Chapter XVII. Solid State Chemistry - Part A

In this Chapter we are going to learn how to deal with a gigantic number of orbitals - roughly Avogadro's number (6.02 x 10²³)! So how are we going to do this? We will take advantage of two factors:

1. Translational symmetry associated with the unit cell

2. Symmetry within the unit cell

3. Perturbation theory ideas within and between unit cells.

But before we get into too many details, we need to have a quick look at a special MO method - Hückel theory.

A. Hückel Theory - the basic tenents are:

1. Use the LCAOMO approach

a. We can proceed exactly as before, except this time using only those p orbitals on adjacent atoms which overlap in a p fashion.

b. We are ignoring all of the filled level interactions, and taking the s bond frame as being fixed.

c. Every adjacent (connected) atom pair has an overlap - S, and a resonance integral b, ie all C-C distances are equal. The starting orbital energy for a carbon p AO is taken as a.

d. No normalization is used. So for ethylene:

2. Remeber that from the top a p AO looks like an s AO:

So the AOs of cyclic polyenes are:

a. The number of nodes increases with each increase in orbital energy.

b. The lowest orbital contains no nodes.

c. The lowest orbital is nondegenerate. The next orbitals on up come in degenerate pairs, except in systems with an even number of atoms, where the highest orbital is nondegenerate. The degeneracy comes from group theory. Look at the C_n group tables. There is an irreducible representation, A, with characters of all +1 so this will lead to the fully symmetric representation. For n=even there is a irreducible representation of B symmtery. This is the totally antisymmetric combination. All others are e sets!

So what will we get when there are ~6 x 10²³ MOs?

y₁ = c₀+c₁+c₂+c₃ +c₄+c₅...+c_N-3+c _N-2+c_N-1+c_N

y_N = c₀-c₁+c₂-c₃ +c₄-c₅...+c_N-3- c_N-2+c_N-1-c_N

and the degenerae pair right at the middle are:

y_N/2 = c₀+c₁-c₂-c₃ +c₄+c₅...+c_N-3+c _N-2-c_N-1-c_N

y_N/2+1 = c₀-c₁-c₂+c₃ +c₄-c₅...+c_N-3-c _N-2-c_N-1+c_N

there is an alternative way to represent them by taking plus and minus combinations:

y_N/2 = 2c₀-2c₂+2c₄...+2c _N-3-2c_N-1 (plus combination)

y_N/2+1 = 2c₁-2c₃+2c₅...+2c _N-2-2c_N (minus combination)

3. Returning to the Hückel approximation the energy of an orbital for a cyclic polyene is given by:

here j runs from 0, ±1, ±2 to N/2 (or [N-1]/2 if N = odd) and so the energy of the lowest MO, y₁, is (j=0) a+2b and at the highest MO, y_N, it is a-2b (b is defined as being negative). Let us recast this equation by defining an index, k, where

We'll get back to what the variable a is in a minute.

Now the Hückel equation for the energy is:

Where k runs from 0 to ±p/a. Let us show these results pictorially. We may plot out the energy levels of, for example, C₅H₅ as shown below using the first expression for the energy. The allowed values of j are 0, ±1, and ±2.

An analogous plot for a ring containing 15 atoms is shown on the right side. Here j runs from 0 through ±1, ±2, and so on to ±7. Finally a diagram for the infinite system is shown below. Now k runs from 0 through -/a or j from 0 through +(N-1)/2 where N is very large, just as in the finite case.

The wavefunction associated with this set of energies is given in a rather simple way:

Here

n = an index for the unit cell (n = 0, 1, 2, 3...N)

a = the lattice spacing

k = the wave vector, which again runs from 0 to ±p/a

Lets take k=0 and p/a:

B. One-dimensional bands

These solutions have exactly the right properties that we want from our previous analysis of a large cyclic molecule. The basic equation we have used is called a Bloch function. We need to be a little more careful with our indices for the general situation. Let us define a positon r from the origin of a coordinate system in a crystal and so the radial portion of our starting AO is c(r). But we also need to have a vector, defined as R, which translates c(r) from one unit cell to another. So our generalized case is

And the corresponding Bloch function is written as:

In this simple one-dimensional case R = na. Returning to our solutions of the energy, e, as a function of the wave vector, k, with the Hückel approximation, notice that the total range of k is from -p/a to +p/a. This is called the first Brillouin zone. Any value - say 3p/a or -7p/a is already covered in this range. However, it is also true that any value of -k = +k. So we only need to plot 1/2 of the energy solutions. This is called an e versus k or band plot.

We can also define a density of states (DOS), n(e), which gives the number of states within a small energy interval around energy e. Then for one band

since a band can be occupied by two electrons per unit cell. The DOS plot looks like

For the one-dimensional case n(e) is known to be inversely proportional to the slope of the e versus k curve

How about a p AO interacting in a sigma way?

Now the band runs from top to bottom, left to right. We can always write down the solution of k = 0 and k = p/a for a few unit cells. The way the orbitals interact between unit cells, ie bonding or antibonding will determine which way the band runs.

C. The hydrogen chain

We have used the Hückel approach for p AOs as a way to see how periodic boundary conditions come about. This problem is topologically identical to a chain of hydrogen s orbitals. In general what we have left out of the picture so far are three aspects:

1. In a realistic calculation overlap is included explicitly and so the antibonding orbitals are destalized more than the bonding ones are stabilized.

2. The band spreads itself out at a particular energy. This is set by what kind of orbital is used, ie it is determined by overlap and electronegativity factors within the unit cell.

3. The magnitude of band dispersion is determined by how much overlap there is between the orbitals between unit cells.

As an example shown on the next page is a calculation for a hydrogen chain where the H-H distance is set at 1.3, 1.7 and 3.0 Å. Recall that the overlap, S, between s AO is µ e^-r so the band width is much smaller, the band is almost flat, when the length of the unit cell is enlarged to 3.0Å. The DOS plots for these same distances are also shown on the next page. Were the band completely flat, then it would be a single peak.

The one last analytical tool that we need is the crystal orbital overlap population, COOP. Recall for a molecule the overlap population for all MOs occupied with 2 electrons is

COOP is the overlap population weighted by the density of states. The H-H COOP for the hydrogen chain is:

Our choice of a unit cell was arbitrary. Suppose we chose two p AOs on a carbon chain or hydrogen s AOs in a unit cell:

The DOS stays exactly the same as before but the band structure plot does change. Now the are two AOs per unit cell so there are two MOs per unit cell and these two MOs spread out to form two bands. This is called band folding.

We might want to double or triple the unit cell to study some distortion or electronegativity change, ie:

when the bond alternation distortion is carried out the resulting band structure and DOS plots are shown on the next page by the dashed lines.

notice that a band gap is created right around the k = p/a points

For the electronegativity perturbation:

Again a band gap is created. Notice here that the electronegativity perturbation can be used to mix the upper and lower bands around the k = 0 point. At k = p/a, there is no mixing since the AO coefficients are on nonadjacent atoms. The perturbation results have the e⁽¹⁾ and e⁽²⁾ corrections both negative for the lower band whereas e⁽²⁾ for the upper band is positive.

We are now in a position to look at any 1-D problem!

Here are some other concepts and nomenclature used in the solid state field:

D. Polyacetylene

Problms to consider:

1. It has alternating C-C bond lengths

2. It is an insulator

3. Upon doping with Br₂, it becomes conducting.

for k = p/a

The real band and DOS plots along with the projection of the pz AO (dashed line) for polyacetylene in the delocalized versus alternating geometries is shown on the next page.

E. The sulfur chain

We are going to try to figure out how a 1-D chain of sulfur atoms might look. Our unit cell is just one S atom with its s and three p AOs. therefore, we should get 4 bands:

What we actually get from a calculation is:

There is actually more detail here, the s and p_z bands undergo an avoided crossing:

The insert shows what happens for an analogous carbon chain. the s/p mixing is even larger so the p_z band actually rises in energy at low k values before it falls. This intermixing is also reflected in the COOP curve; The top of what starts as the

s band should have a negative S-S overlap population, but in fact it is positive. Likewise the bottom of the p band should be positve S-S overlap population and it instead is negative. This is in keeping with the We will come back to what the structure of elemental sulfur really is in the next chapter.