We know that the structure of $\ce{CH4}$ is tetrahedral, and not planar tetragonal, as a tetrahedron enables maximum distribution of $\ce{H}$-atoms in space, as they repulse from each other. This is proved by the fact that each $\ce{H-C-H}$ bond angle is equal in magnitude.

In the case of $\ce{PCl5}$, why should the shape be trigonal bipyramidal? Obviously the top and bottom $\ce{Cl}$-atoms, above and below the plane subtend $90^\circ$ with each of the $3$ planar $\ce{Cl}$-atoms; whilst planar ones subtend $120^\circ$ in between each other. This means all bond angles aren't equal. Why so? Shouldn't there be regular distribution of $\ce{Cl}$-atoms in space?

To my opinion, the XY-plane distribution of $\ce{Cl}$-atoms should be such that each $\ce{Cl-P}$ is $72^\circ$ apart from the other $\ce{Cl-P}$ bond (atoms would not be in the same plane at all. But we should see that the orthogonal projection of each on XY-plane subtends equal angles, i.e. $72^\circ$). Same for YZ-plane. Hence there is regular distribution of atoms along all planes. Now it is to be calculated what the ultimate structure results in.

Sorry, I don't have the necessary tools/knowledge to represent my hypothesis. Else could've just modelled this figure and shown a representative model.

  • 4
    $\begingroup$ The figure you described is simply not there. People with necessary tools have thought of this before. $\endgroup$ Nov 12, 2019 at 14:16
  • $\begingroup$ See, there are two ways to get anything done: either do it yourself, or have someone do it for you. Those who are qualified to do it for you are telling you that it can't be done. Can you do it yourself? $\endgroup$ Nov 12, 2019 at 14:33
  • 5
    $\begingroup$ math.stackexchange.com/questions/185441/… $\endgroup$ Nov 12, 2019 at 14:39
  • 1
    $\begingroup$ Is your proposal in accordance with experimental results? $\endgroup$
    – aventurin
    Nov 12, 2019 at 23:49

2 Answers 2


The trigonal bipyramidal shape of $\ce{PCl5}$ is indeed the lowest energy conformation.

I have had a very hard time figuring out what your description of the structure meant and I am still not convinced I have actually understood correctly. However I was trying to model it in my head I either arrived at a variety of different angles, or at least two of them were equal to $72^\circ$, or both of these cases.

The only structure $\ce{ML5}$, where all $\ce{L-M-L}$ bond angles are equal is a regular pentagon, i.e. all ligands are in one plane, with $\angle(\ce{L-M-L})=72^\circ$, e.g. $\ce{[XeF5]-}$.

Another possibility for (almost all) equal bond angles is a square pyramid, where $\angle(\ce{L-M-L})=90^\circ$, e.g. $\ce{ClF5}$. Note however, that there are also other angles, so not all bond angles are equal, $\angle(\ce{L-M-L})=180^\circ$.

There is no better solution to arrange 5 point 'uniformly' on a sphere than the trigonal bipyramid. This is also known as the Thompson Problem, which is addressed in the linked question on Mathematics.se, provided by orthocresol: 5 Points uniformly placed on a sphere.
Another point of view would be to ask the question whether there is a way to put 5 points on the surface of the sphere so that they are indistinguishable, which is also answered there. Spoiler warning: there isn't.

You can dive deeper into the matter. For example for $\mathbb{R}^n$ there can only be $n+1$ points equally equidistant to each other. In our perceived reality $n=3$, which makes the tetrahedron the most complex molecule with this property. See on mathematics.se: Proof maximum number of equidistant points for a given dimension (duplicate), Prove that $n+2$ points in $\mathbb{R}^n$ cannot all be at a unit distance from each other (duplicate), $n$ points can be equidistant from each other only in dimensions $\ge n-1$? (Which are all duplicates, but still have answers attached to them.)

And then there are only five Platonic solids, none of which has five vertices.

Furthermore, I would like to add to all of that another word of caution: Molecules are flexible. The trigonal bipyramidal arrangement of ligands just happens to be the low energy conformation for a variety of reasons (considering the necessary and useful approximations, e.g. Born-Oppenheimer approximation).
The molecule (and other analogues) are special though, as the undergo pseudorotation according to the Berry Mechanism (see Wikipedia).

Side note:

We know that the structure of $\ce{CH4}$ is tetrahedral, and not planar tetragonal, as a tetrahedron enables maximum distribution of $\ce{H}$-atoms in space, as they repulse from each other. This is proved by the fact that each $\ce{H-C-H}$ bond angle is equal in magnitude.

This is only half the truth. It is certainly true that protons repel each other, more importantly though is that electrons repel each other, too. The most important part of which is also built into the very simple VSEPR theory. While this theory has its uses, there is a lot of criticism attached to it and one should be aware of that. I shall stress again at this point that VSEPR theory is unrelated to orbital hybridisation; there are quite a few textbooks which do make this post-hoc rationalisation, which can be misleading and dangerous.


The shape of any molecule may be obtained by applying the VSEPR model (Valence Shell Electron Pair Repulsion). In this theory the angle between the bonds must be as large as possible. With your idea, the angles between P-Cl bonds are equal to 72°. It is not very much. In the bipyramidal model, the angles are 90° and 120°. This is much better than with your geometry.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.