Atomic Structure — A/L Chemistry

How do I practise Atomic Structure?

1 question with a worked solution

Check before you answer: Quantum Numbers & Electronic Configuration: Writing correct ground-state electronic configurations using s, p, d, f notation and applying the Pauli exclusion principle, Aufbau principle, and Hund's rule (especially anomalies like Cr: [Ar] 3d⁵ 4s¹ and Cu: [Ar] 3d¹⁰ 4s¹).

1. The first five successive ionization energies of an element X are 738, 1450, 7730, 10540, and 13630 kJ mol⁻¹. Deduce the periodic group of element X and the chemical formula of its stable chloride.
   1. A) Group 13, with formula XCl₃
   2. B) Group 2, with formula XCl₂
   3. C) Group 1, with formula XCl
   4. D) Group 14, with formula XCl₄
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   Correct answer: B) Group 2, with formula XCl₂

   A sharp jump in ionization energy occurs between IE₂ (1450 kJ/mol) and IE₃ (7730 kJ/mol), indicating that the first two electrons are in the valence shell and the third electron is removed from a stable inner noble gas core. Thus, element X belongs to Group 2 (alkaline earth metals) and forms a divalent cation X²⁺, producing the chloride XCl₂.

   Show worked solution

   Step 1: Analyze successive ionization energies: IE₁ = 738, IE₂ = 1450, IE₃ = 7730 kJ mol⁻¹. Step 2: Calculate the ratio IE₃/IE₂ = 7730/1450 ≈ 5.33. This large ratio (>5) indicates the removal of an electron from an inner complete shell. Step 3: Therefore, element X possesses 2 valence electrons and belongs to Group 2 (e.g., Magnesium). Step 4: A Group 2 metal forms an X²⁺ ion, which combines with Cl⁻ ions in a 1:2 ratio to give XCl₂. Answer: Group 2, formula XCl₂.
   Modelled on: the A/L Chemistry examination pattern

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Where does Atomic Structure fit in the syllabus?

Grade 12 · Term 1