Chemistry - PERIODIC TABLE Question with Solution | TestHub

The ionic radius generally decreases across a period and increases down a group. For isoelectronic species, the ionic radius decreases with increasing nuclear charge.

Comparing the given ions:

${\mathrm{Al}}^{3+}$: 10 electrons, 13 protons

${\mathrm{Ba}}^{2+}$: 54 electrons, 56 protons

${\mathrm{K}}^{+}$: 18 electrons, 19 protons

${\mathrm{Mg}}^{2+}$: 10 electrons, 12 protons

${\mathrm{Al}}^{3+}$ and ${\mathrm{Mg}}^{2+}$ are isoelectronic (10 electrons). ${\mathrm{Al}}^{3+}$ has a higher nuclear charge (13 protons vs 12 protons), pulling the electrons closer, resulting in a smaller ionic radius.

${\mathrm{K}}^{+}$ has 18 electrons, which is a larger electron shell than ${\mathrm{Al}}^{3+}$ and ${\mathrm{Mg}}^{2+}$.

${\mathrm{Ba}}^{2+}$ has 54 electrons, a much larger electron shell.

Therefore, ${\mathrm{Al}}^{3+}$ has the smallest ionic radius.

The final answer is $\boxed{A{l}^{3+}}$.