A) \[{{K}_{4}}[Fe{{(CN)}_{6}}]\]
B) \[N{{a}_{4}}[Fe{{(CN)}_{6}}]\]
C) \[F{{e}_{4}}{{[Fe{{(CN)}_{6}}]}_{3}}\]
D) None of these
Correct Answer: C
Solution :
\[\text{F}{{\text{e}}^{\text{2+}}}\]ions are oxidized to \[F{{e}^{3+}}\]which reacts with sodium ferrocyanide to give ferri ferrocyanide. \[4F{{e}^{3+}}+3N{{a}_{4}}[Fe{{(CN)}_{6}}]\to \underset{\text{Prussian}\,\text{blue}}{\mathop{F{{e}_{4}}{{[Fe{{(CN)}_{6}}]}_{3}}}}\,+12\,N{{a}^{+}}\] Hence, the correct option is (c).
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