I understand that $\ce{CdCl2}$ has a rhombohedral unit cell as shown in this book. However I often find it being referred to as a "CCP analogue" of $\ce{CdI2}$. Presumably it is the chlorides which form a cubic lattice; how would you draw the cubic unit cell ?
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Cubic close packing is another (confusing) name for face-centered cubic. So, to visualize these structures, you start with a hexagonal 2D layer of the chlorine or iodine. Now start stacking those layers, ABAB... for hcp (iodine), or ABCABC... for fcc (chlorine). Finally, stick the Cd in between alternate layers of the chlorine/iodine. The net result for CdCl2, is that the unit cell is indeed rhombohedral, since there is one Cd layer for every two Cl layers, so it is stretched in that direction.
answered Apr 10, 2015 at 19:59
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$\begingroup$ Why does the ABCABC stacking correspond to a FCC lattice though ? I can't see the right orientation in a 3d structure model (chemtube3d.com/solidstate/_CdCl2%28final%29.htm). $\endgroup$– J. LSCommented Apr 10, 2015 at 20:31
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1$\begingroup$ Orient the live 3D model so you are looking perpendicular to a face on the long side of the indicated rhombohedron. Now the green layers (at least on my browser) are the close packed planes of chlorine. They repeat in a pattern of A-B-C.. Now, go find a picture of an fcc metal. You will note that if you look along the body diagonal you see an atom on the bottom point, the atoms on the faces of the bottom sides, the atoms on the faces of the top sides, and finally come back to the same position with the top vertex atom. These are the close packed planes stacked up as A-B-C-A. $\endgroup$ Commented Apr 10, 2015 at 21:57
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$\begingroup$ I can see the alternation but not where the fcc unit cells of Cl are... $\endgroup$– J. LSCommented Apr 13, 2015 at 8:43
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1$\begingroup$ Again, compare with a simple fcc metal such as copper. Turn the traditional unit cell so that the body diagonal is pointing up. Take the bottom vertex as an atom on a hexagonal plane. The atoms at the center of the lower faces are also a hexagonal pattern, and are the B position. Finally, the atoms at the center of the upper faces are yet another hexagonal plane which line up with neither A or B, so they are C. The atom at the top vertex is back to A, and it repeats from there. For FCC, the stacking is along the 111 direction. $\endgroup$ Commented Apr 13, 2015 at 13:44
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$\begingroup$ Horray, I can see it now ! Thank you so much. $\endgroup$– J. LSCommented Apr 13, 2015 at 13:56