A Level Chemistry Transition Metal Complex Colours PDF

A LEVEL CHEMISTRY 3.2.5 TRANSITION METALS COPPER COMPLEXES [Cu(H2O)6]2+ [CuCl4]2- 2+...

A LEVEL CHEMISTRY 3.2.5 TRANSITION METALS COPPER COMPLEXES [Cu(H2O)6]2+ [CuCl4]2- 2+ 2- H2O Cl : : H2O : :OH2 + conc. HCl(aq) Cu + H2O(l) Cu L.S. Cl : : H2O: : OH2 Cl : H2O Cl : Blue Solution Yellow-Green Solution NOTE: Coordination number is 4, + NaOH(aq) not 6 here as the Cl- + excess NH3(aq) ligands are large & or charged. Fewer can fit A-B L.S. around the metal ion. + NH3(aq) drop by drop [Cu(H2O)4(OH)2] [Cu(NH3)4(H2O)2]2+ OH H2O 2+ : : H2O : :OH2 H3N : :NH3 + excess NH3(aq) Cu Cu L.S. : H2O: OH2 H3N: : NH3 : OH : H2O Blue Precipitate Deep Blue Solution A-B = Acid-base reaction L.S. = ligand substitution reaction AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.2.5 TRANSITION METALS IRON II COMPLEXES [Fe(H2O)6]2+ 2+ H2O : H2O : :OH2 + conc. HCl(aq) No Further Change / Reaction Fe : H2O: OH2 : H2O Green Solution + NaOH(aq) or A-B + NH3(aq) drop by drop [Fe(H2O)4(OH)2] OH : H2O : :OH2 + excess NH3(aq) No Further Change / Fe Reaction : H2O: OH2 : OH Green Precipitate NOTE: This green precipitate may turn brown as some Fe2+ may be oxidised to Fe3+. A-B = Acid-base reaction L.S. = ligand substitution reaction AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.2.5 TRANSITION METALS IRON III COMPLEXES [Fe(H2O)6]3+ [FeCl4]- 3+ - H2O Cl : : H2O : :OH2 + conc. HCl(aq) Fe + H2O(l) Fe Cl : : H2O: L.S. : OH2 Cl : H2O Cl : Purple Solution Yellow Solution NOTE: Coordination number is 4, + NaOH(aq) not 6 here as the Cl- ligands are large & or charged. Fewer can fit A-B around the metal ion. + NH3(aq) drop by drop [Fe(H2O)3(OH)3] OH : H2O : : OH + excess NH3(aq) No Further Change / Fe Reaction : H2O: OH2 : OH Brown Precipitate A-B = Acid-base reaction L.S. = ligand substitution reaction AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.2.5 TRANSITION METALS FORMATION OF COLOURED IONS Transition metal complexes are coloured. The origin of this colour arises from the fact that some wavelengths (colours) of white light (ROYGBIV) are absorbed by the molecule while others are reflected. e.g. for a blue complex, only blue wavelengths are reflected and all others are absorbed The explanation behind this is based on the fact that the transition metal ions have partially filled d-orbitals. e.g. Cu2+ 1s2 2s2 2p6 3s2 3p6 4s0 3d9 ↼ ↼ ↼ 1. Coordinately bonded ΔE ligands cause the five 3d orbitals to split, with two of ↼↼ ↼ ↼ ↼ ↼ ↼ ↼ ↼ ↼ ↼ ↼ ↼ ↼ ↼ these having slightly more energy than the other three. 2. Specific wavelengths of visible light match the ↼↼ ↼ energy difference (ΔE) between the split orbitals. These are then absorbed, and causes an electron(s) ΔE to be “promoted” from their ground state to the ↼ ↼ ↼ ↼ ↼↼ higher, partially filled d-orbital. 3. The wavelengths that are not absorbed are reflected. This is the colour we see! In the case of Cu2+, only blue is reflected. Why are different complexes different colours? The size of ΔE can vary in different complexes. This causes them to have different colours. ΔE varies depending on three major factors: 1. The oxidation state of the central metal ion 2. The type of ligand bonded 3. The coordination number of the complex ion. As transition metal complexes are coloured, we can use a simple colorimeter to measure the concentration of ions in solution. Details on this can be found in the A Level required practical book. AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.2.5 TRANSITION METALS Why are some d-block NOT coloured? We learned earlier that Sc and Zn are not considered transition metals as their ions do not have partially filled d-orbitals. This is the reason behind why they do not have coloured complexes. Here are the most commonly examined examples of this and the explanations behind them… Sc3+ has empty d-orbitals so there are no ΔE X electrons to make the d-orbital transition. ↼ ↼ ↼ ↼ Zn2+ has full d-orbitals so there is no room for the electrons to make the d-orbital ΔE X transition. ↼ ↼ ↼ ↼ ↼ ↼ ↼ ↼ ↼ ↼ Cu+ (Copper 1+) also has full d-orbitals so there is no room for the electrons to make ΔE X the d-orbital transition. ↼ ↼ ↼ ↼ ↼ ↼ If electrons are not promoted from their ground state, no wavelengths of light are absorbed. This means all wavelengths are reflected by the complex and it appears white. AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.2.5 TRANSITION METALS ΔE CALCULATIONS Of course, there are associated calculations with all this! Energy (ΔE) and Frequency (f) are linked by the following expression: ΔE = hf ΔE = Difference in energy levels (J) h = Planck’s Constant (6.63x10-34 J.s-1) f = Frequency (Hz) In exam questions they often give you wavelength (𝜆) instead of frequency to find ΔE. In which case you can use the following expression to find the frequency, THEN find ΔE. 𝜆 c f= f = Frequency (Hz) 𝜆 = wavelength (m) c = speed of light (3.00x108 m.s-1) Be careful! The ΔE that you calculate will be very small. This is because it is the difference in energy required to promote one electron in one atom! (J.atom-1) In order to convert this to kJ.mol-1, you must /1000 and multiply by Avogadro’s constant. How To Tackle ΔE Calculations AQA www.chemistrycoach.co.uk © scidekick ltd 2024