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In the question I actually got I had to name the above compound, and I know how to name it but I don't know how to find the oxidation number of the above compound(which has to be mentioned in the IUPAC name). I actually know how to find the oxidation number in complexes but here both the ions are complex ions which makes it confusing. If someone needs the name its Hexaamminecobalt(?) trioxalatochromate(?)

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  • $\begingroup$ So, what you propose to be the name? $\endgroup$ – Mithoron Nov 21 at 16:11
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$\ce{Co}$ is known to have two possible oxidation numbers in solution : +II and +III. Both make complexes with 6 $\ce{NH3}$. If $\ce{Co}$ is at +II, the charge of the complex is $2$+, as $\ce{NH3}$ is not charged. If $\ce{Co}$ is at +III, the charge of the complex is $3$+ for the same reason.

For chromium, it is similar, as $\ce{Cr}$ may have oxidation number +II and +III. But the ligand oxalate is charged $-2$. The complex of $\ce{Cr}$ with three oxalate ions must compensate the charges of the cobalt complex (+$2$ or +$3$). So the charge of the chromium complex has to be -$2$ or -$3$. In other words, the complex ion is either $\ce{[Cr(C2O4)3]^{2-}}$ or $\ce{[Cr(C2O4)3]^{3-}}$. Let's compare the two possibilities, and the oxidation numbers of $\ce{Cr}$ in these two structures.

To get a complex $\ce{[Cr(C2O4)3]^{2-}}$, $\ce{Cr}$ oxidation number should be equal to $x$, with $x + 3(-2) = -2$. So $x$ = $+4$ or +IV. And unfortunately, this oxidation number IV does not exist for $\ce{Cr}$ in solution. This structure has to be abandoned,

On the other hand, to get a complex $\ce{[Cr(C2O4)3]^{3−}, Cr}$ must have an the oxidation number $y$ given by : $y +3(-2) = +3$. So $y$ = $+3$ = +III, and this is possible and even welcome. It corresponds to one of the requirements developed previously. As a final result, the complex ion $\ce{[Cr(C2O4)3]^{3-}}$ has a charge $3-$ that can be coupled with $\ce{Co(NH3)6^{3+}}$.

As a consequence, both Co and Cr are at oxidation number +III in the original structure $\ce{[Co(NH3)6][Cr(C2O4)3]}$

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    $\begingroup$ You could just say the charge of the ligands adds up to negative six, so the cations have to have a total of six charges as well. $\endgroup$ – Karsten Theis Nov 22 at 15:09

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