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In Graphene, each carbon use 3 electrons to form sp2 bonding with neighboring, and in a unit cell, there are 2 carbon atoms, so at least these 6 electrons contribute to 6 valence bands.

Then my question is, why the two Pz orbital electrons in each unit cell consist one conduction and one valence band, but not both valence bands?

thanks

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The $sp^2$ bonds can be regarded as approximately localised because each bond only involves two carbon atoms. I suppose they will form bands in graphene, but very narrow ones. Anyhow, when two $sp^2$ orbitals interact you get a bonding and an antibonding state. Since each carbon atom contributes one electron you fill the bonding state and leave the anti bonding state empty. This is just your regular covalent bond as in diamond.

By contrast, the remaining $p_z$ orbital interacts with three nearest neighbours, so you'd get (this is all getting a bit arm-waving now!) four bonding and four anti-bonding orbitals. But only four electrons are involved, so only two of the bonding orbitals are filled leaving two empty. When you add in next-nearest neighbours, then next-next-nearest etc the multiple splittings of the bonding orbitals forms an energy band that is only half full. This is your conduction band.

The anti-bonding orbitals will form an empty band some distance above the conduction band. However I believe the bands overlap in graphene so there isn't a band gap.

Needless to say, the electronic structure of graphene is more complicated than the above description suggests, and you shouldn't take the molecular orbital analogy too seriously.

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  • $\begingroup$ As you have commented " four bonding and four anti-bonding orbitals. But only four electrons are involved, so only two of the bonding orbitals are filled leaving two empty", you said there are 4 electrons and 8 orbitals in total, then why you draw the conclusion that only two of the bonding orbitals are filled. By the way, can you explain more about the relation between Pz orbital and bonding anti-bonding orbitals? Thanks! $\endgroup$
    – Lorniper
    Commented Jul 18, 2013 at 12:55
  • $\begingroup$ @user4532: Consider just nearest neighbour interactions: with two C atoms the $p_z$ orbitals split into two MOs. The two electrons go into the bonding MO - so far so good. Bring up a third C atom and the two MOIs split again to give four MOs, but now we have only three electrons. Bring up a fourth C atom and the MOs split again to give eight MOs, each of which can hold two electrons, but there are only four electrons. Hence only two of the eight MOs are full. $\endgroup$
    – John Rennie
    Commented Jul 18, 2013 at 15:52
  • $\begingroup$ See en.wikipedia.org/wiki/Pi-bond for more on how the $p_z$ atomic orbitals form the bonding $\pi$ and anti-bonding $\pi^*$ molecular orbitals. $\endgroup$
    – John Rennie
    Commented Jul 18, 2013 at 15:55
  • $\begingroup$ thank you! so there will be 16-4=12 conduction bands totally? $\endgroup$
    – Lorniper
    Commented Jul 19, 2013 at 7:09
  • $\begingroup$ You're taking my analogy too literally. Remember that there are an effectively infinite number of carbon atoms, not just four. The result is that the molecular orbitals broaden out into a band. The bonding MOs form a partially filled conduction band while the anti-bonding orbitals form an empty valence band. This happens in all solids, however in graphene the two bands overlap. I think you should put the effort into reading around the literature on graphene as I think my simple analogy may be misleading you. $\endgroup$
    – John Rennie
    Commented Jul 19, 2013 at 7:25

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