I am quoting a rule on resonance from Organic Chemistry by T.W. Graham Solomons, Craig B. Fryhle, 12th edition, page 25.

  1. Structures in which all the atoms have a complete valence shell of electrons (i.e., the noble gas structure) are more stable.

I tried to apply to the following.

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According to this rule structure 4 must be most stable. But structure 4 is not aromatic, while structure 1 is aromatic. Since aromatic species are more stable then nonaromatic species, structure 1 must be more stable.

My question is which of the structures is more stable?

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    $\begingroup$ One resonance structure simply can't be (or not be) aromatic, much like an island inhabited only by Tom Hanks can't be democratic. It is a collective feature. $\endgroup$ Aug 21, 2019 at 11:52
  • 4
    $\begingroup$ There is no more (or less) stable resonance structure. Only the complete set describes the molecule, they simply do not exist on their own. See: What is resonance, and are resonance structures real? The question you pose is flawed and cannot be answered. You might be able to find the contribution of the different conformations though. $\endgroup$ Aug 21, 2019 at 12:41
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    $\begingroup$ Better question would be to ask, "Which cannonical structure contributes more to the resonance hybrid? " $\endgroup$ May 13, 2021 at 6:05
  • $\begingroup$ From my studies, I would have chosen number 4 because it has the greastest number of bonds (covalent), which carries more weight than the precense and placement of charge. My text (Organic Chemistry 7th ed. by Francis Carey) doesn't cite aromaticity as a main rule for determining resonance contributions. $\endgroup$
    – Uchuuko
    Oct 10, 2021 at 17:03

1 Answer 1


There are few things you should keep in mind while deciding comparative stability of resonance structures:

  • Non-polar structures are more stable than dipolar.

  • Resonance structures with a greater number of covalent bonds are more stable than those with lesser number. Thus, nonpolar structure of buta-1,3-diene is more stable than any other resonating structures.

  • Resonance structures with similar charges on adjacent atoms are insignificant due to electrostatic repulsion and thus are unstable.

  • Structures with positive charge on electropositive element and negative charge on electronegative element are more stable.

  • Structures in which charge is delocalised are more stable than those in which there is charge separation.

  • Structures with complete octet are more stable.

  • All the atoms in a molecule taking part in resonance should be co-planar. because this enables effective overlap of p-orbitals and delocalization of electrons.

But when you have to decide among these properties itself, you can use the following observational facts:

  • Although during comaprison, there may be a positive charge on the more electronegative atom(O in this case) with complete octet, but it will be preferred over a structure with an electropositive atom ( C in this case ) having positive charge and incomplete octet.

  • But when number of covalent bonds are same, positive charge on C atom with incomplete octet is more preferred than positive charge on more electronegative O atom.

  • A stable neutral structure is a major contributor of a resonance hybrid.

An important note: when you are comparing the stability of the resonance structures , you are actually trying to figure out the major contributor to the resonance hybrid (that which best describes it). Technically, all the resonance structures are contributors to the same true structure of the ion. Since all of your mentioned structures represent the same species, they cannot have have different energies, and therefore they cannot have different stabilities.

Your question is actually asking "which contributor is more important in describing the structure and behavior of the hybrid?"

So fact number 1 and 2 makes it very clear that structure 4 in this case will be the most stable.

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    $\begingroup$ I can only reiterate: There is no such thing as a more (or less) stable resonance structure. They simply do not exist on their own and are only a tool for approximating the electronic bonding situation; only a complete set would actually do that. This answer is unfortunately wrong. $\endgroup$ Aug 22, 2019 at 13:54
  • $\begingroup$ what's a complete set ?.. It must be the pack of all the possible resonance structures of a single compound..in this case it looks complete.. $\endgroup$
    – Fancy-Toon
    Aug 22, 2019 at 15:27
  • $\begingroup$ A complete set is all possible resonance structures. The one in the question is by far not complete, it is missing the charge separated structures. The only part in which it could be considered complete is the pi system. $\endgroup$ Aug 22, 2019 at 15:51
  • $\begingroup$ but don't you think it's not possible to make charge separated structures here .. I mean the "O" atom should be pi bonded in such a case ... how can a negative charge develop over the "O" atom..?? $\endgroup$
    – Fancy-Toon
    Aug 22, 2019 at 16:23
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    $\begingroup$ With charge separated structures I mean those representing ionic bonds, i.e. in the way like $\ce{[H-CH3 <-> H+\quad^-CH3]}$. These do have non-negligible contributions and are necessary for complete description. What you describe in your answer is approximated beyond the useful, even if you disregard that resonance structures do not exist. $\endgroup$ Aug 23, 2019 at 9:48

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