The problems are on the transition metals worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Transition metals are d-block elements that form ions with a partially filled d subshell. These worked answers correct the 20 problems of the Chemistry 11-12 worksheet on transition metals, from electron configurations to identify solution P: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the reasoning and the result in bold. Check configurations and charges first, then ratios in titrations, and finally significant figures and units.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.2.5
Problem 1 Electron configurations
1. [Ar] 3d⁶ 4s².2. Fe²⁺: [Ar] 3d⁶; Fe³⁺: [Ar] 3d⁵ (4s electrons leave first).
3. Cu: [Ar] 3d¹⁰ 4s¹.
Problem 2 Is it a transition element?
Sc: no, Sc³⁺ is 3d⁰. Ti: yes, Ti³⁺ is 3d¹. Zn: no, Zn²⁺ is 3d¹⁰ (full). Cu: yes, Cu²⁺ is 3d⁹ (partly filled).Problem 3 Match the colours
[Cu(H₂O)₆]²⁺ pale blue; [Fe(H₂O)₆]²⁺ pale green; [Co(H₂O)₆]²⁺ pink; MnO₄⁻ purple.Problem 4 Oxidation state and coordination number
[Fe(CN)₆]⁴⁻: +2, 6; [Cr(H₂O)₆]³⁺: +3, 6; [CuCl₄]²⁻: +2, 4; [Ag(NH₃)₂]⁺: +1, 2.Problem 5 Classify the ligands
H₂O, Cl⁻ and NH₃ are monodentate; ethanedioate is bidentate (two O donors); EDTA⁴⁻ is hexadentate.Problem 6 Writing complex formulae
1. [Ni(en)₃]²⁺; each en forms 2 bonds so the coordination number is 6.2. [Cr(NH₃)₄Cl₂]⁺: +3 + 2(−1) = +1.
3. [Cr(NH₃)₄Cl₂]Cl.
Problem 7 Colour and energy gap
1. λ = hc ÷ ΔE = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (3.98 × 10⁻¹⁹) = 5.00 × 10⁻⁷ m = 500 nm.2. Blue-green is absorbed, so the solution looks pink-red.
3. A 3d electron is promoted from the lower to the higher set of d orbitals.
Problem 8 Dichromate titration
1. n = 0.01860 × 0.0150 = 2.79 × 10⁻⁴ mol.2. n(Fe²⁺) = 6 × 2.79 × 10⁻⁴ = 1.674 × 10⁻³ mol; c = 1.674 × 10⁻³ ÷ 0.0250 = 0.0670 mol dm⁻³.
Problem 9 A homogeneous catalyst
1. 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂ and 2Fe²⁺ + S₂O₈²⁻ → 2Fe³⁺ + 2SO₄²⁻.2. Adding and cancelling 2Fe³⁺ and 2Fe²⁺ gives S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂.
3. Two negative ions repel one another, so collisions are rarely successful; the Fe²⁺/Fe³⁺ steps involve oppositely charged ions.
Problem 10 Contact process
1. SO₂ + V₂O₅ → SO₃ + V₂O₄; 2V₂O₄ + O₂ → 2V₂O₅.2. Vanadium goes +5 → +4.
3. V₂O₅ is regenerated in the second step, so it appears on both sides and is unchanged overall.
Problem 11 Copper and ammonia
1. A pale blue precipitate forms first, which then dissolves in excess ammonia to give a deep blue solution.2. Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s); [Cu(H₂O)₆]²⁺ + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O.
Problem 12 Copper and chloride
1. Left: +2 + 4(−1) = −2; right: −2.2. Chloride ions are larger than water molecules, so only four fit around Cu²⁺: 6 → 4.
3. Adding water shifts the equilibrium left, so the solution turns blue again.
Problem 13 Manganese oxidation states
MnO₄⁻ +7 (purple); MnO₄²⁻ +6 (green); MnO₂ +4 (brown solid); Mn²⁺ +2 (pale pink).Problem 14 Trends across the first row
1. Highest vanadium (1910 °C); lowest zinc (420 °C).2. 1910 − 420 = 1490 °C.
3. Many delocalised 3d and 4s electrons give strong metallic bonding, except zinc whose 3d is full and does not contribute.
Problem 15 Spot the errors
1. Chromium is [Ar] 3d⁵ 4s¹, a half-filled d subshell is stable.2. Zinc is not a transition metal: Zn²⁺ is 3d¹⁰.
3. Iron loses its 4s electrons first, giving [Ar] 3d⁶.
Problem 16 Chromium chemistry
1. Cr in [Cr(OH)₆]³⁻ is +3; in CrO₄²⁻ and Cr₂O₇²⁻ +6.2. Chromium stays +6 in both, so there is no electron transfer.
3. OH⁻ removes H⁺, shifting the equilibrium left, so the orange solution turns yellow.
Problem 17 Cisplatin
1. Ligands are neutral NH₃ and two Cl⁻, so platinum is +2.2. cis-trans (geometric) isomerism.
3. In the cis form the two Cl⁻ are adjacent, so the new bonds can form close together; in the trans form they are on opposite sides, too far apart to bind to the same DNA site.
Problem 18 Tablet analysis
1. n(MnO₄⁻) = 0.02160 × 0.0100 = 2.16 × 10⁻⁴ mol; n(Fe²⁺) = 5 × 2.16 × 10⁻⁴ = 1.08 × 10⁻³ mol.2. ×10 = 1.08 × 10⁻² mol.
3. Mass = 1.08 × 10⁻² × 151.9 = 1.640 g; % = 1.640 ÷ 2.00 × 100 = 82.0 %.
Problem 19 The chelate effect
1. 4 particles on the left become 7 on the right, so disorder increases: ΔS is positive.2. ΔG = −12 − (298 × 0.120) = −12 − 35.8 = −47.8 kJ mol⁻¹.
3. ΔG is negative, so the substitution is feasible.
Problem 20 Identify solution P
1. Fe²⁺ (pale green, green hydroxide that oxidises) and sulfate (white BaSO₄).2. Fe²⁺(aq) + 2OH⁻(aq) → Fe(OH)₂(s).
3. 4Fe(OH)₂ + O₂ + 2H₂O → 4Fe(OH)₃.
