So the origin of my question is from a model known as VSEPR which helps you predict the shapes of molecules.

According to this model, the bonds (or electron groups) arrange themselves in space in such a way that they are at maximum angles to their adjacents and therefore electrons experiencing minimum repulsion. few shapes of molecules

So my query arises from the fact that when arranging 5 bonds in space around an atom, due to some asymmetry they aren't actually arranged with maximum angles from each other and hence two bonds are at shorter angles and 3 are at larger. But if I were to arrange them at equal angles, what would the spatial arrangement be and what possibly that angle could be?

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    $\begingroup$ Maybe it's impossible to arrange five bonds with all pairs at equal angles in three dimensions, just as you can't arrange four bonds in the plane with each pair at equal angles. $\endgroup$ Apr 20 at 5:26
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    $\begingroup$ You can't arrange them at equal angles, period. This is math. There are things that we don't know, and then there are things that we do know, and they can't be changed by investigating further, and this thing is one of those we know for sure. $\endgroup$ Apr 20 at 9:15
  • $\begingroup$ @IvanNeretin Aristotle taught that regular tedrahedrons can tile space. It took 1800 years for someone to realize he was mistaken. $\endgroup$
    – Dan
    Apr 21 at 4:28
  • $\begingroup$ @Dan, Aristotle didn't give a proof. $\endgroup$ Apr 21 at 5:37
  • $\begingroup$ @Dan Aristotle thought that spiders have six legs. Much as we respect him for his foundational work in logic and other fields, the dude held some weird views. $\endgroup$ Apr 21 at 18:46

3 Answers 3


You can't do better than the trigonal bipyramidal geometry of (say) phosphorus pentafluoride, where the inter-bond angles are either $90^\circ$ or $120^\circ$. Any small rotation of one of these bonds will result either in a reduction of one of the $90^\circ$ inter-bond angles or in a reduction of one of the $120^\circ$ angles without any increase in a $90^\circ$ angle.

We will prove that there are no five rays from a point in euclidean 3D space $\mathrm E_3$ at equal angles to each other. To do this, we consider the general case of $n$ dimensions, treat $\mathrm E_n$ as a vector space, and represent the rays by equal-length vectors. Thus, let $\pmb u_i\in\mathrm E_n\;(i=1,...,n+2)$. We say that the set $\{\pmb u_1,...,\pmb u_{n+2}\}$ is hyperequal in $\mathrm E_n$ if there are constants $a_n>0$ and $b_n<a_n$ such that $\pmb u_i\cdot\pmb u_i=a_n$ and $\pmb u_i\cdot\pmb u_j=b_n$ for all unequal $i,j=1,...,n+2$.

Lemma If there is a hyperequal set in $\mathrm E_n$, then there is one in $\mathrm E_{n-1}$.

Proof. We can decompose $\mathrm E_n$ as $\mathrm E_1\oplus\mathrm E_{n-1}$, where $\mathrm E_1$ is spanned by $\pmb u_{n+2}$ and $\mathrm E_{n-1}$ includes the span of the $n+1$ vectors $$\pmb v_i:=\pmb u_i-\frac{b_n}{a_n}\pmb u_{n+2}\quad(i=1,...,n+1),$$ where $a_n$ and $b_n$ are defined as before. It is straightforward to check that, if $\{\pmb u_1,...,\pmb u_{n+2}\}$ is hyperequal in $\mathrm E_n$, then $\{\pmb v_1,...,\pmb v_{n+1}\}$ is hyperequal in (a version of) $\mathrm E_{n-1}.\quad\square$

From the lemma, it follows that, if there is a hyperequal set in $\mathrm E_3$, then there is one in $\mathrm E_2$, and therefore in $\mathrm E_1$. Thus, if so, then there are real numbers $a,b,c$ such that $a^2=b^2=c^2=a_1>0$ and $bc=ca=ab=b_1<a_1$. Consequently, since $a$, $b$, and $c$ are nozero, they must all be equal. But then $b_1=a_1$: a contradiction. Hence there is no hyperequal set in $\mathrm E_3$.

  • $\begingroup$ If we dig into the chemistry of that structure, the axial bonds are longer and equitorial bonds are spherical, due this asymmetry they are at different angles But if we forget the asymmetry of the nature of these bonds, there has to be a way of arranging them at equal angles in space $\endgroup$ Apr 20 at 6:03
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    $\begingroup$ @HarshChaudhari : No. The length of the bonds is immaterial. There is no way that five rays can radiate, each at equal angles to its neighbours, from a point in 3D space. Similarly, there is no way to arrange three points on a line, four points on a plane, or five points in 3D space equidistant from each other. $\endgroup$ Apr 20 at 6:23
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    $\begingroup$ @JohnBentin "no way that five rays can radiate, each at equal angles to its neighbours, from a point in 3D space." Is there a proof of this? $\endgroup$ Apr 20 at 15:50
  • $\begingroup$ @An_Elephant en.wikipedia.org/wiki/Point_groups_in_three_dimensions $\endgroup$ Apr 20 at 19:36
  • $\begingroup$ While the approach in the answer may work to demonstrate that the arrangement is locally optimum, it doesn't (necessarily) follow that there isn't some other arrangement that's more globally optimum. (One which, starting from trigonal bipyramidal, could only be reached by continuous angular deformation through a state with a < 90 angle.) -- That arrangement doesn't exist, but this argument doesn't demonstrate that, though. $\endgroup$
    – R.M.
    Apr 20 at 22:45

We have a known arrangement of five points on a unit sphere (the five points where the "sticks" of the molecular model, or the extensions of those "sticks", intersect the unit sphere) called a "trigonal bipyramidal" arrangement. In this arrangement, the smallest angle between a pair of points is $90$ degrees.

Can we have an arrangement of five points where the minimum angle is more than $90$ degrees? That is, can we have a minimum angle of $90 + \varepsilon$ degrees where $\varepsilon > 0$? Let's try.

Start by placing two points on a unit sphere, separated by $90 + \varepsilon$ degrees, so as to establish the minimum angle between two points.

Set up a spherical coordinate system so that the two given points are on the equator, one point has longitude $0$ degrees, and the other point has longitude $90 + \varepsilon$ degrees.

Now paint every point on the sphere that is within an angle $90 + \varepsilon$ degrees from either of the two given points. You will then have painted every point that is not within the quadrant from $-90$ degrees longitude to $-180$ degrees longitude. You will have painted some points inside the quadrant as well.

The remaining points must all be in the remaining unpainted region. Otherwise they would be too close to the two points we've already identified.

Consider a point inside the unpainted region. If we put that point on the equator and paint every point within $90 + \varepsilon$ degrees of that point, we'll have painted the entire quadrant from $-90$ degrees longitude to $-180$ degrees longitude (plus other points) and there will be no place remaining to put any other points.

So let's suppose we put a third point in the unpainted region "above" the equator. This point is somewhere in the octant between $-90$ degrees and $-180$ degrees longitude and above the equator. Since any two points in an octant are at most $90$ degrees apart, when we then paint every point within $90 + \varepsilon$ degrees of this third point, we will have painted all the unpainted points above the equator (as well as at least some of the unpainted points below the equator). If any points remain unpainted, they are all within the octant of the sphere between $-90$ degrees and $-180$ degrees longitude and below the equator.

We still have two points left to place. But within one octant of the sphere, there are no two points more than $90$ degrees apart.

Things are no better if we put the third point "below" the equator.

So there is no arrangement of five points where the minimum angle between any two points is greater than $90$ degrees.


Stated another way, you want to find a convex, regular polyhedron with five faces. The angles between the centers of any two neighboring faces would all be equal. However, there is no such polyhedron. There are only 5 convex, regular polyhedra, the platonic solids, and the have 4, 6, 8, 12, and 20 faces.

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    $\begingroup$ There are also arrangements of 1, 2 and 3 points on a sphere such that their pairwise angles are all equal (and maximal), but the corresponding "generalized polyhedra" are degenerate, as the points are all coplanar. Also, strictly speaking only these and the tetrahedral arrangement of 4 points have all the angles between the points equal — for the other platonic solids, that only holds for the centers of adjacent faces. (And if that's good enough, the Catalan solids also qualify. There's no 5-sided Catalan solid either, though.) $\endgroup$ Apr 20 at 16:40

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