Chemistry Q56 – JEE-MAIN 2025

Home Chemistry ElectrochemistryQuestion #401

Correct order of limiting molar conductivity for cations in water at

298 K

is:

Clue 1 of 3: Conceptual Principle

Limiting molar conductivity depends on the effective size of the hydrated ion and its charge. Smaller hydrated ions and higher charge generally lead to higher conductivity.

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Ninja Strategy•Special Case and General Trends

Recognize the unique high conductivity of H+ due to proton hopping, then apply the general trend that for simple ions, smaller hydrated radius and higher charge lead to higher conductivity, keeping in mind the relationship between ionic radius and hydrated radius.

🧠 Full Solution Path

Step 1: Identify Special Case for H+ Ion✦ Active

The $H^{+}$ ion (proton) has an exceptionally high limiting molar conductivity in water due to the Grotthuss mechanism (proton hopping), where the proton effectively 'jumps' from one water molecule to another. This mechanism allows for much faster charge transfer than the physical movement of other hydrated ions. Therefore, $H^{+}$ will have the highest conductivity among the given ions.

Step 2: Compare Other Cations based on Hydrated Size and Charge○ Expand

For other metal cations, limiting molar conductivity is inversely related to the size of the hydrated ion. Smaller ionic radius leads to stronger hydration and thus a larger hydrated radius, which decreases mobility. However, higher charge also increases the driving force for movement. Considering standard values at $298 K$ :

Λ_{m}^{0} (H^{+}) \approx 349.8 {S cm}^{2} {mol}^{- 1}

Λ_{m}^{0} (C a^{2 +}) \approx 119.0 {S cm}^{2} {mol}^{- 1}

Λ_{m}^{0} (M g^{2 +}) \approx 106.0 {S cm}^{2} {mol}^{- 1}

Λ_{m}^{0} (K^{+}) \approx 73.5 {S cm}^{2} {mol}^{- 1}

Λ_{m}^{0} (N a^{+}) \approx 50.1 {S cm}^{2} {mol}^{- 1}

💡 Teacher's Secret Hint

Remember that for alkali and alkaline earth metals, smaller ionic radius means larger hydrated radius due to stronger electrostatic interaction with water molecules, leading to lower mobility. However, the higher charge of divalent ions like $C a^{2 +}$ and $M g^{2 +}$ can compensate for their larger hydrated size compared to monovalent ions like $K^{+}$ and $N a^{+}$ , resulting in higher conductivity.

Step 3: Determine the Correct Order○ Expand

Based on the conductivities, the decreasing order is $H^{+} > C a^{2 +} > M g^{2 +} > K^{+} > N a^{+}$ . This matches option 4.

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Chemistry Question 56 – JEE-MAIN 2025

Clue 1 of 3: Conceptual Principle

🧠 Full Solution Path