Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Ligang H2O NH3 Ti h.aqua No info OH ppt OH negative complex (i.e. soluble) Cl- V h. aqua No info Cr - amphoteric h.aqua Mn h.aqua plae pink. Fe Co Fe3+ = yellow Fe3+ = pale green pink h.aqua h. ammine ## after hydroxide ppte forms first No amine complex No amine complex Cr(OH)3 a green ppte, then goes to hexaammine Mn(OH)2 a sandy or buff Goes dark over time as MnO2 slowly forms. Fe(OH)2 = dirty green h.aqua h. ammine h.aqua h. ammine Cu h.aqua sometimes shown as tetra aqua as axial H2O’s are further away than equatorial H2O’s. Writing it as hexaaqua is accepted. [Cu(NH3)4]2+ Can also show as [Cu(NH3)4(H2O)2]2+ H2O in axial positions TiCl4 reacts w. H2O like SiCl4 does TiO2 formed Zn - amphoteric h.aqua ** [Zn(NH3)4]2+ Zn(OH)2 Fe(OH)3 = orange/brown [Cr(OH)6]3- green solution [Zn(OH)4]2- [FeCl4]- Fe = +III [FeCl4]2- [CoCl4]2blue Tetrahedral [Cu(Cl)4]2yellow Tetrahedral [ZnCl4]2- [Cu(CN)2]- [Zn(CN)4]2- Fe = +II Tetrahedral F fluoro CN (cyano) [Fe(CN)6]4i.e normal ‘hexa’ [Co(CN)6]4- Fe(CO)5 unusual pentavalent complex Fe in zero oxidation state [Co(Cl)2(NH + 3) 4] CO (carbonyl) Other info Ni [Cr(CO)6] Cr in zero oxidation state [Cr(Cl)3(NH3)3] Cr in +III O.S. CrO42- yellow tetrhedra’ – stable in alkaline solution Cr2O72- stable in acidic solution. Tetrahedral about each Cr, bonded to one common O atom [Ni(CN)4]2square planar [Ni(CO)4](l) Used to extract Ni from ore. ** George Facer “Make the Grade” says Zn forms [Zn(H2O)4]2+ but G.I.Brown’s “Introduction to Inorganic Chem” (2nd Edition) p391 says it forms the hexaaqua ion. ## George Facer “Make the Grade” says Cr(OH)3 remains on addition of excess NH3. I think that is wrong. I think it forms a hexaammine complex. Actually the learning requirements for transition metals is quite small. Test for Fe3+ ions: Add –SCN i.e. thiocyanide ions. The yellow solution goes red. One water ligand is replaced. [Fe(SCN)(H2O)5]+ = blood red Test for Fe2+ ions: [Fe(H2O)6]2+ (aq) + [Fe(CN)6]3- [Fe[Fe(CN)6]]- Prussian blue or Fe(III)4[Fe(II)(CN)6]3 This blue was the colour used in old ‘imprint paper’ but was replaced by carbon imprint paper. CuCl4]2- = yellow, but often appears green as some blue hexaaqua complex remains yellow and blue look green. Same problem with Vanadium, Problem with cobalt: (a) = [Co(CN)6]4- , (b) = [Co(NH3)6]2+ , (c) = [Co(H2O)6]3+ [Co(CN)6]3- + e- [Co(CN)6]4- (a) E = -0.80 V. (a) will revert back to [Co(CN)6]3- as (a) is reducing and so is oxidized in air 3+ 2+ [Co(NH3)6] + e- [Co(NH3)6] (b) E = +0.10 V (b) will revert back to [Co(NH3)6]3+ as (b) is reducing (w.r.t. O2 in air) so is oxidized in air [Co(H2O)6]3+(c) + e- [Co(H2O)6]2+ E = +1.82 V – (c), the 3+ ion (Co also in +III) state will go on to produce [Co(H2O)6]2+ . It is quite a strong oxidizing agent. (c) will actually react with the surrounding water. O 2 is produced. A blue/green precipitate of Co(OH)2 will slowly turn into [Co(NH3)6]3+ over time. Note about Copper: Copper I oxide has 2 Cu’s in the formula. Cu2O - red Cupper II oxide has 1 Cu in the formula: CuO - black Cu+ ions are not stable in solution. If Cu+ is produced in a REDOX reaction, it usually precipitates out. This may take time. Catalysts: The catalyst example I gave was with V and Fe but the Fe wasn’t discussed as in the haber process, rather in reaction with S 2O82(peroxydisulphate ions) ions and sulphate ions with Iodide ions (I-) and Iodine (I2). Deprotonation: When reacting with dil NH3 , deprotonation occurs as ammonia solutions are basic. NH 3 + H2O NH4+ + OH-. The presence of the OH- in ammonia solution deprotonates the H2O ligand. I advise you if showing this reaction, to bring the resulting product to the neutral hydroxide compound e.g. [Cu(H2O)6]2+(aq) + dilute ammonia [Cu(OH)2] or [Cu(OH)2(H2O)4] or Cu(OH)2 Is better than [Cu(H2O)6]2+ + dilute ammonia [Cu(OH)(H2O)5]+ The reason why the top line is better is because you are much less likely to spend less time with trying to write the correct number of ligands and resulting charge. And less likely to make a mistake. If however the question said something like: aq. Cu 2+ ions have a small amount of dil NH3 added to it and no ppte is seen. Give an equation describing this reaction, then of course, you MUST write down the second equation! This is rare however, as in real life, it seems the OH- isn’t shared evenly so some ppte’s out while other Cu2+ remains. The Cu2+ precipitating out iwill be the ones closest to the OH- ions as they are dropped into the solution. The entropy increase by forming solid probably accounts for some Cu 2+ being greedy for OH- and not sharing it. Ligand Ag Pt Pd Al amphoteric Pb amphoteric H2O NH3 hexaaqua hexaaqua hexaaqua hexaaqua hexaaqua [Ag(NH3)2]+ [Pt(NH3l)4]2+ OH ppt AgOH (ppt) linear d.n.f Al(OH)3 ppte deprotonation only Al(OH)3 ppte Pb(OH)2 ppte deprotonation only Pb(OH)2 ppte [Al(OH)6]3ppte [Pb(OH)4]2- OH negative complex Cl- CN (cyano) AgCl (ppt) [Pt(Cl)4]2- [Pd(Cl)4]2- [Ag(CN)2]- Pt(NH3)2(Cl)2 cis-platin an anti-cancer drug. Stops ‘unzipping’ of DNA. Square planar Can get [Cu(NH3)2]2+ but not necessary to know this for A-level. Nice definition of a complex ion: Complex ion: A molecule or ion bonded via a dative covalent bond to a metal ion. It’s likely you will be asked something about the nature of colour in transition metal complexes. E.g. Account for the fact [Cu(NH3)4]2+ complexes are a blue colour yet [Co(H2O)6]2+ are pale pink. Questions like this are typically worth about 4-5 marks. You must first describe why they have a colour to begin with then why that colour is blue (the energy of the blue wavelength doesn’t correspond to d-d transition energy, so is unabsorbed) and then why the Co complex is pink and then why they differ. Be prepared for quick Q’s such as: Other than forming coloured complexes, give 2 properties of transition metals.