As Fullerene's have an empty shell inside their structure how they withstand the pressure from the outside?

As a sidetone, are there any methods for calculating the force that the "vacuum" exerts on the outside? At some point as Fullerene's get too big I imagine that the internal force becomes overwhelmed, so this could give different perspective on this question already on the website:

What is the maximum amount of carbon atoms that can comprise a fullerene?

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    $\begingroup$ Fullerene isn't a balloon, it's like empty ball of steel - why would a ball of steel "collapse", whatever it would really mean? $\endgroup$
    – Mithoron
    Dec 16, 2021 at 0:15
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    $\begingroup$ I'm baffled by your question. 1 bar of pressure difference won't crush a steel sphere, much less a fullerene, for which you need hundreds of GPa! $\endgroup$
    – Mithoron
    Dec 16, 2021 at 1:04
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    $\begingroup$ Mithoron, I would suggest for you to see the practical limitations of a system having a vacuum. There is a reason why, as mentioned above, no vacuum airships are functional. The thickness requiered to prevent compression and buckling from the exterior atmosphere would negate the density difference needed for buoyancy. A fullerene is a one-atom layer molecule, not centimetres of steel. Big magnitude difference. $\endgroup$ Dec 16, 2021 at 5:00
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    $\begingroup$ The shape of the fullerene $\ce{C_{60}}$ is close to the one of a sphere. Forces acting from the outside of this body, initially directed inward of this body, are deflected laterally (like e.g., gothic arches, cupolas; submarine vessels) as pressure and tension. So the question might be closer to civil engineering/architecture (limits of traction, torsion, shear and bending like for I-beams used in construction of bridges, and rebars), or vector algebra/finite element analysis (depending on the level of detail if one want's to compute it ahead of time), than chemistry only. $\endgroup$
    – Buttonwood
    Dec 16, 2021 at 8:59
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    $\begingroup$ Pressure is a pretty meaningless term at the level of single molecules. The real question here is how does a fullerene not deform when interacting with other molecules, perhaps with significant kinetic energy. $\endgroup$
    – matt_black
    Dec 16, 2021 at 11:39

1 Answer 1


In addition to my take, based on what seems qualitatively reasonable and somewhat easy to explain without much technicalities, I have tried to organise the various comments that followed the question.

This answer includes two aspects, and assume a single molecule (it seems what the question refer to), and for some parts a single molecule immersed in a fluid, so the pressure is hydrostatic.

One general aspect is that chemical bonds can be relatively hard to be broken by elongation but compressing them past a certain point is almost impossible. To figure out why it is so, there is no really need for quantum physics or to invoke the Pauli's principle, as for Coulomb repulsion between the electrons in the outer shells suffices, at least qualitatively.

Even for a force coaxial with the bond, if a breaking would occur during compression it will be a kind of slipping or shearing of the bound atoms. Think of the situation in which you try to push together two magnets by the same pole, it should give a good picture.

At this point, you may ask why compression along bonds is relevant to the discussion given that pressure exerts forces which are orthogonal to them, in the current case. One point is that if the atoms are bound in a two-dimensional array, a departure from their equilibrium position in the plane is never bending alone, but geometry implies that bond lengths are altered as well. Typically for a flat array this will be elongation (if you buckle a rubber sheet, the elongation would be radial around the pushing point, to give an example). But keep reading...

The second aspect is due to the particular shape of fullerene, the latter being what has prompted you to pose the question (indeed similar ones can be posed about the structural integrity of other molecules or covalent solids, as for the latter contain "empty space", too). Objects with a high radius of curvature are effectively redirecting and distributing the forces perpendicular to the tangent towards the latter. This property is well know from ancient times and found application in the construction of archvolts, cupolae, pressurised tanks or vacuum containers as well as the light and resistant domes by Buckminster Fuller (the name Fullerenes is after him, as the molecules and the domes show remarkable similarity). We have seen that bending bonds of a surface implies stretching. However, for a spherical surface, applying outside hydrostatic pressure equates to the simultaneous compression of all bonds.

Now, it should be clear that the two aspects work together, and a buckyball is more resistant than an hypothetical sheet having the same bond strengths but lacking curvature. The more closest would be a graphene, or a monoatomic layer of graphite.

So, more you push on a fullerene, it gets harder and harder. While in principle a P leading to its implosion - whatever it means, perhaps it is better to call it rupture - could be reached, its value would be very high, certainly order of magnitude bigger than atmospheric P.

The paper at https://arxiv.org/pdf/cond-mat/0610007 gives an estimated figure of 8000 GPa (!) for the critical pressure at which one molecule of C60 fullerene becomes unstable. Rather than on the value, it is worth noting that at sufficiently high pressure the coordination number of the carbon atoms changes even upon "symmetric", hydrostatic compression. The molecule cannot simply shrink but loose its identity as fullerene.

This points again to the tremendous strength of repulsion forces, so that in short and pictorially, we could say that "there is even not that much space for the molecule to implodes in".

It should be clear that, while the shape of fullerenes makes this type of questions almost natural, the scale involved as well as considering the behaviour of a single molecule makes them devoid of sensible, or at least accessible, consequences,with perhaps the exception of atomic force spectroscopy.

The fact that the interior of fullerene is vacuum it does not mean much. What count is the difference in pressure, and we handle fullerenes at about 1 atmosphere. So, from this point of view, it is perhaps more surprising that my vacuum dessicator doesn't implode.

The question is more interesting if taken, as in the linked paper, or as I take it, for how much external pressure fullerene can sustain.

The case of an inflating fullerene (also treated in the linked paper) is even less sensible. There is nothing we can do by suction, nor a kind of needle to directly inflate it. Perhaps one way to pump the fullerene from inside would be some optically pumped endohedral fullerene. As people on optics do quite some magic, never say never...

Besides theoretical work as that linked above, much work is done on fullerenes as very little but macroscopic solid samples. Under high pressure they are known to undergo oligo- and polymerisation, or transitions leading to graphitic or diamond-like phases, depending also on the nature of the sample and temperature. Data are usually attained in the range 5 - 10 GPa. These pressures, as well the higher achievable with diamond anvil cells (about 500 GPa) are already millions times the atmospheric pressure and still one or two orders of magnitude smaller than the figure mentioned above.

Further note. The original question contained reference to virtual particles, which I have removed. A part of it perhaps remains where OP refers to "force that vacuum exerts to the outside". There is not such a force. What might be interesting and connected to the stability of higher term fullerenes is to ascertain if a smaller curvature is so detrimental to the mechanical strength that the cage is indeed unable to sustain pressure. Given the discussion and values mentioned above I would be astonished of such a big difference. But the idea is intriguing.


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