This the structure of Dolomite.STRUCTURE OF DOLOMITE

Can someone explain this structure to me?( like what are the blue, yellow etc atoms are? why are they arranged this way?)

Any help would be appreciated.


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  • 2
    $\begingroup$ There are quite a few levels of abstraction between this picture and the real compound. It would be good to know which level you don't understand. To begin with, do you know which spheres corresponds to carbon atoms? $\endgroup$ – Ivan Neretin Feb 15 '18 at 11:45
  • $\begingroup$ First, the tag molecular-structure is out of place. Second, strictly speaking, crystal structure is an infinite periodic three-dimensional lattice and cannot be depicted with a single image. What you show here is a unit cell. $\endgroup$ – andselisk Feb 15 '18 at 13:36
  • $\begingroup$ What do you want explaining? $\endgroup$ – bon Feb 15 '18 at 17:35

I don't really like the picture but let's try to do this with that. So, you probably know about the NaCl lattice? In there sodium is surrounded by six chlorides, which forms an octahedron. Chloride has six sodium around it to form another octahedron. And you have probably heard that also other compounds crystallize in this lattice, let's say sodium bromide (so increasing the anion in size) or potassium chloride. It does not change the crystal structure.

Now, what about more complicated units. Let's deform that spherical anion to an elongated one. What about calcium carbide? The C-C unit with its triple bond is unlikely cleaved during the crystallization, so we can count the $C_2^2-$ unit as one part. This forms an elongated cell but still, it is a NaCl-type lattice.

If we make it flat instead we end up with something like carbonate for example. As you may no the carbonate unit is flat, it's basically a triangle with the carbon in the center. If we see this unit as one part then we can just form another NaCl-lattice. And as the jump from sodium chloride to potassium chloride was possible, we can assume that if calcium carbonate crystallizes in a deformed NaCl-lattice, then magnesium carbonate will do the same.

And the final step is to combine them now, instead of separating the two phases, 1/2 of the calcium-ions are just exchanged by magnesium to form the dolomite-structure.

So you could say that carbonate forms a cubic packing where 50% of the octahedral sites are occupied by calcium and the 50% by magnesium for example. All carbonate-units are parallel to each other therefore creating one direction in which they do not require that much space causing the deformation of the cell.

The metals are coordinated by (and I had to load this in my solid state software as well to see it) six oxygen on the carbonates, or if you take the center carbon of each carbonate unit also by six carbonates, which forms an octahedron again. On the other hand, each of these center carbons is coordinated by three oxygen in a triangle or if you count the counter ions by six metals in the shape of an octahedron.

In your structure you can start off by looking at the dark atoms, which are the carbons, and connect them to the red ones, so can see it will always form triangles. For all the others it's hard to see in this picture. If you want I can also highlight you the different polyhedra in my software so you can see it better. Because this is just a unit cell so it is cut at some positions creating the feeling that some of the connections are wrong. You'd have to watch several unit cells aligned to see the full structure of it. But if you imagined the yellow and the blue ones to be the same you'd end up with the Calcium carbonate or magnesium carbonate structure.

EDIT: I also added the crystal structure now: enter image description here

EDIT II: Because there were still some problems I tried to further explain the structure a bit. I don't know how others see this structure but I like this layer explanation and it's also the one you can find in many books. So please just look through the new attachment and tell me if this helps you more.

enter image description here

  • $\begingroup$ I am unable to understand how are we combining both of them? And from where are 1/2 calcium atoms are being replaced by magnesium ions? $\endgroup$ – Piano Land Feb 16 '18 at 16:51
  • $\begingroup$ Well if you read my picture from the left to the right I started off with all metals being the same and being blue, then I seperated them into Mg and Ca by adding the green color as well. In a simplified view you can say that you get a layer-structure, where you have a layer Mg, then a layer carbonate, a layer Ca then carbonate, Mg, carbonate, Ca, .... $\endgroup$ – Justanotherchemist Feb 16 '18 at 20:28
  • $\begingroup$ In the second last diagram, in the upper right carbonate ion, the oxygen atom has formed 3 bonds( with carbon, magnesium and calcium) how is that possible? $\endgroup$ – Piano Land Feb 17 '18 at 18:10
  • $\begingroup$ Well in solid compounds this is like dissolving something in water. NaCl would dissociate and you'd expect the Chloride to be encapsulated by water and not just one water. It's the same here but just a solid type of matrix or solvent if you prefer to imagine it like that. There are layers of positive charge above and below the oxide, so it will certainly 'feel' something and that connection to carbon is a covalent 'real' bond anyways. $\endgroup$ – Justanotherchemist Feb 22 '18 at 10:22
  • $\begingroup$ To draw in bonds helps to see why for example some things are not well centered by kind of move to the opposite charge. And it also helps to organize everything in your mind. It's the same as for complex compounds when we talk about ligands and polyhedra forming around an ion. $\endgroup$ – Justanotherchemist Feb 22 '18 at 10:24

Here is my understanding of it:

1) The black (carbon) and red (oxygen) atoms form the triangular carbonate ($\ce{CO_3}$) units.

2) Each oxygen atom is bonded to a gold magnesium atom and a blue calcium atom.

3) Look carefully at the bonds to each calcium or magnesium atom. The bonds shown fit onto the vertices of a regular octahedron, we see a pair of oppositely directed metal-oxygen bonds at each metal atom. The metal atoms are in fact octahedrally coordinated to oxygen.

So why do we not see all six metal-oxygen bonds at each metal atom? The answer is that the oxygen ends of the "missing" bonds are in adjacent unit cells. I would have drawn those bonds through the faces of the cell, even if I have to leave off the oxygen atoms so that the bonds appear to end in "thin air". But having the "thin air" bonds makes it easier to piece multiple cells together so that you see the octahedral coordination come together.

  • $\begingroup$ Well that's basically it. For the coordinations you can simply look at my picture, I just took the published data on dolomite and drew in the octahedra. And yes, you cut of at the borders. What you see is the unit cell, the repeating unit. You'd have to look at a supercell to see more connections, I could add that but the structure is already quite big and it becomes quite hard to see anything. $\endgroup$ – Justanotherchemist Feb 16 '18 at 20:25

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