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Question #44594

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The Question

Consider a 3d-transition metal $M$ that exhibits the following properties at $298 \text{ K}$ : I. Its standard reduction potential $E^\circ_{M^{2+}/M}$ is anomalously more negative than the element immediately preceding it in the periodic table. II. The hexaaqua complex of $M^{2+}$ is faintly colored, exhibiting an extremely low molar extinction coefficient ( $\epsilon < 1 \text{ L mol}^{-1} \text{ cm}^{-1}$ ) in the visible region. III. The third ionization enthalpy ( $\Delta_i H_3$ ) of $M$ is exceptionally high compared to the element immediately following it.

Which of the following statements regarding $M$ and its compounds is INCORRECT?

Options

A.
The anomalous $E^\circ_{M^{2+}/M}$ value is primarily driven by the exceptionally high (more exothermic) hydration enthalpy of the $M^{2+}$ ion.
B.
The faint color of $[M(H_2O)_6]^{2+}$ is a consequence of the d-d transitions being both Laporte-forbidden and spin-forbidden.
C.
In the presence of excess cyanide ions, $M^{2+}$ forms an octahedral complex with a spin-only magnetic moment of approximately $1.73 \text{ BM}$ .
D.
The metal $M$ has a lower enthalpy of atomization compared to the elements immediately preceding and succeeding it in the 3d series.

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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 d-d Transitions|E° (M²⁺/M) Anomalies|Crystal Field Splitting (Octahedral)

Why This Question Is Tricky

Common Mistakes

The question tests the interplay between thermodynamic stability and spectroscopic properties of 3d transition metals.

Students often assume that a highly stable oxidation state (like $Mn^{2+}$ ) must be driven by a highly exothermic hydration enthalpy.

However, the 'Hidden Variable' here is the Crystal Field Stabilization Energy (CFSE).

For a high-spin $d^5$ complex, CFSE is zero, making the hydration enthalpy less exothermic than expected.

The true drivers for the anomalous $E^\circ$ are the low enthalpy of atomization (due to weak metallic bonding from stable half-filled orbitals) and the low ionization enthalpies.

Furthermore, it tests the selection rules for UV-Vis spectroscopy, linking the thermodynamic $d^5$ stability to the spin-forbidden nature of its electronic transitions.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Coordination Compounds - Class 12
Concept	Color and d-d Transitions
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free