I may be a rudimentary fact or something really "not-that-hard-to-figure-out", but

In inorganic chemistry it is common to consider a single value of the electronegativity to be valid for most "normal" situations. While this approach has the advantage of simplicity, it is clear that the electronegativity of an element is not an invariable atomic property and, in particular, increases with the oxidation state of the element. Wikipedia

Really? That means for example $\ce{C}$ in $\ce{CO3^{2-}}$ is more electronegative than a "bare" C atom? Why is that so?

Interestingly, I'm not able to find relevant info in the Net (and the wikipedia citation that should have hopefully got me results is behind a big bad paywall), so am leaving "reference-request" for the case of additional study.


1 Answer 1


Yes absolutely, electronegativities are hardly static values when you start combining elements and forming ions or molecules.

Electronegativity is simply a measure of an ability of an atom to attract electrons in a covalent framework.

So keeping in mind that like charges repel (electrons repel electrons) ... it follows that elements in a high oxidation state are more electronegative - i.e. more able to attract and stabilize electrons. High oxidation state elements have essentially been "robbed" of their electrons and likely bear some degree of partial positive charge ... this is "good" for attracting and stabilizing nearby electrons.

Conversely consider a lone pair - which is defined to have an electronegativity of 0. Lone pairs cannot stabilize other electrons and don't attract other electrons to themselves ... the electron-electron repulsion would be too great.

  • $\begingroup$ Thanks an amount of Avogadro's number, and would you mind adding something (paper, page or whatever) relevant as for a more thorough studying? $\endgroup$
    – M.A.R.
    Feb 5, 2015 at 22:01
  • 1
    $\begingroup$ books.google.com/… $\endgroup$
    – Dissenter
    Feb 5, 2015 at 22:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.