According to general opinion, one can expect that the lattice parameter will get increased when the smaller atom is substituted with larger one. This is true for atoms of different types. However, the ionic radius, valance state and electron affinity are the different factors that decide bond length. Here electron affinity is strongly depending on the types of cation and the anion in a crystal and their valance state. If a system has high electron affinity, it reduces the bond length and hence decreases in lattice parameter.

Moreover, one should do careful determination of lattice parameter before and after substitution, probably using some structure refinement software. This may give you correct value of parameters and hence the solution to your problem.

According to Vegard's law increase of size of anu atom or ion being a part of a crystal leads to directly proportional increase of lattice constants. All the same, cation it or anion.

Do both cations have the same charge?

What is the percentage of the "larger cations" that were substituted in the lattice?

Can the "new" cation interact strongly with the original atoms and decrease the bond length between the atoms?

I agree with previous remarks. If the substituting cation has a greater valence state, there can be clustering of these cations accompanied by the loss of smaller cations and the vacancies in the metal sublattice will result. Then a decrease in the unit cell size may follow.

Titov is correct according to Vegards Law which assumes a one for one substitution. However there are deviations to Vegards law when there are very strong or very weak bonding interactions. For example many layered structures exhibit weak interactions along one direction (typically normal to the layer direction) and large lattice parameters and very strong in plane interactions with smaller than predicted lattice parameters.

The actual lattice parameter of a crystal is determined by the equilibrium positions of the atoms relative to each other once all of the relevant forces are balanced. Atomic radii are not always well defined and rigid, since quantum mechanics tells us that the electron wavefunction actually extends to infinity, but its probability amplitude is well defined mathematically and drops off rapidly as you move away from the atom.

One important aspect to consider is the coordination number of the atoms in question, since the effective size of an atom changes depending on how many nearest neighbors it is bonded to. I suspect that the answer to your question may have something to do with that.

There are a set of things you need to take into account, and not just the simple number of atomic radii.

1) There are a dozen different definitions of atomic radii, and when you work in a crystal you should you crystal radii to start with.

1b) Crystal radii are both valence AND coordination dependent (check Shannon radii table for example)

--> So does your substituent and original ion have the same valence/coordination ?

2) If the substituents have a different valence for the coordination of the site than the original, this will lead to the formation of vacancies in the anions (eg O vacancies in doped CeO2) these vacancies will lead to a lattice expansion (because Ce goes from 4+ to 3+ nearby the vacancy)

3) What will your substituent atoms do? Cluster, diffuse homogeneously, stay stuck at the surface of your crystals? This may lead to all kinds of reconstructions, and may result in your cell parameters changeing asymmetrically.

4) are you sure the atoms are not substituted interstitially?

For many of these aspects, if uncertain or unknown, theoretical calculations may be of use in clarifying the situation.

As you can see quite some complicated things can be happening. I have done theoretical work in this area with regard to doped CeO2, maybe this can be of use for you:

http://onlinelibrary.wiley.com/doi/10.1111/jace.12650/abstract

http://journals.aps.org/prb/abstract/10.1103/PhysRevB.84.054110

http://www.sciencedirect.com/science/article/pii/S0169433212000451

http://pubs.rsc.org/en/Content/ArticleLanding/2012/JM/c2jm15752g#!divAbstract

Danny

I agree firmly to Titov. However, as mentioned by Fernandez, the charge of the introduced cataion also matters. If the charge of the introduced cation differs from the charge of the cation of host lattice, use of Vegard´s law will not give you the correct lattice constant.

The answer to the question is given in the problem description itself as Vegard's law. However, an other problem which reads opposite to this behaviour is mentioned in the same description. That needs to be addressed.

If the crystallographic structure has two or more lattice sites, the size of the ions with same valence state occupying those different sites differ according to Shannon. Also, some ions change their valence too in the synthesis. In such case, it may be erroneous to mention that ion with smaller ionic radius is substituted by an ion with larger ionic radius without knowing their specific site preferences and valence states. For example, if an ion changes its valence to higher level while occupying the preferred lattice site, the lattice constant decreases even though its ionic radius is larger in the original valence state. Further, if the substituted ions crystallize into additional phases, then also the lattice constant of the major phase decreases. More explanations in this regard are only structure specific and can not be generalized.

When a cation or anion replace or substitute in the host atom, the size of cation/anion and which atom is replacing matter to estimate the lattice parameters.

Suppose, If same ionic radius atoms replaces to each other, no effect will come on lattice parameters. Lattice parameters change due to dissimilar ionic radius, which causes to generate the lattice strain too.

You should check if there is a change in the oxidation state of that ion in the crystal , especially if it companies with phase transformation , or if there is defect in the lattice

Most possible reasons for this could be:

1. The substituted cation has different valence state(possibly higher that the former one) and thus cause enhanced bonding strength and shorter bond length. Meanwhile, due to the charge uneutrality, vacancies may be introduced, which could also increase the lattice parameters.

2. The substituted cation entered different sites, or the substituted cation and the former ones formed some kind of superstructure, such as cation ordering. However, all those should cause variation of crystal structures more or less.

XRD Rietveld analysis should be possible to explain the phenomena in detail.

It is possible that the substitution of small ions by large ions did not happen. Maybe the large ions replace other sites instead of the sites of small ions.

Change of valence state, additional phase formation and defect structures due to strain on such substitutions are some of the reasons for decrease in lattice constant.

Lattice constant is decided by its constituent atoms/(cat)ions. One may thus assume that size of the atom/ion decides the size of a lattice. Hence, a bigger ion/atom ought to occupy more space than a smaller one and consequently replacement of a bigger ion/atom for smaller one results in enhanced lattice parameter.

To judge about the influence of substituents on the lattice parameter I think it wise first to get assured that the solution is in the equilibrium state. The matter is that sometimes after a rather prolonged ceramic procedure you do not obtain the distribution of cations which can be called equilibrium, i.e. meeting the minimum of mixing energy. The second: some kinds of bismuth niobates (Bi3NbO7) have fluorite type structure, however niobium has octahedral surrounding with oxygen - I still cannot understand how it happens, but its parameter must be lower.

Thank you all for the responses. As some of you wanted to clarify, in my system a 2+ cation with larger ionic radius was substituting a 3+ cation. In fact the original lattice has two different cations occupying the same wyckoff site, one 3+ and another 4+ cations. But the 4+ cations are way too smaller to be substituted by this substituent cation. That's why I assumed the 2+ cation does go to the site of 3+ cation (not quite sure whether I'm right). The trend of decrease in lattice parameter with substituent concentration was almost linear too.

According to general opinion, one can expect that the lattice parameter will get increased when the smaller atom is substituted with larger one. This is true for atoms of different types. However, the ionic radius, valance state and electron affinity are the different factors that decide bond length. Here electron affinity is strongly depending on the types of cation and the anion in a crystal and their valance state. If a system has high electron affinity, it reduces the bond length and hence decreases in lattice parameter.

Moreover, one should do careful determination of lattice parameter before and after substitution, probably using some structure refinement software. This may give you correct value of parameters and hence the solution to your problem.

For the same element ionic radius of a cation 2+ more than cationic radius of a cation of 3+ as the first has more than electrons. So linear increase of lattice constant at substitution 3+ -> 2+ seems quite natural.

The answer from Rangasami Chinnusamy is quite a reasonable explanation.