P63/mmc structure

[boris runakov]

Hello Everyone !
I am trying to construct Tb₂CuIn_{3  .}

The parameters that I have are : a = b= 4.7028 , c = 7.3790 .
Positions for Tb : 1(a) , x: 0 , y:0 , z: 0.250
Cu : 2(d) , x:0.333 , y : 0.667 , z:0.02960
In : 2(d) , x:0,333 , y: 0.667, z :0.4560

Can anyone tell me how I can visualize stis structure? What are the structure parameters for the atoms that I have to enter?
Thank you in advance

[Philip Howie]

Dear Boris,

Tb2CuIn3 is based on the CaIn2 structure, with Tb occupying the Ca sites and Cu and In disordered across the In sites.  In the space group P63/mmc, the positions x,y,z and x,y,(1/2-z) are equivalent.  Therefore, you'll see that the Cu and In positions you've listed are very nearly identical.

Create a new structure and in the 'Unit Cell' tab, set its space group to 194 (i.e. P63/mmc) and the a and c lattice parameters to the values you have.  Next, go to the 'Structure Parameters' tab and create atoms at the required locations.  Set the site occupancy ('Occ.') to 0.25 for the Cu atom and 0.75 for the In atom.  The remaining atoms in the unit cell are automatically generated by the symmetry operations of the group P63/mmc.

At this point, if you click 'Ok', you'll see the Tb atoms and 3/4 of an In atom at each site.  The smaller Cu atoms are unfortunately hidden 'inside' the In atoms.  If, instead, you adjust the position of one or other of the atoms so that they overlap exactly, VESTA will recognise them as sharing the same site.  You'll see a hybrid of 3/4 of an In atom and 1/4 of a Cu atom at each site.

I hope this helps,

Philip

[boris runakov]

Great !

So many thanks for your reply . I shall try it and post back

Boris

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[boris runakov]

By the way , "Tb2CuIn3 is based on the CaIn2 structure, with Tb occupying the Ca sites and Cu and In disordered across the In sites.  In the space group P63/mmc, the positions x,y,z and x,y,(1/2-z) are equivalent." , where do you find this information?

[Philip Howie]

The information about the Tb2CuIn3 structure I found by looking it up in the Inorganic Crystal Structure Database, hosted at http://icsd.cds.rsc.org/  This gives the lattice parameters and atomic coordinates for each compound, as well as listing a 'prototype' for each structure - in this case, CaIn2, which was the first such compound to be studied and described.

Everything you need to know about the space group can be found on the space group table.  See, for example, http://img.chem.ucl.ac.uk/sgp/large/sgp.htm  If you find the entry for group 194 (P63/mmc), you'll see a graphical representation of all of the symmetry operations within the group and a list of equivalent positions; number 16 is the one I quoted to you.

Best,

Philip

[boris runakov]

So i tried what you supposed in the first post and I have attached my result . Is this correct ?

Shouldn't I get something like this http://img.chem.ucl.ac.uk/sgp/large/194az1.htm ??

Is number 16 the only equivalent position ?

I really appreciate your help !

Best regards

Boris

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[Philip Howie]

Your image looks correct.  You'll see that certain sites contain 75% In and 25% Cu.

All 24 positions listed in the space group table are equivalent, but they're not necessarily distinct!  The red symbols represent points which do not lie on any of the symmetry elements; there are 24 of them inside the unit cell (since each appears at two heights) and an atom at one position would be reproduced at the other 23 by the operation of the symmetry elements of the group P63/mmc.

However, the more symmetry elements pass through a site, the fewer distinct sites are generated.  An mirror plane near an atom generates a second atom an equal distance on the other side of the plane; a mirror plane passing THROUGH an atom does not generate any new atoms.

Let's take your Cu site at (1/3, 2/3, 0.03) as position 1.  Position 2 is then (-2/3, -1/3, 0.03) and position 3 is (1/3, -1/3, 0.03); these are identical to position 1, since crystal periodicity allows us to add 1 to any coordinate without changing the location of the site.  It's not until we get to position 4 that we find a 'new' atom at (-1/3, -2/3, 5.03) = (2/3, 1/3, 5.03).  If you go through the rest of the list, you'll find that each position will map onto one of the four that are displayed in your image.

I hope that helps,

Philip

[boris runakov]

Dear Philip ,

I am sorry for my late reply , there is too much work for me right now .

Please give me some time to read your post thoroughly and post back !

As always , I appreciate your effort

Regards

Boris

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[boris runakov]

I was thinking that the atoms should be places similar to the attachment!

Thank you for your time !

Boris

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[Philip Howie]

The space group table - and the diagram you've attached - refer to a set of symmetry elements we call P63/mmc, NOT to the atom positions in any particular crystal structure.  Indeed, there are many different crystal structures with symmetry P63/mmc but completely different atom positions.

Are you familiar with a children's toy called the kaleidoscope?  The symmetry elements are the mirrors of the kaleidoscope; the atoms are the coloured beads, which can be placed anywhere but which are then reflected and multiplied by the mirrors.  The layout of the mirrors doesn't tell us where we have to put the beads, but once we choose where to put a bead, it's multiplied and appears at several other positions.  If the bead moves, the reflected images also move.

The red markers show low-symmetry points which are related to each other by the symmetry elements.  If we put an atom at one such point, it must also appear at every other related point, because of the symmetry operations that act on it.  If we move one atom, all of its twins must also move, to maintain the symmetry of the crystal.

Imagine placing a kaleidoscope bead in front of a mirror.  It will be reflected in the mirror, and we'll see it twice.  Now imagine moving the bead toward the mirror; the image also moves toward the mirror, from the other side.  When the bead is located exactly at the plane of the mirror, it will meet its reflection and suddenly we'll only see one bead.

Exactly the same thing happens to atoms that are located at positions of high symmetry.  Instead of seeing 24 copies, we only see 4.

If you want to convince yourself of this effect, you can experiment by opening VESTA and moving one of the atoms slightly from its position - you'll see the number of atoms changes as you move away from a high symmetry site.

Hope that helps,

Philip

[boris runakov]

Ok let me try to experiment with what you suggested and post back!

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[boris runakov]

Once again I am sorry for my late reply !

Please can you tell me what one of the red positions might be ?

 Also where is the origin in this space group diagram so I can start from there and find these 24 different positions and visualize them ?

[Philip Howie]

Boris,

I feel that this discussion has gone far beyond the original scope!  I believe that the latest version of the Tb2CuIn3 structure you showed me is correct.  If you wish, you can verify this by comparing it with the structure displayed in the ICSD or with other published data.

If you want to understand the spacegroup tables, I suggest you start from the beginning and make a proper study of symmetry.  Make sure you're happy with the idea that a symmetry element - whether it's a mirror plane, a rotation axis, the periodic symmetry of a crystal, an inversion centre or anything else - acts on a point to create one or more points which are equivalent by symmetry.

Next, understand that a spacegroup is a group (in the mathematical sense) of symmetry elements.  That is, the combination of any two (or more) symmetry elements must be another symmetry element which is also in the group.  There are 230 possible spacegroups, each of which represents a self-consistent arrangement of symmetry elements.

Within each spacegroup, we can calculate the effect of each symmetry element on the point (x,y,z).  The result is the list of equivalent positions found on the spacegroup table.  For each of the equivalent positions listed, we can find a symmetry element that maps it onto (x,y,z).

Notice that I haven't spoken of atoms yet.  What I've called a 'point' is a purely mathematical idea - a location in space that is equivalent to other locations.  If we now place an atom in the crystal, it must be acted on by the symmetry elements in the same way, and in that way the small number of unique atom positions listed in, for example, the ICSD can be used to generate a whole crystal structure.  Of course, it's possible that two symmetry elements might map to the same point - that's fine, and in this case the number of equivalent atoms will be less than the number of equivalent positions for the general point (x,y,z).

As I said above, I recommend you find some more detailed instruction than I can provide in this forum.  Find someone closer to you who is an expert in these matters, or read a good textbook on the subject.  Christopher Hammond's book 'The Basics of Crystallography and Diffraction' is one I can recommend.  Good luck!

Philip