How to find the oxidation number of Co and Cr in [Co(NH3)6] [Cr(C2O4)3]?
$\ce{Co}$ 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 $\ce{[Co(NH3)6][Cr(C2O4)3]}$