Chemistry - How to tell which species has the highest ionization energy?

Solution 1:

Like most chemical conundrums you need to simplify the problem using some knowledge of chemistry.

First there are two electron configurations. $\ce{Ne, Na+}$ and $\ce{Mg^{2+}}$ all have the configuration ${1s^2 2s^22p^6}$. $\ce{Ar, K+}$ and $\ce{Ca^{2+}}$ all have the configuration ${1s^2 2s^22p^6 3s^2 3p^6}$.

For $\ce{Ne, Na+}$ and $\ce{Mg^{2+}}$, $\ce{Mg}$ has the highest atomic number so of these three, it would be most difficult to ionize $\ce{Mg^{2+}}$.

For $\ce{Ar, K+}$ and $\ce{Ca^{2+}}$, $\ce{Ca}$ has the highest atomic number so of these three, it would be most difficult to ionize $\ce{Ca^{2+}}$.

Now it is just necessary to compare the ionization of $\ce{Mg^{2+}}$ to $\ce{Ca^{2+}}$. Since the $2p$ orbitals are closer to the nucleus than the $3p$ orbitals, it will be harder to ionize $\ce{Mg^{2+}}$ than $\ce{Ca^{2+}}$.

Solution 2:

Axiom 1: Ionisation energy increases from left to right along a period.

Axiom 2: Ionisation energy decreases from top to bottom along a group.

Both of these are often or mostly correct. However, there is more. Remember that the electron you are extracting carries a negative charge ($\ce{e-}$). So:

Axiom 3: Ionisation energy increases dramatically for every positive charge introduced.

And also, to state the concept of core versus valence electrons:

Axiom 4: The first ionisation energy of a core electron will be dramatically more than that of a valence electron.

In your case, you have six atoms to compare which all have core electrons only (if you count the fully populated shell of the noble gases as core electrons which isn’t strictly correct but close enough). Thus we do not need to consider axiom 4. Axiom 3 tells us, that $\ce{Mg^2+}$ and $\ce{Ca^2+}$ are the only contestants for maximum ionisation energy since they both have the highest charge. And comparing the two according to axiom 2, we arrive at $\ce{Mg^2+}$ having the highest ionisation energy of these atoms.