Questions on colour in transition metal complexes and colorimetry

Question 1

Describe how a colorimeter can be used to find the concentration of Cu²⁺ ions in a solution, and explain why excess ammonia is added to every solution before it is measured.

Accepted Answer

Indicative content

make a series of standard solutions of known Cu²⁺ concentration by diluting a stock solution, each with the same excess of ammonia
choose a filter of the complementary colour, orange or red, the light that the deep blue complex absorbs most strongly
zero the colorimeter with a cuvette of the solvent alone, so that the cuvette and solvent contribute no absorbance
measure the absorbance of each standard and plot a calibration curve of absorbance against concentration, with a line of best fit
measure the unknown with the same filter and cuvette, and read its concentration from the curve, within the range of the standards
ammonia forms deep blue [Cu(NH₃)₄(H₂O)₂]²⁺, which absorbs far more strongly than the aqua ion, so absorbances are larger and percentage uncertainties smaller

Levels

Level 3 (5–6 marks) the standards with the same excess of ammonia, the complementary filter, zeroing with the solvent, the calibration curve and reading the unknown within its range are all present, and the ammonia is explained through the much stronger absorbance of the deep blue complex and the smaller percentage uncertainty
Level 2 (3–4 marks) the calibration curve from standards, the filter and the reading of the unknown are described; zeroing, keeping the unknown inside the range of the standards, or the reason for the ammonia is left out
Level 1 (1–2 marks) absorbance is compared with known solutions; the ammonia is typically said to be there to precipitate the copper, and calibration and filter are largely missing

Question 2

State what happens to a d electron in a complex ion when the ion absorbs visible light, and state the equation that links the energy absorbed with the frequency of the light, giving the meaning of each symbol.

Accepted Answer

B1 the d electron is promoted from its ground state to an excited state, a higher-energy d orbital B1 the energy absorbed equals ΔE, the gap between the split d orbitals B1 the equation is ΔE = hν; allow ΔE = hc/λ with λ defined B1 h is the Planck constant and ν the frequency of the light absorbed, in Hz or s⁻¹

Question 3

Explain why a solution containing [Cu(H₂O)₆]²⁺ ions is blue.

Accepted Answer

B1 the water ligands split the 3d orbitals of the Cu²⁺ ion into two groups of different energy B1 light of energy ΔE is absorbed as a d electron is promoted to the higher group B1 the light absorbed is at the orange and red end of the visible spectrum B1 the remaining light, which appears blue, is transmitted

Question 4

Calculate the frequency of the light absorbed by a complex whose split d orbitals are 2.6 × 10⁻¹⁹ J apart, and the energy absorbed per mole of complex in kJ mol⁻¹. h = 6.63 × 10⁻³⁴ J s and L = 6.022 × 10²³ mol⁻¹.

Accepted Answer

M1 ν = ΔE ÷ h = 2.6 × 10⁻¹⁹ ÷ 6.63 × 10⁻³⁴ A1 ν = 3.9 × 10¹⁴ Hz, to two significant figures M1 energy per mole = 2.6 × 10⁻¹⁹ × 6.022 × 10²³ = 1.57 × 10⁵ J, divided by 1000 A1 1.6 × 10² kJ mol⁻¹, to two significant figures

Question 5

Calculate the mass of nickel in 100.0 cm³ of an unknown solution whose absorbance is 0.204. The absorbances of four nickel(II) standards are given in the table. Ar of Ni = 58.7.

Accepted Answer

M1 the standards lie on a straight line through the origin with gradient 0.340 ÷ 0.040 = 8.5 dm³ mol⁻¹ M1 concentration of the unknown = 0.204 ÷ 8.5 = 0.0240 mol dm⁻³; allow a value read from a plotted graph M1 moles of Ni²⁺ = 0.0240 × 0.1000 = 2.40 × 10⁻³ mol A1 mass = 2.40 × 10⁻³ × 58.7 = 0.141 g, to three significant figures

Question 6

Calculate the energy gap ΔE between the split d orbitals of a complex that absorbs light most strongly at a wavelength of 500 nm. h = 6.63 × 10⁻³⁴ J s and c = 3.00 × 10⁸ m s⁻¹.

Accepted Answer

M1 λ = 500 × 10⁻⁹ m = 5.00 × 10⁻⁷ m M1 ΔE = hc/λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ ÷ 5.00 × 10⁻⁷ A1 ΔE = 3.98 × 10⁻¹⁹ J, to three significant figures

Question 7

Explain why solutions of zinc sulfate are colourless.

Accepted Answer

B1 Zn²⁺ has the configuration [Ar] 3d¹⁰ B1 the 3d subshell is full, so there is no vacancy in the upper d orbitals and no d electron can be promoted B1 no visible light is absorbed, so all wavelengths are transmitted and the solution is colourless

Question 8

Give three changes to a complex ion that alter ΔE and so alter its colour, with an example of each.

Accepted Answer

B1 a change of oxidation state of the metal, as in [Fe(H₂O)₆]²⁺ to [Fe(H₂O)₆]³⁺ B1 a change of ligand, as in [Cu(H₂O)₆]²⁺ to [Cu(NH₃)₄(H₂O)₂]²⁺ B1 a change of coordination number and shape, as in [Co(H₂O)₆]²⁺ to [CoCl₄]²⁻

Question 9

Explain why adding excess ammonia to a pale blue solution of [Cu(H₂O)₆]²⁺ produces a deep blue solution.

Accepted Answer

B1 ammonia replaces four of the water ligands, giving [Cu(NH₃)₄(H₂O)₂]²⁺ B1 the new ligand splits the d orbitals by a different (larger) amount, so ΔE changes B1 light of a different frequency is absorbed, so a different colour is transmitted

Question 10

Suggest why a student's result is unreliable when her unknown solution gives an absorbance of 0.92 but her highest standard gives only 0.60, and suggest what she should do.

Accepted Answer

B1 the unknown lies outside the range of the standards, so its concentration can only be found by extrapolation B1 the line may not stay straight at high absorbance, so the extrapolated value may be wrong B1 dilute the unknown by a known factor until its absorbance lies within the standards, then multiply the concentration read off by that factor

Question 11

Compare [Co(H₂O)₆]²⁺ and [CoCl₄]²⁻ in terms of shape, d-orbital splitting and colour.

Accepted Answer

B1 [Co(H₂O)₆]²⁺ is octahedral with coordination number 6, while [CoCl₄]²⁻ is tetrahedral with coordination number 4 B1 the different ligand and shape give a smaller ΔE in [CoCl₄]²⁻, so it absorbs light of longer wavelength B1 [Co(H₂O)₆]²⁺ is a pink solution and [CoCl₄]²⁻ a blue solution

Question 12

Explain why the solution of a complex looks colourless when the complex has a partly filled d subshell but absorbs only at 300 nm.

Accepted Answer

B1 300 nm is in the ultraviolet, outside the visible range B1 so ΔE is too large to correspond to any visible wavelength and no visible light is absorbed B1 all visible wavelengths are transmitted, so the solution appears colourless

Question 13

Evaluate a student's conclusion that the hexaaqua ions of Fe³⁺ and Mn²⁺ must be the same colour because both ions have the configuration [Ar] 3d⁵.

Accepted Answer

B1 the number of d electrons is the same, but the metal and its charge are different (+3 iron against +2 manganese) B1 a different nuclear charge and ion charge change how strongly the ligands interact, so ΔE is different B1 different wavelengths are absorbed, so the colours differ and the conclusion is wrong

Question 14

Explain why, in an octahedral complex, the 3dz² and 3dx²−y² orbitals are raised in energy more than the 3dxy, 3dxz and 3dyz orbitals.

Accepted Answer

B1 the six ligands approach the metal ion along the x, y and z axes B1 the lobes of 3dz² and 3dx²−y² lie along the axes, pointing at the ligands' lone pairs, so they are repelled more strongly B1 the lobes of 3dxy, 3dxz and 3dyz lie between the axes, away from the ligands, so they are repelled less and lie lower

Question 15

Calculate the wavelength, in nm, of the light absorbed by a complex for which ΔE is 186 kJ mol⁻¹. h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹ and L = 6.022 × 10²³ mol⁻¹.

Accepted Answer

M1 ΔE for one ion = 1.86 × 10⁵ J mol⁻¹ ÷ 6.022 × 10²³ = 3.089 × 10⁻¹⁹ J M1 λ = hc ÷ ΔE = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ ÷ 3.089 × 10⁻¹⁹ = 6.44 × 10⁻⁷ m A1 λ = 644 nm, to three significant figures

Question 16

Suggest the colour of filter to use when measuring the concentration of pink [Co(H₂O)₆]²⁺ ions in a colorimeter, and explain your choice.

Accepted Answer

B1 a green filter should be used B1 green is the complementary colour of pink, the colour the solution absorbs most strongly, so the absorbance changes most with concentration

Question 17

Suggest why copper(I) chloride is a white solid but copper(II) chloride is coloured.

Accepted Answer

B1 Cu⁺ is [Ar] 3d¹⁰, a full d subshell, so no d electron can be promoted and no visible light is absorbed B1 Cu²⁺ is [Ar] 3d⁹, with a vacancy in the upper d orbitals, so a d electron can absorb visible light and be promoted

Question 18

Suggest why the purple colour of the manganate(VII) ion cannot be explained by a d electron moving between split d orbitals.

Accepted Answer

B1 manganese in MnO₄⁻ is +7, [Ar] 3d⁰, so it has no d electrons to promote B1 the colour must come from another process, an electron moving from oxygen to the metal (charge transfer)

Concentration of Ni²⁺ / mol dm⁻³	Absorbance
0.010	0.085
0.020	0.170
0.030	0.255
0.040	0.340