# CH3. Intro to Solids Lattice geometries Common structures

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

Stacked 2D hexagonal arrays
B C

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

Close-packing of polymer microspheres

hcp vs ccp (AB)n hcp (ABC)n ccp Also close-packed: (ABAC)n (ABCB)n
Not close packed: (AAB)n (ABA)n Why not ? (ACB)n

Unit cells for hcp and fcc
Unit cells, replicated and translated, will generate the full lattice Hexagonal cell = hcp Cubic cell ccp = fcc

Generating lattices lattice point CN Distance from origin (a units)
(½,½,0) 12 0.71 (1,0,0) 6 1.00 (½,½,1) 24 1.22 (1,1,0) 1.41 (3/2,½,0) 1.58 Etc…

Oh and Td sites in ccp rOh: a = 2rs + 2rOh a / √ 2 = 2rs
fcc lattice showing some Oh and Td sites rOh: a = 2rs + 2rOh a / √ 2 = 2rs rOh / rs = 0.414 4 spheres / cell 4 Oh sites / cell 8 Td sites / cell

Ionic radii are related to coordination number

Element Structures at STP
(ABCB)n

Ti phase transitions RT → 882°C hcp 882 → 1667° bcc
→ ° liquid → gas

Classes of Alloys Substitutional Interstitial intermetallic

Some alloys Alloy Composition 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

A few stainless steels Chemical Composition % (Max unless noted)
Mn P S Si Cr Ni Mo N 410 0.15 1.00 0.040 0.030 0.500 430 0.12 1.000 0.75 304 0.08 2.00 0.045 316 2205 0.02 0.17

Zintl phases KGe

NaCl (rocksalt) Cl Na A c B a C b
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 Cl Na B C A a b c

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

Other fcc anion arrays

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

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. B C A a b c

Sphalerite

Semiconductor lattices based on diamond / sphalerite
Group 14: C, Si, Ge, a-Sn, SiC 3-5 structures: cubic-BN, AlN, AlP, GaAs, InP, InAs, InSb, GaP,… 2-6 structures: BeS, ZnS, ZnSe, CdS, CdSe, HgS… 1-7 structures: CuCl, AgI

Structure Maps incr. radius, polarizability more ionic more covalent

Lattices with hcp anion arrays

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

LiTiS2 (AcBc’)n Ti S Li

LDH structures Mg(OH)2 (brucite) MgxAl1-x(OH)2.An

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.

Rutile TiO2-x and SiO2

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.

Perovskite (CaTiO3) An ordered AA’BX3 perovskite
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 p e0) (ZA ZB / d) e = e- charge = x C e0 = vac. permittivity = x 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) (1/4) +…] = (e2/4pe0)(ZAZB/d) (2 ln2)

Madelung constants Generalizing the equation for 3D ionic solids, we have: Eel = (e2 / 4 p e0) * (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:
Madelung constants 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) 0.874 sphalerite 1.638 (4,4) 0.819 wurtzite 1.641 0.821 fluorite 2.519 (8,4) AB2 0.840 rutile 2.408 (6,3) 0.803

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 Å) Vrep mimics a step function for hard sphere compression (0 where d > hard sphere radius, very large where d < radius)

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‘: E = -DHL = (e2/4pe0) (NAZAZB/d0) (1 - d*/d0) (Born-Meyer equation) Note sign conventions !!!

Further refinements Eel’ include higher order terms
Evdw NC’’r-6 instantaneous polarization EZPE Nhno lattice vibrations For NaCl: Etotal = Eel’ + Erep + Evdw + EZPE 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 = kJÅ/mol (nZAZB / d0) (1 - d*/d0) with d0 in Å

Kapustinskii eqn Using the average 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 Solve to get DHL(B-H) ½ D0 Ea I -DHL DHsub
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 = /2 (240) (350) ≈ 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)) = = 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.

(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.

Thermochemical Radii Example: DHL(BH) for Cs2SO4 is 1658 kJ/mol
Use the Kap eqn: DHL = 1658 = 1210(6/d0)( /d0) solve for d0 = 4.00 Å Look up r (Cs+) = 1.67 Å r (SO42-) ≈ = 2.33 Å The tabulated value is 2.30 Å (an avg for several salts)

Predictive applications
O2 (g) + PtF6 (l) → O2PtF6 (s) Neil Bartlett (1960); side-reaction in preparing PtF6 Ea(PtF6) = 787 kJ/mol. Compare Ea(F) = 328 I(Xe) ≈ I(O2), so Xe+PtF6-(s) may be stable if DHL is similar. Bartlett reported the first noble gas compound in 1962. O2(g) → O2+(g) + e kJ/mol e- + PtF6(g) → PtF6-(g) O2+(g) + PtF6-(g) → O2PtF6(s) * O2(g) + PtF6(g) → O2PtF6(s) ≈ * Estimated from the Kap eqn

Some consequences of DHL
Ion exchange / displacement Thermal / redox stabilities Solubilities

Exchange / Displacement
Large ion salt + small ion salt is better than two salts with large and small ions combined. Example: Salt DHL sum CsF 750 NaI kJ/mol CsI 620 NaF This can help predict some reactions like displacements, ion exchange, thermal stability.

Thermal stability of metal carbonates
An important industrial reaction involves the thermolysis of metal carbonates to form metal oxides according to: MCO3 (s) → MO (s) + CO2 (g) DG must be negative for the reaction to proceed. At the lowest reaction temp: DG = 0 and Tmin = DH / DS DS is positive because gas is liberated. As T increases, DG becomes more negative (i.e. the reaction becomes more favorable). DS depends mainly on DS0{CO2(g)} and is almost independent of M.

Thermal stability of metal carbonates
MCO3 (s) → MO (s) + CO2 (g) Tmin almost directly proportional to DH. DHL favors formation of the oxide (smaller anion) for smaller cations. So Tmin for carbonates should increase with cation size.

Solubility MX (s) --> M + (aq) + X - (aq)
DS is positive, so a negative DH is not always required for a spontaneous rxn. But DH is usually related to solubility. Use a B-H analysis to evaluate the energy terms that contribute to dissolution: MX(s) → M+(g) + X-(g) DHlat M+(g) + n L → ML'n+(aq) DHsolv, M X-(g) + m L → XL'm-(aq) DHsolv, X L'n + L'n → (n + m) L DH L-L MX(s) → M+(aq) + X-(aq) DHsolution, MX Driving force for dissolution is ion solvation, but this must compensate for the loss of lattice enthalpy. LiClO4 and LiSO3CF3 deliquesce (absorb water from air and dissolve) due to dominance of DHsolv

Solubility The energy balance favors solvation for large-small ion combinations, salts of ions with similar sizes are often less soluble.

Solubility Some aqueous solubilities at 25°C: DHsolution solubility
salt (kJ/mol) (g /100 g H2O) LiF LiCl LiI MgF MgO DHL terms dominate when ions have higher charges; these salts are usually less soluble.

Orbitals and Bands

Band and DOS diagrams

s vs T

Intrinsic Semiconductors
s = n q m s = conductivity n = carrier density q = carrier charge m = carrier mobility P = electron population ≈ e-(Eg)/2kT Eg C 5.5 eV Si 1.1 eV Ge 0.7 eV GaAs 1.4 eV

Bandgap vs Dc

Arrhenius relation Arrhenius relation: s = s0 e-Eg/2kT ln s 1 / T
Slope = -Eg/2k

Extrinsic Semiconductors
n-type p-type n-type example: P-doped Si p-type example: B-doped Si