Transition Metals

Chemistry, HSSC-Punjab Second Year. This chapter is split into 40 topics and has 1,045 practice questions. The topics covered are listed below, followed by 10 free sample questions from the chapter.

Topics in this chapter

1. Permanganate and iron(II) titrations 25 questions
2. Atomic and ionic radii of 3d elements (Figure 18.1) 30 questions
3. Ligand exchange reactions of copper(II) 28 questions
4. Variable oxidation states: the reason 24 questions
5. Cisplatin and the polarity of geometrical isomers 23 questions
6. Feasibility of redox reactions from E values 28 questions
7. Stability constant and its expression 22 questions
8. Anomalous configurations of chromium and copper 22 questions
9. Physical properties of the 3d transition metals 22 questions
10. Energy, frequency and wavelength in d-d transitions 25 questions
11. Finding the oxidation state of the metal in a complex 28 questions
12. Types and names of ligands 27 questions
13. d orbitals and crystal field splitting 25 questions
14. Mixed practice on 3d elements and complexes (1) 20 questions
15. Exercise and Quick Check style questions 28 questions
16. Geometrical (cis-trans) isomerism in complexes 26 questions
17. Metal complexes: central ion, ligand and dative bond 25 questions
18. Chelate effect and ligand displacement 22 questions
19. Monodentate, bidentate and polydentate ligands 26 questions
20. Mixed practice on 3d elements and complexes (2) 20 questions
21. Transition metals as catalysts 23 questions
22. Unpaired electrons and magnetic behaviour 30 questions
23. Common oxidation states of 3d elements (Table 18.2) 30 questions
24. Calculating oxidation states of transition metals 30 questions
25. Ligand exchange reactions of cobalt(II) 27 questions
26. Other redox reactions of transition metal ions 20 questions
27. Configurations of M2+ and M3+ ions 30 questions
28. Electronic configurations of 3d transition atoms 30 questions
29. Definition and position of transition elements 24 questions
30. Colour of complexes and d-d transitions 30 questions
31. Effect of ligands on splitting and colour 25 questions
32. Melting and boiling points of 3d elements (Figure 18.1) 30 questions
33. Permanganate and ethanedioate reaction 28 questions
34. Copper(II) and iodide: iodine liberated and thiosulphate titration 24 questions
35. Coordination number, sphere and charge of ligands 28 questions
36. Optical isomerism in complexes 25 questions
37. Calculations with stability constants 25 questions
38. Geometries of transition metal complexes 30 questions
39. Predicting the formula and charge of complex ions 30 questions
40. Naming coordination complexes 30 questions

Sample questions: Transition Metals

10 sample questions with the correct answer and an explanation. Try to answer each one before reading the explanation.

Q1. In the general structure [M(L)x] of a complex ion, the overall charge is shown

• A. only as a name and never written in the formula
• B. as a coefficient written before the brackets
• C. as a subscript inside the square brackets
• D. as a superscript outside the square brackets(correct answer)

Answer: D.The book writes the charge of the complex ion as a superscript after the closing bracket, with x the number of ligands.

Q2. A smaller ΔE in tetrahedral complexes explains why they often absorb

• A. lower energy light, e.g. in the red(correct answer)
• B. no light
• C. gamma rays
• D. higher energy light, e.g. in the ultraviolet

Answer: A.Smaller splitting means lower-energy photons are absorbed, giving blue/green colours as in [CoCl₄]²⁻ (standard knowledge).

Q3. In the book's MnO₄⁻/Fe²⁺ example, the overall E°cell is correctly found by

• A. multiplying the two E° values
• B. subtracting E°(Fe³⁺/Fe²⁺) from E°(MnO₄⁻/Mn²⁺)(correct answer)
• C. adding the E° values of the two couples directly
• D. dividing the larger E° by the smaller

Answer: B.E°cell = 1.52 - 0.77 = +0.75 V. (The book adds the reversed sign E° = +0.77 to obtain 2.29 V because it prints E°(Fe³⁺/Fe²⁺) as −0.77; the standard value is +0.77 V.)

Q4. The balanced equation for the reaction is

• A. 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O(correct answer)
• B. MnO₄⁻ + C₂O₄²⁻ + 8H⁺ → Mn²⁺ + 2CO₂ + 4H₂O
• C. 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 5Mn²⁺ + 2CO₂ + 8H₂O
• D. 2MnO₄⁻ + 5C₂O₄²⁻ + 8H⁺ → 2Mn²⁺ + 10CO₂ + 4H₂O

Answer: A.Combining 2 × (MnO₄⁻ + 8H+ + 5e-) with 5 × (C₂O₄²⁻ → 2CO₂ + 2e-) gives 16 H⁺ in total (the book's printed 8H+ is a typographical slip).

Q5. The colour change at the end-point when KMnO₄ is added to hot acidified oxalate is from

• A. blue to colourless
• B. yellow to green
• C. colourless to the first permanent pale pink(correct answer)
• D. pink to colourless

Answer: C.MnO₄⁻ (purple) is decolourised until all oxalate has been used; the first excess gives a permanent pink (standard knowledge).

Q6. 25.0 cm³ of Fe²⁺ solution is titrated with 0.02 mol/dm³ KMnO₄; 22.5 cm³ is needed. The concentration of Fe²⁺ is

• A. 0.0900 mol/dm³(correct answer)
• B. 0.4500 mol/dm³
• C. 0.1800 mol/dm³
• D. 0.0180 mol/dm³

Answer: A.n(MnO₄⁻) = 22.5/1000 × 0.02 = 4.50 × 10⁻⁴ mol; n(Fe²⁺) = 5 × 4.50 × 10⁻⁴ = 2.25 × 10⁻³ mol; c = 2.25 × 10⁻³/0.025 = 0.0900 mol/dm³.

Q7. 20.0 cm³ of Fe²⁺ solution is titrated with 0.02 mol/dm³ KMnO₄; 16.0 cm³ is needed. The concentration of Fe²⁺ is

• A. 0.0800 mol/dm³(correct answer)
• B. 0.4000 mol/dm³
• C. 0.0160 mol/dm³
• D. 0.1600 mol/dm³

Answer: A.n(MnO₄⁻) = 16.0/1000 × 0.02 = 3.20 × 10⁻⁴ mol; n(Fe²⁺) = 5 × 3.20 × 10⁻⁴ = 1.60 × 10⁻³ mol; c = 1.60 × 10⁻³/0.02 = 0.0800 mol/dm³.

Q8. 25.0 cm³ of Fe²⁺ solution is titrated with 0.015 mol/dm³ KMnO₄; 28.0 cm³ is needed. The concentration of Fe²⁺ is

• A. 0.0840 mol/dm³(correct answer)
• B. 0.4200 mol/dm³
• C. 0.0168 mol/dm³
• D. 0.1680 mol/dm³

Answer: A.n(MnO₄⁻) = 28.0/1000 × 0.015 = 4.20 × 10⁻⁴ mol; n(Fe²⁺) = 5 × 4.20 × 10⁻⁴ = 2.10 × 10⁻³ mol; c = 2.10 × 10⁻³/0.025 = 0.0840 mol/dm³.

Q9. 10.0 cm³ of Fe²⁺ solution is titrated with 0.02 mol/dm³ KMnO₄; 18.4 cm³ is needed. The concentration of Fe²⁺ is

• A. 0.9200 mol/dm³
• B. 0.3680 mol/dm³
• C. 0.1840 mol/dm³(correct answer)
• D. 0.0368 mol/dm³

Answer: C.n(MnO₄⁻) = 18.4/1000 × 0.02 = 3.68 × 10⁻⁴ mol; n(Fe²⁺) = 5 × 3.68 × 10⁻⁴ = 1.84 × 10⁻³ mol; c = 1.84 × 10⁻³/0.01 = 0.1840 mol/dm³.

Q10. 50.0 cm³ of Fe²⁺ solution is titrated with 0.01 mol/dm³ KMnO₄; 13.5 cm³ is needed. The concentration of Fe²⁺ is

• A. 0.0027 mol/dm³
• B. 0.0675 mol/dm³
• C. 0.0135 mol/dm³(correct answer)
• D. 0.0270 mol/dm³

Answer: C.n(MnO₄⁻) = 13.5/1000 × 0.01 = 1.35 × 10⁻⁴ mol; n(Fe²⁺) = 5 × 1.35 × 10⁻⁴ = 6.75 × 10⁻⁴ mol; c = 6.75 × 10⁻⁴/0.05 = 0.0135 mol/dm³.