1. I understand that we need to supply energy to counter the nuclear attraction when we remove electrons, and that is the reason why ionization energy is endothermic. However, why does an atom release energy when we add an electron to it?

  2. With respect to ionization energy, why is energy required to remove electrons of elements like $\ce{Be}$, $\ce{Li}$ that naturally want to lose their electrons?

  3. My teacher told me that elements with fully filled and half filled highest occupied orbitals have positive electron affinity (endothermic). Why is this true?


3 Answers 3

  1. The energy of an electron in a vacuum is zero. If it can attach (however weakly), the energy is gained and the process is therefore exothermic. It is not an atom that releases energy, it is the whole system.

  2. Exactly opposite of previous. The atoms you mentioned gladly release electrons to other atoms, but not to vacuum (which is how ionization energy is defined).

  3. In some cases, adding of extra electron to neutral atom is not an exothermic process. See Periodic table of electron affinities. For example, a helium atom has filled shell with $n=1$. The next electron would be accepted to the shell $n=2$ which has much higher energy, and therefore is not stable. For the half-filled orbitals, similar reasoning applies, but spicing it with Hund's rule.

  • $\begingroup$ I don't get it. 1) If an electron in vacuum attaches itself to the atom, energy is gained by the electron? why? and doesn't "gain" mean endothermic? are you saying the energy lost by the atom goes into the electron that attaches itself?3) don't you mean "exothermic" instead of "endothermic"? just to clarify endothermic- gain and exothermic -loss right? everything else you said makes sense. $\endgroup$
    – SMcCK
    Commented Jun 1, 2015 at 18:40
  • $\begingroup$ Few words of thermodynamics... you describe the reaction $\ce{A + e^- -> A^-}$. If the system gains energy, the reaction is exotermic. Lowering the energy corresponds to more stable system. Not atom, not electron, the whole system. I hope I do not use the energy stuff too vaguely to confuse you. $\endgroup$
    – ssavec
    Commented Jun 1, 2015 at 21:29
  • $\begingroup$ i am rather confused. Please do forgive me but i am new to this. $\endgroup$
    – SMcCK
    Commented Jun 2, 2015 at 4:00
  • 1
    $\begingroup$ So the reaction is exothermic and so energy is gained by the whole system? But why does the reaction have to be exothermic? How is it exothermic? What happens for it to be exothermic? And does the system become stable again by releasing the energy it got from the reaction? $\endgroup$
    – SMcCK
    Commented Jun 2, 2015 at 4:11
  • 1
    $\begingroup$ Let's call classical mechanics to help. You drop 1kg weight from 1m to the ground. The system is the Earth and the 1kg weight. System goes to the state with lowest energy (guess which), process is spontaneous, energy is released, as long as we think only about gravitational interaction. The system gained these 10J which would finally change to heat. So the process is exothermic. $\endgroup$
    – ssavec
    Commented Jun 2, 2015 at 5:23

Looking at the comment thread above, I believe a bit more elaboration can help, so here it is

Why is electron gain generally exothermic?

Think of a unit positive sphere and a unit negative sphere. When is their system most stable? It is when the spheres are closest to each other. Why is that position stable? Because if we slightly disturb those spheres to take them farther from each other, they'll again come back to their original position, due to their attractive forces.

This stable position is an energy minima for this sphere system. When a system goes from a higher energy state to a lower energy state, it loses energy.

The same is the case with the system of the negative electron and the positive nucleus pair. The energy of an electron in vacuum (isolation from the nucleus) is zero (because it cannot feel the nucleus' attractive pull). But when it occupies a definite energy level in the atom, the energy of the system reduces. The extra energy is then released, which is why you find that the electron gain is generally exothermic.

Why is energy required to remove electrons of elements like $\ce{Be, Li}$ that naturally want to lose their electrons?

I assume you mean to say that $\ce{Na}$ easily loses electrons in presence of, say $\ce{Br}$ (assume $\ce{NaBr}$ ionic), even though $\ce{Na}$'s ionization energy is positive. That is because the formation of $\ce{NaBr}$, or any ionic compound in general, involves a number of steps, each involving a definite energy change. Ionization of the sodium atom is just one of those several steps!

This is called the Born-Haber cycle. The general idea is that you start with solid sodium and bromine liquid, and eventually form solid sodium bromide. The net enthalpy change for this entire process is negative, or exothermic, hence it is favorable. The positive ionization enthalpy is compensated by the other processes which release energy.

However, in vacuum, the only process that occurs with the isolated (other ions are absent) gaseous sodium atom is to remove an electron. That process - which defines ionization energy - is still endothermic, because you need to oppose the nucleus' attractive pull.

Elements with fully filled and half filled highest occupied orbitals have positive electron affinity (endothermic). Why is this true?

That is true because elements with fully filled and half filled highest occupied orbitals are stable (you can read why that's true here). Adding an electron to this atom would make it unstable, thereby increasing the atom's energy. Now, don't get me wrong, when you add an electron to the atom: the energy of the nucleus-electron pair still decreases, but the rise in energy due to the instability is much more, causing a net increase in the energy of the atom.

Hence, in such cases, the electron gain process is endothermic.


When two oppositely charged particles are brought closer together under a Coulombic potential, the potential energy of the pair decreases. However, conservation of energy requires that this potential energy be converted into some other form (total energy is conserved). In a simple classical description we say that it is converted into kinetic energy, ie $-\Delta E_{Coulomb} = \Delta E_{kin}$, but the total energy remains constant, unless some of it is transferred to some other body. In the quantum mechanical description of atoms, electrons occupy discrete states, each state characterized by a total electronic energy (with potential and kinetic contributions) and a distance distribution between electron and nucleus (which can be used to compute an average distance). Transitions can occur between states through the exchange of discrete amounts of energy (quanta) with other bodies.

An ion formed by a neutral atom and an electron bound at a large separation is in a high energy state relative to the ground state, the relative energy being nearly the ionization energy of the ion. This is a highly unstable state because the slightest perturbation can drive electron and nucleus apart into an unbound (free) state. For the ion to settle into a lower energy (more stable) bound electronic state, corresponding to a smaller average distance between electron and nucleus, it has to release energy. It might do this radiatively (emitting photons) or through collisions with other atoms, dissipating the energy as heat. Another way of seeing this is that the newly formed ion has an excess of energy that it can give away to colder atoms in order to relax into a more stable lower energy state. QM dictates what are the most stable arrangements of electrons about nuclei, and only particular arrangements are possible. It is largely a geometric problem involving a balance between electron-electron repulsions, electron-nuclear attraction, the wave nature of matter evident particularly at small scales, and odd effects such as Pauli exclusion (the impossibility of two electrons having identical properties).

Considering now your list in more detail:

  1. It doesn't always do so. Sometimes it's impossible. Not all combinations are stable. Only when the electron affinity is consistent with electron attraction is a stable combination possible. This is largely addressed above. Assuming it is possible to form a stable atom or ion, the combined particles have to shed energy to the surroundings, otherwise they risk falling apart again. When it gives off energy the combination relaxes from an excited state to a lower energy state. It might only shed some energy and remain in a highly excited (reactive) state, however.

  2. This is addressed in another answer to your post. The argument is somewhat circular. Higher energy is less stable. Basically you are creating an unstable (excited) system when you attempt to remove an electron from a stable atom, so you have to add energy to do so.

  3. This is addressed by the hand-wavy QM argument at the end of my dissertation. The combination of charge attraction and repulsion with the wave nature of subatomic particles and the Pauli exclusion principle lead to particular particle combinations and geometric arrangements being stable. This is the subject of a more rigorous QM course.


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