Question

In a ferric salt on adding KCN a Prussian blue is obtained which is:
A. ${K_3}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$
B. $F{e_3}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$
C. $FeS{O_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$
D. $F{e_4}{\left[ {Fe{{\left( {CN} \right)}_6}} \right]_3}$

Answer 1

We know that Prussian blue is formed by the oxidation of ferrous ferrocyanide salts. It appears as a dark blue pigment. It contains two iron atoms in different oxidation states and a negatively charged hexacyanoferrate ion. Prussian blue is mostly used in paints, used as an antidote from poisoning of heavy metals and radioactive isotopes of cesium. The other name of Prussian blue is Berlin blue. We know that Prussian blue is formed by the oxidation of ferrous ferrocyanide salts. It appears as a dark blue pigment.

Complete step by step answer:
We have to know that the coordination complex is a compound that consists of a central atom (ion) that is metallic and is known as the coordination centre. It is enclosed by array bound molecules (or) ions called ligands (or) complexing agents. Transition metals are coordination complexes. A coordination complex whose centre comprises a metal atom is called a metal complex of d block element.
Firstly, we have to react with iron with cyanide to form ${\left[ {Fe\left( {C{N_6}} \right)} \right]^{4 - }}$ . We can write the equation as,
$F{e^{2 + }} + 6C{N^ - } \to {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }}$
The product ${\left[ {Fe\left( {C{N_6}} \right)} \right]^{4 - }}$ again reacts with iron to form ${\left[ {F{e_4}\left( {C{N_6}} \right)} \right]_3}$ . We can write the equation as,
$3{\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }} + 4F{e^{3 + }} \to F{e_4}{\left[ {Fe\left( {C{N_6}} \right)} \right]_3}$
The name of the compound ${\left[ {F{e_4}\left( {C{N_6}} \right)} \right]_3}$ is iron (III) hexacyanoferrate (II).
So, when we react iron (III) chloride with potassium ferrocyanide ${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$ , we get the product iron (III) hexacyanoferrate (II) $F{e_4}{\left[ {Fe{{\left( {CN} \right)}_6}} \right]_3}$ and potassium chloride. We can write the chemical equation as,
$3{K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right] + 4FeC{l_3} \to F{e_4}{\left[ {Fe\left( {C{N_6}} \right)} \right]_3} + 12KCl$
The product $F{e_4}{\left[ {Fe{{\left( {CN} \right)}_6}} \right]_3}$ is also called ferric ferrocyanide and it appears as Prussian blue (or) Berlin Blue.
The oxidation state of iron in the compound $F{e_4}{\left[ {Fe{{\left( {CN} \right)}_6}} \right]_3}$ is +3 and +2. So, the IUPAC name of this compound is Iron (II,III) hexacyanoferrate (II,III).
When we add ferric chloride over potassium ferrocyanide, it produces a white precipitate that will turn into blue. We can write this reaction as,
${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right] + FeC{l_3} \to KFe\left[ {Fe\left( {C{N_6}} \right)} \right] + 3KCl$
This is normal, since being only a small amount of ferric chloride, and not all potassium is displaced.

Therefore, the option D is correct.

Note: In the Lassaigne’s test for the identification of nitrogen in organic compounds, we can notice the appearance of a blue coloured compound which is because of ferric ferrocyanide.
In this test, when we boil sodium fused extract with ferrous sulfate, and later acidify it with concentrated sulfuric acid, we can confirm the presence of nitrogen due to the appearance of Prussian blue coloured complex ferric ferrocyanide. We can write the chemical reaction as,
$FeS{O_4} + 2NaOH \to Fe{\left( {OH} \right)_2} + N{a_2}S{O_4}$
$Fe{\left( {OH} \right)_2} + 6NaCN \to N{a_4}Fe{\left( {CN} \right)_6} + 2NaOH$
$3N{a_4}Fe{\left( {CN} \right)_6} + 4FeC{l_3} \to F{e_4}{\left[ {Fe{{\left( {CN} \right)}_6}} \right]_3} + 12NaCl$ .