(This is a question similar to this and this, but none of the answers to those two questions answer my question.)

I've seen many places explain the dipole moment of ozone from the formal charges of the O atoms. However, I don't quite feel comfortable with this explanation since I feel like formal charges are not actual charges; they're more like a bookkeeping convention.

From this question, it seems like we should take into account the lone pair on the central O when summing the dipole moments. Thus, shouldn't the central O be the negative end of the dipole (since it has the lone pair)? But why does this contradict the prediction made by formal charges?

In general, to find the overall polarity of a molecule, when is it appropriate to sum up all the dipole moments (including those from the lone pairs), and when should we use formal charge instead?

  • $\begingroup$ The formal charges are a part of the resonance depiction of ozone, which is a representation of the electronic structure so they are a completely acceptable way of rationalizing the dipole moment. $\endgroup$ – bon Apr 17 '16 at 21:24
  • $\begingroup$ Ozone is not a symmetric molecule: though bent, it's not bent enough to be equilateral, there is a "middle one". The central O is in rather a crappy position for its self-respect as an electronegative atom: it is having electrons pulled from it in both directions. It has a bit of a +ve charge, at least when compared to its neighbours at either end. You can think of it in terms of resonance (easy, but not really physical), or by what they tell you (the orbitals are depleted near the middle). Either way: not many electrons there. As that middle O also sticks out, there will be a dipole moment. $\endgroup$ – Dan Sheppard Apr 17 '16 at 22:05
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    $\begingroup$ Other oxygens have their lone pairs as well. $\endgroup$ – Ivan Neretin Apr 17 '16 at 22:21

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