Chemistry of Silver (Ag) - Solved Practice Questions

Multiple Choice Questions (MCQs)

These questions are designed to test your understanding of key concepts related to Silver (Ag). Choose the most appropriate answer from the given options.

Question 1

Which of the following processes is primarily used for the extraction of silver from its sulphide ore? (A) Bayer’s process (B) Hall-Heroult process (C) MacArthur-Forrest cyanide process (D) Froth flotation process

Solution: (C) Explanation: The MacArthur-Forrest cyanide process is a hydrometallurgical method used for the extraction of precious metals like silver and gold from their low-grade ores. In this process, the finely powdered sulphide ore of silver (e.g., argentite, Ag₂S) is leached with a dilute solution of sodium cyanide (NaCN) or potassium cyanide (KCN) in the presence of air (oxygen). This forms a soluble dicyanoargentate(I) complex, from which silver is later precipitated using zinc dust.

The reactions involved are:

1. Leaching: Ag₂S(s) + 4NaCN(aq) + ½O₂(g) + H₂O(l) → 2NaAg(CN)₂ + 2NaOH(aq) + S(s)
2. Reduction (by displacement): 2NaAg(CN)₂ + Zn(s) → Na₂Zn(CN)₄ + 2Ag(s)

Bayer’s process is for aluminium, Hall-Heroult for aluminium electrolysis, and Froth flotation is for concentration of sulphide ores, not extraction of the metal itself.

Question 2

When silver reacts with dilute nitric acid, the major gaseous product formed is: (A) NO₂ (B) N₂O (C) NO (D) NH₄NO₃

Solution: (C) Explanation: The reaction of silver with nitric acid depends on the concentration of the acid.

• With dilute nitric acid, silver is oxidized to silver nitrate, and nitric oxide (NO) gas is produced. 3Ag(s) + 4HNO₃(dilute) → 3AgNO₃(aq) + NO(g) + 2H₂O(l)
• With concentrated nitric acid, silver also forms silver nitrate, but nitrogen dioxide (NO₂) gas is produced. Ag(s) + 2HNO₃(conc.) → AgNO₃(aq) + NO₂(g) + H₂O(l)

Therefore, for dilute nitric acid, the major gaseous product is NO.

Question 3

Among the silver halides (AgF, AgCl, AgBr, AgI), which one is most soluble in water? (A) AgF (B) AgCl (C) AgBr (D) AgI

Solution: (A) Explanation: Silver fluoride (AgF) is significantly soluble in water, whereas AgCl, AgBr, and AgI are sparingly soluble. The solubility of ionic compounds in water depends on a balance between lattice energy and hydration energy.

• AgF: Fluoride ion (F⁻) is much smaller than other halide ions (Cl⁻, Br⁻, I⁻). This leads to a higher charge density on F⁻ and thus a much stronger hydration energy for AgF. Although AgF also has high lattice energy due to the small size of F⁻, its hydration energy is sufficiently high to overcome the lattice energy, making it soluble.
• AgCl, AgBr, AgI: As the size of the halide ion increases from Cl⁻ to I⁻, the lattice energy of the silver halide decreases, but the hydration energy also decreases even more significantly (as the larger ions are less hydrated). Additionally, the increasing covalent character (due to Fajan’s rules – large anion, small cation) from AgCl to AgI further reduces their solubility in polar solvents like water.

Assertion-Reason Questions

In these questions, a statement of Assertion (A) is given followed by a statement of Reason (R). Mark the correct option out of the choices given below: (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. (E) Both A and R are false.

Question 1

Assertion (A): Silver is not considered a true transition element. Reason (R): Silver has a completely filled d-subshell in its ground state as well as in its common +1 oxidation state.

Solution: (A) Explanation: Assertion (A): The electronic configuration of silver (Ag) is [Kr] 4d¹⁰ 5s¹. In its most common and stable oxidation state, Ag⁺, the configuration is [Kr] 4d¹⁰. According to the IUPAC definition, a transition element is an element which has incompletely filled d-orbitals in its ground state or in any of its common oxidation states. Since Ag⁺ has a completely filled 4d subshell, it does not strictly meet this definition. Therefore, the assertion is true. Reason (R): As stated, the ground state configuration of Ag is 4d¹⁰ 5s¹ and the common +1 oxidation state (Ag⁺) is 4d¹⁰. Both have completely filled d-subshells. Therefore, the reason is true. The reason correctly explains why silver is not considered a true transition element according to the strict definition. Hence, (A) is the correct answer.

Question 2

Assertion (A): Silver bromide is extensively used in black and white photography. Reason (R): Silver bromide undergoes photochemical decomposition upon exposure to light.

Solution: (A) Explanation: Assertion (A): Silver bromide (AgBr) is indeed a key component of photographic emulsions used in black and white photography due to its high photosensitivity. Therefore, the assertion is true. Reason (R): AgBr crystals embedded in gelatin are sensitive to light. When exposed to light, AgBr undergoes a photochemical reaction, decomposing into fine particles of metallic silver (Ag) and bromine (Br₂). This reaction is: 2AgBr(s) → 2Ag(s) + Br₂(g) The metallic silver particles form a “latent image,” which is then chemically developed to produce the visible image. This property is crucial for its use in photography. Therefore, the reason is true. The reason directly explains the utility of AgBr in photography. Hence, (A) is the correct answer.

Short Answer Questions

Answer the following questions concisely and accurately.

Question 1

Explain why silver chloride (AgCl) is sparingly soluble in water but readily soluble in aqueous ammonia. Write the relevant chemical equations.

Model Answer: Silver chloride (AgCl) is sparingly soluble in water primarily due to its high lattice energy, which is the energy required to break apart the ionic lattice. This energy is not sufficiently overcome by the hydration energy released when the Ag⁺ and Cl⁻ ions interact with water molecules.

However, AgCl readily dissolves in aqueous ammonia (NH₃(aq)) due to the formation of a stable, soluble complex ion, diamminesilver(I) ion, [Ag(NH₃)₂]⁺. The ammonia molecules act as ligands, coordinating with the Ag⁺ ion to form this complex. The formation of this highly stable complex significantly lowers the concentration of free Ag⁺ ions in the solution, effectively shifting the dissolution equilibrium of AgCl to the right, according to Le Chatelier’s principle. This drives more AgCl to dissolve to replenish the Ag⁺ ions, until virtually all of the AgCl has dissolved.

Chemical Equations:

1. Sparingly soluble in water: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) (Ksp is very low, ~1.8 × 10⁻¹⁰)

2. Readily soluble in aqueous ammonia: AgCl(s) + 2NH₃(aq) → [Ag(NH₃)₂]⁺(aq) + Cl⁻(aq)

Question 2

Describe the phenomenon of tarnishing of silver articles and write the balanced chemical equation for the process.

Model Answer: Tarnishing of silver is a common phenomenon where the surface of silver articles, such as jewelry, cutlery, or coins, loses its characteristic luster and turns dull, often with a dark grey or black discoloration. This occurs when silver metal reacts with certain substances present in the atmosphere.

The primary cause of silver tarnishing is its reaction with trace amounts of hydrogen sulfide (H₂S) gas, which is a component of air pollution and is also naturally released from some biological processes. Sulfur-containing compounds can also contribute. Silver reacts with hydrogen sulfide to form a thin layer of silver sulfide (Ag₂S) on its surface. Silver sulfide is a black compound, which gives tarnished silver its characteristic darkened appearance. Unlike rust on iron, which flakes off, the silver sulfide layer adheres to the silver surface.

Balanced Chemical Equation: 2Ag(s) + H₂S(g) → Ag₂S(s) + H₂(g)

High-Order Thinking Skills (HOTS) Question

Question 1

Silver (Ag) predominantly exhibits a +1 oxidation state, while its group neighbors, Copper (Cu) and Gold (Au), also commonly show higher oxidation states (+2 for Cu, +3 for Au). Furthermore, silver’s catalytic activity is comparatively limited despite being a d-block element. Discuss the underlying electronic structure reasons for these observations.

Detailed Chemical Explanation:

1. Predominant +1 Oxidation State of Silver:

• Electronic Configuration: Silver has the electronic configuration [Kr] 4d¹⁰ 5s¹. The +1 oxidation state is achieved by the loss of the single 5s electron.
• Stability of Ag⁺: The resulting Ag⁺ ion has a stable, completely filled 4d¹⁰ electronic configuration. This filled d-subshell is energetically very stable, making it difficult to remove further electrons from the 4d orbitals. The energy required to remove a 4d electron from Ag⁺ is significantly higher than for elements like Cu.
• Comparison with Cu and Au:
  • Copper (Cu): [Ar] 3d¹⁰ 4s¹. It commonly shows +1 (Cu⁺, 3d¹⁰) and +2 (Cu²⁺, 3d⁹) oxidation states. The energy difference between the 3d and 4s orbitals is smaller than between 4d and 5s in Ag, making the removal of an additional 3d electron to form Cu²⁺ more feasible. The d-orbital is also not as deeply penetrating, making it more accessible.
  • Gold (Au): [Xe] 4f¹⁴ 5d¹⁰ 6s¹. Gold exhibits common oxidation states of +1 and +3. The prevalence of the +3 state is largely attributed to relativistic effects. These effects cause the 6s orbital to contract and stabilize, while the 5d orbitals expand and destabilize, making the 5d electrons more accessible for bonding and promoting higher oxidation states. Additionally, the lanthanide contraction influences the energy levels.

2. Limited Catalytic Activity of Silver:

• Mechanism of Transition Metal Catalysis: Transition metals are excellent catalysts because they typically possess:
  • Variable Oxidation States: To facilitate electron transfer in reaction mechanisms.
  • Partially Filled d-orbitals: To allow for temporary bonding with reactants through vacant d-orbitals, forming intermediate complexes.
  • Large Surface Area: For heterogeneous catalysis.
• Silver’s Constraint: In its most common and stable +1 oxidation state, silver has a completely filled 4d¹⁰ configuration. This means:
  • Lack of Vacant d-orbitals: There are no easily accessible vacant d-orbitals to form strong intermediate bonds with reactant molecules or to participate effectively in electron transfer mechanisms that involve changes in d-electron count.
  • Limited Variable Oxidation States: While Ag can achieve higher oxidation states (+2, +3), these are less stable and require stronger conditions or specific stabilizing ligands compared to many other transition metals. The inability to easily interconvert between multiple stable oxidation states through d-orbital participation restricts its role in many catalytic cycles.
• Contrast: Elements like Fe, Pt, Pd, and V have incompletely filled d-orbitals in their common oxidation states (e.g., Fe²⁺/Fe³⁺, V²⁺/V³⁺/V⁴⁺/V⁵⁺). This allows them to readily form stable intermediate complexes and undergo facile electron transfer, which are crucial steps in many catalytic processes.
• Specific Catalysis: While silver does exhibit some catalytic activity (e.g., in the oxidation of ethylene to ethylene oxide), its overall range and efficiency in catalytic processes are generally more limited compared to other transition metals that possess partially filled d-orbitals. Silver’s behavior is often termed “pseudo-transition” in this context.