Chemistry of Gold (Au) - Solved Practice Questions

Multiple Choice Questions (MCQs)

Question 1

What is the most common and stable oxidation state of gold (Au) in its compounds?

A) +1 B) +2 C) +3 D) +4

Correct Answer: C) +3

Explanation: Gold primarily exhibits +1 and +3 oxidation states. While Au(I) compounds are stable (e.g., AuCN), Au(III) is often more common and stable in aqueous solutions and many solid compounds like AuCl₃ and H[AuCl₄].

Question 2

Gold is known for its inertness. Which of the following reagents can dissolve gold?

A) Concentrated HCl B) Concentrated HNO₃ C) Aqua Regia D) Dilute H₂SO₄

Correct Answer: C) Aqua Regia

Explanation: Gold is a noble metal and does not react with single acids like concentrated HCl or concentrated HNO₃. Aqua Regia, a mixture of 3 parts concentrated HCl and 1 part concentrated HNO₃, dissolves gold by a synergistic action. Nitric acid acts as an oxidizing agent, forming Au³⁺ ions, and hydrochloric acid forms the stable tetrachloroaurate(III) complex, [AuCl₄]⁻, which shifts the equilibrium forward, promoting dissolution.

Question 3

In the extraction of gold by the cyanide process (MacArthur-Forrest process), gold is converted into which of the following complexes?

A) [AuCl₄]⁻ B) [Au(CN)₂]⁻ C) [Au(OH)₄]⁻ D) [Au(NH₃)₄]³⁺

Correct Answer: B) [Au(CN)₂]⁻

Explanation: The MacArthur-Forrest cyanide process involves dissolving finely crushed gold ore in a dilute solution of sodium or potassium cyanide in the presence of air (oxygen). This forms a soluble dicyanoaurate(I) complex, [Au(CN)₂]⁻. The reaction is: 4Au(s) + 8CN⁻(aq) + O₂(g) + 2H₂O(l) → 4[Au(CN)₂]⁻(aq) + 4OH⁻(aq)

Assertion-Reason Questions

Directions: In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as: A) Both A and R are true and R is the correct explanation of A. B) Both A and R are true but R is NOT the correct explanation of A. C) A is true but R is false. D) A is false but R is true.

Question 4

Assertion (A): Gold is a noble metal and does not corrode easily. Reason (R): Gold has a very high positive standard electrode potential (E°Au³⁺/Au = +1.50 V).

Correct Answer: A) Both A and R are true and R is the correct explanation of A.

Explanation: A high positive standard electrode potential indicates a strong tendency for a metal to resist oxidation and remain in its metallic state. This makes gold highly resistant to oxidation, corrosion, and reaction with most acids, classifying it as a noble metal. Therefore, the reason correctly explains the assertion.

Question 5

Assertion (A): Gold appears yellow, unlike silver which is white. Reason (R): The yellow color of gold is due to relativistic effects on its electron orbitals, leading to absorption of blue light.

Correct Answer: A) Both A and R are true and R is the correct explanation of A.

Explanation: Gold’s unique yellow color is indeed attributed to relativistic effects. For heavy elements like gold, relativistic effects cause the 6s orbital to contract and lower in energy, while the 5d orbitals are stabilized. This reduces the energy gap between the 5d and 6s/6p bands, allowing for the absorption of higher energy photons (blue and violet light) from the visible spectrum. The complementary yellow and red light are then reflected, giving gold its characteristic golden hue. Silver, being lighter, does not experience these effects to the same extent, leading to a larger energy gap and absorption of only UV light, reflecting all visible light (appearing white).

Short Answer Questions

Question 6

Explain why gold dissolves in aqua regia but not in nitric acid alone.

Model Answer: Gold does not dissolve in nitric acid alone because it is a noble metal with a very high positive standard electrode potential (E°Au³⁺/Au = +1.50 V). This high potential indicates its strong resistance to oxidation by H⁺ ions or nitrates.

However, gold dissolves in aqua regia (a 3:1 mixture of concentrated HCl and concentrated HNO₃) due to a synergistic action:

Oxidation by Nitric Acid: Nitric acid acts as a powerful oxidizing agent, oxidizing metallic gold (Au⁰) to gold(III) ions (Au³⁺): Au(s) + 4H⁺(aq) + 3NO₃⁻(aq) → Au³⁺(aq) + 3NO₂(g) + 2H₂O(l) (Simplified reaction)
Complex Formation by Hydrochloric Acid: Simultaneously, hydrochloric acid reacts with the nascent Au³⁺ ions to form the highly stable tetrachloroaurate(III) complex anion, [AuCl₄]⁻: Au³⁺(aq) + 4Cl⁻(aq) → [AuCl₄]⁻(aq)

The formation of the stable [AuCl₄]⁻ complex significantly reduces the concentration of free Au³⁺ ions in solution. According to Le Chatelier’s principle, this removal of Au³⁺ ions shifts the equilibrium of the initial oxidation reaction to the right, promoting the continuous dissolution of more gold. The overall reaction can be represented as: Au(s) + 3HCl(aq) + HNO₃(aq) → HAuCl₄ + NO(g) + 2H₂O(l)

Question 7

Describe the role of cyanide and oxygen in the MacArthur-Forrest cyanide process for gold extraction.

Model Answer: In the MacArthur-Forrest cyanide process for gold extraction, finely crushed gold ore is leached with a dilute solution of sodium or potassium cyanide in the presence of oxygen. Both cyanide and oxygen play crucial roles:

Role of Cyanide (CN⁻): Cyanide acts as a complexing agent. It reacts with metallic gold to form a highly stable, soluble dicyanoaurate(I) complex, [Au(CN)₂]⁻. This complex formation is vital because it allows gold, which is otherwise insoluble in water and most reagents, to dissolve in an aqueous solution. The stability of the complex shifts the equilibrium towards the dissolution of gold.
Role of Oxygen (O₂): Oxygen acts as an oxidizing agent. It oxidizes metallic gold (Au⁰) to the Au(I) oxidation state, which is necessary for the gold to react with cyanide ions and form the complex. Without the presence of oxygen, the oxidation step would not occur, and gold would not dissolve. Oxygen essentially facilitates the conversion of solid gold into its ionic form.

The overall chemical reaction demonstrating their combined role is: 4Au(s) + 8CN⁻(aq) + O₂(g) + 2H₂O(l) → 4[Au(CN)₂]⁻(aq) + 4OH⁻(aq) This process effectively solubilizes gold from its ore, allowing for its subsequent recovery by reduction (e.g., using zinc).

High-Order Thinking Skills (HOTS) Question

Question 8

Most transition metals exhibit variable oxidation states and form colored compounds. Gold (Au), a d-block element, primarily shows +1 and +3 oxidation states and forms stable complexes. Explain why gold often appears to be less reactive than other transition metals like iron or copper, and how its electronic configuration contributes to its properties.

Model Answer: Gold’s unique properties, including its relative inertness compared to other transition metals like iron or copper, and its characteristic electronic configuration, are rooted in several factors:

Electronic Configuration and Ionization Energy: Gold has the electronic configuration [Xe] 4f¹⁴ 5d¹⁰ 6s¹.
- High First Ionization Energy: Gold possesses a very high first ionization energy. This is due to its compact nuclear charge (Z=79) and the relativistic contraction of its 6s orbital. The relativistic effect makes the 6s electrons more tightly bound to the nucleus, requiring a significant amount of energy to remove them. This high ionization energy contributes to its reluctance to lose electrons and undergo oxidation.
- Stable d¹⁰ Configuration: Gold has a completely filled 5d¹⁰ subshell. This stable configuration contributes to its generally lower reactivity compared to transition metals with partially filled d-orbitals that readily participate in metallic bonding and chemical reactions.
High Standard Electrode Potential: Gold has a very high positive standard electrode potential (E°Au³⁺/Au = +1.50 V; E°Au⁺/Au = +1.69 V). This indicates that gold has a strong tendency to remain in its metallic, unoxidized state rather than forming ions. In comparison, elements like iron (E°Fe²⁺/Fe = -0.44 V) and copper (E°Cu²⁺/Cu = +0.34 V) have much lower or negative electrode potentials, indicating a greater tendency to lose electrons and react.
Relativistic Effects: For gold, being a very heavy element, relativistic effects become significant:
- Contraction of 6s Orbital: The electrons in the 6s orbital move at speeds high enough for relativistic mass increase, causing the orbital to contract and become significantly lower in energy. This makes the 6s electrons exceptionally difficult to remove.
- Stabilization of 5d Orbital: The 5d orbitals are also somewhat stabilized due to increased shielding from the contracted s-electrons. These relativistic effects lead to a larger energy gap between the occupied d-orbitals and the unoccupied s/p orbitals, affecting its chemical behavior and contributing to its relative inertness.
Complex Formation Tendency: Despite its inertness to direct oxidation by many reagents, gold readily forms stable complexes, particularly with soft ligands like cyanide (CN⁻) and halides (Cl⁻). The formation of these stable complexes (e.g., [Au(CN)₂]⁻, [AuCl₄]⁻) is crucial for its dissolution in specific reagents like aqua regia or cyanide solutions. The high stability of these complexes shifts the reaction equilibrium, effectively ‘pulling’ Au ions into solution and overcoming its inherent inertness.

In summary, gold’s exceptional inertness stems from its high ionization energy, stable d¹⁰ electronic configuration, and crucially, significant relativistic effects that tightly bind its valence electrons, manifesting as a very high positive standard electrode potential. While inert to many reagents, its ability to form highly stable complexes allows it to react under specific conditions.