Why does $\ce{Ca}$ have a negative electron affinity, i.e. energy is released when it gets an electron (ca. $\pu{-2 kJ/mol}$), when $\ce{Be}$ and $\ce{Mg}$ have positive values?

I know that the electron will be settled in the $\ce{3d}$ orbital which has a lower principal quantum number than $\ce{4s}$, but still, why would energy be released?


Two sign conventions: (1) the more common one states that a positive electron affinity value represents energy release when an electron is added to an atom; (2) the other states that a negative electron affinity represents a release of energy.


Recommended values for these electron affinities, in the units commonly employed in introductory texts and with the sign convention used here, are 2.37, 5.03, and 13.95 kJ/mol for Ca, Sr, and Ba, respectively. The endothermic electron affinities often quoted for Be and Mg are also too large and should be reported simply as ">0". An argument for a return to the original sign convention for the electron affinity is presented in this paper.


OP: I know that the electron will be settled in the 3d orbital

Actually, as explained in the above article, the electron occupies 4p.

OP: but still, why would energy be released?

Very little energy is released and the lifetime of Ca- is on the microsecond scale in practical experiments. The reference Contributions to the electron affinity of calcium and scandium may be the best source for theoretical models that do and don't predict a slight energy release. According to Atomic negative ions: structure, dynamics and collisions, core electrons need to be considered in addition to the valence electrons in calculating the stability of Ca-. Early calculations that considered only valence electrons predicted that Ca- was unstable. Configuration-interaction study of differential correlation energies in Ca-, Ca, and Ca+ presents detailed calculations of energy levels.

  • $\begingroup$ Can't access it. $\endgroup$ – EJC Nov 10 '14 at 18:57
  • $\begingroup$ can you access sciencedirect.com/science/article/pii/S0370157304000316 ? $\endgroup$ – DavePhD Nov 10 '14 at 19:07
  • $\begingroup$ I can, but I couldn't find a satisfying explanation (probably there isn't one). $\endgroup$ – EJC Nov 10 '14 at 19:27
  • $\begingroup$ try this one: journals.aps.org/pra/abstract/10.1103/PhysRevA.62.052518 $\endgroup$ – DavePhD Nov 10 '14 at 19:32
  • $\begingroup$ I read that they relativistic effects i.e. electrons have bigger speed, lower mass, the orbital contracts, the electrons spend more time near the nucleus and have lower energies. Did I figure it out correctly, or is the explanation different? $\endgroup$ – EJC Nov 10 '14 at 19:44

Please verify your data.

$\ce{Be}$ has negative electron affinity (precisely -2.4 eV).

Also calcium has positive electron affinity (precisely 0.0245 eV).

Check the Wikipedia page Electron Affinity Values

I assume you got confused between electron affinity and gain enthalpy. Electron gain enthalpy is negative of electron affinity.

EDIT 1: DAVE PhD, you misunderstood me. Electron affinity has absolute definite values. Only one convention is used. Electron affinity can assume both positive and negative values. But beryllium has negative electron affinity, by any convention you can't assume it to be positive. You can even check the Wikipedia page.

  • 1
    $\begingroup$ There are two conventions. I used the one where negative electron affinities represent stabilization. In your case, the positive electron affinities represent stabilization of the system. $\endgroup$ – EJC Nov 7 '14 at 15:00
  • $\begingroup$ Nit picking, but those values can't be "precise" and exact because they're not definitions. $\endgroup$ – Zhe Feb 5 '17 at 15:26

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