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

As written, you have an ${\left[F e {\left(C \equiv N\right)}_{6}\right]}^{- 4}$ complex ion.
Calcium metal typically forms the $C {a}^{2 +}$ ion. If this is a real compound (and not a typo), you thus have a $F e \left(I I +\right)$ centre. The charge of the complex as written is of course zero. Why?