) = 2326 • DHL(B-H value) = 2526

Unit cell volume relation • Note that d*/d0 is a small term for most salts, so (1 - d*/d0) ≈ 1, • Then for a series of salts with the same ionic charges and formula units: DHL ≈ 1 / d0 • For cubic structures: DHL ≈ 1 / V1/3 where V is the unit cell volume

DHLvs V-1/3 for cubic lattices V1/3 is proportional to lattice E for cubic structures. V is easily obtained by powder diffraction.

Born – Haber cycle ½ D0 DHf {KCl(s)} = DH {K(s) + ½Cl2(g) → KCl(s)} Ea I DHf {KCl(s)} = DHsub(K) + I(K) + ½ D0(Cl2) – Ea(Cl) - DHL -DHL DHsub All enthalpies are measurable except DHL Solve to get DHL(B-H) -DHf

Is MgCl3 stable ? DHf = DHat,Mg + 3/2 D0(Cl2) + I(1)Mg + I(2)Mg + I(3)Mg - 3 Ea(Cl) - DHL = 151 + 3/2 (240) + 737 + 1451 + 7733 - 3 (350) - 5200 ≈ + 4000 kJ/mol • DHL is from the Kapustinskii eqn, using d0 from MgCl2 • The large positive DHf means it is not stable. • I(3) is very large, there are no known stable compounds containing Mg3+. Energies required to remove core electrons are not compensated by other energy terms.

Entropic contributions DG = DH - TDS Example: Mg(s) + Cl2(g) → MgCl2(s) • DS sign is usually obvious from phase changes. DS is negative (unfavorable) here due to conversion of gaseous reactant into solid product. • Using tabulated values for molar entropies: DS0rxn = DS0(MgCl2(s)) - DS0(Mg(s)) - DS0(Cl2(g)) = 89.6 - 32.7 - 223.0 = -166 J/Kmol -TDS at 300 K ≈ + 50 ; at 600 K ≈ +100 kJ/mol • Compare with DHf {MgCl2(s)} = -640 kJ/mol • DS term is usually a corrective term at moderate temperatures. At high T it can dominate.

Thermochemical Radii • What are the radii of polyatomic ions ? (Ex: CO32-, SO42-, PF6-, B(C6H6)-, N(Et)4+) • If DHL is known from B-H cycle, use B-M or Kap eqn to determine d0. • If one ion is not complex, the complex ion “radius” can be calculated from: d0 = rcation + ranion • Tabulated thermochemical radii are averages from several salts containing the complex ion. • This method can be especially useful when for ions with unknown structure, or low symmetry.