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Question #44201

Difficulty: 8/10 | 🟢 Free

The Question

In the context of Molecular Orbital (MO) Theory, the energy level ordering of homonuclear diatomic molecules is dictated by the extent of $s-p$ mixing, which depends on the $\Delta E(ns, np)$ energy gap. Consider the Group 15 diatomic species \ce{N2} and \ce{Bi2} in their ground states. Taking into account relativistic effects and $d/f$ -orbital contractions for heavier elements, identify the symmetry of the Highest Occupied Molecular Orbital (HOMO) for \ce{N2} and \ce{Bi2}, respectively (assuming standard $\Lambda-S$ symmetry labels).

Options

A.
$\sigma_g(2p)$ and $\pi_u(6p)$
B.
$\sigma_g(2p)$ and $\sigma_g(6p)$
C.
$\pi_u(2p)$ and $\sigma_g(6p)$
D.
$\pi_u(2p)$ and $\pi_u(6p)$

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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|d-orbital contraction

Why This Question Is Tricky

Common Mistakes

Students typically memorize that all Group 15 diatomics (like N2) exhibit s-p mixing, leading to a HOMO of sigma_g symmetry.

The trap lies in the hidden variable of relativistic effects and d/f-orbital contraction in Period 6 elements (Bismuth).

This strongly stabilizes the 6s orbital, drastically increasing the ns-np energy gap.

Consequently, s-p mixing is entirely suppressed in Bi2, causing the MO diagram to revert to the 'O2/F2' non-mixed ordering, making the HOMO pi_u instead.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Chemical Bonding and Molecular Structure - Class 11 Part 1
Concept	Molecular Orbital Mixing
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free