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I teach an MCAT course in chemistry. I like to explain VSEPR by saying.. first, imagine arranging electron pairs around the central atom so they are maximally distant from each other, and uniformly arranged. If there are two pairs, they'll just be across from each other. Three? You get a triangle, and so on.

This rhetorical device fails with five pairs. Five is strange because the electron pairs are most definitely not equidistant from each other, with the equatorial pairs being 120 degrees from each other, but 90 degrees from the axial electron pairs.

My question: is this a simplifying geometry for high school / college classes, and the geometry is actually more equally arranged in 3D space among the pairs? Alternatively, if 'trigonal bipyramidal' is actually the geometry, why? What induces the asymmetry? And does that imply that molecules with the same formula and connectivity, but with an atom equatorially versus axially located correspond to distinct stereoisomers?

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    $\begingroup$ Welcome to chemistry.se! Let me just quickly point out the obvious: VSEPR-theory is based on solely based on an observation, that many molecules behave like this. There is absolutely no theoretical fundament behind it whatsoever. In this sense it is a oversimplification for chemistry beginners. But since your question is quite broad I am currently not able to provide a complete answer. $\endgroup$ – Martin - マーチン Oct 2 '14 at 8:36
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    $\begingroup$ As a simple example, take PF5. The axial and equatorial positions are different, but exchange rapidly through Berry pseudorotation. So in short, there is asymmetry (stemming from orbital theory), and the molecule does not like it. $\endgroup$ – ssavec Oct 2 '14 at 10:57
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Yes, 5-coordinate compounds are always a bit confusing.

The best way I've answered this question about VSEPR for students is to do a little demo with balloons. Blow up balloons (i.e., electron pairs) and tie them around a central knot. It will definitely give a trigonal bipyramidal shape - it has nothing to do with chemistry, and everything to do with "fitting 5 things around a central point."

I disagree with @Martin that there's no theoretical basis to the shapes. The theory is purely geometric.

There are multiple possible shapes, including square pyramidal, but these are higher in energy because you have more ~90-degree angles. As discussed in the comments, if all 5 atoms are in a plane, the angles are 72 degrees, which is clearly bad. You could in principal have the top of an octahedron, in which case all angles are 90 degrees, but that's also clearly worse. (In practice, square pyramidal structures have the central atom slightly above the plane, increasing some angles slightly beyond 90.)

So I do the demo with the balloons, talk about square pyramidal and how this alternative is clearly slightly higher in energy.

The Berry pseudorotation, as mentioned above, does convert between the shapes, and it's known that at room temperature the axial and equatorial positions scramble quickly.

Update:

I decided to do some quick calculations using Avogadro and MOPAC on $\ce{PF5}$ in square planar versus trigonal bipyramidal using the PM7 semiempirical method. This took longer to write than to run the calculations (seconds).

Here's the optimized square pyramidal geometry:

square pyramidal PF5

The phosphorous isn't in the plane, it's a bit above, to increase four of the bond angles and minimize electron-pair repulsion. Note that the largest angle is ~103 degrees, but the F-P-F angles in the basal plane are all ~87 degrees.

Here's the trigonal bipyramidal geometry:

trigonal bipyramidal PF5

Now the angles are exactly what we expect from VSEPR. We see 90 degree angles between the equatorial plane and the axial F atoms. And the F-P-F in the plane are all 120 degrees.

Just from the angles we can guess that trigonal bipyramidal should be lower in energy.

For reference, the difference in energy between these two geometries using PM7 is ~2.76 kcal/mol. So there's not a huge difference in energy, but it's there.

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  • $\begingroup$ Hmm.. "There is simply no way geometrically to create 5 completely equal angles." A planar pentagon satisfies this! $\endgroup$ – vector07 Oct 2 '14 at 16:20
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    $\begingroup$ On the contrary, if the five electron pairs are arranged in a planar pentagon around the central atom, the angles between the non-adjacent pairs would be 144 degrees while the angle between adjacent pairs would be 72. The initial statement was poorly worded, but a planar pentagon doesn't solve the problem. $\endgroup$ – Julie Thomas Oct 2 '14 at 16:40
  • $\begingroup$ @JulieThomas Ahhhh right, good point. Here's my thought experiment, and why I have trouble with the inability to get 5 equal angles. Suppose you arrange the 5 in 3D randomly. There are clearly some arrangements which are less evenly spaced than others. But if some are less, and others more evenly spaced, then there must exist a maximally distant arrangement (or set of such arrangements). So, is 'trigonal bipyramidal' that arrangement? Can it be shown mathematically? It seems if that's the case, there's something more than just spatial arrangement to it. $\endgroup$ – vector07 Oct 2 '14 at 17:08
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    $\begingroup$ @vector07 The mathematical proof is, in fact, Geoff's example. The the most stable (i.e., lowest energy) configuration is the trigonal bipyramidal with the square pyramidal just a wee bit less stable (higher in energy). Those calculations are the result of minimizing the repulsion among all the electrons arranged around the central atom. It is, in effect, a spatial arrangement, because the repulsion between electrons makes it conceptually appear as though they take up space. That's why the balloon analogy works as a visual aid. $\endgroup$ – Julie Thomas Oct 2 '14 at 17:49
  • $\begingroup$ @vector07 Yes, trigonal bipyramidal is exactly the structure that provides a maximally distant arrangement of 5 things around a central point. The reason it's not intuitive is because we don't frequently encounter 5 things around a central point except in chemistry (or the balloon demo). $\endgroup$ – Geoff Hutchison Oct 2 '14 at 18:22
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When I studied this in school this year, I too found it confusing as trigonal bipyramidal geometry is asymmetric. I did some research and found out that in n dimensions maximum of n+1 vectors can be equidistant from each other(like trigonal planar and tetrahedral).(see https://math.stackexchange.com/questions/714711/how-to-find-n1-equidistant-vectors-on-an-n-sphere ) So for 5 vectors, it would require 4 dimensions, which would make a 5 cell( https://en.wikipedia.org/wiki/5-cell#Projections ). Its projection on 3 dimensions gives us a trigonal bipyramidal. I think that it can be proved mathematically too using the fact that it is the projection of 5 cell on 3 dimensions.

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