# Structural isomers of hexaammineiron(III) hexacyanidocobaltate(III)

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.

• 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 May 24 '20 at 7:28
• We prefer to not use MathJax in the title field due to issues it gives rise to; see here for details. May 24 '20 at 7:30
• @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 ;) May 24 '20 at 9:20
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• @andselisk Thank you for that! I will keep that in mind in the future! May 24 '20 at 14:28

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.
• @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. May 24 '20 at 17:32