QN.17 Based on the molecular orbital theory, which of the following statements with respect to N₂, O₂ and O₂⁺ is correct?
Option (b) is correct. Molecular orbital theory reveals distinct bonding behaviors in N₂, O₂, and their ions due to differences in orbital energy ordering and electron removal sites.
Bond Orders
N₂ has 14 valence electrons in the configuration KK (σ2s)²(σ2s)²(π2p)⁴(σ2p)², yielding a bond order of (8-2)/2 = 3. N₂⁺ loses one electron from the bonding σ2p orbital (13 electrons), reducing bond order to (7-2)/2 = 2.5. O₂ (16 electrons) follows KK (σ2s)²(σ2s)²(σ2p)²(π2p)⁴(π2p)² with bond order (8-4)/2 = 2, while O₂⁺ (15 electrons) removes one from antibonding π2p, increasing it to (8-3)/2 = 2.5.
Unpaired Electrons
N₂⁺ has one unpaired electron in the σ2p bonding orbital. O₂ contains two unpaired electrons, one each in the degenerate π*2p antibonding orbitals, explaining its paramagnetism.
Bond Lengths
Higher bond order correlates with shorter bonds. N₂⁺ (bond order 2.5) has a longer N-N bond than N₂ (3). O₂⁺ (2.5) features a shorter O-O bond than O₂ (2), around 116 pm vs. 121 pm.
Option Analysis
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(a): Bond order of N₂⁺ (2.5) is lower than N₂ (3), but O₂ (2) is lower than O₂⁺ (2.5)—opposite for N₂. Incorrect.
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(b): Matches exactly—N₂⁺ unpaired in σ (bonding), O₂ in π* (antibonding). Correct.
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(c): N₂⁺ bond longer (not shorter) than N₂; O₂⁺ shorter than O₂. Incorrect.
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(d): N₂ (3) > N₂⁺ (2.5), but O₂ (2) < O₂⁺ (2.5). Incorrect.
Nitrogen (N₂) and oxygen (O₂) molecules showcase molecular orbital theory principles, where atomic orbitals form bonding/antibonding MOs, determining bond order as (N_b – N_a)/2. Key phrase “molecular orbital theory N₂ O₂ O₂⁺” highlights differences: N₂/ions use π2p below σ2p ordering (up to N), while O₂/ions switch to σ2p below π2p due to s-p mixing. This affects electron removal in cations—bonding orbital for N₂⁺ (lowers bond order), antibonding π* for O₂⁺ (raises it)—core to CSIR NET questions.
Key Comparisons
| Species | Valence Electrons | Key MO Loss/Gain | Bond Order | Unpaired e⁻ Orbital | Bond Length Trend |
|---|---|---|---|---|---|
| N₂ | 14 | – | 3 | None | Shortest |
| N₂⁺ | 13 | σ2p (bonding) | 2.5 | σ2p | Longer than N₂ |
| O₂ | 16 | – | 2 | π*2p (2 e⁻) | 121 pm |
| O₂⁺ | 15 | π*2p (antibonding) | 2.5 | π*2p | Shorter, 116 pm |
Higher bond order strengthens bonds, inverting trends between N and O systems for competitive exams like CSIR NET.
