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According to this ChemistryLibretexts website and this StackExchange answer, 1st electron affinites are negative and 2nd electron affinities are positive.

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However, according to this Study.com website, 1st electron affinity (energy released) is positive.

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Both contradicting websites address the point that when electrons are added to a neutral atom, energy is released. But the only problem is that one website says that when this exothermic reaction occurs, the 1st electron affinity is negative, while the other states that the 1st electron affinity is positive. Which one is right?

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  • $\begingroup$ en.wikipedia.org/wiki/Electron_affinity tells it all $\endgroup$ – Mithoron Dec 18 '16 at 22:31
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    $\begingroup$ Well, I edited your question to look better. Don't use super sized font for writing paragraph. It is only used in headings. Looks like you were shouting... $\endgroup$ – Nilay Ghosh Dec 19 '16 at 2:43
  • $\begingroup$ I also don't get what you're saying about that question on SE. The question clearly states that the first EA is positive and the second one negative. $\endgroup$ – orthocresol Dec 19 '16 at 3:44
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Well, the first electron that is taken up any atom other than a noble gas atoms or light atoms with half filled or full filled orbitals, is taken up with a release of energy, so ∆E should be negative. For the other atoms, they are already in the preferred electronic configuration, so they reluctantly take up electrons, tHus having a positive ∆E. In case of the second electron acceptance, there is repulsion of the electron by the negative charge gained by the atom, so it depends upon the situation. If the second electron acceptance leads to a stabler electronic configuration, like a noble gas, then the second ∆E may also be negative.

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  • $\begingroup$ >Well, the first electron that is taken up any atom other than a noble gas atom, is taken up with a release of energy || revisit electron affinity of II group elements. $\endgroup$ – permeakra Dec 19 '16 at 5:52
  • $\begingroup$ @permeakra so mostly atoms with half filled and full filled orbitals are reluctant to accept electrons, aren't they? $\endgroup$ – Mrityunjay Gupta Dec 19 '16 at 6:00
  • $\begingroup$ yep. Large atomic size do help to lift the limitation, but still. $\endgroup$ – permeakra Dec 19 '16 at 6:17
  • $\begingroup$ Yes, exactly the answer I was looking for! Thank you $\endgroup$ – Aniket Dec 19 '16 at 19:04
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This is sadly, one of those matters of "convention" that chemistry is so plagued by.

For most neutral atoms, the magnitude of $\Delta H$ for the following reaction is negative (i.e. energy is released upon the addition of an electron).

$$\ce{A (g) + e- (g) -> A- (g)} \tag{1}$$

The value of $\Delta U$ for the reaction is, in some places, called the "electron affinity". In other places, $\Delta H$ for the reaction is called the "electron-gain enthalpy".[1]

The convention that is more common (as far as I know) is to define the electron affinity as the negative of $\Delta U$ for the above reaction. This means that electron affinities, as defined in this way, are usually positive. This convention is adopted by most major introductory inorganic chemistry textbooks and is also what I learnt in undergrad. However, if you are supposed to use a different convention for your examinations etc., then do so.

The moral of the story is: in every book you read, always double check what is the sign convention. It can and will differ from book to book.


[1] To make things even more confusing, $\Delta H$ (the change in enthalpy) and $\Delta U$ (the change in internal energy) is used in different contexts. The difference between the two is given by

$$\Delta H(T) = \Delta U(T) + \frac{5}{2}RT$$

which may be understood using Kirchhoff's law since $\Delta H(0) = \Delta U$. (Alternatively there is a derivation in Housecroft & Sharpe's Inorganic Chemistry 4th ed., p 25.) Since $5RT/2$ is only roughly $6~\mathrm{kJ~mol^{-1}}$ at $298~\mathrm{K}$, this contribution is usually negligible compared to the usual magnitudes of electron affinities.

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  • $\begingroup$ Is that difference between dH and dU is only accurate for ideal gases? $\endgroup$ – electronpusher Dec 20 '16 at 0:57
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    $\begingroup$ @electronpusher Yes it is for ideal gases. I should've mentioned. $\endgroup$ – orthocresol Dec 20 '16 at 3:07
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  1. When an electron is being added to a fully filled or half-filled shell, that electron would encounter repulsion due to the stability that already existed in the atom and therefore one would have to input more energy making the EA value positive.

  2. However, when the ingoing electron doesn't encounter any repulsion, then energy would rather be given out making the EA value negative. This refers to any other case apart from the one in (1).

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