JEE Chemistry 2023 Question with Solution

Given: The ions are $\mathrm{Be}^{2+}$, $\mathrm{Ba}^{2+}$, $\mathrm{Ca}^{2+}$ and $\mathrm{Sr}^{2+}$.

Find: Which ion has the highest hydration enthalpy.

Hydration enthalpy depends on the charge density of the ion. Highly charged and smaller-sized metal ions have higher hydration enthalpy.

All the given ions have the same charge, namely $+2$. Therefore, hydration enthalpy will be greater for the ion with the smallest size.

Down the alkaline earth metal group, ionic size increases, so hydration enthalpy decreases. Hence the order is:

Therefore, the ion having the highest hydration enthalpy is $\mathrm{Be}^{2+}$. The correct option is A.

Given: All ions belong to alkaline earth metals and each has charge $+2$.

Find: The ion with maximum hydration enthalpy.

Hydration enthalpy is the energy released when a gaseous ion gets surrounded by water molecules to form a hydrated ion. Its magnitude increases with increasing charge density.

This can be expressed qualitatively as:

Since the charge is the same for all given ions, we compare only their sizes:

As we move down the group from beryllium to barium, ionic size increases. So the smallest ion, $\mathrm{Be}^{2+}$, has the greatest hydration enthalpy.

Hence, $\mathrm{Be}^{2+}$ shows the highest hydration enthalpy, so the correct option is A.