Molecular Orbitals
What about homonuclear diatomics of the next row of the Periodic Table? We will start with F2 and work backwards, for reasons which will become clear shortly. We now have available on each atom three 2p orbitals; in addition we have 2s which will interact with each other, as did 1s for di-hydrogen, forming s and s*. Conventionally, we define the 'special' direction joining F to F as the z-axis. It is only useful to consider overlap of orbitals where wave functions match. (+ to + or - to - to form bonding interactions - remember these signs do NOT represent charges) or totally mis-match (+ to - to form antibonding interactions).
A partial match, partial mis-match (e.g. sideways s to p) is net non-bonding, and not permitted.
It is essential to match angular nodal planes. The two 2px and the two 2py orbitals each overlap to form bonding and antibonding interactions, but these are now pi (p) molecular orbitals: one pair is illustrated, the other occupies the direction perpendicular to the plane of the diagram.
We can illustrate all these interactions on a molecular orbital, energy level diagram which shows the ss and ss* molecular orbitals already filled with the first four electrons.
We put the valence orbitals of the two atoms (in this case the n=2 orbitals) to the left and to the right of an energy level diagram (in which potential energy increases upwards), and fill in the gap in between with the molecular orbitals produced by their overlaps. Initially, we can keep the *s and *p orbitals separate, although, since both are * in type, they could mix (hybridise). We now add the remaining ten ( [2 x7] -4) valence electrons of F2 according to Hund's Rule .
For the purpose of calculating bond order, the doubly occupied ss and ss* cancel out (just like in He2). Similarly, the two doubly-occupied p and the two doubly-occupied p* orbitals cancel out, leaving a doubly occupied sp and thus a single bond between fluorines in F2. Using the same diagram, Ne2 would have all molecular orbitals doubly occupied, leading to (essentially) zero bond energy and monatomic Ne, not Ne2 (just like He2).
If you consider O2, there are two fewer electrons than in F2, thus the two p* orbitals (which are degenerate) will each be singly occupied, with electrons of parallel spin (Hund's Rule): thus O2 is paramagnetic . Since the two p* orbitals are each only singly occupied, only half the p-bond energy is cancelled, leaving one net s-bond (as in F2) and the energy equivalent of one net p-bond: thus O2 can be considered to have a double bond, although, as you see, it is not quite as simple as that!
The molecule N2 would lack the two electrons in the p* orbitals and would therefore retain the full two p-bonds, along with the s-bond, and be triple-bonded. Since it no longer has unpaired electrons, it would be diamagnetic . However, there is now the need to amend the diagram somewhat. In the case of F2 and O2, the energy gap between 2s and 2p orbitals (specifically the 2pz) was sufficiently large for it to be justifiable to ignore any mixing between these orbitals, both of which form s-bonding molecular orbitals. From di-nitrogen downwards to the transient species C2 and B2, the omission of such mixing leads to an error in the final energy level diagram, although it does not affect the bond order of N2. The effect of this on the final energy level diagram is to cause ss and sp (which mix) to 'push apart' in energy: likewise ss* and sp* mix and 'push apart'. Thus ss is lowered and sp* is raised at the energetic extremities of the diagram, which has no effect on the order of the molecular orbitals: similarly ss* is lowered, but only towards an already falling ss. The only significant change occurs when the sp is raised sufficiently to lie above the doubly degenerate p-bonding molecular orbitals. Thus the highest occupied molecular orbital (HOMO) in N2 is actually the sp orbital.
