Molecular Orbital Theory

Question

The bond order of O₂⁻ (superoxide ion) calculated using molecular orbital theory is:

RankGuru · Accepted Answer

1. **Count Total Electrons**: O₂ has $2 	imes 8 = 16$ electrons. O₂⁻ has one extra electron, so total = **17 electrons**.
2. **Fill Molecular Orbitals** (increasing energy order for O₂): σ1s², σ*1s², σ2s², σ*2s², σ2p², π2p⁴, π*2p³.
3. **Count Bonding and Antibonding Electrons**:
   - Bonding electrons (in σ2s, σ2p, π2p): $2 + 2 + 4 = 8$ (excluding core 1s which cancel).
   - Considering only valence MOs: Nb = 8, Na = 5 (σ*2s² + π*2p³).
   Wait — let's recount properly from valence shell:
   - σ2s² (bonding) = 2
   - σ*2s² (antibonding) = 2
   - σ2p² (bonding) = 2
   - π2p⁴ (bonding) = 4
   - π*2p³ (antibonding) = 3
   - Nb = 2 + 2 + 4 = 8, Na = 2 + 3 = 5.
4. **Calculate Bond Order**: $BO = \frac{N_b - N_a}{2} = \frac{8 - 5}{2} = 1.5$.
5. **Physical Significance**: Bond order 1.5 indicates a bond strength intermediate between a single and double bond. O₂⁻ is paramagnetic with one unpaired electron in the π* orbitals.

Molecular Orbital Theory

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