Find the attached paper.

Thank you

The crystal structure suffer from an effect as if pressure were applied, chemical pressure. Twinning may result with the large cation found on the twinning plane in a higher coordination.

3 posibilities.

1. Large ions do not incorporate into the lattice replacing smaler ions.

2. Large ions replace the smaller ions, causing the bigger the lattice parameters if the structure remain the same.

3. The structure will reconstruct as result of ion size effects, producing a different structure.

The ABO3 perovskite structures usually adopt various elements of different ionic radius in their crystal lattices. So, the crystal structure will not significantly change or distorted  in ABO3 within the limit of tolerance factor. Of course depending on the substituting elements their properties might be drastically varied. 

I propose you to see my thesis, you may see the structural evolutions in R2-xCexCuO4 with R = Gd, Eu, Nd, Pr with high precisions.

Hi, may I request for the copy of your thesis? Thank you.

Hi. What parts of my thesis do you want?

t = (RA + R0) / (square root(2) x (RB + R0))

In this equation RA, RB and R0 denote the ionic radii of the A-site cation, B-site cation and oxygen ion size respectively. A tolerance, t, value of 0.89 or less indicates a monoclinic or rhombic structure, 0.89 to 1.01 an orthorhombic or tetragonal structure, 1.01 to 1.07 cubic, and greater than 1.07 hexagonal structure.

Dear Shielo,

In fact, several ways can be reached with you system.

First of all, if your substitution can't be achieved, mainly due to size, you may have phase speciation. That's usual way to release tension on the insoluble system. However, if you can incorporate some amount of your cation in the structure, following your description, your crystal lattice will need some small arrangements to release the internal tension. Some crystal structures have the ability to overcome the excess of tension with distortion. It will change your symmetry operators, that may results in another space group for your material, as already point out in the answers you received. The amount incorporated will be lesser than you expected. Higher substitutions usually result in heavy lattice distortions, and sometimes local phase formations, given rise to solid solutions (same material with slight differences in chemical composition) or a more stable second phase precipitation.

Both situations can be easily probed with X-ray diffraction experiment.

Nevertheless, it is important to get some references about your system to have a good idea about the behavior under isotopic substitution.

Say for a crystal having two X atoms in the unit cell and one is replaced by Y atom. won't the symmetry get reduced (change in space group) if we replace an atom with different one ?

Dear Sharad,

In the usual way, if the replacement happens with a lower radius atom one, the lattice should have its parameters changed. Usually, one slight contraction. If the symmetry would or not change, depends on the original space group. If the space group is already of lower symmetry, it tends to keep the lower symmetry. If the atomic radius is higher, the tendency is to slightly increase the lattice parameters. Of course, we are talking about a simple and direct substitution. In some very rare cases, the symmetry would get higher, nevertheless, just keep symmetry is by far, the usual. The lattice substitution is controlled by some limits of size changes. 15 % in difference are the accepted range. Higher than that, undertake large changes and/or nothing happens. A second phase just segregates and no change, nor doping effect is achieved.