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Question #44236

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The Question

Consider the thermodynamic parameters governing the standard reduction potentials ( $E^\circ_{M^{2+}/M}$ ) of the 3d transition metal series at $298\text{ K}$ . The Born-Haber cycle for the half-cell reaction $M^{2+}(aq) + 2e^- \rightarrow M(s)$ involves the enthalpy of atomization ( $\Delta_{a}H^\circ$ ), ionization enthalpies ( $IE_1 + IE_2$ ), and hydration enthalpy ( $\Delta_{hyd}H^\circ$ ). Which of the following statements is \textbf{INCORRECT}?

Options

A.
The positive $E^\circ$ value for $Cu^{2+}/Cu$ ( $+0.34\text{ V}$ ) arises because the sum of its $\Delta_{a}H^\circ$ and $(IE_1 + IE_2)$ is exceptionally high and is not adequately compensated by its $\Delta_{hyd}H^\circ$ , making the oxidation process thermodynamically non-spontaneous relative to the Standard Hydrogen Electrode.
B.
The $E^\circ$ value for $Ni^{2+}/Ni$ ( $-0.25\text{ V}$ ) deviates from the general less-negative trend of the late 3d series primarily due to the exceptionally high negative $\Delta_{hyd}H^\circ$ of the $Ni^{2+}$ ion, which thermodynamically favors its oxidized state.
C.
If the aqueous medium is replaced by a solvent with a significantly lower dielectric constant, the absolute standard reduction potential ( $E^\circ_{abs}$ ) of $Cu^{2+}/Cu$ will become less positive because the decreased solvation energy makes the reduction of $Cu^{2+}$ thermodynamically less favorable.
D.
The pronounced negative dips in the $E^\circ_{M^{2+}/M}$ trend for Manganese and Zinc are predominantly driven by the lower $(IE_1 + IE_2)$ values associated with the formation of exchange-energy stabilized $d^5$ and fully-filled $d^{10}$ configurations, respectively.

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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

Bond Enthalpy Pitfall|E° (M²⁺/M) Anomalies

Why This Question Is Tricky

Common Mistakes

The question tests the student's mastery of the Born-Haber cycle applied to electrode potentials and the specific thermodynamic trap of solvent effects.

Students often rely on the heuristic that 'less solvation = less reactivity', leading them to incorrectly assume that a lower dielectric constant decreases the reduction potential.

However, reduction involves the destruction of the solvated ion; destabilizing the reactant ( $Cu^{2+}$ ) actually increases the thermodynamic driving force for its reduction to solid copper, thereby making the absolute reduction potential more positive.

This exposes a critical misunderstanding of state functions and relative energy levels.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Thermodynamics - Class 11 Part 1
Concept	Bond Enthalpy Pitfall
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free