I suppose this is simply the same effect as found for Al2O3 when you add e.g. Cr or Ti making the typical ruby or sapphir gemstones. At these high temperature, you have inserted Ce atoms on regular lattice points of either Sr or Ti, resulting in the color change. The more Ce you add, the more lattice point Ce atoms occupy.

It is always a case in cerium doping as follows:

To start with the calcination, it forms Ce(III, 4f^1)) oxide [Ce2O3; colorless but absorbs in uv]. With the increasing intensity of calcination, further oxidation to its most stable Ce(IV, 4f^0)) [ CeO2; orange red / yellow ] takes place. Depending upon the % of Ce(III) / Ce(IV) obtained which occupy the interstial sites in the defect stcucture,the color intensities vary to give different colors.

No, mr. Sehgal. I have doped with Ce(IV) and then calcined it. That means that before calcination the dopant is in (IV) state. I have used CeO2 as a dopant which is light yellow in color. and also Ce is occupying Sr sites, not interstitial sites. It might be possible that Ce(IV) get reduced to Ce(III).

I have made another sample having 0 mol % CeO2 doped in SrTiO3 [i.e. pure SrTiO3] and then calcined it at same temperature, then the again the color of sample get changed to light brown. I want to know the chemistry behind color change due to calcination.

[A] Ce(III) occupy lattice sites and not the interstitial sites.

[B] It has well been reported [J. Am. Ceram. Soc., 1–8 (2011)] that:

(i)Ti(iV) of the perovskite [SrTiO3;ABO3] gets reduced to Ti(III) THOUGH THIS IS NOT CLEARLY EXPLAINED in the paper.

(ii) As the lattice parameters of the dopped sample was lower than the undopped SrTiO3 sample,it suggests that the majorirty of the cerium is Ce(III) substituting on the A site

.B site substitution will result in an in INCREASE IN LATTICE PARAMETERS and

A site substitution by Ce(IV) should not be expected due to the large difference in the relative ionic radii of Sr(II) and Ce(IV).AGAIN REDUCTION OF Ce(IV) TO Ce(III) is NOT EXPLAINED IN THE PAPER.

[C]I, very humbly, put forth two arguments to explain both the above points which were left unexplained as:

(i)There occurs MLCT [Metal Ligand charge transfer]to reduce it to Ce(III) which occupy some A sites

(ii) The excess negative charge reduces Ti(IV) to Ti(III).

Ti(IV) +e -----------Ti(III).

Pls. treat it as a humble academic discussion which is open to constructive criticism.

Pls read

here occurs MLCT [Metal Ligand charge transfer]to reduce it to Ce(III) which occupy some A sites

As

There occurs LMCT[Ligand metal charge transfer] from the O(-2) to Ce(IV) to reduce it to Ce(III) which occupy some A sites

It means that the color change is due to reduction of Ti(IV) to Ti(III). Because in another sample having 0 mol % of ceria doped, again the color is changing.

Initially, before calcination the color of sample was white. [ See the attached file, which is showing the sample after calcination] I am calcining the sample in air.

The color change is due to the electronic exchange produced by the presence of the the different oxidation states of cerium and titanium see P. Day Introduction to mixed valence chemistry in Mixed Valence Compounds Eds D.B. Brown, D. Ridel or Mixed valence: origins and developements P. Day N.S. Hush, R.J.H. Clark Phylosofical Trans of the Royal Society A 2008 366.

I think high temp calcination change crystallinity which effect refractive index due to which color is changed.