noc19-cy16-Lecture 46 - Closed - Packed Structures and Voids

[Music] [Music] now I will start the tenth week of this course and in this week you will you will see the common crystal structures so so so it will be it'll be you will look at various crystal structures of elements of simple binary compounds and a few oxide compounds and these are the common crystal structures that you encounter in lot of materials so in a sense this will be fairly descriptive and it will be it will be nice to see how some of the concepts that we have talked about so far in this course they actually apply in the calculation of in the in the actual structures of of different materials so so it will be in some sense it will be it will be rewarding because all this work we have done about structures and symmetry and packing and so on will actually be applied to see how to understand the structures of of different crystals and I'll stop this first lecture of the tenth week with a discussion on on closed packing and voids and as as I had said earlier that a lot of a lot of crystal structure can be understood in terms of thinking of atoms as spheres which pack into space okay so week 10 in the first lecture we look at close packed structures and voids so what are close packed structures as we as we already learnt the FCC or the cubic close packed and HCP that is a hexagonal close-packed or close packed structures and they have a packing fraction of 0.74 and that is the highest possible packing fraction if you if you have spheres okay so if you if your atoms are perfect spheres then this is a highest packing fraction that you can get the body centered cubic has a packing fraction of 0.68 and the simple cubic has a flat packing fraction of 0.5 - this is the monatomic BCC crystal okay and similarly this is a monatomic simple cubic crystal okay and as I had already said that this idea of packing assumes that the atoms are spheres and intuitively we feel that the most efficient packing should be preferred because they would like to pack very closely to each other okay so so you might expect that most of the structures you'll see are FCC and HCP but the fact is that atoms are not spheres okay and there are orbitals okay so the the bonding between atoms goes through orbitals and there is a directionality in the bonding so what we'll see is that even though you would expect most of the of the atomic crystals to be FCC or HCP we will find that significant number of them are BCC and other structures so let's let's look at the crystals of elements so so I'm just listing some elements and they're crystal structures so if you look at the cubic close packed or the FCC you have copper silver aluminum gold nickel lead these are some of the some of the metals that that form a cubic close packed the hexagonal close-packed you have barium magnesium zinc cadmium titanium zirconium these are just some of the examples for BCC body centered cubic you have iron iron iron actually also forms an FCC but at much higher temperatures at more than thousand Kelvin chromium molybdenum tungsten tantalum barium these all from body centered cubic structures simple cubic there is only one example of an element that crystallizes in simple Kubik that is polonium there is a hexagonal structure like for example graphite carbon in the form of graphite crystallizes in a hexagonal structure there are several tetrahedral materials we have already seen the diamond structure of silicon or diamond carbon or germanium there are several others okay then there are some elements that actually crystallize in a rhombohedral structure okay with a space group r3m so some of these examples are arsenic antimony and bismuth so so this is actually you know you know the trigonal crystal system can have either the trigonal or the hexagonal and you can have the hexagonal or the rhombohedral okay so so so this comes in the rhombohedral crystal system okay arsenic antimony and bismuth are some examples of this okay so what you notice is that there is there is no clear trend okay no clear trend okay you cannot say that heavier elements for prefer certain structures or something of that we notice that arsenic arsenic antimony and bismuth okay are in the same same period okay so so so there is some some similarity for example carbon silicon germanium they are in the fourth row of the period the the fourth period and they they also have tetrahedral structure similarly the rhombohedral elements are in the are in the in the in the fifth period okay and so so there is some some trends okay some trends with with periodic table okay some but it's but it's not overall there is no clear trend okay and for example you there is no clear understanding of why iron is BCC whereas copper is FCC okay and really these are just these are just the crystal structures that are observed and it turns out that the electronic structure of these of these elements makes them prefer a BCC structure make Sion prefer a BCC whereas it makes copper prefer of prefer an FCC now these are some of the crystals of elements now one of the very useful concepts when you go to crystals of crystals of multi element compounds for example binary compounds okay what we like to think is that we like to visualize these elements as forming some crystal structure and then and then the other element okay occupies some voids in that structure okay so we'll come to the concept of voids in the cubic close packed structure okay and now here here I'm showing I'm showing two layers okay of a cubic close pack structure okay so the black black circles this black circle represents the bottom layer and the next layer on top is represented by blue so these are atoms that form a cubic close packed array okay I'm just showing two layers now the difference between hexagonal and cubic close packed comes in the third layer in the hexagonal close-packed the third layer is on top of the first layer on top of this black layer whereas in the cubic close packed it's actually it's actually in the it's it's directly on top of these this green green hexagons okay so that's where the third layer would be but but but we are just looking at these two layers okay and what this these two layers if you just look at these two layers you know we can immediately identify location of voids in cubic close packed structures okay so what do we notice is that is that this black circle okay this black circle okay it is now now imagine that it is at the center it is at the center of one two three and four the the three black the three black circles and this and this blue circle so it is right at the center of these four and clearly it is in a tetrahedral void so all these black all these black filled circles they correspond to tetrahedral voids okay now there are other tetrahedral voids which are indicated by these red circles okay again very similar to the black circles okay the only difference is that in these red circles okay so there is one they are directly on top of this black atom and they are in the middle of these three blue blue atoms okay so so in a sense the orientation of the tetrahedron around the red and black circles is inverted okay so so so with respect to the black filled circles okay the orientation of the tetrahedron has three three bonds going down and one bond coming up whereas the red one has one bond coming it has three bonds coming up and one bond going down okay again please excuse the fact that I haven't drawn these exactly okay so so I so actually this the location of some of these dots is not precise but you can clearly make out where they should be so so we can immediately what this tells you is that these tetrahedral voids are going to be there in any cubic close packed structure whether it is whether it is HCP or FCC okay and similarly the octahedral voids okay they will be located okay they will be located right here at the center of these three so so if you look at this green hexagons okay now is equidistant from these three blue circles and these three black circles okay so clearly it is equidistant from these six circles okay and this is going to be a octahedral void okay again again I should emphasize that this is actually below below the blue blue atom okay so it is it is in between in between the blue layer and the black layer okay so the black layer black black open circles are the bottom layer on top of that or the blue open circles okay and in between those two right at this Center right here there is a octahedral void there is an octahedral void and the end due to periodicity it is there in all these locations so what this shows okay is that you can immediately identify the tetrahedral and octahedral voids in both and hexagonal close-packed and a cubic close packed okay as I said a cubic close packed the third layer will be directly on top of these of these hexagons in a hexagonal close-packed the third layer will be directly on top of these these black black circles okay however let's I mean in so also we should keep in mind that for cubic close packed the closed pack direction is actually the 1 1 1 direction of the crystal that means that means the this this screen ok screen is the 1 1 1 plane okay so that is the closed pack direction okay and and you can you can easily take your cubic close packed or an FCC structure and see this see this packing arrangement okay and what it says is that you know whether you have a you have an FCC or an HCP you you can identify the location of the voids okay and for the and so this is very useful okay and and if you go back to the lectures where we discussed the location of voids and HCP you can go back and confirm that these were indeed the location of the voids now what is interesting about these voids is is that these voids okay are spaces where you could in principle fit atoms but if you wanted to fit an atom in a void you need to know the size of the void okay and again so we are going to do some calculations from the with respect to the size of the voids now let me emphasize one point which might not have been clear okay is the following that you know you know we usually show the show the FCC structure in the form of a cube and so we usually show it in the form of a cube and we show the and we show the location of the voids okay so the tetrahedral voids as you as you are familiar will be located along the body diagonals along each of the body diagonals so there are eight tetrahedral voids located along each of the body diagonals I am NOT I am not focusing on the sign of these tetrahedral voids okay but you know that there are eight of these located along the body diagonals okay these are what I want to emphasize that these are exactly the same same as shown here okay there is no difference between what is shown in this in this in this diagram and in this figure okay so the tetrahedral voids exactly the the same locations that are described here this is just a different way of of looking at the structure in fact in fact you are looking at it from the 1 1 1 direction ok which is which can be shown as let's see if we take this as the origin X why is it so we can take this as the 1 1 1 direction as a 1 1 1 play so actually the screen it corresponds to this this plane ok and if you look if you look in this plane then you will get you'll see exactly the structures that we are seeing here ok and now if you go to the layer below this plane ok then you'll see the packing arrangement that is senior and UK and and then you can immediately identify the tetrahedral voids and you can show that what you see in this figure is the same as what you see here similarly for the octahedral voids octahedral voids you know are located at the at the edge centers and the body centers so there's one right at the body center that's an octahedral void I'll use the green colour and located at each of the edge centers ok and again this is exactly the same that you see in the other figure ok and until you until you draw it and convince yourself ok you want you you won't be sure of this ok so these two are two different ways to look at the cubic close packed structure now let's calculate the radius of these voids now to calculate the radius of the voids okay I will start with the I'll start with the cubic close packed but but but is the same for the HCP so so if you look at a cubic close packed now for the for the tetrahedral void for the tetrahedral void we know that we have so now the tetrahedral void which along is located along the body diagonal and it's located at 1 by 4 at 1/4 of the way along the body diagonal okay so I'll just show tetrahedral void here and along the body diagonal okay so this is a corner and let's say this is a tetrahedral void what you have is something like this you have you have your sphere which is the atom and you have the void and you know that you know that this distance is a root 3 by 4 and that is equal to R of the atom plus R of the void and now our of the atom this again again let me remind that R is the radius of the atom and a is the size of the cell okay now so R of white is equal to a root 3 by 4 - now our R of the atom our of the atom can be related to the to the lattice parameter because along the along the face face diagonal okay you have you have a root 2 equal to 4 R okay so R is equal to here root 2 by 4 okay so then the I can write this as here root 2 by 4 so you get you get the size of the tetrahedral void a times root 3 minus root 2 by 4 ok so this is the radius of that at idle right in terms of in terms of this in terms of the cell parameter a okay now now this is the radius of a tetrahedral void okay in a perfect assumes atoms form a perfect FCC so in the sense in the sense we we impose this condition okay which is only true if the blue atoms form a perfect FCC okay and then and then we used you use this condition to find the size of the tetrahedral void okay now what I mean is that if you actually squeeze an atom okay then the then the FCC might get distorted a bit okay but but still for in a perfect FCC this is the size of the void okay now interestingly you can you can ask what is the ratio of the radius of the void to the radius of the atom okay so this is is root 3 minus root 2 divided by root 2 okay so this is the ratio of radius of the void to the radius of the atoms so all I did was take the radius of the wild divided by the radius of the atom so so this is for the tetrahedral void let me put a T in front next to the void so this for the tetrahedral void you can do for the octahedral void now for the octahedral void okay again again you can just go you can just go along the along the edge okay and you can you can easily calculate that so for the octahedral void you have you have radius of octahedral void plus radius of atom this is equal to a by 2 and so you can you can immediately see that the radius of octahedral void is equal to a by 2 minus a root 2 by 4 okay and further you can you can you can also do the same you can you can divide by a root 2 by 4 to get the ratio of radius of octahedral void divided by radius of atom this is 2 minus root 2 by root 2 so so so we can work these things out and these are these are useful relations you can do the same for hexagonal okay you can if you want you can start with the HCP structure and you will get essentially the same result okay so so the size of the voids okay so what this says is that you can fit another atom of this size okay but now if you calculate these fractions okay so this is this is of the order of okay so this is this is only about this is only this is less than 25% okay so what this says is that the atom that you can fit in the tetrahedral void has to be extremely small okay it has to be extremely small okay if it has to fit and if it has to fit in the tetrahedral void and to keep the structure intact and clearly that seems very unreasonable okay in fact in fact we will see structures like the zinc blende okay where the atoms fit in the tetrahedral void but the size is a lot more okay size is not the ratio of sizes is not this small okay so that's what I meant by saying that you know this is in a perfect FCC but there can be distortions similarly this is about you and this is this is about this is much higher okay so this is nearly about about point four three for two okay and yes yeah I mean this is significantly larger okay but still it is much smaller than what you would expect okay so this is about the size of the voids okay now let's look at something that that had we sort of skipped when we did the hexagonal closed pack structure and this has to do with an ideal hexagonal close-packed structure now what do I mean by an ideal hexagonal closed pack structure now let me let me draw the hexagonal close-packed structure in the in the in the in the way that you are used to seeing okay so this is where you show a hexagonal structure okay and what you have is you have these atoms here and you have the next layer okay which the layer below will have three three atoms and then you have another layer below where you have six atoms six or rather okay and now the reason I am showing it this way is the following that that this distance in a hexagonal in a high HCP this is called C okay and this lattice parameter is called a okay now if you take a unit cell of this okay what you will see is the following that let me let me draw it in the following way so you have so you have a B C this angle is 120 degrees okay and now now the question is if you have an ideal hex HCP okay then it turns out that there are some further constraints okay C has to be related to a okay C is not completely independent of a okay so so this side is a and this side is C so there are two two parameters okay so so you have eight equal to B and gamma equal to 120 alpha equal to beta equal to 90 okay and in an ideal in an ideal HCP okay now there is going to be one other atom that is located right here okay inside this inside this inside the cell okay it is not it is not right at the center okay it has to be it has to form it is located in it is located in this in this it is located at the center of of these three atoms okay so what this tells you is the following that this distance okay in an ideal ideal HCP should also be a okay so if it was perfectly closed packed then this distance this diagonal distance should also be a this distance let me show it in the red okay so this is from the from the from this corner to the center of this atom okay so that distance should also be a okay and it turns out that this so you immediately have a relation now now you know that this is related to C in this way okay and so so you will get a relation between C and a okay now let's work out that relation now so so you can let's see the location of this atom okay and for this let me use the planer representation so in the planer representation okay I have okay so so I have I have these four atoms located here okay and then I have the this center atom is located at height C by two along this okay so so let me these are all at Z equal to 0 set equal to zero and this is located it z equal to half and you know that this distance is a Z is Z equal to half or C by two more like C by 2 okay so now so now let's look at the coordinate of this atom so the coordinate of of this atom in blue this equal to so let me let me make a choice of the origin I will choose I will choose II this is my beam so the coordinate of that is clearly the x-coordinate is 0 it is located right here so the x-coordinate is 0 the y-coordinate okay y-coordinate now now my core in order to find the y-coordinate you have to find this distance and this distance can be found out from this angle okay so so so this distance can be found out from this angle so it is a by 2 divided by cosine of this angle now cosine of this angle is root 3 by 2 okay this is this is a 30-degree angle so the cosine of that is root 3 by 2 so so this is equal to a a divided by root 3 okay so the y-coordinate is a by root three and the Z coordinate is C by two okay so so now now if you take if you take a distance from the origin of this point and the origin is located at one of these origin is located right here okay so now if we take a distance from origin okay then you can immediately find the Khans you can you can say that a square by three plus C square by four okay that is that is the distance of this blue point from the origin okay this should be equal to a by 2 square a by 2 is square is sorry it should be a square and this immediately implies that C square by 4 is equal to 2 a square by 3 or C by a is equal to square root of 8 by 3 and this works out about 1 point 6 3 okay so so this is the this is the ideal ratio for a hexagonal close-packed so if the atoms were really close packed then the C by a ratio would be exactly 1 point 6 3 in a hexagonal close-packed structure okay now let's look at an ideal cubic close packed structure as I had already said in an ideal cubic close packed structure there the face centers okay should be touching each other so let me again draw this so so so we just lower we just take the planar representation okay now it should be such so so in an ideal cubic close pack structure you should have now now you have only you have only one parameter a so a root two okay so this this distance which is a root two should be equal to four times the radius of your atom okay and so this is an ideal cubic close pack structure in the sense the atoms in blue are forming a perfect closed pack structure so interestingly the so unlike unlike the hexagonal closed pack structure where we where we had a parameter C okay in this case there is no additional parameter okay and again again this is just a convenience of notation I mean if we had if we had directly imposed the condition for ideal hexagonal close-packed structure in that case also we would not have any independent parameter C C would be equal to a into root 8 by 3 okay so with this I will I will conclude this lecture I will conclude this first lecture so here we have seen the idea how the idea of closed packing can be used to understand the sizes of voids and also and also we saw the structures of simple elements now in the in the next lecture I will look at and look at binary compounds and see how these can be understood in terms of in terms of close packed structures and atoms in voids okay and we have already seen this in several examples like the most common example that we see the rock salt and and the cesium chloride okay or the or the zinc blende structure okay so these are all these all can be understood in terms of one of the elements forming a closed pack structure and the other elemental avoid so we look at this in more detail in the next lecture thank you [Music] [Music] [Music]