ch3. intro to solidssites.science.oregonstate.edu/chemistry/courses/ch411... · 2011-09-15 ·...
TRANSCRIPT
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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