Molecular Orbital Theory
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?
Select the correct option:
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About This Question
- Subject
- chemistry
- Chapter
- chemical bonding and molecular structure
- Topic
- molecular orbital theory
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
He2isunstable(bondorder0),He2+isstable(bondorder0.5)
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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