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Question #44265

Difficulty: 8/10 | 🟢 Free

The Question

A 3d-block metal MM forms an octahedral aqua complex [M(H2O)6]3+[M(H_2O)_6]^{3+} which exhibits an extremely weak d-d transition spectrum because all such transitions are both Laporte and spin-forbidden. The metal MM forms a sulfate salt MSO4MSO_4. When an aqueous solution of MSO4MSO_4 is treated with an excess of KCNKCN, a stable octahedral complex YY is formed. Assume a hypothetical scenario where the ionization of MM to M2+M^{2+} incorrectly occurs by removing electrons from the 3d3d subshell before the 4s4s subshell. Based on this hypothetical electronic configuration, what would be the predicted spin-only magnetic moment (μhypo\mu_{hypo}) of complex YY? Also, determine the true Crystal Field Stabilization Energy (CFSE) of complex YY (ignoring pairing energy PP).

Options

  • A.

    μhypo=2.83 BM\mu_{hypo} = 2.83 \text{ BM}, True CFSE = −2.4Δo-2.4 \Delta_o

  • B.

    μhypo=4.90 BM\mu_{hypo} = 4.90 \text{ BM}, True CFSE = −0.4Δo-0.4 \Delta_o

  • C.

    μhypo=0.00 BM\mu_{hypo} = 0.00 \text{ BM}, True CFSE = −2.4Δo-2.4 \Delta_o

  • D.

    μhypo=3.87 BM\mu_{hypo} = 3.87 \text{ BM}, True CFSE = −2.0Δo-2.0 \Delta_o

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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|d-block Variable Oxidation States|Color and d-d Transitions

Why This Question Is Tricky

Common Mistakes

This problem tests the energy crossover between 4s and 3d orbitals during ionization.

Students often memorize the Aufbau principle (4s fills before 3d) but fail to realize that upon ionization, 4s electrons are removed first because the 3d orbitals drop below 4s in energy once occupied.

The 'trap' forces students to apply Crystal Field Theory (CFT) to a hypothetical anomalous electronic configuration (4s23d44s^2 3d^4).

They must recognize that even in this hypothetical state, the strong field ligand CN−CN^- forces the 3d43d^4 electrons to pair up in the t2gt_{2g} level, yielding n=2n=2 instead of n=4n=4.

Additionally, it requires identifying Iron via the spectroscopic hallmark of a completely spin-forbidden d-d spectrum in a high-spin d5d^5 system.

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