Explanation
The molecular orbital configuration of O₂ (16 e⁻) molecule will be,
(σ1s)² (σ1s*)² (σ2s)² (σ2s*)² (σ2pz)² (π2px)² = (π2py)² (π2px*)¹ = (π2py*)¹.
The molecular orbital configuration of O₂²⁺ (14 e⁻) molecule will be,
(σ1s)² (σ1s*)² (σ2s)² (σ2s*)² (σ2pz)² (π2px)² = (π2py)²
Thus, we can see that, removing an electron from O₂²⁺ will require more energy as compared to removing an electron from O₂ because in the case of O₂²⁺, the electron to be removed is paired but in O₂ two electrons are unpaired.
Hence, option A is correct. The order of first ionization potential is O₂²⁺ > O₂ .
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