3

So this is easy to calculate in 2 dimensions, if you have a circle represented by 3 points the angle between any two consecutive points and the spheres centre is simply $\frac{2\pi}{n}$. I basically would like to know if there is a version of the same thing in 3d for spheres.

The reason why I need to know this is because I am trying to calculate the bonding angles of different molecules. For example, methane has a tetrahedral shape and whilst I could just go look up the H-C-H bond angle, it would be better to have a general formula. This is also how I realised it was actually a math problem not just a chemistry problem.

Jonathan
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  • This looks like a though one. What do you mean exactly by "equidistant"? – Daniel Robert-Nicoud Sep 13 '17 at 16:43
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    Evidently, by "equidistant" you mean that, in 2D they form a regular n-agon inscribed in the circumference. But in 3D you are limited to the five platonic solids, if you are imposing full symmetry. Otherwise you shall specify how the points are .."equidistant" – G Cab Sep 13 '17 at 16:59
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    You can find the central angle of platonic solids in this link: https://en.wikipedia.org/wiki/Table_of_polyhedron_dihedral_angles –  Sep 13 '17 at 17:13
  • @GCab That's kind of my point. A possibility is to ask about configurations minimizing potential energy for $n$ charged particles constrained on a sphere. – Daniel Robert-Nicoud Sep 13 '17 at 17:31
  • that's a well known problem. see e.g this page – G Cab Sep 13 '17 at 23:44
  • See https://math.stackexchange.com/questions/1710388/why-is-arccos-1-3-the-optimal-angle-between-bonds-in-a-methane-ch4-molecule seems relevant; the answer points out that even for water, the story is more complicated, and the two hydrogen atoms do not end up "equidistant" on a sphere around the oxygen atom. – David K Sep 14 '17 at 12:47
  • Yes you're right, I am aware of this and I'm basically running a rough computational chemistry simulation in which I'm implementing a very rough version of VSEPR (ignoring proper orbitals for the purposes of what I'm doing). So in essence lone pairs will be treated as if they were other bonds. So in H2O the bond angle would be $\frac{2\pi}{3}$ which is inaccurate but enough for what I'm doing. – Jonathan Sep 14 '17 at 12:53
  • @Daniel only three of the five Platonic solids give minimal electrostatic repulsion. The cube is a saddle point, same for the dodecahedron; the minimum electrostatic potential for eight or twenty vertices uses lower-symmetry polyhedra with triangular faces. – Oscar Lanzi Sep 16 '17 at 00:52

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