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In the process of extraction of gold, roasted gold ore \(+\mathrm{CN}^{-}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{O}_{2}}{\longrightarrow}[\mathrm{X}]+[\mathrm{Y}]+\mathrm{Zn}\) \(+\mathrm{Au},[\mathrm{X}]\) and \([\mathrm{Y}]\) are (a) \([\mathrm{X}]=\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-},[\mathrm{Y}]=\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{-2}\) (b) \([\mathrm{X}]=\left[\mathrm{Au}(\mathrm{CN})_{4}\right]^{-3},[\mathrm{Y}]=\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{-2}\) (c) \([\mathrm{X}]=\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-},[\mathrm{Y}]=\left[\mathrm{Zn}(\mathrm{CN})_{6}\right]^{-4}\) (d) \([\mathrm{X}]=\left[\mathrm{Au}(\mathrm{CN})_{4}\right]^{-},[\mathrm{Y}]=\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{-2}\)

Short Answer

Expert verified
The correct option is (a): \([X] = [\text{Au(CN)}_2]^-, [Y] = [\text{Zn(CN)}_4]^{2-}\).

Step by step solution

01

Understanding the Question

The problem involves the chemical process of extracting gold using cyanide solution and zinc. The primary part of the process is to identify the complex ions \([X]\) and \([Y]\), formed as intermediates in the given equation.
02

Identify Gold Complex

During the cyanidation process, gold forms a soluble complex with cyanide ions. This complex is typically \([\text{Au(CN)}_2]^-\), as the gold coordination number in cyanide complexes is generally 2. Thus, \([X]\) should be \([\text{Au(CN)}_2]^-\).
03

Identify Zinc Complex

In gold extraction, zinc displaces gold from its cyanide complex, forming a zinc complex. Zinc usually forms a tetrahedral complex with cyanide, \([\text{Zn(CN)}_4]^{2-}\), due to its 2+ oxidation state and the tetrahedral geometry of the complex. This suggests \([Y]\) should be \([\text{Zn(CN)}_4]^{2-}\).
04

Choosing the Correct Option

Based on the identified complexes, \([X] = [\text{Au(CN)}_2]^-\) and \([Y] = [\text{Zn(CN)}_4]^{2-}\). Comparing with the options given: (a) \([X] = [\text{Au(CN)}_2]^-, [Y] = [\text{Zn(CN)}_4]^{2-}\). This matches our findings, so (a) is the correct answer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cyanidation Process
The cyanidation process is a widely used method for extracting gold from its ores. This process involves mixing the finely ground gold ore with a dilute solution of sodium cyanide. The cyanide solution dissolves the gold, facilitating its separation from the rest of the material. This technique exploits a simple chemical principle: gold can form a soluble complex with the cyanide ion. The equation often used to depict this is:\[ \text{4 Au} + 8\text{CN}^- + \text{2 H}_2\text{O} + \text{O}_2 \rightarrow 4 [\text{Au(CN)}_2]^- + 4 \text{OH}^- \]Gold is transformed into a stable and soluble form, making it easier to extract and process. This extraction method offers efficiency but also poses potential environmental risks due to the use of cyanide.
Complex Ion Formation
Complex ion formation is a key aspect of the gold extraction process. When gold interacts with cyanide ions, it forms a complex ion, which is a crucial step in cyanidation. Gold, being a transition metal, can coordinate with several ligands, like cyanide, to form complex ions.
For gold in the cyanidation process, the complex formed is \([\text{Au(CN)}_2]^-\), where the gold ion (\(\text{Au}^+\)) is stabilized by coordination with two cyanide ions. This process enhances the solubility of gold, enabling its separation and recovery from ores. These complexes are stable enough to remain in solution, making it possible to filter out other unwanted materials during processing.The stability of these complex ions depends on the particular metal and ligands involved, influencing the efficiency of the extraction process.
Zinc Displacement Reaction
Zinc displacement is an important step that follows the cyanidation process during gold extraction. Once gold has been dissolved in the cyanide solution and forms the complex \([\text{Au(CN)}_2]^-\), zinc is added to displace the gold from this complex. This is known as the zinc displacement or cementation reaction.
The reaction can be represented as:\[ \text{[Au(CN)_2]}^- + \text{Zn} \rightarrow \text{Zn(CN)_4}^{2-} + \text{Au} \]
This process leverages zinc's higher reactivity compared to gold. Zinc replaces gold due to its ability to lose electrons more readily, which allows the gold to precipitate out of the solution as solid metal, while zinc forms its complex ion \([\text{Zn(CN)}_4]^{2-}\).This step is crucial in recovering pure gold from the solution and is much more cost-effective than using other metals for displacement. The efficiency and simplicity of the zinc displacement reaction make it an enduring technique in the field of metallurgy.

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Most popular questions from this chapter

The number of unpaired electrons present in \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) and \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) are, respectively (a) 0,0 (b) 0,4 (c) 1,2 (d) 2,4Among the following four sets, the one having the same geometry and same magnetic property for both complexes is (a) \(\left[\mathrm{Co}\left(\mathrm{CN}_{6}\right)\right]^{3-}\) and \(\left[\mathrm{CoF}_{6}\right]^{3-}\) (b) \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}\) and \(\left[\mathrm{MnF}_{6}\right]^{3-}\) (c) \([\mathrm{Cr}(\mathrm{CN})]^{3-}\) and \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) (d) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) and \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\)

The soldiers of Napolean's army, while on Alps during freezing winter, suffered a serious problem as regards to the tin buttons of their uniforms. White metallic tin buttons got converted to grey powder. This transformation is related to (a) an interaction with nitrogen of the air at very low temperatures (b) a change in the crystalline structure of tin (c) an interaction with water vapour contained in the humid air (d) a change in the partial pressure of oxygen in the air.

The aqueous solution of the following salts will be coloured in the case of (a) \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (b) \(\mathrm{LiNO}_{3}\) (c) \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) (d) potash alum

Which of the following statement is correct with reference to the ferrous and ferric ions? 1\. \(\mathrm{Fe}^{3+}\) gives brown colour with potassium ferricyanide 2\. \(\mathrm{Fe}^{2+}\) gives blue precipitate with potassium ferricyanide 3\. \(\mathrm{Fe}^{3+}\) gives red colour with potassium thiocyanate 4\. \(\mathrm{Fe}^{2+}\) gives brown colour with ammonium thiocyanate (a) 1,4 (b) 1,2 (c) 2,3 (d) all of these

The transition elements with some exceptions can show a large number of oxidation states. The various oxidation states are related to the electronic configuration of their atoms. The variable oxidation states of a transition metal is due to the involvement of \((\mathrm{n}-1) \mathrm{d}\) and outer \(\mathrm{ns}\)-electrons. For the first five elements of 3 d-transition series. The minimum oxidation state is equal to the number of electrons in 4s shell and the maximum oxidation state is equal to the sum of \(4 \mathrm{~s}\) and \(3 \mathrm{~d}\)-electrons. The relative stability of various oxidation state of a given element can be explained on the basis of stability of \(\mathrm{d}^{0}, \mathrm{~d}^{5}\) and \(\mathrm{d}^{10}\) configurations. In 3 d-series, the maximum oxidation state is shown by (a) \(\mathrm{Fe}(26)\) (b) Mn (atomic no.: 25 ) (c) \(\mathrm{Cr}(24)\) (d) Sc (atomic no: 21 )

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