The enthalpy of hydration measures the energy released when gaseous ions interact with water molecules. A higher (more negative) value indicates stronger interactions.
For octahedral complexes like these, with a constant charge (+2), the hydration enthalpy depends mainly on the ionic radius of the central metal ion. Smaller ions have higher charge density, leading to stronger hydration.
Let's examine the ionic radii (in picometers, pm) for the relevant +2 ions:
The order of ionic radii is: V2+ < Mn2+ < Cr2+ < Ca2+.
The metal ion with the smallest radius, V2+, possesses the highest charge density.
Therefore, $[V(H_2O)_6]^{2+}$, featuring V2+, exhibits the strongest ion-dipole interactions with water and thus has the highest enthalpy of hydration among the given options.
Consider the figure given below, where M is a metal and L is a monodentate ligand. The $\sigma$-bonding ligand group orbital (LGO) having same symmetry with $d_{z^2}$ orbital of M in the octahedral coordination geometry is
Among the given platinum(II) complexes, the one that is thermally the most unstable is