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Isolation of Elements question

2019 · Shift 2 · Q7
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Isolation of Elements question

2019 · Shift 2 · Q7

JEE AdvancedChemistryIsolation of ElementsMultiple correct+4 / −1
The cyanide process of gold extraction involves leaching out gold from its ore with CN −-− in the presence of Q in water to form R. Subsequently, R is treated with T to obtain Au and Z. Choose the correct option(s).
  1. A
    Q is O2O_2O2​
  2. B
    Z is [Zn(CN)4{}_44​]2−{}^{2-}2−
  3. C
    T is Zn
  4. D
    R is [Au(CN)4{}_44​]−{}^-−
View written solutionFree

Correct answer: A, B, C

The problem describes the MacArthur-Forrest cyanide process for the extraction of gold. We need to identify the reagents and products involved in the two main steps of this process.

Step 1: Leaching of Gold

The first step involves leaching the gold ore with a dilute solution of sodium cyanide (NaCN) or potassium cyanide (KCN). This is an oxidation process where gold is converted into a soluble complex. An oxidizing agent is required for this reaction, which is typically oxygen from the air.

The balanced chemical equation for this step is: 4Au(s)+8CN−(aq)+O2(g)+2H2O(l)→4[Au(CN)2]−(aq)+4OH−(aq)4 \text{Au(s)} + 8 \text{CN}^-(\text{aq}) + \text{O}_2(\text{g}) + 2 \text{H}_2\text{O(l)} \rightarrow 4 [\text{Au(CN)}_2]^-(\text{aq}) + 4 \text{OH}^-(\text{aq})4Au(s)+8CN−(aq)+O2​(g)+2H2​O(l)→4[Au(CN)2​]−(aq)+4OH−(aq)

According to the question, gold is leached with CN−^-− in the presence of Q to form R.

  • Comparing this with the reaction, the substance present along with CN−^-− is oxygen. Therefore, Q is O2_22​.
  • The soluble complex formed is the dicyanoaurate(I) ion. Therefore, R is [Au(CN)2_22​]−^-−.

Step 2: Precipitation of Gold

The second step is the recovery of gold from the soluble complex solution. This is done by a displacement reaction using a more electropositive metal, typically zinc (Zn). Zinc reduces the gold complex, causing metallic gold to precipitate out.

The balanced chemical equation for this step is: 2[Au(CN)2]−(aq)+Zn(s)→[Zn(CN)4]2−(aq)+2Au(s)2 [\text{Au(CN)}_2]^-(\text{aq}) + \text{Zn(s)} \rightarrow [\text{Zn(CN)}_4]^{2-}(\text{aq}) + 2 \text{Au(s)}2[Au(CN)2​]−(aq)+Zn(s)→[Zn(CN)4​]2−(aq)+2Au(s)

According to the question, R is treated with T to obtain Au and Z.

  • We identified R as [Au(CN)2]−[\text{Au(CN)}_2]^-[Au(CN)2​]−. It is treated with zinc metal. Therefore, T is Zn.
  • The products are metallic gold (Au) and a soluble zinc complex. Therefore, Z is [Zn(CN)4_44​]2−^{2-}2−.

Step 3: Evaluating the Options

Now we evaluate each option based on our findings:

  • A: Q is O2_22​. This is correct. Oxygen acts as the oxidizing agent in the leaching step.
  • B: Z is [Zn(CN)4{}_44​]2−{}^{2-}2−. This is correct. Zinc forms the stable tetracyanozincate(II) complex.
  • C: T is Zn. This is correct. Zinc metal is the reducing agent used to displace gold.
  • D: R is [Au(CN)4{}_44​]−^-−. This is incorrect. The stable complex formed by Gold(I) with cyanide is the linear dicyanoaurate(I) ion, [Au(CN)2]−[\text{Au(CN)}_2]^-[Au(CN)2​]−. The coordination number is 2, not 4.

Therefore, the correct options are A, B, and C.

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