Question

(A) When ${\text{Fe}}{({\text{CN}})_2}$​ solution is treated with ${\text{KCN}}$solution, species formed does not give tests of ${\text{F}}{{\text{e}}^{2 + }}$ and ${\text{C}}{{\text{N}}^ - }$.
(B) When ${{\text{K}}_2}{\text{S}}{{\text{O}}_4}$​ solution is treated with ${\text{A}}{{\text{l}}_2}{{\text{(S}}{{\text{O}}_4})_3}$​ solution, species formed gives tests of ${K^ + },{\text{A}}{{\text{l}}^{3 + }}{\text{ and SO}}_4^{2 - }$.
Species formed in experiment A does not give test of ${\text{F}}{{\text{e}}^{2 + }}$ and ${\text{C}}{{\text{N}}^ - }$It is due to formation of________ .
A. ${{\text{K}}_2}{\text{[Fe}}{({\text{CN}})_4}]$
B. ${{\text{K}}_3}{\text{[Fe}}{({\text{CN}})_5}]$
C. ${{\text{K}}_3}{\text{[Fe}}{({\text{CN}})_6}]$
D. ${{\text{K}}_4}{\text{[Fe}}{({\text{CN}})_6}]$

Answer 1

The test is not shown by the ferrous and cyanide ions due to the formation of complex. The complex formed by iron with cyanide is octahedral with an oxidation state of iron as same as that was in ionic state.

Complete step by step answer:
When ferrocyanide that is ${\text{Fe}}{({\text{CN}})_2}$ is treated with potassium cyanide that is ${\text{KCN}}$ they form a coordination complex as ${{\text{K}}_4}{\text{[Fe}}{({\text{CN}})_6}]$. The complex formed is octahedral. Octahedral complexes are those complexes in which the coordination number of metal is 6. Coordination number is the number of ligands that are directly attached with the metal ion in a coordination sphere. The coordination sphere remains intact and does not dissociate into the ions. Hence instead of ferrous and cyanide ions, ${{\text{[Fe}}{({\text{CN}})_6}]^{4 - }}$ ions will be there in the solution.
Hence no test will be shown due to the formation of the complex ${{\text{K}}_4}{\text{[Fe}}{({\text{CN}})_6}]$.
If we look at equation B, when potassium sulphate that is ${{\text{K}}_2}{\text{S}}{{\text{O}}_4}$ is treated with ${\text{A}}{{\text{l}}_2}{{\text{(S}}{{\text{O}}_4})_3}$ a double salt is formed. The formula of double salt formed is ${{\text{K}}_2}{\text{S}}{{\text{O}}_4}\,.{\text{ A}}{{\text{l}}_2}{{\text{(S}}{{\text{O}}_4})_3}$. A double salt completely dissociates into the ions unlike the complexes. Hence in the solution potassium, aluminium and sulphate ions will be there and so they will give the positive test. A double salt is a salt that contains more than one type of cation or anion.

Hence the correct option is D.

Note:
${{\text{K}}_4}{\text{[Fe}}{({\text{CN}})_6}]$ is known as potassium ferrocyanide. The IUPAC name of ${{\text{K}}_4}{\text{[Fe}}{({\text{CN}})_6}]$ is potassium hexacyanoferrate. It is used as an anti-caking agent in salts, especially table salt and road salt. It is non toxic and does not give cyanide upon dissociation.