I'll note that I'm a student in a rather introductory course, and today we learned about molecular geometries.

I'm curious as to why certain geometries (say, $\ce{CH4}$, tetrahedral) are uniform (evenly distributed in a sphere) while other geometries, say "$\ce{SbCl5}$," (5 bonds to the central atom) are less-uniformly distributed - two of the atoms are linear and 3 of them of trigonal planar.

Is this just a simplification for teaching introductory students, or is this how the molecule actually behaves? I would think that to achieve "equilibrium" from the $\ce{Cl}$ atoms repelling each other, the distribution of the $\ce{Cl}$ atoms would be perfectly normal around the sphere.

  • $\begingroup$ Duplicate: A more general version of this question, and one of the two cases asked about in this question. $\endgroup$
    – hBy2Py
    Feb 16, 2017 at 1:14
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    $\begingroup$ The short answer, summarizing from the answers here, is that it's not geometrically possible to arrange five points in a mutually equidistant fashion around a sphere. $\endgroup$
    – hBy2Py
    Feb 16, 2017 at 1:15
  • 1
    $\begingroup$ You could have five ligands equidistant from their nearest neighbors in a regular pentagon. But that is not generally a favorable structure. $\endgroup$ Feb 16, 2017 at 1:31

1 Answer 1


You may want to look into VSEPR theory. To a good first order approximation you can think of electron pairs and external atoms on a central atom as taking up about the same space. This explains why water's bond angle (104 degrees) is very close to the 109.5 degrees exact equidistant angle associated with methane, or $\ce{AlCl4+}$.

Your example with $\ce{SbCl5}$ is a good example where the theory gets a bit more complicated than simply maximizing distance between steric centers. The bonds to the chlorines are not equal. And yes this is how the chemical would look: see the Wikipedia article on orbital hybridisation.

Now, as your example has 5 identical groups, inter-conversion can occur, where the axial group "switches" with an equatorial group. But at any moment there will be two distinct chemical environments around the compound.

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    $\begingroup$ This doesn't explain why a uniform distribution of five groups is impossible. $\endgroup$ Sep 24, 2018 at 18:33

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