Ionization energy comparison between K+ and Cl-

Question

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.

Answer 1

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.

Answer 2

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.