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Question #44585

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The Question

In the 3d transition series, metal M1M_1 exhibits an anomalously negative standard reduction potential (EM2+/M∘E^\circ_{M^{2+}/M}) primarily due to its exceptionally high negative hydration enthalpy. Metal M2M_2 also shows a sharp negative discontinuity in its EM2+/M∘E^\circ_{M^{2+}/M} trend, but this is driven by the low sum of its first two ionization enthalpies associated with a stable half-filled dd-subshell.

Aqueous solutions of M12+M_1^{2+} and M22+M_2^{2+} are treated with excess \ce{KCN(aq)} under an inert atmosphere at 298 K to form the major complex ions PP and QQ, respectively.

Which of the following correctly identifies the spin-only magnetic moment (μ\mu) of PP, the spin-only magnetic moment of QQ, and the Crystal Field Stabilization Energy (CFSE) of QQ? (Assume Δo>P\Delta_o > P for octahedral complexes. Express CFSE in terms of the octahedral splitting energy Δo\Delta_o and the mean pairing energy PP).

Options

  • A.

    μP=0\mu_P = 0 BM, μQ=1.73\mu_Q = 1.73 BM, CFSEQ=−2.0Δo+2P\text{CFSE}_Q = -2.0 \Delta_o + 2P

  • B.

    μP=2.83\mu_P = 2.83 BM, μQ=5.92\mu_Q = 5.92 BM, CFSEQ=0Δo\text{CFSE}_Q = 0 \Delta_o

  • C.

    μP=0\mu_P = 0 BM, μQ=1.73\mu_Q = 1.73 BM, CFSEQ=−2.0Δo+3P\text{CFSE}_Q = -2.0 \Delta_o + 3P

  • D.

    μP=2.83\mu_P = 2.83 BM, μQ=1.73\mu_Q = 1.73 BM, CFSEQ=−2.0Δo+2P\text{CFSE}_Q = -2.0 \Delta_o + 2P

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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

Spin-Only Magnetic Moment|E° (M²⁺/M) Anomalies|Crystal Field Splitting (Octahedral)

Why This Question Is Tricky

Common Mistakes

This question tests the thermodynamic origins of E∘E^\circ anomalies in the 3d series and their intersection with Coordination Chemistry.

Students often memorize the E∘E^\circ trend but fail to distinguish between the hydration energy driver (Ni) and the ionization energy driver (Mn).

The 'Lethal Collision' occurs when predicting the complexes: students frequently assume Ni2+Ni^{2+} forms an octahedral \ce{[Ni(CN)6]^{4-}} complex (yielding μ=2.83\mu = 2.83 BM) instead of the sterically and electronically favored square planar \ce{[Ni(CN)4]^{2-}} (μ=0\mu = 0 BM).

Furthermore, they must correctly calculate the pairing energy contribution to the CFSE of the low-spin d5d^5 \ce{[Mn(CN)6]^{4-}} complex by subtracting the zero pairs of the free Mn2+Mn^{2+} ion from the two pairs in the t2g5t_{2g}^5 configuration.

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Question Information

SubjectChemistry
ChapterCoordination Compounds - Class 12
ConceptSpin-Only Magnetic Moment
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree