A) ferricyanide
B) ferrocyanide
C) hydrogen cyanide
D) sodium cyanide
Correct Answer: B
Solution :
\[\text{F}{{\text{e}}^{\text{3+}}}\]solution gives Prussian blue with \[{{K}_{4}}[Fe{{(CN)}_{6}}],\]potassium ferrocyanide. \[4\overset{\text{III}}{\mathop{Fe}}\,C{{l}_{3}}+3{{K}_{4}}[\overset{\text{II}}{\mathop{F}}\,e{{(CN)}_{6}}]\xrightarrow{{}}\underset{\begin{smallmatrix} \text{ferric}\,\text{ferrocyanide} \\ \,\,\,\,\,\,\,\,(Prussian\,blue) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,12\,KCl \end{smallmatrix}}{\mathop{{{\overset{\text{III}}{\mathop{Fe}}\,}_{4}}{{[\overset{\text{II}}{\mathop{Fe}}\,{{(CN)}_{6}}]}_{3}}}}\,\]
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