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Find the number of structural isomers of $\ce{[Fe(NH_3)_6][Co(CN)_6]}$

This was a question solved by my teacher as follows:

The $\ce{CN}$ will show linkage isomerism while $\ce{NH_3}$ and $\ce{CN}$ will show coordination isomerism.

We will now consider cases of coordination isomerism and then use linkage isomerism.

  • $\ce{[Fe(NH_3)_6][Co(CN)_6]}\rightarrow 7$ isomers(due to linkage isomerism of $\ce{CN}$)
  • $\ce{[Fe(NH_3)_5(CN)][Co(CN)_5(NH_3)]}\rightarrow 2\times 6=12$
  • $\ce{[Fe(NH_3)_4(CN)_2][Co(CN)_4(NH_3)_2]}\rightarrow 3\times 5=15$
  • $\ce{[Fe(NH_3)_3(CN)_3][Co(CN)_3(NH_3)_3]}\rightarrow$ not counted.

The further $3$ cases are similar to the first $3$ and will result in the same number of isomers. Hence the answer is $2\times(7+12+15)=68$.

Why is $\ce{[Fe(NH_3)_3(CN)_3][Co(CN)_3(NH_3)_3]}$ "not counted"?

I feel that isomers due to $\ce{CN}$ showing linkage isomerism are perfectly valid. Please correct me if I'm wrong and sorry if the question is too trivial.

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  • $\begingroup$ Firstly, there is a typo in your solution. It should be 2*(7+12+15)=68. Also I feel that the compound which you mentioned has to be counted but only once.(in the sense that it wouldn't be inside the *2. And just 4*4=16 isomers in the case. Giving a total of 68+16=84 isomers $\endgroup$ – user600016 May 24 at 7:28
  • $\begingroup$ We prefer to not use MathJax in the title field due to issues it gives rise to; see here for details. $\endgroup$ – Zenix May 24 at 7:30
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    $\begingroup$ @user600016 You are right, it's not the case in MSE, rather we generally edit the titles and add MathJax. Here, \ce{^} paves problem, which is not the case in MSE ;) $\endgroup$ – Zenix May 24 at 9:20
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    $\begingroup$ @AryanSonwatikar Rule of thumb for the same chemical compound appearing both in title and in the body: whenever possible, use chemical names in the title and chemical formulas in the body. This way you not only avoid problems with the search engines incorrectly parsing MathJax, but also gain better search coverage for the future visitors, sort of kindergarten-level CEOing. $\endgroup$ – andselisk May 24 at 14:16
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    $\begingroup$ @andselisk Thank you for that! I will keep that in mind in the future! $\endgroup$ – AryanSonwatikar May 24 at 14:28
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I have found the answer:

In the following case: $$\ce{[Fe(NH_3)_3(CN)_3][Co(CN)_3(NH_3)_3]}$$ Both the constituent entities are neutral and hence will exist independently, and thus will not form isomers of the original compound.

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  • $\begingroup$ Pfft, it still could be a compound not necessarily a mixture. CN ligands are bridging, so most of reasoning in all this makes little practical sense. $\endgroup$ – Mithoron May 24 at 15:57
  • $\begingroup$ @Mithoron Could you elaborate as to how $\ce{CN}$ ligands being a bridging states that the above will exist as a compound? Please keep in in mind I'm only 16 and have just started learning Coordination Chemistry- all I have learnt is nomenclature and Isomerism; no CFT and so on. $\endgroup$ – AryanSonwatikar May 24 at 17:32
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    $\begingroup$ Well if the two metallic centers were connected (Fe-CN-Co) then they wouldn't exist independently. Of course formula would be different, but I'm talking about practical aspect here, maybe pointlessly as such isomer counting has little to do with realism and more with training in 3D thinking. $\endgroup$ – Mithoron May 24 at 17:47
  • $\begingroup$ @Mithoron I guess that's what my teacher had in mind. This was just after we had been explained all the types of isomerizations. $\endgroup$ – AryanSonwatikar May 25 at 2:33

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