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28.1 General physical and chemical properties

9701_s24_qp_43_qu_002 (1/2)

Question

• Define transition element.

Answer

• An element which forms one or more stable ions with incomplete d orbitals.

Notes

• Nothing To See Here!

9701_m24_qp_42_qu_003 (1/2)

Question

• Complete the shorthand electronic configurations of Fe and Fe³⁺.

Answer

• Fe: [Ar] 3d⁶ 4s²
• Fe³⁺: [Ar] 3d⁵

Notes

• This was generally answered well. Common errors included 3d⁸ (for Fe) and 4s² 3d³ (for Fe³⁺).

9701_w25_qp_43_qu_005 (1/2)

Question

• Complete the table to show the total number of unpaired electrons in the 3d and 4s orbitals of an isolated gaseous Cu atom and a Cu²⁺ ion.

Answer

• Cu: 3d = 0, 4s = 1
• Cu²⁺: 3d = 1, 4s = 0

Notes

• Nothing To See Here!

9701_s24_qp_42_qu_004 (1/2)

Question

• Explain why transition elements have variable oxidation states.

Answer

• The (3)d and (4)s sub-shells/orbitals are close/similar in energy.

Notes

• Nothing To See Here!

9701_w24_qp_43_qu_004 (1/2)

Question

• Explain why transition elements form complex ions.

Answer

• Empty / vacant (d) orbitals are energetically accessible.
• OR empty / vacant (d) orbitals can form dative bonds with ligands.

Notes

• Many candidates could explain that transition elements form complex ions because they have empty d orbitals that are energetically accessible. Some described this due to having variable oxidation states or partially filled d orbitals (which is incorrect).

9701_s24_qp_43_qu_002 (1/2)

Question

• Explain why transition elements can form complex ions.

Answer

• They have vacant d orbitals that are energetically accessible.

Notes

• This was answered well by many candidates. The distinction between ‘partly filled d-orbitals’ (used to define transition elements) and ‘empty d-orbitals’ (used to explain complex formation) was important.

9701_w24_qp_42_qu_001 (1/2)

Question

• Platinum is a transition element. Explain why transition elements behave as catalysts.

Answer Any two from:

• They have variable oxidation states.
• They have vacant / empty / unfilled d orbitals.
• They can form dative bonds / can accept electrons.

Notes

• This question discriminated well. A full explanation was required. Partial answers such as ‘has vacant d-orbitals’ did not score.

9701_w23_qp_43_qu_005 (2/4)

Question

• Explain why transition elements behave as catalysts.

Answer

• They have more than one stable oxidation state.
• Empty/vacant d orbitals are energetically accessible OR empty/vacant d orbitals can form dative bonds with ligands.

Notes

• Most candidates suggested that variable oxidation states were needed for transition elements to act as catalysts. Less candidates were able to recall that vacant d-orbitals that are energetically accessible (to accept lone pairs from ligands) are also required.

9701_w24_qp_42_qu_001 (2/4)

Question

• Platinum acts as a heterogeneous catalyst in the removal of nitrogen dioxide, NO₂, from the exhaust gases of car engines.
• Describe the mode of action of a platinum catalyst in this process.

Answer

• Adsorption of reactants by Pt.
• Bonds of reactants weaken.
• Desorption of products.

Notes

• This question was answered well with the terms ‘adsorption’ and ‘desorption’ being used correctly. Many candidates stated, ‘the bonds between reactants are weakened’. This did not score, as it implies the weakening of intermolecular forces. The statement ‘the bonds within NO₂ molecules are weakened’ is a much better description.

9701_w24_qp_42_qu_001 (1/2)

Question

• NO₂ acts as a homogeneous catalyst in the oxidation of atmospheric sulfur dioxide, SO₂.
• Write equations for the two reactions that occur.

Answer

• NO₂ + SO₂ → NO + SO₃
• 2NO + O₂ → 2NO₂

Notes

• Many unbalanced equations, and equations with SO, were seen.

9701_s24_qp_43_qu_002 (1/2)

Question

• Define degenerate d orbitals.

Answer

• Orbitals of the same energy.

Notes

• Most candidates gave a correct definition. Some candidates wrongly gave a definition of ‘non-degenerate’. No credit was given for ‘orbitals of similar energy’.

9701_w24_qp_42_qu_004 (1/2)

Question

• Define the term non-degenerate d-orbitals.

Answer

• Not of the same energy / have different energy.

Notes

• This mark was awarded to most candidates. Some candidates appear to have misread the question, resulting in their answer ‘d-orbitals that have the same energy.’

9701_s25_qp_44_qu_001 (1/2)

Question

• Sketch the shape of a 3dz² orbital.

Answer

• Correct sketch showing two lobes along the z-axis and a ring (doughnut shape) centered in the xy-plane around the nucleus.

Notes

• Many accurately drawn correct answers were seen. However, sometimes it was difficult to distinguish between the ‘doughnut’ shaped electron cloud and an s-orbital in between a lobe.

9701_s24_qp_43_qu_002 (1/2)

Question

• Sketch the shape of a 3dxy orbital.

Answer

• Correct sketch showing four lobes situated exactly between the x and y axes.

Notes

• Most candidates could sketch the shape. The commonest error was to draw the lobes along the x and y axes, rather than between them.

28.2 General characteristic chemical properties

9701_m23_qp_42_qu_004 (2/4)

Question

• Define monodentate ligand.

Answer

• A species that uses one lone pair of electrons…
• …that forms a single dative covalent bond to a central metal atom / ion.

Notes

• Nothing To See Here!

9701_s24_qp_42_qu_004 (2/4)

Question

• Explain what is meant by a bidentate ligand.

Answer

• A species with two lone pairs (of electrons)…
• …that forms dative (covalent) / co-ordinate bonds to a (central) metal atom / ion.

Notes

• Nothing To See Here!

9701_w23_qp_43_qu_006 (1/2)

Question

• The molecule H₂NCH₂CH₂NHCH₂CH₂NH₂ is a tridentate ligand. Suggest the meaning of tridentate ligand.

Answer

• (A ligand that) donates 3 lone pairs to a central metal atom / ion OR forms 3 dative bonds to a central metal atom / ion.

Notes

• Most candidates gave a correct definition for a tridentate ligand.

9701_s23_qp_43_qu_001 (2/4)

Question

• EDTA⁴⁻ is a polydentate ligand. Explain what is meant by a polydentate ligand.

Answer

• (A species) that donates more than two lone pairs…
• …to form dative / coordinate bonds to a metal atom or ion.

Notes

• Stronger candidates had learnt the standard definition of a polydentate ligand.

9701_w25_qp_43_qu_002 (1/2)

Question

• Define complex ion.

Answer

• A species or ion formed by a central metal atom/ion AND surrounded by/bonded to (one or more) ligands.

Notes

• Nothing To See Here!

9701_s23_qp_43_qu_001 (1/2)

Question

• Explain what is meant by coordination number.

Answer

• The number of co-ordinate bonds being formed by the metal ion.

Notes

• Nothing To See Here!

9701_w24_qp_43_qu_004 (2/4)

Question

• Complexes P, Q, R, and S have tetrahedral, square planar, octahedral, and linear geometries respectively.
• Identify the size of the bond angle in degrees for each.

Answer

• P (tetrahedral): 109.5°
• Q (square planar): 90° (or 180°)
• R (octahedral): 90° (or 180°)
• S (linear): 180°

Notes

• Most candidates gave the correct bond angles. Common errors included P 104° and 120°; Q 109.5°; R 45°.

9701_s24_qp_43_qu_002 (2/4)

Question

• Name the shape of the complex ion [Ag(NH₃)₂]⁺.
• State the bond angle for H-N-Ag and for N-Ag-N.

Answer

• Shape: Linear.
• Bond angle for N-Ag-N = 180°.
• Bond angle for H-N-Ag = 109.5°.

Notes

• Many candidates gained at least one mark. The deduction that the H-N-Ag bond angle is equal to or close to 109.5° was found difficult.

9701_w25_qp_42_qu_005 (4/8)

Question

• An excess of concentrated HCl is added to a solution containing [Co(H₂O)₆]²⁺.
• Describe the colour change observed and the state of the cobalt-containing product.
• Write an equation for the reaction and name the type of reaction.

Answer

• Colour change: pink to blue. State: solution.
• Equation: [Co(H₂O)₆]²⁺ + 4Cl⁻ → [CoCl₄]²⁻ + 6H₂O
• Type of reaction: Ligand exchange.

Notes

• Nothing To See Here!

9701_s25_qp_43_qu_005 (4/8)

Question

• State the observations and write equations when aqueous copper(II) sulfate reacts separately with:

1. a few drops of NH₃(aq)
2. an excess of NH₃(aq)

Answer

• 1. Pale blue precipitate forms. [Cu(H₂O)₆]²⁺ + 2OH⁻ → Cu(OH)₂(H₂O)₄ + 2H₂O
• 2. Deep/dark blue solution forms. [Cu(H₂O)₆]²⁺ + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O

Notes

• Nothing To See Here!

9701_s24_qp_43_qu_002 (2/4)

Question

• Tollens’ reagent contains [Ag(NH₃)₂]OH, which can be prepared in a two-step process: Step 1: Aqueous NaOH is added to aqueous AgNO₃ to form Ag₂O. Step 2: Aqueous NH₃ is added to Ag₂O to form [Ag(NH₃)₂]OH.
• Construct equations for each step.

Answer

• Step 1: 2Ag⁺ + 2OH⁻ → Ag₂O + H₂O
• Step 2: Ag₂O + 4NH₃ + H₂O → 2[Ag(NH₃)₂]⁺ + 2OH⁻

Notes

• Many candidates found these equations particularly difficult.

9701_s24_qp_43_qu_002 (2/4)

Question

• An electrochemical cell uses Ag₂O as the positive electrode and Zn as the negative electrode immersed in an alkaline electrolyte.
• The overall cell reaction is: Ag₂O + Zn + H₂O → 2Ag + Zn(OH)₂
• Complete the half-equation for the reaction at each electrode.

Answer

• Positive electrode: Ag₂O + H₂O + 2e⁻ → 2Ag + 2OH⁻
• Negative electrode: Zn + 2OH⁻ → Zn(OH)₂ + 2e⁻

Notes

• This was found to be very difficult by many candidates. Many half-equations were seen that did not involve Ag⁺ gaining electrons at the positive electrode, or Zn losing electrons at the negative electrode; such answers gained no credit.

9701_w24_qp_42_qu_001 (3/6)

Question

• SO₂ dissolves in water, forming H₂SO₃. H₂SO₃ can be oxidised to SO₄²⁻ under acidic conditions ( = +0.17 V for the reversed reduction).
• Four more half-equations are given: H₃BO₃ + 3H⁺ + 3e⁻ ⇌ B + 3H₂O ( = –0.73 V) BiO⁺ + 2H⁺ + 3e⁻ ⇌ Bi + H₂O ( = +0.28 V) S + 2H⁺ + 2e⁻ ⇌ H₂S ( = +0.14 V) Sb + 3H⁺ + 3e⁻ ⇌ SbH₃ ( = –0.51 V)
• Select the oxidising agent that could oxidise H₂SO₃ to SO₄²⁻. Write an equation and calculate .

Answer

• Oxidising agent: BiO⁺ (since its > +0.17 V).
• Equation: 3H₂SO₃ + 2BiO⁺ + H₂O → 3SO₄²⁻ + 8H⁺ + 2Bi
• = +0.28 - 0.17 = +0.11 V

Notes

• This was a difficult question. Candidates who were able to deduce that BiO⁺ is the only oxidising agent on the list strong enough to oxidise H₂SO₃ to SO₄²⁻ often went on to score all three marks.

9701_w25_qp_43_qu_005 (3/6)

Question

• An impure sample of a vanadium(V) compound (mass 0.250 g) containing VO₃⁻ ions is reduced entirely to V²⁺ using excess Zn.
• The resulting solution is titrated with 0.0750 mol dm⁻³ MnO₄⁻, reaching the end-point at 22.5 cm³.
• The reaction is: 3MnO₄⁻ + 5V²⁺ + 3H₂O → 3Mn²⁺ + 5VO₃⁻ + 6H⁺
• Calculate the percentage by mass of vanadium in the sample.

Answer

• Moles of MnO₄⁻ = 0.0750 × 0.0225 = 1.6875 × 10⁻³ mol
• Moles of V²⁺ = (5/3) × 1.6875 × 10⁻³ = 2.8125 × 10⁻³ mol
• Mass of V = 50.9 × 2.8125 × 10⁻³ = 0.143 g
• % of V = (0.143 / 0.250) × 100 = 57.3%

Notes

• Nothing To See Here!

9701_w23_qp_43_qu_005 (3/6)

Question

• A 0.567 g piece of copper alloy is dissolved in acid, yielding 100.0 cm³ of solution containing Cu²⁺.
• A 25.0 cm³ sample is reacted with excess KI: 2Cu²⁺ + 4I⁻ → 2CuI + I₂
• The liberated I₂ is titrated with 20.10 cm³ of 0.0200 mol dm⁻³ S₂O₃²⁻: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻
• Calculate the percentage of copper in the alloy.

Answer

• Moles of I₂ in 25 cm³ = 0.5 × 0.0200 × 0.0201 = 2.01 × 10⁻⁴ mol
• Moles of Cu²⁺ in 25 cm³ = 2 × 2.01 × 10⁻⁴ = 4.02 × 10⁻⁴ mol
• Total moles of Cu in 100 cm³ = 4.02 × 10⁻⁴ × 4 = 1.61 × 10⁻³ mol
• Mass of Cu = 1.61 × 10⁻³ × 63.5 = 0.102 g
• % of Cu = (0.102 / 0.567) × 100 = 18.0%

Notes

• This calculation was often very well managed. Many candidates explained clearly what each step meant. It was much easier to award marks in such cases.

28.3 Colour of complexes

9701_m24_qp_42_qu_003 (3/6)

Question

• Explain why transition element complexes are coloured.

Answer

• The d orbital(s) split into different energy levels / d-d splitting occurs.
• Electron(s) are promoted / excited to a higher energy d orbital.
• Light is absorbed AND the colour seen is complementary to the frequency/wavelength absorbed.

Notes

• The splitting of d orbitals; the absorption of photons of visible light; the excitation of electrons and the observed colour being complementary to the colour absorbed were all described clearly.

9701_s23_qp_43_qu_001 (2/4)

Question

• A solution of Cr³⁺(aq) and a solution of Fe³⁺(aq) have different colours. Explain why.

Answer

• They have a different energy gap (ΔE) between the split d-orbitals.
• Therefore, a different frequency / wavelength (of light) is absorbed.

Notes

• Stronger candidates gave answers that included ‘the two complexes have different ΔE’ since ΔE is conventionally used for the size of the energy gap between split d-orbitals.

9701_w24_qp_43_qu_004 (2/4)

Question

• Co²⁺ ions form two different complexes, G and M, which have different ligands.
• Explain why G and M have different colours.

Answer

• They have a different ΔE OR d-d energy gap.
• Therefore, a different frequency / wavelength / energy from visible light is absorbed.

Notes

• Those who did not gain full marks referred to light being emitted or answered with an explanation about how colour arises in transition metal complexes rather than explaining why these complexes had different colours.

9701_s24_qp_42_qu_004 (2/4)

Question

• Solutions containing the [Ag(NH₃)₂]⁺ complex are colourless. Explain why.

Answer

• The Ag⁺ ion has a full / complete d-subshell (d¹⁰).
• Therefore, no d-d transitions are possible OR no d electrons can be promoted/excited.

Notes

• Candidates found this question challenging. Common errors included stating Ag⁺ has a full d-orbital and there was no d-d splitting.

9701_w23_qp_43_qu_005 (2/4)

Question

• Suggest why an aqueous solution of Cu²⁺ is coloured but solid CuI is white.

Answer

• Cu²⁺ is d⁹ / its d sub-shell is not full AND Cu⁺ is d¹⁰ / its d sub-shell is full.
• d-d transitions / d electron promotion is not possible in Cu⁺ (CuI).

Notes

• This question proved difficult for some candidates. Many did not reference that Cu²⁺ is d⁹ / had partially filled d subshell and Cu⁺ is d¹⁰ / full d subshell.

9701_w25_qp_43_qu_005 (2/4)

Question

• The 3d orbitals in an isolated Cu²⁺ ion are degenerate.
• Describe the relative energies of the 3d orbitals in an isolated Cu²⁺ ion compared to Cu²⁺ in a tetrahedral complex.

Answer

• Isolated Cu²⁺ ion: all five 3d orbitals are at the same energy level.
• Tetrahedral complex: orbitals split into three at a higher energy level and two at a lower energy level.
• The energy of all five d orbitals in the complex is higher than all d orbitals in the isolated ion.

Notes

• Nothing To See Here!

28.4 Stereoisomerism in transition element complexes

9701_s23_qp_42_qu_002 (1/2)

Question

• State two types of stereoisomerism shown by transition element complexes.

Answer

• Geometrical (cis-trans) isomerism.
• Optical isomerism.

Notes

• Nothing To See Here!

9701_s25_qp_44_qu_001 (3/6)

Question

• The complex [Cu(H₂O)₂(en)₂]²⁺ exists as stereoisomers. Draw three-dimensional diagrams for the three isomers.

Answer

• Draw the cis-isomer and its non-superimposable mirror image (these represent the two optical isomers).
• Draw the trans-isomer (with the two H₂O ligands opposite each other).

Notes

• Nothing To See Here!

9701_s24_qp_42_qu_004 (3/6)

Question

• Ruthenium(III) forms an octahedral complex [Ru(dpys)₂Cl₂]⁺ with the bidentate ligand dpys.
• Draw three-dimensional diagrams to show the three different stereoisomers.

Answer

• Draw the two optical isomers of the cis-form (where the two Cl⁻ ligands are at 90° to each other, acting as non-superimposable mirror images).
• Draw the trans-form (where the two Cl⁻ ligands are at 180° to each other).

Notes

• Many accurately drawn correct answers were seen, using correct conventions for the three-dimensional bonds of wedged and dashed bonds. The most common errors were diagrams without any 3-D bonds or diagrams in which 3-D bonds were drawn but where the complex could not be octahedral.

9701_s23_qp_42_qu_002 (3/6)

Question

• The complex [Cr(OCH₂CH₂NH₂)₃]⁻ contains an unsymmetrical bidentate ligand and exists as four stereoisomers.
• Describe how to generate the four isomers.

Answer

• Draw one facial (fac) isomer and its non-superimposable mirror image (two optical isomers).
• Draw one meridional (mer) isomer and its non-superimposable mirror image (two optical isomers).

Notes

• This was found to be a difficult question. The three nitrogen atoms in the complex can be ‘meridial’, where one N–Cr–N bond angle is 180°, or ‘facial’, where all N–Cr–N bond angles are 90°. The securest way to generate the four isomers is to draw one meridial isomer and its mirror image, and one facial isomer and its mirror image.

9701_w23_qp_42_qu_006 (1/2)

Question

• The complex Au(dien)Cl₃ contains the tridentate ligand ‘dien’.
• The dien ligand forms three bonds to the gold ion which all lie in the same plane.
• Describe the 3D structure of this complex.

Answer

• It is an octahedral complex where the three bonds from the ‘dien’ ligand sit in the same plane (e.g., equatorial/meridional), and the three Cl⁻ ligands occupy the remaining positions.

Notes

• This mark was rarely awarded. Comparatively few candidates appreciated the importance of the information that ‘The dien ligand forms three bonds to the gold ion… these three bonds lie in the same plane.’

9701_w23_qp_42_qu_006 (1/2)

Question

• Draw the three-dimensional structure of Ni(PH₃)₂Cl₂. (Note: The complex does not show isomerism).

Answer

• A tetrahedral structure with 3D wedges and dashes.

Notes

• This discriminated well. The commonest error was to think that Ni(PH₃)₂Cl₂ is square planar. If this were correct the complex would show cis-trans isomerism, however it states it does not.

9701_s25_qp_44_qu_001 (1/2)

Question

• Identify one isomer of [Cu(H₂O)₂(en)₂]²⁺ that is polar and explain your answer.

Answer

• Either of the cis-isomers is polar.
• Because the dipoles / polar bonds / partial charges do not cancel out (the molecule is asymmetric).

Notes

• Nothing To See Here!

9701_s24_qp_42_qu_004 (1/2)

Question

• Deduce which stereoisomer of [Ru(dpys)₂Cl₂]⁺ is non-polar. Explain your answer.

Answer

• The trans-isomer.
• Because the dipoles cancel / partial charges cancel.

Notes

• This proved challenging for candidates. Many did not describe dipoles (partial charges) cancelling but stated that bonds or polarities would cancel.

28.5 Stability constants,

9701_w23_qp_42_qu_004 (1/2)

Question

• Define stability constant, .

Answer

• The equilibrium constant for the formation of a complex in a solvent / from its constituents.

Notes

• Nothing To See Here!

9701_m25_qp_42_qu_003 (1/2)

Question

• Write an expression for the stability constant, , of the complex [Fe(H₂O)₅SCN]²⁺.

Answer

Notes

• Nothing To See Here!

9701_w23_qp_43_qu_005 (1/2)

Question

• Write an expression for the of [Cu(CN)₄]³⁻.
• In a solution, [CN⁻] = 0.0010 mol dm⁻³ and [[Cu(CN)₄]³⁻] = 0.0010 mol dm⁻³.
• Calculate [Cu⁺(aq)] if = 2.0 × 10²⁷.

Answer

• [Cu⁺] = = 5.0 × 10⁻¹⁹ mol dm⁻³

Notes

• Nothing To See Here!

9701_w24_qp_43_qu_004 (1/2)

Question

• Cadmium forms complex ion X, [Cd(NH₃)₄]²⁺. Adding CN⁻ replaces all NH₃ to form complex Y, [Cd(CN)₄]²⁻, which exists in a much higher concentration.
• A third complex Z is [CdCl₄]²⁻. Adding NaCl to X yields very little Z.
• Write the three complexes X, Y, and Z in order of increasing stability constant, .

Answer

• Z (least stable) < X < Y (most stable).

Notes

• This question was answered well.

9701_m25_qp_42_qu_003 (2/4)

Question

• A few drops of KF(aq) are added to a solution of [Fe(H₂O)₆]³⁺ (violet), turning it to [Fe(H₂O)₅F]²⁺ (colourless, = 2.4 × 10⁵).
• Then, a few drops of KSCN(aq) are added (the SCN⁻ complex is red, = 1.4 × 10²).
• Describe the observations and explain them using values.

Answer

• Observation: The solution turns from violet to colourless, AND then remains colourless (no change upon adding KSCN).
• Explanation: The of the F⁻ complex is greater than the of the SCN⁻ complex, meaning the F⁻ complex is more stable and no ligand exchange occurs.

Notes

• Nothing To See Here!

9701_s23_qp_43_qu_001 (1/2)

Question

• An aqueous solution of [CaEDTA]²⁻ is added to a solution containing equal concentrations of Cr³⁺, Fe³⁺, and Pb²⁺.
• Predict the relative concentrations of [CrEDTA]⁻, [FeEDTA]⁻, and [PbEDTA]²⁻ in the resulting equilibrium mixture.
• (Given values: Ca = 5.0 × 10¹⁰, Cr = 2.5 × 10²³, Fe = 1.3 × 10²⁵, Pb = 1.1 × 10¹⁸)

Answer

• Highest concentration: [FeEDTA]⁻ > [CrEDTA]⁻ > [PbEDTA]²⁻ : Lowest concentration.
• The [FeEDTA]⁻ complex has the highest value so it is the most stable.

Notes

• Nothing To See Here!