Chemistry of Aluminum (Al) Practice Questions

Multiple Choice Questions (MCQs)

Question 1

Which of the following statements about Aluminum is INCORRECT? A) Aluminum is an amphoteric metal. B) Aluminum reacts with dilute HCl to produce hydrogen gas. C) Aluminum oxide (Al₂O₃) is basic in nature. D) Aluminum forms a passive layer of oxide on its surface.

Solution: The correct answer is C) Aluminum oxide (Al₂O₃) is basic in nature.

Explanation: A) Aluminum is indeed an amphoteric metal, meaning it reacts with both acids and bases. B) Aluminum reacts with dilute HCl: 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g). C) Aluminum oxide (Al₂O₃) is amphoteric, not basic. It reacts with acids to form salts and with bases to form aluminates. For example, Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O and Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O. D) Aluminum readily forms a thin, tough, and impervious layer of aluminum oxide (Al₂O₃) on its surface, which protects the underlying metal from further corrosion. This is known as passivation.

Question 2

In the electrolytic extraction of aluminum using the Hall-Héroult process, what is the primary role of cryolite (Na₃AlF₆)? A) To act as an oxidizing agent. B) To lower the melting point of alumina (Al₂O₃). C) To increase the melting point of alumina (Al₂O₃). D) To serve as the main source of aluminum.

Solution: The correct answer is B) To lower the melting point of alumina (Al₂O₃).

Explanation: Alumina (Al₂O₃) has a very high melting point (over 2000 °C). In the Hall-Héroult process, cryolite (Na₃AlF₆) is added to molten alumina primarily to:

1. Lower the melting point of the electrolyte mixture to about 950-1000 °C, making the process economically viable.
2. Increase the electrical conductivity of the electrolyte.
3. Dissolve alumina effectively. Cryolite itself does not serve as the main source of aluminum; alumina is the primary ore. It does not act as an oxidizing agent.

Question 3

Aluminum shows a diagonal relationship with which element, exhibiting some similar properties? A) Boron B) Gallium C) Magnesium D) Beryllium

Solution: The correct answer is D) Beryllium.

Explanation: Diagonal relationship is observed between elements of period 2 and period 3, belonging to diagonally opposite groups. Aluminum (Group 13, Period 3) exhibits a diagonal relationship with Beryllium (Group 2, Period 2). Similarities include:

• Both form covalent compounds.
• Their oxides (Al₂O₃ and BeO) and hydroxides (Al(OH)₃ and Be(OH)₂) are amphoteric.
• They form complex fluorides, e.g., [AlF₆]³⁻ and [BeF₄]²⁻.
• Both metals are passive towards nitric acid.
• Both react with strong bases to form complex anions (aluminates and beryllates).

Assertion-Reason Questions

Question 1

Assertion (A): Aluminum vessels should not be washed with washing soda (sodium carbonate). Reason (R): Washing soda is alkaline and reacts with aluminum to form sodium aluminate and hydrogen gas.

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.

Solution: The correct answer is A) Both A and R are true and R is the correct explanation of A.

Explanation: Aluminum is an amphoteric metal. Washing soda (Na₂CO₃) hydrolyzes in water to produce a strong base (NaOH) and a weak acid (H₂CO₃), making its solution alkaline. Na₂CO₃ + 2H₂O ⇌ 2NaOH + H₂CO₃ Aluminum reacts with the strong base (NaOH) as follows: 2Al(s) + 2NaOH(aq) + 6H₂O(l) → 2Na[Al(OH)₄](aq) + 3H₂(g) (Sodium tetrahydroxoaluminate(III)) This reaction corrodes the aluminum vessel, leading to pitting and damage. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Question 2

Assertion (A): Anhydrous AlCl₃ is a Lewis acid. Reason (R): Anhydrous AlCl₃ is an electron-deficient molecule.

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.

Solution: The correct answer is A) Both A and R are true and R is the correct explanation of A.

Explanation: In anhydrous aluminum chloride (AlCl₃), the central aluminum atom is bonded to three chlorine atoms. Aluminum is in Group 13 and has three valence electrons. After forming three covalent bonds, the aluminum atom has only six electrons in its valence shell (3 electron pairs). It needs two more electrons to complete its octet. According to Lewis theory, a Lewis acid is an electron pair acceptor. Since AlCl₃ is electron-deficient, it readily accepts an electron pair to achieve a stable octet, thus acting as a Lewis acid. For example, AlCl₃ + Cl⁻ → [AlCl₄]⁻. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Short Answer Questions

Question 1

Explain why aluminum forms [Al(OH)₄]⁻ when treated with strong bases, but boron, also in Group 13, does not typically form [B(OH)₄]⁻ under similar conditions.

Model Answer: Aluminum forms the tetrahydroxoaluminate(III) complex, [Al(OH)₄]⁻, due to its amphoteric nature and its ability to expand its coordination number. Aluminum, being a larger atom than boron, has empty d-orbitals in its valence shell, which can be utilized for accommodating additional electron pairs from the hydroxide ligands. This allows aluminum to achieve a coordination number of 4 and form the stable [Al(OH)₄]⁻ complex.

Boron, on the other hand, is a much smaller atom and lacks accessible d-orbitals in its valence shell. Its maximum covalency is restricted to four due to the absence of d-orbitals. While it can form [B(OH)₄]⁻ (tetrahydroxoborate) in some specific conditions, it generally exists as boric acid, B(OH)₃, which is a weak Lewis acid. Boron’s strong tendency to achieve sp² hybridization and form planar structures, combined with steric hindrance and electron deficiency, makes the formation of [B(OH)₄]⁻ less favorable and less common than [Al(OH)₄]⁻ for aluminum. Boric acid exists as B(OH)₃ which functions as a Lewis acid by accepting an OH⁻ ion rather than donating H⁺.

Question 2

Describe the role of cryolite (Na₃AlF₆) in the electrolytic extraction of aluminum (Hall-Héroult process).

Model Answer: In the Hall-Héroult process for the electrolytic extraction of aluminum from alumina (Al₂O₃), cryolite (Na₃AlF₆) plays several crucial roles:

1. Lowering the Melting Point: Pure alumina has a very high melting point (over 2000 °C). By dissolving alumina in molten cryolite, the melting point of the mixture is significantly lowered to about 950-1000 °C. This reduction in temperature makes the industrial process economically feasible, as less energy is required to maintain the molten state.

2. Increasing Electrical Conductivity: Molten alumina itself is a poor conductor of electricity. Cryolite, when molten, dissociates into ions (Na⁺, AlF₆³⁻, AlF₄⁻), which significantly increases the electrical conductivity of the electrolyte. This facilitates the efficient passage of current required for the electrolysis.

3. Solvent for Alumina: Cryolite acts as an excellent solvent for alumina. Alumina dissolves in molten cryolite to form complex ions like AlF₆³⁻ and AlOF₂⁻, which are then reduced at the cathode. This ensures a continuous supply of aluminum ions for the electrochemical reduction process.

In summary, cryolite is essential for lowering the operating temperature, enhancing electrical conductivity, and dissolving alumina, thereby enabling the practical and efficient production of aluminum metal.

High-Order Thinking Skills (HOTS) Question

Aluminum is a highly reactive metal according to its position in the reactivity series. However, aluminum articles do not corrode easily even when exposed to air and moisture. Explain this phenomenon and provide relevant chemical equations to demonstrate aluminum’s amphoteric nature.

Detailed Chemical Explanation:

Aluminum is indeed a highly reactive metal with a standard electrode potential of E° = -1.66 V, indicating a strong tendency to lose electrons. However, its excellent resistance to corrosion, despite exposure to air and moisture, is attributed to the phenomenon of passivation.

1. Formation of a Protective Oxide Layer: When aluminum is exposed to air, it immediately reacts with atmospheric oxygen to form a thin, tough, non-porous, and strongly adherent layer of aluminum oxide (Al₂O₃) on its surface. 4Al(s) + 3O₂(g) → 2Al₂O₃(s) This oxide layer acts as a protective barrier, preventing further reaction of the underlying aluminum metal with oxygen, moisture, and other corrosive agents. This passive layer is very stable and difficult to remove mechanically.

2. Amphoteric Nature of Aluminum Oxide: The protective Al₂O₃ layer is amphoteric, meaning it can react with both strong acids and strong bases. This property allows for controlled dissolution in specific environments but generally contributes to its stability in neutral conditions.

   • Reaction with Acids: Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l) or, with stronger acids like sulfuric acid: Al₂O₃(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂O(l)

   • Reaction with Bases: Al₂O₃(s) + 2NaOH(aq) + 3H₂O(l) → 2Na[Al(OH)₄](aq) (Sodium tetrahydroxoaluminate(III)) This reaction demonstrates that while the oxide layer provides protection, prolonged exposure to strong bases will cause it to dissolve, exposing the underlying metal to corrosion. Similarly, very strong acids can also dissolve the layer.

In summary, the high reactivity of aluminum is masked by the rapid formation of a stable, impervious, and self-healing passive oxide layer (Al₂O₃). This layer prevents further chemical attack, making aluminum highly resistant to corrosion under most environmental conditions. The amphoteric nature of this oxide layer dictates its behavior in acidic and basic media.