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In my book, it is written:

Resonance energy of a system involving monopolar resonance structures is greater than that involving bipolar resonance structures.

The book cites example:

Carboxylate ion which contains negative monopolar charge is in fact more stable than the corresponding carboxylic acid which contains both negative and positive charge.

But what's the reason?? What is going on?? Please help me with a math-free explanation.

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  • $\begingroup$ This sounds weird to me, and googling for "monopolar resonance structures" gives only this page (besides antenna construction). $\endgroup$ – ssavec Dec 2 '14 at 6:06
  • $\begingroup$ @user36790 Could you cite the source of the quote, too, please. $\endgroup$ – Martin - マーチン Dec 2 '14 at 7:55
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    $\begingroup$ In addition to Max's good answer, it is a complete logical fallacy to go into "this resonance contributor is better than that resonance contributor". Neither contributor is a real thing. However, you can talk about "which contributor more closely resembles the true structure of the hybrid". $\endgroup$ – Ben Norris Dec 2 '14 at 13:32
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In the diagram below, the 'bipolar resonance structure' refers to the resonance structure with both positive and negative partial charges (left), while the monopolar resonance structure refers to the anionic form in the right. I don't think this terminology is appropriate, and I have never encountered it before. I will refer to the structures as 'protonated' (left) and 'deprotonated' (right).

enter image description here

The diagram above illustrates the scenario. While it is true that the resonance stabilization is greater for the deprotonated form, the protonated form is still overall more stable. The book is inaccurate in that regard.

The "resonance stabilization" refers to electrons being donated into empty orbitals within the molecule. In this case, there is a lone pair in the hydroxide oxygen atom being donated into the antibonding pi orbital of the C = O. A way to rationalize the effect is that, when the proton is attached to the oxygen, you have both the H and the O pulling on the electrons. When the H is removed the electrons will more readily flow into the C = O pi*. Since both O atoms are equivalent, you end up forming an orbital that looks like an extended pi bond.

The stabilization due to the resonance of the deprotonated form is of greater magnitude than the stabilization due to resonance of the protonated form. However, the proton-oxygen bond is more stabilizing than the resonance, so the protonated form is overall more stable.

Diagram soure: http://chemwiki.ucdavis.edu/Organic_Chemistry/Carboxylic_Acids/Properties_of_Carboxylic_Acids/Physical_Properties_of_Carboxylic_Acids

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