CHEM 548: Materials Chemistry
Lecture 17: Band Structures of Graphene, Primitive Cubic, and FCC Crystals
Mon 11/02/2026 · Meeting 19
Reading: Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry
Today we work out the band structures of graphene, of primitive cubic polonium, and of the face-centered cubic crystals aluminum, diamond, silicon and germanium.
Learning Objectives: Graphene
Hexagonal reciprocal lattice construct reciprocal lattice vectors for non-orthogonal hexagonal systems and identify the first Brillouin zone
Special k-points locate Gamma, M, K, and K’ and convert between reciprocal lattice and Cartesian coordinates
Pi bands distinguish sigma and pi band contributions and apply orbital orthogonality at high-symmetry points
Crystal orbitals predict the bonding character of π orbitals at Gamma, M, and K from phase relationships
Semi-metallic behavior explain Dirac points and linear dispersion as unique features of graphene’s electronic structure
Before we leave two-dimensional materials, it would be a pity not to talk about graphene, which might be the most famous two-dimensional material. By the end you should be able to construct the reciprocal lattice vectors for a hexagonal lattice, where the axes are not orthogonal, and to identify its first Brillouin zone. You should be able to locate gamma, M, K and K prime in that zone. You should be able to separate the pi bands from the sigma bands, and to predict whether the pi crystal orbitals are bonding or antibonding at gamma, M and K from the phases of the orbitals. And you should be able to explain why the bands touch at the Dirac points, and why that makes graphene a semimetal.
Reciprocal Space: 2D Hexagonal Lattice
y x a b a* b* real-space lattice
$\boldsymbol{T} = u\boldsymbol{a} + v\boldsymbol{b}$
$a = b$, $\gamma = 120^{\circ}$ $u$, $v$ = integers
Reciprocal space lattice vectors
$\boldsymbol{a}^{*}$ perpendicular to $\boldsymbol{b}$
$\boldsymbol{a}^{*}\cdot\boldsymbol{a} = 2\pi$
$\boldsymbol{b}^{*}$ perpendicular to $\boldsymbol{a}$
$\boldsymbol{b}^{*}\cdot\boldsymbol{b} = 2\pi$
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry , Fig. 6.13 (§6.4.1, p. 217)
To describe the band structure of graphene we have to deal with axes that are no longer orthogonal, because graphene has a hexagonal lattice. So let's start by working out the reciprocal lattice for a hexagonal system. On the left is the real space lattice. The vectors a and b are equal in length, and the angle between them is 120 degrees.
[advance: add the reciprocal-vector criteria]
The rules for the reciprocal lattice vectors are the same as before. A star has to be perpendicular to b, so in this drawing a star points straight down. B star has to be perpendicular to a, so it points off at an angle. The two reciprocal lattice vectors end up at 60 degrees to each other.
The Reciprocal Lattice and First Brillouin Zone
y x a b real-space lattice
$\boldsymbol{a} = \dfrac{\sqrt{3}a}{2}\hat{x} - \dfrac{a}{2}\hat{y}$ $\boldsymbol{b} = a\hat{y}$
y x a* b* reciprocal-space lattice
first Brillouin zone
$\boldsymbol{a}^{*} = \dfrac{4\pi}{\sqrt{3}a}\hat{x}$ $\boldsymbol{b}^{*} = \dfrac{2\pi}{\sqrt{3}a}\hat{x} + \dfrac{2\pi}{a}\hat{y}$
$a$ = length of real-space unit cell vector
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry , Fig. 6.13 (§6.4.1, p. 217)
If we draw the whole lattice of reciprocal space points, it looks like this. It still has hexagonal symmetry, but the lattice vectors now point in somewhat different directions.
[advance: add the vector formulas and the first Brillouin zone]
The first Brillouin zone is the set of points closer to the lattice point at the origin than to any other point. To find it we draw lines from the origin to all of the nearest points and bisect each of those lines, and we get a hexagon. In Cartesian coordinates b lies along y, and its length is just a, the length of the lattice vector. A has the same length, but it does not point along x. We have a 30 60 right triangle, so its x component is root 3 over 2 times a and its y component is minus one half a. A dot a star and b dot b star both have to be 2 pi, and that gives the reciprocal vectors written here. I will leave it to you to confirm the dot products.
Special Points in a 2D Hexagonal Lattice
label coordinates Γ $0\,\boldsymbol{a}^{*} + 0\,\boldsymbol{b}^{*}$ M $(1/2)\,\boldsymbol{a}^{*} + 0\,\boldsymbol{b}^{*}$ K $(1/3)\,\boldsymbol{a}^{*} + (1/3)\,\boldsymbol{b}^{*}$ K′ $(2/3)\,\boldsymbol{a}^{*} - (1/3)\,\boldsymbol{b}^{*}$
label wave vector (Cartesian) Γ $0\,k_x + 0\,k_y$ M $(2\pi/\sqrt{3}a)\,k_x + 0\,k_y$ K $(2\pi/\sqrt{3}a)\,k_x + (2\pi/3a)\,k_y$ K′ $(2\pi/\sqrt{3}a)\,k_x - (2\pi/3a)\,k_y$
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry , Fig. 6.15 (§6.4.1, p. 218)
Now the special points in the first Brillouin zone of a hexagonal lattice. At the center is gamma, where the k vector is zero. M is at the center of one of the hexagon's edges, one half a star and no b star. K and K prime are at the corners of the hexagon. To get to K we go one third of a star plus one third of b star, and to get to K prime we go two thirds of a star minus one third of b star. These are the points where we will look at the energies and the crystal orbitals of graphene. The Cartesian coordinates are written below, but we will not pay much attention to them.
Graphene
Plane group p6mm
C on Wyckoff site 2b 1/3, 2/3; 2/3, 1/3
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry , Fig. 6.18 (§6.5.1, p. 221)
Everything up to this point is generic for any crystal with a two-dimensional hexagonal lattice. Now let's look at graphene itself. Graphene is a two-dimensional honeycomb of fused benzene rings, or you can think of it as chicken wire. The unit cell contains two carbon atoms, on the Wyckoff 2b site of the plane group p6mm.
Graphene: Band Structure
The 2p z orbitals are orthogonal to the other orbitals at all points in the Brillouin zone. This gives 2 pi bands and 6 sigma bands in the band structure.
At Γ the C 2s does not mix by symmetry with C 2p x and 2p y . σ1 is purely C 2s bonding at Γ, while σ2 and σ3 are antibonding 2p at Γ.
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry , Fig. 6.18 (§6.5.1, p. 221)
The atomic orbitals on carbon that we care about are the 2s, 2p x, 2p y and 2p z. Two carbons per unit cell give eight bands, and here I show the five lowest. The three lowest at M and at K are sigma bands. At gamma the 2s is orthogonal to the 2p orbitals, so the bottom crystal orbital is purely 2s and bonding, and above it the 2p x and 2p y bands are degenerate. We will see this again and again, with s bonding and p antibonding at gamma. But the sigma bands are not the ones we are interested in. Just like in benzene, the key features come from the pi overlap of the 2p z orbitals, perpendicular to the sheet. The 2p z is orthogonal to the other three everywhere in the zone, so we only have to consider the pi overlap between 2p z orbitals.
Graphene π Crystal Orbitals at Γ, M, and K
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry , Fig. 6.19 (§6.5.1, p. 222)
Let's start with gamma, which is always the easiest. We have two bands because we have two p z orbitals in the unit cell, one on each carbon. So the basis for the lower band is the pi bonding molecular orbital, and the basis for the upper band is the pi antibonding molecular orbital. Here we are looking down on the lobe of each p orbital coming up out of the plane. At gamma each carbon has pi bonding interactions with its three nearest neighbors, so this is the lowest energy crystal orbital. In the pi antibonding orbital each carbon has antibonding interactions with its three nearest neighbors, so this is the most antibonding crystal orbital in the whole first Brillouin zone, and it's at a much higher energy.
[advance: move the crystal-orbital pictures to the M point]
Now let's go to M. M is one half a star and zero b star, so in real space the phase of the orbital changes every time we move one unit cell along a, but not when we move along b. Starting from the bonding molecular orbital, each carbon now has two pi bonding interactions with its neighbors and one pi antibonding interaction, so the net result is a weakly bonding crystal orbital. We change the phases the same way for the antibonding band, and now each carbon has two antibonding interactions and one bonding interaction, so this crystal orbital is weakly pi antibonding. At M there is still a fair energy separation between the two bands.
[advance: move the crystal-orbital pictures to the K point]
At K we have one third of a lattice vector in both directions, so the phase of the orbital comes back to where it started once every three unit cells. That makes this orbital a lot harder to visualize, and here I am plotting only the real part of the wave function. If we start on one carbon, the coefficient of the p z orbital on its three neighbors is zero, so this is a nonbonding orbital, and the same is true for both the pi and the pi star band. So at K these two bands touch precisely. Each carbon has four valence electrons, so the unit cell has eight, enough to fill four bands. We fill the three sigma bands and the pi band, and we leave the pi star band empty.
Dirac Points and Semi-Metallic Graphene
At K and K′ the π and π* touch (become degenerate) these are called Dirac points
As we move away from K and K′ the energies of the π and π* bands change linearly with k . This is unusual and gives graphene special properties.
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry , Fig. 10.30 (§10.6.1, p. 446)
Here is a three-dimensional plot of the pi and pi star bands of graphene. The plane is the first Brillouin zone, with gamma at the center, where the two bands are far apart. At K and at K prime the bands touch, so around the edge of the hexagon there are six points where they just touch, called Dirac points. In a metal the Fermi level cuts through a band, and in a semiconductor there is a gap. In graphene there is no gap, but the Fermi level does not cut through a band either, so graphene is a perfect semimetal. If we dope it, by chemical means or with an electric field, the Fermi level moves down or up, and either way graphene becomes a metallic conductor. And away from the Dirac points the energy of each band changes linearly with k. That is very unusual, and we come back to it in chapter 10.
Check Your Understanding: π and π* Orbitals
In the graphene band structure, which atomic orbitals are primarily responsible for forming the π (pi) and π* (pi-star) bands near the Fermi level?
A. 2px and 2py orbitals
B. All 2s and 2p orbitals combined
C. 2pz orbitals
D. 2s orbitals
Correct: C. The 2pz orbitals are perpendicular to the graphene plane and form the π-system, which dictates the electronic properties near the Fermi level.
Question: in the graphene band structure, which atomic orbitals form the pi and pi star bands near the Fermi level?
[advance: reveal the answer]
The answer is C, the 2p z orbitals. They are perpendicular to the graphene plane, so they form the pi system, and the pi system sets the electronic properties near the Fermi level. Next we will go to three dimensions.
Learning Objectives: 3D Band Structures
Reciprocal lattice vectors construct them in 3D with cross products and identify Gamma, X, M, and R in primitive cubic Brillouin zones
Four bands predict s- and p-orbital dispersion from nearest-neighbor bonding patterns at high-symmetry points
P orbital directionality explain the band splitting it causes and which bands remain degenerate at specific k-points
Sigma and pi overlap rationalize the subtle energy variations between k-points
Fermi level place it from the valence electron count and predict Peierls distortions in half-filled p-band systems
We are now ready for three-dimensional crystals, and we start with about the simplest one there is, the primitive cubic structure of alpha polonium. By the end you should be able to construct the reciprocal lattice vectors in three dimensions from cross products, and to locate gamma, X, M and R in the primitive cubic Brillouin zone. You should be able to predict how the s and p bands change between those points from the bonding and antibonding interactions with the nearest neighbors. You should be able to explain how the direction of the p orbitals splits the p bands, and how the weaker pi overlap causes the smaller changes in energy. And you should be able to place the Fermi level from the electron count, and to predict a Peierls distortion when the p bands are half filled.
Reciprocal Space Vectors in 3D
Real space lattice
$r = x\boldsymbol{a} + y\boldsymbol{b} + z\boldsymbol{c}$
Reciprocal space lattice
$k = m\boldsymbol{a}^{*} + n\boldsymbol{b}^{*} + o\boldsymbol{c}^{*}$
$\boldsymbol{a}^{*} = \dfrac{2\pi}{V}(\boldsymbol{b}\times\boldsymbol{c})$
$\boldsymbol{b}^{*} = \dfrac{2\pi}{V}(\boldsymbol{c}\times\boldsymbol{a})$
$\boldsymbol{c}^{*} = \dfrac{2\pi}{V}(\boldsymbol{a}\times\boldsymbol{b})$
As always, the first question is how to define the first Brillouin zone, which contains the k vectors. In three dimensions we have three real space lattice vectors, a, b and c, and the fractional coordinates x, y and z that we have used to describe crystals. In reciprocal space there are three reciprocal lattice vectors, a star, b star and c star. A star is the cross product of b and c, times 2 pi divided by the volume of the unit cell, and b star and c star follow the same pattern. The cross product of two vectors is perpendicular to both of them, so if the real space lattice vectors are orthogonal, the reciprocal lattice vectors are also orthogonal and point in the same directions. We will not derive these, but it is good to know where they come from.
First Brillouin Zone: Primitive Cubic
kx ky kz Γ X M R
label coordinates Γ $0\,\boldsymbol{a}^{*} + 0\,\boldsymbol{b}^{*} + 0\,\boldsymbol{c}^{*}$ X $(1/2)\,\boldsymbol{a}^{*} + 0\,\boldsymbol{b}^{*} + 0\,\boldsymbol{c}^{*}$ M $(1/2)\,\boldsymbol{a}^{*} + (1/2)\,\boldsymbol{b}^{*} + 0\,\boldsymbol{c}^{*}$ R $(1/2)\,\boldsymbol{a}^{*} + (1/2)\,\boldsymbol{b}^{*} + (1/2)\,\boldsymbol{c}^{*}$
label wave vector (Cartesian) Γ $0\,k_x + 0\,k_y + 0\,k_z$ X $(\pi/a)\,k_x + 0\,k_y + 0\,k_z$ M $(\pi/a)\,k_x + (\pi/a)\,k_y + 0\,k_z$ R $(\pi/a)\,k_x + (\pi/a)\,k_y + (\pi/a)\,k_z$
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry , Fig. 6.16 (§6.4.2, p. 219)
A primitive cubic lattice is particularly easy, because its reciprocal lattice is also primitive cubic, and the first Brillouin zone is a cube. Gamma is at zero zero zero, the center of the cube. X is one half of a star, and M is one half of a star and b star, the same special points we had in the square lattice. The one new point is R, one half of all three reciprocal lattice vectors. Now let's see how these points change the phases of the orbitals in a real material.
Band Structure of a Primitive Cubic Metal (α -Po): the 6s Band
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry , Fig. 6.25 (§6.6.1, p. 227)
The simplest material in three dimensions is a primitive cubic metal, and the one element that takes this structure is alpha polonium. This band structure is a DFT calculation, which is what we will use for the rest of this chapter, and by convention the zero of energy is at the Fermi level. That is different from the extended Huckel calculations we have used so far. There is one polonium atom per unit cell, and polonium is a p block metal with a 6s and three 6p orbitals, so there are four bands. Without knowing anything about the overlap, you can recognize that the lowest band comes from the 6s orbital and the upper three come from the 6p orbitals.
[advance: show the 6s crystal orbital at Γ]
Let's look at the overlap of the 6s orbitals first. At gamma every atom has the same phase from one unit cell to the next. The basis is just the 6s orbital on polonium, so we get six nearest neighbor bonding interactions, and that is the bottom of the 6s band.
[advance: add the 6s crystal orbital at X]
At X the phase changes as we move along x, but not along y or z. So of the six nearest neighbors, the two along x are now antibonding and the other four are still bonding. The crystal orbital is still net bonding, but it is higher in energy than at gamma.
[advance: add the 6s crystal orbital at M]
At M the phase changes every time we move one unit cell in either x or y. Now the four neighbors in the x y plane are antibonding, and only the two along z are still bonding. This is more antibonding than bonding, so it is higher in energy still.
[advance: add the 6s crystal orbital at R]
Finally, at R the phase changes every time we move one unit cell in any of the three directions, so all six nearest neighbor interactions are antibonding, and this is the top of the 6s band. We could have predicted this band from what we learned with the hydrogen chains and sheets in one and two dimensions.
The $6p_x$ Band
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry , Fig. 6.25 (§6.6.1, p. 227)
Now let's consider the 6p orbitals, starting with the 6p x. At gamma, just as in the fluorine chain, the sigma interaction between the p orbitals is antibonding, so this is at a relatively high energy. At gamma all three p orbitals are degenerate, and they all have this sigma antibonding interaction.
[advance: add the 6px crystal orbital at X]
At X the phase changes when we move one unit cell along x, so the sigma interactions along x have become bonding, and the energy should drop.
[advance: add the 6px crystal orbital at M]
At M the phase changes along both x and y. The orbitals on the back face of the cell now have a different phase than they had at X, but the sigma interactions along x are still bonding.
[advance: add the 6px crystal orbital at R]
At R the phase changes every time we move one unit cell in any direction, and the sigma interactions along x are still bonding. So which of the three p bands is the 6p x band? Going from gamma to X turned sigma antibonding into sigma bonding, which should give a large drop in energy, and only band one drops from gamma to X. So band one is the 6p x band. To see why it is lowest at X and then rises a little to M and again to R, we need the weaker pi interactions. At gamma and at X the pi interactions with the four sideways neighbors are bonding. At M the pi interaction along z is still bonding but the one along y is antibonding, so the net is nonbonding. At R the pi interactions in both directions are antibonding, and that causes the small rise from X to M to R.
The $6p_z$ Band
Band 3 is the 6pz band: σ* at Γ, X, and M, and only band 3 does that.
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry , Fig. 6.25 (§6.6.1, p. 227)
The other p bands do not sit on top of the 6p x band, and we can see why if we look at the overlap in the 6p z band.
[advance: show the 6pz crystal orbital at Γ]
At gamma the 6p z orbitals have sigma antibonding character along z and pi bonding character sideways, same as the 6p x.
[advance: add the 6pz crystal orbital at X]
At X, just as before, the phase changes for the orbitals one unit cell over in the x direction, so all the orbitals on this right-hand face have the opposite phase. But because these are p z orbitals, the sigma interactions along z are still antibonding everywhere, so the 6p z band goes up instead of down in energy from gamma to X.
[advance: add the 6pz crystal orbital at M]
At M the phase changes in both the x and the y direction. The sigma interactions along z are still antibonding, and now the pi interactions sideways are antibonding in both directions.
[advance: add the 6pz crystal orbital at R]
At R the phase changes every time we move one unit cell in any direction, giving sigma bonding and pi antibonding, same as the 6p x at R. So at gamma and at R all three p orbitals are degenerate, but at the other points they are not, because the p orbitals point in different directions.
[advance: hold the full picture while we pick out the band]
So which of the three bands is the 6p z? It is band three, because at gamma, X and M it is sigma antibonding, so it stays at a relatively high energy, and only band three does that. It goes up from gamma to X to M because the pi interactions go from bonding to nonbonding to antibonding, and then it drops sharply at R.
Assembling Polonium's Band Structure
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry , Fig. 6.25 (§6.6.1, p. 227)
Now we can put the band structure of polonium together. We did not look at the 6p y overlap, but by elimination this band is the 6p y band. Notice again that at gamma the s orbitals have a bonding overlap and the p orbitals an antibonding overlap, with a large energy separation between them. Now the Fermi level. Polonium is in group 16, so it has six valence electrons. Two fill the 6s band, and four go into the three 6p bands, so the p bands are two thirds filled, and that puts the Fermi level where it is. If we move one element to the left, to bismuth, there are three p electrons, so the p bands would be half filled. We saw with the hydrogen chain that a half-filled band is susceptible to a distortion, which in one dimension is called a Peierls distortion, and in bismuth metal we see something like a Peierls distortion.
Peierls Distortion in 3D
α-polonium (cubic)p bands 2/3 filled
bismuth (rhombohedral)p bands 1/2 filled
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry , Fig. 6.26 (§6.6.1, p. 228)
Bismuth has a rhombohedral structure, but it is not far from primitive cubic. Each bismuth atom has three short contacts to its neighbors, the solid lines, at 3.07 angstroms, and three longer contacts, the dashed lines, at 3.53 angstroms. The p bands are half filled, so this distortion lowers the energy of the occupied 6p bands at the expense of the empty 6p bands.
The Band Structure of Bismuth
S. Golin, "Band Structure of Bismuth: Pseudopotential Approach", Phys. Rev. 166, 646 (1968), Fig. 2
This is a calculated band structure of bismuth. The distortion splits bands that were degenerate in the cubic structure and opens gaps where the bands would have crossed near the Fermi level. One final note on polonium and bismuth. These are very heavy elements, so there is a lot of spin-orbit coupling, and if we include it in the calculations it changes them substantially. It removes the degeneracy of the 6p orbitals at gamma and at R. That is something to keep in mind for real materials. Next we will look at face-centered cubic crystals.
Learning Objectives: FCC Band Structures
FCC Brillouin zone describe the first Brillouin zone structure for face-centered cubic lattices
Aluminum analyze the electronic band structure of an FCC metal
Diamond structure examine the band structures of carbon, silicon, and germanium
Bands and gaps distinguish valence bands, conduction bands, and band gaps in semiconductors
Group 14 trend explain the progression from insulator to semiconductor to metal down the group
Orbital overlap connect bonding and anti-bonding states to metallic versus semiconducting behavior
Polonium is great for the concepts, but you are unlikely to come across it in real life. So we stay with p block elements that adopt cubic structures, but now with a face-centered cubic Bravais lattice, and we look at some representative metals and semiconductors. By the end you should be able to describe the first Brillouin zone of a face-centered cubic lattice, and to analyze the band structure of aluminum, a face-centered cubic metal. You should be able to read the band structures of carbon, silicon and germanium in the diamond structure, and to tell the valence bands, the conduction bands and the band gap apart. And you should be able to explain why group 14 goes from an insulator to semiconductors to a metal, from the bonding and antibonding states and the orbital overlap.
First Brillouin Zone: Face-Centered Cubic
kx ky kz Γ X W K L
label coordinates Γ $0\,\boldsymbol{a}^{*} + 0\,\boldsymbol{b}^{*} + 0\,\boldsymbol{c}^{*}$ X $0\,\boldsymbol{a}^{*} + (1/2)\,\boldsymbol{b}^{*} + (1/2)\,\boldsymbol{c}^{*}$ W $(1/4)\,\boldsymbol{a}^{*} + (3/4)\,\boldsymbol{b}^{*} + (1/2)\,\boldsymbol{c}^{*}$ L $(1/2)\,\boldsymbol{a}^{*} + (1/2)\,\boldsymbol{b}^{*} + (1/2)\,\boldsymbol{c}^{*}$
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry , Fig. 6.16 (§6.4.2, p. 219)
First we need the first Brillouin zone, and for a face-centered cubic lattice it looks a lot more complicated than for simple cubic. That is because it corresponds to the primitive unit cell. The cubic centered cell has four atoms, and the dark lines show the primitive cell, a rhombohedron with equal edges and 60 degrees between them. The cubic cell would give four copies of every atom, so the primitive cell with one atom makes more sense. Putting those primitive vectors into the reciprocal lattice formula gives a body-centered cubic reciprocal lattice, and the Wigner-Seitz cell of that lattice is the truncated octahedron. Gamma is, as always, at the center. L is the face-centered cubic equivalent of R, one half of all three reciprocal lattice vectors, so the phase changes from one unit cell to the next in every direction.
Aluminum
Face-centered cubic elements do not get simpler than a cubic close-packed metal, so let's look at aluminum. One aluminum atom per primitive cell gives four bands, one 3s and three 3p. At gamma the 3s crystal orbital is in phase everywhere, so all twelve neighbors of each aluminum are bonding, and this is the lowest crystal orbital in the structure. The p orbitals change phase across their nodal plane, so at gamma all three are degenerate and antibonding, the highest in the zone. At other points s and p can mix, and the crystal orbitals fall in between. The bands are very wide, about 24 to 25 electron volts. Aluminum has three valence electrons, so the bands are three eighths filled and the Fermi level cuts through them. That is a recipe for delocalized electronic conductivity.
Diamond
Space Group $Fd\bar{3}m$
$a$ 3.57 Å
atoms/unit cell 8
C–C distance 1.54 Å
Now let's go on to diamond. We have already looked at graphene, the two-dimensional version of graphite, and the band structure of graphite is quite similar to that of graphene. What happens in the sp3 form of carbon? Diamond still has a face-centered cubic lattice, with carbon at the corners and on each face of the unit cell, just like aluminum. But there are also four carbons inside the cell, at the centers of alternating octants, so there are eight carbons in the centered cell. The primitive cell is four times smaller and holds two carbons.
Diamond: Band Structure
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry , Fig. 6.27 (§6.6.2, p. 229)
The eight corners in the pictures count as one carbon and the atom inside is the second, so we have eight bands. The lower four come from the bonding combination of the two carbons, and as in aluminum, s and p do not mix at gamma, so there is a fully bonding 2s crystal orbital and a 2p crystal orbital much higher. In that 2p state every bond is sigma bonding but pi antibonding. The upper four bands come from the antibonding combination, one s and three p. Eight valence electrons per cell fill the bottom four, and then there is a gap of 5.5 electron volts, so we think of diamond as an insulator. The filled, mostly bonding states are the valence band, and the empty, mostly antibonding states are the conduction band. The valence band top is at gamma and the conduction band bottom is between gamma and X, so the gap is indirect.
Diamond-like Group 14 Elements
Element Bond distance (Å) Band gap (eV) C 1.54 5.5 Si 2.35 1.1 Ge 2.45 0.7 Sn 2.81 0.1
13 14 15 16 17 18 B C N O F Ne Al Si P S Cl Ar Ga Ge As Se Br Kr In Sn Sb Te I Xe Tl Pb Bi Po At Rn
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry , §6.6.3 (p. 230)
What happens as we move down group 14? Silicon and germanium also have the diamond structure. The bonds get longer, as expected, and the band gap drops a lot. Silicon and germanium have gaps around one electron volt, which suits them to many electronic devices. Gray tin also has the diamond structure, with longer bonds still and a gap of only a tenth of an electron volt, so it is practically a semimetal. White tin is tetragonal and much closer to a metal, and lead is a close-packed metal. We see this change from insulator to semiconductor to metal across much of the p block, so let's look at the band structures to understand it from a bonding point of view.
Silicon and Germanium
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry , Fig. 6.28 (§6.6.3, p. 230)
Silicon and germanium have band structures like diamond, both with indirect gaps. Look at the s star band at gamma, the antibonding combination of the two s orbitals. In carbon it was well above the antibonding p bands. In silicon it is just above them, and in germanium it has dropped well below them, close to the top of the valence band. The atoms get bigger down the group, so the bonds get longer, 1.54 angstroms in diamond and 2.45 in germanium, and neighboring orbitals overlap less. So the bonding states get a little less bonding, the antibonding states get a lot less antibonding, and the gap shrinks. In lead, s star would fall below the bonding p states, so antibonding states would fill first and the diamond structure would be unstable. So falling orbital overlap drives the change from insulator to semiconductor to metal.
