3
$\begingroup$

I am currently studying the textbook Physics of Photonics Devices, Second Edition, by Shun Lien Chuang. In a section discussing the basic concepts of semiconductor band and bonding diagrams, the author gives the following description:

The basic idea is that for a semiconductor, such as $\ce{GaAs}$ or $\ce{InP}$, many interesting optical properties occur near the band edges. For example, Table 1.1 shows part of the periodic table with many of the elements that are important for semiconductors, including group IV, III-V, and II-VI compounds. For a III-V compound semiconductor such as $\ce{GaAs}$, the gallium ($\ce{Ga}$) and arsenic ($\ce{As}$) atoms form a zinc-blende structure, which consists of two interpenetrating face-centered-cubic lattices, one made of gallium atoms and the other made of arsenic atoms (Fig. 1.1).

enter image description here

This seems to be the relevant Wikipedia article for zinc-blende (Zincblende) structure.

I was interested in understanding the relevance of this structure to semiconductors, so I did further research. I then came across this website, which says the following:

Here are the two most important crystal structures for semiconductors.

They are often referred to by the historical names "Zinc blende" from the German "Zinkblende" $= \ce{\alpha-ZnS}$, a rather ubiquitous mineral. The name "Sphalerite" also comes form the German: "Sphalerit", which, as was the custom of the time, stems from the Greek "sphaleros" meaning treacherous or malicious because it is easy to confuse it with other minerals.

Wurtzite was and is the name of the $\ce{\beta-ZnS}$ modification - the hexagonal high-temperature variant. The named after the French chemist C. A. Wurtz (* 1817, † 1884), which gives us an idea of how old those names are.

Zinc blende or sphalerite or Diamond structure

enter image description here

Wurtzite or hexagonal structure

Red lines are not showing bonds.

enter image description here

However, this website does not actually explain why these are the two most important crystal structures for semiconductors. I was wondering if someone would please take the time to explain this.

EDIT:

Thinking about Justanotherchemist's comment and Oscar Lanzi's answer, I wonder if all group $\ce{III-V}$ semiconductors have these $\ce{ZnS}$ structures? Because, if so, then that could very well be the answer to my question here, since, as I understand it, group $\ce{III-V}$ semiconductors are the most important semiconductors for practical purposes? As Oscar Lanzi said in his answer, not all semiconductors have a $\ce{ZnS}$ structure, so the answer cannot be that they are the most important because all semiconductors have a $\ce{ZnS}$ structure.

$\endgroup$
  • 1
    $\begingroup$ I don't know if the structure has something to do with it's ability to be semiconducting but these III-V structures adopt the ZnS structures due to something we call 'Grimm-Sommerfeld' rule. Same applies for I-VII as well in for example AgI. $\endgroup$ – Justanotherchemist Jan 28 at 11:52
  • $\begingroup$ @Justanotherchemist Interesting. Thanks for the information. $\endgroup$ – The Pointer Jan 28 at 11:57
  • $\begingroup$ @Justanotherchemist Thinking about your comment and Oscar Lanzi's answer, I wonder if all group $\ce{III-V}$ semiconductors have these $\ce{ZnS}$ structures? Because, if so, then that could very well be the answer to my question here, since, as I understand it, group $\ce{III-V}$ semiconductors are the most important semiconductors for practical purposes? As Oscar Lanzi said in his answer, not all semiconductors have a $\ce{ZnS}$ structure, so the answer cannot be that they are the most important because all semiconductors have a $\ce{ZnS}$ structure. $\endgroup$ – The Pointer Jan 31 at 15:40
2
+50
$\begingroup$

As suggested I can try to summarize the basic idea in an answer although I wouldn't say it solves the bounty. I'm also unsure why you even set the bounty for it.

For this I'll quote the book that I found a few days ago. As I own the missing pages now as well the answer seems a bit more elaborate then before. As the book is in German I will translate the text.

The formation of the band-structure of typical element semi-conductors like diamond (C), Si and Ge was already shown in chapter 7.3. By mixing s- and p-wavefunctions we'll get a tetrahedral bonding orbital ($sp^3$), that will lead to a splitting into a bonding and an anti-bonding orbital in the range of the equilibrium-bonding distance. The bonding orbitals form the valence band and the anti-bonding orbitals form the conducting band. The distribution of the four s- and p-electrons into bonding and anti-bonding orbitals leads to a completely filled valence band while the conducting band remains empty. The result is therefore an insulator like diamond or, if the energy gap between the valence and conducting band is smaller, to semi-conductors like Si or Ge.

I'll skip the next pages that deal with the formation of the band-structure and the symmetry aspects as I have no experience in these topics and continue with the $III-V$-semi-conductors.

As the formation of $sp^3$ hybrid orbitals in the chemical bonding of Si and Ge seem to be the a substancial part of the semi-conducting properties it seems reasonable that other materials with a tetrahedral crystal structure, i.e. $sp^3$ hybridisation, should show semi-conducting properties as well.

At this point I will switch to basic solid state chemistry as the book doesn't elaborate this part well enough. As solid state chemist I often have examples like these with my students. A while ago we were discussing the silver-mobility in high-temperature $AgI$, which shows quite some conductitiy through defects and the low temperature modification. At this point I would typically ask for reasons why $AgI$ might be problematic. Typical answer that I expect are:

  • HSAB-concept with soft $Ag^+$ and soft $I^-$
  • A covalent three-dimensional tetrahedral network with ZnS-structure

For the second part they either check these facts before we talk about it or they can use the things they have learned in lectures. One of them being the Grimm-Sommerfeld rule. Basically, compounds, that form between two elements to the left and to the right of another element will likely adopt the structure of said element. So for a group $IV$ element like the silicon structure, it can be $III-V$ or $II-VI$ or $I-VII$. It's one of these rules that are taught in the very first semester but people forget about it although it's super useful for compounds you don't know.

Therefore my idea would be here that if, as the book mentions, $sp^3$ hybridisations and covalent, tetrahedral structures as the main source for the semi-conducting properties it's only natural that $III-V$ compounds show a similar effect. Those bands are basically molecular orbitals. If you start from a di-atomic molecule and increase the amount of bonding partners you will get more and more orbitals that move closer and closer together until you have bands. If let's say Ge has a certain band-gap and you change to GaAs, or course the single contributions to the HOMO and LUMO and later to the bands will be different but not that far away from Ge, so we expect a comparable band-gap.

Source: H. Ibach, H. Lüth, Festkörperphysik: Einführung in die Grundlagen, 7th edition, Springer, 2008.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for taking the time to post this excellent answer! So just to clarify something that I wasn't clear on from the textbook, but that seems to be implied: Group III-V compound semiconductors all form a zinc-blende structure? $\endgroup$ – The Pointer Feb 3 at 18:28
  • $\begingroup$ That's a good question, at least hat high pressure they will go into the NaCl structure. Some lead compounds might have problems as the lone-pair will require some space. There is an entry somewhere here where I explained how the PbO structure is created from the CaF2 structure. There it should become obvious why no ZnS is formed. $\endgroup$ – Justanotherchemist Feb 3 at 19:51
  • $\begingroup$ But the author of the website says that the two $\ce{ZnS}$ crystal structures are the most important for semiconductors. And we are told that silicon and germanium semiconductors form the $\ce{ZnS}$ structure, and that they have $\ce{sp^3}$ hybridization. But I'm struggling to understand what the connection is between these two. The former seems to be a statement about all semiconductors, whereas the latter seems to be a statement about silicon and germanium semiconductors. $\endgroup$ – The Pointer Feb 3 at 20:07
  • 1
    $\begingroup$ Well it's ZnS crystallizing in the Silicon structure rather than the other way but yeah. I have to guess here. The most common ones are in fact all ZnS-type. There may be some that do not form this structure (although I cannot find any) but still this wouldn't matter. The bands form from a combination of the two elements orbitals. If the energy gap is within a certain range it's semi-conductive, if they overlap and delocalize it's conductive, if the gap is really high it's an insulator. Hence other structures besides ZnS can be semi-conducting as well. $\endgroup$ – Justanotherchemist Feb 3 at 21:02
  • $\begingroup$ Interesting. Thank you for taking the time to do all of this research and explain it to me. This has been very illuminating. $\endgroup$ – The Pointer Feb 3 at 21:47
1
$\begingroup$

Zincblende and wurtzite structures are not the only ones that afford semiconducting materials. This short list of narrow-gap semiconductors involves compounds with a variety of crystal structures. Aside from the fact that several examples clearly have different structures because they do not have 1:1 stoichiometry, the examples lead sulfide (PbS) and lead selenide (PbSe) have halite structures.

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Interesting, but that doesn't tell us why, if at all, these are the "two most important crystal structures for semiconductors". $\endgroup$ – The Pointer Jan 31 at 14:26
  • $\begingroup$ It may have nothing to do with semiconduction as such. It may have more to do with what has been applied in practice. $\endgroup$ – Oscar Lanzi Jan 31 at 14:55
  • 1
    $\begingroup$ Unfortunately I'm just a solid state chemist and no physicist. I can only tell you that most things like I-VII, II-VI, III-V will adopt either the hexagonal Wurzite or the cubic Sphalerite type ZnS structure. It was mentioned in a book somewhere that in Silicon and Germanium the sp³-hybrids formed in these structure might actually be responsible for the semiconducting properties. Unfortunately Google Books does not show me the rest of the pages. $\endgroup$ – Justanotherchemist Feb 1 at 16:13
  • 1
    $\begingroup$ They also argumented that as most of these III-V compounds like GaAs or so are somewhat covalent we can still think of sp³-hybrids but with different band gaps. That might actually be the answer here. Your original IV structure like Silicon has interesting properties. If you change the elements but remain similar bonding situations the different orbital energies overlapping will lead to new band structures. $\endgroup$ – Justanotherchemist Feb 1 at 16:15
  • 1
    $\begingroup$ You can see something similar with Graphite. Graphite can be seen as big system of connected benzene rings. Which means there are orbitals all pointing orthogonally where current can be transported. If we switch to BN however, though it's the same structure, the nitrogen will draw more of the electron density and the thing turns white and non-conductive. $\endgroup$ – Justanotherchemist Feb 1 at 16:18

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.