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Is there any difference between electronegativities of isotopes of the same element. For example consider the above compound. Will all the resonance structures be equivalent?

Also if I place the above compound in water (which will act as a Bronsted acid), which carbon will get the H atom?

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    $\begingroup$ As for "equivalence", they certainly aren't, because one resonance structure has a negative charge on 14C and four others have a negative charge on 12C. I assume what you meant was whether they contribute equally? $\endgroup$ – orthocresol May 13 at 11:13
  • $\begingroup$ An equally intriguing question could be how the resonance hybrid might be represented in this case. $\endgroup$ – William R. Ebenezer May 13 at 11:23
  • $\begingroup$ @orthocresol Can you please explain why a negative charge on a $\ce{_{14} C}$ would be different then a negative charge on a $\ce{_{12} C}$. Is there a difference of electronegativities (hence the question)? $\endgroup$ – gauri agrawal May 13 at 13:38
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    $\begingroup$ It's really just a wording nitpick. 14C isn't the same thing as 12C; thus a negative charge on 14C isn't the same thing as a negative charge on 12C. Asking whether the resonance structure are "equivalent" has the answer "no", because they aren't the same thing. Chemically speaking, it may of course be that the difference is so small so as to be completely negligible. I am guessing that is the point of your question. But we don't describe that as "equivalence". We'd describe it by saying that they contribute (almost) equally. $\endgroup$ – orthocresol May 13 at 14:11
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    $\begingroup$ @orthocresol , the background of this question was what happens when we place this compound in water. Where would the hydrogen get placed in the acid-base reaction? This would be based on which resonating structure would be more stable, in turn giving different carbons a different negative charge. My guess was that maybe it was due to electronegativities, but I may be wrong. $\endgroup$ – gauri agrawal May 13 at 14:41
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Yes it is there due to the increased size of nucleus. But, the difference in the electronegativity of the two isotopes would be so small so you can easily neglect it. The resonating structure will be different in C14 and C12. The reason is C14 is radioactive (very less) so it may cause difference between the chemical properties of the two resonating structure.

Sorry for the English.

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  • $\begingroup$ so where will hydrogen be added if we react this with water $\endgroup$ – gauri agrawal May 14 at 4:08
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A tentative reasoning could be made on the basis of charge density.

More number of neutrons means larger size of nucleus, but with the same number of protons. Hence, +ve charge density will get reduced, and hence the electronegativity.

But these comparisons can be made on numbers only, as discussed in the comments.

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  • $\begingroup$ Intuitively it's not clear how extra neutral will impact electronegativity. Will it make protons more sparse in the nucleus? Or vice versa? I assume denser protons would decrease electronegativity while sparser protons would increase it? $\endgroup$ – Stanislav Bashkyrtsev Jun 14 at 10:09
  • $\begingroup$ @StanislavBashkyrtsev, I guess I thought it the other way. But, here's what I think: more number of neutrons will decrease charge density and hence the $\ce{Z_{eff}}$, which in turn makes it easy for the atom to lose electrons, and therefore make it, relative to its isotope, a good e-donor. Hence, reduces its electronegativity. Am I missing something? $\endgroup$ – Rahul Verma Jun 14 at 15:07

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