In this lesson, we're going to talk about structures in which we have two ions or two different atoms per lattice point. What I've illustrated up here is the Cesium Chloride Structure. And when you first look at the cesium chloride structure on the left, what it appears to be is body center cubic. However, we need to go back and consider exactly what we mean by the Concept of a Lattice Point. Remember if it's body-centered cubic, then we have two equivalent lattice points. All those that lie on the corners and the one that lies wholly in the interior of the unit cell. Now, if you stand at a lattice point and you move to another lattice point, the environment that you see around that new lattice point has to be exactly the same as the environment that you saw in the first. However, in the case of the cesium chloride structure, when you're standing on the corner, what you see is the open circle whereas when you're sitting at the open circle, what you see is that filled in black circle. So consequently, those are two different environments, so rather than referring to this as a Body Center Cubic unit, what we do is we refer to it as a Simple Cubic Structure in which we have two ions, one cesium ion and one chlorine ion. And those two ions move together as we translate the unit cell in the X, Y, and Z directions. Now there's an alternative way that we can look at the cesium chloride structure. And we can look at it in terms of having two different simple cubic lattices that interpenetrate one another. So for example, if we look at the filled in circles, we're going to refer to those as the B ions and the open circles as the A ions. And each of those on the corners has eight corner positions in one eighth therefore a total of one lattice point for the blue. And one lattice point to the associated black. And so consequently we're dealing with two ions per lattice point. Now, as I say, cesium chloride is the characteristic that we usually refer to these structures by, by calling them cesium chloride. But it turns out that there are other systems. For example, in the metallurgy community, when we look at copper zinc alloys, which are the bases of brasses, what we have is a structure where when the material goes through a particular type of transformation, the copper sits at one of those positions and the zinc sits at the other position. So once again, it's a simple cubic structure. Now what we can do is also because we have the relationship between the radius and the body diagonal of the unit cell, we can actually therefore go through and make a calculation for the density for example. So here is the cesium chloride structure. We have two different atoms or two different ions. In this case, one at the positions 0,0,0. And the other one at one-half, one-half, one-half. Now what we need to do is to turn our attention to what is the diameter or what is the radius of these two ions? So what we're going to do is look up the values and it turns out that we find the value for the cesium and it's written in terms of nanometers. We now look for the radius of the chlorine and so now we have little r and big R and we can come up with the radius ratio, of course. And we can also come up with a relationship with a0 in r. Now in this particular case because we have two different ions, a cesium and a chlorine, we want those two ions to touch one another. So they're going to actually determine the dimensions of the unit cell as we described in an earlier module. So what I can then do is, based upon the fact that I am looking at a structure of cesium chloride. I can look at the number of ions per unit cell times the mass of the ions and in this case, I have two. One of them is cesium and one of them is chlorine. So then I need to come up with the atomic mass of each of those ions. And once I know the atomic mass, I know that I have one ion per unit cell and I can then calculate the mass of that particular structure of cesium chloride in the unit cell. Knowing the relationship between r and a0, I can come up with the value for a0. And once I have a0, I now am able to calculate the density of cesium chloride. So we go through these steps and each one of those then becomes highlighted. And ultimately we come up with the answer of 3.99 grams per centimeter cubed. We can turn our attention to another common structure, namely that of Sodium Chloride. This time sodium chloride is, again, it's 2 ions per lattice point, but now what we're going to be dealing with is a face centered cubic structure. So in this particular case we have a sodium ion and a chlorine ion. And they're located at two types of positions. One of them is on the corner positions, and the other ones are on the face centering operations. So we can look at the unit cells in alternate ways. We can either put the black filled in circles as our corner position, or alternatively we could put the blue filled in positions as our corner. When we look at the structure, where I have the blue chosen as my alternate, I see that I have the sodium ion and the chlorine ion together. The last structure that we want to look at and examine in this particular lesson is the structure called Zinc Blend, or Zinc Sulfide. And here we have two different ions per lattice point, we have a zinc ion and a sulfur ion and they're coupled together. And of course what has to happen, because of the material has to have the stoichiometry of the compound, what we see is once we make a calculation of how many sulfur atoms, how many zinc atoms. We know that there has to be an equal number of those. So it turns out then, since we're dealing with FCC and two ions per lattice point, we now have a total of four zinc and four sulfur, giving us eight ions per unit cell. In the next lesson we'll be looking at slightly more complicated structures. But again, what we want to do in this case is to illustrate some very important points regarding the development of crystalline materials. Thank you.
