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Question #44588

Difficulty: 8/10 | 🟢 Free

The Question

In the qualitative analysis of nitrate ions, a brown ring of an octahedral coordination complex $X$ is formed at the junction of the liquids when freshly prepared \ce{FeSO4(aq)} and concentrated \ce{H2SO4} are added. Spectroscopic studies reveal that the nitrosyl ligand in $X$ is non-innocent; it undergoes an internal redox process by transferring one electron to the central iron ion. Given that complex $X$ maintains a high-spin configuration, determine the spin-only magnetic moment ( $\mu_{eff}$ ) and the Crystal Field Stabilization Energy (CFSE) of the central metal ion in $X$ . (Ignore pairing energy; $\Delta_o$ represents the octahedral crystal field splitting energy).

Options

A.
$3.87 \text{ BM}, -0.8 \Delta_o$
B.
$4.90 \text{ BM}, -0.4 \Delta_o$
C.
$5.92 \text{ BM}, 0.0 \Delta_o$
D.
$1.73 \text{ BM}, -1.8 \Delta_o$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Crystal Field Splitting (Octahedral)|Spin-Only Magnetic Moment|Oxidation State Traps

Why This Question Is Tricky

Common Mistakes

This question tests the 'Oxidation State Trap' via a non-innocent ligand.

Students often assume the central metal retains its initial oxidation state (Fe2+, d6) and blindly calculate CFSE and magnetic moment based on that assumption.

By introducing the internal redox mechanism where NO coordinates as NO+ and reduces Fe to Fe+, the d-electron count shifts to d7.

Failing to account for this hidden variable leads to incorrect orbital filling and erroneous CFSE values.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Coordination Compounds - Class 12
Concept	Crystal Field Splitting (Octahedral)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free