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Question:

Which of the following options represents the correct bond order?

O2⁻>O2>O₂⁺

O2⁻>O2<O₂⁺

O2⁻O₂⁺

O2⁻<O2<O₂⁺

Solution:

Bond Order = (Bonding electrons - Antibonding electrons) / 2
The electronic configuration of the O₂ containing 16 electrons can be written as:
1σ²₁s 1σ²₁s 2σ²₂s 2σ²₂s 2pz² 2px² 2py¹ 2py¹
Bond Order = (8 - 4) / 2 = 2
The electronic configuration of the O₂⁺ containing 15 electrons can be written as:
1σ²₁s 1σ²₁s 2σ²₂s 2σ²₂s 2pz² 2px² 2py¹
Bond Order = (8 - 3) / 2 = 2.5
The electronic configuration of the O₂⁻ ion containing 17 electrons can be written as:
1σ²₁s 1σ²₁s 2σ²₂s 2σ²₂s 2pz² 2px² 2py² 2py¹
Bond Order = (8 - 5) / 2 = 1.5
O₂⁻ contains 3 electrons in the anti-bonding orbital. It requires less energy to remove an electron from the anti-bonding orbital.
Hence, the correct bond order in the following species is O₂⁺ > O₂ > O₂⁻.
Therefore option D is correct.