Ozone is a resonating structure, is it because of its resonance and bent structure that it is a polar molecule? Or Is it because of the formal charges whereby the formal charges determine the direction of the dipole moment. If yes, how does this work with resonating structures. Initially, I thought that resonating structures automatically are polar, however, the azide ion is a clear counter-example. I am curious, how one deals with such a case, should we take each resonating structure by itself and consider the superposition in the end.

Ultimately, my question is how can you explain ozone's polarity without formal charges because on a wikipedia page, they explain it using formal charges. If that's the case, shouldn't resonance be considered. If yes, how?

  • $\begingroup$ Concerning your ultimate point, presumably the "polar" you mean is the dipole moment (correct me if I am wrong). N$_3^+$ is linear. Its eigenstates are parity eigenstates. Therefore it has no dipole. O$_3$ is V-shape. The only possibility to let O$_3$ has vanishing dipole is to let its electronic contribution cancels exactly with its nuclear part. It could only happen in very special cases, e.g. the total density is a perfect superposition atomic density. $\endgroup$
    – user26143
    Jun 16, 2014 at 10:48
  • $\begingroup$ This is highly unlikely. Since covalent bond will produce superposition of wavefunction, thus deviates electronic density. $\endgroup$
    – user26143
    Jun 16, 2014 at 11:58

2 Answers 2


When talking about resonance it is important to know that this is only a concept to visualise certain bonding features. It is very important to understand, that the actual bonding situation is a mixture of all possible resonance structures. The Molecule itself exists at all times as this mixture, hence neither of these resonance representations can be observed. One can regard it also as a superposition of these structures.

Resonance structures had to be introduced into the Lewis formula context, when it was obvious, that one formula can no longer describe a bonding pattern remotely accurate. While the Lewis formalism is great for many things, as it keeps chemistry visualisable and simple, it unfortunately has its boundaries. Only for molecules with very simple bonding patterns it is somehow possible to guess some physical properties (qualitatively).

Formal charges are also only a concept, or a result from the limited depiction of the Lewis representation - hence the name formal. They do not usually represent any physical charge. However, sometimes, the physical charge will coincide with the formal charge. Therefore it is usually not possible to derive any actual dipole moment from them.

Charge separation within molecules depend of the geometric arrangements of the nuclei and the electrons around them, i.e. the electron density. It is a physical property which can be observed and also predicted by means of quantum chemical considerations. Molecular Orbital Theory is a very fundamental concept which allows to derive these properties (at least on a qualitative level).

The bent arrangement of ozone is a result of its electronic structure. And the dipole moment is a result coming from that.

Upon the extension of the question and some comments:

Like Thomij already stated, the formal charges are only for bookkeeping. They always have an integer value. A partial charge, however, is a physical observable, resulting from the difference from the electron density and the field of the nuclei.

In all cases the overall formal charge has to add up to the overall physical charge, which can also be calculated from all partial charges. These partial charges can also be calculated from quantum chemical calculations, since you are approximately solving for the optimal electron density at a certain nuclear arrangement.

There is absolutely no possibility to predict a dipole moment without knowing the electron density (or wavefunction) at a given nuclear geometry. A Lewis formula can be generated from these facts, showing the simplest but still well matching bond description. This representation has to sometimes include formal charges. The linked wikipedia page does not employ this concept to explain the polarity, they rather use partial charges.

The partial charges in ozone result from a superposition of the molecular orbitals (click here to enlarge). These are the valence orbitals obtained from BP86/cc-VPDZ:
valence orbitals of ozone

Especially the HOMO (very right) and the HOMO-3 (4th from right) and HOMO -6 (4th from left) contribute massively to a slight charge separation in ozone.

  • $\begingroup$ I disagree; overall formal charge represents a real physical quantity - the overall charge on the molecule. $\endgroup$
    – Dissenter
    Jun 15, 2014 at 16:07
  • 2
    $\begingroup$ @dissenter The formal charges must add up to the overall charge, but the overall charge is not the same thing as a formal charge. Formal charge is a bookkeeping method for assigning charges to individual atoms simply as a way of visualizing electron density distribution. $\endgroup$
    – thomij
    Jun 15, 2014 at 16:51
  • $\begingroup$ What's the difference between overall charge and overall formal charge? Numerically, they're equivalent right? I think what Martin and you are getting at is that formal charge =! partial charge (which is a lot more useful). $\endgroup$
    – Dissenter
    Jun 15, 2014 at 16:51
  • $\begingroup$ Thanks for your answer, but I don't think it answers my question. $\endgroup$
    – user29568
    Jun 15, 2014 at 17:24
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    $\begingroup$ @Kexanone Indeed the partial charge is not an observable, but please do not confuse ill-defined with unambiguous. Many different schemes exist, this is also true, but most of them are well-defined. There is just not one unambiguous way to do it. But I kind of get where you are coming from and this answer could probably should be improved. $\endgroup$ Jan 11, 2022 at 23:34

Yes, ozone has a dipole moment because it is bent and has resonance structures that involve formal charge separation. Resonance can affect the polarity and dipole moment of a molecule, but only if at least one of the resonance structures involves formal charge separation, and even then there are cases where the overall molecule will not have a dipole moment. Some examples will help. Let's start with benzene. Resonance structures can be drawn for benzene, but none of them involve charge separation, therefore benzene is not polar and does not have a dipole moment. So just being able to draw resonance structures does not mean the molecule must be polar and have a dipole moment. Next let's examine cyclohexanone. In this case two resonance structures can be drawn and one of them involves charge separation. Therefor, we would expect this molecule to be polar and have a dipole moment. For 1,4-cyclohexanedione we can also draw resonance structures with formal charge separation, however in this case the symmetry of the molecule causes the local polarization at each carbonyl to cancel and overall, the molecule has no dipole moment.

enter image description here

  • $\begingroup$ By formal charge separation ,do you mean difference? Then, does that mean that the net dipole vector is bigger if the difference is bigger? $\endgroup$
    – user29568
    Jun 15, 2014 at 17:30
  • $\begingroup$ In some of the resonance structures I've drawn there is a "plus" and a "minus" charge - that is called "formal charge separation". The further apart the charges and the more that resonance structure contributes to the overall "true" picture of the molecule, the bigger the net dipole. $\endgroup$
    – ron
    Jun 15, 2014 at 17:35
  • $\begingroup$ So formal charge and formal charge seperation are two different things? $\endgroup$
    – user29568
    Jun 15, 2014 at 17:44
  • $\begingroup$ Not really, I could say "the resonance structure shows the formal charges" or I could say "the resonance structure shows the formal charge separation." $\endgroup$
    – ron
    Jun 15, 2014 at 17:56

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