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$$\begin{array}{cc}\hline \textit{Formula of Complex} & {\textit{Rewritten Formula (showing the}\\\textit{ coordinating ligands)}}\\ \hline \ce{Pt(NH3)6Cl4} & \ce{[Ph(NH3)6]Cl4}\\ \ce{Pt(NH3)5Cl4} & \ce{[Ph(NH3)5Cl]Cl3}\\ \ce{Pt(NH3)4Cl4} & \ce{[Ph(NH3)4Cl2]Cl2}\\ \ce{Pt(NH3)3Cl4} & \ce{[Ph(NH3)3Cl3]Cl}\\ \ce{Pt(NH3)2Cl4} & \ce{[Ph(NH3)2Cl4]}\\ \hline \end{array}$$

How is it determined that all the $\ce{NH3}$ go in the brackets with $\ce{Pt}$ and then how ever many $\ce{Cl}$ can fit in with them? Why not all the $\ce{Cl}$ and then just some of the $\ce{NH3}$?

And is there any way to tell the $\ce{Pt}$ should have 6 ligands other then just memorizing it?

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    $\begingroup$ You simply need to have zero total charge and NH3 can't act as counterion as it's neutral. You'd need to know much more to predict coordination numbers with ease. $\endgroup$ – Mithoron Apr 18 '16 at 21:17
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    $\begingroup$ It doesn't always get six ligands. $\endgroup$ – SendersReagent Apr 19 '16 at 1:55
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There is quite a bit behind the table that you are either expected to know or expected to derive. I hope the latter, so here’s the way to go.

First, consider the sum formulae. Ammonia is a neutral entity, platinum is most likely positively charged which leaves chlorine to be negatively charged (as a chloride ion). Since we always have four chlorides, and they all carry a single negative charge, the platinum must be $\ce{Pt^4+}$. Since it is in the 10th group, it will have six remaining valence electrons.

Considering the 18-electron-guideline,[1] this calls for six ligands in an octahedral ligand-sphere. Considering platinum to adopt a low-spin conformation (the standard for 4d and 5d metals) would allow maximum LFSE stabilisation,[2] giving us a lot of indications to assume an octahedral configuration.[3]

The next thing to know is that ammonia is generally a stronger-field ligand when compared to chloride, meaning that in the presence of more ammine ligands the LFSE is even larger. This makes it reasonable to assume a maximum of ammine ligands before any remaining ligand positions are filled up by chloride.

You could have arrived here differently by assuming that anything not forming part of the cationic (or neutral) platinum complex must be a counterion and thus chloride. But that may fail especially in the presence of water due to water of crystallisation. While I have never seen it in person, I am sure that ammonia of crystallisation will exist, too.

As I kind-of explained in the previous paragraph, having more chlorides form part of the complex and additional ammonia of crystallisation sounds possible, but due to what I wrote in the paragraph above it just seems less likely. For this platinum series, all compounds were probably isolated and analysed by X-ray diffraction which would prove unanimously which structure is correct; later the correct structures were rationalised by the scheme shown.


Notes

[1]: I just don’t want to use the word rule as it is too strong.

[2]: $24~\mathrm{Dq}$, in case you want to calculate it.

[3]: No, platinum doesn’t always in every configuration take six ligands. Platinum(II) is well-known to take only four in a square-planar configuration, e.g. cisplatin or cis-diamminedichloridoplatinum(II). But for a given oxidation state of a metal, there is usually one preferred coordination geometry or a small set of preferred geometries.

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  • $\begingroup$ Thanks a lot for the in depth answer. And I'n not expected to know or derive any of it. At least not at the time being. Despite people assuming it is, and tagging re-tagging as 'homework' it actually isn't. I was just reviewing topics from last year, when we went over coordination compounds very briefly, found that table and some problems online, and became interested in finding out more. Again, thanks for the explanation, and note 2 as well. $\endgroup$ – Caesium-133 Apr 19 '16 at 21:13
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    $\begingroup$ @Caesium-133 Oh, the tag homework doesn’t necessarily mean it is an assignment given by your teacher/professor at school/uni/whatever that you are expected to solve. It has a much larger scope that is poorly reflected in its name and we have a heated Chemistry Meta-debate as to which name it should get, what questions it should be applied to etc. As the current unchanged policy stands, the question is a great example of an acceptable homework question. $\endgroup$ – Jan Apr 20 '16 at 11:22
  • $\begingroup$ Ahh ok, I see. That's interesting. $\endgroup$ – Caesium-133 Apr 20 '16 at 14:00

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