What is the charge on the complex ion in #Ca_2[Fe(CN)_6]#?

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1 Answer
anor277
Jul 29, 2016

As written, you have an #[Fe(C-=N)_6]^(-4)# complex ion.

Explanation:

Calcium metal typically forms the #Ca^(2+)# ion. If this is a real compound (and not a typo), you thus have a #Fe(II+)# centre. The charge of the complex as written is of course zero. Why?

I would check your source.

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