Transcript Slide 1

Lectures 21-22
Solid state materials. Electronic structure and conductivity
1) Band theory
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•
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The electronic structure of solids can also be described by MO theory.
A solid can be considered as a supermolecule.
One mole of atoms (NA), each with X orbitals in the valence shell contributes X moles of
atomic orbitals producing X moles of MO’s.
Consider qualitatively bonding between N metal atoms of ns1 configuration (Li, Na etc)
arranged in a chain; N = 2, 4, NA. Assume that X=1 for simplicity.
• In the case of N ~ NA atoms they form not bonds but bands.
• The band appearing in the bonding region is called valence band. The antibonding
region is called conduction band.
• In the case of metals the valence and conduction bands are immediately adjacent.
N = 2 Li atoms
4 Li atoms
NA Li atoms
conduction band



F
Fermi level

valence band
2) Band theory. Insulators, semiconductors, conductors
•
If we apply now an electrostatic potential to a conductor, the population of the energy
levels will tend to change and electrons will be able to flow using empty adjacent
conduction band.
F
negative potential
•
no potential
positive potential
In the case of insulators and semiconductors, the energy gap between the valence and
conduction bands is more or less significant; electrons cannot easily get into the
conduction band and cannot move along the sample; thermal or photo-energy is needed
to bring some electrons to the conduction band.
Bandgap
T (intrinsic
conductivity)
Bandgap
F
or h
(photoconductivity)
F
Semiconductors
Insulators
F
Bandgap,
eV
Conductivity,
W-1cm-1
C
6.02
< 10-18
Si
1.09
5·10-6
Ge
0.72
0.02
Sn
0.07
104
3) Crystal Orbital theory
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The band structure of a crystalline material of virtually any complexity can be found through
the application of the MO theory for solid state materials (Crystal Orbital theory).
•
One of the ways to model a real (finite size) crystal is by using cyclic boundary conditions
assuming that a chain of bound atoms forms a very large ring.
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It turns out that the energy levels in a cyclic molecule composed of N hydrogen atoms look
as shown below.
E
Energy levels of resulting MO's are indicated with
energy level of
isolated s-orbitals
N=3
N=4
N=5
N=6
N=7
4) Crystal orbitals (Bloch functions)
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If we have N hydrogen atoms with atomic wave
functions fm (m = 1 … N) related by symmetry and
spaced at distance a, we can get N MO’s yn (n = N/2, …, 0, …, N/2) which are called Bloch functions.
For the n-th crystal orbital, yn, we will have:
E, eV
E, eV
a=3A a=2A a=1A
a=1A
2pn
2pn 

y n   cos(
m)  i sin(
m) m
N
N

m 1 
N
When n changes from 0 to N/2, variable k =
2pn/(aN)) (wave vector) changes from 0 to p/a and
the type of the MO changes from the completely
bonding y0 to the completely antibonding yN/2:
a
k=0
0
-10
k=p/a
-20
y0 = f1+f2+f3+...
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yN/2 = -f1+f2-f3+...
Energy levels of the resulting set of MO’s (band
structure) can be described with help of continuous
functions E and density of states dn/dE (DOS)
k
0
p/a 0
k
p/a 0
k
DOS
p/a
5) Bonding in solids: Crystal Orbital Overlap Population
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A common way to analyze bonding in solids is by
calculating and analyzing the crystal orbital
overlap population (COOP).
E, eV
E
k=p/a
20
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COOP is defined in the same way as the bond
order is defined in MO theory of molecules.
For any two atoms i and j COOP(i-j) = S2cicjSij
(Sij is the overlap integral for two atomic
wavefunctions; summation should be performed
for all pairs of overlapping orbitals of atoms i and
j). A negative value of COOP means antibonding
situation while a positive value is characteristic
for bonding.
Sij < 0
10
0
a
-10
-20
COOP
k
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For the chain of hydrogen atoms the lower half of
the band is bonding while the upper half is
antibonding (see diagram on the right).
0
0
p/a
k=0
Sij > 0
6) Simplified picture of bonding in crystalline metals
E
E
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Using crystal orbital
theory we can
rationalize the wellknown fact that the
metals with highest
melting points are those
belonging to 6th and 7th
groups (see diagram
below).
s, p - band
14 e's
p
s
12-13 e's
s, p - band
6-7 e's
d
d - band
d - band
Melting point, oC
3950
3450
2950
2450
1950
1450
950
450
-50
Sc Ti
V Cr Mn Fe Co Ni Cu Zn Ga
Y
Zr Nb Mo Tc Ru Rh Pd Ag Cd In
La Hf Ta W Re Os
Ir
Pt Au Hg Tl
0
+
COOP
7) The Peierls distortion
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When working with highly symmetrical structures one has to be cautious.
Highly symmetrical structures with not completely filled degenerate or near-degenerate levels are a
subject to distortions which lower the symmetry and the energy of the system (Peierls distortion).
Diagrams on the left and in the center show how we can form bands for polymeric dihydrogen (MO) with twice larger four-atomic unit 2a and then distort the polymer to produce an array of
dihydrogen molecules (the diagram on the right).
Similarly an infinite polyene -HC=HC-HC=HC-… polyacetylene will have alternating HC-HC and
HC=HC bonds due to the Peierls distortion. Because of the large band gap it will behave not as a
conductor but as an semiconductor.
E H2 or HC CH
E
E
2a
2a
a

F
F

0
p/2a
k
0
p/2a
k
8) Band structure of one dimensional polymers: a stack of PtII
square planar complexes
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In some cases one dimensional consideration is sufficient for a satisfactory analysis of
band structure of solids – one dimensional polymers. For example, we can get a
satisfactory description of bonding and conductivity of K2[Pt(CN)4Clx] (x = 0 … 0.3) using
just one-dimensional model of crystal.
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The complexes K2[Pt(CN)4Clx] (x = 0 … 0.3) have Pt(CN)4 – squares stacked one above
another with Pt-Pt separation of 3.3 (x = 0) or 2.7-3.3 Ǻ (0 < x < 0.3).
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Purely PtII complex (x = 0 in the formula above) is an insulator while oxidized
cyanoplatinates are low-dimensional conductors.
a
2a
z
arrangement of
anions in K 2[Pt(CN) 4]
arrangement of
anions in
K2[Pt(CN) 4Clx]
9) Forming bands: Principles
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To predict a qualitative band structure of
stacked [Pt(CN)4]2-, we will consider [PtH4]2as a model.
We will need for this analysis a MO diagram
of PtL4.
Each of the monomer’s MOs generates a
band when we form a polymer. We can
analyze all MO’s one by one and then
combine all bands together.
To get an idea about bands width use the
rule which states that better orbital overlap
will produce a wider band (>p>d):
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E
a2u
eu
p
pz
s
dx2-y2
a1g
d
z
pz
a

x
y

d
dx2-y2
z
4 L -GO's
Pt
b1g
b2g
a1g
eg
dz2
dxy
dyz
dxz

dz2
dxy
dyz
dxz
d
p

eu
b1g
a1g
10) How bands behave
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To learn, how the “frontier” bands will run (“up” or “down”)
let’s write corresponding Bloch functions for frontier
x
orbitals, pz, and all d-orbitals, for k = 0 and k = p/a.
a
k=0
k=0
pz
dz2
k=p/a
E
z
pz
y
E
runs down
k
k=p/a
p/a
0
E
E
k=0
dxz
k=p/a
E
dz2
runs up
k=0
dxy
k=p/a
E
k
0
p/a
11) Band structure of a stacked [PtH4]2•
The predicted band structure of a stacked [PtH4]2- in the center match well a calculated
diagram on the right.
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With band structure or DOS diagram in hands we can answer the questions: 1) why oxidized
K2[Pt(CN)4Clx] (x>0) is a conductor and 2) why Pt-Pt distance shortens as x increases.
E
x
y
E
pz
z
pz

dx2-y2
d
dx2-y2
dz2
dz2
dxy
dyz
dxz
d
p
dyz
dxz

dxy
k
0
p/a
12) Bonding and conductivity in stacked [PtH4]2-
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Conductivity. The Fermi level of stacked [PtH4]2is on the top of the z2-band since the monomer
HOMO is dz2 orbital. The conduction band is pzband which is almost 3 eV higher in energy.
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When the z2-band is completely filled (case of
PtIIL4), no conductivity is expected / observed. For
partially oxidized materials z2–band is filled only
partially and we expect and observe conductivity.
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Bonding. In solids like in molecules if bonding
and antibonding MO’s are completely filled, the
net bonding is zero.
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For partially oxidized materials K2[Pt(CN)4Clx] (x =
0 … 0.3) z2–band is partially empty and we
observe (dz2-dz2) bonding between Pt atoms.