# Homework Help: Packing fraction in ionic structure

1. Oct 19, 2014

### CAF123

The problem statement, all variables and given/known data
The CsCl structure is stable when the ratio of the smaller ion radius to larger ion radius is $r > 0.73$. Why is this so? Show that the packing fraction, $p$, is given by $$p = \frac{\sqrt{3} \pi}{2} \frac{1+r^3}{(1+r)^3}$$

2. Relevant equations
volume of sphere, volume of structure

3. The attempt at a solution

I have obtained the correct answer for $p$ but I am not sure how to explain why $r > 0.73$ but I have some calculations. The CsCl structure has a Cl in the middle of a cube with 8 Cs around it (so like a BCC structure and the unit cell is BCC cubic). Suppose that along the edge of a cube, the two adjacent Cs ions are touching. Then $a = 2R_1$ where $a$ is the side of the cube and $R_1$ is the radius of a Cs ion. The face diagonal is then $h = 2\sqrt{2}R_1$, from which we find the cube diagonal to be $2\sqrt{3}R_1$. Now if we further suppose that the Cl ion is touching two Cs from either side (so close packing of spheres as much as possible) then we have $2R_1 + 2R_2 = 2 \sqrt{3}R_1$ where $R_2$ is the radius of Cl. Rearranging for $R_2$ gives $R_2 = (\sqrt{3} - 1)R_1 \approx 0.73 R_1$. So I seemed to have extracted the correct number (0.73) but not the correct ratio. (i.e I wanted to derive $R_1 = 0.73 R_2$), Cl is certainly bigger than Cs. So what went wrong here? Actually I have found the same calculation in this doc (bottom of first page), but nothing was said about it.

http://depts.washington.edu/chemcrs/bulkdisk/chem364A_spr05/notes_Lecture-Note-6-7.pdf [Broken]1.

Many thanks.

Last edited by a moderator: May 7, 2017
2. Oct 19, 2014

### ehild

Think: The Cs+ ions attract the Cl- ions, but the Cs+ repulses the other Cs+ and also the Cl- ions repulse each other. To be stable, the positive and negative ions have to be as close as possible, they need to touch each other, but the ions of the same kind do not.

The vacancy in the middle of a cube made of spheres at the corners which touch each other can accept a smaller sphere. If the spheres at the corners have radius R, the vacancy is enough for a sphere of r=0.73R. r is the smaller ion in that sense.
Cs+ and Cl- have very similar ionic radii, 174 pm and 181 pm, respectively. In the CsCl structure, 8 Cs ions surround one Cl and also 8 Cl ions surround one Cs. The central ion touches the ions on the edges, but the ions on the edges do not touch.

ehild

3. Oct 20, 2014

### CAF123

Hi ehild,
So the stable configuration is one in which the oppositely charged ions are as close as possible. So this means in the CsCl structure, the ions touch along the cube diagonal. If we take the unit cell of CsCl to be composed of a single Cl in the middle of the cube with 8 Cs ions around it on the nodes of the cube then, if the ions don't touch along the edge, then the Cs ions do not touch along the edge. If this is the case, then wouldn't that mean that $a$, the length of a side of the cube, is not equal to $2R_1$ as I initially wrote down? How does this reconcile with the derivation in the document I showed? (which as far as I can tell, does suppose that $a = 2R_1$ to derive the face diagonal to be $2\sqrt{2}R_1$ and so on..)
Thanks

4. Oct 20, 2014

### ehild

R1 is used to determine the size of the vacancy if the spheres on the edges touch each other. They have radius R1. The spheres are just spheres, not any of the ions. If the central sphere has radius of R2=0.73 R1 then all spheres touch each other. It is not so in the CsCl structure: the central atom is big. It touches the ions at the corners, which do not touch each other.

ehild

5. Oct 20, 2014

### CAF123

I see, so in the case where all the spheres touch each other the central sphere has radius 0.73R1. (ie it is smaller than the spheres at the nodes). But in the CsCl structure, the Cl ion is bigger than the Cs ion, so how does the calculation in the OP help show that the stable configuration of the structure is when the ratio of smaller ion to bigger ion is greater than 0.73?
Thanks.

6. Oct 20, 2014

### ehild

The calculation in the OP means that the radius of the central ion must be at least 0.73 times the radius of the ions at the vertexes in order the positive ion touches the negative one. It does not matter what ions they are. To get a stable configuration the positive ions must be in contact with the closest negative ones. That can happen when the central ion is bigger than 0.73 times the one on the vertex.
In case of CsCl, the cell parameter is not twice the radius of the CS ion. You get the cell parameter knowing that the body diagonal is 2(RCs +RCl).

Last edited: Oct 20, 2014
7. Oct 20, 2014

### CAF123

The thing I am not understanding is that I derived (I think) for the ions to be touching on both the edges and on the diagonal the radius of the central ion to be R2 = 0.73R1. But I am not seeing how this relates to the case here (the CsCl structure). In that case, for the atoms to be in a stable configuration, the positive and negative ions are touching along the cube diagonal and not on the edges and the central ion (Cl) is bigger than the nodal ions (Cs). Sorry if you explained already.

8. Oct 20, 2014

### ehild

The problem you solved is pure Geometry. That is the other statement, that the CsCl structure is stable if the negative ions touch the positive ones.
The CsCl structure does not mean the crystal CsCl. It only means 8 positive ions at the vertices of a cube surrounding a negative one in the middle. Forget about Cs and Cl altogether or which of them is bigger. It is irrelevant for the problem.

Last edited: Oct 20, 2014
9. Oct 20, 2014

### CAF123

Ok so I suppose the nodal ions have radius R1 and that the central ion has radius R2. Then in the OP, I derived that R2 = 0.73R1 which means the central ion is smaller than the nodal ones. I guess my difficulty is making sense of this result when it is applied to CsCl structure, since there the central ion is bigger than the nodal ones. I am not even sure if I can apply this result to CsCl since in that structure, the ions don't touch along the edges as assumed in the OP. What I believe I want to show in the problem is that for a stable configuration (i.e one in which the positive and negative ions touch) I must have, in the same notation as in the beginning of this message, R1/R2 > 0.73).

10. Oct 21, 2014

### ehild

I give up. We are going in circles. I think you have not read my posts, or my English was so poor that you could not understand it.

You need to know what you have derived and what conclusion you drew from your derivation.

Again:
In caesium chloride, the Cs ions do not touch. But Cs and Cl ions do.
They have to, so as the attraction between them overcomes the repulsion between like ions.
If the central ion was smaller then 0.71 R1 it would not touch the nodal ions.

See http://en.wikipedia.org/wiki/Caesium_chloride.

Last edited: Oct 21, 2014
11. Oct 21, 2014

### CAF123

;( sorry
But I think I understand now, equivalently we could take a Cs ion in the middle surrounded by 8 Cl ions, making the calculation directly relevant. Can I ask if you are familiar at all with the Sommerfeld model in condensed matter? I made another thread but it hasn't attracted a lot of attention.

12. Oct 26, 2014

### TKBonny

Hi CAF123, i was wondernig how you found p? i am struggling to manage it!

13. Oct 26, 2014

### CAF123

Hi ehild, I have a final question about the derivation of $p$. I managed to derive $p$ but my work used the strict equality $r=0.73$. I think this is valid because it is the instantaneous ratio of radii before which the structure becomes stable. I am just a bit hesitant because by using this value for $r$, I am assuming that all spheres touch,(not the case in CsCl) so despite the fact I get the correct result for $p$, the reasoning is not so clear. Thanks.

14. Oct 26, 2014

### TKBonny

CAF123, could you help me with the method of finding p?

15. Oct 26, 2014

### CAF123

hey TKBonny, sure. Although I am not sure how the forum system works in this case, usually you have to give your attempt and explain your difficulties.

16. Oct 26, 2014

### TKBonny

I based the ionic structure as a BCC lattice but i dont see where the 1+r^3 comes in

17. Oct 26, 2014

### ehild

The question was "The CsCl structure is stable when the ratio of the smaller ion radius to larger ion radius is r>0.73. Why is this so?"
You calculated the radius of the central ion with the condition that all nearest ions touch each other. In CsCl, the radius of the central atom is greater than the radius of a corner ion. So r>0.73 holds.
If you read my posts you also find the answers why is that condition needed for stability.
I do not know how you derived the general relation for p, when the nodal atoms do not touch. You have not shown it.

Last edited: Oct 26, 2014
18. Oct 26, 2014

### ehild

What is the question exactly?
CsCl is not a BCC lattice. It is simple cubic with a two-atomic base. One kind of atoms (ions ) are at the corners of a cube, the other kind ones are at the centre.

19. Oct 27, 2014

### CAF123

yeah sorry, the body diagonal satisfies $2(R_1 + R_2) = \sqrt{3}a$ where $a$ is the lattice parameter. I then went to sub in $R_1 = 0.73R_2$ but this was a completely unnecessary step because later I convert back to $r$. So my derivation is fine I just need to delete this line.

20. Oct 27, 2014

### ehild

I see only the derivation when you assumed that all ions touch each other.

What would happen if the radius of the central ion was smaller than 0.73 R1? For example, R2=0.6R1.
As $\sqrt{3}a = 2(R_1+R_2) = 3.2 R_1$ the lattice parameter (side of the cubic cell) would be 1.85 R1, less than 2R1. So it would not be enough distance for two spheres of radius R_1 beside each other.