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

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A

Stacked 2D hexagonal arrays

BC

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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

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Close-packing of polymer microspheres

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hcp vs ccp

Also close-packed: (ABAC)n (ABCB)n

Not close packed: (AAB)n (ABA)n

Why not ? (ACB)n

(AB)n hcp (ABC)n ccp

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Unit cells for hcp and fcc

Hexagonal cell = hcp

Cubic cell

ccp = fcc

Unit cells, replicated and translated, will generate the full lattice

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Generating latticeslattice 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) 12 1.41

(3/2,½,0) 24 1.58

Etc…

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Oh and Td sites in ccp

rOh:a = 2rs + 2rOh

a / √ 2 = 2rs rOh / rs = 0.414

4 spheres / cell

4 Oh sites / cell

8 Td sites / cell

fcc lattice showing some Oh and Td sites

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Ionic radii are related to coordination number

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Element Structures at STP

(ABCB)n

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Ti phase transitions

RT → 882°C hcp 882 → 1667° bcc1667 → 3285° liquid3285 → gas

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Classes of Alloys

(a) Substitutional(b) Interstitial(c) intermetallic

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Some alloys

Alloy Composition Cu, Ni anyCu and Ni are ccp, r(Cu) = 1.28, r(Ni) = 1.25 ÅCast iron Fe, C (2+ %), Mn, Si r(Fe) = 1.26, r(C) = 0.77Stainless Steels Fe, Cr, Ni, C … Brass CuZn () = bcc r(Zn) = 1.37, hcp

substitutional

interstitial

intermetallic

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A few stainless steels

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

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Zintl phases

KGe

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NaCl (rocksalt)

Cl

Na

B

C

A

a

b

c

– 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

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CaC2

Tetragonal distortion of rocksalt structure (a = b ≠ c)

Complex anion also decreases (lowers) symmetry

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Other fcc anion arrays

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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

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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.

BC

Aa

b

c

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Sphalerite

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Semiconductor lattices based on diamond / sphalerite

• Group 14: C, Si, Ge, 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

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Structure Maps

more covalent

more ionic

incr

. rad

ius,

pol

ariz

abili

ty

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Lattices with hcp anion arrays

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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

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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

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LiTiS2

(AcBc’)n

Ti

S

Li

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LDH structures

Mg(OH)2 (brucite)

MgxAl1-x(OH)2.An

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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.

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Rutile

TiO2-x and SiO2

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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

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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.

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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.

An ordered AA’BX3 perovskite

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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.

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Charged spheres

For 2 spherical ions in contact, the electrostatic interaction energy is:

Eel = (e2 / 4 0) (ZA ZB / d)

e = e- charge = 1.602 x 10-19 C

0 = 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

Assumes a uniform charge distribution (unpolarizable ions). With softer ions, higher order terms (d-2, d-3, ...) can be included.

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Consider an infinite linear chain of alternating cations and anions with charges +e or –e

The electrostatic terms are:Eel = (e2/40)(ZAZB/ d) [2(1) - 2(1/2) + 2(1/3) - 2(1/4) +…]

= (e2/40)(ZAZB/d) (2 ln2)

Infinite linear chains

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Madelung constants

Generalizing the equation for 3D ionic solids, we have:

Eel = (e2 / 4 0) * (ZA ZB / d) * A

where A is called the Madelung constant and is determined by the lattice geometry

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Madelung constantsSome values for A and A / n:

lattice A CN stoich A / nCsCl 1.763 (8,8) AB 0.882NaCl 1.748 (6,6) AB 0.874

sphalerite 1.638 (4,4) AB 0.819wurtzite 1.641 (4,4) AB 0.821fluorite 2.519 (8,4) AB2

0.840

rutile 2.408 (6,3) AB20.803

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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)

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Born-Meyer eqn The total interaction energy, E: E = Eel + Erepulsive

= (e2 / 40)(NAZAZB /d) + NC'e-d/d*

Since E has a single minimum d, set dE/dd = 0 and solve for C‘:

E = -HL = (e2/40) (NAZAZB/d0) (1 - d*/d0)

(Born-Meyer equation)

Note sign conventions !!!

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Further refinements

• Eel’ include higher order terms

• Evdw NC’’r-6 instantaneous polarization

• EZPE Nhlattice vibrations

For NaCl:Etotal = Eel

’ + Erep + Evdw + EZPE

-859 + 99 - 12 + 7 kJ/mol

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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:

HL = -1210 kJÅ/mol (nZAZB / d0) (1 - d*/d0)

with d0 in Å

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Kapustinskii eqn Using the average A / n value decreases the

accuracy of calculated E’s. Use only when lattice structure is unknown.

HL (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.

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Example:

Use the Kapustiskii eqn to estimate HL for MgCl2

1. ZA = +2, ZB = -1, n = 3

2. r(Mg2+) CN 8 = 1.03 Å r(Cl-) CN 6 = 1.67 Å3. d0 ≈ r+ + r- ≈ 2.7 Å

4. HL(Kap calc) = 2350 kJ/mol

5. HL(best calc) = 2326

6. HL(B-H value) = 2526

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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:HL ≈ 1 / d0

For cubic structures:HL ≈ 1 / V1/3

where V is the unit cell volume

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HLvs V-1/3 for cubic lattices

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

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Born – Haber cycle

Hf {KCl(s)} = Hsub(K) + I(K) + ½ D0(Cl2) – Ea(Cl) - HL

All enthalpies are measurable except HL

Solve to get HL(B-H)

HL

Ea

Hsub

½ D0

I

Hf

Hf {KCl(s)} = H {K(s) + ½Cl2(g) → KCl(s)}

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Is MgCl3 stable ?

HL is from the Kapustinskii eqn, using d0 from MgCl2

The large positive Hf 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.

Hf = Hat,Mg + 3/2 D0(Cl2) + I(1)Mg + I(2)Mg + I(3)Mg - 3 Ea(Cl) - HL

= 151 + 3/2 (240) + 737 + 1451 + 7733 - 3 (350) - 5200

≈ + 4000 kJ/mol

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Entropic contributionsG = H - TS Example: Mg(s) + Cl2(g) → MgCl2(s) S sign is usually obvious from phase changes. S is

negative (unfavorable) here due to conversion of gaseous reactant into solid product.

Using tabulated values for molar entropies:S0

rxn = S0(MgCl2(s)) - S0(Mg(s)) - S0(Cl2(g)) = 89.6 - 32.7 - 223.0

= -166 J/Kmol-TS at 300 K ≈ + 50 ; at 600 K ≈ +100 kJ/mol Compare with Hf {MgCl2(s)} = -640 kJ/mol S term is usually a corrective term at moderate

temperatures. At high T it can dominate.

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Thermochemical Radii What are the radii of polyatomic ions ? (Ex: CO3

2-, SO42-, PF6

-, B(C6H6)-, N(Et)4+)

If HL 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.

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Thermochemical RadiiExample:

HL(BH) for Cs2SO4 is 1658 kJ/molUse the Kap eqn:HL = 1658 = 1210(6/d0)(1-0.345/d0) solve for d0 = 4.00 ÅLook up r (Cs+) = 1.67 Å

r (SO42-) ≈ 4.00 - 1.67 = 2.33 Å

The tabulated value is 2.30 Å (anavg for several salts)

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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

HL is similar. Bartlett reported the first noble gas compound in 1962.

O2(g) → O2+(g) + e- + 1164 kJ/mol

e- + PtF6(g) → PtF6-(g) - 787

O2+(g) + PtF6

-(g) → O2PtF6(s) - 470*O2(g) + PtF6(g) → O2PtF6(s) ≈ - 93

* Estimated from the Kap eqn

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Some consequences of HL

• Ion exchange / displacement• Thermal / redox stabilities• Solubilities

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Exchange / Displacement Large ion salt + small ion salt is better than

two salts with large and small ions combined.Example: Salt HL sum

CsF 750NaI 705 1455 kJ/molCsI 620NaF 926 1546

This can help predict some reactions like displacements, ion exchange, thermal stability.

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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)

G must be negative for the reaction to proceed. At the lowest reaction temp:

G = 0 and Tmin = H / S S is positive because gas is liberated. As T

increases, G becomes more negative (i.e. the reaction becomes more favorable). S depends mainly on S0{CO2(g)} and is almost independent of M.

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Thermal stability of metal carbonates

MCO3 (s) → MO (s) + CO2 (g) Tmin almost directly proportional to H. HL favors formation of the oxide

(smaller anion) for smaller cations. So Tmin for carbonates should

increase with cation size.

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SolubilityMX (s) --> M + (aq) + X - (aq)

S is positive, so a negativeH is not always required for a spontaneous rxn. But H 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) Hlat M+(g) + n L → ML'n+(aq) Hsolv, M

X-(g) + m L → XL'm-(aq) Hsolv, X L'n + L'n → (n + m) L H L-L

MX(s) → M+(aq) + X-(aq) Hsolution, 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 Hsolv

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Solubility

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

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Solubility

HL terms dominate when ions have higher charges; these salts are usually less soluble.

Some aqueous solubilities at 25°C: Hsolution solubility

salt (kJ/mol) (g /100 g H2O)

LiF + 5 0.3 LiCl - 37 70 LiI - 63 180 MgF2 0.0076

MgO 0.00062

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Orbitals and Bands

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Band and DOS diagrams

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vs T

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Intrinsic Semiconductors

= n q = conductivityn = carrier densityq = carrier charge = carrier mobility

P = electron population ≈ e-(Eg)/2kT

Eg

C 5.5 eVSi 1.1 eVGe 0.7 eV

GaAs 1.4 eV

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Bandgap vs

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Arrhenius relation

ln

1 / T

Slope = -Eg/2k

Arrhenius relation:

e-Eg/2kT

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Extrinsic Semiconductorsn-type p-type

p-type example:

B-doped Si

n-type example:

P-doped Si