Question

Reddish brown chocolate precipitate is formed by mixing solutions containing:
a.) $C{{u}^{2+}}$ and ${{[Fe{{(CN)}_{6}}]}^{3-}}$ ions
b.) $C{{u}^{2+}}$ and ${{[Fe{{(CN)}_{6}}]}^{4-}}$ ions
c.) $P{{b}^{2+}}$ and $S{{O}_{4}}^{2-}$ ions
d.) $P{{b}^{2+}}$ and ${{I}^{-}}$ ions

Answer 1

To solve this type of question first we should understand each and every reaction given in the options and the type of product formed after the reaction. The reddish brown precipitate will be of copper Ferrocyanide in which copper will be present in +2 oxidation state.

Complete Solution :
Potassium Ferrocyanide is used to precipitate reddish brown copper Ferrocyanide in +2 oxidation state in the solution. The reaction involves id mentioned below:
$C{{u}^{2+}}(aq)+{{[Fe{{(CN)}_{6}}]}^{4-}}(aq)\to C{{u}_{2}}[Fe{{(CN)}_{6}}](s)$

The balanced chemical reaction is mentioned below:
$2C{{u}^{2+}}(aq)+{{[Fe{{(CN)}_{6}}]}^{4-}}(aq)\to C{{u}_{2}}[Fe{{(CN)}_{6}}](s)$

Hence, the correct answer is option (B) i.e. reddish brown chocolate precipitate is formed by mixing solutions containing $C{{u}^{2+}}$ and ${{[Fe{{(CN)}_{6}}]}^{4-}}$ ions.

Additional information:
Copper Ferrocyanide precipitates as an inorganic material which acts as the best semipermeable membrane because copper ferrocyanide contains pores that are big enough for the water molecule to pass through it but the pores are two smaller for the particles which are larger in size than the water molecules. Hence, the heavier molecules do not pass through this solution.
So, the correct answer is “Option B”.

Note: The compound ${{[Fe{{(CN)}_{6}}]}^{4-}}$ is an ionic compound it is usually found in the form of either potassium salts or sodium salts of this complex in the laboratory along with the yellow color solution. It is less toxic than other cyanide complexes because the cyanide ligand present in this compound is not in free state. If the cyanide will be present in free state, then this compound would be highly toxic as it contains 6 cyanide atoms.