# What's the electronic configuration of Nickel and Oxygen in RNiO3 (R=Rare-Earth)?

Nickel's electronic ground state is $\mathrm{3d^8 4s^2}$ (there's a dispute on this, but let's not worry about it for now). Oxygen's is $\mathrm{2s^2 2p^4}$. In a Rare-Earth based Nickelate with chemical formula $\mathrm{RNiO_3}$, if we assume simple ionic bonds, then what should the electronic state be?

Literature indicate that the "classical" case is $\mathrm{Ni^{3+} O^{2-}}$ corresponding to $\mathrm{3d^7, 2p^6}$. But this means that Oxygen has gained 2*3=6 electrons. But how can this be possible if Nickel has donated only 3 electrons?

As an example, let's look at $\ce{LaNiO3}$. Each oxygen has an oxidation state of -2, nickel has an oxidation state of +3 and lanthanum has an oxidation state of +3. Overall the molecule is neutral. $$\ce{La(+3)~ plus~ Ni(+3)~plus~ 3~oxygens (-6)~=~0}$$
• Thanks. In this configuration: $Ni^{2+} O^{-}$, why do we write $3d^8 \underline{L}$? What does it mean? – student1 Oct 19 '14 at 21:09
• I'm not familiar with this notation, is there a specific web page that you are referring to? Perhaps it represents nickel being $\ce{3d^8}$ after giving up two electrons and becoming $\ce{Ni^2}$, and the L represents any rare earth lanthanide ion. – ron Oct 19 '14 at 21:29