How to determine the oxidation states of the metal atoms when both cation and anion are coordination complexes?
For example, how to determine the oxidation states of $\ce{Pt}$ in the following compound?:
$[\ce{Pt(NH3)4}][\ce{PtCl6}]$
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$\begingroup$ You just have two different oxidation states. Pt in the cation is +2, Pt in the anion is +4. Saying that the average oxidation state is +3 is technically correct, but isn't really useful (imo) and can potentially be misleading. $\endgroup$ – orthocresol♦ Sep 15 '16 at 13:47
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2$\begingroup$ Related: IUPAC name of a salt made of both cationic and anionic complexes $\endgroup$ – user7951 Sep 15 '16 at 14:23
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There is something called an average oxidation state, which is typically not very meaningful chemically, but can (usually) easily be calculated by just knowing the structural formula. In your case, since chlorine is the most electronegative element (save nitrogen), you can assume it to be in $\mathrm{-I}$ state while nitrogen of course would have $\mathrm{-III}$ and hydrogen $\mathrm{+I}$. Thus to get platinum’s average oxidation state just sum up to the compound’s charge:
$$\begin{align}2 x + 4 \cdot (-3) + 3\cdot 4 \cdot (+1) + 6 \cdot (-1) = 0\\ 2x - 12 + 12 - 6 = 0\\ 2x -6 = 0\\ 2x = 6\\ x = 3\end{align}$$
Which gives you an average oxidation state of $\mathrm{+III}$.
But as I said before, average oxidation states do not carry any physical or chemical meaning and can only serve as exercises to students and pupils. Take for example tetrathionate’s ($\ce{S4O6^2-}$) average sulphur oxidation state of $+\frac{5}{2}$.
Instead, one should always attempt to allocate per-atom oxidation states you cannot do that without knowing the charge of each of those complexes, and you need to do it per complex; ideally by drawing it, heterolytically cleaving all bonds (or homolytically for homoatom bonds) and then counting electrons. In the case of your compound — tetraamminplatinum(II) hexachloridoplatinate(IV) — you should arrive at the numbers in brackets in the IUPAC name I gave you, knowing that the cation has $2+$ charge and the anion $2-$.
