Concept:Hydration energy is the amount of energy released when one mole of gaseous ions dissolves in water. It is heavily dependent on the "charge density" of the ion.
Formula:$$ \text{Hydration Energy} \propto \frac{\text{Charge of Ion}}{\text{Size of Ion}} $$
Solution:- To find an ion with a greater hydration energy than \( \text{Mg}^{2+} \), we need an ion with a higher charge, a smaller size, or both.
- \( \text{Al}^{3+} \) has a much higher charge (+3 compared to +2) and a smaller ionic radius than \( \text{Mg}^{2+} \).
- This results in a massively higher charge density for Aluminum, allowing it to attract water molecules much more violently, releasing far more hydration energy.
Why other options are incorrect:- \( \text{Na}^+, \text{Li}^+ \): Have only a +1 charge, resulting in drastically lower charge densities.
- \( \text{Ca}^{2+} \): Has the same charge as Mg, but is physically larger (lower down Group 2), giving it a lower charge density and lower hydration energy.
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