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Question #44875

Difficulty: 8/10 | 🟢 Free

The Question

Transition metals in their highest oxidation states exhibit unique optical and chemical properties governed by charge transfer and relativistic effects. Which of the following statements regarding these high oxidation state species is \textbf{INCORRECT}?

Options

  • A.

    \ce{MnO4^-} and \ce{Cr2O7^{2-}} are intensely colored due to Laporte-allowed Ligand-to-Metal Charge Transfer (LMCT).

  • B.

    The LMCT energy gap follows the order \ce{MnO4^-} < \ce{TcO4^-} < \ce{ReO4^-}, rendering \ce{ReO4^-} colorless.

  • C.

    5d high oxidation state fluorides like \ce{PtF6} are stabilized by d-orbital contraction and relativistic effects.

  • D.

    \ce{PtF6} is a colorless, diamagnetic d0d^0 complex due to the extreme electronegativity of fluorine.

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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

Color and Charge Transfer|d-orbital contraction|Stability of High O.S. Fluorides/Oxides

Why This Question Is Tricky

Common Mistakes

Students often falsely conflate high oxidation states with a d0 configuration and subsequently assume that d0 species must be colorless.

While KMnO4 breaks the 'd0 is colorless' rule via LMCT, PtF6 breaks the 'high O.S. means d0' assumption.

Platinum (Group 10) in the +6 oxidation state retains a d4 configuration.

Furthermore, the question tests the understanding that the stability of extreme oxidation states in the 5d series is driven by relativistic effects and d-orbital contraction, which students frequently misinterpret as a reason for colorlessness rather than enhanced covalent bonding.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterD and F Block Elements - Class 12
ConceptColor and Charge Transfer
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree