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I am not now nor have I ever been a chemist.

I'm practicing by drawing molecules in the Chemspider structure search tool. Today I drew buckminsterfullerene, in order also to test my 3D spatial awareness. I chose to draw the double bonds on the hexagonal rings only - the resulting symmetry making this the intuitive choice, I thought.

Chemspider recognised my sketch as C60 right enough, but in its own 3D rendering has the double bonds placed differently, with some in the pentagonal rings, so of course I run to Google to see what is up with that, and learn the following from Wikipedia :

C60 tends to avoid having double bonds in the pentagonal rings, which makes electron delocalization poor, and results in C60 not being "superaromatic".

So according to this my intuition was correct, but I guess Chemspider just has a different version stored for some reason?

So:

Q0: Any guess as to WHY Chemspider has different configuration?

Q1: Is the extract above actually true? (If you can imagine for a moment a world where Wikipedia might not be 100% accurate and authoritative)

Q2: If so, why? As I said, the more maximally symmetric arrangement seems intuitive. I guess the charge of the mutually repelling (repulsive?) electrons would be more evenly distributed this way, and so this is how it would self-organise, like a soap bubble.

Q2: In the context of the extract, does electron delocalization being 'poor' mean that there is a lot of it (this is my instinct), or not so much?

Q3: Either way, why is the degree of delocalization contingent upon where the double bonds are? My first guess would be that the positively charged nuclei in the hexagons keep the electrons in a less vice-like grip than in the more tightly convex pentagons.

Q4: Are the bonds static? Once you have a C60 molecule with its bonds a certain way, do they stay that way, or do they jump about arbitrarily?

Q5: Does it matter? Does the geometry of the bonds affect the chemistry, or indeed physics, of C60 that much?

If your answer is 'go read a book/paper on C60' then that's fine, if you could recommend a good one for a beginner. There's a lot out there.

Cheers MD.

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closed as too broad by Mithoron, airhuff, Jan, Nilay Ghosh, bon Oct 21 '17 at 16:27

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Well, you have talent for finding dumber lines in Wikipedia, or just bad luck? ;) $\endgroup$ – Mithoron Oct 20 '17 at 18:32
  • $\begingroup$ That's imo a common misuse of the term "delocalised bond", which really is a 0/1 statement. pi electrons however delocalise even when you only have a chain of conjugated double bonds, where each double bond is perfectly fixed and cannot change unless you do actual, practical chemistry to it. $\endgroup$ – Karl Oct 20 '17 at 19:00
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All carbon atoms in fullerene are part of one "double bond", as you draw it. In quantum chemical reality, the structure of the fullerene is made up by fixed "single" bonds between those 60 sp2 hybridised carbon atoms, with the residual thirty pairs on pi electrons being somewhat delocalised.

In a sheet of graphene, the pi electrons are delocalised, and every bond lying between two six-rings gets the same share in the electron cloud.

In fullerene, they are as delocalised, but not equally distributed. The bonds between two six-rings are "more equal" (more aromatic) than those between five- and six-rings, and grab themselves a larger part of the electron density.

Problem of the five-rings is that the bond angles are not 120°, as they should be for a proper sp2 hybridised carbon atom. They become more sp2.4 or something, which let's the pi electrons move out of there a bit.

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