My textbook says removing an electron always requires an input of heat. That makes sense because of coulombic force between electron and nucleus. But it also says that the process is endothermic always.

Why is that? In the case of Na, after the input of "activation energy", isn't Na+ more stable than Na? So therefore it would be exothermic?

Furthermore, this is not always true for electron affinity like it is for ionization. Adding an electron to Helium is an endothermic process because He- is less stable than He.* But it's exothermic for Cl and Cl-.

*Not totally sure about that. Contradicting information on the internet. Some sources say electron affinity of He is 0. Others say its negative (thus endothermic).

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    $\begingroup$ You can't compare $\ce{Na}$ versus $\ce{Na+}$. The ion is missing an electron, and since $E=mc^{2}$, it makes no sense to compare things that don't have the same amount of matter. That also should hint at why it takes energy to remove an electron. Since $\ce{Na}$ and $\ce{Na+ + e-}$ are identical, except in the latter case, you put two charges farther apart. $\endgroup$
    – Zhe
    Jul 23, 2020 at 20:34
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    $\begingroup$ Ionization is always an endothermic process as bound electrons have lower energy than free electrons. It is very analogical to astronomical bodies bound by gravity of the central body. Note that ionization can be a part of more complex overall process that can be ( even highly ) exothermic. $\endgroup$
    – Poutnik
    Jul 24, 2020 at 9:59
  • $\begingroup$ See chemistry.stackexchange.com/questions/146481/… $\endgroup$
    – Ian Bush
    Apr 13, 2022 at 9:39

2 Answers 2


My textbook says removing an electron always requires an input of heat. Better to say it needs input of energy, which can come e.g. from thermal motion, ongoing reaction or absorption of a photon.

Ionization (of neutral atoms or positive ions) is always an endothermic process as bound electrons have lower energy than free electrons. It is very analogical to astronomical bodies bound by gravity of the central body. Note that ionization can be a part of more complex overall process that can be ( even highly ) exothermic.

$\ce{Na+(g)}$ + $\ce{e-(g)}$ are not more stable than $\ce{Na(g)}$ by itself. In fact the opposite. $\ce{Na+}$ is more stable than $\ce{Na}$ if and only if there follows a process making all the operation thermodynamically favourable, like production of solid salt or hydrated $\ce{Na+}$ ion.

Imagine high temperature gas that consists of $\ce{Na}$ and $\ce{Cl}$ atoms. Chlorine afinity to electron is lower than sodium ionization energy. So reaction $\ce{Na(g) + Cl(g) -> Na+(g) + Cl-(g)}$ is thermodynamically unfavoured. It becomes favoured and exothermic, if ions pack together to form ionic pairs and later solid ionic lattice. Then coulombic energy released by bounding ions together overcomes unfavoured process of forming independent ions in gaseous phase.

Helium electron affinity is negative, estimated byWikipedia: Electron affinity data page as $\pu{-0.5(2) eV}$ ($\pu{-0.5 \pm 0.2 eV}$). Note that (from the link):

Negative electron affinities can be used in those cases where electron capture requires energy, i.e. when capture can occur only if the impinging electron has a kinetic energy large enough to excite a resonance of the atom-plus-electron system. Conversely electron removal from the anion formed in this way releases energy, which is carried out by the freed electron as kinetic energy. Negative ions formed in these cases are always unstable. They may have lifetimes of the order of microseconds to milliseconds, and invariably autodetach after some time.

There are more elements with negative electron affinity than just noble gases. Most of them are metals/metaloids ($\ce{Be}$, $\mathrm{Mg}$, $\ce{Mn}$, $\ce{Zn}$, $\ce{Cd}$, $\ce{Yb}$, $\ce{Hg}$, $\ce{Pu}$), sharing some electron configuration patterns, but suprisingly also $\ce{N}$ ($\pu{-0.07 eV}$).

  • $\begingroup$ "Electron affinity can be defined in two equivalent ways. First, as the energy that is released by adding an electron to an isolated gaseous atom. The second (reverse) definition is that electron affinity is the energy required to remove an electron from a singly charged gaseous negative ion. The latter can be regarded as the ionization energy of the –1 ion or the zeroeth ionization energy. Either convention can be used." $\endgroup$
    – user119245
    Apr 14, 2022 at 4:33
  • $\begingroup$ goldbook.iupac.org/terms/view/E01977 $\endgroup$
    – user119245
    Apr 14, 2022 at 4:40
  • $\begingroup$ @TomHardy That is known, but the former way is usually preferred. $\endgroup$
    – Poutnik
    Apr 14, 2022 at 5:38
  • $\begingroup$ If the latter definition was used, the electron affinity of N would've still been -0.07eV, right? Or would it change from -0.07eV to +0.07eV? $\endgroup$
    – user119245
    Apr 14, 2022 at 5:44
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    $\begingroup$ @TomHardy The same. It is like energy released by falling an item from the shelf on the floor versus energy required for lifting the item from the floor to the shelf. The same too. $\endgroup$
    – Poutnik
    Apr 14, 2022 at 5:48

All nonradioactive, isolated atoms are stable and do not emit electrons. If one is excited thermally by collision or photolytically by absorption of a photon with sufficient energy, the photoelectric effect, an electron can or will be removed from the atom to give a positive ion and either a negative ion or a free electron. In most chemical reactions the charge separation is not complete, and the electrons involved are attracted to both nuclei in a bonding orbital, a covalent bond. Removal of an electron from a neutral moiety always requires energy, the quantum mechanical bond energy and the simple energy in electrostatically separating a positive and negative charge.

The case of negatively charged ions is more complicated! An atom or molecule with a half empty orbital or an empty orbital of similar energy that is not shielded from the nucleus can add an electron to that orbital with a release of energy. This process is best examined in isolated atoms and can be extended to molecular ions and solids. [Careful here signs are confusing, positive Electron Affinities, EA, usually mean energy is released in forming the negative ion, some sources treat EA as an energy and give energy release a negative sign. Key is H and F release energy in forming H- and F-.] An ion with a release of energy in formation is endothermic in ionization. In the removal of an electron once the electron is out of the valence orbital the electrostatic force is electrons repelling an electron. This repulsion increases the KE of the fleeing electron increasing its energy. If the original binding energy to the nucleus is very low and a negative ion is formed it is possible that its decay could be exothermic. This is explained in the following link[that is also quoted in a previous answer to this question.] Such a negative ion would have a short lifetime. https://en.wikipedia.org/wiki/Electron_affinity_(data_page)

A second EA for an atom such as O- + e- = O= is invariably exothermic even tho the empty orbital is available because it involves bringing two negative charges together. This sets up a conundrum because oxide ions do exist in metallic oxides so there must be a stabilizing mechanism. These are usually said to be charge transfer between the negative ion and the positive ions surrounding. Compounds such as cesium oxide, Cs2O, have very low electron ionization potentials. There might be a situation where the loss of an electron might be exothermic in such a compound.


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