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EEE2 14-15 transition elements p.1 A2 Module F325: Equilibria, Energetics and Elements EEE 2: Transition elements For the elements up to Ca the 3d orbitals are higher in energy than the 4s orbital. Therefore, after argon (element 18), the 4s orbital is filled: Ca has electron configuration [Ar] 4s2. From scandium on, the 3d orbitals are filled, until they have ten electrons at zinc. The term “d-block elements” refers to those elements in which this d-subshell is filling (Sc–Zn), but the term “transition elements” is used for d-block elements that form one or more stable ions with a partially filled d-subshell. This excludes Sc and Zn, since their only common oxidation states are Sc3+ (3d0) and Zn2+ (3d10). This distinction is made because the main features of the chemistry of the transition elements depend largely on this partially filled d-subshell. You will be expected to use your Periodic Table to deduce the electronic configurations of atoms and ions. Remember that: (i) the stability of the half-filled sub-shell means that d5 and d10 configurations are particularly stable e.g. Cr is 1s2 2s2 2p6 3s2 3p6 3d5 4s1 (not 3d4 4s2); (ii) the 3d is written before the 4s; (iii) in ions the 4s electrons are always lost first: Fe2+ is 1s2 2s2 2p6 3s2 3p6 3d6. Remember: the 4s fills before the 3d and is emptied before the 3d. Properties of transition elements (i) Variable oxidation state A Group 2 element like Mg shows only one oxidation state (+2); transition elements, however, normally show a range of different oxidation states e.g. Fe2+ and Fe3+; Cu+ and Cu2+. Generally transition metals show a fairly steady increase in successive ionisation energies until all the 4s and 3d electrons have been removed, when there is a sharp jump as the closed [Ar] shell is broken into. Generally the +2 oxidation will be common as this corresponds to the removal of the element’s two 4s electrons. Higher oxidation states than +2 mean that 3d electrons (close in energy to the 4s) have also been removed. Copper is the only transition metal to form a significant number of compounds in the +1 state because the one 4s electron can be removed to leave a stable 3d 10 configuration. However, even here the +1 state is very easy to oxidise, with the result that copper(I) compounds are only stable if they are in a complex (see below), or if they are insoluble. (ii) Coloured ions Unlike the elements of, for example, Groups 1 and 2, which exhibit white solids and colourless solutions, transition elements show a wide variety of colours in their compounds. When ligands (see below) pack around a metal ion, the d-orbitals no longer have exactly the same energies. If they are partially filled, it is possible for an electron to jump from a lower-energy d-orbital to an unoccupied higher-energy d-orbital. These “d–d transitions” are of an energy corresponding to absorption in the visible region, and so the compound appears to be coloured. The colour is that of the light which is not absorbed: e.g. copper(II) ions look blue because they absorb red light. The energies of d-orbitals, and so the colour of the complex, are very sensitive to the ligands present. EEE2 14-15 transition elements p.2 Colours of some common complexes of iron and copper: Fe(H2O)62+ very pale green; Fe(OH)2 dirty grey/green Fe(H2O)63+ red/brown Fe(OH)3 red/brown (rust) Fe(H2O)5SCN2+ blood red CuI, CuCl, Cu2SO4 all white (d10 is full, so no d–d transitions possible) Cu(H2O)62+ blue CuCl42– yellow/green Cu(NH3)4(H2O)22+ deep blue (iii) Catalytic behaviour A wide range of industrial catalysts consist of transition elements or their compounds. They work through many different mechanisms, of which two will be mentioned here. In the Haber process, Fe2O3 is reduced to finely divided iron metal, and this provides a surface for the nitrogen and hydrogen to react (heterogeneous catalysis – different phase). It is likely that the low energy empty orbitals on the iron can be used to accept co-ordinate bonds, and so absorb the gases temporarily and allow short-lived intermediates to be formed. Homogeneous catalysis – in which the catalyst is in the same phase as the reactants – often involves a reversible change in oxidation states by a d-block metal. Here Fe(OH)3 is insoluble, but the decomposition probably takes place via the small amounts which remain in solution. The Fe3+ can oxidise the hydrogen peroxide, changing to Fe2+, and then another hydrogen peroxide molecule can oxidise it back to Fe3+ again: 2 Fe3+ + H2O2 2 Fe2+ + 2 H+ + H2O2 2 Fe2+ + 2H+ + O2 2 Fe3+ + 2 H2O Other examples of catalysts in industry (all heterogeneous) include Ni in hydrogenation of oils to make margarine, V2O5 in the contact process (oxidation of SO2 to SO3 to make sulfuric acid), and platinum + rhodium in catalytic converters in cars. EEE2 14-15 transition elements p.3 (iv) Complexes Ions of transition elements are able to accept the donation of several lone pairs into their outer shells. The lone pairs will come from species able to donate lone pairs of electrons into coordinate bonds – we call such species ligands in this context. Examples of ligands include H2O, CN-, NH3 and Cl-. A complex ion is one in which a central positive ion is surrounded by ligands, which are co-ordinately bonded to it; e.g. Cr(H2O)63+, Fe(CN)64–. The transition elements are not unique in forming complexes (there are small numbers formed by metals in groups 2, 3 and 4), but they form a much wider range than other elements. This is because the transition metal ions are small and polarising, since their nuclei are poorly shielded, and so they attract ligands strongly. The most common type of complex ion by far is the aqua-ion, where the transition element ion is surrounded by water ligands (usually six, because this is the number able to fit around the ion). Ligands and Complexes The co-ordination number of a metal ion in a complex is the number of atoms that are coordinately bonded to the metal ion. This can be 2 (rarely), e.g. CuCl2–, Ag(CN)2–; or 4, e.g. CuCl42–; or most commonly 6, e.g. Fe(H2O)63+. A chelating ligand can co-ordinate from more than one atom at once, creating a ring structure. A bidentate ligand is one which can co-ordinate from two positions at once, in the same molecule or ion. e.g. O H2N NH2 H2C CH2 ethane-1,2-diamine O C C O O ethanedioate "en" "ox" (old name: ethylenediamine) (old name: oxalate) You are expected to be able to predict the formula and charge of a complex. For this you need to know : (a) the charge on the metal ion; (b) the charge on the ligand; (c) the likely co-ordination number [C.N.]. Examples: Fe2+, with CN–, C.N. 6: gives Fe(CN)64– Cr3+, with C2O42– (bidentate), C.N. 6: gives Cr(C2O4)33– Conversely, in Fe(C2O4)34–, the oxidation state of Fe is given by x – 6 = –4; x = +2. EEE2 14-15 transition elements p.4 Shapes of 4- and 6-co-ordinate complexes (a) 4-co-ordinate Complexes with four ligand positions can be either tetrahedral or square planar. Simple repulsion of ligands would give the former, but in some complexes (especially d8 complexes of Ni-Pd-Pt) interaction with the d-electrons means that the square-planar arrangement is more stable. We shall treat the shape as something to be determined experimentally. A square-planar shape can give rise to cis-trans isomers in complexes like ML2Q2 (where L and Q are different ligands). Comparison with carbon compounds like CH2Br2 reminds us that no cis-trans possibilities occur in a tetrahedral shape: it follows that, if a metal complex of formula ML2Q2 shows cis-trans isomerism, this is proof that it must be square planar. 2- Cl H3N Co Ni Cl Cl Cl H3N Cl tetrahedral e.g. CoCl42- Cl square-planar e.g. cis-Ni(NH3)2Cl2 The cis-isomer of PtCl2(NH3)2, called cis-platin, is effective at treating certain types of cancer, by binding to DNA (in place of the two Cl– ions), thus distorting the shape of DNA and inhibiting replication and transcription in cell production. Trans-platin has no effect, confirming that the shape is critical in enabling the cis-isomer to bind to the DNA. Cis-platin also inhibits repair processes by blocking proteins. The drug is highly toxic, and must be administered with much water and diuretic drugs to limit damage to the kidneys. (b) 6-co-ordinate Complexes with six ligand positions are octahedral [though occasionally distorted, as in some copper(II) complexes, such as Cu(NH3)4(H2O)22+ ]. Octahedral complexes of the type ML4Q2 can show cis-trans isomerism (the two Q ligands can be axial, or opposite, each other, or can be adjacent. If they have three bidentate ligands, as in Ni(NH2CH2CH2NH2)32+, they also show optical isomerism. A complex of this type is chiral, with the three bidentate ligands acting like propeller blades — the propeller can be left-handed or right-handed, like a corkscrew. NH2 NH2 = CH2 CH2 NH2 NH2 NH2 Ni NH2 NH2 Ni NH2 NH2 NH2 NH2 NH2 mirror EEE2 14-15 transition elements p.5 When prepared from symmetrical material (e.g. by adding 1,2-diaminoethane to Ni(H2O)62+) a 50:50 racemic mixture of the two enantiomers results. If the enantiomers are separated (e.g. by crystal picking), each rotates the plane of polarised light in the opposite direction to the other. Reactions of Complex Ions One set of ligands may be displaced by others that form stronger co-ordinate bonds, or which are present in very high concentration – this is called ligand substitution: When copper(II) sulfate solution is treated with dilute aqueous ammonia, the solution starts blue because of the Cu(H2O)62+ ion. It first forms a pale blue precipitate of Cu(OH)2, and then this dissolves to give a deep blue coloured solution, containing the Cu(NH3)4(H2O)22+ ion. [N.B. the hydroxide is formed first because ammonia solution is alkaline – see below for the effect of alkaline solutions on transition element ions] When copper(II) sulfate solution is treated with concentrated hydrochloric acid (or sodium chloride solution), the solution starts blue because of the Cu(H2O)62+ ion. As Cl– ions are added they displace water molecules, forming CuCl42–, which is yellow. The colour changes from blue through lime-green to yellow-green, and becomes more intensely coloured despite the dilution: Cu(H2O)62+ + 4Cl– ⇌ CuCl42– + 6H2O blue yellow Another example: in solution Co(H2O)62+ ions are pink. When concentrated hydrochloric acid is added to this pink solution, the solution eventually becomes deep blue, as a ligand substitution reaction takes place, and the co-ordination number changes from 6 (octahedral) to 4 (tetrahedral): Co(H2O)62+ + 4Cl– ⇌ CoCl42– + 6H2O pink deep blue The reason the CN changes from 6 to 4 is that the Cl- ligands are larger and negatively charged, so only 4 of them can fit around the Co2+ ion, as opposed to 6 water ligands. Ligand substitution is also involved in iron’s role in haemoglobin. Each haemoglobin molecule contains four haem ligands, which are tetradentate – i.e. they can make four coordinate bonds to an ion in their centre. The ion in question is Fe2+. Oxygen molecules can form (weak) coordinate bonds to the Fe2+ ions, and this is how they are transported around the blood stream. However, carbon monoxide molecules can also act as ligands and they form stronger coordinate bonds to the Fe2+ than O2, so as CO gets into the blood stream, it stops the haemoglobin molecules from transporting oxygen and the person may die of suffocation. EEE2 14-15 transition elements p.6 The Stability Constant, Kstab A ligand substitution reaction can be viewed as an equilibrium and as such can have an equilibrium constant assigned to it – in this case called Kstab, as it shows the stability of the new complex formed relative to the original complex. The stability constant of a complex ion is the equilibrium constant for its formation from its constituent ions in a particular solvent. e.g. in a generalized substitution reaction: M(H2O)62+ + 6X- ⇌ MX64- + 6H2O The equilibrium constant will be represented by the expression Kstab = [MX64-] / [M(H2O)62+] [X-]6 [H2O] is omitted because in aqueous solution the concentration of water is essentially constant. Clearly the more the equilibrium is to the right, the more stable the new complex in question and the larger the Kstab value. Thus complexes with the largest Kstab values are considered the most stable. Reaction with Alkalis (NaOH or dilute NH3) Reacting a transition element ion with an alkali such as NaOH or dilute NH3 (which can also effectively provide OH- ions) will result in the formation of a coloured precipitate: (a) Cu2+(aq), a blue solution, gives a mid/light-blue precipitate: Cu2+(aq) + 2OH–(aq) Cu(OH)2(s) 2+ (b) Fe (aq), a very pale green solution, gives a dirty grey/green precipitate: Fe2+(aq) + 2OH–(aq) Fe(OH)2(s) 3+ (c) Fe (aq), a yellow solution, gives a brown (rust-colour) precipitate: Fe3+(aq) + 3OH–(aq) Fe(OH)3(s) 2+ (d) Co (aq), a pink solution, gives a blue precipitate: Co2+(aq) + 2OH–(aq) Co(OH)2(s) Redox systems Because transition elements have variable oxidation states, they will frequently be involved in redox processes in which they switch between these oxidation states. Example 1: Observation: an Fe2+ solution starts very pale green. Dilute sulfuric acid is added, then purple potassium manganate(VII) solution dropwise. The yellow colour of the Fe 3+ begins to appear, and at EEE2 14-15 transition elements p.7 the end point the whole solution turns pink, due to the excess of KMnO4. The half equations: MnO4- Fe2+ Fe3+ + e+ 8H+ + 5e- Mn2+ 4H2O Overall: MnO4- + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2O This reaction could be used in a redox titration, e.g.: Calculation: 5.00g of impure FeSO4.7H2O (Mr = 278) is weighed out and made up to 250 cm3 in a standard flask. 25.0cm3 aliquots of this solution are titrated with 0.0200 mol dm –3 KMnO4 solution, and the mean titre is 14.8cm3. Calculate the % purity of the iron(II) sulfate. Steps: (a) Find moles of MnO4–: amount of MnO4– in 14.8cm3 = 0.02 x 14.8/1000 = 0.000296 mol (b) Use chemical equation (above) to find no. of moles of Fe2+ from equation, 1 mol of MnO4– reacts with 5 mole Fe2+ therefore 0.000296 mol of MnO4– reacts with 5×0.000296 = 0.00148 mol Fe2+ (c) Convert to mass of pure FeSO4.7H2O in 250cm3 amount of Fe2+ in 250cm3 solution = 10 × 0.00148 = 0.0148mol mass of pure FeSO4.7H2O in 250cm3 = 0.0148 × 278 = 4.11g (d) Find percentage purity percentage purity = ( 4.11 / 5.00 ) x 100 = 82.2% Example 2: A copper coin weighing 0.99g is reacted with concentrated nitric acid, producing a solution of Cu 2+ ions which is made up to 250cm3 using distilled water. A 25cm3 portion of the solution is reacted with an excess of potassium iodide solution, producing a white precipitate of CuI and a brown solution of I2. This I2 solution is titrated against 0.1 mol/dm3 sodium thiosulfate solution, requiring 15.00cm3 to reach the end point (noted when starch indicator turns colourless from black/blue). The relevant equations are shown below. What is the percentage by mass of copper in the coin? 2Cu2+(aq) + 4I–(aq) 2CuI(s) + I2(s) I2(aq) + 2S2O32–(aq) S4O62–(aq) + 2I–(aq) (a) Find moles of S2O32-: amount of S2O32- in 15.00cm3 = 0.1 x 15/1000 = 0.0015 (b) Use second equation above to find moles of I2: from equation, 2 moles S2O32- require 1 mole of I2; therefore, 0.0015 moles S2O32- required 0.00075 moles of I2 (c) Use first equation above to find moles of Cu2+: EEE2 14-15 transition elements p.8 from equation, 1 mole of I2 is produced by 2 moles of Cu2+; therefore, 0.00075 moles of I2 is produced by 0.0015 moles of Cu2+ (d) Find original number of moles of Cu2+: 0.0015 moles of Cu2+ in 25cm3 portion; therefore, in 250cm3 solution there are 0.015 moles of Cu2+ (e) Convert to mass of copper: mass of 0.015 moles of copper = 0.015 x 63.5 = 0.9525g (f) Find percentage purity: percentage purity by mass of copper in 0.99g coin = (0.9525 / 0.99) x 100 = 96.2%