Question

Turnbull’s blue and Prussian’s blue respectively are:
$\text{KF}{{\text{e}}^{\text{II}}}\left[ \text{F}{{\text{e}}^{\text{III}}}{{\left( \text{CN} \right)}_{6}} \right]\text{ , KF}{{\text{e}}^{\text{III}}}\left[ \text{F}{{\text{e}}^{\text{II}}}{{\left( \text{CN} \right)}_{6}} \right]$
State whether the given statement is true or false.

Answer 1

Turnbull’s blue is produced when a ferrous salt reacts with ferricyanide. On the other hand, the Prussian’s blue is formed when a ferric ion is treated with a ferrocyanide.

Complete step by step solution: We all have heard of the term Prussian’s blue during Lassaigne’s test to detect the presence of nitrogen in an organic compound. It is a deep blue-colored complex formed when a ferrocyanide complex compound reacts with iron in a $+3$ oxidation state or we can say an aqueous solution of salt-containing ferric ions.
The reaction involved is given below:

$$\text{F}{{\text{e}}^{\text{III}}}\text{(aq)}+{{\text{K}}_{4}}\left[ \text{F}{{\text{e}}^{\text{II}}}{{\left( \text{CN} \right)}_{6}} \right]\text{(aq)}\to \text{KF}{{\text{e}}^{\text{III}}}\left[ \text{F}{{\text{e}}^{\text{II}}}{{\left( \text{CN} \right)}_{6}} \right]\text{(s) (Prussian }\\!\\!'\\!\\!\text{ s blue)} \\\ \text{Ferric salt}+\text{Potassium ferrocyanide}\to \text{Potassium ferric-ferrocyanide} \\\$$

Here, the iron outside the coordination sphere is in a $+3$ oxidation state, and the iron inside the coordination sphere is in a $+2$ oxidation state.
Now Turnbull's blue has the same chemical composition as that of Prussian’s blue. The only difference is the reactants involved in their formation. When an aqueous solution of $\text{F}{{\text{e}}^{2+}}$ salt is treated with a ferricyanide compound, the blue-colored precipitate formed is named Turnbull’s blue.
The reaction involved is given below:

$$\text{F}{{\text{e}}^{\text{II}}}\text{(aq)}+{{\text{K}}_{4}}\left[ \text{F}{{\text{e}}^{\text{III}}}{{\left( \text{CN} \right)}_{6}} \right]\text{(aq)}\to \text{KF}{{\text{e}}^{\text{II}}}\left[ \text{F}{{\text{e}}^{\text{III}}}{{\left( \text{CN} \right)}_{6}} \right]\text{(s) (Turnbull }\\!\\!'\\!\\!\text{ s blue)} \\\ \text{Ferrous salt}+\text{Potassium ferricyanide}\to \text{Potassium ferro-ferricyanide} \\\$$

Here, the iron outside the coordination sphere is in a $+2$ oxidation state, and the iron inside the coordination sphere is in a $+3$ oxidation state.
So, Turnbull’s blue and Prussian’s blue respectively are:
$\text{KF}{{\text{e}}^{\text{II}}}\left[ \text{F}{{\text{e}}^{\text{III}}}{{\left( \text{CN} \right)}_{6}} \right]\text{ , KF}{{\text{e}}^{\text{III}}}\left[ \text{F}{{\text{e}}^{\text{II}}}{{\left( \text{CN} \right)}_{6}} \right]$
Hence, the given statement is true.

Note: The Prussian’s blue and Turnbull’s blue are identical compounds. The reason for the blue color in both of these compounds is the intervalence charge transfer of electrons from $\text{F}{{\text{e}}^{2+}}$ to $\text{F}{{\text{e}}^{3+}}$.