In the lanthanoid series, the +3 oxidation state is the most...

The stability of an oxidation state in aqueous solution is determined by the overall enthalpy change of the process $\ln (s) \rightarrow Ln^{3+}(aq) + 3e^{-}$.

This process involves three steps:
1. Sublimation/Atomization: $\ln (s) \rightarrow \ln (g)$ (requires $\Delta_{atom}H$)
2. Ionization: $\ln (g) \rightarrow Ln^{3+}(g) + 3e^{-}$ (requires $IE_{1} + IE_{2} + IE_{3}$)
3. Hydration: $Ln^{3+}(g) \rightarrow Ln^{3+}(aq)$ (releases $\Delta_{hyd}H$)

The sum of the first three ionization enthalpies is very high and endothermic. However, the $Ln^{3+}$ ions have a high charge (+3) and a relatively small ionic radius, which results in a very large negative hydration enthalpy. This highly exothermic hydration process releases enough energy to compensate for the high ionization enthalpies and the enthalpy of atomization, making the $+3$ oxidation state the most stable in aqueous solutions.

The $4f$ orbitals are deeply buried and do not interact significantly with ligands, so crystal field stabilization energy is negligible for lanthanoids.

Answer: Large negative hydration enthalpy of $Ln^{3+}$ ions