# Questions in MO theory

Τhe electrons in the antibonding MO are the unpaired electrons of the molecule? E.g. $$\ce{O2}$$, by writing down the diagram for the MO we notice that there are only 2 electrons in the top antibonding orbitals thus 2 unpaired electrons, yet by writing down the Lewis structure, the molecule turns out to have 0 unpaired electrons.. So, what do the electrons in the anti bonding orbitals do if they don't represent the unpaired electrons of the molecule?

• Refrain from merging multiple questions into one. Dec 25, 2019 at 17:30
• For second part, how did you get two $\sigma$ bonds? Dec 25, 2019 at 17:46

The antibonding orbital doesn't necessarily tell you the number of unpaired electrons - it's coincidental that for O$$_2$$, the HOMO (highest occupied molecular orbital) is a (1$$\pi_g)^2$$ orbital - or, if you prefer to write it like you have for N$$_2$$, ($$\pi ^1 _{2px} = \pi ^1 _{2py}$$). $$\pi$$ orbitals are doubly degenerate, so the lowest energy configuration is for the 2 electrons to occupy different orbitals (one in each of the 2 $$\pi$$ orbitals). This means they're both unpaired, as shown in this MO diagram of O$$_2$$ from Wikipedia. For an example of an MO diagram where there are unpaired electrons in an orbital which is not antibonding, try drawing the B$$_2$$ MO diagram (remember that s-p mixing occurs).
In terms of what they do, antibonding electrons destabilise the bond, "cancelling out" the energetically favourable effect of bonding electrons - try drawing an MO diagram for He$$_2$$ or Ne$$_2$$. It should be apparent why they don't form stable molecules.
As far as N$$_2$$ goes, I think you may have neglected to take into account the effect of the $$\sigma ^* _{2s}$$ electrons. Remember that antibonding electrons cancel out bonding electrons.