California Institute of Technology

Inter-Office Memo

To David Shoemaker

From Linus Pauling

Aug. 26, 1949

Some predicted structures, including AB₁₂ and AB₆

I think that it should be possible to show the existence of compounds AB₁₂ and AB₆, and perhaps of certain other compounds, corresponding to superstructures on cubic closest packing, some of the other structures being superstructures on hexagonal closest packing.

Let us first take the case in which atom A is surrounded by 12 atoms of B, the group AB₁₂ being equivalent to one another, and with all 13 atoms occupying points of either the cubic or hexagonal closest packed arrangement, or some similar arrangement. Let us find a structure in which the group AB₁₂ are crystallographically equivalent and all of the points of the closest packed arrangement are occupied.

We have 13 atoms. The number 13 can be represented as 1²+3x2², and hence a close packed layer can be divided into units containing 13 atoms. It is seen on examining these units that they correspond to an atom A with 6 atoms B arranged around it at each lattice point, and 2 triangular groups of 3 atoms B in addition. The orientations of the triangles are such as to cause the layers above and below to be arranged in the way corresponding to hexagonal closest packed. Accordingly the unit of structure is rhombohedral, containing 1 AB₁₂. The length of the corresponding C axis is the same as the body diagonal of cubic closest packing, and the length of the A axis is √13 x the interatomic distance.

I think that it should be possible to show the existence of this rhombohedral structure, in well-annealed solid solutions of atoms A in a metal B which crystallizes in cubic closest packing. It is very interesting to note that many metals that crystallize in cubic closest packing dissolve other metals with the same atomic radius just to the extent corresponding to the formula AB₁₂.

Sometimes the solubility is somewhat less than this, indicating that the activities of the metals in the melt are not such as to permit the ordered arrangement to be achieved, there being instead some extra atoms B occupying interstitial points between the groups AB₁₂.

I propose that you suggest to one of the new graduate students who come to see you at the beginning of the new term (I shall be away on Monday, September 26, but I shall want to see the new students during the next day or so) that he take this as his problem. This would involve the preparation of a number of alloys of composition AB₁₂, perhaps including CdAu₁₂, NaPb₁₂, SiCu₁₂, SnCu₁₂, ZnCu₁₂, MgCu₁₂, HgCu₁₂, SnPt₁₂, MnCu₁₂, SnNi₁₂, HgPb₁₂, HgTl₁₂ (kept above the β-α transition of thallium), etc. It might well be necessary to anneal the alloys in a number of different ways, and to test for superstructures in each case. Perhaps the most effective way would be to reduce the temperature of a furnace very slowly from a high enough temperature to room temperature, to permit superstructure to be formed no matter what the temperature is at which the procedure would occur.

In case that an alloy of the proper composition does not form a super-structure AB₁₂, or in case it is not possible to dissolve this much A in the phase, it might be worth while to carry on experiments to show the complexes AB₁₂ in the alloy. This might be done with, say, Cu₁₂Hg; a radial distribution might prove that mercury atoms are not present adjacent to one another, nor even with ligands shared. However, this would be a more difficult problem, applicable only in a few cases, and I think that it should be possible to demonstrate the existence of a superstructure.

In the case of metals with hexagonal close packing, I believe that there is no way to achieve the formula AB₁₂ with independent AB₁₂ groups. However, if there are extra atoms B I think that an arrangement based on hexagonal close packing can be constructed - I have not attempted to do so. Probably a compound AB₁₃ based on hexagonal closest packing could exist with AB₁₂ groups equivalent to one another, and an additional interstitial B atom. It might be worth while to examine the lithium-magnesium system, to see if the compound LiMg₁₃ exists.

The next case of interest is that in which atom A is surrounded by 12 atoms B, each of which is shared with one other AB₁₂ group. The composition of such a phase would be AB₆. Since the number 7 can be represented as 2² + 3 x 1², and since the atoms A with a hexagon of 6 atoms B around the corresponding unit in the hexagonal plane are so related to one another as to permit their corners to be connected by 3 triangles, a structure satisfying our requirements that the complexes AB₁₂ be equivalent can be formulated. It is based on a rhombohedral lattice, the corresponding hexagonal unit having its c axis equal to the body diagonal of the close packed cube and the a axis being equal to √7 x the interatomic distance.

I suggest that reported compounds with formula AB₆ be prepared and investigated to see if they correspond to this interesting rhombohedral structure. Reported compounds include WNi₆, NaHg₆, LiHg₆ (?), CeCu₆, CsHg₆, and a few others. I do not think that CeCu₆ is interesting in this respect, because the atoms involved do not have the same atomic radii, but most of the others are worth preparing and investigating (anyway I think that Knop and the people in Norway have already studied CeCu₆). In addition, the student working on this problem might well prepare a number of solid solutions of A in B, with B having the cubic close packed arrangement, and then anneal them and look for superstructure lines. The composition should correspond to the formula AB₆. Among the compounds that might be sought for in this way are MhCu₆, HgPb₆ (superstructure lines perhaps wouldn't show up) FeZr₆, HgTl₆, LiMg₆. I think that the last three at room temperature are based on hexagonal C.P.

And so it might be better to investigate the compositions AB₇.

Linus Pauling