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Question #44241

Difficulty: 8/10 | 🟢 Free

The Question

The chelate effect is classically responsible for the enhanced stability of complexes containing polydentate ligands. Consider the ligand substitution reaction of a d10d^{10} Cadmium(II) complex at 300 K300 \text{ K}:

\ce[Cd(NH3)4]2+(aq)+2en(aq)<=>[Cd(en)2]2+(aq)+4NH3(aq)\ce{[Cd(NH3)4]^{2+}(aq) + 2en(aq) <=> [Cd(en)2]^{2+}(aq) + 4NH3(aq)}

The thermodynamic parameters for the formation of these complexes from \ce[Cd(H2O)6]2+\ce{[Cd(H2O)6]^{2+}} at 300 K300 \text{ K} are provided below: Overall formation constant (log⁡β4\log \beta_4) for \ce[Cd(NH3)4]2+=7.0\ce{[Cd(NH3)4]^{2+}} = 7.0 Overall formation constant (log⁡β2\log \beta_2) for \ce[Cd(en)2]2+=10.5\ce{[Cd(en)2]^{2+}} = 10.5 Standard enthalpy of complexation (ΔHc∘\Delta H^\circ_c) for \ce[Cd(NH3)4]2+=−53.0 kJ mol−1\ce{[Cd(NH3)4]^{2+}} = -53.0 \text{ kJ mol}^{-1} Standard enthalpy of complexation (ΔHc∘\Delta H^\circ_c) for \ce[Cd(en)2]2+=−56.0 kJ mol−1\ce{[Cd(en)2]^{2+}} = -56.0 \text{ kJ mol}^{-1}

Evaluate the standard entropy change (ΔSsub∘\Delta S^\circ_{sub}) for this substitution reaction and identify the dominant thermodynamic driving force. (Given: 2.303RT=5.75 kJ mol−12.303 RT = 5.75 \text{ kJ mol}^{-1} at 300 K300 \text{ K})

Options

  • A.

    +57.1 \text{ J K}^{-1} \text{ mol}^{-1}; \text{Entropy-driven}

  • B.

    +77.1 \text{ J K}^{-1} \text{ mol}^{-1}; \text{Entropy-driven}

  • C.

    +17.1 \text{ J K}^{-1} \text{ mol}^{-1}; \text{Enthalpy-driven}

  • D.

    -77.1 \text{ J K}^{-1} \text{ mol}^{-1}; \text{Enthalpy-driven}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Spontaneity Criteria|Ligand Denticity Variations|Complex Ion Oxidation States

Why This Question Is Tricky

Common Mistakes

This problem tests the quantitative thermodynamic basis of the Chelate Effect.

Students often memorize that chelating ligands form more stable complexes but incorrectly attribute this to stronger metal-ligand bonds (enthalpy).

By using a d10d^{10} Cd2+Cd^{2+} system where Crystal Field Stabilization Energy (CFSE) is zero, the σ\sigma-bond strengths of Cd−NH3Cd-NH_3 and Cd−enCd-en are nearly identical, isolating entropy as the true driving force.

The trap lies in assuming the larger formation constant (log⁡β\log \beta) implies a highly exothermic reaction.

Distractor B catches sign errors in Hess's Law (ΔHsub\Delta H_{sub}), Distractor C catches unit conversion failures where TΔST\Delta S is not divided by TT, and Distractor D catches reversed subtraction of ΔG\Delta G values.

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Question Information

SubjectChemistry
ChapterThermodynamics - Class 11 Part 1
ConceptSpontaneity Criteria
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree