Question

The central metal ion in Turnbull's blue complex is

• A. $F{e}^{2+}$
• B. $F{e}^{3+}$
• C. $F{e}^{4+}$
• D. Both (a) and (b)

Answer 1

The correct option is D Both (a) and (b)

The Turnbull's blue complex is $F{e}_3\left[Fe\right(CN{)}_6{]}_2$. It is prepared by reaction of $FeC{l}_2$ with $K_3\left[Fe\right(CN{)}_6]$.

$3FeC{l}_2+2K_3\left[Fe\right(CN{)}_6]\to F{e}_3[Fe(CN{)}_6{]}_2+6KCl$​
The complex $F{e}_3\left[Fe\right(CN{)}_6{]}_2$ dissociate as $3F{e}^{2+}\text{and}2\left[Fe\right(CN{)}_6{]}^{3-}$
Thus, oxidation state of Fe in coordination entity is +3.
It can also be represented as $KFe\left[Fe\right(CN{)}_6]$ where central metal atom is in +2 oxidation state.
Hence, option (d) is correct.

The colour of the complex is due to the Metal-Metal charge transfer.