General properties
General properties of transition metals transition metal characteristics of elements Ti Cu arise from an incomplete d sub-level in 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 ions. these characteristics include •formation of coloured ions, •variable oxidation state •catalytic activity. •Complex ion formation, Why is Sc not a transition metal? Sc can only form a +3 ion. In this ion the Sc3+ has an empty d orbital and so does not meet the criteria of having an incomplete d orbital in one of its ions. Typical Properties of Transition metals The existence of more than one oxidation state for each element in its compounds For example iron commonly forms +2 and +3, chromium commonly forms +2,+3 and +6 see page 5 of this guide for more detail The formation of coloured ions There are characteristic colours for each transition metal element and the colours can vary in the different oxidation state. See the rest of the guide for many examples The catalytic behaviour of the elements and their compounds and their importance in the manufacture of chemicals by industry Iron is used as a catalyst in the Haber process to produce ammonia. Manganese dioxide MnO2 catalyses the decomposition of hydrogen peroxide. Vandandium pentoxide V2O5 catalyses the contact process . See 3.2.2 reaction rates for some more detail
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5.3.1 Transition elements
Properties (a) the electron configuration of atoms and ions of the d-block elements of Period 4 (Sc–Zn), given the atomic number and charge (see also 2.2.1 d) Learners should use sub-shell notation e.g. for Fe: 1s22s22p63s23p63d64s2. (b) the elements Ti–Cu as transition elements i.e. d-block elements that have an ion with an incomplete d-sub-shell (c) illustration, using at least two transition elements, of: (i) the existence of more than one oxidation state for each element in its compounds (see also 5.3.1 k) (ii) the formation of coloured ions (see also 5.3.1 h, j–k) (iii) the catalytic behaviour of the elements and their compounds and their importance in the manufacture of chemicals by industry (see 3.2.2 d) No detail of how colour arises required. Practical examples of catalytic behaviour include: Cu2+ for reaction of Zn with acids; MnO2 for decomposition of H2O2. No detail of catalytic processes required. HSW9 Benefits of reduced energy usage; risks from toxicity of many transition metals.
Ligands and complex formation
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 unidentate (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). A complex with bidentate ligands e.g. [Cr(NH2CH2CH2NH2 )3 ]3+ It has a coordination number of 6 Cu(H2O)6 2+ + 3NH2CH2CH2NH2 [Cu(NH2CH2CH2NH2 )3 ]2+ + 6H2O Cu(H2O)6 2+ + 3C2O4 2- [Cu(C2O4 )3 ]4- + 6H2O Cu(H2O)6 2+ + EDTA4- [Cu(EDTA)]2- + 6H2O Equations to show formation of bidentate and 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 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 ]+ , [Ag(S2O3 )2 ]3- and [Ag(CN)2 ] – (all colourless). [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-Ni(NH3)2Cl2 trans-Ni(NH3)2Cl2. Optical isomerism. 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 The Pt(II) complex cisplatin is used as an anticancer drug. cisplatin transplatin The cisplatin version only works as two chloride ions are displaced and the molecule joins on to the DNA. In doing this it stops the replication of cancerous cells. It can also prevent the replication of healthy cells by bonding on to healthy DNA which may lead to unwanted side effects like hair loss. In the body one Cl ligand is subsituted by a water molecule Pt(NH3 )2Cl2 + H2O [Pt(NH3 )2Cl(H2O)]+ + Cl– Be able to apply your knowledge of bonding to given information in the question to explain how it bonds to DNA molecule- generally a combination of dative covalent bonding and hydrogen bonding Platin binds to DNA of cancer cells and stops cancer cells dividing. Cisplatin
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5.3.1 Transition elements
Ligands and complex ions (d) explanation and use of the term ligand in terms of coordinate (dative covalent) bonding to a metal ion or metal, including bidentate ligands Examples should include: monodentate: H2O, Cl – and NH3 bidentate: NH2CH2CH2NH2 (‘en’). In exams, other ligands could be introduced. (e) use of the terms complex ion and coordination number and examples of complexes with: (i) six-fold coordination with an octahedral shape (ii) four-fold coordination with either a planar or tetrahedral shape (see also 2.2.2 g–h) M4.1, M4.2 Examples: Octahedral: many hexaaqua complexes, e.g. [Cu(H2O)6] 2+, [Fe(H2O)6] 3+ Tetrahedral: many tetrachloro complexes, e.g. CuCl 4 2– and CoCl 4 2– Square planar: complexes of Pt, e.g. platin: Pt(NH3) 2Cl 2 (see also 5.3.1 g). (f) types of stereoisomerism shown by complexes, including those associated with bidentate and multidentate ligands: (i) cis–trans isomerism e.g. Pt(NH3) 2Cl 2 (see also 4.1.3 c–d) (ii) optical isomerism e.g. [Ni(NH2CH2CH2NH2) 3] 2+ (see also 6.2.2 c) M4.1, M4.2, M4.3 Learners should be able to draw 3-D diagrams to illustrate stereoisomerism. HSW8 (g) use of cis-platin as an anti-cancer drug and its action by binding to DNA preventing cell division HSW9 Benefits of chemotherapy; risks from unpleasant side effects.
Ligand substitution and precipitation reactions
Ligand substitution Reaction with excess NH3 With excess NH3 ligand substitution reactions occur with several transition aqueous ions. The ligands NH3 and H2O are similar in size and are uncharged. Ligand exchange occurs without change of co-ordination number for Co and Cr Note: This substitution in the case with Cu is incomplete as not all the water molecules are substituted. deep blue solution [Cu(H2O)6 ]2+ (aq) + 4NH3 (aq) [Cu(NH3 )4 (H2O)2 ]2+ (aq) + 4H2O (l) blue solution [Co(H2O)6 ]2+ (aq) + 6NH3 (aq) [Co(NH3 )6 ]2+ (aq) + 6H2O (l) Be able to write equations for other metal ions given information about the complex formed. No need to learn colours for these other ions. 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. [Cu(H2O)6 ]2+ + 4Cl- [CuCl4 ]2- + 6H2O [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 blue solution blue solution yellow/green solution pink solution. Biological Complexes Fe(II) in haemoglobin enables oxygen to be transported in the blood . Haem is an iron(II) complex with a multidentate ligand. O2 bonds to Fe2+ ions in the Haemoglobin and when required the O2 is released. 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 prevents the oxygen attaching to the haemoglobin. With CO, the stability constant is greater than with complex in O2. Precipitation Reactions with sodium hydroxide and ammonia The bases OH- and ammonia when in limited amounts form the same hydroxide precipitates. [Cu(H2O)6 ]2+ (aq) + 2OH- (aq) Cu(H2O)4 (OH)2 (s) + 2H2O (l) [Mn(H2O)6 ]2+ (aq) + 2OH- (aq) Mn(H2O)4 (OH)2 (s) + 2H2O (l) [Fe(H2O)6 ]3+ (aq) + 3OH- (aq) Fe(H2O)3 (OH)3 (s) + 3H2O (l) These reactions are classed as precipitation reactions [Fe(H2O)6 ]2+ (aq) + 2OH- (aq) Fe(H2O)4 (OH)2 (s) + 2H2O (l) Pale brown ppt green ppt brown ppt Cu2+ (aq) + 2OH- (aq) Cu(OH)2 (s) Mn2+ (aq) + 2OH- (aq) Mn(OH)2 (s) Fe 2+ (aq) + 2OH- (aq) Fe(OH)2 (s) Fe3+ (aq) + 3OH- (aq) Fe(OH)3 (s) blue ppt blue ppt brown ppt Blue solution Blue solution Very pale pink solution green solution green solution Yellow/brown solution Yellow/brown solution Very pale pink solution Pale brown ppt green ppt Reaction with excess OH- With excess NaOH, the Cr hydroxide dissolves. Cr becomes [Cr(OH)6 ]3- (aq) green solution •This hydroxides is classed as amphoteric because it can react with alkali to give a solution and react with acid to form the aqueous salt Cr(H2O)3 (OH)3 (s) + 3OH- (aq ) [Cr(OH)6 ]3- (aq) + 3H2O(l) Cr(H2O)3 (OH)3 (s) + 3H+ (aq ) [Cr(H2O)6 ]3+ (aq) [Mn(H2O)6 ]2+ (aq) + 2NH3 (aq) Mn(H2O)4 (OH)2 (s) + 2NH4+ (aq) [Fe(H2O)6 ]3+ (aq) + 3NH3 (aq) Fe(H2O)3 (OH)3 (s) + 3NH4+ (aq) With ammonia when added in limited amounts the same hydroxide precipitates form. The ammonia acts as a base, removes a proton from the aqueous complex and becomes the ammonium ion. Reaction with excess NH3 With excess NH3 ligand exchange reactions occur with Cu and Cr, and their hydroxide precipitates dissolve in excess ammonia The ligands NH3 and H2O are similar in size and are uncharged. Ligand exchange occurs without change of co-ordination number for Cr Cr becomes [Cr(NH3 )6 ]3+ purple solution This substitution may, however, be incomplete as in the case with Cu Cu becomes [Cu(NH3 )4 (H2O)2 ]2+ deep blue solution Cr(OH)3 (H2O)3(s) + 6NH3 (aq) [Cr(NH3 )6 ]3+ (aq) + 3H2O(l) + 3OH- (aq) Cu(OH)2 (H2O)4(s) + 4NH3 (aq) [Cu(NH3 )4 (H2O)2 ]2+ (aq) + 2H2O (l) + 2OH- (aq)
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5.3.1 Transition elements
Ligand substitution (h) ligand substitution reactions and the accompanying colour changes in the formation of: (i) [Cu(NH3) 4(H2O)2] 2+ and [CuCl 4] 2– from [Cu(H2O)6] 2+ (ii) [Cr(NH3) 6] 3+ from [Cr(H2O)6] 3+ (see also 5.3.1 j) Complexed formulae should be used in ligand substitution equations. (i) explanation of the biochemical importance of iron in haemoglobin, including ligand substitution involving O2 and CO Precipitation reactions (j) reactions, including ionic equations, and the accompanying colour changes of aqueous Cu2+, Fe2+, Fe3+, Mn2+ and Cr3+ with aqueous sodium hydroxide and aqueous ammonia, including: (i) precipitation reactions (ii) complex formation with excess aqueous sodium hydroxide and aqueous ammonia For precipitation, non-complexed formulae or complexed formulae, are acceptable e.g. Cu2+(aq) or [Cu(H2O)6] 2+; Cu(OH)2(s) or Cu(OH)2(H2O)4. With excess NaOH, only Cr(OH)3 reacts further forming [Cr(OH)6] 3–. With excess NH3, only Cr(OH)3 and Cu(OH)2 react forming [Cr(NH3) 6] 3+ and [Cu(NH3) 4(H2O)2] 2+ respectively (see also 5.3.1 h).
Variable oxidation states
Variable Oxidation States Transition elements show variable oxidation states When transition metals form ions they lose the 4s electrons before the 3d Iron (II) Oxidation Fe2+ (green solution) can be easily oxidised to Fe3+ (brown solution) by various oxidising agents. We commonly use potassium manganate (VII), although oxygen in the air will bring about the change MnO4 -(aq) + 8H+ (aq) + 5Fe2+ (aq) Mn2+ (aq) + 4H2O (l) + 5Fe3+ (aq) Purple colourless Iron (III) Reduction Fe3+ (brown solution) can be reduced to Fe2+ (green solution) by various reducing agents. We commonly use potassium iodide solution. The brown colour of the iodine formed can mask the colour change of the Iron. 2Fe3+ (aq) + 2I- (aq) I2 (aq) + 2Fe2+. Reducing Chromium Cr3+ (green) and then Cr2+ (blue) are formed by reduction of Cr2O7 2- (orange) by the strong reducing agent zinc in (HCl) acid solution. Fe2+ is a less strong reducing agent and will only reduce the dichromate to Cr3+ Cr2O7 2- + 14H+ + 6Fe2+ 2Cr3+ + 7H2O + 6 Fe3+ Orange green The Fe2+ and Cr2O7 2- in acid solution reaction can be used as a quantitative redox titration. This does not need an indicator Oxidising Chromium When transition metals in low oxidation states are in alkaline solution they are more easily oxidised than when in acidic solution [Cr(H2O)6 ]3+ (aq) [Cr(OH)6 ]3- (aq) excessNaOH Acidified Not easy to oxidise alkaline easier to oxidise It is easier to remove an electron from a negatively charged ion Alkaline chromium(III) can be oxidised by using oxidising agents such as hydrogen peroxide to the (yellow solution) chromate ion. Cr(OH)6 3- (aq) H2O2 CrO4 2- (aq) Green solution Reduction :H2O2 + 2e- 2OH- Oxidation: [Cr(OH)6 ]3- + 2OH- CrO4 2- + 3e- + 4H2O 2 [Cr(OH)6 ]3- + 3H2O2 2CrO4 2- +2OH- yellow + 8H2O solution. Reduction of Cu2+ to Cu+ Cu2+ (blue solution) can be reduced to Cu+ (colourless solution) by various reducing agents. We commonly use potassium iodide solution. 2Cu2+ (aq) + 2I- (aq) I2 (aq) + 2Cu+ (aq) Disproportionation of copper(I) ions Copper(I) ions when reacting with sulphuric acid will disproportionate to Cu2+ and Cu metal 2Cu+ Cu + Cu2+ Cu+ (aq) + e− Cu(s) Eo = +0.52 V Cu2+ (aq) + e− Cu+ (aq) Eo = +0.15 V So Eo cell = 0.52 − 0.15 = +0.37 V As Eo Cu+ /Cu > Eo Cu2+/Cu+ and Ecell has a positive value of +0.37V , Cu+ disproportionates from +1 oxidation state to 0 in Cu and +2 in Cu2+
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5.3.1 Transition elements
Redox reactions (k) redox reactions and accompanying colour changes for: (i) interconversions between Fe2+ and Fe3+ (ii) interconversions between Cr3+ and Cr2O7 2– (iii) reduction of Cu2+ to Cu+ and disproportionation of Cu+ to Cu2+ and Cu Fe2+ can be oxidised with H+/MnO4 – and Fe3+ reduced with I–, Cr3+ can be oxidised with H2O2/ OH– and Cr2O7 2– reduced with Zn/H+, Cu2+ can be reduced with I–. In aqueous conditions, Cu+ readily disproportionates. Learners will not be required to recall equations but may be required to construct and interpret redox equations using relevant half-equations and oxidation numbers (see 5.2.3 b–c). (l) interpretation and prediction of unfamiliar reactions including ligand substitution, precipitation, redox.
Credits: Neil Goalby