Why does nature "prefer" symmetry? Why is symmetry energy-lowering? I keep hearing statements along the lines of this tossed around by chemists. Here are some examples:



IBrF2 - van der Waals repulsions - and number of unique bond angles (The more symmetric conformation of the molecule is probably preferred).


1) Are any of the above statements true? If so, can you provide example(s)?

2) I think the statement is false at least to some extent. Consider cyclobutadiene ... it's not a square; it's a rectangle.

  • $\begingroup$ $H_2S_2O_5$ is not symmetric. Myth busted! $\endgroup$
    – Rohinb97
    Commented Feb 15, 2015 at 9:07
  • $\begingroup$ @Rohinb97 what's the significance of the molecule you named? $\endgroup$
    – Dissenter
    Commented Feb 16, 2015 at 4:50
  • $\begingroup$ Well symmetry isn't favored so much. Nature does like it, but it's not everywhere. $\endgroup$
    – Rohinb97
    Commented Feb 16, 2015 at 14:02

3 Answers 3


Symmetry is preferred if it lowers energy, but it does not always.

My high school chemistry teacher had a good rule of thumb. A chemical system will [almost] always tend towards the lowest energy configuration. (I insert "almost," because there are some cases where this isn't true - usually because of entropy.)

So yes, many things are symmetric and this usually lowers the energy.

You raise a good counter-example. Cyclobutadiene is not square. A square cyclobutadiene would be a di-radical, so it undergoes symmetry breaking (i.e., a Jahn-Teller distortion) and becomes a rectangle.

enter image description here

We see here that the rectangular distortion has lower total energy - one orbital goes down in energy and becomes doubly-occupied. The other goes up in energy and is now empty.

So what about benzene? Why is it symmetric, vs. a double-single-double.. bond alternation?

Well, in this case, the symmetric (aromatic) version is lower in energy. (Figure, Chemwiki)

enter image description here


The Jahn-Teller theorem says almost opposite:

any non-linear molecular system in a degenerate electronic state will be unstable and will undergo distortion to form a system of lower symmetry and lower energy, thereby removing the degeneracy


Why does nature "prefer" symmetry?

Why is symmetry energy-lowering?

Are any of the above statements true?

Things are either "true" or "false". The condition is binary, there is no "somewhat true". In that sense, the above statements are false. There are many, many non-symmetric things in nature.

  • The human heart is not symmetrically positioned within the human body.
  • The narwhal's tusk is helically grooved and projects from the left side of the jaw.
  • You are not symmetric with the person standing next to you.
  • Like people, most snowflakes are different one from another (see here).
  • The staggered conformation ($\ce{D_{3d}}$, lower symmetry) of ethane is lower in energy than the eclipsed conformation ($\ce{D_{3h}}$, higher symmetry).
  • Cyclooctatetraene exists in a tub shaped conformation with 4-fold symmetry, rather than planar with 8-fold symmetry.
  • Cyclobutane exists in a puckered $\ce{D_{2d}}$ geometry rather than a planar geometry with 4-fold symmetry.
  • Cyclohexane exists in a chair conformation with a 3-fold axis of symmetry rather than a planar configuration with 6-fold symmetry.

From these examples we see that higher symmetry is not always preferred. In those situations where a symmetric arrangement is preferred it is likely due to

  • It requires less (DNA) code to create identical arms or arrangements on a living structure
  • Electrostatic repulsions are often minimized by creating structures with higher symmetry (e.g. methane is tetrahedral rather than planar).

In those situations where the symmetric arrangement is not preferred evolutionary advantages or steric\electrostatic repulsions, hyperconjugation (ethane conformation), etc., outweigh other factors.

  • $\begingroup$ "The staggered conformation (lower symmetry) of ethane is lower in energy than the eclipsed conformation (higher symmetry)." - actually, the symmetry group is just as big in either case. There is a 3-fold rotational symmetry with a mirror symmetry parallel to the axis in either case, and while the eclipsed conformation has an extra mirror symmetry that the staggered conformation doesn't have, the staggered conformation has an extra point symmetry. Moreover point symmetries are usually better at distributing points along a sphere - of all platonic solids, only the tetrahedron doesn't have one. $\endgroup$ Commented Feb 15, 2015 at 12:34
  • $\begingroup$ @JanDvorak staggered ethane is $\ce{D_{3d}~ (C_3 ~+~ 3C_2~+~3\sigma_{\mathrm{v}})}$; eclipsed ethane is $\ce{D_{3h} ~ (C_3 ~+~ 3C_2~+~3\sigma_{\mathrm{v}}~+~ \sigma_{h})}$, the latter containing the extra $\ce{\sigma_{h}}$ symmetry plane - because of this, I've always seen $\ce{D_{nh}}$ listed as higher symmetry than $\ce{D_{nv}}$ $\endgroup$
    – ron
    Commented Feb 15, 2015 at 14:56
  • 3
    $\begingroup$ D3d also has i (inversion) $\endgroup$
    – DavePhD
    Commented Feb 16, 2015 at 3:23

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