Bond Parameters

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The correct order of bond dissociation enthalpy for the oxygen species O₂⁺, O₂, and O₂⁻ is:

RankGuru · Accepted Answer

1. **Calculate Bond Orders Using MOT**:
   - O₂⁺ (15 e⁻): Configuration ends at ...π2p⁴, π*2p¹. Bond order = $\frac{10 - 5}{2} = 2.5$.
   - O₂ (16 e⁻): Configuration ends at ...π2p⁴, π*2p². Bond order = $\frac{10 - 6}{2} = 2.0$.
   - O₂⁻ (17 e⁻): Configuration ends at ...π2p⁴, π*2p³. Bond order = $\frac{10 - 7}{2} = 1.5$.
2. **Relate Bond Order to Bond Dissociation Enthalpy**: Higher bond order implies a stronger bond, which requires more energy to break. Therefore, bond dissociation enthalpy follows the same trend as bond order.
3. **Order**: O₂⁺ (BO = 2.5) > O₂ (BO = 2.0) > O₂⁻ (BO = 1.5).
4. **Experimental Values**: O₂⁺ ≈ 643 kJ mol⁻¹, O₂ ≈ 498 kJ mol⁻¹, O₂⁻ ≈ 395 kJ mol⁻¹, confirming the MOT prediction.
5. **Physical Insight**: Each additional electron enters an antibonding π* orbital, weakening the bond progressively.

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