Bond polarity, as far as I understand, is a measure of the degree to which shared electron density is distorted, and thus solely depends on the electronegativity difference.

Up until now, I had learnt that the dipole moment (charge on each atom x separation) is used as a measure of the polarity. However, for molecules with a formal charge like CO, even though the shared electron density is distorted towards the more electronegative atom, the dipole moment points in the opposite direction. For other molecules like Ozone, even though there is approximately no distortion of bonding electron density, the formal charges alone result in a dipole moment.

For a general molecule, am I supposed to take into account the formal charges for bond and thus molecule polarity, and it it even a measure of degree of distortion/ionic character at that point?

  • $\begingroup$ it all depends on whether you are talking about the polarity of a bond or the molecule $\endgroup$ – Nicolas Feb 20 at 7:02
  • $\begingroup$ At the end of the day, my major use of bond polarity is to find the overall molecular polarity. Could you elaborate on why there’s a difference? $\endgroup$ – OVERWOOTCH Feb 20 at 7:06
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    $\begingroup$ polarity is a vector quantity: it is not a simple addition but a vector addition. For example, for the molecule of $ \ce {CO_2} $, each bond $ \ce {C-O} $ is polarized but as the molecule is linear, the moments of the 2 bonds are compensated: the bonds are polarized but the molecule does not 'is not $\endgroup$ – Nicolas Feb 20 at 7:08
  • $\begingroup$ I am aware that polyatomic molecule polarity depends on bond polarity and molecular geometry. my question is regarding formal charge and has nothing to do with thus $\endgroup$ – OVERWOOTCH Feb 20 at 7:10
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    $\begingroup$ If there is formal charge, there is a corresponding vector. Just sum it to all the other el. dip. moments due to elemental electronegativity etc. Of course it does. The extent of the effect depends on the weight of the limiting form with charge separation, in a resonance frane. $\endgroup$ – Alchimista Feb 20 at 10:30

When you compute formal charges you split bonding electrons evenly between bonded atoms. This does not account for differences in electronegativity. It is similar to the difference between formal charge and oxidation state, as well explained in the Wikipedia.

A better picture (closer to the real electronic distribution) is, as Pauling would probably suggest, intermediate to the two, assuming a hybrid between possible Lewis structures (covalent and ionic).

In any case, when considering bond polarity, electronegativity overrules formal charge. On the other hand inductive effects should not be dismissed.

The OP presented an excellent counterexample: CO. Here the carbon has a very slight negative charge relative to oxygen, as predicted based on the computation of formal charges.

I am not an expert on the development and application of electronegativity scales, but suffice it to say that these scales are (1) empirical and (2) provide useful guidelines. The original Pauling formulation of electronegativity cannot account for all possible bonding schemes. Case in point is the modification of the electronegativity scale to account for orbital hybridization, the following data via the Wikipedia:

Hybridization χ (Pauling)1
C(sp3) 2.3
C(sp2) 2.6
C(sp) 3.1
'generic' C 2.5

Based on this modified scale the electronegativity of oxygen, 3.44, is no longer much larger than that of the carbon atom in carbon monoxide.


  1. Data found in Wikipedia, original reference: Fleming, Ian (2009). Molecular orbitals and organic chemical reactions (Student ed.). Chichester, West Sussex, U.K.: Wiley.
  • $\begingroup$ What I’m trying to ask if I should take into account the formal charges in addition to the electronegativity when finding the bond polarity. Is the triple bond in CO polarised towards carbon or oxygen? The carbon appears to have a partial negative charge despite the lower electronegativity $\endgroup$ – OVERWOOTCH Feb 20 at 9:23
  • $\begingroup$ See eg en.wikipedia.org/wiki/Inductive_effect $\endgroup$ – Buck Thorn Feb 20 at 9:28
  • $\begingroup$ @OVERWOOTCH You're right, CO is a strange beast. $\endgroup$ – Buck Thorn Feb 20 at 9:34
  • $\begingroup$ I do not see why CO should be called counterexample. It is consistent with your answer, at least it seems so to me. OP you appear to dissect bonds. You can of course think about how the bond are polarized in a structure with formal charges. But the resulting polarity will be determined also by the separated charges. $\endgroup$ – Alchimista Feb 20 at 10:46
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    $\begingroup$ @BuckThorn right. In my comment to OP under the question I wanted to mentioned right C as faced from different directions. I skipped because I was afraid to makes thing unclear, in a comment. The main point is that whatever analized the el dip moment would be the sum or the various components we - correctly or not - search for. $\endgroup$ – Alchimista Feb 20 at 12:39

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