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I've been working through some questions online (self study, not homework) predicting the various forms for molecules. One of the questions involved the following molecule (see image below). It had four structures listed in total, but the two below have me baffled. They appear to be identical if rotated by 180 degrees along the horizontal axis.

Please explain the difference between these two structures.

enter image description here

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    $\begingroup$ They are the same as they are superimposable by rotation and reflection and the energy of the molecule cannot change by simply doing this. $\endgroup$
    – porphyrin
    Commented Jul 15, 2017 at 8:03
  • $\begingroup$ @porphyrin Do you mean that resonance structures must have different energies? $\endgroup$
    – andselisk
    Commented Jul 15, 2017 at 8:19
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    $\begingroup$ @andselisk no ! exactly the opposite, because they are related by simply rotating etc they must have the same energy $\endgroup$
    – porphyrin
    Commented Jul 15, 2017 at 8:23
  • $\begingroup$ @porphyrin Ahh! Now we are on the same page, sorry, I misunderstood your comment. $\endgroup$
    – andselisk
    Commented Jul 15, 2017 at 8:23
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    $\begingroup$ @porphyrin In resonance structures, the atoms are regarded as fixed, so the two structure are not the same. Born-Oppenheimer appoximation. Of course the two have the same energy, and if you regarded them as actual molecules, they'd be identical. $\endgroup$
    – Karl
    Commented Jul 15, 2017 at 9:08

2 Answers 2

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Resonance structures do not describe the conformation (=energy state) of a molecule, but are didactic. They show how you can just flip schemactic $\pi$ electrons around the structure, without breaking any connection. The bond lenghts and angles are constant in the actual molecule, at values somewhere between those indicated by the resonance structures.

Between resonance structures, you don't look at a possible symmetry of the location of atoms, but regard them as unique and fixed (see also https://en.wikipedia.org/wiki/Born-Oppenheimer_approximation). Resonance structures can have different energy or be symmetric (=conformationally identical) to each other. If one resonance structure has higher energy than the others, that is just a hint that the actual structure might by more closely described by the other resonance structures.

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  • $\begingroup$ The B-O approximation is not relevant here, and the structures are more than didactic. In a variational approach we would consider various covalent and ionic structures, the true energy will lie below any of these and the superposition corresponds to a 'species of resonance' (Coulson 'Valence') as a result of which energy is lowered. The amount of lowering below the lowest component energy is the resonance energy. $\endgroup$
    – porphyrin
    Commented Jul 17, 2017 at 7:08
  • $\begingroup$ @porphyrin You regard the atoms as fixed and wonder what electronic structure(s) could fit. How is that not B-O? $\endgroup$
    – Karl
    Commented Jul 18, 2017 at 5:25
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While drawing resonance structures, we keep the orientation of the molecule same and assume that the bonds have changed their positions. This is done to pictorially show the movement of electrons. The two structures given are 'equivalent' since they can be superimposed but not the same since the double bond is now between two different atoms.

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    $\begingroup$ Oh, ok. Let me see if I've understood correctly. So with the benzene ring, even though both structures are the 'same' under rotation, they are different if you fix the orientation. Fixing this makes it possible to write two sensible forms of the ring that, when considered together, show all the C-C bonds are equivalent. Is that right? $\endgroup$
    – Guest
    Commented Jul 15, 2017 at 8:15
  • $\begingroup$ So the two 'different' resonance forms in my question show that all the bonds in the ring are 'equivalent' in the same sense the C-C benzene bonds are 'equivalent'? And this would be difficult to communicate without fixing the orientation? $\endgroup$
    – Guest
    Commented Jul 15, 2017 at 8:17
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    $\begingroup$ This answer is ambigous, on the border to incomplete or just plain wrong. $\endgroup$
    – Karl
    Commented Jul 15, 2017 at 8:28
  • $\begingroup$ @Guest You are right in saying that the orientation is kept fixed to show the movement of electrons/bonds. However, unlike benzene, not $ all $ bonds in the compound you have given are equivalent, only those opposite to each other about an axis passing through the B and N atoms are. Overall the two structures are 'equivalent' but not the same. $\endgroup$
    – User
    Commented Jul 15, 2017 at 10:04

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