Band formation in diamond Why is it that in the band diagram of diamond does the sp3 hybridised orbital split into two forming valence band and conduction band?
I thought that these sp3 hybridised orbital forms 4n bonding and 4n antibonding orbital accommodating 8n electrons but it turn out that valence band accommodates 4n and conduction band accommodates 4n electrons....
 A: (Starting from the molecular orbital picture.)
If you check up on the role of anti-bonding states/orbitals e.g., here you'll note that these states "increase" the energy of the atomic pair. This means, it is less stable than two separate atoms. In contrast, bonding states lower the energy (=increase stability). Looking at the primitive picture of a band diagram as a solid state equivalent of molecular states, the energy of the system is obtained by summing the energies of the contributing parts (the electrons for our current purpose). The most stable configuration is the one with the lowest energy, which means, the energy of the electrons should be as low as possible. Ergo, filling up from the bottom, we create maximally filled orbitals.
Pointing our attention to C (and diamond), then the 2s2p orbitals contain 4 electrons per C. Of the 8 sp3 hybridized orbitals[**], 4 are occupied with 8 electrons (2 electrons per orbital as usual, 1 from the center C and 1 from the C it binds to), and 4 are unoccupied (the anti-bonding states). The former will have an energy below the Fermi-level, while the latter have an energy above the Fermi-level.
This however, is still a very simplified picture. It only explains that only half the bands should be occupied. What their energies should be is a more complex story. Also the nature of the actual band-gap is beyond the scope of what can be explained with molecular orbitals.
[**] Note that there is a conservation of orbitals/states. The 8 states are a reorganization of 4 states of the central C, and 4x1 state of the surrounding 4 C atoms.
