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Question #44267

Difficulty: 8/10 | 🟢 Free

The Question

Consider the reaction of the pale green complex \ce{[Fe(H2O)6]^{2+}} (Complex A\mathbf{A}) with excess 1,10-phenanthroline (phen) in an aqueous medium at 298 K to form the intensely red complex B\mathbf{B}. Let n1n_1 and n2n_2 represent the number of d-electrons with spin quantum numbers ms=+1/2m_s = +1/2 and ms=−1/2m_s = -1/2, respectively, in the metal's valence shell for Complex B\mathbf{B}. Which of the following correctly identifies the value of ∣n1−n2∣|n_1 - n_2|, the magnetic behavior, and the primary origin of the intense red color for Complex B\mathbf{B}?

Options

  • A.

    ∣n1−n2∣=0|n_1 - n_2| = 0, Diamagnetic, MLCT (t2g→π∗t_{2g} \rightarrow \pi^*)

  • B.

    ∣n1−n2∣=0|n_1 - n_2| = 0, Diamagnetic, Laporte-allowed d−dd-d transition

  • C.

    ∣n1−n2∣=2|n_1 - n_2| = 2, Paramagnetic, LMCT (n→egn \rightarrow e_g)

  • D.

    ∣n1−n2∣=0|n_1 - n_2| = 0, Diamagnetic, Spin-allowed d−dd-d transition

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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

Pauli Exclusion Principle|Ligand Denticity Variations|Color and d-d Transitions

Why This Question Is Tricky

Common Mistakes

The question tests the intersection of Crystal Field Theory, the Pauli Exclusion Principle, and electronic spectroscopy.

The 'Hidden Variable' or 'Exception' is the origin of the intense color.

Students often default to d−dd-d transitions for transition metal colors (Option D).

Some might mistakenly think the loss of perfect OhO_h symmetry makes it fully Laporte-allowed (Option B).

However, for complexes with π\pi-acceptor ligands like phenanthroline, the intense color is due to MLCT.

Furthermore, the question rigorously tests the understanding that while the ligand field strength (Δo>P\Delta_o > P) provides the thermodynamic driving force for pairing, it is the Pauli Exclusion Principle that strictly enforces the opposite spins (∣n1−n2∣=0|n_1 - n_2| = 0) when electrons share the t2gt_{2g} spatial orbitals.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterStructure of Atom - Class 11 Part 1
ConceptPauli Exclusion Principle
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree