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Question #44259

Difficulty: 8/10 | 🟢 Free

The Question

A 3d-block transition metal $M$ possesses an anomalous ground-state electronic configuration to maximize exchange energy, having a fully filled penultimate d-subshell. In an aqueous acidic medium at 298 K, its monovalent cation $M^+$ undergoes a spontaneous disproportionation reaction to yield a solid precipitate and a colored cation $X$ . Let $n_r$ be the number of radial nodes in the orbital from which the first ionization of $M$ occurs. Let $E$ be the equivalent weight of $M^+$ in this reaction (where $M_0$ is the molar mass of $M^+$ ), and $\mu$ be the spin-only magnetic moment of $X$ in Bohr Magnetons (BM). Which of the following correctly represents the set $(n_r, E, \mu)$ ?

Options

A.
$(3, 2M_0, \sqrt{3})$
B.
$(0, \frac{M_0}{2}, \sqrt{3})$
C.
$(3, \frac{M_0}{2}, \sqrt{8})$
D.
$(0, 2M_0, 0)$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Aufbau Principle Exceptions|Equivalent Concept in Redox|Disproportionation Constraints

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of atomic structure (Aufbau exceptions and radial nodes) with redox stoichiometry.

Students often incorrectly assume the first electron is removed from the 3d subshell for transition metals, leading to an incorrect radial node count of 0.

Furthermore, calculating the equivalent weight of a disproportionating species is a classic trap; students frequently use $E = M_0 / (n_{ox} + n_{red}) = M_0/2$ instead of the correct derivation $E = M_0/n_{ox} + M_0/n_{red} = 2M_0$ , failing to account for the stoichiometric ratio of electron transfer.

Quick Facts About This Question

Question Information

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