Problem. I've come up with the strange example of the third energy of ionization of both $\pu{Mg}$ and $\pu{Al}$, the standard logic that is to be applied on any problem of "which element has more energy of ionization", should be guided by how that element it's located on the periodic table, that way we can find it's periodic characteristics, like nuclear charge, and all of them.

My issue. The intuitive way to think about it, is that since $\pu{Al}$ is more to the right than $\pu{Mg}$ in the periodic table, it should be enough to prove that with all the periodic characteristics, Al has more energy of ionization than Mg. But what happens is all the way around actually, same thing happens with the first energy of ionization, which made me very confused.

What I tried to think to resolve this whole mess. I then thought that it has to happen due to another factor, and then I looked at the electronic configuration: $$\pu{Mg:[Ne]} \ 3s^2$$ $$\pu{Al:[Ne]} \ 3s^2\ 3p^1$$ and took a look at the ions $\pu{Mg+3 and Al+3}$, since I think that's where I have to look if we are investigating the 3rd energy of ionization, if this is wrong, please do let me know: $$\pu{Mg+3:[He]}\ 2s^2\ 2p^5$$ $$\pu{Al+3:[Ne]} $$ so then I thought that in these last ions, $\pu{Mg+3 is quite more inestable than Al+3, since Mg+3 has lost one full orbital while Al+3 has it full}$. And then that would kind of make sense intuitively? Since if one ion is more stable, it would be harder to take away one electron from it, hence the energy of ionization would be higher from the stable one, right?

Is this logic correct or is it flawed? And, if it's flawed, then how is it that Al has more energy of ionization than Mg? What should my general strategy to find which element has more energy of ionization be? Since just looking at the orientation on the periodic table doesn't seem to work always, like in this one.

  • 1
    $\begingroup$ Logic itself is misleading, if not based on domain specific data, facts, rules and laws. Vast majority of things is not deducible from first principles. $\endgroup$
    – Poutnik
    Oct 12, 2021 at 12:31
  • $\begingroup$ It would be higher because it is already charged. Your question confuses me a bit. But in principle you should be able to close thermodynamical cycles and put everything in the right order. That the latter is what you expect or something else is a different story. $\endgroup$
    – Alchimista
    Oct 12, 2021 at 13:30
  • $\begingroup$ "if this is wrong, please do let me know" - this is indeed wrong: to assess the third ionization energy you need to look at Al2+, not 3+. Apart from that, you are on the right track - see my answer below. $\endgroup$
    – Pallas
    Dec 9, 2021 at 20:03

3 Answers 3


Rounded atomic ionization energy values:

Element 1st IE[eV] 2nd IE 3rd IE 4th IE
Mg 7.646 15.04 80.14 109.3
Al 5.986 18.83 28.45 120.0


If energies are grouped according to orbitals:

Element 3p[eV] 3s 3s 2p 2p
Mg 7.646 15.04 80.14 109.3
Al 5.986 18.83 28.45 120.0

The respective orbital related ionization energies are greater for aluminium because of smaller respective ion sizes and greater effective ( screened ) nucleus charge.

Stability of ions has questionable meanings, as it is usually related to spontaneous changes, If one insists, it could be considered by 2 independent ways:

  • The energy released by capturing an electron, or all missing ones . ( A bigger one means thermodynamically less stable.
  • The energy needed for ionization of yet another electron.( A bigger means more TD stable.

While both values are related, the eventual stability order would depend on the chosen criteria.


When comparing the 3rd ionization energy of $\pu{Mg}$ and $\pu{Al}$, you are right to look at their electron configurations to assess their relative magnitudes. However, the ions you need to be comparing are $\pu{Mg^{2+}}$ and $\pu{Al^{2+}}$, not the three-plus ones (the first ionization energy relates to $\pu{Al}$, the second to $\pu{Al^{+}}$ and the third to $\pu{Al^{2+}}$).

When you write down the electron configurations

$$\pu{Mg^{2+}:[Ne]}$$ $$\pu{Al^{2+}:[Ne]} \ 3s^1$$

you find that you are comparing the ionization energy of a noble-gas-like ion with an ion that has noble-gas-plus-one-electron. It is well known that the noble gases are the most stable configurations possible, i.e. that they have the highest ionization energies among their neighbors (and are thus least reactive, whence the term "noble").

So, you may conclude that the third ionization energy of $\pu{Mg}$ is higher than that of $\pu{Al}$. With the same reasoning, you'll find that the fourth ionization energy of of $\pu{Mg}$ is lower than that of $\pu{Al}$, since in that case it is $\pu{Al^{3+}}$ that has the noble gas electron configuration.


The notion of relative stability of an ion is a non-sense when applied to comparing two different ions of the same element or two equally charged ions of two different elements. Production of $\ce{Mg^{3+}}$ requires a much higher energy ($7.73$ MJ/mol) than the $2.74$ MJ/mol necessary for producing $\ce{Al^{3+}}$. It does not imply that one of these ions is more or less stable than the other one.


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