P-Block Elements JEE 2019: Gold Extraction Explained

What is the answer to the p-block elements JEE 2019 gold-extraction question?

A, B and D are correct in this p-block elements JEE 2019 question: oxygen oxidises gold during cyanide leaching, and zinc then recovers metallic gold. The source is JEE Advanced 2019, Paper 2, Chemistry. “p-Block Elements” is the supplied question-bank chapter tag.

Gold from its ore enters an aqueous cyanide solution in the presence of reagent Q, producing soluble species R. Treatment of R with reagent T recovers metallic gold and forms product Z. This is a multiple-correct question, so check each option independently.

A: Z is the zinc complex:

[Z n (C N)_{4}]^{2 -}

B: T is zinc.
C: R is the gold complex:

[A u (C N)_{4}]^{-}

D: Q is oxygen.

The question bank labels it medium difficulty and gives a 120-second target. These are question-bank benchmarks, not official exam standards. Use the two reactions below rather than memorising four separate identities.

How do cyanide and oxygen convert gold into the soluble complex?

Oxygen is the oxidising agent, and the soluble product is dicyanoaurate(I). Cyanide acts as the ligand that stabilises this dissolved gold(I) species:

[A u (C N)_{2}]^{-}

Each cyanide ligand carries one negative charge. The gold oxidation state therefore follows from the product’s charge:

x + 2 (- 1) = - 1 \Rightarrow x = + 1

Metallic gold starts at oxidation state zero. Each gold atom therefore loses one electron, giving the oxidation half-reaction:

A u + 2 {C N}^{-} \to [A u (C N)_{2}]^{-} + e^{-}

Oxygen accepts those electrons. In the aqueous alkaline balance, its reduction half-reaction is:

O_{2} + 2 H_{2} O + 4 e^{-} \to 4 {O H}^{-}

Multiply the gold half-reaction by four and add the oxygen half-reaction. Cancelling four electrons gives the official leaching equation:

4 A u + 8 {C N}^{-} + O_{2} + 2 H_{2} O \to 4 [A u (C N)_{2}]^{-} + 4 {O H}^{-}

Check atoms and charge before assigning Q and R. Cyanide groups remain intact, so counting them also checks carbon and nitrogen.

Gold: four atoms on each side.
Cyanide: eight groups on each side.
Oxygen and hydrogen: four atoms of each on each side.
Charge: eight negative charges on each side.

The official reaction therefore identifies:

Q = O_{2}, R = [A u (C N)_{2}]^{-}

D is correct and C is incorrect. The soluble intermediate has two cyanide ligands, not the four stated in C.

How does adding zinc recover metallic gold?

Zinc displaces gold because zinc is more electropositive. Zinc is oxidised, while gold(I) in the dissolved complex is reduced to metallic gold. One zinc atom supplies the two electrons needed to recover two gold atoms.

Reverse the earlier gold oxidation half-reaction to show gold recovery. Each complex ion accepts one electron:

[A u (C N)_{2}]^{-} + e^{-} \to A u + 2 {C N}^{-}

Zinc supplies electrons while forming its cyanide complex. Its oxidation half-reaction is:

Z n + 4 {C N}^{-} \to [Z n (C N)_{4}]^{2 -} + 2 e^{-}

Double the gold reduction half-reaction, then add the zinc half-reaction. Cancel two electrons and four free cyanide ions to obtain the official displacement equation:

2 [A u (C N)_{2}]^{-} + Z n \to [Z n (C N)_{4}]^{2 -} + 2 A u

Both sides contain two gold atoms, one zinc atom and four cyanide groups. The total charge is two negative charges on each side.

Check zinc’s oxidation state in the product:

x + 4 (- 1) = - 2 \Rightarrow x = + 2

Zinc starts at oxidation state zero, so this confirms the loss of two electrons. The reagent and product are:

T = Z n, Z = [Z n (C N)_{4}]^{2 -}

B correctly identifies the displacement reagent. A correctly identifies its soluble product.

Which options should you mark after checking both reactions?

Select A, B and D; reject C. The two reactions give a separate chemical reason for each verdict:

A, correct: the zinc product is:

Z = [Z n (C N)_{4}]^{2 -}

B, correct: zinc is the displacement reagent T that recovers metallic gold.
C, incorrect: leaching produces:

[A u (C N)_{2}]^{-} rather than [A u (C N)_{4}]^{-}

D, correct: oxygen oxidises gold during leaching.

This is a multiple-correct selection, not a choice between the leaching and displacement steps. Final selection: A, B and D.

Why is option C wrong even though its complex has a valid charge?

C assigns gold(III), whereas the official leaching reaction produces gold(I). The method error is recalling a four-cyanide gold complex, or copying the zinc product’s ligand count, without checking the reaction. The gold and zinc products need not have the same number of ligands.

For the complex proposed in C:

[A u (C N)_{4}]^{-} : x + 4 (- 1) = - 1 \Rightarrow x = + 3

For the actual leaching product:

[A u (C N)_{2}]^{-} : x + 2 (- 1) = - 1 \Rightarrow x = + 1

Charge arithmetic diagnoses the difference, but cannot alone predict which complex forms. The reaction chemistry supplies that information. The four-cyanide gold(III) complex is not an impossible species; it is the wrong intermediate here.

Establish the product from the reaction first, then check its oxidation state. Practise that distinction with Coordination Compounds Practice Questions JEE: 6 Solved.

Can you solve three related cyanidation practice questions?

Use reaction coefficients for quantities and electron transfer for reagent roles. These are original practice questions, not additional verified JEE PYQs. They stay within cyanidation under the supplied p-Block Elements topic.

The numerical drills assume ideal reactions, without side reactions or reagent excess. Their answers are stoichiometric requirements, not practical reagent doses.

How many moles of oxygen are required to leach 2 mol of gold?

The requirement is 0.5 mol of oxygen. The leaching equation uses one mole of oxygen for every four moles of gold, so divide the given gold amount by four: $4 A u : 1 O_{2}$

n (O_{2}) = \frac{2}{4} = 0.5 m o l

How many moles of zinc recover 3 mol of gold from dicyanoaurate(I)?

The requirement is 1.5 mol of zinc. One zinc atom releases two electrons, enough to reduce two gold(I) centres: $1 Z n : 2 A u$

n (Z n) = \frac{3}{2} = 1.5 m o l

Does cyanide act as the oxidising agent during leaching?

No. Oxygen accepts electrons and is the oxidising agent. Cyanide is the ligand stabilising the soluble gold(I) complex, not the electron acceptor.

Before marking a cyanidation answer, write both balanced reactions and label the electron donor and acceptor in each. Then check the complex formulas against those reactions.

Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).

Frequently asked questions

What is the answer to the JEE 2019 gold-extraction question?

A, B and D are correct in the JEE Advanced 2019 Paper 2 Chemistry question. Oxygen is reagent Q, zinc is reagent T, and the zinc product Z is [Zn(CN)₄]²⁻. Option C is incorrect because the soluble gold intermediate is [Au(CN)₂]⁻.

Why is option C wrong in the gold-extraction question?

Option C gives [Au(CN)₄]⁻, which contains gold in the +3 oxidation state. The leaching reaction instead produces [Au(CN)₂]⁻, containing gold in the +1 oxidation state. The proposed complex is not an impossible species; it is the wrong intermediate for this reaction.

What is the role of oxygen in cyanide leaching of gold?

Oxygen accepts electrons and acts as the oxidising agent, allowing metallic gold to become gold(I). Cyanide acts as the ligand that stabilises the soluble [Au(CN)₂]⁻ complex, not as the oxidising agent.

Why is zinc used to recover gold from cyanide solution?

Zinc is more electropositive than gold and reduces gold(I) in the dissolved complex to metallic gold. Zinc is oxidised to the +2 state and forms [Zn(CN)₄]²⁻. One zinc atom supplies two electrons, so one mole of zinc recovers two moles of gold in the ideal reaction.