Transcript Document

CH3. Intro to Solids
Lattice geometries
Common structures
Lattice energies
Born-Haber model
Thermodynamic effects
Electronic structure
1
Stacked 2D hexagonal arrays
A
B
C
2
Packing efficiency
“…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, onehundred degrees
Fahrenheit, or, better still,
… one-hundred-and-thirty
degrees.”
Kurt Vonnegut, Jr.
Cat’s Cradle
• 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
3
Close-packing of polymer microspheres
4
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
5
Unit cells for hcp and fcc
Unit cells,
replicated and
translated, will
generate the
full lattice
Hexagonal
cell = hcp
Cubic cell
ccp = fcc
6
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…
7
Oh and Td sites in ccp
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
8
Ionic radii are related to
coordination number
9
Element Structures at STP
(ABCB)n
10
Ti phase transitions
RT
882
1667
3285
→ 882°C
→ 1667°
→ 3285°
→
hcp
bcc
liquid
gas
11
Classes of Alloys
(a) Substitutional
(b) Interstitial
(c) intermetallic
12
Some alloys
substitutional
interstitial
intermetallic
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
13
A few stainless steels
Chemical Composition %
(Max unless noted)
Stainless
C
Mn
P
S
Si
Cr
Ni
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
2205
0.02 2.00 0.045 0.030 1.000 22.00-23.00
5.50-6.00
Mo
N
2.00-3.00
3.00-3.50 0.17
14
Zintl phases
KGe
15
NaCl (rocksalt)
–
–
Cl
–
–
Na
b
C
a
B
c
A
–
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
16
CaC2
Tetragonal
distortion of rocksalt
structure (a = b ≠ c)
Complex anion also
decreases (lowers)
symmetry
17
Other fcc anion arrays
18
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
19
Sphalerite (ZnS)


c
B
A
a
b

C

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.
20
Sphalerite
21
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
22
incr. radius, polarizability
Structure Maps
more ionic
more covalent
23
Lattices with hcp anion arrays
24
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
25
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
26
LiTiS2
S
(AcBc’)n
Ti
Li
27
LDH structures
Mg(OH)2 (brucite)
MgxAl1-x(OH)2.An
28
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 edgesharing 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.


29
Rutile
TiO2-x and SiO2
30
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


31
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.
32
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.
33
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.
34
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 = 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
35
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)
36
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
37
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)
0.840
rutile
2.408
(6,3)
AB2
AB2
0.803
38
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)
39
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‘:
Note sign conventions
!!!
E = -DHL = (e2/4pe0) (NAZAZB/d0) (1 - d*/d0)
(Born-Meyer equation)
40
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
41
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
42
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.
43
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
44
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
45
DHLvs V-1/3 for cubic lattices
V1/3 is
proportional to
lattice E for cubic
structures. V is
easily obtained
by powder
diffraction.
46
Born – Haber cycle
½ D0
Ea
DHf {KCl(s)} =
DHsub(K) + I(K) + ½ D0(Cl2)
– Ea(Cl) - DHL
I
DHsub
-DHf
DHf {KCl(s)} =
DH {K(s) + ½Cl2(g) → KCl(s)}
-DHL
All enthalpies are
measurable except DHL
Solve to get DHL(B-H)
47
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.
48
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.

49
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.
50
Thermochemical Radii
Example:
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 Å
r (SO42-) ≈ 4.00 - 1.67 = 2.33 Å
The tabulated value is 2.30 Å (an
avg for several salts)
51
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

* Estimated from
the Kap eqn
O2(g)
→ O2+(g) + ee- + PtF6(g)
→ PtF6-(g)
O2+(g) + PtF6-(g) → O2PtF6(s)
O2(g) + PtF6(g) → O2PtF6(s)
+ 1164 kJ/mol
- 787
- 470*
≈ - 93
 I(Xe) ≈ I(O2), so Xe+PtF6-(s) may be
stable if DHL is similar. Bartlett reported
the first noble gas compound in 1962.
52
Some consequences of DHL
• Ion exchange / displacement
• Thermal / redox stabilities
• Solubilities
53
Exchange / Displacement

Large ion salt + small ion salt is better than
two salts with large and small ions
combined.
Example:
Salt DHL sum
CsF
NaI
CsI
NaF

750
705
620
926
1455 kJ/mol
1546
This can help predict some reactions like
displacements, ion exchange, thermal
stability.
54
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.
55
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.
56
Solubility
MX (s) --> M + (aq) + X - (aq)
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
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
57
Solubility
The energy balance
favors solvation for
large-small ion
combinations, salts of
ions with similar sizes
are often less soluble.
58
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
59
Orbitals and Bands
60
Band and DOS diagrams
61
s vs T
62
Intrinsic Semiconductors
s =nqm
s
n
q
m
= conductivity
= carrier density
= carrier charge
= carrier mobility
P = electron population
≈ e-(Eg)/2kT
Eg
C
5.5 eV
Si
1.1 eV
Ge
0.7 eV
GaAs 1.463eV
Bandgap vs Dc
64
Arrhenius relation
Arrhenius relation:
s = s0 e-Eg/2kT
Slope = -Eg/2k
ln s
1/T
65
Extrinsic Semiconductors
n-type
p-type
n-type example:
P-doped Si
p-type example:
B-doped Si
66