Is the position of the central Tin atom as shown in the first figure ($\ce{SnO2}$) drawn offset towards the oxygen to the right for a reason? The oxygen also seems to reciprocate and move towards the tin. Usually the rutile structure is shown as symmetric with the metal the center. I'm showing a GIF alternating images of $\ce{SnO2}$ and $\ce{RuO2}$, with complete image source information following.

note: I'm interested to know about the locations of the atoms within a unit cell as determined by crystallographic information - not one's impression of the drawing. I'd like to know if the offset shown in the drawing is there to represent an actual asymmetry, or if it's just an error.

enter image description here

Figure 1. Tin Dioxide (Tin IV oxide), also found as Cassiterite, image from Rutile unit cell 3D balls.

Figure 2. Ruthenium(IV) oxide, image from Ruthenium(IV) oxide unit cell 3D vdW.

Figure 3. Rutile structure, image from Rutile structure

enter image description here Fig. 1

enter image description here Fig. 2

enter image description here Fig. 3

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    $\begingroup$ I can't find any information which suggests that the Sn atom at the centre is not at (0.5, 0.5, 0.5). The unit cell parameters are slightly different, and the oxygens are shifted by a very tiny bit (the oxygens are at (x,x,0), (1-x,1-x,0), (0.5+x, 0.5-x, 0.5), (0.5-x,0.5+x,0.5), and x = 0.305 for TiO2 and 0.307 for SnO2) but nothing on the tin atoms. $\endgroup$ Commented May 2, 2017 at 12:20
  • $\begingroup$ @orthocresol "nothing" means (0,0,0), (0.5, 0.5, 0.5), or nothing as in no mention of the location of the tins? I'm asking this to personally learn more on the subject, and it seems that there should be some sort of symmetry argument against (0,0,0), (0.5, 0.6, 0.6) for a crude example - the unit cell would have a net direction and the material might show particular optical or dielectric behavior. Answer that included a non-paywalled way to see the numbers you are seeing would be greatly appreciated! Also there's still this question... chemistry.stackexchange.com/q/73254/16035 $\endgroup$
    – uhoh
    Commented May 2, 2017 at 12:36
  • $\begingroup$ @orthocresol Also I realized that that the unique motion of the one oxygen atom on the right could not be explained with the simple expression in your comment, so I have a hunch you've got the answer. $\endgroup$
    – uhoh
    Commented May 2, 2017 at 12:38
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    $\begingroup$ No mention of the location of the tins, so I expect it to be the usual (0,0,0) + (0.5,0.5,0.5). I don't know much about it, I just flipped through the books that I had. It's in A.R. West "Solid State Chemistry and Applications" $\endgroup$ Commented May 2, 2017 at 12:47
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    $\begingroup$ The tin is not offset. Likely the images have been distorted. $\endgroup$
    – Jon Custer
    Commented May 2, 2017 at 13:53

1 Answer 1


From http://www.crystallography.net/cod/1000062.html one can see that Sn sits in the middle, without any displacement, and mimics the classic Rutile-type structure:

enter image description here

I checked original Wikimedia image, and according to metadata

CIF retrieved from The American Mineralogist Crystal Structure Database

is no longer available. I also failed to discover this source:

X-ray crystallographic data from: R. W. G. Wyckoff (1963) Second edition. Interscience Publishers, New York, New York. Crystal Structures 1, 239-444

Ideally, if you have *.cif file, it might be worth checking occupancies for the atoms you are interested in. Probably there is a 50-50 disordered Sn atom about center of unit cell, and you only observe a single site. As long as this is not the case, I would rely on the available COD data from 1971 instead.

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    $\begingroup$ Thanks very much for your post! Each figure in my question is numbered, and the captions contain the word "from" followed by the link to the original image. I did not draw any figures myself. Can you help me understand your comment about "50-50 disordered Sn atom"? Does this mean that the atom will in fact be more likely to be found on one side or the other rather than the center? Would this show up crystallographically as an elongated charge distribution for that atom? $\endgroup$
    – uhoh
    Commented Jun 26, 2017 at 5:26
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    $\begingroup$ I slightly edited my answer to clarify my point of view. By "50-50 disordered Sn atom" I suggested that there might be statistically disordered Sn atoms, most likely with 50% probability, either due to poor quality of crystals, or experimental technique, or just because of different phase. Anyway, this recent publication reveals that this is most likely not the case. I'm also not sure what you mean by "elongated charge distribution" (dipole?), in crystallography this entity is usually called "thermal ellipsoid" (also see ORTEP). $\endgroup$
    – andselisk
    Commented Jun 26, 2017 at 5:37
  • $\begingroup$ Oh, my ellipsoid was part of my trying to understand the 50-50. I thought that you were suggesting that in 50% of the unit cells within a physical crystal the center Sn would be on one side, and in 50% of them it would be on the other side of the center, and so this might show up as a non-spherical electron distribution when modeling diffraction data. But it looks like that's not what you meant. Since I didn't make these drawings, can you adjust the sentence "I'm not sure what data you used to draw this structure." to perhaps "...what data was originally used..."? Thanks! $\endgroup$
    – uhoh
    Commented Jun 26, 2017 at 5:44
  • $\begingroup$ I edited my answer a bit more, please let me know if you are not happy with the new one. Regarding disorder: you are right, "when modeling diffraction data" it really is sometimes looking as an elongated ellipsoid, and we embrace it as a mathematical model for representation of experimental data; whereas in reality it might be two (or more) separate atoms with a set of own coordinates each statistically distributed among all the unit cells, so that we see their average electron density. $\endgroup$
    – andselisk
    Commented Jun 26, 2017 at 5:56
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    $\begingroup$ No prob at all, I'm very glad I could help. $\endgroup$
    – andselisk
    Commented Jun 26, 2017 at 6:12

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