JEE Chemistry 2025 Question with Solution

Given: The order of limiting molar conductivity of cations in water at $298 \, \text{K}$ is to be identified.

Find: The correct decreasing order among the given ions.

From the extracted solution, the limiting molar conductivities are:

$$\lambda^0_{H^+} = 349.8 \, \text{S cm}^2 \, \text{mol}^{-1}$$ $$\lambda^0_{Na^+} = 50.11 \, \text{S cm}^2 \, \text{mol}^{-1}$$ $$\lambda^0_{K^+} = 73.52 \, \text{S cm}^2 \, \text{mol}^{-1}$$ $$\lambda^0_{Ca^{2+}} = 119 \, \text{S cm}^2 \, \text{mol}^{-1}$$ $$\lambda^0_{Mg^{2+}} = 106.12 \, \text{S cm}^2 \, \text{mol}^{-1}$$

Therefore, the decreasing order is:

$$H^+ > Ca^{2+} > Mg^{2+} > K^+ > Na^+$$

Matching with the given options, the correct option is B.

Given: Limiting molar conductivity depends on ionic mobility in water.

Find: Which listed order is correct.

The solution states that smaller and highly charged ions are strongly hydrated in water. Stronger hydration lowers mobility. However, $H^+$ shows exceptionally high conductivity because it moves through the Grotthuss proton-hopping mechanism.

Also, between $Ca^{2+}$ and $Mg^{2+}$, calcium ion has higher limiting molar conductivity than magnesium ion because $Mg^{2+}$ is more strongly hydrated and hence less mobile.

Thus, the correct order is:

$$H^+ > Ca^{2+} > Mg^{2+} > K^+$$

Therefore, the correct option is B.