Use Fajan's rule

The covalent character in an ionic compound increases when:

• the cation is small and highly polarizing,
• the anion is large and highly polarizable.

Here, the compounds are alkaline earth metal halides: $MX_2 \, , \quad X = Cl, Br, I$

Compare the alkaline earth metal cations

The cations are: $Ca^{2+},\ Sr^{2+},\ Mg^{2+},\ Be^{2+}$

Among these, the size order is approximately: $Be^{2+} < Mg^{2+} < Ca^{2+} < Sr^{2+}$

Since all have charge $+2$, the smallest cation has the highest polarizing power.

Therefore, $Be^{2+}$ has the maximum tendency to distort the electron cloud of halide ions and hence forms the most covalent halide.

Check the options

• A: $CaX_2$ — less covalent than beryllium halides
• B: $SrX_2$ — even larger cation, hence more ionic
• C: $MgX_2$ — shows some covalent character, but less than beryllium halides
• D: $BeX_2$ — maximum covalent character among the given alkaline earth metal halides

Conclusion

The covalent alkaline earth metal halide is: $\boxed{BeX_2}$

Comparison with stored answer

Stored correct answer: D

Our derived answer is also D, so they agree.