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Question #44689

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The Question

The isoelectronic diatomic species \ce{N2}, \ce{CO}, and \ce{NO+} all exhibit a formal bond order of 3 according to standard homonuclear and heteronuclear Molecular Orbital (MO) theory. However, their effective bond orders—and consequently their stretching frequencies ( $\nu$ )—deviate significantly when subjected to different electronic environments, such as formal charge variations, $\pi$ -backbonding, and resonance delocalization.

Consider the effective \ce{N-O} or \ce{C-O} bond orders in the following species: (I) \ce{NO+} (II) Free \ce{CO} (III) \ce{[Ti(CO)6]^{2-}} (IV) \ce{CO3^{2-}}

Which of the following options correctly represents the decreasing order of their effective bond orders, and accurately identifies the underlying electronic rationale?

Options

A.
I > II > III > IV; Strong $d\pi \rightarrow \pi^*$ back-donation from \ce{Ti(-II)} populates the \ce{CO} antibonding MO, reducing its bond order, while \ce{CO3^{2-}} has a fractional bond order of 1.33 due to resonance.
B.
II > I > III > IV; The formal positive charge on \ce{NO+} withdraws electron density from the bonding MOs, reducing its effective bond order compared to neutral \ce{CO}.
C.
I > II > IV > III; Extreme back-donation from the highly electron-rich \ce{Ti(-II)} center reduces the \ce{C-O} bond order below the 1.33 value of the resonance-stabilized \ce{CO3^{2-}}.
D.
I > III > II > IV; Synergistic $\sigma$ -donation from \ce{CO} to \ce{Ti(-II)} removes electron density from the slightly antibonding $5\sigma$ orbital, increasing the \ce{C-O} bond order above free \ce{CO}.

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Isotopes-Isobars-Isotones-Isoelectronic|Bond Order in Resonance Hybrids|Resonance Energy and Stability

Why This Question Is Tricky

Common Mistakes

This problem tests the limitations of simple MO theory by forcing students to compare the effective bond orders of isoelectronic species (NO+ vs CO) and species in complex environments (metal carbonyls vs resonance hybrids).

The core trap is twofold: (1) Assuming extreme pi-backbonding in a highly reduced metal center like Ti(-II) reduces the C-O bond order below that of a resonance-stabilized carbonate ion (BO=1.33), and (2) Misapplying the synergistic bonding model by assuming sigma-donation from the 5-sigma orbital dominates over pi-backbonding in electron-rich complexes.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Structure of Atom - Class 11 Part 1
Concept	Isotopes-Isobars-Isotones-Isoelectronic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free