Molecular Orbital Theory
According to Molecular Orbital Theory, the bond order of O_2^{2-} (peroxide ion) and whether it is paramagnetic or diamagnetic are respectively:
Select the correct option:
Solution
1, diamagnetic
Molecular Orbital Theory (MOT) builds molecular orbitals by linear combination of atomic orbitals. For O_2^{2-} (peroxide ion), oxygen has 8 electrons, so two oxygen atoms contribute 16 electrons, and the 2^- charge adds 2 more, giving 18 total electrons. The MO filling sequence for the second-period homonuclear diatomics is: \sigma(1s)^2, \sigma^(1s)^2, \sigma(2s)^2, \sigma^(2s)^2, \sigma(2p_z)^2, \pi(2p_x)^2, \pi(2p_y)^2, \pi^(2p_x)^2, \pi^(2p_y)^2. With 18 electrons, both \pi^(2p_x) and \pi^(2p_y) are completely filled (two electrons each). Bond order = (bonding electrons - antibonding electrons)/2 = (10 - 8)/2 = 1. Since all 18 electrons are paired (no unpaired electrons remain), O_2^{2-} is diamagnetic. Option '2, paramagnetic' describes neutral O_2 (16 electrons, two unpaired electrons in degenerate \pi^* orbitals). Option '1, paramagnetic' would imply unpaired electrons at bond order 1, which contradicts the completely filled \pi^* orbitals. Option '2, diamagnetic' overestimates the bond order for 18 electrons. This is a frequently tested NCERT MOT concept in JEE. Plausibility check: bond order of 1 is consistent with the known O-O single bond in H_2O_2, where the peroxide unit also has bond order 1.
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About This Question
- Subject
- chemistry
- Chapter
- chemical bonding and molecular structure
- Topic
- molecular orbital theory
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
1, diamagnetic
Molecular Orbital Theory (MOT) builds molecular orbitals by linear combination of atomic orbitals. For O_2^{2-} (peroxide ion), oxygen has 8 electrons, so two oxygen atoms contribute 16 electrons, and the 2^- charge adds 2 more, giving 18 total electrons. The MO filling sequence for the second-period homonuclear diatomics is: \sigma(1s)^2, \sigma^(1s)^2, \sigma(2s)^2, \sigma^(2s)^2, \sigma(2p_z)^2, \pi(2p_x)^2, \pi(2p_y)^2, \pi^(2p_x)^2, \pi^(2p_y)^2. With 18 electrons, both \pi^(2p_x) and \pi^(2p_y) are completely filled (two electrons each). Bond order = (bonding electrons - antibonding electrons)/2 = (10 - 8)/2 = 1. Since all 18 electrons are paired (no unpaired electrons remain), O_2^{2-} is diamagnetic. Option '2, paramagnetic' describes neutral O_2 (16 electrons, two unpaired electrons in degenerate \pi^* orbitals). Option '1, paramagnetic' would imply unpaired electrons at bond order 1, which contradicts the completely filled \pi^* orbitals. Option '2, diamagnetic' overestimates the bond order for 18 electrons. This is a frequently tested NCERT MOT concept in JEE. Plausibility check: bond order of 1 is consistent with the known O-O single bond in H_2O_2, where the peroxide unit also has bond order 1.
This medium 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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