A) \[{{K}_{4}}\left[ Fe{{\left( CN \right)}_{6}} \right]\]
B) \[{{H}_{2}}S\]
C) \[N{{H}_{4}}CNS\]
D) \[KCNS\]
Correct Answer: A
Solution :
\[FeC{{l}_{3}}+{{K}_{4}}\left[ Fe{{(CN)}_{6}} \right]\to \underset{\text{Ferri ferrocyanide (Blue)}}{\mathop{F{{e}_{4}}{{\left[ Fe{{(CN)}_{6}} \right]}_{3}}}}\,\] \[2FeC{{l}_{3}}+3{{H}_{2}}S\to F{{e}_{2}}{{S}_{3}}+6HCl\] \[3N{{H}_{4}}CNS+FeC{{l}_{3}}\underset{(\text{Blood}\ \text{red})}{\mathop{\to Fe{{(CNS)}_{3}}+}}\,\]\[3N{{H}_{4}}Cl\] \[FeC{{l}_{3}}+3KCNS\to \underset{(\text{Blood red)}}{\mathop{Fe{{(CNS)}_{3}}}}\,\]+3KCl
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