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Question #44249

Difficulty: 8/10 | 🟢 Free

The Question

The radial probability distribution curve for the valence d-electrons of a transition metal $M$ exhibits exactly 2 radial nodes. $M$ reacts with aqueous $HCl$ to form a stable homoleptic octahedral complex $[MCl_6]^{2-}$ , in which the metal retains a $d^4$ electronic configuration. Assuming standard Crystal Field Theory, what is the calculated spin-only magnetic moment ( $\mu_s$ ) of this complex, and how does the experimentally observed effective magnetic moment ( $\mu_{eff}$ ) at 298 K compare to $\mu_s$ due to the intrinsic properties of the metal's valence orbitals?

Options

A.
$2.83 \text{ BM}, \mu_{eff} < \mu_s$
B.
$4.90 \text{ BM}, \mu_{eff} \approx \mu_s$
C.
$2.83 \text{ BM}, \mu_{eff} > \mu_s$
D.
$4.90 \text{ BM}, \mu_{eff} > \mu_s$

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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

Radial Probability Distribution|Spin-Only Magnetic Moment|d-orbital contraction

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Hidden Variable' of relativistic effects and the lanthanoid contraction in 5d transition metals.

Students often fall into the trap of treating $Cl^-$ as a weak-field ligand to predict a high-spin $d^4$ complex ( $\mu_s = 4.90$ BM).

Even if they correctly identify the low-spin state, they may fail to account for the massive spin-orbit coupling in 5d metals.

For a $d^4$ low-spin system, the unquenched orbital angular momentum and positive spin-orbit coupling constant lead to a $J=0$ ground state, drastically quenching the magnetic moment below the spin-only prediction.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Structure of Atom - Class 11 Part 1
Concept	Radial Probability Distribution
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free