Out of all halides of cesium (${\mathrm{Cs}}^{+}$):

Here, $\mathrm{CsF}$ (maximum lattice energy), $\mathrm{CsI}$(maximum covalent character) and  $\mathrm{CsF}$(has maximum covalent character). Because F is the most electronegative halogen, and because Cs and F have a strong electrostatic affinity, $\mathrm{CsF}$ has the greatest lattice energy.
The amount of polarization increases with the charge on the cation. Ion size shrinks as follows: Since ${\mathrm{Na}}^{+}>{\mathrm{Mg}}^{+2}>{\mathrm{Al}}^{+3}>{\mathrm{Si}}^{+4}$and ${\mathrm{Cl}}^{-}$ ion have the same size in all of these compounds, ${\mathrm{Si}}^{+4}$ will exhibit stronger polarization. ${\mathrm{SiCl}}_4$ will exhibit the most covalent nature as a result.
$\mathrm{CsF}$ has a stronger ionic bond due to the higher difference in electronegativity. According to Fajan's rule, the anion in $\mathrm{CsF}$ is much smaller than the cation. Ionic bonding will be quite strong and covalent character will be very low.
Hence, the option D is correct.