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I was looking at the molecule of $\ce{SnI4}$ and how its structure looks like in its crystal lattice: enter image description here

Wikipedia.de says its crystal lattice is of the fcc (ccp) type (for the iodine atoms), which means that there are eight tetrahedral sites, 1/8 of which are occupied by the tin. I understand the calculation behind it and how it leads to the formula of $\ce{SnI4}$, but I don't quite understand how one can deduce it is of the fcc type just by looking at its crystal lattice, as I'm quite bad at visualizing 3-dimensional structures. How can I visualize it properly?

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    $\begingroup$ pubs.acs.org/doi/pdf/10.1021/ja01657a014 has the atom positions listed. Since the iodine atoms are not all in equivalent positions I find it hard to believe Wikipedia's assertion. $\endgroup$
    – Jon Custer
    May 24, 2023 at 19:41
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    $\begingroup$ You can't and arent't supposed to deduce the crystal system just by looking at the picture. Also, it is not really fcc. $\endgroup$ May 24, 2023 at 20:05
  • $\begingroup$ Wiki says it’s cubically close packed.. is that the same as fcc in this case though? $\endgroup$
    – Mäßige
    May 25, 2023 at 7:03

2 Answers 2

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How can I visualize it properly?

Here is a visualization showing the iodine atoms as large spheres, and the tin atoms as tiny transparent spheres. The colors for iodine are chosen by layers along the c-axis facing the viewers, with the top and bottom layers in black, and the other three layers in yellow, purple and green. In one view, you can see that iodine atoms are roughly arranged on a grid with four grid lengths fitting into the unit cell.

enter image description here

Coordinates from https://legacy.materialsproject.org/materials/mp-23182/#

When you tilt the unit cell and look at the center of the unit cell, you can see that the yellow, purple and green layers show an approximate FCC pattern with a "unit cell" that has half the length of the real unit cell. The symmetry is broken by the tin atoms, explaining why the iodine atoms are not exactly on the smaller lattice and why we can't choose a smaller unit cell (because the tin atoms would be in different positions of the 8 smaller unit cells).

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  • $\begingroup$ Now this summarizes nicely why it is not a true fcc, and also why calling it so is not entirely untrue either. $\endgroup$ May 25, 2023 at 13:26
  • $\begingroup$ Or, maybe it is fcc. My rendering identifies $D_{3d}$ symmetry, but that can be embedded into a cubic structure (your original reference identifies the crystal system as cubic rather than rhombohedral). The five green atoms, five yellow atoms and four purple atoms centered inside the box boundaries seem to form a face-centered cube. $\endgroup$ May 25, 2023 at 14:20
  • $\begingroup$ @OscarLanzi They do form a face-centered cube, but only approximately. On the other hand, the cubic symmetry of the actual unit cell is exact, as it should be. $\endgroup$
    – Karsten
    May 25, 2023 at 14:21
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Whether the structure is fcc is disputed in the comments, but the picture shown gives a hint of at least rhombohedral ($D_{3d}$) symmetry. The two molecules at the corners of the parallelopiped are arranged in a way that shows improper sixfold rotational symmetry (they are invariant under a 120° rotation and also an inversion through the center), and the count of six remaining molecules also conforms with this symmetry.

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  • $\begingroup$ Well wiki says that’s it’s cubically closed packed.. I always understood it as a synonym to fcc. Is that wrong? $\endgroup$
    – Mäßige
    May 25, 2023 at 7:01
  • $\begingroup$ I wouldn't even call this an error. A metaphor, maybe. $\endgroup$ May 25, 2023 at 10:09
  • $\begingroup$ @Mäßige I updated the German Wikipedia article to say "Dies entspricht ungefähr einer kubisch-dichtesten Kugelpackung von Iod-Atomen, in der 1/8 aller tetraedrischen Lücken mit Zinn-Atomen besetzt sind." $\endgroup$
    – Karsten
    May 25, 2023 at 13:38
  • $\begingroup$ Okay thank you. However I am unsure if fcc is really the same as cubic close packed because here it is said that fcc is only =cubic close packed if it is regarding metals; non-metals (for example diamond) form fcc, but are not cubic close packed. Maybe you can spot why but I also have trouble understanding it: en.wikipedia.org/wiki/Cubic_crystal_system $\endgroup$
    – Mäßige
    May 25, 2023 at 18:37
  • $\begingroup$ @Mäßige Quoted from Diamond article in Wikipedia: "The diamond lattice can be viewed as a pair of intersecting face-centered cubic lattices, with each separated by 0.25 of the width of the unit cell in each dimension." The space group is #227, whereas the space group for closest packing is #224. The coordination number for diamond is 4, while it is 12 for fcc, so they must be different. $\endgroup$
    – Karsten
    May 25, 2023 at 19:42

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