Okay, this will just remind me to start recording, so don't forget some announcements. Yes, so quiz one is available, and again, this will only be, for this time will be full credit if you attempt it. It is a timed quiz, so there's five questions. You have seven minutes to complete it. It's very similar to the homework questions. And this is what the quizzes will be like for the remaining quizzes as well. And it's only available today, so you have until midnight tonight to start and complete this quiz in seven minutes. Homework two, I'll post that today. It'll be due next Wednesday at noon. It has 20 questions. I think I might actually, I still have to edit one of the questions, but that'll be available this afternoon, this evening. Next week, we have a new TA in charge of quiz section that will be SID. And homework two is due next week, and again, there'll be another quiz on the same day in the similar format. And then the week after that is your first midterm. So I will not be doing a quiz section, but during quiz section, I believe this is the case, you'll be taking your midterm. And we still don't know the format. It's either going to be on canvas or through, I mean, it's definitely going to be on canvas, but it might be similar to the quiz where you have like an hour to complete it and whatnot. You also have homework due that day as well. Okay. And then make sure you're on top of the reading. Like I said, the summer quarter things are going by a bit quickly. So, you know, read ahead for next week. This is the reading assignments for next week. Also, I've established my office hours on Monday, after class on Monday, every now and then I might offer some additional office hours and I'll send an announcement out. Okay. So Dr. Brush today talked about crystals, and I wanted to talk more about that crystallography. Of course, the study of the atomic arrangement in solids. So you have a crystalline solid where it has a long range order, and then opposed to an amorphous solids known as a glass material where it still has short range order, right? So you have silicon, silicon still tetrahedrally coordinated to oxygen, but there's no long range order in an amorphous solid. All right. So this is an example of quartz versus silica. In quartz and silica, they have the same chemistry, the same components, silicon and oxygen, but one is crystalline, one is amorphous. And of course, as you know, silica makes up most glass is a major component of like window glass and other glasses. Like drinking glasses. So here's another example of crystals. These three crystals look very similar, right? They have the same sort of shape to them. And although they're different colors. So this is corundum, ruby and sapphire. And does anyone know the atomic, excuse me, the chemical formula for these three crystals? Or one of these three crystals? Okay. The chemical formula for corundum, ruby and sapphire is aluminum oxide. Let me, oops, move my mouse. Aluminum oxide. For all three of these are made out of the same base material, which is aluminum oxide, a crystal, crystalline aluminum oxide. Now for ruby and sapphire is interesting to make the color, it gets doped, doping. That means we're adding some kind of impurity to the crystal. In this case, for ruby, we're adding about one to three percent by weight, I think it's weight, not atomic percent, chromium three plus, and we're substituting it with aluminum inside the crystal structure. So the crystal structure is still the same, just some of the atoms have been replaced with chromium. And that makes the crystal red. And sapphire, you do the same thing with titanium and iron, you have to have both in order to make it blue color. So that's a little bit of interesting. But the point is that they have the same crystal structure. And that's what gives it this macroscopic shape. And as Dr. Brush had mentioned, there's polycrystals. Polycrystalline material have many smaller crystals. And the boundaries of these crystals are called grain boundaries. So each one of these crystallites has its atomic arrangement, this crystal structure, but they might be in different orientations next to each other. Here's an example of a polycrystalline material. Not only is it polycrystalline, but it's polycomposite material. You could consider it a composite, right? It has multiple minerals, multiple different chemical formula, excuse me, chemical compounds that make up this granite. So quartz, mica, feldspar. Here's a little quiz I like to give this to. I do a lot of outreach for K through 12. So I'm going to give you guys the same kind of quiz. Which one of these materials contains crystals? And there's one of these materials that does not contain crystals. So in the chat where you can shout it out, which one of these do you think does not contain crystals? We have crystal wear, copper metal, graphite, polyethylene. All right, it looks like most of you guys are saying the polyethylene jug does not contain crystals. And that is incorrect. Yeah, it's a trick question. The crystal wear is the only material on this list that does not contain crystals. Crystal wear is actually glass. It's just called crystal wear because it has facets, it's cut, and it looks like crystals. Where, you know, this is something I learned, I didn't know until like high school, I took a material science camp here at the UW that metals contain crystals and that kind of blew my mind that metals and everything almost is made up of crystals. Graphite, of course, has a crystal structure. We talked about that on the first week. The polymer, and this is where most people say it doesn't have contained crystals, polymers do will be kind of semi-crystalline where their polymer chains will arrange in a periodic fashion with long range order. And so that is the polymers do contain crystals, crystal structure. Crystal wear is a specific type of glass. Does anyone know? No, the K through 12 students also all said the polymer. Some of them say the copper metal too. So it's all right. The crystal where does anyone know what type of glass crystal wear is or what the chemical what the major what what's the chemical formula that makes crystal wear unique and doesn't look like anyone knows. So crystal wear boron silicate. Nope, not quite not, but it's not a borosilicate glass like Pyrex is a borosilicate glass. Crystal wear is a lead glass. So, you know, if you have any fine crystal in your house, it contains a bit of lead oxide. And the reason why does anyone know why? Well, I guess you might not know because you didn't know the first question. The reason why crystal wear uses lead oxide is that the lead oxide component in glass, adding lead oxide to this glass increases the index of refraction. All right, the refractive index. So it makes it so the light that bounces in bounces out at a higher angle. And so it makes it optically attractive, right? Like diamond has a very high index of refraction. Cubic zirconia has a very high index of refraction. That's why we use it as a substitute for diamond and jewelry. Crystal lead glass also has a fairly high index of refraction. The other reason why lead glass is used for crystal wear is that it makes it easy to machine. It makes it softer. It's easier to cut these facets into the glass to make it, as we call it, a crystal. But in fact, it's actually amorphous. Does it make it melt into molds? Yes. But related to the manufacturer ability, it makes it easier to cut to make these facets. And there's actually been some studies that look at, you know, different types of beverages, acidic beverages like wine, and how much lead is actually leached out of the wine.   So I should probably check in how safe it actually is. Anyways, just and here's another example of crystals in real life. So you might have seen these guard rails or like light posts, lamp posts, hand rails that are zinc galvanized, right? So they're made out of steel, but they're dip coated in zinc metal. And to act as a corrosion protection for the steel or iron. And the zinc makes these very beautiful crystals in the shape and size of these crystals. This is called a spangle, right? You can have different types of spangle. And depending on the different additives you put into the zinc alloy, you can make them smaller or bigger for different optical properties. But it's mainly for corrosion prevention. So next time you're at the street corner and you take a look at if the lamp post is zinc galvanized, you can see spot out these giant crystals. So metals, the element, pure elements have a variety of different crystal structures. And we're going to look at some of the more common crystal structures that Dr. Brush has already covered. But you can see how the crystal structures follow these different chemical groups. So obviously, how the atoms arrange depends on the electron configuration in the chemistry of the atoms. So the common ones are the face centered cubic, also known as cubic close packing.  And then BCC is another common iron at room temperature is BCC. And the alkali metals are also BCC. And then hexagonal close pack is also the common element or crystal structure we talked about, titanium specifically is when I think of HCP, I always think of titanium metal. Yes, some of the crystal structures will change at different temperatures. Iron is the obvious example of that at a higher temperature, I forget which what temp is like 700 and something degrees Celsius, 724, something like that, it changes from BCC into FCC. And then at even higher temperature, it changes back to VCC. Another interesting thing about the crystal structure of iron is when we start talking about alloys, so we're substituting some of the iron atoms for some other atoms, specifically stainless steel is made of a nickel iron alloy, there's a large percentage of nickel. So if you put a large percentage of nickel, substitute with the iron, what happens is your crystal structure changes, the crystal structure of stainless steel is FCC, not BCC. All right, let's move on. So this, we're coming to our first question, first problem is unit cell calculations. I'm going to have you guys do these number of problems for each unit cell. The first is the number of atoms in the unit cell. So in this example is just the simple cubic lattice. And as Dr. Brush had said, polonium is the only single element that forms a simple cubic lattice. All right, the crystal structure is simple cubic. And we'll talk about this a bit more. The crystal structure is made up of a lattice points and a basis. In this case, it's just a single atom. All right. And so the number of atoms in each unit cell you need to calculate for simple cubic is rather easy. Each corner, there's eight corners and each corner has one eighth of an atom. So there's one atom per unit cell. Okay. And then you want to find out the coordination number. So this is the number of nearest neighbors for any atom. They're all, they're all, they're nonunique, right? They're all the same. So if you take one atom here and find its nearest neighbors, and you say, you have to kind of imagine what does the unit cell above and to the right and in front of it look like. So if we take this corner atom, there's these three here, one, two, three, and there's also going to be three. There's one above, one to the right and one in front. So the coordination number is six, which makes an octahedral coordination. Okay. Then last parameter, A. So that's the dimension of the unit cell. Okay. And in terms of radius. So a simple cubic is very easy. It's just the two radii. So A equals two R. And then also the atomic packing factor, it's going to be the volume of atoms. So four thirds pi R cubed times the number of atoms in the unit cell divided by the volume of the unit cell. So be in this case, a cubed, which is two R cubed. Okay. So a simple cubic is not very dense. It's a 50% atomic packing factor. All right. I wanted to do this we're going to do it for body center cubic and face center cubic, but not hexagonal. It's a little bit more difficult. I wanted to do this, break you guys into breakout rooms. I haven't really prepared it. And then I'll give you about five or so minutes to do it. We're going to start with the face center cubic. And then I after five or so minutes, I'll call out on a group for the answer. But so when I break you out into breakout rooms, I think you're going to lose this page. So what I suggest you doing is if you're on your computer, if you open up like PowerPoint or Word, you can quickly copy this, this slide. If you're on Windows, the shortcut for a screenshot is just Windows shift S. And then it'll allow you to make a screenshot. I don't know if you can see what I see because my free Mr. Green, this gets frozen. And then when you select the area you want a screenshot, it copies it to the clipboard. So for example, I like to do everything in PowerPoint. So yeah, there it is. So that way you have it when I break you out in the breakout room. So I'll go back here real quick. This will be a little bit of an experiment for me because I haven't really done this before. Let's see, we're going to have, let's try to get four participants per room, about three to four participants per room. All right, looks like everyone's back now. So we're going to go over the answers. I'm going to pick on people from different groups to give their answer. Now, if you don't have an answer, that's okay, just say you didn't come up with an answer or whatever excuse you want to come up with, but it's totally fine. So let's start with from, let's see, room one, if you're there, can you answer the first question? How many atoms fit in the in one FCC unit cells? Count the number of nearest neighbors or neighbors that are touching. So below it, directly below it, we should have four, right? We have one and two, we can see two here, but there's also two in the back that we can't see, then along the equator here, we have these four that we can see here. And then directly above it, there's going to be another four. All right, so four times three is 12. And this makes this symmetry called a cubo octahedron. All right, so if you have your central atom, and then this is the nearest neighbors around it makes 12 nearest neighbors. All right, how about the next question, last parameter, a in terms of radius r. So in this diagram I drew here is just of the face of the FCC unit cell. And we're looking for a in terms of r, the radius of the atom. So here we can draw the shape, we have a triangle where the, what is this, the hypotenuse, man, it's been a long time since trigonometry, is equal to four r. All right, and then the legs of the triangle are a, and we can use, what is that Pythagorean's theorem, a squared plus a b squared equals c squared, and then you can solve for a in terms of r. So a equals two r times a squared of two. So if you know the radius of any atom, okay, you can calculate, and you know it's FCC crystal structure, then you can calculate its lattice parameter a. Okay, and then from that you can use it to calculate the density of the material, if you know it's molar mass. 0.74 is correct, yeah. So you just take the volume of the atoms in the unit cell. So we have four atoms times the volume of a sphere divided by the volume of the unit cell. And if you've calculated the parameter, last parameter, just a squared, or cubed, excuse me, a cubed, so it's that value cubed, 0.74. So that's an important number to memorize, 0.74 for FCC, 0.74. Also 12 is an important number, and four. So these, those are the important numbers. Okay, let's move on. And we're kind of, it takes a little bit of time to go through all this. So I think if it's okay with you guys, we're going to just not break out into breakout rooms for this next one. So I can move on to the next content. But the same thing for BCC. So I'll go through it with you, or we can just answer out loud. So how many atoms fit in one? This is a typo here. It should say one BCC unit cell. How many atoms are inside one BCC unit cell? Anyone want to shout out? Should be two. Yes, that is two. So how about the coordination number? Eight. Eight is correct, right? So just take this one out on the center, and it's got eight corners surrounding it. So it makes a cubic geometry. How about the last parameter? This one's a bit more time consuming. So I'll just go through it. So here I've drawn a shape, drawn a plane that cuts from edge to edge, right down the center of this cube. And I've drawn out the two dimensional projection of that plane of the atoms. And you'll see that there's a continuous three atoms down the center here. So that's equal to four r, four times the radius. And then the edge is the last parameter a. So that's what we're looking for in terms of r. And then I've also drawn this triangle, which is the base of the square. And so I've kind of unfolded it out into this two dimensional projection, kind of like origami here. And it has two last parameters as well, these two edges here. So again, through Pyrethagorean's theorem, a squared plus a squared equals c squared, a squared b squared equals c squared. So that makes this edge length here along the diagonal, a times the square root of two. So now we have our two legs, a and b, and that equals c. So again, Pyrethagorean's theorem. And you get a in terms of r. So a equals four r over square root of three. How about atomic packing factor? Does anyone know the answer on the top of their head? Yeah, 0.683 says that's correct, 0.68. So it's the same calculation, you have the information you need. But like I said, for for cubic, for FCC and VCC and HCP, you should memorize these numbers, because they might come up in a test or something, and you don't have to calculate it. Yeah, two atoms per unit cell, eight nearest neighbors, the coordination, and 0.68, 68%. Hexagonal close pack, not going to do any calculations, really, but it's just to visualize how the atoms are packed, right? So here's again, a two dimensional projection of just this first layer here, and then I move up to the layer above it. Okay, so this is that J atom here on the layer above it, and then the next layer on top of that. Okay, so you get the parameter, last parameter a is just two r, and there's two atoms per this unit cell and the smaller unit cell, but the expanded unit cell, the hexagonal unit cell, there's six atoms. But what is important, like Dr. Brush had mentioned in class, is the similarity between HCP and FCC. They are both close packed lattices, and they both have the same atomic packing factor, right, which was 0.74. I get that one, right? I always forget. I haven't memorized it myself. Yeah, 0.74. Just double checking. Okay, so they have the same packing factor, and like Dr. Brush had mentioned, the only difference is the, you know, they have these close packed planes of atoms, and the layering sequence is different where HCP is ABA, and FCC is ABC. And I've tried to visualize this further for the FCC, so you can try to see where the cubic structure comes from in this orientation, right? So you can see where this is the FCC cube.  Okay, and then the coordination is the same for both of them, but what is different is the symmetry of that, the geometry, right? So for HCP, which is ABA, it has this weird shape, triangular orthobiopola. You don't need to know the names of anything, but you do need to know that the coordination is 12. And so this is that visualization of the 12, and then FCC, because it's ABC, it has a different geometry, cubo-octahedron, and this is important, and I think the mechanical engineers in the group will find this interesting. You know, even though they have the same coordination, the same atomic packing factor, but because the difference in the stacking between HCP and FCC, their mechanical properties are much different in general. For metals that are FCC crystal structure, that's aluminum, copper, nickel, gold, they are very malleable metals, but metals that are HCP, primarily titanium, are not very malleable. Okay, and that's because of the crystal structure, the arrangement of these layers. To make a metal malleable means we're we're we're plastically deforming the material, and means we're physically having the atoms slide past each other when we pull the metal apart. All right, and so the question is how easy is it for atoms to slide past each other between these two different crystal structures, and it depends on the number of directions that the plane that we're talking about, a plane of atom, like an A plane, B plane, that can slide. Okay, so I try to visualize this. This is a little bit beyond the scope of the class, although I think we might talk a little bit more about it when we discuss mechanical properties, but at this point you don't really need to know, but so I've drawn out a plane of atoms. All right, this is a closed packed plane. For FCC and HCP, they both have these same closed packed planes, but FCC has more than one of the same closed packed planes. So I'm going to draw another one here. Here's another plane of closed packed atoms. Okay, the same shape, the same arrangement of atoms, where HCP does not have that arrangement. You can't make another one of these hexagons in this geometry. Okay, and then actually FCC has two more, so it has a total of four closed packed planes. Okay, and so later on you'll talk about slip systems. These are the systems of how atoms can slide past each other. All right, and so FCC has many more slip systems than HCP does. HCP only has three slip systems. FCC has 12. That's why FCC metals are in general are more malleable than HCP. Okay. All right, let's move on. So yeah, the equation is to know APF, atomic factor factor, the volume of the atoms divided by unit cell. This will also come up. The density, I think you've already probably seen it in last homework, is of course mass over volume, but when you're calculating using the atoms and the crystal structure, you can use this equation here. N is the number of atoms in the unit cell. A is the molar mass of the atom, or atomic mass of the atom. N is Avogadro's number. Vc is the volume of the unit cell. Okay, so as long as you know those parameters like the A and R, then you should be able to calculate density if you're given a molar mass. And you should also know the number of atoms in the unit cell. N is the number of atoms in the unit cell. So we calculated that for FCC, it was four atoms, BCC it was two atoms, and so on. A is the molar mass. So it's going to be in grams per mole, okay. N sub A is just Avogadro's number. What was that? 6.022 times 10 to 23. What is that? Atoms per molecule, just atoms, right? Per mole. Atoms per mole, excuse me. So then that will cancel out the moles, and you're left with grams per atom times the number of atoms divided by Vc is the volume of the unit cell. So you're left with, you know, the mass of the atoms in the unit cell divided by the volume of the unit cell. And that's density. Okay, so we're going to move on and talk about some symmetry of crystal structures. And Dr. Brush had briefly mentioned this, the Bravais lattices. So like he said, all crystals can be arranged or categorized into one of these 14 Bravais lattices, okay. And so the ones that we typically deal with are the cubic system because the symmetry is easy. Mathematically, it's easy to work with. But again, there's different lattice types. And these are just lattices, okay. These are not crystal structures. Lattices are just a set of points in space. It doesn't mean where the, I mean, atoms can be put onto these lattice points or a group of atoms can be put onto these lattice points. And then there's a bit of complexity, you know, especially in the hexagonal, so I just screen clip this from Wikipedia if you look up like crystal structure or something, you know, between hexagonal, the crystal family system and the lattice system and trigonal hexagonal. It's not, you know, you don't need to know it's not too important. But we'll talk a little bit when we start looking at hexagonal closed packed in which system it's in. But what this is showing is the Bravais lattice. And it's just a way to categorize the symmetries of these crystals. In a graduate level courses, we talk about space group symmetries, which are just further ways of defining the symmetry of the different crystals and so on. And so what I want to emphasize, and a lot of students miss this, and even I've even had some professors that have incorrectly said this. But I want to emphasize that the crystal structure is made up of the lattice plus a basis. And this oftentimes is also called the motif. So I use these interchangeably motif and basis. So the lattice, like I said, is just a group of points in space. It doesn't represent, you know, the crystal structure physically is just a group of points. So FCC is a lattice is a Bravais lattice. And for example, gold, the metal has the FCC crystal structure. So in this case, the lattice and crystal structure have the same name. Okay, but it's not the same. It's not necessary for other crystals. Here's another example. BCC of iron has a BCC crystal structure. And then inert gases, this is methane, and other noble gases, if you cool it down to solidify them, generally form in the FCC crystal structure. So in this case, the basis is a group of atoms. It's a molecule. And if you cool it down enough, it forms the FCC crystal structure. But then it can get more complex. So here's an example. This is the cesium chloride crystal structure. So what do you think the base or excuse me, what do you think the lattice is of this crystal structure? What is the lattice? And specifically, what is the Bravais lattice? Right? So it's going to be one of these cubic bang BCC because it's incorrect. And this is the point I'm trying to make the lattice in for this crystal structure is not BCC. It's actually simple cubic. Okay, so I have drawn this out, the motif is cesium chloride, it's a group of two atoms. Okay, so that's the motif there. There's the simple cubic lattice points. Okay, so if you were to take this motif, let's take this chlorine ion with the cesium attached, you move, you translate it to each one of these lattice points. That's how you expand to this cesium chloride crystal structure. So the crystal structure is not BCC. That's why I've heard some professors say, oh, this is BCC. No, it's not BCC. BCC would be if the atom in the center is the same as the atoms at these lattice points. But the atom in the center is not the same. Okay, the lattice structure is simple cubic. The Bravais lattice is simple cubic, it has a motif of two atoms. So if you translate this motif to all of these different lattice points, that's how you build up the crystal. And the crystal structure is called cesium chloride. It's the other like cesium bromide, cesium iodide, I think they all make the cesium chloride crystal structure. So does that make sense? Why does the central atom not count? So in BCC, and I should have included this, I had a picture of what it would look like if it was BCC. In BCC, all the atoms are the same. Okay, like iron is BCC. Okay, but in this case, the atom in the center is not the same as the atom on the corner. Yeah, it's simple cubic of cesium chloride, like of the cesium chloride molecule, I guess you could say. All right, so this is what I wanted to emphasize, that this is not the BCC structure. And I have some more examples that should make it clear. So here are a set of different crystal structures and their respective names. So sodium chloride is called the rock salt crystal structure. It's also called the halite crystal structure is the same thing. So what is the lattice? What is the Bravais lattice for sodium chloride? What do you guys think? All right, it's not BCC this time. This time it's FCC. Yeah. And the motif, what do you think the motif is for FCC? And you can see it here, just look at these blue atoms. Which atom is the blue atom? It is smaller, so it's going to be the cation. Typically cations are smaller. We're giving away electrons. So you see the sodium makes this FCC lattice here. So what is the basis? Yeah, so sodium chloride is right, two atoms. So sodium chloride is the basis. So if you took this sodium, attached to this chlorine, even if you just took the middle or center of mass and you put it at each one of these lattice points, you're going to construct the sodium chloride crystal structure. Okay, how about calcium fluoride or fluoride crystal structure? What's the lattice? Yeah, this one's also FCC. And how about the basis for this one? Yeah, so this one was you see, here's the FCC lattice right here. Was this the calcium? Makes that FCC structure. All right, I'm going to move on. The motif for this is just a group of three atoms, calcium and two fluorines. So if you were to take calcium and two fluorines and translated it to every lattice point in the FCC, you're going to make the fluoride structure. So if you took, say, these three atoms right here, calcium and two fluorines, you translate this over to the next one, you're going to build up the symmetry. And there shouldn't be any overlapping. When you translate it, there's no duplicates. That wouldn't work. They're all unique. Okay, now we get to titanium, the HCP. So I want to emphasize this as well because a lot of students get this incorrect. HCP is a crystal structure. It is not a Bravais lattice. If we go back to the Bravais lattices, and this will give away the next answer, hexagonal doesn't have face center, doesn't have body center. There's only simple hexagonal is the only Bravais lattice for hexagonal. So the crystal structure is HCP. What is the lattice I already gave away? It's hexagonal. The motif is actually two titanium atoms. So that's the point I want to make that lattice points and crystal structure are different. Okay, they're not the same lattice. It's just a set of points in space. And depends on what you put onto those points in space, and that can build up. And that's how we get so many different types of crystal structures, not just 14. There's way more than 14 crystal structures, but there's only 14 Bravais lattices. Okay. Yeah, simple hexagonal. Let's move on. So here's a question. And I don't remember if Brush got to this in his lecture, but it's definitely part of the reading, I believe. Here's an example of an ionic solid, magnesium oxide. So magnesium is cation 2 plus oxygen is anion 2 minus it the ionic solid. So oxygen arranges itself into an FCC anion lattice. So here are the lattice points for the oxygen. So you can assume that oxygen is going to be on each one of these points. Okay. The question is, where will the magnesium go? And this deals with what we call interstitials. Interstitials are the spaces in between. So in FCC, there are two different types of interstitials. There is this red one here is the octahedral interstitial, interstitial, right? That's made up from the faces. And then on the corners is the tetrahedral interstitial. So for each unit cell, whether it should be eight, there should be eight octahedral, excuse me, there should be eight tetrahedral interstitial sites. And I believe there's also, let's see, there's one on each face, seven. No, it's each edge. So there's what, one, two, three, four, that's eight, that's 12, 13. Oh, I forgot to divide by two. Anyways, the question is, where will magnesium go? Will magnesium go into the octahedral sites or will go into the tetrahedral sites? Does anyone know how to go about solving the problem? What do we have to look at? And I believe this was part of the reading for this week. Yeah, the radii. So we look at the atomic radii. And what do we look about?  Make sure they don't overlap. I think there's a, it's a more general rule to determine whether it's octahedrally coordinated or tetrahedrally coordinated. So I'll go into that. So this is from Callister. It might be chapter 14, actually, which is part of your reading assignment. And you, what you do is take the ratio of the cation to the anion radius. Okay. And it's a general rule of thumb. If they fall in one of these ranges, that will be the coordination of the anion or the cation to the anion. And remember, I usually the cation is the smaller anion. Excuse me. I'm getting my words messed up. Cation is usually the smaller ion because you're giving away electrons. Your electron shell is getting smaller. The anion is usually the bigger ion because it's receiving electrons. It's less, it's more loosely bounded to that atom. And so in this case, you just take the ratio between the two. Okay. And so the ratio between magnesium and oxygen 0.51. So that falls into here coordination number six, which is an octahedral. All right. So magnesium is going to fit into the octahedrals, not into the tetrahedrals. It's more energetically favorable. And that's why these numbers are determined. It's which coordination makes it more energetically favorable. If you have a too small of an atom inside the octahedral is not very energetically stable. You can think of like a small atom rattling around in a cage, it rather be too big than it than too small. So this is what that looks like. If we take magnesium, put it at each of these edges and inside the center. And then if we expand this octahedral, this is the next door neighbor octahedral interstitial site. I think there was another question I had. Yeah, what's the crystal structure of magnesium oxide? What do you guys think? And you should be able to just look at this and know right away what the crystal structure and it's one of the crystals that we already talked about. What is the crystal structure? And what I'm looking for is the name of the crystal structure. Right. So not BCC. BCC is only a crystal structure if the entire material is the same composition like iron. It's all made of iron. That's the only time BCC can be a crystal structure. Otherwise, it's a lattice. Yeah, you're right. Halite is correct. Sodium chloride crystal structure or rock salt. This is the halite rock salt or sodium chloride crystal structure. All right. It's the same as what we saw before. I won't go all the way back. Okay. Let's go into Miller indices and crystallographic directions. So if we want to describe the range or the plane of atoms inside a unit cell or perhaps what the atoms look like at the surface of different single crystals, we use what's called Miller indices. Here's an example of the spinel crystal structure. This is iron 304. It's also iron 23 oxide because it has mixed of iron 2 plus and iron 3 plus. It's called magnetite. It makes the spinel crystal structure and these nanoparticles of that spinel and macro particles, crystals having the same shape. And the shape is determined, of course, by the crystal structure. But the question is, you know, how can you describe the surface of these crystals? And we use Miller indices. Miller indices have this abbreviation Hkl. And these are integer values. So here's another example of some kind of nanocrystal, I forget, but it describes the different surfaces of this crystal with the Miller indices. And then there's also you should distinguish it from the direction. Where are the crystal structures such as halite found? I don't quite don't quite understand the question. Where are crystal structures such as halite found? You mean like on earth? Oh, the like a list of crystal structures. I am pretty sure chapter three. If not chapter three, it's chapter 14, I believe was part of the reading. There's a small reading assignment chapter 14. I think that talks about ceramic crystals 14 or 12. Oh, did I mix that up? It might be 12. Then if I'm, I don't have it right in front of me. So I'm not sure. Maybe 12 not 14. The chapter is about ceramic crystals or ionic crystals, I believe, and it should give you a different list of crystal structures. And then the direction are just vectors, but you need to make sure that you're using the correct notation. If you're talking about a plane in space for Miller indices is using parentheses or sometimes these curly brackets, we're talking about a family of planes. And if it's a direction, you use the square brackets or these little, what do you call those? I don't know, carrots, arrows, whatever, for family of directions. So I have a couple of problems, but I think we'll just go through them because we don't have much time. And like I said, don't feel obligated to stay. I this recording will be posted this evening once it finishes compiling. So for crystallographic directions, all right, using the square brackets is very simple, just vector math, right? So you take the final coordinate point in real space and minus the initial coordinate point. Or excuse me, this should be of lattice space because this, this, you know, here I show X, Y, and Z, and to think of three dimensions. But that's only correct if your unit cell is orthogonal, okay, if it has these right angles. But as you know, some unit cells don't have right angles at the origin. So it's not really correct to say X, Y, and Z. But we're talking about lattice space. The vector math will still be the same. So you should use these unit vectors instead of X, Y, and Z. But in these examples, they're all cubic. So it's all the same. But like in this example, you know, just doing the vector math, finding the end end coordinate, the beginning coordinate, and that will give you and you should multiply the fractions out. So in this example, the end coordinate is one and three quarters zero, because there's no, there's no change in the Z dimension. Initials the origin. So you need to do is multiply by four to remove fractions. Okay, so there's no fractions in the Miller notation for directions or for planes. So this direction will be called brackets four three zero. Okay, I had some questions, I was going to have you guys do yourself. And but because there's not enough time. And I'm pretty sure your homework deals with questions like this as well. But here's I'll just go through this. So this direction to point a, right, the coordinates are one half one half one half, all right, and it starts at the origin. So then if we multiply out the fractions, so we multiply everything by two, this is going to be one one one, right, and brackets one one one is the one one one direction. This one is not starting at the origin. Okay, so you have to kind of do the math here. So we start in, you can always move this to the origin to make it easier. And that's what I'll try to visualize. So in the x direction, we're moving over one third, we're going from two thirds to one. Okay, and then in the z y direction, we're moving one and the z direction, we're moving negative one. For negatives in Miller notation, negatives are written with a bar over the number. Okay, so we need to multiply out the fraction. So this was what I say one third for the x direction. So we're going to multiply everything by three. So it's going to be turned into three, and then one in the y direction, and then one, excuse me, I'm sorry, I messed that up. It's going to be one, three, three bar. Okay, one, three, three bar. Okay, three bar means negative. And then the last one, we're going from this corner here, removing positive one in the x direction, negative one half in the y direction, and negative one in the z direction. Okay, we remove the fraction. So we're going to multiply everything by two. So it's going to be two, one bar, one bar. Let's see. Oh, excuse me, did I do that wrong? Two, one bar, two bar. Yeah, I forgot. Okay, now for Miller indices, talking about planes, it's a bit more complex. It's not as simple. There's a, we need to change it into what we call reciprocal space, but still not too complicated. A lot of your homework will be, will have some of these problems. In fact, this might even be a problem from your homework. Let's start with B, for example. So we want to find out what the Miller indices of plane B is. B is easier than A, so we'll start with B. So what you need to do is from the origin, find the intercepts of this plane with the different axes. All right, so this B plane intercepts x-axis at one half. Okay, it intercepts the y-axis at one half, and intercepts the z-axis at one. Okay, and then the next step is to take the reciprocal. So that's what this, this is how it's different than the regular vector notation. We're going to take the reciprocal of this, these intercepts, so becomes two, two, one. Reciprocal of one is one. And then if there's any fractions, you multiply them out. And then the final notation is two, two, one for this plane. That's how we describe this plane. For, for A, it's a bit more complex because A intersects our origin. So we need to move our origin. We can't have it intersect the origin. We need to move the origin in order to calculate it. So if we move the origin to this position here, so we consider this zero, zero, zero. Sorry. In the x-axis, intercepts at two thirds. The y-axis, now we're going to negative, intersects at one. And the z-axis intercepts at one half. Let me see. There's a question. How do we know when to write two, two, one that we mean the plane and not the vector? Oh yeah. So the vector is always going to be, this is the square brackets.  But when it's a plane, we use the, the parentheses. So that's how you know. So moving, going back, let's see where was I, the, the z, the y-direction is negative one. And the z direction is one half. Right here's, excuse me, the z-intercept is one half. Then take the reciprocal, so two thirds becomes three halves. Negative one is still negative one. One half becomes two. And then you need to multiply out the refractions. So multiply everything out by two. So this becomes three, two bar, four. So that is the Miller indices for this plane A. There's another question I missed. How do you know the right order of the numbers? Are we talking about, Daniel, this was your question, the right order of the numbers. Are we talking about the directions here? So it's always going to be like x, y, z for the planes, the right order. It's, again, the, when we're talking about intercepts, it'll be x, y, z is always in that order. And like I said, x, y, z is a bit misleading, especially if we, if we're changing into a, this is supposed to be lattice space and not real space. But for cubic, the lattice space is the same as real space. But if you're, if your unit cell, you know, is like trigonal or monoclinic is going to have different angles, the procedure is all the same. Yeah. So here's some problems that we'll go over. I think there was a typo here. This one half is by itself. It shouldn't be there. So this, we're going to find the Miller indices of these different planes of the cubic unit cell. So this one here is a bit different because we don't have, it doesn't intersect, right? In fact, you could expand this plane and intersect way up here. But instead what we'll do is we'll move the origin, we're going to move the origin to this top corner here. And so in the x direction, it's negative one. In the y direction is negative one. And the z direction is negative one. So this, this Miller indices of this is going to be one bar, one bar, one bar. Let's see. Yeah, that's right. Okay. And this plane is equivalent to one, one, one. All right. One bar, one bar, one bar is equivalent to one, one, one. It's also equivalent to two, two, two, and so on. All right. But so one, this, you know, the correct answer would be one bar, one bar, one bar. The next one, let's see. So this problem is a bit different than the other ones because it never intersects the y axis. All right. This plane never intersects the y axis. So that in that case, for y will be zero. Let's start with x, though. x is going to be one, y is zero, because it never intersects y. And then z is one half. So remember, we need to take the reciprocal. So we start one zero one half, take the reciprocal is one zero two. And then there's no fractions. So that should be the answer one zero two is the middle end to see for this plane. This next one is a bit more difficult. Because again, the intersection with the x axis is not in the view of this unit cell. So you would have to extrapolate this plane down. So if this is one half here, and this is one to one half, and then continue down, then it's going to end up intersecting x at two. So x is two, y is one, and z is one. And this was starting at the origin again. Okay, so it's two one one. Oh, I got that wrong. What did I have? Oh, I forgot to take the reciprocal. Yeah. So it's two one one before the reciprocal. But remember, we need to take the reciprocal reciprocal. So two turns into one half. So then it's one half, one one, then multiply out the fractions, it becomes two one one, excuse me, one, two, two. Okay, so I got that one wrong, because I forgot to do the reciprocal. So always remember for planes, you need to take the reciprocal before the final answer. Okay. And then here's another question. Find the one one one plane of this unit cell, this FCC unit cell, and then also draw the 2d atomic arrangement of that one one one plane. Now there's a question on your homework that's very similar to this. And I want to, you know, it's it's the homework said it was a hard, a difficult question, but it could be very easy if you know this trick, if you know this, I guess. So one one one, of course, the X intercept is one here, the Y is one here, and Z is one here. So that makes this triangular plane. Okay. And then if we were looking at the atoms that intersect this plane and to draw out those atoms, you know, you get the closed pack structure, the closed packed plane. So for FCC unit cells, one one one is the closed packed plane of atoms. Whenever you see this hexagonal symmetry, right, this hexagonal symmetry of atoms like this, this closed packed plane, you should know right away that's either it's either the one zero zero plane of HCP, one zero zero of HCP, or a one one one plane of FCC. Alright, so there's a question on your homework that might seem at first a bit difficult because it wants you like to calculate something in order to find it. It gives you like an image of this. And you should know right away when you see this, oh, that's the one one one plane of FCC, or the one zero zero plane of HCP. But the question made it obvious as FCC. Okay. The same thing for BCC find the one one zero plane and then draw the atomic arrangement of that. So this one one one zero intercept on the x axis one here and the intercept on the y axis is one here. And there's no intersect on the zero on Z. So it's just going to be a slice along the diagonal here. Okay, is the one one zero plane. And if we were to draw out the atoms on this plane, this is what it looks like. Remember, we talked previously what the unit cell last parameters were. And so this is the atomic arrangement of the one one zero plane for BCC. So if you had a iron, for example, iron is a BCC metal. If you had a single crystal of iron, for example, if you had a single crystal of iron and it was cut along this one one zero plane so that the surface of the single crystal was one one zero at the surface, the atoms would be arranged in this the pattern. Alright, so that's what that means there. Okay. I think next I'm going to talk about x-ray diffraction, but we've gone a bit over time. And Dr. Brush has not yet talked about x-ray diffraction yet. So I think I'm going to end it here and leave that for Dr. Brush and Sid to talk about next time. So I'm going to stop the recording. And if you guys have any last questions, you can go ahead and ask and we can go back if you need to because I did go through it a bit quickly.