So I have encountered a question which looks very suspicious to me.

If you have $\ce{Cl-}$ and $\ce{K+}$, Can you say that $\ce{K+}$ has more ionization energy than $\ce{Cl-}$?

We know for a fact that when you put more electrons it makes the the pulling force of the protons a bit scattered (because of the incremental of the electrons in $\ce{Cl-}$) so the ionization energy of chlorine should go down.

The opposite happens to $\ce{K+}$ when I remove electrons the pulling force of protons is stronger (Because it pulls less electrons) than it used to be so the ionization energy goes up!

So What I am proposing here that we can't just say that $\ce{K+}$ has more ionization energy than $\ce{Cl-}$ just because that happened. $\ce{Cl}$ had a really high ionization energy compared to $\ce{K}$ and the decrement of the ionization wouldn't be that big to make it less than the ionization energy of $\ce{K+}$.

(The answer was that $\ce{K+}$ has more ionization energy than $\ce{Cl-}$ in the exam paper but I questioning that)

So if there is a chart or something for this that would be awesome. I hope I can get some help here.

  • $\begingroup$ Gravity ain't involved here. Are you looking for a chart of ionization energies? If so, they're all over the place - Google it. As it reads now, your question is a little unclear: if you can edit it for clarity to specify what the gist of your question is, that will help you and the community here get an answer (or more) to your query. $\endgroup$ Commented Dec 27, 2015 at 18:58
  • $\begingroup$ Oops didnt mean gravity sorry. I know that there are a lot of charts out there but I cant seem to find any chart that has cl- in it. The question is simply is: Who has a higher Ionization energy k+ or cl-? and explain why. I put my argument up there. So that who will answer the question can check it out. $\endgroup$
    – Biker
    Commented Dec 27, 2015 at 19:17
  • 1
    $\begingroup$ Look at it this way: what is the electronic configuration of $\ce{K^+}$? What is the electronic configuration of $\ce{Cl-}$? $\endgroup$ Commented Dec 27, 2015 at 19:30

2 Answers 2


The general reaction we are looking at when discussion ionisation enthalpy is:

$$\ce{X -> X+ + e-}$$

With the electron removed into space (i.e. totally disappearing). An electron is a negatively charged particle, so removing it from something negative will be easier than removing it from something positive. Thus, you have a very stable $\ce{K+}$ ion and are trying to rip away another electron from this (bad) — and you need to rip it out of a core orbital — all valence orbitals are already empty, the last one when you removed that previous electron to generate $\ce{K+}$ (very bad). A combination of bad and very bad gives us a very high ionisation energy for $\ce{K+}$.

Discussing chloride, first of all we have an anion, so the removal of an electron is generally good for removing excess charge. Second, we again have something that has a full octet — but this time the octet is a valence octet, not a core octet, so removing the electron will be a lot easier. We have a combination of good and relatively good, so the ionisation energy of $\ce{Cl-}$ will be rather low.


The simplest way to approach this is to look at the total number of electrons, which is 18 for both $\ce{Cl-}$ and $\ce{K+}$ (the electron configuration of Argon).

In one case you have 17 protons in the nucleus pulling all these electrons together, in the other 19. It is obvious that the $\ce{K+}$ nucleus will be pulling harder, and hence has the higher ionization energy. You can safely predict that the ionization energy of $\ce{Ar}$ will be in between both values.

Numerically, however, the approximate proportion between the ionization energies is not 17 : 18 : 19, but rather (17-17) : (18-17) : (19-17) due to screening by the remaining electrons. Indeed, ionization energy is mainly driven by $\frac{Z_\rm{eff}}{r}$, where $Z_\rm{eff}$ is the effective (screened) nuclear charge and $r$ the radius. The radius will be similar for $\ce{Cl-}$, $\ce{Ar}$, and $\ce{K+}$, but $Z_\rm{eff}$ is 0, 1, and 2, respectively. If we assume a radius of 0.9, the ionization energies will be roughly 0, 1.1, and 2.2 Rydberg ($\pu{0 eV}$, $\pu{15 eV}$, and $\pu{30 eV}$).

This can be understood by considering that the ionizing electron from $\ce{Cl-}$ does not feel any electrical force as it moves away from the neutral $\ce{Cl}$ atom, it is only bound by electron affinity, which is an order of magnitude lower ($\pu{3.5 eV}$). The ionizing electrons of $\ce{Ar}$ and $\ce{K+}$, on the other hand, are strongly attracted by the positive ion they leave behind.


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