1. Identify electronic configurations of the ions

• $K^+$ : $19 - 1 = 18$ electrons $\Rightarrow [Ar]$
• $Na^+$ : $11 - 1 = 10$ electrons $\Rightarrow [Ne]$
• $Mg^{2+}$ : $12 - 2 = 10$ electrons $\Rightarrow [Ne]$
• $Al^{3+}$ : $13 - 3 = 10$ electrons $\Rightarrow [Ne]$

So, $Na^+$, $Mg^{2+}$ and $Al^{3+}$ are isoelectronic (all have 10 electrons).

2. Compare the isoelectronic ions

For an isoelectronic series, ionic radius decreases as nuclear charge increases, because the same number of electrons are pulled more strongly by more protons.

Nuclear charges are:

• $Na^+$ : $Z=11$
• $Mg^{2+}$ : $Z=12$
• $Al^{3+}$ : $Z=13$

Therefore, $Al^{3+} < Mg^{2+} < Na^+$

3. Compare $K^+$ with the above ions

$K^+$ has 18 electrons and configuration $[Ar]$, so it has one more occupied shell than the $[Ne]$ ions. Hence $K^+$ is larger than $Na^+$, $Mg^{2+}$, and $Al^{3+}$.

Thus, $Al^{3+} < Mg^{2+} < Na^+ < K^+$

4. Match with the options

This corresponds to Option C.

$\boxed{Al^{3+} < Mg^{2+} < Na^+ < K^+}$

5. Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

So they agree.