Spaces Between Atoms - Interstitial Sites

The locations of atoms is clearly important. However, it turns out that in Materials Science and Engineering that the space between the atoms is nearly as important! Let's take a look at how we'd quantify these open spaces in atoms. At this point, we'll just work on identifying and quantifying these sites, and we'll soon use them and how we might use them in creating stronger alloys, predicting ceramic structures, and in materials processing via diffusion.

First, let's just give it a try, based on what you learned in the previous section. We'll try the easiest structure again (simple cubic) and try to investigate the space between - or the interstitial space. Give it a shot in Exercise 5.11.1.

Using the interactive 3D visulalization tool (Section 5.11.6, at the end of the page) really helps with this visualization. Look at the directions on the right (or press "Help" to look at different views, activate different features, or manipulate controls. One can control the viewing of the interstitial sites by clicking (e.g., for simple cubic) the "Cubic On" button on the left. Other interstitial sites like octahedral and tetrahedral are viewable in the FCC and BCC structures.

Interstitial Sites in Other Common Metal Structures

As structures get more complicated, the number, time, and location of the interstitial sites get more complicated. Here, we'll ask you to think about interstitial sites in FCC, and we'll assign the BCC sites on your homework.

Interstitial sites are named based on the their geometry - specifically, the shape that surrounding atoms take. In the simple cubic Section 5.10.9, the atoms surrounding the interstitial at $\frac{1}{2}\frac{1}{2}\frac{1}{2}$ formed a cube - and so we call it a cubic interstitial site. In FCC there are two major types of interstitial sites: the tetrahedral site and the octahedral site. (There's also a third type of site that's typically less important... the triangular site. See if you can find it.) In the tetrahedral site, the surrounding atoms form a tetrahedron, and in the octahedral site the atoms form an octahedron. @ref(CB) shows these shapes within the FCC unit cell.

After observing Figure 5.11.1, you might be worried that you can't remember all these sites! Here is where the beauty of the crystal's Bravais lattice helps us. For FCC we have a cubic-F lattice, and this means that every atoms on on the of the lattice sites has identical surroundings. So - in the FCC structure we really only need to remember (or be able to observe a 2-3 interstitial site positions and we can find them all. Let's try with each site:

FCC Octahedral

We first want to count the octahedral sites in the unit cell. Looking at Figure 5.11.1 we can see that there's one interstitial site fully within the cell at $qrs = \frac{1}{2}\frac{1}{2}\frac{1}{2}$, and then we have one quarter of a interstitial site on each edge of the cube. We can use a slightly adapted version of Eq. 5.10.1 to find the number of sites in the cell:

$$N_{\text{Tot}} = N_{\text{i}} + \frac{N_{\text{f}}}{2}+\frac{N_{\text{e}}}{4}+\frac{N_{\text{c}}}{8}$$

where we've included $N_{\text{e}}$ to be the number of sites on the edge of the cell. So the number of octahedral sites are

$$N_{\text{FCC,Oct}} = 1 + \frac{12}{4} = 4$$

There are four octahedral sites per FCC unit cell, and there are four atoms per FCC unit cell. It seems, then, that due to the properties of the Bravais latties, we could associate one interstitial site with each atom in the cell and predict the positions of all interstitial sites. This is indeed true, so let's make it easy for ourselves: let's find the position of one interstitial site and repeat that site throughout the unit cell. I'll choose the one at the body-center: $qrs = \frac{1}{2}\frac{1}{2}\frac{1}{2}$ and will associate this site with the atom at the $\frac{1}{2}0\frac{1}{2}$ (although any neighboring atom works. This position is visual observable, but how might you calculate a position of the interstitial site if you only know the surrounding atoms? See the solution for Exercise 5.11.1.

So, the position of the body-centered interstitial site is simply the position of the $\frac{1}{2}0\frac{1}{2}$ + $\Delta q \Delta r \Delta s$ where $\Delta q \Delta r \Delta s = 0\frac{1}{2}0$ is the distance between the atom and the interstitial site. This means that there will be an interstitial site at every FCC lattice point $qrs$ position + $0\frac{1}{2}0$.

Size of Sites

All the octahedral sites are at the center of 6 atoms, so we can use the same process as we did in Exercise 5.11.1 to calculate their size. I'll look at the size of the $\frac{1}{2}\frac{1}{2}\frac{1} {2}$ (computing it with respect to the neighboring $\frac{1}{2}0\frac{1}{2}$ atom for simplicity, but the calculation is the same for any site.

$$ \begin{align} d &= \sqrt{(\frac{1}{2}a-\frac{1}{2}a)^2 + (\frac{1}{2}a-0a)^2+(\frac{1}{2}a-\frac{1}{2}a)^2} = r_{\text{Atom}}+r_{\text{Oct}}\\ \frac{1}{2}a &= r_{\text{Atom}}+r_{\text{Oct}}\\ \frac{1}{2} (2\sqrt{2}r_{\text{Atom}}) &= r_{\text{Atom}}+r_{\text{Oct}}\\ \sqrt{2}r_{\text{Atom}} &= r_{\text{Atom}}+r_{\text{Oct}}\\ r_{\text{Oct}} &= (\sqrt{2}-1)r_{\text{Atom}} \\ r_{\text{Oct}} &= 0.414r_{\text{Atom}} \end{align} $$

This means that for any FCC crystal, the maximum size of an atom that you could place in an interstitial site without distorting the structure is $0.414r_{\text{Atom}}$, about 40% of the constituent atom itself. It sounds like this might limit how atoms mix when alloying or induce some sort of structural change if the atom occupying the lattice site isn't the right size. We'll explore that soon.

FCC Tetrahedral

Location of Sites

For the tetrahedral sites we again watt to count the number of site in the unit cell. Figure 5.11.1 shows us the all the tetrahedral sites are fully within the unit cell, and there are 8 of them

$$N_{\text{Tet.}} = 8$$

There are eight tetrahedral sites per FCC unit cell and there are four atoms per FCC unit cell. This means for each atom we need to define two interstitial site positions, but then we have defined them all. Again, we'll look at the $\frac{1}{2}0\frac{1}{2}$ atom. We see there are four interstitial sites nearby, but we'll only need two, I'll choose the two with $s = \frac{1}{4}$: $\frac{1}{4}\frac{1}{4}\frac{1}{4}$ and $\frac{3}{4}\frac{1}{4}\frac{1}{4}$. Again - I'm not showing this explicitly, but how might you find this position using the atoms surrounding these sites?

To find the $\Delta q \Delta r \Delta s$ for each, I just need to find the displacement vector between the interstitial sites and the atom:

$$ \begin{align} \Delta q. \Delta r. \Delta s = q_1-q_0.r_1-r_0.s_1-s_0\\ \Delta q. \Delta r. \Delta s = \frac{1}{4}-\frac{1}{2}.\frac{1}{4}-0.\frac{1}{4}-\frac{1}{2}\\ \Delta q. \Delta r. \Delta s = -\frac{1}{4}.+\frac{1}{4}.-\frac{1}{4}\\ \end{align} $$

and similarly

$$ \begin{align} \Delta q. \Delta r. \Delta s = q_2-q_0.r_2-r_0.s_2-s_0\\ \Delta q. \Delta r. \Delta s = \frac{3}{4}-\frac{1}{2}.\frac{1}{4}-0.\frac{1}{4}-\frac{1}{2}\\ \Delta q. \Delta r. \Delta s = +\frac{1}{4}.+\frac{1}{4}.-\frac{1}{4}\\ \end{align} $$

So, for every atom, all I need to do to reproduce the pattern of interstitial sites is to two associated sites: one at $\Delta q. \Delta r. \Delta s = -\frac{1}{4}.+\frac{1}{4}.-\frac{1}{4}$ and one at $\Delta q. \Delta r. \Delta s +\frac{1}{4}.+\frac{1}{4}.-\frac{1}{4}$. I'd invite you to give it a shot.

Size of Sites

All the tetrahedral sites are at the center of 4 atoms, so we can use the same process as we did in Exercise 5.11.1 to calculate their size. I'll look at the size of the $\frac{1}{4}\frac{1}{4}\frac{1} {4}$ (computing it with respect to the neighboring atom at the origin: $000$. atom for simplicity, but the calculation is the same for any site.

$$ \begin{align} d &= \sqrt{(\frac{1}{4}a-0a)^2 + (\frac{1}{4}a-0a)^2+(\frac{1}{4}a-0a)^2} = r_{\text{Atom}}+r_{\text{Tet}}\\ \frac{\sqrt{3}}{4}a &= r_{\text{Atom}}+r_{\text{Tet}}\\ \frac{\sqrt{3}}{4}(2\sqrt{2}r_{\text{Atom}}) &= r_{\text{Atom}}+r_{\text{Tet}}\\ \frac{\sqrt{6}}{2}r_{\text{Atom}} &= r_{\text{Atom}}+r_{\text{Tet}}\\ \frac{\sqrt{6}-2}{2}r_{\text{Atom}}) &= r_{\text{Tet}}\\ r_{\text{Tet}} &= 0.225r_{\text{Atom}} \end{align} $$

This means that for any FCC crystal, the maximum size of an atom that you could place in a tetrahedral interstitial site without distorting the structure is $0.225r_{\text{Atom}}$, about 23% of the constituent atom itself.

Interstitial Repeating Patterns

To conclude - one last important feature of the interstitial sites is that they can be defined to have their own structure (often called a sublattice) or repeating pattern. Sometimes this is the same as that of the atomic lattice, but sometimes it isn't (i.e. in the case of the FCC tetrahedral sites). The description of this sublattice is especially useful when we start putting atoms into the interstitial sites and we want to describe them independently, as well as when we're considering diffusion of atoms within the interstitial lattice. More on those topics soon.

Try to identify the sublattice for both of the in Question 5.11.2.1.