It is perfectly possible for an atom - particularly on the electronegative end of the periodic table to form negatively-charged ions by attracting an electron, and these species can be stable, requiring (a small but positive amount of) energy to detach the extra electron.

Is the same thing possible with doubly-charged anions? I know hydrogen can form $\mathrm H^{2-}$, but that one has a finite lifetime. Are there elements that can acquire a second extra electron and keep it until they're perturbed?

  • $\begingroup$ Like a sulfide ion? $\endgroup$ – M. Enns Apr 15 '16 at 13:40
  • $\begingroup$ @M.Enns If it is stable by itself in the gas phase, then yes. $\endgroup$ – Emilio Pisanty Apr 15 '16 at 13:58
  • $\begingroup$ This is not something I've thought about before but here's a clue. People sell tests for detecting sulfide ions in air. Kitagawa and Buck so maybe they're out there. $\endgroup$ – M. Enns Apr 15 '16 at 14:11

Assuming I understand you correctly the quantity you refer to is the second electron affinity i.e. the energy absorbed in the gas phase reaction:

$$ X^- + e \rightarrow X^{2-} $$

(it's the energy absorbed because a negative second electron affinity means energy is released)

If so, there are no elements for which the second electron affinity is negative. The nearest is sulphur with a second electron affinity of $+532$ kJ/mol followed by oxygen at $+844$ kJ/mol. So there no elemental $X^{-2}$ anions that are stable in the gas phase.

  • $\begingroup$ Yeah, I'm looking for elements for which $X^-+e\to X^{2-}$ releases energy. $\endgroup$ – Emilio Pisanty Apr 15 '16 at 14:44
  • $\begingroup$ @EmilioPisanty: There are none. $\endgroup$ – John Rennie Apr 15 '16 at 14:46
  • $\begingroup$ OK, that's good to know then. I imagine that things like $S^{2-}$ are only stable in liquid phase because of solvent effects, then? $\endgroup$ – Emilio Pisanty Apr 15 '16 at 14:48
  • $\begingroup$ Yes, and similarly for solids. $\endgroup$ – John Rennie Apr 15 '16 at 14:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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