1. We need the trend in octahedral crystal field splitting, $\Delta_o$, for aqua complexes.

   For a given ligand ($H_2O$ here), $\Delta_o$ generally depends on:

   • oxidation state of the metal: higher oxidation state $\Rightarrow$ larger $\Delta_o$
   • principal quantum number / row of transition metal: typically $3d < 4d < 5d$ for comparable ions, so heavier congeners usually give larger splitting.

2. Compare $[Ti(H_2O)_6]^{3+}$ and $[Ti(H_2O)_6]^{2+}$

   Both have the same metal and same ligand, only oxidation state differs.

   • In $[Ti(H_2O)_6]^{3+}$, metal is $Ti^{3+}$
   • In $[Ti(H_2O)_6]^{2+}$, metal is $Ti^{2+}$

   Since higher oxidation state increases metal-ligand attraction and shortens bond length, we get $\Delta_o\big([Ti(H_2O)_6]^{3+}\big) > \Delta_o\big([Ti(H_2O)_6]^{2+}\big)$

   So any option saying $[Ti(H_2O)_6]^{3+} < [Ti(H_2O)_6]^{2+}$ is incorrect.

   Hence options B and D are eliminated.

3. Compare $[Cr(H_2O)_6]^{2+}$ and $[Mo(H_2O)_6]^{2+}$

   These are both $M^{2+}$ aqua complexes, but:

   • $Cr$ is a $3d$ element
   • $Mo$ is a $4d$ element

   For similar oxidation state and ligand, $4d$ metals usually have larger crystal field splitting than $3d$ metals because their orbitals are more diffuse and overlap better with ligands.

   Therefore, $\Delta_o\big([Mo(H_2O)_6]^{2+}\big) > \Delta_o\big([Cr(H_2O)_6]^{2+}\big)$

   or equivalently, $\Delta_o\big([Cr(H_2O)_6]^{2+}\big) < \Delta_o\big([Mo(H_2O)_6]^{2+}\big)$

4. Match with the options

   We need:

   • $\Delta_o([Cr(H_2O)_6]^{2+}) < \Delta_o([Mo(H_2O)_6]^{2+})$
   • $\Delta_o([Ti(H_2O)_6]^{3+}) > \Delta_o([Ti(H_2O)_6]^{2+})$

   This corresponds to Option C.

5. Final answer

   $\boxed{\text{C}}$

This matches the stored correct answer.