A) \[{{K}_{2}}[Cu{{(CN)}_{4}}]\]
B) \[{{K}_{3}}[Cu{{(CN)}_{4}}]\]
C) \[CuC{{N}_{2}}\]
D) \[Cu\,[K\,Cu{{(CN)}_{4}}]\]
Correct Answer: B
Solution :
\[CuS{{O}_{4}}\] reacts with KCN and gives a white precipitate of cuprous cyanide and cyanogen gas. The cuprous cyanide dissolves in excess of KCN forming \[{{K}_{3}}[Cu{{(CN)}_{4}}]\]. \[CuS{{O}_{4}}+2KCN\xrightarrow{{}}{{K}_{2}}S{{O}_{4}}+Cu{{(CN)}_{2}}\] \[2Cu{{(CN)}_{2}}\xrightarrow{{}}\underset{insoluble}{\mathop{2CuCN}}\,+\underset{cyanogen}{\mathop{CN-CN}}\,\] \[CuCN+3KCN\xrightarrow{{}}{{K}_{3}}\underset{so\operatorname{lub}le}{\mathop{[Cu{{(CN)}_{4}}]}}\,\]
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