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Question #44279

Difficulty: 8/10 | 🟢 Free

The Question

In the framework of advanced Molecular Orbital (MO) theory, the energy ordering and mixing of atomic orbitals dictate the bond characteristics of diatomic species. Consider the isoelectronic molecules \ce{N2} and \ce{CO}. Due to the electronegativity asymmetry in \ce{CO}, the nature of its Highest Occupied Molecular Orbital (HOMO) deviates significantly from the strongly bonding HOMO of \ce{N2}.

Let $b_1, b_2, b_3,$ and $b_4$ represent the formal bond orders of the isolated monocations \ce{CO+}, \ce{N2+}, \ce{NO+}, and \ce{O2+} respectively.

Determine the exact value of the expression: $\Phi = 2b_1 + 4b_2 + b_3 + 2b_4$

Options

A.
25
B.
23
C.
24
D.
27

Check Your Answer

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

Isoelectronic Energy Ordering|Molecular Orbital Mixing|Bond Order in Resonance Hybrids

Why This Question Is Tricky

Common Mistakes

The core trap lies in the 'Chemical Exception' of the \ce{CO+} bond order.

Students heavily rely on the shortcut 14-electron rule (where 14e- yields BO=3, and 13e- yields BO=2.5).

Applying this blindly to \ce{CO} (14e-) -> \ce{CO+} (13e-) yields an incorrect bond order of 2.5.

However, due to the significant electronegativity difference between C and O, the $5\sigma$ HOMO is weakly antibonding.

Ionization removes an electron from this antibonding orbital, thereby increasing the bond order to 3.5.

This question effectively filters students who memorize the electron-counting trick from those who understand the underlying MO energy level diagrams and orbital mixing.

Quick Facts About This Question

Question Information

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