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Question #44197

Difficulty: 8/10 | 🟢 Free

The Question

In Molecular Orbital Theory, diatomic molecules like N2N_2 and COCO exhibit s-p mixing, rendering the HOMO of COCO weakly antibonding (3σ3\sigma) and its LUMO as 2π∗2\pi^*. When COCO acts as a ligand in homoleptic octahedral complexes, synergic bonding dictates the C-O bond order.

Consider the isoelectronic complexes [Cr(CO)6][Cr(CO)_6] and [Mn(CO)6]+[Mn(CO)_6]^+. The standard reduction potentials are ECr2+/Cr∘=−0.91 VE^\circ_{Cr^{2+}/Cr} = -0.91\text{ V} and EMn2+/Mn∘=−1.18 VE^\circ_{Mn^{2+}/Mn} = -1.18\text{ V}. The anomalously negative E∘E^\circ for Mn reflects the profound exchange energy stabilization of its d-orbitals, which persists in the M(I)M(I) state, significantly lowering the energy of its t2gt_{2g} electrons compared to Cr(0)Cr(0).

Given that the C-O stretching frequency (νCO\nu_{CO}) for free CO(g)CO(g) is 2143 cm−12143\text{ cm}^{-1} and for [Cr(CO)6][Cr(CO)_6] is 2000 cm−12000\text{ cm}^{-1}, identify the most chemically accurate νCO\nu_{CO} for [Mn(CO)6]+[Mn(CO)_6]^+ from the given options.

Options

  • A.

    1860 cm−11860\text{ cm}^{-1}

  • B.

    2000 cm−12000\text{ cm}^{-1}

  • C.

    2090 cm−12090\text{ cm}^{-1}

  • D.

    2190 cm−12190\text{ cm}^{-1}

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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

Molecular Orbital Mixing|E° (M²⁺/M) Anomalies

Why This Question Is Tricky

Common Mistakes

Tests the synthesis of MO theory (s-p mixing in CO), Coordination Chemistry (synergic back-bonding), and Periodic Trends (E∘E^\circ anomalies).

Students often assume a linear trend in d-orbital energies across the 3d series.

The trap is failing to recognize that the anomalous stability of Mn's d-orbitals (evidenced by its E∘E^\circ) drastically lowers the t2gt_{2g} energy, widening the HOMO-LUMO gap for back-bonding.

Another trap is assuming back-bonding is completely absent in cations, which would incorrectly lead to a νCO\nu_{CO} higher than free CO (>2143 cm−1>2143\text{ cm}^{-1}) due to pure σ\sigma-donation from the weakly antibonding 3σ3\sigma HOMO.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterChemical Bonding and Molecular Structure - Class 11 Part 1
ConceptMolecular Orbital Mixing
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree