Which of the following is least ionic?

We need to determine which of the given compounds is the least ionic.

We know from Fajans’ rules that ionic character decreases (and covalent character increases) when the cation is small with high charge (high charge density equals high polarizing power), the anion is large and easily polarizable, and the cation adopts a pseudo-noble gas configuration (such as an 18-electron outer shell) instead of a noble gas configuration (8-electron outer shell), since cations with (n-1)d$$^{10}$$ configurations exhibit higher polarizing power.

With these considerations in mind, we examine each compound in turn.

In BaCl$$_2$$ the Ba$$^{2+}$$ ion is large and has the noble gas configuration [Xe], which results in low polarizing power; consequently, BaCl$$_2$$ is highly ionic.

In AgCl, the Ag$$^{+}$$ ion has the electronic configuration [Kr]4d$$^{10}$$, and despite its +1 charge, its filled d-shell provides strong polarizing power; this causes AgCl to exhibit significant covalent character.

In KCl, the K$$^{+}$$ ion has the noble gas configuration [Ar] and a relatively large radius, leading to low charge density and making KCl one of the most ionic compounds.

In CoCl$$_2$$ the Co$$^{2+}$$ ion has the configuration [Ar]3d$$^7$$ and thus some polarizing power due to its +2 charge and incomplete d-shell, but it is not as strongly polarizing as Ag$$^{+}$$; therefore, CoCl$$_2$$ is more ionic than AgCl yet less so than BaCl$$_2$$ and KCl.

Ranking these compounds by ionic character from most to least ionic gives KCl > BaCl$$_2$$ > CoCl$$_2$$ > AgCl.

AgCl is the least ionic compound because Ag$$^{+}$$ has exceptionally high polarizing power according to Fajans’ rules.

The correct answer is Option (2): AgCl.