As the size of Li is smaller than Ca, so when the Ca is replaced by Li volume should decrease ,But in our case a increase in volume is observed. Plz explain with a reference. Thanks.
It does not depend only on the size of the ion but charge is equally important. Ca2+ has greater charge than Li+ causing the negatively charged ligands to bind more tightly to Ca. Hence one sees an increase in volume as Ca is replaced by Li.
There is another explanation
Li doping causes Mn4+ to change to Mn3+ and Mn3+ being large in size causes increase in lattice size.
Dear Narayan Dutt,
As you are preparing La0.7Ca0.3-xLixMnO3. According to the charge neutrality, for x=0, La3+0.7Ca2+0.3Mn3+0.7Mn4+0.3O3 and for x≠0, La3+0.7Ca2+0.3-xLixMn3+0.7-xMn4+0.3+xO3. This shows that doping of Li+ at Ca-site enhances concentration of Mn4+ in the systems. As ionic size of Mn4+ in smaller than Mn3+, this caused suppression in lattice constants. I hope you will get the point.
Dear Dinesh Kumar,
i have asked the reason for increasing volume , not for decreasing. If Mn4+ are increasing, it will lead to decrease of volume.
Dear Narayan Dutt,
It's Ok. Take it easy. In such case, the increment in the volume may be only when Li+ occupy interstitial site and forms nonstoichiometric system. When an ion having larger size causes increase in unit cell volume and becomes nonstoichiometric. For the reference you can follow any book on defects and vacancy theory.
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