# Conduction band and antibonding orbitals

1. Jan 29, 2010

### bearcharge

I'm not from the chemistry major but I need to learn solid state chemistry out of the need of my research. I viewed solid state chemistry lectures online delivered by professor Donald Sadoway from MIT. Particularly on the topic of conductors, insulators, semi-conductors, I have some confusions.

Is it safe to have the following conception?
conductors, insulators and semi-conductors can be distinguished by their respective band gap. This band gap is the least amount of energy required to excite an electron to conduction band level. In energy diagram, this is the gap between the highest level in valence band and lowest level in conduction band. As soon as the electron reaches the conduction band level, it's free to move around.

If the above statement is right, then I have some questions:
what's the relationship between antibonding and conduction band? are they the same or one contains the other? they don't look like the same, because in some cases, the electrons will not only fill the bonding orbitals, but also part of even the whole antibonding orbitals, like in magesium. Then I'm even more confused, because if the energy level above the top of valence band is not 0, then how can electron move freely in that level? For example, in the case of magnesiu, the orbital above valence band should be p orbital, how can electrons move freely here?

As I said, I'm not from the chemistry major, so please forgive me if I made rookie mistakes :)

2. Feb 9, 2010

### ykent

I am not from chemistry either, but I did some research on bonding analysis in solid state system using a downfolding scheme called QUAMBO. So I would like to address some points as far as I know.

3. Feb 10, 2010

### DrDu

The relation between anti-bonding orbitals and the conduction band is not so direct.
A good reading which explains the bonding in solids from the angle you are interested in is the book by Walter A. Harrison, "Electronic structure and the property of solids", Dover publ.

4. Feb 10, 2010

### knghrts17

I agree with the previous post, the Harrison text is a great reference to clarify these issues. To a first approximation, one can assume that the antibonding state and conduction band are equivalent. As a qualitative example of this consider the following:

Consider two hydrogen atoms approaching each other along the x axis. As you bring the two closer together, the s-orbitals of each overlap, resulting in the well known $$\sigma$$ and $$\sigma$$* states. These MO's can also be viewed as symmetric ($$\sigma$$) or antisymmetric ($$\sigma$$*) linear combinations of the 1s orbitals (The so-called LCAO approach). Assume we bring in another hydrogen x axis. This gives rise to
an additional MO - an antisymmetric non-bonding orbital, with an energy situated somewhere between the the symmetric bonding and antibonding orbital. A fourth hydrogen gives an additional orbital, situated below the antibonding orbital and the antisymmetric non-bonding orbital.

No consider you have a chain of N hydrogen atoms. Certainly, there will be a bonding and antibonding state. However, there will also be bands associated with the other N-2 hydrogens. As it happens, these bands will be extremely close together, so close that they can be considered continuous and hence bands.

Its important to note however, that the previous description is purely qualitative. The band diagrams of which you are interested in actually plots energy v. the reciprocal lattice vector k. k is not incorporated in the previous discussion. This comes from wavefunctions composed of Bloch functions used to treat the periodicity inherent in a crystal

Hope this helps

5. Feb 10, 2010

### DrDu

What I meant was the following: In a molecular crystal, like one made up from H_2 molecules, it makes approximately sense to consider the valence and conduction band as made up of bonding and antibonding wavefunctions, respectively. However, in a metal, there are several nearest neighbours which have the same or nearly the same distance from a given metal atom. It is then normally not possible to imagine the band as built up from purely bonding or anti-bonding localized orbitals, because the best localized orbitals still will extend over several atoms and will have partly bonding and partly antibonding character. This is somehow related to the bonding in benzene where the bonds are also delocalized over all the atoms.

6. Feb 10, 2010

### ykent

But in benzene, it is still OK to think of it as pi bond and pi*(anti-bonding) states. I think the language of bonding and antibonding are are related with a representation of valence bond theory. While in molecular orbital theory, people do not talk about bonding and antibonding stuff. Different representation may give same physical result anyway. But they may be adapted for different situations.

7. Feb 10, 2010

### Kracatoan

http://img535.imageshack.us/img535/8942/orbitals.png [Broken]

As this diagram of nanotubes shows, the pi orbitals (the '8' shapes around each point of intersection, these being atoms, are Pi orbitals) can overlap, creating one really big orbital. This conducts electricity because it effectively extends the orbital across the whole material, so instead of the electrons 'hopping' from orbital to orbital, they can move with much more ease becuase it is just like ordinary movement within any other orbital, but on a larger scale.

Last edited by a moderator: May 4, 2017
8. Feb 12, 2010

### bearcharge

Thanks! I think I'm much more clear from your explaination. Indeed, electrons in metals are delocalized, therefore electron orbitals have partly bonding and partly antibonding character. I totally agree with this. Thanks!

9. Feb 12, 2010

### bearcharge

very vivid explaination! Thanks!

Last edited by a moderator: May 4, 2017
10. Feb 22, 2010

### Modey3

bearcharge,

The terms "bonding" and "antibonding" molecular orbitals are routinely used when describing electronic interactions using LCAO (linear combination of atomic orbital) approaches. In reality, these molecular orbitals are solutions to the molecular Schrodinger equations. For example consider the 2s orbital. What distinguishes a bonding from an anti-bonding 2s-orbital is whether that orbital-energy is higher (anti-bonding) or lower (bonding) than some reference orbital-energy, which is the 2s orbital-energy of the isolated atom when considering the formation of HCP Mg. When talking about electron-transport properties what really matter is if there is a band-gap between the highest unoccupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO). Whether an orbital is considered a bonding or anti-bonding orbital is irrelevant.

modey3

11. Feb 22, 2010

### bearcharge

well explained! Thanks!
I think with what Kracatoan had previously explained about the conductivity of pi orbitals, I'm getting a more clear picture about the band gap, conductivity and relative concepts.
Modey3 has explained well about the concept of band-gap and its relevance to conductivity. As for the conductivity of unoccupied molecular orbitals, i.e, its capability of letting electrons move around freely, explaination of the pi orbitals by Kracatoan gives me some idea. I think we can say electrons do move freely in unoccupied molecular orbitals, but the mechanism may be not so easy to illustrate.

12. Feb 22, 2010

### Modey3

bearcharge,

Pi-orbitals are do to the interaction of un-hybridized p(z) orbitals between carbon atoms. They form pi-orbital bands in the x,y directions of the nanotube. During the bonding process, these orbitals have a "weak interaction" and as a result there is a "anti-bonding" pi-orbital-band symmetry just above a "bonding" pi-orbital-band.