According to the MOT diagram there are no unpaired electrons but in an exam I gave the answer key said it has 2 unpaired electrons. I can also find some images on google where it has 2 unpaired electrons.
Which one is correct and why?
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$\begingroup$ @santimirandarp I believe, in the above MO, the BO is 2. $\endgroup$ – William R. Ebenezer May 17 '19 at 12:28
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1$\begingroup$ @Random, peroxide is diamagnetic. So the provided MO is correct. Oxygen ($\ce{O2}$) has 2 unpaired electrons, but $\ce{O2^2-}$ hasn't. $\endgroup$ – William R. Ebenezer May 17 '19 at 12:31
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The MO diagram you show is correct. $O_2^{2-}$ has a bond order of 1 (equivalent to a single bond between the two O atoms) as the antibonding $\pi^*$ orbitals contain 4 electrons and this cancels the $\pi$ orbitals' 4 bonding electrons. See the Mo diagrams for the different $O_2$ species below:
https://www.chem.uci.edu/~lawm/Ch%205%20Solutions.pdf
The Lewis diagram is incorrect. It shows the O atoms with an expanded octet (not allowed for period 2 elements by the rules of Lewis diagrams) and it shows the bond between the two O atoms as a double, rather than the measured single O-O bond length:
http://www.wiredchemist.com/chemistry/data/oxygen-selenium-compounds
It also shows the peroxide ion as having unpaired electrons, when it is actually diamagnetic. It should look like this:
