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In the late 80's and early 90's, Smalley and others made claims that the C60 fullerene bearing icosahedral symmetry was the most spherical molecule known, and perhaps the most spherical that could exist. While it makes sense to me that the smallest member of the 60n^2 isocahedral family of buckyballs would be the most spherical (I believe larger members should tend towards and isocahedron?), are there really no other more spherical geometries that are possible or actually experimentally realized?

I recently took a look at: "Production and isolation of an ellipsoidal C-80 fullerene" by Chun-Ru Wang et. al. (http://pubs.rsc.org/en/content/articlelanding/2000/cc/b000387p/unauth), and it looks like C80 forms an ellipsoid. However, I'm not sure about C80's endohedral counterparts.

Update - To be clearer about what I mean by "more spherical", I'm looking for a family of buckyballs that converges quickly on the volume of a circumscribing sphere as a function of the number of carbon atoms (or alternatively, as a function of the 'k'th smallest member of its symmetry family). For example, from this Mathematica document: ("http://reference.wolfram.com/legacy/v5_2/Demos/Notebooks/BuckyballConstruction.html") we can see that the icosahedral C60 fills ~87% the volume of its circumscribing sphere. If C-80 filled a larger volume of its circumscribing sphere than C60, as the k=1 member of its family of buckyballs, I would consider that "more spherical" than C60. The problem with that assumption, however, is that C60 has 20 fewer carbon atoms than C80.

Update 2 - Another way to measure how "spherical" a buckyball is could involve looking at the distribution of angles between any two adjacent carbon bonds. One would particularly want to look at the global minimum angle(s), being careful to avoid problems with large planar faces that approximate graphene and washout outliers. If the global minimum angle between carbon bonds tended to ~(120 degrees - epsilon), as a function of the kth minimum member of a particular symmetry group, that family of buckyballs would presumably also converge on their circumscribing spheres as a function of the number of carbon atoms.

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    $\begingroup$ When You ask such "question", You shold define a scale for "more or less" spherical. I would close this question because of "chattiness" $\endgroup$ – Georg Jul 19 '11 at 19:29
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    $\begingroup$ I think that if I wanted to make this notion precise, I'd calculate the distance from the center of mass for all of the atoms (specifically, their nuclei, which are small enough to be treated as classical mass points), and compute the standard deviation of these values. The most spherical molecule would have the smallest standard deviation (relative to the mean, of course). Of course, that criterion expresses the idea that all points are at nearly the same radius, not that they're equally distributed over the surface -- for instance, N$_2$ would be perfectly spherical by this criterion! $\endgroup$ – Ted Bunn Jul 19 '11 at 19:49
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    $\begingroup$ @Ted The C60 has 60 chemically identical atoms. For that reason all the atoms lie exactly on that circumscribed sphere. This was reason to tickle user8861 a bit. In the sense of Your definition C60 is perfect, and any question of more or less is in vain. There could be C20 (dodecahedrane C20H20 exists) but C20 has much strain. One should never say never, but C20 will stay a dream. Because only 5-rings and 6-rings are possible constituents such C-molecules C60 will be the only perfect spherical case. $\endgroup$ – Georg Jul 19 '11 at 20:16
  • $\begingroup$ @Ted: looks like a good idea except that (in 2D) it doesn't distinguish regular polygons from the circle. For a more natural description one should also include centers of faces or something. $\endgroup$ – Marek Jul 19 '11 at 20:23
  • $\begingroup$ Georg is right that C$_{60}$ is "perfect" by this criterion. Re Marek's comment: I'm not sure I understand why failure to distinguish regular polygons from circles is a bad thing! Any molecule has a finite number of atoms and hence describes a polygon (or would in 2D). A criterion for "most circular" that gave the highest ratings to regular polygons sounds right to me. $\endgroup$ – Ted Bunn Jul 19 '11 at 21:38
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Since graphene exists, there is no limit to how spherical you can ma ke a Buckyball by angle-measure. You can just take a flat sheet of graphene with a tiny number of defects (or just a tensile stress) and make an enormous sphere, as large as you like. The statement that the icosahedron is the "most spherical" refers to its symmetry group, which is the largest discrete subgroup of SO(3). Any other near-sphere carbon structure would have equal symmetry or less, and the macroscopic graphene balls will have no exact symmetry at all.

LATER EDIT: As per Peter Shor's comment, tensile stress alone can't do it, since the graphine structure embeds in a tiling of the plane by triangles, and you can't tile a sphere by triangles fitting together in 6's without at least some 5-vertices, because of the Euler characteristic triangulation constraints V-E+F=2 / 3F=2E. There must be defects in the lattice, whose local curvature, positive and negative, end up totaling the Euler characteristic. This is certainly possible, but it requires special defect points of a certain density to avoid the tensile stress of any region getting too great, and the problem is more involved than noting the planar limit exists.

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    $\begingroup$ You can't get create a sphere out of a flat surface. A sphere has non-zero curvature. A flat surface has zero curvature. If you try to deform a sheet of graphite into a sphere, you need to use chemistry to see what will happen. How do you get the lines of carbon atoms to match up? $\endgroup$ – Peter Shor Sep 5 '11 at 21:46
  • $\begingroup$ I know you need an Euler's characteristic worth of curvature, but I thought one could get around this using defects and stresses. In thinking about it more, it is clear that there are tilings of the plane that do not form tilings of the sphere with the same number of neighbors, for topological reasons, without any deformation helping (3V-E=6 for a triangular lattice). So it requires defects. I convinced myself that one can always introduce defects which together will form a sphere, but I am not sure about other tilings. The question is involved, but I guess the mathematicians know. $\endgroup$ – Ron Maimon Sep 6 '11 at 3:45

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