Which of the alkaline earth metal halides given below is essentially covalent in nature?

The alkaline earth metals belong to Group 2 of the periodic table and form di-halides of general formula $$MCl_2$$, where $$M^{2+}$$ is the metal cation. Whether a metal chloride is ionic or covalent is decided mainly by the polarising power of the cation.

We first recall Fajan’s rules. These rules state that:

$$\text{Polarising power of a cation} \propto \dfrac{\text{charge on cation}}{(\text{radius of cation})^{2}}.$$

Greater polarising power means that the cation can pull (distort) the electron cloud of the anion more strongly, introducing covalent character in the bond. Smaller highly charged cations thus favour covalent bonding, whereas larger cations of the same charge form predominantly ionic bonds.

Inside the alkaline earth series the charge on every metal ion is the same, $$+2$$, but the ionic radius increases down the group:

$$r(Be^{2+}) \lt r(Mg^{2+}) \lt r(Ca^{2+}) \lt r(Sr^{2+}) \lt r(Ba^{2+}).$$

Because the charge is identical for all five cations, the factor that actually varies is the denominator $$(\text{radius})^{2}$$. A smaller radius makes the denominator smaller, increasing the overall value of the fraction and thereby increasing the polarising power.

Hence

$$\text{Polarising power:}\; Be^{2+} \gt Mg^{2+} \gt Ca^{2+} \gt Sr^{2+} \gt Ba^{2+}.$$

Chloride ion, $$Cl^{-}$$, is moderately large and easily polarisable, so if a cation has enough polarising power, the bond will become covalent. Among the options given, $$Be^{2+}$$ is by far the smallest and therefore the most powerful in polarising $$Cl^{-}$$. This makes $$BeCl_2$$ essentially covalent in nature, whereas $$MgCl_2, CaCl_2$$ and $$SrCl_2$$ remain largely ionic.

So, the compound that is predominantly covalent is $$BeCl_2$$.

Hence, the correct answer is Option A.