2022 May/June · Paper 43 · Q4 (a, b(i), b(ii))7 marksColour & d-orbital splitting · Complex ions: ligands, shapes & coordination number · Isomerism in complexes
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Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Ligand exchange & stability constants · Core · Paper 4 style · 4 marks
(a)Describe what is seen when aqueous ammonia is added dropwise, until in excess, to aqueous copper(II) sulfate.[2]
(b)Write an equation for the formation of the precipitate from [Cu(H₂O)₆]²⁺.[1]
(c)Write an equation for the ligand exchange that forms the deep blue complex from [Cu(H₂O)₆]²⁺.[1]
Show answer
(a) A pale blue precipitate forms first; it dissolves in excess ammonia to give a deep (dark) blue solution.
B1 pale blue precipitate; B1 dissolves to deep blue solution
(c) Chloride ions are larger than water molecules, so fewer fit around the Cu²⁺ ion.
B1
(d) It turns blue again; the equilibrium shifts back to [Cu(H₂O)₆]²⁺ as water is in large excess.
B1
Q3Ligand exchange & stability constants · Standard · Paper 4 style · 5 marks
(a)Define the term stability constant, Kstab, of a complex ion.[2]
(b)Write the expression for Kstab of [Cu(NH₃)₄(H₂O)₂]²⁺ formed from [Cu(H₂O)₆]²⁺, and give its units.[2]
(c)Explain why the concentration of water does not appear in the expression.[1]
Show answer
(a) The equilibrium constant for the formation of the complex ion in a solvent from its constituent ions or molecules (the aqueous metal ion and the ligands).
B1 equilibrium constant; B1 for formation of complex from metal ion and ligands (in water)
(b) Kstab = [Cu(NH₃)₄(H₂O)₂²⁺] / ([Cu(H₂O)₆²⁺][NH₃]⁴); units dm¹² mol⁻⁴.
B1 expression (water omitted); B1 units
(c) Water is the solvent, in large excess, so its concentration is effectively constant.
B1
Q4Ligand exchange & stability constants · Standard · Paper 4 style · 5 marks
lg Kstab values: [CuCl₄]²⁻ 5.6; [Cu(NH₃)₄(H₂O)₂]²⁺ 13.1; [Cu(EDTA)]²⁻ 18.8.
(a)Predict what happens when excess aqueous ammonia is added to a solution of [CuCl₄]²⁻. Explain using the data.[2]
(b)EDTA⁴⁻ solution is added to the product of (a). Write an equation for the reaction and explain why it happens.[2]
(c)Explain why lg Kstab values are often quoted rather than Kstab.[1]
Show answer
(a) The solution turns deep blue: [Cu(NH₃)₄(H₂O)₂]²⁺ forms, because its Kstab is much larger, so NH₃ replaces Cl⁻.
B1 deep blue / ammine complex forms; B1 larger Kstab, more stable
(b) [Cu(NH₃)₄(H₂O)₂]²⁺ + EDTA⁴⁻ → [Cu(EDTA)]²⁻ + 4NH₃ + 2H₂O; [Cu(EDTA)]²⁻ has the largest Kstab, so it is the most stable complex.
B1 equation; B1 explanation
(c) Kstab values are very large and vary over many powers of ten, so logarithms are easier to compare.
B1
Q5Ligand exchange & stability constants · Standard · Paper 1 style · 1 mark
(a)Concentrated hydrochloric acid is added to a pink solution containing [Co(H₂O)₆]²⁺. Which row describes the new complex?[1]
Show answer
(a) A [CoCl₄]²⁻, blue, tetrahedral
Four large Cl⁻ ligands replace six H₂O, giving blue tetrahedral [CoCl₄]²⁻. B wrongly keeps six ligands; Cl⁻ ions are too large to fit six around Co²⁺.
Q6Ligand exchange & stability constants · Standard · Paper 4 style · 5 marks
(a)Aqueous potassium thiocyanate is added to aqueous iron(III) chloride. Write an equation for the formation of the blood-red complex [Fe(SCN)(H₂O)₅]²⁺ from [Fe(H₂O)₆]³⁺.[1]
(b)When aqueous sodium fluoride is added to the blood-red solution, the colour fades to almost colourless as [FeF(H₂O)₅]²⁺ forms. What does this show about the stability constants of the two complexes? Explain.[2]
(c)Write the Kstab expression for [Fe(SCN)(H₂O)₅]²⁺ and give its units.[2]
Show answer
(a) [Fe(H₂O)₆]³⁺ + SCN⁻ ⇌ [Fe(SCN)(H₂O)₅]²⁺ + H₂O
B1
(b) Kstab of [FeF(H₂O)₅]²⁺ is larger than that of [Fe(SCN)(H₂O)₅]²⁺; F⁻ competes for Fe³⁺ and the equilibrium favours the more stable complex.
B1 Kstab(F complex) larger; B1 competing equilibria / ligand exchange to more stable complex
Q8Ligand exchange & stability constants · Challenge · Paper 4 style · 5 marks
Kstab of [Ag(NH₃)₂]⁺ is 1.7 × 10⁷ dm⁶ mol⁻².
(a)Write the expression for Kstab of [Ag(NH₃)₂]⁺.[1]
(b)In a solution, the equilibrium concentration of free ammonia is 0.50 mol/dm³. Calculate the ratio [Ag(NH₃)₂⁺] : [Ag⁺]. Give your answer to 3 significant figures.[2]
(c)Hence calculate the fraction of the silver present as free Ag⁺(aq) ions. Give your answer to 2 significant figures.[1]
(d)Use your answer to explain why silver chloride dissolves in aqueous ammonia.[1]
(d) Free [Ag⁺] becomes so small that [Ag⁺][Cl⁻] falls below the solubility product, so more AgCl dissolves.
B1
Q9Ligand exchange & stability constants · Challenge · Paper 1 style · 1 mark
(a)Carbon monoxide is dangerous because it binds to the Fe²⁺ ion in haemoglobin. Which statement explains this?[1]
Show answer
(a) B The complex with CO has a larger stability constant than the complex with O₂, so the exchange is effectively irreversible
CO replaces O₂ in a ligand exchange; its complex is much more stable (larger Kstab), so O₂ cannot displace it and oxygen transport stops. CO is monodentate and does not oxidise the iron.
Q10Ligand exchange & stability constants · Challenge · Paper 4 style · 5 marks
Equal volumes of solutions are mixed to give initial concentrations [Fe³⁺] = 2.00 × 10⁻³ mol/dm³ and [SCN⁻] = 2.00 × 10⁻³ mol/dm³. Using a colorimeter, the equilibrium concentration of [Fe(SCN)(H₂O)₅]²⁺ is found to be 1.20 × 10⁻³ mol/dm³.
(a)Calculate the equilibrium concentrations of Fe³⁺(aq) and SCN⁻(aq).[1]
(b)Calculate Kstab for [Fe(SCN)(H₂O)₅]²⁺, giving its units. Give your answer to 3 significant figures.[2]
(c)Explain how the colorimeter was used to find the concentration of the complex.[2]
(c) A calibration graph of absorbance against known concentrations of the complex is made (using a complementary filter, blue-green); the absorbance of the equilibrium mixture is read and the concentration taken from the graph.
B1 calibration with standard solutions; B1 read concentration from absorbance (suitable filter)
Q11Complex ions: ligands, shapes & coordination number · Core · Paper 4 style · 4 marks
(a)Define the terms ligand and complex ion.[2]
(b)Define coordination number and give the coordination number of the metal in [Fe(H₂O)₆]³⁺ and in [CuCl₄]²⁻.[2]
Show answer
(a) Ligand: a species with a lone pair of electrons that forms a dative (coordinate) bond to a central metal atom/ion. Complex ion: a central metal ion surrounded by ligands joined by dative bonds, with an overall charge.
1 mark each
(b) The number of dative bonds formed to the central metal ion; 6 and 4.
B1 definition; B1 both values
Q12Complex ions: ligands, shapes & coordination number · Core · Paper 1 style · 1 mark
(a)What are the shape and the coordination number of the complex ion [Ag(NH₃)₂]⁺?[1]
Show answer
(a) A linear, 2
Two NH₃ ligands each form one dative bond, so the coordination number is 2, and the two bonds point in opposite directions (180°). Bent shapes arise from lone pairs on a central atom, not here.
Q13Complex ions: ligands, shapes & coordination number · Core · Paper 4 style · 4 marks
(a)Which of these can act as a ligand? NH₃, CH₄, Cl⁻, NH₄⁺, H₂O, CO[2]
(b)Explain why NH₄⁺ cannot act as a ligand.[1]
(c)State the type of bond a ligand forms with a metal ion and the feature of the ligand that makes it possible.[1]
Show answer
(a) NH₃, Cl⁻, H₂O and CO.
2 marks all four and no others; 1 mark for three correct
(b) The lone pair on nitrogen is used to bond to H⁺, so it has no lone pair to donate.
B1
(c) Dative (coordinate) bond, using a lone pair of electrons.
B1
Q14Complex ions: ligands, shapes & coordination number · Standard · Paper 4 style · 6 marks
(a)Give the oxidation number of the metal in each complex: (i) [Fe(CN)₆]³⁻ (ii) [Cr(H₂O)₄Cl₂]⁺ (iii) [Co(NH₃)₅Cl]²⁺ (iv) [Cu(EDTA)]²⁻ (EDTA⁴⁻).[4]
(b)State the coordination number in (ii) and (iv).[2]
Show answer
(a) (i) +3 (ii) +3 (iii) +3 (iv) +2
1 mark each
(b) (ii) 6; (iv) 6
1 mark each (EDTA⁴⁻ is hexadentate)
Q15Complex ions: ligands, shapes & coordination number · Standard · Paper 4 style · 6 marks
(a)Explain the term bidentate ligand, using ethanedioate, C₂O₄²⁻, as an example.[2]
(b)Give the overall charge, coordination number and shape of the complex formed by Fe³⁺ with three ethanedioate ions.[3]
(c)Name another bidentate ligand.[1]
Show answer
(a) A ligand that forms two dative bonds to the metal ion; ethanedioate donates a lone pair from each of two O⁻ atoms.
B1 two dative bonds; B1 two O atoms with lone pairs
(b) [Fe(C₂O₄)₃]³⁻: charge 3−, coordination number 6, octahedral.
1 mark each
(c) 1,2-diaminoethane (en), H₂NCH₂CH₂NH₂.
B1
Q16Complex ions: ligands, shapes & coordination number · Standard · Paper 4 style · 4 marks
(a)Give the shape and the bond angles in each of: [Pt(NH₃)₂Cl₂], [CoCl₄]²⁻ and [Ni(H₂O)₆]²⁺.[3]
(b)Suggest why most aqua complexes of first-row transition metal ions are octahedral.[1]
(b) Water molecules are small, so six can fit around the ion.
B1
Q17Complex ions: ligands, shapes & coordination number · Standard · Paper 1 style · 1 mark
(a)What is the coordination number of cobalt in [Co(en)₂Cl₂]⁺, where en is H₂NCH₂CH₂NH₂?[1]
Show answer
(a) C 6
Each en is bidentate (2 bonds), so 2 × 2 + 2 Cl⁻ = 6. B counts each en as one bond.
Q18Complex ions: ligands, shapes & coordination number · Challenge · Paper 4 style · 5 marks
Chromium(III) chloride hexahydrate, CrCl₃·6H₂O, exists in forms with different colours. A solution containing 0.0100 mol of the pale green form is treated with excess aqueous silver nitrate at once; 2.87 g of silver chloride (Mr 143.4) is precipitated. Chromium has coordination number 6 in all forms.
(a)Calculate the number of moles of chloride ions precipitated per mole of complex.[2]
(b)Deduce the formula of the complex ion and the full formula of the compound, explaining your reasoning.[2]
(c)State the type of bond between the chloride ligand and the chromium ion.[1]
(b) [Cr(H₂O)₅Cl]²⁺; [Cr(H₂O)₅Cl]Cl₂·H₂O. Only free Cl⁻ ions (not ligands) precipitate, so two Cl⁻ are outside and one is a ligand; coordination number 6 is completed by five H₂O.
B1 complex ion; B1 full formula with reasoning
(c) Dative covalent (coordinate) bond.
B1
Q19Complex ions: ligands, shapes & coordination number · Challenge · Paper 4 style · 6 marks
25.0 cm³ of a solution of Ni²⁺ ions is titrated with 0.0500 mol/dm³ EDTA⁴⁻. 18.60 cm³ is needed. The ions react in a 1 : 1 ratio.
(a)State the coordination number of nickel in [Ni(EDTA)]²⁻ and explain how one EDTA⁴⁻ ion achieves this.[2]
(b)Calculate the concentration of Ni²⁺ in the solution, in mol/dm³, to 3 significant figures.[2]
(c)The solution was made by dissolving a nickel alloy sample in acid and making it up to 250 cm³. Calculate the mass of nickel in the sample, to 3 significant figures. (Ar Ni = 58.7)[2]
Show answer
(a) 6; EDTA⁴⁻ is hexadentate, forming dative bonds from two N atoms and four O⁻ atoms.
M1 n in 250 cm³ = 10 × 0.000930 = 0.00930 mol; A1 × 58.7 = 0.546 g
Q20Isomerism in complexes · Core · Paper 4 style · 5 marks
(a)Describe the two stereoisomers of square planar [Pt(NH₃)₂Cl₂].[2]
(b)State the type of stereoisomerism shown.[1]
(c)The cis isomer, cisplatin, is an anticancer drug. Explain how it acts.[2]
Show answer
(a) Cis: the two Cl ligands are next to each other (90° apart); trans: the two Cl ligands are opposite each other (180° apart).
B1 cis; B1 trans
(b) Geometric (cis/trans) isomerism.
B1
(c) Its two Cl ligands are replaced by nitrogen atoms of guanine bases on DNA; this binds adjacent bases, distorts the DNA and stops the cancer cell dividing.
B1 ligand exchange: binds to N on DNA bases; B1 prevents DNA replication / cell division
Q21Isomerism in complexes · Standard · Paper 4 style · 4 marks
(a)Octahedral [Ni(NH₃)₄(H₂O)₂]²⁺ has two geometric isomers. Describe the arrangement of the water ligands in each.[2]
(b)Explain why [Ni(NH₃)₅(H₂O)]²⁺ has no geometric isomers.[1]
(c)State whether either isomer in (a) is optically active and explain.[1]
Show answer
(a) Cis: two water ligands adjacent (90°); trans: two water ligands opposite (180°).
1 mark each
(b) With only one water ligand, every position is equivalent; all arrangements are identical.
B1
(c) No; each has a plane of symmetry, so its mirror image is superimposable.
B1
Q22Isomerism in complexes · Standard · Paper 1 style · 1 mark
(a)Which complex ion shows optical isomerism?[1]
Show answer
(a) C [Cr(C₂O₄)₃]³⁻
Three bidentate ethanedioate ligands give a chiral octahedral complex. The trans complex has a plane of symmetry; tetrahedral [CuCl₄]²⁻ with four identical ligands is not chiral.
Q23Isomerism in complexes · Standard · Paper 4 style · 4 marks
(a)For each complex, state whether it shows geometric isomerism, optical isomerism, or neither. Give a reason in each case.(i) tetrahedral [Zn(NH₃)₂Cl₂](ii) octahedral [Co(NH₃)₅Cl]²⁺(iii) octahedral [Fe(C₂O₄)₃]³⁻(iv) octahedral [Cr(H₂O)₄Cl₂]⁺[4]
Show answer
(a) (i) neither: in a tetrahedron all positions are adjacent. (ii) neither: only one Cl, all positions equivalent. (iii) optical: three bidentate ligands, non-superimposable mirror images. (iv) geometric: Cl ligands can be cis (90°) or trans (180°).
1 mark each, with reason
Q24Isomerism in complexes · Challenge · Paper 4 style · 6 marks
(b)State the total number of stereoisomers of [Co(en)₂Cl₂]⁺. Explain.[3]
(c)Suggest how the two optical isomers could be distinguished.[1]
Show answer
(a) The three bidentate ligands make a propeller shape with no plane of symmetry; the complex and its mirror image are non-superimposable.
B1 no plane/centre of symmetry; B1 non-superimposable mirror images
(b) Three: the trans isomer (Cl ligands opposite) has a plane of symmetry and is not chiral; the cis isomer is chiral, giving a pair of optical isomers.
B1 three; B1 trans optically inactive; B1 cis exists as two enantiomers
(c) They rotate plane-polarised light in opposite directions.
B1
Q25Isomerism in complexes · Challenge · Paper 4 style · 5 marks
(a)Two different compounds of formula [Pt(NH₃)₂Cl₂] exist. Explain how this shows that the complex is square planar and not tetrahedral.[2]
(b)One isomer has a dipole moment and the other does not. Identify which, and explain.[2]
(c)Suggest which isomer is more soluble in water, and why.[1]
Show answer
(a) A tetrahedral MA₂B₂ complex has all positions adjacent, so only one arrangement exists; two isomers are possible only for square planar (cis and trans).
B1 tetrahedral: one form only; B1 square planar gives cis and trans
(b) Cis is polar: the Pt–Cl bond dipoles do not cancel; trans is non-polar: the bond dipoles are opposite and cancel.
B1 cis polar; B1 dipoles cancel in trans
(c) Cis, because it is polar and interacts more strongly with polar water molecules.
B1
Q26Colour & d-orbital splitting · Core · Paper 4 style · 5 marks
(a)Explain why [Cu(H₂O)₆]²⁺ is coloured.[3]
(b)Explain why [Zn(H₂O)₆]²⁺ and [Sc(H₂O)₆]³⁺ are colourless.[2]
Show answer
(a) Ligands split the d orbitals into two energy levels; an electron absorbs visible light and moves to a higher d orbital (energy ΔE = hf); the light not absorbed (complementary colour, blue) is transmitted/seen.
B1 d orbitals split by ligands; B1 electron promoted absorbing visible light of energy ΔE; B1 remaining colour transmitted
(b) Zn²⁺ is 3d¹⁰ (full: no space to promote an electron); Sc³⁺ is 3d⁰ (no d electrons to promote).
1 mark each
Q27Colour & d-orbital splitting · Standard · Paper 4 style · 5 marks
(a)Name the d orbitals that are raised in energy in an octahedral complex and explain why.[3]
(b)Describe how the splitting in a tetrahedral complex differs from that in an octahedral complex.[2]
Show answer
(a) dx²−y² and dz²; their lobes point along the axes directly at the ligands; electrons in them are repelled more by the ligand lone pairs.
B1 names; B1 lobes along axes / towards ligands; B1 greater repulsion
(b) Reversed: dxy, dxz, dyz are higher and dx²−y², dz² lower; the energy gap is smaller.
B1 reversed pattern (three up, two down); B1 smaller ΔE
Q28Colour & d-orbital splitting · Standard · Paper 1 style · 1 mark
(a)Excess ammonia is added to aqueous [Cu(H₂O)₆]²⁺, forming deep blue [Cu(NH₃)₄(H₂O)₂]²⁺. Which statement is correct?[1]
Show answer
(a) B The d-orbital splitting increases, so light of shorter wavelength is absorbed
NH₃ splits the d orbitals more than H₂O, so ΔE is larger and higher-energy (shorter-wavelength) light is absorbed, deepening the blue. A reverses the trend; the oxidation state stays +2.
Q29Colour & d-orbital splitting · Standard · Paper 4 style · 5 marks
(a)[Co(H₂O)₆]²⁺ is pink but [CoCl₄]²⁻ is blue. Give two reasons why the colours differ.[2]
(b)[Cu(H₂O)₆]²⁺ is pale blue but [Cu(NH₃)₄(H₂O)₂]²⁺ is deep blue, although both contain Cu²⁺ in an octahedral complex. Explain why their colours differ.[2]
(c)State one other factor that affects the colour of a transition element complex.[1]
Show answer
(a) Different ligands (H₂O v Cl⁻) and different coordination number/shape (octahedral v tetrahedral) change the size of ΔE, so a different wavelength is absorbed.
B1 ligand; B1 coordination number/geometry; (both linked to ΔE)
(b) NH₃ ligands split the d orbitals by a different (larger) amount than H₂O, so ΔE is different and light of a different wavelength (frequency) is absorbed; the transmitted colour changes.
B1 different ligands give different ΔE; B1 different wavelength absorbed, so different complementary colour seen
(c) Oxidation state of the metal (or identity of the metal).
B1
Q30Colour & d-orbital splitting · Challenge · Paper 4 style · 5 marks
The complex [Ti(H₂O)₆]³⁺ has one absorption peak in the visible region, at 500 nm.
(a)Calculate the energy gap ΔE between the split d orbitals, in J per ion, to 3 significant figures. (h = 6.63 × 10⁻³⁴ J s; c = 3.00 × 10⁸ m/s)[2]
(b)Convert your answer to kJ/mol, to 3 significant figures. (L = 6.02 × 10²³ mol⁻¹)[1]
(c)Light of 500 nm is green-blue. Suggest the colour of the solution.[1]
(d)Explain why the ion is coloured although titanium has only one d electron in this ion.[1]
