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