Chapter XVIII. Solid State Chemistry - Part B

A. The Pt(CN)₄^2- Story:

1. What is the bonding in this polymer chain.

2, The Pt-Pt distance in K₂Pt(CN)₄^.3H₂ O is 3.48Å

3. The compound is an electrical insulator.

4. Why does this compound form a 1-D chain?

5. The Pt-Pt distance in K₂Pt(CN)₄^.3H₂ O(Br)_0.3 is 2.85Å.

6. This compound is metallic.

Let's start the discussion with trying to figure out what the band structure and DOS plots will look like. This is the compound without the K atoms and H₂O molecules:

It is a 1-D chain. Each Pt(CN)₄^2- unit is rotated. We will not worry about this and the easiest thing is to take each unit as being eclipsed with eachother so the unit cell contains just one Pt(CN)₄^2- unit. The MO's are something that we have seen before:

Taking the five d-based MO's along with p_z, they will spread out into bands because of intercell overlap (particularly for the compound where Br₂ has been added). So the result is

We can easily tell the band width from the type of overlap and which way the bands run by writing down solutions for the two k points.

For the real calculation, we are going to use a simplified model - PtH₄^2- - this will show us the principle patterns in an easier manner. The band structure of PtH₄^2- is

And the DOS with several d - AO projections is

So what happens when Br₂ is added? The shrinking of the Pt-Pt bond lengths and electrical conductivity is easy to explain

So why does the undoped material hold together at all? A good part of it has to do with the formation of hydrogen bonds to the cyano nitrogen atoms from the water molecules of hydration. But look carefully at the COOP curve. It does not show the expected feature of more antibonding at the top of the Pt z² band. This is due to mixing with the z band :at intermediate k points-

B. Electron counting and the Zintl-Klemm concept.

1. Take all valence electrons from the very electropositive elements and give them to the electronegative ones.

2. If the electronegative elements are short to fulfilling the Lewis octet rule, then make bonds between them.

C. Some elementary structures:

1. NaCl

2. S (elemental)

3. BaSi

4. BaSi₂

5. Ba₃Si₄

6. BaLiSi₂ = Ba₂Li₂Si₄

7. SiS₂

8. ZnS, GaS and As

9. Ba₂Si₃Ni

D. Two and Three-Dimensional Structures.

Recall that the Bloch functions were defined as

where R = na.

For a two or three-dimensional structure - in other words where the unit cells are translated in two or three dimensions - this formula must be expanded. Since k is a wave vector, it now needs to be expanded to encorporate the other dimensions. Let us take a very simple structure based

on a cube with coordiantes a, b and c (in this case like the x, y, z coordinates in Cartesian space but the a, b, c unit dimensions do not have to be the same).

And for the first Brillouin zone:

-p/a £ k_a £ p/a

-p/b £ k_b £ p/b

-p/c £ k_c £ p/c

Once more we only need to plot one-half of the values that the k's span since +k = -k. We are going to look at three very simple examples, namely first a 2-D net of hydrogen atoms, secondly a cube of phosphorus atoms and finally the p MOs in graphite.

1. The 2-D hydrogen net. Let the H-H distance be the same in the x direction as it is in the y direction. That stucture is said to be tetragonal (if the x and y directions were not the same then it is an orthorhombic structure).

The translation vectors here are x and y. The corresponding Bloch functions are given by

The part of the Brillouin zone that is unique and needs to be considered is shown below in the dotted area. We can do calculations for each of the (k_x,k_y) given by the points (in practise one does many more) and this will give us the DOS curve.

For the e versus k band structure plot the best one can do is to cover the range of values by going around the high symmetry points of the Brillouin zone. A calculation of the bands for an H-H distance of 1.3Å is given below.

The orbitals can easily be constructed from the Bloch functions:

The density of states plots for H-H distances of 1.5, 1.7 and 2.2Å are shown below. These shapes are very representative for a 2-D band.

2. A cube of phosphorus atoms. Lets take phosphorus or any other group 15 element with a cubic structure. One might think that the s and p bands will have a large dispersion if the P-P distance was about that or perhaps a little larger than a typical single bond. With five valence electrons, the s bands would be full and the p bands would hold three electrons. Certainly this is not going to be a viable solution. Partly filled bands often signal that a distortion will take place. So let's see how this comes about. The first Brillouin zone for a material with a cubic structure is:

The p bands for this are 1/2 filled - there are three out of a possible of six electrons in each unit cell. The p bands using a Hückel approach for both the p and s interactions leads to:

Fermi level

A distortion is going to happen - half of the six bonds are broken:

There is another, equally valid way to look at this structure that we shall see in the next example.

3. Graphite.

Graphite is one of three forms of elemental carbon.

diamond

graphite

This has a hexagonal unit cell which is a little more difficult to deal with. First of all there are a couple of choices how to translate the unit cell (a C₂ unit). We have picked one shown below that is not common.

The translation in the a and b directions takes the solid circle to the open ones. Several unit cells are outlined by the dashed lines.

The unique portion of the Brillouin zone is shown by the shaded area within the hexagon. Normally one would plot from say G to X to K to G. The K point with k_a = 1/3, k_b = 1/3 (in units of 2p/a and 2p/b) has coefficients that are immaginary. To see what happens we will plot from X to G to K to M. Here the M point with k_a = 1/2, k_b = 1/2 is outside the Brillouin zone. In fact it is identical to the X point!

Here is the e versus k plot for the two p bands, p and p*.

The orbitals are drawn out on the next page. Notice that p and p* bands become degenerate at the K point. This is a result of the fact that the p* band at M must be lower in energy than the p band at this point. Notice also that p* at M is identical in energy and form with p at X and p at M is identical to p* at X. So at the K point filled and empty orbitals meet. Graphite is not, however, a good metal.

It is said to be a psuedo-metal. As one can see from the DOS plot below, the Fermi level falls in a region where the density of sates is small, ie just around k values close to K. Also

notice if there are two more electrons in the unit cell the p and p* bands will be completely filled. A distortion just like in ammonia or water is to be expcted. This is another way to show how the black phosphorus structure evolves.

In the next chapter we will look at really big systems where the band structure is quite complex and more of a reliance on the DOS curves and numerical results is necessary.