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MOT configurations of O2 and O2+ are given below. Determine the number of unpaired electrons in each case.

1. O2+: (σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (σ2pz)2 (π2px = π2py)2 (π*2px = π*2py)1
2. O2: (σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (σ2pz)2 (π2px = π2py)2 (π*2px = π*2py)2
3. Number of unpaired electrons = 2, paramagnetic.
4. Number of unpaired electrons = 1, paramagnetic.

Correct answer: Number of unpaired electrons = 1, paramagnetic.

Solution

O2 has 2 unpaired electrons in π* orbitals, making it paramagnetic. O2+ loses one electron, leaving 1 unpaired electron in π* orbitals, also paramagnetic.

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