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If two similar sized atoms have different electronegativity values, the atom with higher electronegativity holds the electrons more tightly, and thus is more stable. Example: hydroxide vs azanide (NH2-).

If two atoms have different sizes (down the periodic table), the larger atom has electrons more spread out and thus is more stable. Example: fluoride vs iodide anion

To me, these two statements are contradictory. One says tight electron cloud means not reacting while the other says spread electron cloud means stability (less repulsion). Is one of the statements wrong? I'm sure I am missing something here. Thank you in advance for your help.

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    $\begingroup$ The word "stable" has several quite different (and indeed sometimes contradictory) meanings. How is fluoride anion less stable than iodide? The latter is much easier to oxidize. But it is "more stable" in a different sense: it is harder to protonate ($\ce{HI}$ is a stronger acid than $\ce{HF}$). $\endgroup$ – Ivan Neretin May 20 '16 at 9:14
  • $\begingroup$ When I mean stability, I mean how well are the electrons kept to the atoms themselves and not shared with their neighbors? I was thinking about repulsion between negative charge of electrons. For specific case, I was thinking about nucleophilicity in protic solvents. With electrons less likely to stay with the atom, the electrons are more likely to attack (electronegativity is small). With larger atom, electrons are more spread out, and there's less repulsion, but larger volume for interaction with other molecules. Somehow larger atoms react less with hydrogen of solvent. $\endgroup$ – Kenny Kim May 20 '16 at 9:38
  • $\begingroup$ I may be confusing between nucleophilicity and basicity as well, because it seems hydrogen bonding is more likely to occur with smaller atoms such as fluoride ions. My thought is that if fluoride ions attract more hydrogens, they would also attract more electrophiles. Of course, somehow basicity and nucleophilicity has inverse correlation in protic solvent for some unknown reason. $\endgroup$ – Kenny Kim May 20 '16 at 9:41
  • $\begingroup$ @IvanNeretin Stability you talk about is achieved in aqueous solution. In anionic gaseous form the stability is different and I- is more stable than F-. $\endgroup$ – shre_sudh_97 May 20 '16 at 11:15
  • $\begingroup$ @IvanNeretin I meant stable as in electron's tendency to stay there. It seemed that I was simply confused on the concepts of electronegativity and electron affinity. $\endgroup$ – Kenny Kim May 20 '16 at 23:57
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I believe that you are mixing up the concepts of electronegativity and of electron affinity. The first statement you present is related to Electronegativity.

If two similar sized atoms have different electronegativity values, the atom with higher electronegativity holds the electrons more tightly, and thus is more stable.

Yes. This is generally true. Electronegativity establishes how well an element can pull electron density towards itself.

The second statement is about electron affinity.

If two atoms have different sizes (down the periodic table), the larger atom has electrons more spread out and thus is more stable. Example: fluoride vs iodide anion

Generally true as well. Flouride anion is less stable than iodide anion due to the repulsion of electrons in the outermost octet of fluoride. But this rule is not absolute as chloride has the highest stability (in the gaseous state) though it is above bromine and iodine.

Edit: Considering some confusion in the comments I shall put this here.

The electronegativity is generally not compared based on inter-electron repulsion because the shared pair doesn't completely enter the orbits of the electronegative atom but is shared and so electron density is between the two atoms and not concentrated on one atom for repulsion to take much effect.

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  • $\begingroup$ The point of my question is that increase in electron density and increase in electron distance both decreases electronegativity and electron affinity. I don't think this is an issue of electronegativity vs electron affinity since the two statements are true for both concepts. $\endgroup$ – Kenny Kim May 20 '16 at 20:11
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    $\begingroup$ Okay, after hours of frustration, I think I understand more of what you mean. Electronegativity does not take into account electron repulsion but electron affinity does. This is because electronegativity is about tendency of electrons to move to one side in a bond (thus somewhat distant from the atom) while electron affinity is about an electron completely becoming part of the atom and thus getting the full effect of electron repulsion. $\endgroup$ – Kenny Kim May 20 '16 at 23:54
  • $\begingroup$ @KennyKim These facts are all predicted by quantum mechanics and confirmed by experiment. It's hard to say just by qualitatively comparing ionic size and charge density. $\endgroup$ – shre_sudh_97 May 24 '16 at 15:32
  • $\begingroup$ @KennyKim You seem to have cleared up confusion. Yes, the electronegativity is generally not compared based on inter-electron repulsion because the shared pair doesn't completely enter the orbits of the electronegative atom but is shared and so electron density is between the two atoms and not concentrated on one atom for repulsion to take much effect. $\endgroup$ – shre_sudh_97 May 24 '16 at 15:34
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The electronegativity stabilizes due to the lower coulombic potential between opposite charges. However, as we go down the periodic table, we are adding electron shells to the atom which dramatically increases its size. This means the negative charge is distributed over a wider space, reducing repulsion among all electrons.

You might say its a competition between attraction to the nucleus and repulsion among electrons, and reducing repulsion among all electrons takes precedence over increasing attraction because of how quickly the volume expands. The key point is that repulsion is mostly mitigated by adding shells, which does not take place in your example of NH2 vs OH.

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