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Question #44292

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The Question

Consider the thermodynamic parameters governing the standard reduction potentials ( $E^\circ_{M^{4+}/M}$ ) of the Group 4 transition metals Zirconium (Zr) and Hafnium (Hf) in aqueous medium at 298 K. The standard reduction potential is determined by the interplay of sublimation enthalpy ( $\Delta_{sub}H^\circ$ ), the sum of the first four ionization enthalpies ( $\sum_{i=1}^{4} IE_i$ ), and the hydration enthalpy ( $\Delta_{hyd}H^\circ$ ). Based on the effects of the Lanthanoid contraction, identify the correct relationship between these parameters for Zr and Hf.

Options

A.
$|\Delta_{hyd}H^\circ_{Zr^{4+}}| \approx |\Delta_{hyd}H^\circ_{Hf^{4+}}| \text{ ; } \sum_{i=1}^{4} IE_{Hf} > \sum_{i=1}^{4} IE_{Zr} \text{ ; } E^\circ_{Hf^{4+}/Hf} > E^\circ_{Zr^{4+}/Zr}$
B.
$|\Delta_{hyd}H^\circ_{Zr^{4+}}| > |\Delta_{hyd}H^\circ_{Hf^{4+}}| \text{ ; } \sum_{i=1}^{4} IE_{Hf} < \sum_{i=1}^{4} IE_{Zr} \text{ ; } E^\circ_{Hf^{4+}/Hf} < E^\circ_{Zr^{4+}/Zr}$
C.
$|\Delta_{hyd}H^\circ_{Zr^{4+}}| \approx |\Delta_{hyd}H^\circ_{Hf^{4+}}| \text{ ; } \sum_{i=1}^{4} IE_{Hf} < \sum_{i=1}^{4} IE_{Zr} \text{ ; } E^\circ_{Hf^{4+}/Hf} < E^\circ_{Zr^{4+}/Zr}$
D.
$|\Delta_{hyd}H^\circ_{Zr^{4+}}| < |\Delta_{hyd}H^\circ_{Hf^{4+}}| \text{ ; } \sum_{i=1}^{4} IE_{Hf} > \sum_{i=1}^{4} IE_{Zr} \text{ ; } E^\circ_{Hf^{4+}/Hf} < E^\circ_{Zr^{4+}/Zr}$

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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

Ionization Enthalpy (IE) Anomalies|Standard Electrode Potential Sign Conventions|Post-Lanthanoid Radii Equality

Why This Question Is Tricky

Common Mistakes

Students often rely on simple vertical periodic trends, assuming that ionization enthalpy strictly decreases down a group, which would make heavier transition metals easier to oxidize (more negative $E^\circ$ ).

The 'Hidden Variable' here is the Lanthanoid Contraction.

The poor shielding of 4f electrons in Hafnium (5d) leads to a higher effective nuclear charge, making its $\sum IE$ anomalously higher than Zirconium (4d).

Because their ionic radii are nearly identical, hydration enthalpy cannot compensate for this high IE.

Consequently, Hf is harder to oxidize, and its standard reduction potential $E^\circ_{Hf^{4+}/Hf}$ is less negative (more noble) than $E^\circ_{Zr^{4+}/Zr}$ .

The trap punishes the blind application of group trends and rewards a rigorous Born-Haber cycle analysis.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Classification of Elements and Periodicity in Properties - Class 11 Part 1
Concept	Ionization Enthalpy (IE) Anomalies
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free