The order of filling of molecular orbitals $\ce{O2}$ by MOT is this :

enter image description here

But from it, how can I deduce that there is one sigma and one pi bond in an oxygen molecule ?

  • $\begingroup$ If I am not mistaken, the $2\sigma_g$ should be higher in energy than $\pi_u^x,\pi_u^y$. $\endgroup$ Commented Nov 18, 2014 at 13:06
  • 1
    $\begingroup$ @martin That's for molecules having electrons less than or equal to 14 , like nitrogen $\endgroup$
    – biogirl
    Commented Nov 18, 2014 at 16:26

1 Answer 1


Overall bond order equals number of electrons that occupy bonding orbitals minus numer of electrons that occupy antibonding orbitals divided by two.

Now you can apply the same concept indivudually:

For s molecular orbitals, both electron pairs "cancel out".

For p molecular orbitals, you have 2 electrons in sigma bonding orbitals and none in sigma antibinding orbitals, which means (2-0)/2 = 1 sigma bond. Also, you have 4 electrons within pi bonding orbitals and only 2 electrons within pi antibonding orbitals, which means (4-2)/2 = 1 pi bond. This yields the result that you expected.

  • $\begingroup$ Actually, there is not a sigma and a pi bond. The MO scheme clearly depicts that - the bonding situation is a little bit more complicated. $\endgroup$ Commented Nov 18, 2014 at 12:37
  • 1
    $\begingroup$ Yeah, right, but that just makes the OP question nonsense. In MO theory there are no individual bonds, just that, molecular orbitals filled with electrons. $\endgroup$ Commented Nov 18, 2014 at 12:47
  • $\begingroup$ @martin I am studying MOT as a part of high school chemistry and this was a question in one of the tests. But actually, are MOT and VBT mutually exclusive theories ? Should I not try to find anything in common ? $\endgroup$
    – biogirl
    Commented Nov 18, 2014 at 16:35
  • $\begingroup$ @biogirl They are not mutually exclusive, quite the opposite is the case. They are complementary. So they are just views from different angles. If you want to know more about this matter I encourage you to ask a new question, it is too extensive to be dealt with in the comments. $\endgroup$ Commented Nov 18, 2014 at 16:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.