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Question #44596

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The Question

An intensely red-colored octahedral complex $\ce{[Fe(phen)3]^2+}$ (where phen = 1,10-phenanthroline) is treated with a standard nitrating mixture (conc. $\ce{HNO3}$ + conc. $\ce{H2SO4}$ ) at 298 K. The reaction is monitored colorimetrically. Instead of yielding the expected electrophilic aromatic substitution (EAS) product on the aromatic rings, the solution rapidly changes to a pale blue color, and no nitrated organic product is isolated upon immediate aqueous workup. Which of the following best explains the observed phenomena?

Options

A.
The nitrating mixture oxidizes the $d^6$ metal center to a $d^5$ low-spin state, quenching the intense MLCT band. The resulting $+3$ charge strongly withdraws electron density from the phenanthroline $\pi$ -system, severely deactivating it towards EAS.
B.
Coordination to $\ce{Fe^{2+}}$ activates the rings towards EAS via strong $d\pi-p\pi$ back-bonding. The color change to blue indicates successful nitration, which shifts the MLCT absorption to a longer wavelength due to the $-\ce{NO2}$ group.
C.
The strong acidic medium protonates the phenanthroline ligands, causing complete dissociation. The pale blue color arises from the formation of $\ce{[Fe(H2O)6]^3+}$ , while the free protonated phenanthroline resists nitration.
D.
The $\ce{NO2+}$ electrophile attacks the metal center directly due to its high electron density, displacing a phenanthroline ligand to form a blue $\ce{[Fe(phen)2(NO2)2]}$ complex, bypassing the sterically hindered aromatic rings.

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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

Color and d-d Transitions|Electrophilic Substitution Control|Crystal Field Splitting (Octahedral)

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of Coordination Chemistry (Spectroscopy/Redox) and Organic Reaction Mechanisms (EAS).

The 'Chemical Exception' is the assumption that metal coordination universally activates aromatic ligands via back-bonding.

Students often overlook that in the presence of strongly oxidizing electrophilic reagents (like nitrating mixtures), the metal center acts as a redox sink.

The resulting high-oxidation-state metal ( $+3$ ) acts as a powerful electron-withdrawing group, overriding any potential $\pi$ -activation and completely deactivating the coordinated aromatic rings towards electrophilic attack.

The color change serves as a diagnostic trap: it is a redox-induced MLCT quenching, not a substituent-induced bathochromic shift.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Coordination Compounds - Class 12
Concept	Color and d-d Transitions
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free