Transition metals

General properties of transition metals. transition metal characteristics of elements Sc  Cu arise from an incomplete d sub-level in atoms or ions Sc 1s22s22p63s23p6 4s23d1 Ti 1s22s22p63s23p6 4s23d2 V 1s22s22p63s23p6 4s23d3 Cr 1s22s22p63s23p6 4s13d5 Mn 1s22s22p63s23p6 4s23d5 Fe 1s22s22p63s23p6 4s23d6 Co 1s22s22p63s23p6 4s23d7 Ni 1s22s22p63s23p6 4s23d8 Cu 1s22s22p63s23p6 4s13d10 Zn 1s22s22p63s23p6 4s23d10 Sc 3+ [Ar] 4s03d0 Ti 3+ [Ar] 4s03d1 V 3+ [Ar] 4s03d2 Cr 3+ [Ar] 4s03d3 Mn 2+ [Ar] 4s03d5 Fe 3+ [Ar] 4s03d5 Co 2+ [Ar] 4s03d7 Ni 2+ [Ar] 4s03d8 Cu 2+ [Ar] 4s03d9 Zn 2+ [Ar] 4s03d10 When forming ions lose 4s before 3d Why is Zn not a transition metal? Zn can only form a +2 ion. In this ion the Zn2+ has a complete d orbital and so does not meet the criteria of having an incomplete d orbital in one of its compounds. these characteristics include •complex formation, •formation of coloured ions, •variable oxidation state •catalytic activity.

Complex formation complex :is a central metal ion surrounded by ligands. ligand.: An atom, ion or molecule which can donate a lone electron pair. Co-ordinate bonding is involved in complex formation. Co-ordinate bonding is when the shared pair of electrons in the covalent bond come from only one of the bonding atoms. Co-ordination number: The number of co-ordinate bonds formed to a central metal ion. Cu OH2 OH2 H2O H2O OH2 OH2 2+ Ligands can be monodentate (e.g. H2O, NH3 and Cl- ) which can form one coordinate bond per ligand or bidentate (e.g. NH2CH2CH2NH2 and ethanedioate ion C2O4 2- ) which have two atoms with lone pairs and can form two coordinate bonds per ligand or multidentate (e.g. EDTA4- which can form six coordinate bonds per ligand)

H2O, NH3 and Cl− can act as monodentate ligands. The ligands NH3 and H2O are similar in size and are uncharged. Exchange of the ligands NH3 and H2O occurs without change of co-ordination number (eg Co2+ and Cu2+). [Co(H2O)6 ]2+ (aq) + 6NH3 (aq)  [Co(NH3 )6 ]2+ (aq) + 6H2O (l) This substitution may, however, be incomplete as in the case with Cu. Cu becomes [Cu(NH3 )4 (H2O)2 ]2+ deep blue solution [Cu(H2O)6 ]2+ (aq) + 4NH3 (aq)  [Cu(NH3 )4 (H2O)2 ]2+ (aq) + 4H2O (l) Reactions with Chloride Ions Addition of a high concentration of chloride ions (from conc HCl or saturated NaCl) to an aqueous ion leads to a ligand substitution reaction. The Clligand is larger than the uncharged H2O and NH3 ligands so therefore ligand exchange can involve a change of co-ordination number. Addition of conc HCl to aqueous ions of Cu and Co lead to a change in coordination number from 6 to 4. [CuCl4 ]2- yellow/green solution [CoCl4 ]2- blue solution [Cu(H2O)6 ]2+ + 4Cl-  [CuCl4 ]2- + 6H2O These are tetrahedral in shape [Co(H2O)6 ]2+ + 4Cl-  [CoCl4 ]2- + 6H2O Be careful: If solid copper chloride (or any other metal) is dissolved in water it forms the aqueous [Cu(H2O)6 ]2+ complex and not the chloride [CuCl4 ]2- complex. [Fe(H2O)6 ]3+ + 4Cl-  [FeCl4 ] – + 6H2O

Ligands can be bidentate (e.g. NH2CH2CH2NH2 and ethanedioate ion C2O4 2- ) which have two atoms with lone pairs and can form two coordinate bonds per ligandCu(H2O)6 2+ + EDTA4- [Cu(EDTA)]2- + 6H2O Equations to show formation of mutidentate complexes N CH2 CH2 N CH2 CH2 CH2 CH2 C C C C O O O -O -O OOO The EDTA4- anion has the formula with six donor sites(4O and 2N) and forms a 1:1 complex with metal(II) ions 3- C C O O C C O O C C O O Cr O O O O O O 3+ H2C NH2 NH2 CH2 H2C NH2 NH2 CH2 CH2 NH2 NH2 Cr CH2 A complex with bidentate ethanedioate ligands e.g. [Cr(C2O4 )3 ]3- Learn the two bidentate ligands mentioned above but it is not necessary to remember the structure of EDTA. There are 3 bidentate ligands in this complex each bonding in twice to the metal ion. N Goalby chemrevise.org 2 Ligands can be multidentate (e.g. EDTA4- which can form six coordinate bonds per ligand). Bidentate Ligands Cu(H2O)6 2+ + 3NH2CH2CH2NH2  [Cu(NH2CH2CH2NH2 )3 ]2+ + 6H2O Cu(H2O)6 2+ + 3C2O4 2- [Cu(C2O4 )3 ]4- + 6H2O Equations to show formation of bidentate complexes Octahedral shape with 90o bond angles Ethane-1-2-diamine is a common bidentate ligand. Ethane-1-2-diamine Ethanedioate Partial substitution of ethanedioate ions may occur when a dilute aqueous solution containing ethanedioate ions is added to a solution containing aqueous copper(II) ions. In this reaction four water molecules are replaced and a new complex is formed. 2- OH2 O O C C O O Cu O O C C O O OH2 Cu(H2O)6 2+ + 2C2O4 2- [Cu(C2O4 )2 (H2O)2 ]2- + 4H2O Multidentate Ligands Oxygen forms a co-ordinate bond to Fe(II) in haemoglobin, enabling oxygen to be transported in the blood. CO is toxic to humans as CO can from a strong coordinate bond with haemoglobin. This is a stronger bond than that made with oxygen and so it replaces the oxygen attaching to the haemoglobin.

Stability of complexesThe substitution of monodentate ligand with a bidentate or a multidentate ligand leads to a more stable complex. This is called the chelate effect [Cu(H2O)6 ]2+ (aq) + EDTA4- (aq)  [Cu (EDTA)]2- (aq) + 6H2O (l) The copper complex ion has changed from having unidentate ligands to a multidentate ligand. In this reaction there is an increase in the entropy because there are more moles of products than reactants (from 2 to 7), creating more disorder. This chelate effect can be explained in terms of a positive entropy change in these reactions as more molecules of products than reactants Free energy ΔG will be negative as ΔS is positive and ΔH is small. The enthalpy change is small as there are similar numbers of bonds in both complexes. The stability of the EDTA complexes has many applications. It can be added to rivers to remove poisonous heavy metal ions as the EDTA complexes are not toxic. It is in many shampoos to remove calcium ions present in hard water, so helping lathering. [Co(NH3 )6 ]2+ + 3NH2CH2CH2NH2  [Co(NH2CH2CH2NH2 )3 ]2+ + 6NH3 This reaction has an increase in entropy because of the increase in moles from 4 to 7 in the reaction. ΔS is positive. Its enthalpy change ΔH is close to zero as the number of dative covalent and type (N to metal coordinate bond) are the same so the energy required to break and make bonds will be the same. Therefore Free energy ΔG will be negative and the complex formed is

Shapes of complex ions transition metal ions commonly form octahedral complexes with small ligands (e.g. H2O and NH3 ). transition metal ions commonly form tetrahedral complexes with larger ligands (e.g.Cl- ). square planar complexes are also formed, e.g. cisplatin Ag+ commonly forms linear complexes e.g. [Ag(NH3 )2 ]+ used as Tollen’s Reagent [Co(NH3 )6 ]2 [Cu(H2O)6 ]2+ [CoCl4 ]2- Isomerism in complex ions Complexes can show two types of stereoisomerism: cis-trans isomerism and optical isomerism Ni NH3 Cl H3N Cl Ni Cl NH3 H3N Cl Cis-[Cr(H2O )4Cl2 ]+ trans-Ni(NH3)2Cl2 Cis-trans isomerism in square planar complexes H3N Ag NH3 Pt + Cl Cl H3N Ni H3N NH3 NH3 H3N H3N NH3 NH3 2+ 2- Cu Cl Cl Cl Cl Complexes with 3 bidentate ligands can form two optical isomers (non-superimposable mirror images). 2+ CH2 CH2 NH2 NH2 CH2 CH2 H2N H2C NH2 H2C NH2 NH2 NiOptical isomerism in octahedral complexes Cis-trans isomerism in octahedral complexes cis–trans isomerism is a special case of E–Z isomerism HCl Cr2+ OH2 OH2 H2O OH2 ClH HCl Cr2+ OH2 OH2 H2O OH2 ClH Cis-Ni(NH3)2Cl2 trans-[Cr(H2O)4Cl2 ]+ + + 2 Ag+ commonly forms linear complexes e.g. [Ag(H2O)2 ]+ [Ag(NH3 )2 ]+ , [Ag(S2O3 )2 ]3- and [Ag(CN)2 ] – All are colourless solutions

Formation of coloured ions Colour changes arise from changes in 1. oxidation state, 2. co-ordination number 3. ligand. Colour arises from electronic transitions from the ground state to excited states: between different d orbitals. A portion of visible light is absorbed to promote d electrons to higher energy levels. The light that is not absorbed is transmitted to give the substance colour. Changing colour Changing a ligand or changing the coordination number will alter the energy split between the d- orbitals, changing E and hence change the frequency of light absorbed. Ligands cause the 5 d orbitals to split into two energy levels. Co(H2O)6 2+ + 4Cl-  [CoCl4 ]2- + 6H2O pink blue [Co(NH3 )6 ]2+ (aq)  [Co(NH3 )6 ]3+ (aq) +e- yellow brown Compounds without colour Scandium is a member of the d block. Its ion (Sc3+) hasn’t got any d electrons left to move around. So there is not an energy transfer equal to that of visible light. In the case of Zn2+ ions and Cu+ ions the d shell is full e.g.3d10 so there is no space for electrons to transfer. Therefore there is not an energy transfer equal to that of visible light. O2 In this equation only oxidation state is changing. Co(H2O)6 2+ + 6 NH3  Co(NH3 )6 2+ + 6H2O Pink yellow brown In this equation both ligand and co- ordination number are changing. In this equation only the ligand is changing. How colour arises 5 Equation to learn! This equation links the colour and frequency of the light absorbed with the energy difference between the split d orbitals. E = hv. v = frequency of light absorbed (unit s-1 or Hz) h= Planck’s constant 6.63 × 10–34 (J s) E = energy difference between split orbitals (J) A solution will appear blue if it absorbs orange light. The energy split in the d orbitals E will be equal to the frequency of orange light(5 x1014 s -1 ) x Planck’s constant E in a blue solution = hv = 6.63 × 10–34 x 5 x1014 = 3.32 × 10–19 J N Goalby chemrevise.org Average energy of d orbitals in field of ligands dz2 dx2 – y 2 dxy dxz dyz E Octahedral complex ion Absorption of visible light is used in spectrometry to determine the concentration of coloured ions. method •Add an appropriate ligand to intensify colour •Make up solutions of known concentration •Measure absorption or transmission •Plot graph of absorption vs concentration •Measure absorption of unknown and compare If visible light of increasing frequency is passed through a sample of a coloured complex ion, some of the light is absorbed. The amount of light absorbed is proportional to the concentration of the absorbing species (and to the distance travelled through the solution). Some complexes have only pale colours and do not absorb light strongly. In these cases a suitable ligand is added to intensify the colour. Spectrophotometry Spectrometers contain a coloured filter. The colour of the filter is chosen to allow the wavelengths of light through that would be most strongly absorbed by the coloured solution.

General trends •Relative stability of +2 state with respect to +3 state increases across the period •Compounds with high oxidation states tend to be oxidising agents e.g MnO4 – •Compounds with low oxidation states are often reducing agents e.g V2+ & Fe2+. Transition elements show variable oxidation states. When transition metals form ions they lose the 4s electrons before the 3d.The redox potential for a transition metal ion changing from a higher to a lower oxidation state is influenced by pH and by the ligand. Vanadium has four main oxidation states VO2+ Oxidation state +5 ( a yellow solution) VO2+ Oxidation state + 4 (a blue solution) V3+ Oxidation state + 3 (a green solution) V2+ Oxidation state + 2 (a violet solution) Vanadium The ion with the V at oxidation state +5 exists as a solid compound in the form of a VO3 – ion, usually as NH4VO3 (ammonium vanadate (V). It is a reasonably strong oxidising agent. Addition of acid to the solid will turn into the yellow solution containing the VO2+ ion. Addition of zinc to the vanadium (V) in acidic solution will reduce the vanadium down through each successive oxidation state, and the colour would successively change from yellow to blue to green to violet [Ag(NH3 )2 ]+ is used in Tollen’s reagent to distinguish between aldehydes and ketones . Aldehydes reduce the silver in the Tollen’s reagent to silver. Red ½ eq: [Ag(NH3 )2 ]+ + e-  Ag +2NH3 Ox ½ eq: CH3CHO + H2O  CH3CO2H + 2H+ + 2e

Manganate Redox Titration The redox titration between Fe2+ with MnO4 – (purple) is a very common exercise. This titration is self indicating because of the significant colour change from reactant to product. MnO4 -(aq) + 8H+ (aq) + 5Fe2+ (aq) Mn2+ (aq) + 4H2O (l) + 5Fe3+ (aq) Purple colourless Choosing correct acid for manganate titrations. The acid is needed to supply the 8H+ ions. Some acids are not suitable as they set up alternative redox reactions and hence make the titration readings inaccurate. Only use dilute sulphuric acid for manganate titrations. Insufficient volumes of sulphuric acid will mean the solution is not acidic enough and MnO2 will be produced instead of Mn2+ . MnO4 -(aq) + 4H+ (aq) + 3e-  MnO2 (s) + 2H2O The brown MnO2 will mask the colour change and lead to a greater (inaccurate) volume of Manganate being used in the titration. Using a weak acid like ethanoic acid would have the same effect as it cannot supply the large amount of hydrogen ions needed (8H+ ). It cannot be conc HCl as the Clions would be oxidised to Cl2 by MnO4 – as the Eo MnO4 -/Mn2+ > Eo Cl2 /Cl- MnO4 -(aq) + 8H+ (aq) + 5e–  Mn2+ (aq) + 4H2O(l) E+1.51V Cl2 (aq) +2e–  2Cl– (aq) E +1.36V This would lead to a greater volume of manganate being used and poisonous Cl2 being produced. It cannot be nitric acid as it is an oxidising agent. It oxidises Fe2+ to Fe3+ as Eo NO3 -/HNO2> Eo Fe3+/Fe2+ NO3 – (aq) + 3H+ (aq) + 2e–  HNO2 (aq) + H2O(l) Eo +0.94V Fe3+ (aq)+e–  Fe2+ (aq) Eo +0.77 V This would lead to a smaller volume of manganate being used. The purple colour of manganate can make it difficult to see the bottom of meniscus in the burette. If the manganate is in the burette then the end point of the titration will be the first permanent pink colour. Colourless  purple

Other useful manganate titrations With hydrogen peroxide Ox H2O2  O2 + 2H+ + 2e- Red MnO4 -(aq) + 8H+ (aq) + 5e-  Mn2+ (aq) + 4H2O Overall 2MnO4 -(aq) + 6H+ (aq) + 5H2O2  5O2 + 2Mn2+ (aq) + 8H2O Ox C2O4 2-  2CO2 + 2e- Red MnO4 -(aq) + 8H+ (aq) + 5e-  Mn2+ (aq) + 4H2O Overall 2MnO4 -(aq) + 16H+ (aq) + 5C2O4 2-(aq) 10CO2 (g) + 2Mn2+(aq) + 8H2O(l) With ethanedioate With Iron (II) ethanedioate both the Fe2+ and the C2O4 2- react with the MnO4 – 1MnO4 – reacts with 5Fe2+ and 2 MnO4 – reacts with 5C2O4 2- MnO4 -(aq) + 8H+ (aq) + 5Fe2+  Mn2+ (aq) + 4H2O + 5Fe3+ 2MnO4 -(aq) + 16H+ (aq) + 5C2O4 2-  10CO2 + 2Mn2+ (aq) + 8H2O So overall 3MnO4 -(aq) + 24H+ (aq) + 5FeC2O4  10CO2 + 3Mn2+ (aq) + 5Fe3+ + 12H2O So overall the ratio is 3 MnO4 – to 5 FeC2O4 The reaction between MnO4 – and C2O4 2- is slow to begin with (as the reaction is between two negative ions). To do as a titration the conical flask can be heated to 60o C to speed up the initial reaction. Be able to perform calculations for these titrations and for others when the reductant and its oxidation product are given. A 2.41g nail made from an alloy containing iron is dissolved in 100cm3 acid. The solution formed contains Fe(II) ions. 10cm3 portions of this solution are titrated with potassium manganate (VII) solution of 0.02M. 9.80cm3 of KMnO4 were needed to react with the solution containing the iron. What is the percentage of Iron by mass in the nail? Manganate titration example MnO4 – (aq) + 8H+ (aq) + 5Fe2+  Mn2+ (aq) + 4H2O + 5Fe3+ Step1 : find moles of KMnO4 moles = conc x vol 0.02 x 9.8/1000 = 1.96×10-4 mol Step 2 : using balanced equation find moles Fe2+ in 10cm3 = moles of KMnO4 x 5 = 9.8×10-4 mol Step 3 : find moles Fe2+ in 100cm3 = 9.8×10-4 mol x 10 = 9.8×10-3 mol Step 4 : find mass of Fe in 9.8×10-3 mol mass= moles x RAM = 9.8×10-3 x 55.8 = 0.547g Step 5 : find % mass %mass = 0.547/2.41 x100 = 22.6% N Goalby chemrevise.org A 1.412 g sample of impure FeC2O4 .2H2O was dissolved in an excess of dilute sulphuric acid and made up to 250 cm3 of solution. 25.0 cm3 of this solution decolourised 23.45 cm3 of a 0.0189 mol dm–3 solution of potassium manganate(VII). What is the percentage by mass of FeC2O4 .2H2O in the original sample? Step1 : find moles of KMnO4 moles = conc x vol 0.0189 x 23.45/1000 = 4.43×10-4 mol Step 2 : using balanced equation find moles FeC2O4 .2H2O in 25cm3 = moles of KMnO4 x 5/3 (see above for ratio) = 7.39×10-4 mol Step 3 : find moles FeC2O4 .2H2O in 250 cm3 = 7.39×10-4 mol x 10 = 7.39×10-3 mol Step 4 : find mass of FeC2O4 .2H2O in 7.39×10-3 mol mass= moles x Mr = 7.39×10-3 x 179.8 = 1.33g Step 5 ; find % mass %mass = 1.33/1.412 x100 =

Transition metals and their compounds can act as heterogeneous and homogeneous catalysts Catalysis A heterogeneous catalyst is in a different phase from the reactants A homogeneous catalyst is in the same phase as the reactants Surface area: Increasing the surface area of a solid catalyst will improve its effectiveness. A support medium is often used to maximise the surface area and minimise the cost (e.g. Rh on a ceramic support in catalytic converters). Heterogeneous catalysts are usually solids whereas the reactants are gaseous or in solution. The reaction occurs at the surface of the catalyst. Heterogeneous catalysis Adsorption of reactants at active sites on the surface may lead to catalytic action. The active site is the place where the reactants adsorb on to the surface of the catalyst. This can result in the bonds within the reactant molecules becoming weaker, or the molecules being held in a more reactive configuration. There will also be a higher concentration of reactants at the solid surface so leading to a higher collision frequency. Transition Metals can use the 3d and 4s e- of atoms on the metal surface to form weak bonds to the reactants. Steps in Heterogeneous Catalysis 1. Reactants form bonds with atoms at active sites on the surface of the catalyst (adsorbed onto the surface) 2. As a result bonds in the reactants are weakened and break 3. New bonds form between the reactants held close together on catalyst surface. 4. This in turn weakens bonds between product and catalyst and product leaves (desorbs). Strength of adsorption The strength of adsorption helps to determine the effectiveness of the catalytic activity. Some metals e.g. W have too strong adsorption and so the products cannot be released. Some metals e.g. Ag have too weak adsorption, and the reactants do not adsorb in high enough concentration. Ni and Pt have about the right strength and are most useful as catalysts. Catalysts increase reaction rates without getting used up. They do this by providing an alternative route with a lower activation energy. V2O5 is used as a catalyst in the Contact Process. Overall equation : 2SO2 + O2  2SO3 step 1 SO2 + V2O5  SO3 + V2O4 step 2 2V2O4 + O2  2V2O5 Examples of heterogeneous catalysts Learn the equations for this mechanism. Note the oxidation number of the vanadium changes and then changes back. It is still classed as a catalyst as it returns to its original form. Cr2O3 catalyst is used in the manufacture of methanol from carbon monoxide and hydrogen. CO + 2H2  CH3OH Fe is used as a catalyst in the Haber Process N2 + 3H2  2NH3 Poisoning Catalysts Catalysts can become poisoned by impurities and consequently have reduced efficiency. Poisoning has a cost implication e.g. poisoning by sulphur in the Haber Process and by lead in catalytic converters in cars means that catalysts lose their efficiency and may need to be replaced. Leaded petrol cannot be used in cars fitted with a catalytic converter since lead strongly adsorbs onto the surface of the catalyst. It is important to ensure the purity of the reactants if poisoning can occur. When catalysts and reactants are in the same phase, the reaction proceeds through an intermediate species. Homogeneous catalysis The intermediate will often have a different oxidation state to the original transition metal. At the end of the reaction the original oxidation state will reoccur. This illustrates importance of variable oxidation states of transition metals in catalysis. N Goalby chemrevise.org Examples of homogeneous catalysts The reaction between I- and S2O8 2- catalysed by Fe2+ overall S2O8 2- + 2I-  2SO4 2- + I2 Catalysed alternative route stage 1 S2O8 2- + 2Fe2+  2SO4 2- + 2Fe3+ stage2 2I- + 2Fe3+  2Fe2+ + I2 The uncatalysed reaction is very slow because the reaction needs a collision between two negative ions. Repulsion between the ions is going to hinder this – meaning high activation energy. Learn these 2 examples and equations carefully Both of the individual stages in the catalysed mechanism involve collision between positive and negative ions and will have lower activation energies. The autocatalysis by Mn2+ in titrations of C2O4 2- with MnO4 – overall 2 MnO4 – + 5 C2O4 2- + 16 H+  2Mn2+ + 10 CO2 + 8 H2O Catalysed alternative route Step 1 4Mn2+ + MnO4 – + 8 H+  5Mn3+ + 4 H2O Step 2 2Mn3+ + C2O4 2-  2Mn2+ + 2 CO2 The initial uncatalysed reaction is slow because the reaction is a collision between two negative ions which repel each other leading to a high activation energy. The Mn2+ ions produced act as an autocatalyst and therefore the reaction starts to speed up because they bring about the alternative reaction route with lower activation energy. The reaction eventually slows as the MnO4 – concentration drops. This is an example of autocatalysis where one of the products of the reaction can catalyse the reaction. Mixture with only reactants showing effect of autocatalysis Mixture with added catalyst Mn2+ at start Following the reaction rate This can be done by removing samples at set times and titrating to work out the concentration of MnO4 – . It could also be done by use of a spectrometer measuring the intensity of the purple colour. This method has the advantage that it does not disrupt the reaction mixture, using up the reactants and it leads to a much quicker determination of concentration. Autocatalytic Reaction between Ethanedioate and Manganate ions Reaction between iodide and persulphate ions Fe3+ ions can also act as the catalyst because the two steps in the catalysed mechanism can occur in any order. 9 Constructing a catalysed mechanism for a reaction Example The following reaction is catalysed by Co2+ ions in an acidic solution. SO3 2– + ½ O2  SO4 2– . Write a mechanism for the catalysed reaction by writing two equations involving Co 2+ and Co3+ ions. Split the full equation into its two half equations SO3 2– + ½ O2  SO4 2– SO3 2– + H2O  SO4 2– + 2H+ + 2e- ½ O2 + 2H+ + 2e-  H2O Add in cobalt to make two new redox equations. Making sure the oxidised cobalt equation is combined with the original reduced half equation and vice versa Co2+  Co3+ + e- Co3+ + e-  Co2+ ½ O2 + 2H+ + 2Co2+  H2O + 2Co3+ 2Co3+ + SO3 2– + H2O  SO4 2– + 2H+ + 2Co2+ Check your two mechanism equations add up to the original full non-catalysed equation.