I am reading Modern Quantum Chemistry by Szabo and Ostlund and on page 62 he says "A minimal basis set for benzene consists of 72 spin orbitals." I tried to understand this number but failed.

Previously he illustrates his concept of a minimal basis set with the Hydrogen molecule $\text{H}_2$, where he starts off with one spacial orbital per atom. Then he combines both orbitals to obtain two properly symmetrized spacial wavefunctions and then multiplies with the two spin states (spin up, spin down) to arrive at a total of four spin orbitals.

In the case of benzene however things get more complicated. We have now 12 atoms that we need to consider and we need to construct properly symmetrized wavefunctions. It is clear that we have more options now, but how do we get to 72?

  • 3
    $\begingroup$ Could Chemistry be a better place for this question? $\endgroup$
    – Danu
    Sep 17, 2014 at 20:38
  • $\begingroup$ I think it might be on topic here, since it is about wavefunctions after all. We'll see what other people say. $\endgroup$
    – David Z
    Sep 17, 2014 at 22:16
  • 1
    $\begingroup$ Note: Chemistry is fine with this question, but so is Physics IMO. $\endgroup$ Sep 18, 2014 at 15:39

1 Answer 1


You have one 1s atomic orbital for each H atom, and one 1s, 2s and three 2p for each C. This makes a total of 36 atomic orbital in the whole molecule, and so you have 72 spin states.

The 2s and 2p's orbitals are going to hibridate giving three 2sp$^2$ orbitals and one 2p orbital, so it has its characteristic $\pi$-delocalized electronic estructure on the ring.


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.