Part I:- Solid Solution and Alloys

 Part I:- Solid Solution and Alloys

In the previous lecture, we talked about packing in metallic solids, and how do atoms pack in these materials? What is the atomic packing factor? What is the coordination number? We looked at the other two types of lattices, which are BCC and Simple Cubic lattices, which are not close-packed. Close packed structures, CCP, HCP, BCC, and simple cubic are not close-packed structures, as a result, we need to know about them.

So now, we will talk about the new topic in this metallic material, which is essentially Interstices. An interstice is void. We can see that, since atoms are spherical in nature, they do not completely fill their space, but there are empty spaces within the lattice. For example, if you see this particular kind of arrangement, you can see I can put an atom here on top. The next layer goes here, and all are the same atoms. The red is only for the sake of illustration, but you can see that within this body, which is made by three atoms at the bottom and one atom on the top, if all the atoms were of same sizes, I can create a body by connecting the centers of all the four atoms, three at the bottom, one at the top, this body is called as a tetrahedron.

So, a tetrahedron is something like that, I will have to draw a little oblique shape, I mean, you have certain space lying at the center of the tetrahedron. So, space which is available at the center of the tetrahedron, you can fit a certain size of the atom there so obviously, that atom is very small. So, the size of an atom that can fit in the space within the tetrahedron is called rtet. Let say, and this rtet is much smaller than the radius of the host atom (r).

So, we will see what the size is, where you can make a smaller atom sit here. This is called a tetrahedral void. So, you have a tetrahedral void here. Similarly, you can have a tetrahedral void underneath. Which is symmetric in nature, which is a regular tetrahedron, you will have a tetrahedral void. Tetrahedral does not need to be regular. We will see in certain cases tetrahedron is not regular, but as long as it has the shape of it is that how many sides does it have? It has three on the top and three at the bottom. It has a total of six sides.

Now, as long as you have six sides, which looks like a tetrahedral, it does not need to be regulated tetrahedral. In this case, let us say 2r, but in some other case, all the dimensions may not be equal. In FCC and be HCP structures, it is a regular tetrahedral. So, this is the tetrahedral void and the other is octahedral void.
 

In this case, you can see that this has six edges and four faces. So, it has eight faces and twelve edges and six corners. And in this case, how many corners you had? You had four corners. This is the description. Now, this octahedra can be found in ABCABC or ABAB kind of packing. The configuration looks something like that. So, you have an atom sitting here, the next layer goes there, and the next layer which comes on top, it does not need to go exactly on top of A, it can go to C. It has six possibilities of 6 coordination. So, this small atom is basically an octahedral atom sitting in the octahedral site.

So, this is my A layer, this is my B layer, and between the A and B layer, I have an octahedral void. How do you visualize this? You can visualize this, so, this octahedral is a little tricky to visualize. So, let me show you first the FCC structure. It is easy to visualize in the case of FCC structure. It is difficult to visualize in the case of the HCP structure. The octahedral void, in this case, is made by connecting the sorry, one atom will be here as well. You connect these phases, this is the octahedral void in case of FCC structure. And the location of the octahedral void is 1/2 1/2 1/2. So, this is the octahedral void that you can format 1/2 1/2 1/2 position. Are there any other octahedral voids present in a structure? No, not on surfaces, on the edges, so this will be one octahedral void, this is another octahedral void. So, how do you form an octahedral here?

This is, you will have an atom sitting here, somewhere here, and the neighboring unit cell. This will connect to that you can again connect this to the atom at the face center. And the one at the bottom face will also be there. So, this will be an octahedral void. So, all the edges will also be octahedral. The center of the edges will be octahedral void locations. So, the octahedral voids in case of FCC are located at 1/2 1/2 1/2,  1/2 0 0, 0 0 1/2,  0 1/2 1/2, and 1/2 1/2 1/2. And since each edge is shared by 4 unit cells, as a result, since there are 12 edges, so, this will contribute to 1, these will contribute to 14∗12=3. But since each is shared by 4 unit cells, 12 divided by 4 is equal to 3. So, you have a total of 4 octahedral voids within the FCC unit cell.

So, and tetrahedral void, on the other hand, in FCC structure lies at the position which I am going to show you now. And if you connect these places with each other, this will make a regular tetrahedral. So, these are tetrahedral voids, and they are located along a body diagonal. So, this position is 1/4 1/4 1/4 type. So, you have four body diagonals, and you have two on each co near each corner. As a result, you have eight tetrahedral voids in an FCC unit cell. Locations are 1/4 1/4 1/4, 1/4 1/4 3/4, 1/4 3/4 1/4, 3/4 1/4 1/4 What else? 1/4 3/4 3/4, 3/4 1/4 3/4, 3/4 3/4 1/4, 3/4 3/4 3/4. These are 8 locations that you will have in FCC unit cell, and octahedral voids are four per unit cell or one per atom because there are four atoms in FCC unit cell. Tetrahedral voids are eight per unit cell or two per atom same is the case with HCP because both are close-packed structures.

Though this is one tetrahedral void, this is another tetrahedral void. So, what is the location of the tetrahedral void? In this case, this tetrahedral void is that 1/3 2/3 1/4, 1/3 2/3 3/4. Where are the other two tetrahedral voids? Now, if you look at that top view of a hexagonal unit cell, this is the top view of the hexagonal unit cell. So, the unit cell is you have one atom sitting in let’s say if this is the if this is my first unit cell, I am saying that one atom is sitting here, the second atom got to be sitting here, and the third item is got to be sitting somewhere here.

So, you can see that these three give you one tetrahedral void, but remember these three atoms also share that a tetrahedral void on the side. They are sharing tetrahedral void of the neighboring unit cell. So, those two neighboring unit cells contribute to the tetrahedral void of this unit cell. So, as a result, you have because of sharing with the neighbors just like FCC, because of sharing with neighboring cells, we have four tetrahedral voids per unit cell, and each unit cell has two atoms. Similarly, the octahedral void is two per unit cell or one per atom.

So, this is how the octahedral void looks like FCC. So, this is the bottom one is the first layer, the lighter one is the second layer, the top layer. So, you can see three atoms at the bottom, three atoms on the top between them; there is an octahedral void. So, this is the square-shaped body this one alright, and this is one apex, and this is the second apex. So, if you connect now, and of course, you can connect have to you will have to connect there as well, and then you connect to here, and you connect to here. So, this is how the octahedral are arranged you can see on the right, these are the various octahedral within the hexagonal unit cell. So, you will have a total of two octahedra is per unit cell.

Let us take the example of FCC unit cell because what will go in them is determine quite a lot quite strongly by the size of this atom, which is dependent upon the size of the octahedral void. The atom which goes in there is somewhere there. Now, of course, atoms are not touching here, if you make a 2D view of this. So, what is this length?

Size of octahedral atom = roct

So, the maximum size of the atom that can fit in the octahedral void is 0.414 of the radius of the host lattice, and the same is true about hexagonal lattice as well. Similarly, if you need to find out the size of the octahedral atom, in case of HCP its easier. In case of the tetrahedral void, since it is at 1/4 1/4 1/4, the distance between this point and the apex.

So, I will draw these three atoms. One at the center tetrahedron is made by connecting these atoms, and our atom is sitting at 14 13 14, and this connects the body-diagonal. So, if this is O, this is A, that distance OA is, diagonal is 𝑎√3, and 

So, this is the maximum size of the atom that can fit in a tetrahedral void without distorting the tetrahedral void. So, the similar size of the atom you can also do tetrahedral void calculation just by drawing a tetrahedron.

So, this is your tetrahedron, for example, and the atom which will fit somewhere here at the center of the tetrahedral, these are all touching atoms there. So, just like you do that calculation for 𝑐/𝑎 Remember we worked out two distances CV and CF. So, you just have to work out this distance from the center to center. So, this distance is equal to let us say this is A, this is B, AB is equal to r + rtet. So, you need to find AB in terms of r using the geometrical methods of a simple tetrahedron.

So, this will again give you the same relation, the octahedral void remains the same, and so, in the case of HCP also rtet is equal to 0.225 r, and roct is equal to 0.414 r. These not the same in the case of BCC. In the case of BCC, since it is not a closed pack structure as a result it has voids. So, this is the octahedral void, but this is not a regular octahedron because all the sides are not equal. what is that tetrahedral void? It is also irregular because you can see that this length is a 𝑎√3/2.

I will leave it to you as homework that how many octahedral and tetrahedral voids in BCC locations, determine the exact location in terms of coordinates and the size of these voids in BCC. So, by it is a fairly straight forward thing to work out because you remember that you have to take the shortest size, you do not have to take the longest size. There will be two sizes, one, so depending upon which distance you take, it will be long and short, the smallest that can fit is the given by the shortest.

So, these are the octahedral and tetrahedral voids that are present in metallic structures, which are useful because the impurities go in them like steel contains carbon. So, as carbon is a smaller atom, it goes to interstices, and that makes the steel stronger than pure iron. So, it is good to know about these interstices, and these interstices are also important when we consider the ionic solids. Because in ionic solids, anions are bigger, cations are smaller and occupy the interstice sites. So, that is where the knowledge of these interstices is very useful. So, in the next lecture, we will talk about the solid solutions and alloys of metals, and once we finish the metal part, we will move on to covalent solids and ionic solids. So, we have spent quite a bit of time on packing of solids the packing fraction coordination number and then the interstices, and this has formed the basis of studying the substitutional interstitial solid solutions and alloys in metallic structures.

For the Previous Lecture Click below

Structure of Metals Packing Co-ordination Interstices 

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