An Introduction to Band Theory, A Molecular Orbital Approach

An Introduction to Band
Theory, A Molecular
Orbital Approach
Chemistry 754
Solid State Chemistry
Dr. Patrick Woodward
Lectures #17-18
References – Band Theory
The material for this lecture comes primarily from the work of
Roald Hoffmann and those who came through his group (Mike
Whangbo,
Whangbo, Jeremy Burdett, Thomas Albright, Tim Hughbanks,
Hughbanks,
etc.). His treatment (and many of the figures used in this
lecture) are nicely summarized in a small book he wrote some
years ago:
“Solids and Surfaces: A chemists view of bonding
in extended structures”
Roald Hoffmann, VCH Publishers, (1988)
See also
“The Electronic Structure and Chemistry of Solids”
P.A. Cox, Oxford University Press, Oxford (1987).
“Chemical Bonding in Solids”
Jeremy K. Burdett, Oxford University Press, Oxford (1995).
1
Relevance of Band Structure
• What is the electronic band structure?
– For an extended solid the band structure plays the
same role that an MO diagram does for a molecule.
• Why it is important to be able to comprehend the
electronic band structure?
– The band structure is the link between crystal
structure, bonding and properties. In particular
details of the band structure are closely linked to the
following properties:
• Electronic conductivity
• Optical properties, including color
• Electronically driven structural distortions
• Mechanical and Magnetic Properties
• Catalytic Activity
MO Diagram ReO66- Octahedron
In the last lecture we discussed the
MO diagram of an octahedron
(shown to the left). How does this
diagram change when we have an
extended structure?
2
Band Structure ReO3
Band Structure
(aka Spaghetti diagram): MO
diagram with translational
symmetry taken into account.
Density of States (DOS)
Integration of the band
structure. Shows the # of
available levels between E and
E+dE
E+dE as dE → 0.
Constructing a DOS Diagram: TiO2
It’s possible to construct a reasonable
approximation of the DOS diagram
from the MO diagram of the building
block.
The energy levels of each block of
bands or “BOB” comes from the MO
diagram (based on electronegativity
and bonding interactions)
The area of each BOB is proportional
to the number of MO’s at that
approximate energy.
The width of each BOB is derived from
the overlap between building blocks
3
Constructing a DOS Diagram: TiO2
Ti eg
Ti-O σ*
Ti t2g
Ti-O π*
Ef
O 2p,
Ti-O bonding
O 2s
The DOS plots for TiO2 (rutile),
rutile), SrTiO3 (perovskite), CaTiO3 (cubic perovskite)
and MgTiO3 (ilmenite
(ilmenite)) are all going to be qualitatively similar, because the
electronic building block is the same in each case, the TiO6 octahedron. The
electropositive cations: such as Sr2+, Ca2+ and Mg2+ only make minor
contributions near the Fermi energy. The main difference will be the width of
the bands (as the octahedra are connected differently in each structure).
H2 Molecule
Chain of 5 H atoms
χ0
E
χ1
χ2
Antibonding
χ1
χ2
χ3
χ4
4
3
Nonbonding
2
1
Bonding
0
# of Nodes
4
Infinite 1D Chain of H atoms
If there are N atoms in the chain
there will be N energy levels and
N electronic states ("MOs"). The
wavefunction for each electronic
state is:
χ0
χ1
χ2
χ3
χ4
k=π/a
Ψk = Σ eiknaχn
Where:
k=π/2a
•a is the lattice constant (spacing
between H atoms),
•n identifies the individual atoms
within the chain,
• χn represents the atomic orbitals
k=0
•k is a quantum # that identifies
the wavefunction and tells us the
phase of the orbitals.
orbitals.
a
Infinite 1D Chain of H atoms
k = π/a
χ0
χ1
χ2
χ3
χ4
k=π/a
Ψπ/a = χ0+(exp{i
+(exp{iπ})χ
+(exp{i2π})χ
})χ1 +(exp{i2π
})χ2
+(exp{i3π
+(exp{i3π})χ
+(exp{i4π})χ
})χ3+(exp{i4π
})χ4+…
Ψπ/2a = χ0 - χ1 + χ2 - χ3 + χ4 +…
k = π/2a
k=π/2a
Ψπ/2a = χ0+(exp{i
+(exp{iπ/2})χ
+(exp{iπ})χ
/2})χ1 +(exp{i
})χ2
+(exp{i3π
+(exp{i3π/2})χ
+(exp{i2π})χ
/2})χ3+(exp{i2π
})χ4+…
Ψπ/2a = χ0 + 0 - χ2 + 0 + χ4 +…
k = 0
Ψ0 = χ0+χ1 +χ2 +χ3 +χ4 +…
k=0
a
k=0 → orbital phase does not change when we translate by a
k=π
k=π/a → orbital phase reverses when we translate by a
5
Band
Structure
Linear H Chain
E(k)
EF
0
k
π/a
•The Fermi energy separates the filled states (E < EF at T = 0 K)
from the empty states (E > EF at T = 0 K). Here it splits the band
(each band can hold 2 electrons)
electrons)
•A 1D chain of H atoms is predicted to be metallic because the
Fermi level cuts a band (there is no gap so it takes only an
infinitesimal energy to excite an electron into an empty state).
•The band runs "uphill" (from 0 to π/a) because the in phase (at k=0)
combination of orbitals is bonding and the out of phase (at k=π
k=π/a) is
antibonding.
antibonding.
Effect of Orbital Overlap
If we reduce the lattice
parameter a it has the following
effects:
•The spatial overlap of the
orbitals increases
•The band becomes more
bonding at k=0
•The band becomes more
antibonding k=π
k=π/a.
•The increased antibonding is
larger than the increased
bonding.
•The bandwidth increases.
•The electron mobility increases.
Wide bands → Good orbital
overlap → High carrier mobility
6
Band Structure: Linear Chain of F
F
F
F
F
(a)
F
F
(b)
F
F
(c)
F
F
(d)
EF
E(k)
EF
EF
0
k
π/a 0
k
π/a 0
EF
k
π/a 0
k
π/a
Which of the following is the correct band structure for
a linear chain of F atoms?
Band Structure: Linear Chain of F
Antibonding 2pz σ∗
Doubly degenerate
EF
Antibonding 2px/2py π∗
Doubly degenerate
Antibonding 2s σ∗
Bonding 2px/2py π
Bonding 2pz σ
Bonding 2s σ
0
k
π/a
7
Band Structure: Linear Chain of F
A more accurate
treatment of the band
structure would show an
avoided crossing
between the 2pz σ and
2s σ * interactions at
k=π
k=π/a. There would be
mixing between these
two bands (creating sphybrid like states)
similar to the 2s/2p σ &
σ* interactions seen in
the MO diagram of N2.
Antibonding 2pz σ∗
EF
Doubly degenerate
Bonding 2px/2py π
Bonding 2s σ
0
k
π/a
Lessons from Linear F Chain
• There are now 4 orbitals in the unit cell (a single F
atom with 1 2s + 3 2p orbitals)
orbitals) giving rise to 4
bands in the band structure.
• The fact that the wavefunction corresponding to a
p-orbital changes sign at the nucleus causes the 2p
σ band to run downhill (opposite of the 2s σ band).
• The reduced spatial overlap of the π interaction
causes the π bands to be narrower than the σ
bands.
• The 2p orbitals start out at a higher energy than
the 2s orbitals (from atomic quantum theory)
8
More than 1 atom in the Unit Cell
The Pierls Distortion
What happens when there is more than one atom in the
unit cell? To illustrate the consequences consider a
distortion of the 1D chain of H atoms which causes the
atoms to dimerize.
dimerize. The consequences will be as follows:
•There are now 2 orbitals in each unit cell, so there will
be 2 bands in the band structure.
•Let’s construct 1 band to be bonding within the unit cell
(HOMO - valence band), and the other to be antibonding
within the unit cell (LUMO - conduction band).
•Translational symmetry and the value of k will take care
of the rest.
Pierls Distortion (H2 Chain)
Conduction Band (LUMO σ*)
χ4 χ5
χ0 χ1
χ2 χ3
Intra
Inter
k=π/a
σ*
σ
k=π/2a
σ*
non
k=0
σ*
σ*
Intra
Inter
k=π/a
σ
σ∗
k=π/2a
σ
non
k=0
σ
σ
Valence Band (HOMO σ)
χ2 χ3
χ4 χ5
χ0 χ1
σ* → σ
Band runs
downhill
σ→σ*
Band runs
uphill
9
E(k)
Band Structure H2
EF
0
k
π/a
Lessons – Pierls Distortion (H2)
•There are two electrons per unit cell so that the valence band is full and
the conduction band empty. Thus EF occupies a position midway between
the two bands and the system is now a semiconductor/insulator.
•The minimum energy gap between bands occurs at k = π/a. Since the
maximum in the VB and the minimum in the CB occur at the same value of k,
the compound is said to be a direct gap semiconductor.
•Compared with the band structure of a linear chain of hydrogen atoms the
bands in this system will be narrower. This is a consequence of the reduced
overlap between molecules.
•The net effect of this distortion is to lower the energy of the filled
states (which originate from bonding MO’
MO’s) while raising the energy of
empty states (antibonding
(antibonding MO’
MO’s). In 1D such a distortion will always be
stable when you have a half filled band (as we did in the H band structure).
Physicists call this a Peierls distortion, it is the solid state equivalent of a
1st order Jahn-Teller distortion.
10
PtH4 a Molecular Chain Compound
Now let’s consider a compound that contains different types of
atoms within the unit cell. The compound A2PtX4 (where A is a +1
cation, such as K+, and X –1 anion, such as Cl-, CN- or H-) forms linear
chains and is a good example to consider. Let’s construct the band
structure of PtH42- (the K+ are very electropositive and serve
primarily as donors of electrons, we will ignore them except when it
comes to counting electrons). The process consists of 3 steps.
Step 1: Consider the MO Diagram of the Building unit (here a PtH4
Square Planar Unit). This gives an idea of the band energy levels.
Step 2: Consider the orbital overlap between units (molecules in
this case). This tells us if the bands run uphill or downhill.
Step 3: Consider the degree of orbital overlap (spatial and
energetic) between units. This gives the band widths.
Linear PtH4 Chain
H
H
H
H
H
H
H
H
Pt
Pt
Pt
Pt
H
H
H
H
H
H
H
H
Step 1: Consider the MO Diagram of the
Building unit (here a PtH4 Square Planar Unit).
This gives an idea of the band energy levels.
11
Linear PtH4 Chain
Step 2: Consider the orbital overlap
between units (molecules in this case). This
tells us if the bands run uphill or downhill.
Linear PtH4 Chain
Step 3: Consider the degree of orbital
overlap (spatial and energetic) between
units. This gives the band widths.
Calculated Band
Structure
12
Geometry and Properties
Consider the salt K2PtH4 (or K2PtCl4)
and predict what would be it’s
optimal geometry and properties.
Which configuration would be more
stable a staggered (PtH4 units
rotated by 45º to each other) or an
eclipsed geometry (PtH4 units
related by a mirror plane)?
Would you expect this material to be
a semiconductor or a metal?
How would the Pt-Pt distance react
if the material were oxidized?
K2Pt(CN)4Clx
K2Pt(CN)4
K2Pt(CN)4Cl0.3
• Pt2+ (d8) - dz2 band is
completely filled
• Pt2.3+ (mixed d7/d8) - dz2
band is partially filled
• Pt-Pt distance = 3.3 A
• Eclipsed geometry (the
Pt-Pt distance is too long
for Pt-Pt π interactions
to matter).
• Semiconductor (Fermi
(Fermi
level falls between dz2
band and the dx2-y2 band)
• Pt-Pt distance = 2.7 A
(electrons have been
removed from Pt-Pt
antibonding states
• Staggered geometry
(minimize dxz-dxz and dyzdyz interactions)
• Conductor (Fermi
(Fermi Level
cuts dz2 band)
13
Two dimensions
Γ = (kx=0, ky=0, kz=0)
M = (π/a, π/a, 0)
R = (π/a, π/a, π/a)
X = (π/a, 0, 0)
Y = (0, π/a, 0)
Z = (0, 0, π/a)
Band Structure of Ba2SnO4
To give a real example of a band structure which is
predominantly two dimensional consider the RuddlesdenPopper compound Ba2SnO4. This compound contains
sheets of corner sharing octahedra. Let’s derive the
band structure for one of these sheets.
y
z
x
14
Constructing the Band Structure
Step 1: Consider the MO Diagram of the building unit. In this
case the building unit is a SnO6 octahedron. The 4d orbitals
of Sn are essentially core orbitals and can be neglected.
The 2s orbitals of oxygen play a role, but for the sake of
simplicity they will be ignored.
Step 2: Consider the orbital overlap between units. In this
case the Sn 5px-O 2px, the Sn 5py-O 2py, and the Sn 5s-O
2px/O 2py interactions need to be considered. The Sn 5pz-O
2pz interactions are essentially isolated on individual building
units and will give rise to flat bands.
Step 3: Consider the degree of orbital overlap. Both the Sn
5s-O 2p and Sn 5p-O 2p interactions have a favorable
spatial overlap. The Sn 5s-O 2p interaction has a better
energetic overlap though.
SnO6 MO Diagram
2t1u
Sn 5p
2a1g
Sn 5s
*
eg
1t1u
eg
t1u
O 2p
a1g
1a1g
*In total there are 3×
3×6-4 = 14 non-bonding O 2p orbitals (neglecting
π-bonding and Sn 4d overlap). Only two are shown here.
15
Sn 5s-O 2p
Valence Band
y
Sn 5s-O 2p
Conduction Band
Majority O 2p character
x
Majority Sn 5s character
y
x
Γ
Γ
O 2p Nonbonding
X
Sn 5s Nonbonding
X
Sn 5s-O 2p
Bonding
M
M
Sn 5px-O 2px
Valence Band
y
x
Γ
Sn 5px-O 2px
Conduction Band
Majority O 2p character
Sn 5p-O 2p
Bonding
X
Majority Sn 5p character
y
x
Γ
Sn 5p-O 2p
Antibonding
X
O 2p Nonbonding
M
Sn 5s-O 2p
Antibonding
Sn 5p Nonbonding
M
16
Sn 5py-O 2py
Valence Band
y
Sn 5py-O 2py
Conduction Band
Majority O 2p character
Sn 5p-O 2p
Bonding
x
Γ
X
Majority Sn 5p character
y
Sn 5p-O 2p
Antibonding
x
Γ
X
Sn 5p Nonbonding
O 2p Nonbonding
M
M
Ba2SnO4: Band Structure
Sn 5pZ-O 2pZ σ∗
Sn 5py-O 2py σ∗
Sn 5px-O 2px σ∗
Sn 5s-O 2p σ∗
Band Gap
EF
O 2p nonbonding
Sn 5py-O 2py σ
Sn 5px-O 2px σ
Sn 5pZ-O 2pZ σ
Sn 5s-O 2p σ
Γ
X
M
Γ
17
3D: The Band Structure ReO3
We can now go full circle and return to the ReO3 spaghetti diagram that
started our discussion. The band structure of this compound is similar in
many ways to the Ba2SnO4 band structure. The main differences are:
•
•
•
Because of the 3-dimensionality all of the Re based bands will be disperse
(only O 2p non-bonding states will give rise to flat bands).
The most important interactions are the Re 5d-orbitals
5d-orbitals,, t2g (π
(π/π*) & eg
(σ/σ*), rather than the Sn 5s & 5p.
The Fermi level cuts the π* bands creating a metallic conductor rather
than a semiconductor.
ReO3 Bandstructure
O 2p 30
nonbonding
30
Re
25
Energy (vs. O 2s)
Overlap at the M-point
kx=ky=π/a, kz=0
O 2p
O
Re MM te
2g
O
20
20
15
15
10
10
Re
X
R
Γ
M
R0
k
3
DOS
π*
π*
O
O
Re
Overlap at the Γ-point
kx=ky=kz=0
X
The points circled in blue are the bonding
counterparts to the antibonding
interactions shown to the right.
W(t2g/π*) ~ 5 eV
W(e
W(eg /σ*) ~ 7 eV
g
EF
25
R
M
ReΓ
O
R0
Re
DOS
k
O
O
6
π*
π
Re
O
Γ point
non-bonding
Re
The eg/σ* band is wider than t2g/π* due to larger orbital overlap (spatial).
18
Summary Band Structures
• What is being plotted? Energy vs. k, where k is the wavevector that
gives the phase of the AO’s as well as the wavelength of the electron
wavefunction (crystal momentum).
• How many lines are there in a band structure diagram? As many as
there are orbitals in the unit cell.
• How is the center of gravity energy level of each band determined?
Usually follows from the MO diagram.
• How do we determine whether a band runs uphill or downhill? By
comparing the orbital overlap at k=0 and k=π
k=π/a.
• How do we distinguish metals from semiconductors and insulators?
The Fermi level cuts a band in a metal, whereas there is a gap between
the filled and empty states in a semiconductor.
• Why are some bands flat and others steep? This depends on the
degree of orbital overlap between building units.
Wide bands → Large intermolecular overlap → delocalized eNarrow bands → Weak intermolecular overlap → localized e-
19