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CH3. Intro to SolidsLattice geometries

Common structures

Lattice energies

Born-Haber model

Thermodynamic effects

Electronic structure

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A

Stacked 2D hexagonal arrays

B

C

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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 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) 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 = 0.4144 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° bcc

1667 → 3285° liquid

3285 → gas

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Classes of Alloys

(a) Substitutional

(b) Interstitial

(c) intermetallic

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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

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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.

B

C

Aa

b

c

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Sphalerite

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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

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Structure Maps

more covalent

more ionic

incr. r

ad

ius, p

ola

riza

bili

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 p e0) (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

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 = (e2/4pe0)(ZAZB/ d) [2(1) - 2(1/2) + 2(1/3) -2(1/4) +…]

= (e2/4pe0)(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 p e0) * (ZA ZB / d) * A

where A is called the Madelung constant

and is determined by the lattice geometry

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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) AB 0.874

sphalerite 1.638 (4,4) AB 0.819

wurtzite 1.641 (4,4) AB 0.821

fluorite 2.519 (8,4) AB20.840

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 / 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 !!!

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Further refinements

• Eel‟ include higher order terms

• Evdw NC‟‟r-6 instantaneous polarization

• EZPE Nhno lattice 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:

DHL = -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.

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.

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Example:

Use the Kapustiskii eqn to estimate DHL for MgCl2

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

2. r(Mg2+) CN 8 = 1.03 Å

r(Cl-) CN 6 = 1.67 Å

3. d0 ≈ r+ + r- ≈ 2.7 Å

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

5. DHL(best calc) = 2326

6. DHL(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:

DHL ≈ 1 / d0

For cubic structures:

DHL ≈ 1 / V1/3

where V is the unit cell volume

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DHLvs 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

DHf {KCl(s)} =

DHsub(K) + I(K) + ½ D0(Cl2)

– Ea(Cl) - DHL

All enthalpies are

measurable except DHL

Solve to get DHL(B-H)

-DHL

Ea

DHsub

½ D0

I

-DHf

DHf {KCl(s)} =

DH {K(s) + ½Cl2(g) → KCl(s)}

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Is MgCl3 stable ?

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.

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

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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.

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Thermochemical Radii

What are the radii of polyatomic ions ?

(Ex: CO32-, SO4

2-, 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.

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Thermochemical RadiiExample:

DHL(BH) for Cs2SO4 is 1658 kJ/mol

Use the Kap eqn:

DHL = 1658 = 1210(6/d0)(1-0.345/d0)

solve for d0 = 4.00 Å

Look up r (Cs+) = 1.67 År (SO4

2-) ≈ 4.00 - 1.67 = 2.33 Å

The tabulated value is 2.30 Å (an

avg 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 DHL 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 DHL

• 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 DHL sum

CsF 750

NaI 705 1455 kJ/mol

CsI 620

NaF 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)

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.

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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.

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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

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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

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

Some aqueous solubilities at 25°C:

DHsolution 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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s vs T

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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

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Bandgap vs Dc

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Arrhenius relation

ln s

1 / T

Slope = -Eg/2k

Arrhenius relation:

s = s0 e-Eg/2kT

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Extrinsic Semiconductors

n-type p-type

p-type example:

B-doped Si

n-type example:

P-doped Si