From doing more Phase ID for XRD on aluminosilicate based coal fly ash samples, I've seen mullite phases appear a few times which have the non-stoichiometric chemical formula $Al_{4+2_x}Si_{2-2x}O_{10-x}$ where $x \approx 0.14 - 0.49$ according to this reference which also details in a good Phase ID methodology for mullites.

My questions are the following:

  1. Why are some solids crystalline instead of being a solid solution; how is the latter formed and why?
  2. How are solid solution chemical formulas identified as opposed to easier-to-understand stoichiometric chemical formulas as seen in crystal structures (e.g. SiO2 in quartz).
  3. What common methods exist to identify the $x$ parameter in solid solution formulas (apart from identifying common ratios between characteristic peaks as mentioned in the reference above).
  • $\begingroup$ Those solutions are crystals. $\endgroup$
    – Jon Custer
    Oct 25, 2022 at 12:51
  • $\begingroup$ Wait, are "solid solutions" synonymous with polycrystalline solids? $\endgroup$
    – Hendrix13
    Oct 25, 2022 at 12:52
  • $\begingroup$ No, not at all. In a bit I’ll get an answer out… $\endgroup$
    – Jon Custer
    Oct 25, 2022 at 12:57

1 Answer 1


Being a solid solution is separate from being a crystal. Some alloys occur with one single stoichiometry (plus/minus defects). Many can accommodate a range of compositions.

Lets start easy: The Au-Cu binary system shows complete solubility across the entire composition range. The crystal structure is fcc, and Au and Cu atoms sit randomly. The lattice parameter will change a bit going across the composition range and that is about it. Similarly the Si-Ge system also shows complete solubility in the diamond cubic crystal structure.

The NaCl-KCl pseudo binary system has a slightly different response across the composition range:

NaCl-KCl binary

Below 800K the fcc-based NaCl (or KCl) crystal structure has a miscibility gap - only so much K in a primarily NaCl (or Na in a primarily KCl) crystal fits. Above 800K you see complete solid solubility, with any amount of Na or K fitting in between the Cl sublattice.

The GaAs-AlAs pseudo binary has complete solid solubility, again with any amount of Al/Ga in the As sublattice. The lattice parameter and the band gap changes across the composition range.

The B2 crystal structure is based on the bcc unit cell, with the corner and center atoms being different sites. The exemplar is CsCl. However, there are many binary metallic systems where the B2 structure shows up, but often with a reasonable composition range. One example is in the Al-Ti system:

Al-Ti system

Here one sees a variety of phases, some line compounds (Ti$_{3}$Al$_{5}$ for example), but many with wide composition ranges. These include, notably, quite reasonable solubilities of Al in both hexagonal and bcc Titanium as a random solution with the Al sitting on hcp or bcc lattice sites. In addition, the light blue B2 $\beta_{0}$ phase is an ordered bcc phase, with the Al preferentially occurring on one of the B2 sites. This additional symmetry of the crystal will be clear in diffraction patterns.

The sublattices do not all need to be completely full. One example is in the Ti-N system: Ti-N binary system

Again one sees that the hcp and bcc Ti phases can accommodate some amount of nitrogen in them - however these N atoms are on interstitial sites not in place of Ti. There are also three stoichiometric compounds (line compounds). Finally there is the TiN phase, which is an fcc Ti lattice with varying amounts of nitrogen on tetrahedral sites. This is modeled as a two-lattice system with either N or a vacancy on the one one site, and fcc Ti on the other. You can only get up to a 1:1 stoichiometry range, then you run out of the space on the (N,Va) sublattice.

In the BaO-CaO system, there is limited solid solubility in the halite (BaO) or periclase (CaO) phases. BaO-SrO behaves much like NaCl-KCl. Al$_{2}$O$_{3}$-BaO only shows line compounds.

For many mineral oxides (including the ones you are looking at, but basically about a million discovered minerals in nature), one has an oxygen sublattice with octohedral and tetrahedral sites (or more complex oxygen lattices but the point remains), and various metals will prefer to sit on one site or the other. In some cases this is a strong preference, in other cases some will sit on either site without caring much. Sometimes this requires vacancies on one sublattice or the other (well, really multiple sublattices), charged defects, or whatnot. Once you get oxygen and Al, Fe, Si, and maybe a few other impurities, well, the sky is the limit.

Your particular case of mullites (see [Journal of the European Ceramic Society][4] for example) shows one set of AlO$_{6}$ octahedra, with "cross-linking" of tetrahedral chains of AlO$_{4}$ or SiO$_{4}$ connecting them. The composition is Al$_{4+2x}$Si$_{2-2x}$O$_{10-x}$ showing you can put some Al in place of the Si, but you also have to get rid of some of the oxygens for charge compensation. A quite complex structure all in all.

[4]: https://www.sciencedirect.com/science/article/pii/S0955221907002634#:~:text=The%20crystal%20structure%20of%20mullite,%2CSi)O4%20tetrahedra.

  • $\begingroup$ Thanks for the very diligent answer! Before I can fully understand if your answer solves my question I think I need to spend more time understanding some background here in your answer. Could you please advise what types of diagrams you refer to so I can do some reading along with some sources? Thanks so much, I'll return here and mark it as officially answered (although I've +1'd) once I'm sure of the answers :) $\endgroup$
    – Hendrix13
    Oct 26, 2022 at 12:36
  • 1
    $\begingroup$ Those are binary phase diagrams (some are pseudo-binary). Porter and Easterling have a classic book called Phase Transformations in Metals and Alloys that gives the basic background to them in the first few chapters. Sorry, I assumed (bad of me) that someone doing phase identification in ceramics would have done a course somewhere on phase diagrams. $\endgroup$
    – Jon Custer
    Oct 26, 2022 at 12:48
  • $\begingroup$ All good, it's a reasonable assumption to make, I'm still a beginner in some ways which is why I hadn't heard of them (although I've seen them before quite a bit). $\endgroup$
    – Hendrix13
    Oct 26, 2022 at 13:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.