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The linked answers do attempt to clarify on 'stability' but I wanted an answer with specific connection to resonance.

For example, the ethanoate ion can exist in many forms but, as we know from data, it exists in a form that is a hybrid of all the possible forms. Why is this? I have heard teachers say that this is since that form has 'less energy' or is 'more stable.' I understand - from the linked answer- that these terms are open to interpretation. The criteria my teacher used was that in each of the contributing mesomers, the electrons are localised. But in the hybrid, they are delocalised. Now I understand this results in more 'stability' in some sense of the word.

But is that only criteria ? Should we not also consider the electronegativities of the atoms involved, for example? Basically, why is the hybrid form the form in which the structure exists?

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  • $\begingroup$ I point you to my A in chemistry.stackexchange.com/questions/111005/… though all the answers are correct I think that is the simpler. Lower the energy, the stable it is. Resonance is a recurrent question. Of course you are right. Resonance isn't stability per se. $\endgroup$ – Alchimista Mar 25 '19 at 14:10
  • $\begingroup$ Stability is a meaningless word if used alone in chemistry, organic chemistry textbooks make the case even worse. The key question is "stability with respect to what?" We need a reference. For example we say this iron block is heavy. This statement is scientifically of no use. Is it heavy with respect to a sand grain or a 70 kg human being? Please see my response in the post suggested above by Harsh Wasnik. $\endgroup$ – M. Farooq Mar 25 '19 at 14:10
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    $\begingroup$ With regard to resonance, see this question: What is resonance, and are resonance structures real? It is important to understand, that the structures with localised electron pairs are hypothetical cases, they do not exist; the expectation value of the (electronic) energy of such hypothetical structures would be higher in energy than the 'observed' structure; or in lab-jargon: less stable/unstable. By reversing the argument, the observed structure is more stable. This of course is not really true. $\endgroup$ – Martin - マーチン Mar 26 '19 at 13:16
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    $\begingroup$ @Sal_99 — The thing between "real-world" chemistry vs quantum chemistry is that, in the real-world, we need some tricks to determine electron-densities in a molecule, resonance... without having to spend 3-week calculations on a super computer. And we have difference ways to explain how we ended with a given conclusion. In the real-world chemistry, it is easier to consider a molecule as a weighted "mix" of all of its mesomers. Though, in quantum chemistry, it would just be: "the calculations show that one species is lower in energy". $\endgroup$ – SteffX Mar 26 '19 at 16:51
  • $\begingroup$ "But is that only criteria ? Should we not also consider the electronegativities of the atoms involved, for example? Basically, why is the hybrid form the form in which the structure exists?". Note that comparing mesomers nothing change in atom composition and configuration. Whatever changes, e.g. electronegativity of a moiety, it will be reflected in the stability of that specific mesomer, thus contributing to determine the importance ("weight") of it in the final hybrid description. Having resonance or not isn't a criterion to compare totally unrelated molecule. It is that sometimes a singl $\endgroup$ – Alchimista Mar 27 '19 at 9:08

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