1
$\begingroup$

Throughout this whole question, I will be referring solely to single element solids. According to band theory, ns and np bands are close enough in energy to overlap and create one band with 8N states available to fill. This band then splits into two 4N bands when the interatomic distance between atoms in the solid increases.

Source: https://csclub.uwaterloo.ca/~matedesc/figures/diamond-crystal-spacing.png

This model explains why diamond is an insulator. A single carbon atom has four valence electrons, so N carbon atoms will have 4N of them. These electrons fill the lower of the 4N bands (in this case, the valence band) completely, meaning that diamond cannot conduct electricity at normal conditions.

However, I don't understand why elements such as boron, nitrogen, oxygen and fluorine aren't able to. In boron, there would be 3N filled states with N empty states still remaining. In nitrogen, there would be 5N states and the higher of the 4N bands would start to fill, leaving another 3N empty states for conduction. The same goes for all non-metallic elements which aren't in the group 14 or 18. Why, then, are these regarded as semi-conductors or insulators even though they have unfilled states in their valence bands, which is a condition for conductivity?

$\endgroup$

3 Answers 3

1
$\begingroup$

The partially filled band explanation of conductor versus insulator behaviour is not convincing. For example both solid an amorphous silicon are semiconductors, while amorphous silicon has no translation symmetry.

A better description of the underlying physics is the Hubbard model. It basically states that conductors have delocalized valence orbitals and insulators have localised ones. Whether localisation or delocalisation is happening depends on the ratio of on-site repulsion to hopping matrix elements. An important factor is the number of nearest neighbors. For example, solid silicon is a semiconductor while liquid silicon is a metal.

https://en.m.wikipedia.org/wiki/Hubbard_model

$\endgroup$
1
  • $\begingroup$ Plus the whole crystal structure (and implicit band structure) needs to be considered. Boron in particular has decidedly weird crystal structures with scads (50-100) atoms in a unit cell, very different from things like Cu or Al. $\endgroup$
    – Jon Custer
    Aug 4, 2020 at 18:47
1
$\begingroup$

The key is the amount of energy needed to bump an electron out of its orbital and make it mobile.

In a conductor, the electrons are already there, typically delocalised in a lower energy state (i.e. freeing them up actually releases energy).

In a semiconductor there is a small energy gap between the valence and conduction bands.

Pure elements such as boron, nitrogen, oxygen and fluorine are covalently bonded into molecules, with no free electrons and no crystal lattice in which "holes" might form. The energy (band gap, if you will) required to ionise an atom is comparatively large. That is enough to ensure the substance is in a gaseous state which, when ionised, is known as a plasma. Such plasmas occur in fluorescent "neon" tubes, lightning and so on.

Note that certain compounds of these elements, such as the III-V compound boron nitride, do form crystalline semiconductors.

$\endgroup$
-1
$\begingroup$

ITs not about fully or partially filled orbital or band as you say, But rather its about how much energy is needed to cause conductance and we call the energy gap between conductive and valence band. If the gap is large more energy is needed to ionize or conduct.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.