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Molecular Orbital Theory

Hardchemistry

When comparing the stability of He_2 versus He_2^+ using Molecular Orbital Theory, which statement is correct and what are their respective bond orders?

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About This Question

Subject
chemistry
Chapter
chemical bonding and molecular structure
Topic
molecular orbital theory
Difficulty
Hard
Year
2025
Tags
molecular orbital theoryHe2 bond orderantibonding orbitalnoble gas diatomicsMO stability

Solution

Correct Answer:

Molecular Orbital Theory predicts the stability of diatomic molecules by comparing the number of electrons in bonding versus antibonding orbitals. For He_2: each He atom has 2 electrons, giving 4 total electrons. These fill \sigma(1s)^2 \sigma^(1s)^2. Bond order = (2 bonding - 2 antibonding)/2 = 0/2 = 0. A bond order of 0 means no net bonding interaction exists, so He_2 cannot exist as a stable molecule. This is consistent with the fact that helium is a noble gas with no tendency to form covalent bonds. For He_2^+: this cation has 3 electrons (4 from two He atoms minus 1 for the positive charge). Electron configuration: \sigma(1s)^2 \sigma^(1s)^1. Bond order = (2 - 1)/2 = 0.5. A positive bond order means there is a net stabilising interaction, so He_2^+ can exist as a transient species with one electron in the antibonding orbital but more electrons in the bonding orbital. Option 1 wrongly assigns bond order 1 to He_2 and 0 to He_2^+, reversing reality. Option 3 incorrectly claims both have bond order 0; He_2^+ does have bond order 0.5. Option 4 reverses the bond orders. This question tests a nuanced MOT concept examined in JEE Advanced. Plausibility: He_2^+ has been experimentally detected in discharge tubes, confirming bond order > 0, while He_2 has never been observed as a stable molecule.

This hard difficulty chemistry question is from the chapter chemical bonding and molecular structure, covering the topic of molecular orbital theory. It appeared in the 2025 exam.

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