Does symmetry lower the energy of molecules? 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:

https://physics.stackexchange.com/questions/135671/nature-favours-symmetry

https://books.google.com/books?id=WkRy8b0F2HwC&amp;pg=PA153&amp;dq=nature+favors+symmetry&amp;hl=en&amp;sa=X&amp;ei=V__fVPvpGsuxggSptIG4AQ&amp;ved=0CDAQ6AEwAw

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

So: 

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.
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.



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)