JEE Chemistry 2023 Question with Solution

Given: The ions are $\mathrm{K}^+$, $\mathrm{Rb}^+$, $\mathrm{Mg}^{2+}$, $\mathrm{Cs}^+$, and $\mathrm{Ca}^{2+}$.

Find: The correct decreasing order of hydration enthalpy.

Hydration enthalpy depends on the charge density of an ion, which depends on its charge and ionic radius. Smaller ions with higher charges have higher hydration enthalpies because they interact more strongly with water molecules.

For alkali metal ions:

$$\mathrm{K}^+ > \mathrm{Rb}^+ > \mathrm{Cs}^+$$

So, $\mathrm{A > B > D}$.

For alkaline earth metal ions:

$$\mathrm{Mg}^{2+} > \mathrm{Ca}^{2+}$$

So, $\mathrm{C > E}$.

Combining both trends:

$$\mathrm{Mg}^{2+} > \mathrm{Ca}^{2+} > \mathrm{K}^+ > \mathrm{Rb}^+ > \mathrm{Cs}^+$$

That is,

$$\mathrm{C > E > A > B > D}$$

Therefore, the correct option is D.

Given: Hydration enthalpy increases with increasing charge density.

Find: Which listed sequence matches the correct order.

Ions with higher positive charge and smaller size attract water molecules more strongly, so their hydration enthalpy is greater.

1. Between $\mathrm{Mg}^{2+}$ and $\mathrm{Ca}^{2+}$, the ion $\mathrm{Mg}^{2+}$ is smaller and has the same charge, so:

$$\mathrm{Mg}^{2+} > \mathrm{Ca}^{2+}$$

2. Among $\mathrm{K}^+$, $\mathrm{Rb}^+$, and $\mathrm{Cs}^+$, size increases down the group, so hydration enthalpy decreases:

$$\mathrm{K}^+ > \mathrm{Rb}^+ > \mathrm{Cs}^+$$

3. The $2+$ ions have much greater hydration enthalpy than the $1+$ ions because of their larger charge and higher charge density.

Hence the overall order is:

$$\mathrm{Mg}^{2+} > \mathrm{Ca}^{2+} > \mathrm{K}^+ > \mathrm{Rb}^+ > \mathrm{Cs}^+$$

So the matching option is $\mathrm{C > E > A > B > D}$, which is option D.