Solutions to Unit Cell Problems

You might like to look at theMaking Matter website for illustrations of the 7 2210 structures - and many more.

1. Use the radius ratio rules to predict the unit cell types of:

a. KBr b. BeO c. ZnS d. RbBr e. MgO

Rules predict: CsCl > 0.73 < NaCl > 0.41 < ZnS > 0.23

a. KBr r⁺/r^- = 0.68 NaCl structure.

a. BeO r⁺/r^- = 0.21 Predicted too small for ZnS. (See below.)

a. ZnS r⁺/r^- = 0.39 ZnS structure.

a. RbBr r⁺/r^- = 0.76 CsCl structure. (CsBr also)

a. MgO r⁺/r^- = 0.45 NaCl structure.

At least one of these predictions does not agree with reality. Which is that? Suggest an explanation?

You would have had to do a bit of research to find this. BeO has the wurtzite structure, (the hexagonal form of ZnS). It is fairly common for very small cations to be found in holes which are apparently too large for them. The explanation is, of course, that there would be a fair bit of covalent character in the bonds of such compounds, and the hard charged sphere model on which the radius ratio rules are based no longer holds.

I meant to put RbBr not CsBr! This is also "wrong", it has the NaCl structure and this is more difficult to explain. It's just an example of a situation where the rules break down. The ratio is quite close to the threshold where one might expect a bit of uncertainty.

2. The length of the side of FeO, which adopts the NaCl structure, is 0.444 nm. What is the radius of the Fe⁺² ion?

In the NaCl structure the ions touch along an edge.

I.e. 2.r⁺ + 2.r^- = L ==> r⁺ = (L - 2.r^-)/2

==> r⁺ = 0.076 nm

3. What is the length of side of the unit cell of CaF₂?

N.B. sqrt = square root, cbrt = cube root.

The point of contact here is along an internal diagonal with the F^- ions centered 1/4 & 3/4 the way down. So the length of the internal diagonal is 4(r⁺ + r^-), and this is sqrt(3) times the cell edge.

I.e. L = 4(r⁺ + r^-)/sqrt(3) ==> L = 0.524 nm

4. Assuming that cations and anions in sodium chloride are in contact what is the minimum separation between the chloride ions?

The closest the Cl^- ions get is along the face diagonals. Let the "gap" between them be g, then the length of a face diagonal is (4.r^- + 2.g) and also sqrt(2).2(r⁺ + r^-) since cations and anions touch along an edge.

Solving for g: g = [sqrt(2).2(r⁺ + r^-) - 4.r^-]/2 = 0.033 nm

5. The density of sodium, which crystallises with the BCC structure, is 0.97 g.cm^-3. What is the radius of the sodium atom?

The BCC structure has atoms in contact along an internal diagonal. If r = radius of Na atom and L = unit cell edge:-

If d = density, M = atomic mass and N = Avogadro's constant:

Combining the two:

      Li+  0.068   Be+2  0.030    O-2 0.146      F-   0.133
      Na+  0.098   Mg+2  0.065    S-2 0.190      Cl-  0.181
      K+   0.133   Ca+2  0.094                   Br-  0.196 
      Rb+  0.148   Zn+2  0.074                   I-   0.216
      

(C) H.J Clase 1999 - Updated 1999.02.17