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Question #44262
The Question
Consider the standard electrode potentials and of the 3d transition metals at , along with their thermodynamic parameters. Which of the following statements are CORRECT?
I. The value of \ce{Mn} () is more negative than that of \ce{Cr} () because both the sublimation enthalpy and the sum of the first two ionization enthalpies are lower for \ce{Mn} than for \ce{Cr}E^\circ(M^{2+}/M)-0.25\text{ V}(IE_1 + IE_2)E^\circ(M^{2+}/M)-0.76\text{ V}+0.34\text{ V} is significantly more negative than that of \ce{Cu^{2+}_{(aq)}}E^\circ(M^{3+}/M^{2+})+1.57\text{ V}+0.77\text{ V}.
Options
- A.
I, II, and IV only
- B.
I, II, and III only
- C.
II, III, and IV only
- D.
I and IV only
Check Your Answer
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
Hund’s Rule of Maximum Multiplicity|Screening Effect in Inner Shells|E° (M²⁺/M) Anomalies
Why This Question Is Tricky
Common Mistakes
This question tests the student's ability to deconstruct the thermodynamic components (Born-Haber cycle) governing standard electrode potentials.
A major trap is assuming that Mn has a higher (IE1 + IE2) than Cr due to its stable 4s2 3d5 configuration, overlooking that Cr's IE2 involves breaking a half-filled 3d5 core.
Another trap is assuming Zn's anomalous potential is driven by a high hydration energy of the d10 ion, whereas Cu2+ actually has a more negative hydration energy due to Jahn-Teller distortion.
Finally, it rigorously tests the quantum mechanical origin (exchange energy) of the 3rd ionization enthalpy differences between Mn and Fe.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Structure of Atom - Class 11 Part 1 |
| Concept | Hund’s Rule of Maximum Multiplicity |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |