The space group information can be used to research the fractional atomic coordinates. This information can be written in a format that can be visualized using software's such as VMD, AtomEye, Ovito and many others. In the visualization scene, you can create a slice along the (111) plane and measure the planar density. 

Hope that helps

Thank you Ilaksh Adlakha for answer. Can you please send me the link of software (free version) ?

Since, for my research I use Ovito here is the link attach. you can easily find the others by performing a google search for VMD or AtomEye. 

http://ovito.org/index.php

Thanks a lot Ilaksh for sharing!

Crystalmaker does this, but it stacks unit cells, so you get the unit cell multiplied by 1, 2 etc.

Dear colleague,

                                      to give a correct answer to your question, one should follow this path:

-  first you have to isolate a slice of thickness d111, wghich is allowed by the space group P21/c;

- to do that, a projection of the crystal structure along the direction [1-10] is needed

-then you will find  a series of parallel and equivalent slices of thickness d111

- the atomic density is then obtained by dividing the number of atoms in the slice by the area of the 2D cell parallel to the slice.

You can repeat this operation for other slices and make a comparison among them.

Using software is good, but using the reasoning would be better...

All the best

Dino

Thank you Dino Sir for your reply.

Hey another easy way is to get a cif file from for instance crystallography.net (search => elements Zr and O and number of distinct elements 2 to 2) and then visualize with VESTA (http://jp-minerals.org/vesta/en/). Inside, click objects => boundaries and type 1 1 1 (disable the symmetry operations else it'll cut all 111 -111 etc planes at the same time. At the top you can also expand the unit cell to see a bigger surface while the original cell is always shown by the box (make sure you're basing 111 on the right cell), there may for instance be a primitive and conventional cif file.

First I have to say that (111) is a lattice plane, i.e. it only contains lattice points and this plane only rarely contains atoms. Actually, both have nothing to do with each other since the lattice point contains/represents all atomes since it is a primitive lattice. In other words, if one lattice point represents 12 atoms then they all belong to (111). Of course, I can imagine what you are interested in. You want look, how many atom layers are parallel formed to (111). But also this is relateted to the lattice points since you need to say how many parallel (111) exist in a unit cell (only in case you have stacked layers which are presented by one lattice point the result can be different). It also depends on the centering, i.e. I-centered lattices would have 6 parallel-stacked layers instead of three  3 (111) planes as in your case. That means, in your case (P lattice) each lattice plane (111) contains in average 4 atoms. However, I am not clear what do you want extract from this information. 

Hello Gert!

I want to find out molar interface area (m^2/mol) of all low index plane for thermodynamic point of view. And for that I should divide it by number of atoms sitting in plane.

Thank you for reply.

Dear Aditi Thanki,

May I ask for what purpose you are studying this?

Most people use the easy to study cubic modification of Zirconia. My PhD thesis long time ago was on monoclinic ZrO2, produced by skull-melting technique; please see the three uploaded papers.

https://www.researchgate.net/publication/233859018_Critical_Tensile_Stresses_for_Microcracking_in_Monoclinic_Zirconia

https://www.researchgate.net/publication/41699361_Fractal_Aspects_of_the_Martensitic_Transformation_in_Zirconia

https://www.researchgate.net/publication/222373071_TEM_studies_on_phases_and_phase_stabilities_of_zirconia_ceramics

There are two main experimental difficulties with monoclinic ZrO2: 1) It contains many twins. Each removed symmetry element will appear as a set of twins. Furthermore, due to the 81 degree angle, it is not single phase anymore, a small amount (5%) of orthorhombic phase together with micro-cracks appear.

2) Strictly speaking, when you leave the easy-to-study cubic crystals, Miller indices of the plane normal and that of crystallographic directions are not parallel anymore. Hope your crystallographic software considers this.

Regards and good luck W. Wunderlich

Article Critical Tensile Stresses for Microcracking in Monoclinic Zirconia

Article Fractal Aspects of the Martensitic Transformation in Zirconia

Article TEM Studies on Phase Stabilities of Zirconia Ceramics

Hello Wilfried Wunderlich!

I want to find out molar interface area (m^2/mol) of all low index plane for thermodynamic point of view. And for that I should divide it by number of atoms sitting in plane.

There is always relaxation going on at interfaces or surfaces. There is no infinite small layer only a slit with a certain wideness, where you can count atoms, especially in crystal structure with low symmetry.

again, from my point of view it is a quite fundamental answer. You know how many atomd belong to one lattice point. For a general plane (hkl) the number of lattice planes intersection the unit cell is h+k+l. This means you only have the devide the number of atoms by the number h+k+l. The only exclusion would be - as already mentioned, you have stackings of layers described by one lattice points (what is not the case in Baddeleyite).

An orderly array of motifs (lattice point) is known as a crystalline lattice. In the case of Baddeleyite (monoclinic ZrO2) you know it's Z value, I.e., the number of motifs in its unit cell. Then the required answer for the (111) plane is 3Z / 3 = Z.

Dear Aditi,

                      the symbol  111  you used to indicate a lattice plane has a  meaning only when you work with lattice points.  But you are interested to the density of atoms (number of atoms per unit area). In your case the physical quantity which is important is the number of atoms (per unit 2D cell) contained in a slice of thickness d111 . This is a thickness which is allowed by the extinction rules of the space group P21/c. 

Hence, try to made a projection of your structure  along the direction [ -1 1 0 ]. Thus you will find the slices of thickness d111  looking the structure along the direction [-101] or [0-11]. Finally, the density you are interested in is equal to the number of atom (within the slice) divided by the area of the 2D cell made by the vectors [ -1 1 0 ] and [-101].

Comparison should be made with the slice d020, parallel to the 010 plane and  allowed by the extiction rules. One can obtain it from a projection along the [001] direction; in this case the number of atoms (contained in one half of the 3D unit cell) should be divided by the area determined by the vectors [001]  and [100].

@D. Aquilano: Doesn't describe your projection the image I posted 2 weeks ago?