
In the process of extraction of gold, \[Roasted\text{ }gold\text{ }ore+C{{N}^{-}}+{{H}_{2}}O\xrightarrow{{{O}_{2}}}[X]+O{{H}^{-}}\]
[X]+Zn\[\to \][Y]+Au
Identify the complexes [X] and [Y].
(1) X = \[{{[Au{{(CN)}_{2}}]}^{-}}\] , Y = \[{{[Zn{{(CN)}_{4}}]}^{2-}}\]
(2) X = \[{{[Au{{(CN)}_{4}}]}^{3-}}\] , Y = \[{{[Zn{{(CN)}_{4}}]}^{2-}}\]
(3) X = \[[Au(CN)_{2}^{-}]\] , Y = \[{{[Zn{{(CN)}_{6}}]}^{4-}}\]
(4) X = \[{{[Zn{{(CN)}_{4}}]}^{-}}\] , Y = \[{{[Zn{{(CN)}_{4}}]}^{2-}}\]
Answer
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Hint: Gold ore contains impurities like sulfide sulfur, organic carbon, and arsenic. Roasting of Gold ore is a process, to remove the impurities present in it to get pure Gold (Au). Removal of these impurities is necessary to attain high gold recovery (>90%) by cyanide leaching process.
Complete step by step answer:
By using the cyanide leaching process gold can be purified.
It is a two-step process.
In the first step roasted gold ore is going to leach with cyanide ion in presence of water.
In the second step the gold in combination with cyanide is going to react with zinc to get pure gold.
Step-1:
Roasted Gold ore is going to leach with cyanide ion in presence of water.
The chemical reactions are as follows.
\[Roasted\text{ }gold\text{ }ore+C{{N}^{-}}+{{H}_{2}}O\xrightarrow{{{O}_{2}}}2\underset{X}{\mathop{{{[Au{{(CN)}_{2}}]}^{-}}}}\,\]
In the above reaction roasted gold ore is going to react with Cyanide ion, and water in presence of oxygen gives two moles of \[[Au(CN)_{2}^{-}]\].
Step-2:
The formed complex \[[Au(CN)_{2}^{-}]\]in step-1 reacts with zinc and gives pure Gold (Au) and a complex of zinc.
The chemical reactions are as follows.
\[2{{[Au{{(CN)}_{2}}]}^{-}}+Zn\to \underset{Y}{\mathop{{{[Zn{{(CN)}_{4}}]}^{2-}}}}\,+2Au\]
Therefore here X = \[{{[Au{{(CN)}_{2}}]}^{-}}\]and Y = \[{{[Zn{{(CN)}_{4}}]}^{2-}}\].
So, the correct option is A.
Note: The process of purifying gold from roasted gold ore by using Cyanide ion is called Macarthur-Forrest process. By using this method we can extract silver and gold from their ores. This process was designed in 1887 by the scientists John S. MacArthur and Robert W.
Complete step by step answer:
By using the cyanide leaching process gold can be purified.
It is a two-step process.
In the first step roasted gold ore is going to leach with cyanide ion in presence of water.
In the second step the gold in combination with cyanide is going to react with zinc to get pure gold.
Step-1:
Roasted Gold ore is going to leach with cyanide ion in presence of water.
The chemical reactions are as follows.
\[Roasted\text{ }gold\text{ }ore+C{{N}^{-}}+{{H}_{2}}O\xrightarrow{{{O}_{2}}}2\underset{X}{\mathop{{{[Au{{(CN)}_{2}}]}^{-}}}}\,\]
In the above reaction roasted gold ore is going to react with Cyanide ion, and water in presence of oxygen gives two moles of \[[Au(CN)_{2}^{-}]\].
Step-2:
The formed complex \[[Au(CN)_{2}^{-}]\]in step-1 reacts with zinc and gives pure Gold (Au) and a complex of zinc.
The chemical reactions are as follows.
\[2{{[Au{{(CN)}_{2}}]}^{-}}+Zn\to \underset{Y}{\mathop{{{[Zn{{(CN)}_{4}}]}^{2-}}}}\,+2Au\]
Therefore here X = \[{{[Au{{(CN)}_{2}}]}^{-}}\]and Y = \[{{[Zn{{(CN)}_{4}}]}^{2-}}\].
So, the correct option is A.
Note: The process of purifying gold from roasted gold ore by using Cyanide ion is called Macarthur-Forrest process. By using this method we can extract silver and gold from their ores. This process was designed in 1887 by the scientists John S. MacArthur and Robert W.
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