Identify the smallest ion among the following:
(A) $ A{l^{3 + }} $
(B) $ N{a^ + } $
(C) $ B{a^{2 + }} $
(D) $ M{g^{2 + }} $

Hint: Determine the atomic numbers of the parent atom of each of these ions. Compare their ionic sizes based on the extent of nuclear charge experienced by each ion, their position in the periodic table and the magnitude of charge present on them.

Complete answer:
The size of an ion depends upon the position of the parent atom in the periodic table. We know that the size of atoms increases when descending a group but decreases with moving from left to right in a period. But the position of the element in the periodic table alone is not sufficient to compare the size of ions.
Cations are formed by losing electrons and are always smaller in size than their parent atoms. A neutral atom has an equal number of electrons and protons (inside the nucleus) which balances the charge, but the loss of electrons by ions makes the number of protons higher than the electrons. This inequality results in a strong nuclear force of attraction experienced by the electrons.
Aluminium, sodium and magnesium belong to the same period and have atomic numbers $ 13,11{\text{ and }}12 $ respectively. But once they transform into cations $ A{l^{3 + }} $ , $ N{a^ + } $ and $ M{g^{2 + }} $ , they become isoelectronic species.
The comparison of ionic sizes of isoelectronic species is done according to their charge and not the position of their parent atom in the periodic table. The highest charged cation loses the maximum number of electrons and experiences the strongest nuclear force of attraction that results in the smallest molecule.
Aluminium cation has the highest charge of $ + 3 $ and is the smallest among the three isoelectronic species. Since, barium belongs to the sixth period i.e. occupies a position much lower in its group making the parent atom too large to be compared with smaller atoms like aluminium, sodium and magnesium. Thus $ B{a^{2 + }} $ is large enough to be compared with the remaining ions
Hence, the correct option is $ (1) $ $ A{l^{3 + }} $ is the smallest ion among the given options.

Note:
Isoelectronic species are those species that contain the same number of electrons. Since aluminium loses three electrons, magnesium loses two electrons and sodium loses a single electron the number of electrons present in all these ions becomes $ 10 $ while the number of protons remains the same.