
Which of the following will give $(F{e^{3 + }})$ ions in the solution?
A. ${[Fe{(CN)_6}]^{3 - }}$
B. ${[Fe{(CN)_6}]^{4 - }}$
C. $N{H_4}{(S{O_4})_2}.FeS{O_4}.6{H_2}O$
D. $F{e_2}{(S{O_4})_3}$
Answer
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Hint: ${[Fe{(CN)_6}]^{3 - }}$ and ${[Fe{(CN)_6}]^{4 - }}$ are the complex ions. $N{H_4}{(S{O_4})_2}.FeS{O_4}.6{H_2}O$ is a double salt. $F{e_2}{(S{O_4})_3}$ also breaks into individual ions and will give $F{e^{3 + }}$ ions in the solution.
Complete step by step answer:
${[Fe{(CN)_6}]^{3 - }}$, also called ferricyanide is an anion and Ferrocyanide, ${[Fe{(CN)_6}]^{4 - }}$ are complex ions. They retain their identity in the solution. Hence, they will not give $F{e^{3 + }}$ ions in the solution.
$F{e^{2 + }} + {[Fe{(CN)_6}]^{3 - }} \to F{e^{3 + }} + {[Fe{(CN)_6}]^{4 - }}$
In this reaction ferricyanide ions ${[Fe{(CN)_6}]^{3 - }}$ oxidise iron (II) to iron (III) forming ferrocyanide ions, ${[Fe{(CN)_6}]^{4 - }}$
The third compound, $N{H_4}{(S{O_4})_2}.FeS{O_4}.6{H_2}O$ or Mohr salt is a double salt. It contains only ferrous ions
In the solution, it breaks into individual ions, but it will give $F{e^{2 + }}$ ions in the solution.
Similarly, $F{e_2}{(S{O_4})_3}$, Ferric sulphate is a chemical compound which is a yellow coloured salt and is soluble in water. Iron (III) sulphate is most often generated as a solution rather than being isolated as a solid. It is an extremely strong acidic compound. It also breaks into individual ions. Hence, they will give $F{e^{3 + }}$ ions in the solution.
Hence, only $F{e_2}{(S{O_4})_3}$ will give
$F{e^{3 + }}$ ions in the solution.
Therefore, the correct answer is option (D).
Note: Ferrocyanide, ${[Fe{(CN)_6}]^{4 - }}$ is a diamagnetic species. Although many salts of cyanides are toxic but ferrocyanide ${[Fe{(CN)_6}]^{4 - }}$ and ferricyanide ${[Fe{(CN)_6}]^{3 - }}$ are less toxic because they do not release free cyanide. The treatment of ferrocyanide ${[Fe{(CN)_6}]^{4 - }}$ and ferricyanide ${[Fe{(CN)_6}]^{3 - }}$ salts will give an intense coloured pigment Prussian blue which is sometimes called ferric ferrocyanide ${[Fe{(CN)_6}]^{4 - }}$ and ferrous ferricyanide ${[Fe{(CN)_6}]^{3 - }}$. They both cannot pass through plasma membranes.
Complete step by step answer:
${[Fe{(CN)_6}]^{3 - }}$, also called ferricyanide is an anion and Ferrocyanide, ${[Fe{(CN)_6}]^{4 - }}$ are complex ions. They retain their identity in the solution. Hence, they will not give $F{e^{3 + }}$ ions in the solution.
$F{e^{2 + }} + {[Fe{(CN)_6}]^{3 - }} \to F{e^{3 + }} + {[Fe{(CN)_6}]^{4 - }}$
In this reaction ferricyanide ions ${[Fe{(CN)_6}]^{3 - }}$ oxidise iron (II) to iron (III) forming ferrocyanide ions, ${[Fe{(CN)_6}]^{4 - }}$
The third compound, $N{H_4}{(S{O_4})_2}.FeS{O_4}.6{H_2}O$ or Mohr salt is a double salt. It contains only ferrous ions
In the solution, it breaks into individual ions, but it will give $F{e^{2 + }}$ ions in the solution.
Similarly, $F{e_2}{(S{O_4})_3}$, Ferric sulphate is a chemical compound which is a yellow coloured salt and is soluble in water. Iron (III) sulphate is most often generated as a solution rather than being isolated as a solid. It is an extremely strong acidic compound. It also breaks into individual ions. Hence, they will give $F{e^{3 + }}$ ions in the solution.
Hence, only $F{e_2}{(S{O_4})_3}$ will give
$F{e^{3 + }}$ ions in the solution.
Therefore, the correct answer is option (D).
Note: Ferrocyanide, ${[Fe{(CN)_6}]^{4 - }}$ is a diamagnetic species. Although many salts of cyanides are toxic but ferrocyanide ${[Fe{(CN)_6}]^{4 - }}$ and ferricyanide ${[Fe{(CN)_6}]^{3 - }}$ are less toxic because they do not release free cyanide. The treatment of ferrocyanide ${[Fe{(CN)_6}]^{4 - }}$ and ferricyanide ${[Fe{(CN)_6}]^{3 - }}$ salts will give an intense coloured pigment Prussian blue which is sometimes called ferric ferrocyanide ${[Fe{(CN)_6}]^{4 - }}$ and ferrous ferricyanide ${[Fe{(CN)_6}]^{3 - }}$. They both cannot pass through plasma membranes.
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