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$\ce{C60}$ has a pentagon-hexagon system. Why can't one make it from a hexagon-heptagon system, or a pentagon-heptagon system for that matter?

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  • $\begingroup$ It is possible to build carbon skeletons of other geometric forms (though they usually contain hydrogen as well). But they are not exactly easy to make. Dodecahedrane, for example, has a carbon skeleton with just pentagons, derivatives of cubane (just squares) are known. And they don't have 60 carbons. $\endgroup$ – matt_black Sep 25 '16 at 11:15
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Well, let's see. This is not quite a chemical question, though I wouldn't recommend moving it to Math.SE either, because they might have hard time recognizing what's $\ce{C60}$.

Now, let's imagine we're considering locally planar conjugated carbon structures. Each carbon has to reserve one $p$ orbital for $\pi$-bonding, which leaves it with 3 bonds. Now, the bonds tend to go in cycles; let's confine ourselves to 5-, 6-, and 7-membered cycles, because otherwise we'll have too much of angular strain (not that other cycles are totally impossible, though). So in effect, we are considering graphs consisting of pentagons, hexagons, and heptagons, with 3 edges meeting at each vertex. Let's say the numbers of these n-gons are $n_5, n_6$, and $n_7$, correspondingly.

That's when old Euler kicks in. Polyhedra must meet a certain relation between the number of faces (F), edges (E), and vertices (V): $$V - E + F = 2$$ OK, what is the number of faces? Why, that's quite simple: $F=n_5+n_6+n_7$. Then what is the number of vertices? Apparently, each pentagonal face has five of them, and so on, so it would be $5n_5+6n_6+7n_7$. But wait, we've counted each atom three times (for it belongs to three polygons), hence we must divide by 3: $V=(5n_5+6n_6+7n_7)/3$. By similar reasoning, the number of edges would be $5n_5+6n_6+7n_7$, if not for the fact that each edge belongs to two faces and thus is counted twice, hence $E=(5n_5+6n_6+7n_7)/2$.

All in all, $$ V - E + F = 2\\ {5n_5+6n_6+7n_7\over3} - {5n_5+6n_6+7n_7\over2} + n_5+n_6+n_7 = 2\\ -{5n_5+6n_6+7n_7\over6} + n_5+n_6+n_7 = 2\\ {n_5-n_7\over6} = 2\\ n_5-n_7 = 12 $$

That's pretty much the size of it. You can do more or less whatever you want, as long as you have twelve more pentagons than heptagons. There is no going around it.

Now chemistry kicks in, too, and brings its own limitations. An $sp^2$ carbon wants to have its bonds in one plane, at $120^\circ$ angles. You can convince it to budge, but only so much. If you strain it more, the system becomes unstable. It turns out that two pentagons meeting at one carbon is already too much (not for an $sp^3$ carbon, though; see dodecahedrane). That's why you can't have constructions made of pentagons alone (or pentagons and heptagons, for that matter). Geometry would allow that, but chemistry wouldn't.

Let's see how that translates to actual shapes. Several fused hexagons give us coronene; it's planar. Coronene

(It is tilted sideways to give it a 3D feel. There are hydrogens attached on the outer edge; I omitted them for clarity.)

Extend it further, and you'll have an infinite flat sheet; that would be graphene.

Throw in a pentagon, and here's corannulene; it's bowl-shaped. Corannulene

Extend that pattern, and you'll have fullerene. Fullerene

Use a heptagon, and get circulene:

Circulene

See that? It's bent indeed, but it's bent the wrong way. It's a saddle-shaped bend. An example of negative Gaussian curvature, so to say. You can't wrap it to a sphere, no matter how hard you try.

Can we extend this further? Well, we kinda can (on paper, that is). Moreover, it can be done in a number of ways, some leading to magnificent structures (purely hypothetical, of course). Big stuff

Upd. It was brought to my attention that according to some studies, so-called carbon nanofoam might be a real example of structures of the latter kind. I'm not quite convinced (to me, until you have an X-ray diffraction study, everything is just hand-waving), but let's mention that anyway.

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    $\begingroup$ I was wondering whether anyone would look at this from a geometrical standpoint, and your answer is even more pleasant to read than I expected! I shall briefly point out Stone-Wales defects and carbon nanofoam as chemically-relevant examples of fused heptagons in carbon structures. $\endgroup$ – Nicolau Saker Neto Sep 25 '16 at 11:22
  • $\begingroup$ Thought of including the Stone-Wales thing at some point or another, but it didn't quite fit anywhere, and the answer is long enough already (by my standards, that is), so let's save it for another day. $\endgroup$ – Ivan Neretin Sep 25 '16 at 12:06
  • $\begingroup$ Really, really great answer! $\endgroup$ – Jan Sep 25 '16 at 19:02

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