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I often see that the most stable monoatomic ion of nitrogen is N$^{3-}$ (for example on Khan Academy), and I remember being taught something similar, along the lines of atoms wanting complete octets.

However, the $1^\text{st}$, $2^\text{nd}$, and $3^\text{rd}$ electron affinities of nitrogen are, according to [1]:

Nitrogen (kJ/mol)

$-7$, $+673$, $+1070$

The positive (endothermic) $2^\text{nd}$ electron affinity implies that the anions with negative charge greater than one are inherently unstable, in the sense that if I bring an electron near an N$^-$ in the vacuum of outer space, then they would not want to bind. And if I had an N$^{2-}$ atom, out in space, then it would spontaneously decompose to N$^-$ and an electron.

My interpretation of "most stable ion" is that it should not involve any stabilising complexation with a (variable) cationic component.

Is N$^{3-}$ the most stable monoatomic ion? Is there some additional completing octet-related favourability that compensates? I would have thought that if there were, that that would turn up in the energy measurement. If I am tutoring people in chemistry, is there a better way to describe the drive of nitrogen to complete its octet, or should the way we teach the non-metals be revamped by actually looking at the electron affinities for each? (Oxygen for example, has negative both first and second electron affinities.)

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    $\begingroup$ By your definition and the data you provide N- is the most stable monoatomic ion of nitrogen in that it is the only ion which will not spontaneously decompose to the uncharged atom and an electron. However that is all but irrelevant for chemistry - it refers to a completely isolated gas phase nitrogen atom. The higher charged ions are stablised in the solid and liquid phase by the favourable electrostatic potential due to the rest of the lattice, and a the greater gain in energy due to the higher charge offsets that lost due to the electron affinity. I guess you would call this "complexation". $\endgroup$
    – Ian Bush
    Oct 9, 2022 at 7:56
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    $\begingroup$ There are no known stable 2- free ions. $\endgroup$
    – Jon Custer
    Oct 9, 2022 at 15:09
  • $\begingroup$ Nitride in the liquid phase? $\endgroup$
    – jimchmst
    Oct 9, 2022 at 19:26
  • $\begingroup$ Wikipedia gives a melting point in vacuum for aluminium nitride. On the other hand not totally convinced of the ionic model in that case, but it shouldn't be a total disaster $\endgroup$
    – Ian Bush
    Oct 9, 2022 at 20:39
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    $\begingroup$ @IanBush There is big difference between free and lattice-bound ions, with the latter stabilized by the released lattice energy. Even isolated Na+(g) + Cl-(g) are energetically unfavoured, unless at least bound in a ionic pair. $\endgroup$
    – Poutnik
    Oct 10, 2022 at 9:17

1 Answer 1

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The correct answer is:

Both nitrogen(-III) and nitrogen(-I) are unstable (with respect to electron ejection) per se, and exist in ionic crystals only due to the lattice energies outweighing the energy for spontaneous electron ejection.

Then,

The former becomes more stable due to the valence orbitals being fully filled, while the latter ends up taking away electrons from the cation(which is, vide supra, endothermic per se, but is compensated for by the growth of lattice energies).

Note: An even bigger difference occurs between oxygen(-I), which is stable per se, and oxygen(-II), which is unstable per se. The latter can be stabilised by lattic energies in crystals, while the former ends up taking electrons away from the cation by the same mechanism as above.

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  • $\begingroup$ So what is meant when we are taught (I was taught this in multiple schools/universities) that atoms “want” to get their nearest noble gas configuration (ex. O wants to form O$^{2-}$)? Is this just plain wrong? $\endgroup$
    – Furrier Transform
    Oct 10, 2022 at 20:07
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    $\begingroup$ @FurrierTransform that holds in MOLECULES or SOLIDS, not IN VACUO. $\endgroup$
    – Kanghun Kim
    Oct 11, 2022 at 1:31
  • $\begingroup$ But surely that depends on the molecule or solid and is a properly of the differences in electronegativity of the pair of atoms (or more for a conjugated system), and not a property of an individual element. So what I am hearing is that what we were taught is just plain wrong. $\endgroup$
    – Furrier Transform
    Oct 13, 2022 at 4:29
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    $\begingroup$ "Both nitrogen(-III) and nitrogen(-I) are unstable (with respect to electron ejection) per se" Doesn't the 1st electron affinity of nitrogen indicate that nitrogen(-I) is stable? $\endgroup$
    – Buck Thorn ♦
    Oct 16, 2022 at 10:57
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    $\begingroup$ The values in the OQ have different signs for the three consecutive EA's of nitrogen, which is not possible as all three are actually endothermic. $\endgroup$
    – Kanghun Kim
    Oct 17, 2022 at 2:18

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