..this is a layered perovskite. Actually it is a (four layers i.e. n=4) Aurivillius phase:

(Bi2O2)(An-1 Bn O3n+1)

From this fact (and assuming full and no cross site occupancy) you can assign some valence the atoms:

Sr(+2) - > +2

Bi(+3) -> +12

Ti(+4) -> +16

O(-2) -> -30

If you want to do some calculation, just use the bond valence method (look for it and for Brown). As far as I remember there is both an XRD and a neutron structure refinement of the phase (I can find the refs in my files if you need them), so you can easily calculate the valence using the bond valence idea.

Elements have typical oxidation numbers. Most have several, but often one is strongly prevalent. You need to look it up in Wikipedia or a chemistry book. I have an annotated periodic table for that purpose in my office.

It depends a bit on the nature of the compound which oxidation number is the right one for an element. In your case, an oxide ceramic, it is rather simple. Such metal oxides usually behave a lot like an ionic crystal, that means formal oxidation number and actual charge more or less match.

Now look at the oxides of the components in Wikipedia or a chemistry book. You'll find SrO, Bi2O3 and TiO2. O is mostly (-II) in metal oxides, with some peroxide exceptions such as BaO2, where it is (-I).

The sum of oxidation numbers in a compound has to be zero. For an ion, the sum has to be the charge of the ion.

That means Sr is (+II), Bi is (+III), Ti is (+IV) if you apply this rule to the oxides mentioned above.

If you now look at your compound, you can do the following calculation:

2 + 4*3 + 4*4 - 15*2 = 0

Perfect! Everything balances. A rather normal oxide. The key here is to know or look up typical oxides and oxidation numbers for all elements involved. If you have an element with many different oxidation numbers, like Ru or Mn, you can infer the oxidation number by assigning all other oxidation numbers in the chemical formula and working out which oxidation number you need to balance the equation. In such cases it is quite often a mix, and interesting electronic properties emerge. (La,Sr)MnO3 would be a textbook example for that.

If I extend the above answer especially in view of manganites example as referred by Dieter Weber, the valence state of Bi and Ti can be(2,3) and (3,4), respectively at the expense of oxygen stoichiometry. These concepts are very helpful for te understanding of conduction mechanism and the origin of polarisation.

As Dieter said, elements have 'more suitable' valence or in a proper way, more stable oxidation states. In case of Sr is always +2 (I think alkaly earth elements don't have any other valence). But as Muhammad Nadeem said the case of Bi is different. It usually has +3 valence but exist compounds with +2 and +4 valencies (Ba2Bi2O6) have Bi 'dismutated in +2 and +4. And the case of Ti is even harder it can be +2 +3 +4 or even (in highly reduction conditions) +1.

Once we know this, is very important to say that more stable oxidation state of an element is HIGHLY affected by some parameter such as Temperature synthesis. Besides if you 'quench' your sample (for example) is very probably that the oxidation states may vary.

So, in my opinion you need to do some calculations as the well accepted Bond-Valence method, which is mainly due to the structure of your material (with some formulaes that you should check) and yo can obtain it from a Rietveld refinement.

The best way is to use spectroscopies (XANES (on a Synchrotron) or maybe XPS) also selective chemical titrations should be a good first step, and for sure try to complement the results between this two techniques.

Is not trivial at all to know the formal valency of elements in a compound and many people use to asign wrongly the valence by using just the 'tipical oxidation state of elements'

Even more, O will probably not be exactly 14 per formulae unit...

@Alejandro, if you look that deeply into the electronic structure of a compound, the concepts of valence and oxidation number cease to make sense. Orbitals overlap and form bands with different energy levels. You'll have to look at the band structure (resp. go beyond the simple band model) and see where the Fermi level is etc. I think that in this simple case, one can safely say that Bi is (III) and Ti is (IV) and there's 15 O per formula unit. Any consideration that goes beyond that should not refer to valency or oxidation number, but consider the true electronic structure and defect chemistry of the compound.

@Dieter For sure you must go under bands, we are working on solids. But take a look to the question, it sais 'How to calculate valency of individual atoms in SrBi4Ti4O15 compounds? I am not a phisiscist so maybe i am wrong, but to the best of my knowledge the bands occupancy will be directly related with the number of electrons of the orbitals which participate in the band. (I.e oxidation state of an element).

I have a little experience on this and as a chemist i promise you that depending of the synthesis route the non stoichiometry is VERY VERY common. Is true that this material seems to be quite simple to elucidate valency, but in my opinion you should proof it by any way...

Yes, I also second Alejandro Gomez Perez, the valence state directly depend upon synthesis. In some materials, where we have perfect stoichiometry, even then one may have different domains where oxygen can be deficient and balanced by other domains where oxygen is excess, especially in high temperature synthesis.

@Alejandro, for sure you have to consider such things! What I wanted to say is that the valency and oxidation number model does not apply well to such questions. This model works very well for simple ionic crystals, for typical salts. Complex oxides with interesting properties are not ionic crystals, they are semiconductors or metals in the electronic sense. I know that chemists like to use this model also for "interesting" complex oxides. But it has no predictive power anymore.

You can somehow rationalize your results by assigning fractional oxidation numbers and such, but you cannot predict relevant properties with it. I happen to know both the "physics" and the "chemistry" side of oxides, and I can tell you: If you get fractional or "odd" oxidation numbers, you have to stop using that model! Use the band model in simple cases, and get a huge headache if you have to go beyond it.

There is actually no proper theory anymore for most interesting complex oxides. Think high-Tc superconductors and complex magnetic materials, (La,Sr)MnO3! Such oxides with complex electronic properties are currently an active field of research. Very complicated.

Yes @Dieter, know i think i get your point.

But i still in disagree with you, but just a little.

If you only want to have a little look on the valence state as an information to compare with other references, is very common talk about ''Average oxidation state of an element''. I work a lot with the pair Co+2/Co+3. In this case, in general the more Co+3 the higher electronic conductivity you (have except if charge ordering phenomena is present for sure). So at this point and mainly semi-quantitatively is very usefull to know the percent of each ox-state you have in order to compare.

But for sure that you are in reason when you say that undergo in bands theory could be a big mess... But anyway, a good selective titration in adition to EDS, EELS, XPS and if posible XANES experiments is a good starting point for a good band model (if you finally want it as i think is the final goal,don't you think?

My apologise if i was hard in the answer @Dieter, was not my idea!

@Alejandro, don't worry, I know that my answers are just as hard! I enjoy such discussions and sometimes I get carried away a bit... Sorry if I sounded too hard myself!

I think that your last post sums it up nicely. This entire discussion reflects the complexity of the subject!

Completly agree Dieter, Nice discussion, hope to see you again overhere!

I am coming a bit late to contribute much to the content - in fact, everything important seems to be included. I think that this discussion is a nice example for the useful application of RG!

How I find valency of the molecule is as follow:

1. Identify the major valency of all of the element.

2. cross- check with the electrical neutrality of molecule.

3.Here oxygen is the only negative ion O(-2)X15=-30

4.Positive ionsTi (4+)X4=16 ,Sr(+2)X1=4,Bi(+3)X4=12 overall +30.

5. In this case, it is balanced.

http://www.vaxasoftware.com/doc_eduen/qui/valencia.pdf

http://edtech2.boisestate.edu/lindabennett1/502/compounds%20and%20naming/charge%20balancing.html