Determining # of Antibonding Orbitals & More?

Question

I'm really confused.

The # of MO are obtained by counting the atomic orbitals; for every atomic orbital combined, one MO results.

So here's my question (s):

1) To determine the # of MO, which atomic orbitals do we count? The valence atomic orbitals? The atomic orbitals with only 1 e-?

2) Once we have the # of MO, how do we determine how many electrons are in either a bonding or antibonding orbital?

For instance, how would I determine that there are 3 bonding pairs and 1 antibonding pair in C2?

I'm really confused, any help would be appreciated. Thank you.

Answer 1

1) Only the valence orbitals matter, and all the valence orbitals count (more complicated version: all valence orbitals which are of the correct symmetry to hybridize to form MO. If you don't know what that means then you're in a lower level class and it doesn't matter). # of atomic orbitals = # of molecular orbitals, and the molecular orbitals may be bonding, nonbonding, or antibonding. Not depending on the particular class you're taking your teacher may or may not want you to draw the non-valence orbitals, but they don't matter at all.

2) You start filling all the MOs with electrons, starting with the lowest energy and moving up.

3) You wouldn't determine that for C2 because that's not correct. The bonding and antibonding orbitals formed from the 2s atomic orbitals would both be filled (net bond order of zero) the lowest bonding (sigma) MO from the 2p AO would be filled, and each of the higher bonding (pi) MO from the 2p AO would have one electron. None of the antibonding orbitals from the 2p AO would be filled.

This predicts that C2 would have 2 unpaired electrons, whereas a Lewis structure would predict a quadruple bond between the carbons. One guess as to which one is right....