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For the case of metals, we observed that metals often have nearest-neighbours in excess of the maximum valency (for example, 8 for Li, which has only one valence electron) and that metals display great conductivity. On this basis, it was suggested that electrons are delocalised over the solid, and we used LCAO (linear combination of AO) over all the valence orbitals to form a band - the 2s band in this case.

The above is from section two in MIT's solid state lecture notes here. They talk about the above two points suggesting an atom in a metal lattice can interact with a large number of others. This then leads to thinking in terms of the orbitals of the "system" or the entire "macromolecule"

However, for non-metallic solids, like diamond for example, bonds are viewed as sp3 localised bonds and the number of nearest neighbours equals how many covalent bonds it can form, and it has poor conductivity as well. The poor conductivity is often explained by saying the band gap($E_g$) to the conduction band is large. But I don't understand the meaning of forming a band in this context, when the assumption we used of delocalised electrons over the entire solid doesn't hold.

Also, the description for the formation of the valence and conduction band for diamond is given in terms of sp3 orbitals combining to the form the bands. Why is hybdridisation (from VBT/VSEPR) being brought in to the picture here?

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Existence of band structure is the result of Bloch theorem, which predicts that electron spectrum in a periodic potential consists of continuous patches, states being label by a continuous number/vector $\mathbf{k}$ and a discrete band index $n$. Thus, the existence of bans is the consequence of crystal structure.

Excess of valency comes into play when distinguishing metals and semiconductors - in metals it results in the last band typically not completely filled, which means that there are mobile electrons present. On the other hand, in diamond structure, characteristic of Carbon, Silicon, and Germanium, all the bonds are saturated and we can distinguish last fully filled band (valence band) and the first empty band (conduction band). Similar situation takes place for close elements, as in GaAs or AlGaAs - which are combinations of elements of valencies three and five.

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    $\begingroup$ But why is assuming electron delocalisation justified for nonmetals. For metals, for example, there are those two points that suggest that electrons are delocalised over the material. $\endgroup$
    – xasthor
    Commented Feb 20, 2023 at 11:34
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    $\begingroup$ As I have already pointed out, electron de-localization is not the reason for band structure existence. It is the reason for the band not being fully filled. This is the same as saying that electrons are freely moving with the metal. Talk of de-localization is a hand-waving way of describing it, not a rigorous argument. Another way to look at it is metallic bond vs. covalent bonds in chemistry. $\endgroup$
    – Roger V.
    Commented Feb 20, 2023 at 11:55
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    $\begingroup$ Well, when we are linearly combining valence orbitals of all the atoms in the solid, the resultant would be a set of orbitals that extend throughout the solid, correct? (Implying delocalisation). Like in the case of MOT, valence orbitals interact to give bonding and anti-bonding orbitals that delocalise the electrons in them over the molecule. $\endgroup$
    – xasthor
    Commented Feb 20, 2023 at 12:10
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    $\begingroup$ Yes, electrons and holes can be though of as plane waves extending through a solid. But if the band is filled, they cannot change state - they cannot be accelerated by electric field. The problem here is mixing different descriptions of solids, chemical bonds, conductance, etc. $\endgroup$
    – Roger V.
    Commented Feb 20, 2023 at 12:54
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    $\begingroup$ Hm. Okay, let's leave conduction aside for a second. Just in terms of electronic structure, for example for diamond we view the covalent bonds as being localised. But for metals, electrons are considered to be delocalized over the solid, like a "gas" (I'll add support for this in the next comment). In this case it makes sense to think in terms of orbitals of the ensemble or entire system. But in the case of diamond, the bonds are fairly localised, and electrons in the vicinity of one atom or in a sigma bonding orbital don't really interact with those in one twenty atoms away $\endgroup$
    – xasthor
    Commented Feb 20, 2023 at 15:13

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