Questions on ligands, the shapes of complexes, isomerism and ligand substitution

Question 1

Explain, with equations, what is seen when excess concentrated ammonia and, separately, concentrated hydrochloric acid are added to copper(II) sulfate solution, and why the two products have different shapes.

Accepted Answer

Indicative content

[Cu(H₂O)₆]²⁺(aq) + 4NH₃(aq) → [Cu(NH₃)₄(H₂O)₂]²⁺(aq) + 4H₂O(l), with only four of the six water ligands replaced
the blue solution becomes a deep blue solution; a pale blue precipitate may form first, before the ammonia is in excess
ammonia and water are similar in size and both uncharged, so the coordination number stays at 6 and the complex stays octahedral
[Cu(H₂O)₆]²⁺(aq) + 4Cl⁻(aq) ⇌ [CuCl₄]²⁻(aq) + 6H₂O(l), with all six water ligands replaced by four chloride ions
the blue solution turns yellow-green (accept green, then yellow), and adding water reverses the change, so it is an equilibrium
chloride ions are larger than water molecules and repel one another, so only four fit round Cu²⁺: coordination number 4, tetrahedral

Levels

Level 3 (5–6 marks) both equations are correct with the right number of ligands replaced, both observations are given, and the two shapes are explained from the size and charge of the ligands: ammonia replacing water without changing the coordination number, and the larger chloride ions fitting only four round the copper
Level 2 (3–4 marks) both observations and one equation are right and the shapes are named; the second equation has the wrong number of ligands or does not balance, or the shapes are not tied to ligand size
Level 1 (1–2 marks) one colour change or one complex is given; six ammonia ligands are often shown replacing all six waters, and the change of shape is not explained

Question 2

Give the shape and one bond angle of each of these complexes: [Cu(H₂O)₆]²⁺, [CuCl₄]²⁻, cisplatin [Pt(NH₃)₂Cl₂] and [Ag(NH₃)₂]⁺.

Accepted Answer

B1 [Cu(H₂O)₆]²⁺ is octahedral, with bond angles of 90° B1 [CuCl₄]²⁻ is tetrahedral, with Cl–Cu–Cl bond angles of 109.5°; allow 109° B1 cisplatin is square planar, with bond angles of 90° B1 [Ag(NH₃)₂]⁺ is linear, with an N–Ag–N bond angle of 180°

Question 3

Explain why [Co(NH₃)₆]²⁺ reacts with ethane-1,2-diamine to form [Co(H₂NCH₂CH₂NH₂)₃]²⁺.

Accepted Answer

B1 four particles (one complex and three diamine molecules) become seven (one complex and six ammonia molecules) B1 the number of particles increases, so the entropy increases and ΔS is positive B1 ΔH is close to zero, because six Co–N bonds are broken and six similar Co–N bonds are formed B1 so ΔG = ΔH − TΔS is negative and the substitution is feasible

Question 4

Explain how cisplatin prevents cancer cells from dividing.

Accepted Answer

B1 inside the cell, where the chloride concentration is low, the two chloride ligands are replaced by water B1 the platinum bonds to nitrogen atoms on guanine bases in DNA B1 it bonds to two guanine bases next to each other on the same strand, so the DNA bends B1 the DNA cannot be copied (replication is prevented), so the cell cannot divide and dies

Question 5

Evaluate the use of cisplatin to treat cancer.

Accepted Answer

B1 benefit: it is highly effective against some cancers, with survival from testicular cancer above 90% B1 drawback: it also acts on healthy rapidly dividing cells, causing hair loss and damage to the lining of the gut B1 drawback: it can damage the kidneys, so the dose must be limited B1 judgement: its use is justified where the benefit outweighs the risk, with the lowest effective dose and the patient's informed consent

Question 6

Calculate the percentage yield of tetraamminecopper(II) sulfate, [Cu(NH₃)₄]SO₄·H₂O (Mr 245.6), from a student's results for core practical 12, given in the table. One mole of copper(II) sulfate-5-water, CuSO₄·5H₂O (Mr 249.6), gives at most one mole of product.

Accepted Answer

M1 by difference, 3.15 g of copper(II) sulfate-5-water was used and 2.37 g of product was collected M1 moles of CuSO₄·5H₂O = 3.15 ÷ 249.6 = 1.262 × 10⁻² mol, which is also the maximum moles of product M1 theoretical mass of product = 1.262 × 10⁻² × 245.6 = 3.10 g; allow error carried forward A1 percentage yield = 2.37 ÷ 3.10 × 100 = 76.5%, to three significant figures

Question 7

Describe what happens when excess concentrated ammonia is added to a solution containing [Co(H₂O)₆]²⁺. Include an equation.

Accepted Answer

B1 [Co(H₂O)₆]²⁺ + 6NH₃ → [Co(NH₃)₆]²⁺ + 6H₂O B1 the pink solution becomes a pale straw-coloured solution, which darkens on standing in air B1 the coordination number stays at 6 and the complex stays octahedral, because NH₃ and H₂O are similar in size and both uncharged

Question 8

Calculate ΔG at 298 K for the reaction [Cu(H₂O)₆]²⁺ + EDTA⁴⁻ → [Cu(EDTA)]²⁻ + 6H₂O, for which ΔH = −4.0 kJ mol⁻¹ and ΔS = +210 J K⁻¹ mol⁻¹.

Accepted Answer

M1 TΔS = 298 × 0.210 = 62.6 kJ mol⁻¹, with ΔS converted to kJ K⁻¹ mol⁻¹ M1 ΔG = ΔH − TΔS = −4.0 − 62.6 A1 ΔG = −67 kJ mol⁻¹, to two significant figures

Question 9

Calculate the percentage by mass of platinum in cisplatin, [Pt(NH₃)₂Cl₂]. Ar: Pt 195.1, N 14.0, H 1.0, Cl 35.5.

Accepted Answer

M1 Mr = 195.1 + 2(17.0) + 2(35.5) = 300.1 M1 percentage = 195.1 ÷ 300.1 × 100 A1 65.0%, to three significant figures

Question 10

Describe the difference between the cis and trans isomers of [Co(NH₃)₄Cl₂]⁺.

Accepted Answer

B1 both are octahedral, with four ammonia and two chloride ligands round a cobalt(III) ion B1 in the cis isomer the two chloride ligands are on adjacent positions, 90° apart B1 in the trans isomer the two chloride ligands are on opposite positions, 180° apart

Question 11

Explain why [Ni(H₂NCH₂CH₂NH₂)₃]²⁺ exists as a pair of optical isomers.

Accepted Answer

B1 three bidentate ligands occupy the six positions of an octahedral complex B1 the ligands can be arranged in two ways that are mirror images of each other B1 the mirror images are non-superimposable, so they are optical isomers that rotate plane-polarised light in opposite directions

Question 12

Suggest the cobalt complexes responsible for the colours of silica gel sachets, which contain a little cobalt(II) chloride and are blue when dry and pink when they have absorbed water, and suggest why heating the gel turns it blue again.

Accepted Answer

B1 the blue form is the tetrahedral [CoCl₄]²⁻ complex, with chloride ligands B1 absorbed water replaces the chloride ligands to give the pink octahedral [Co(H₂O)₆]²⁺ B1 the change to [CoCl₄]²⁻ is endothermic, so heating, which also drives off water, moves the equilibrium back to the blue complex

Question 13

Suggest the formula, shape and coordination number of the complex formed when excess ethanedioate ions are added to a solution of [Fe(H₂O)₆]³⁺, and suggest why the substitution is favoured.

Accepted Answer

B1 the complex is [Fe(C₂O₄)₃]³⁻ B1 it is octahedral with coordination number 6, each ethanedioate forming two coordinate bonds B1 four particles become seven, so the entropy increases and the substitution is favoured (chelate effect)

Question 14

Explain why the cis isomer of [Pt(NH₃)₂Cl₂] is polar but the trans isomer is not.

Accepted Answer

B1 the Pt–Cl and Pt–N bonds are polar, each with a bond dipole B1 in the trans isomer identical bonds are opposite each other, 180° apart, so the dipoles cancel and there is no overall dipole B1 in the cis isomer both chlorine atoms are on the same side, so the dipoles do not cancel and the molecule is polar

Question 15

Calculate the ratio of [Ni(en)₃]²⁺ to [Ni(H₂O)₆]²⁺ in a solution in which the free ethane-1,2-diamine concentration is 0.010 mol dm⁻³. log₁₀ Kstab for [Ni(en)₃]²⁺ is 18.3.

Accepted Answer

M1 Kstab = antilog 18.3 = 2.0 × 10¹⁸ dm⁹ mol⁻³ M1 ratio = Kstab × [en]³ = 2.0 × 10¹⁸ × (0.010)³ A1 ratio = 2.0 × 10¹², to two significant figures

Question 16

Describe what is seen when sodium hydroxide solution is added until in excess to a solution of [Co(H₂O)₆]²⁺, and the mixture is then left in air. Include an equation.

Accepted Answer

B1 [Co(H₂O)₆]²⁺ + 2OH⁻ → [Co(H₂O)₄(OH)₂] + 2H₂O; allow Co(OH)₂ as the product B1 the pink solution gives a blue precipitate, which does not dissolve in excess sodium hydroxide B1 on standing in air the precipitate turns pink and then darkens to brown as cobalt(II) is oxidised to cobalt(III)

Question 17

Explain why ethanol is added to the deep blue solution when tetraamminecopper(II) sulfate is prepared, why the crystals are washed with ethanol rather than water, and why the yield is below 100%.

Accepted Answer

B1 the complex salt is much less soluble in ethanol than in water, so adding ethanol makes it crystallise out of the solution B1 water would dissolve some of the crystals; cold ethanol removes the remaining solution without dissolving the product B1 some complex stays dissolved in the water and ethanol, and some is lost on the filter paper and in transfer; ignore 'human error'

Question 18

Suggest why [Pt(NH₃)₂Cl₂] has cis and trans isomers but [Zn(NH₃)₂Cl₂] does not.

Accepted Answer

B1 the platinum complex is square planar, so the two chloride ligands can be adjacent (90°) or opposite (180°) B1 the zinc complex is tetrahedral, where every position is adjacent to every other, so only one arrangement exists

Question 19

Explain why a yellow-green solution of [CuCl₄]²⁻ turns blue when it is diluted with water.

Accepted Answer

B1 [Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O is an equilibrium B1 adding water lowers the chloride concentration and raises the water concentration, so the position moves left and blue [Cu(H₂O)₆]²⁺ forms

Question 20

Suggest why cisplatin is given to patients as the pure cis isomer and not as a mixture with the trans isomer.

Accepted Answer

B1 the trans isomer cannot bind to two neighbouring guanine bases, so it has no anticancer effect B1 it can still bind to other molecules in the body and cause side effects, so a mixture adds toxicity with no benefit

Measurement	Mass / g
Weighing boat and copper(II) sulfate-5-water	4.62
Empty weighing boat	1.47
Filter paper and dry crystals	3.21
Filter paper alone	0.84