 Solution

Write $\mathrm{Me}=\mathrm{C}{\mathrm{H}}_3$. The alternating-vertex cubane A is $\mathrm{L}{\mathrm{i}}_4\mathrm{M}{\mathrm{e}}_4$; replacing one or two methyl vertices gives $\mathrm{B}=\mathrm{L}{\mathrm{i}}_4\mathrm{M}{\mathrm{e}}_3\mathrm{X}$ and $\mathrm{C}=\mathrm{L}{\mathrm{i}}_4\mathrm{M}{\mathrm{e}}_2{\mathrm{X}}_2$. Balanced redistributions areThe smaller aggregate has composition $\mathrm{D}=\mathrm{L}{\mathrm{i}}_2\mathrm{MeX}$, since the conventional masses $2(7)+15+80=109$ or $2(7)+15+127=156$ match the supplied values. Thus $\mathrm{A}+4LiX\rightleftharpoons 4D$ is another balanced redistribution.

For the structures, place Li on one tetrahedral set of cube corners and Me/X on the complementary set, joining adjacent unlike vertices. A contains four Me vertices, B three Me and one X, and C two of each. D is a four-membered alternating Li-Me-Li-X ring. The quoted relative masses use conventional elemental masses; a fully ${}^6$Li-enriched D would be approximately two mass units lighter, without changing its composition.