1. Use Fajan’s rule

   Covalent character in an ionic compound increases when:

   • the cation is small and has high charge,
   • the cation has high polarising power,
   • the anion is large and easily polarised.

   Here, the anion is the same in all compounds: $Cl^-$.
   So we only compare the polarising power of the cations: $Sn^{2+},\ Al^{3+},\ Mg^{2+},\ Fe^{2+}$

2. Compare the cations

   • $Mg^{2+}$: small, but charge is only $+2$.
   • $Fe^{2+}$: transition metal ion, can show some extra polarisation, but still only $+2$.
   • $Sn^{2+}$: larger ion, so its polarising power is less than a small highly charged ion.
   • $Al^{3+}$: very small size and high charge $+3$, so it has maximum polarising power.

3. Conclusion from Fajan’s rule

   Since $Al^{3+}$ has the greatest polarising power, it distorts the electron cloud of $Cl^-$ the most, giving the compound the maximum covalent character.

4. Check options

   • A: $SnCl_2$ — less covalent than $AlCl_3$
   • B: $AlCl_3$ — most covalent
   • C: $MgCl_2$ — more ionic
   • D: $FeCl_2$ — less covalent than $AlCl_3$

Therefore, the correct answer is: $\boxed{B\; (AlCl_3)}$