Transition Element Chemistry EASE 2 PDF

24.1 What is a transition element? The transition elements are found in the d block of the Periodic Table, between Groups 2 and 13. However, not all d-block elements are classified as transition elements. IMPORTANT A transition element is a d-block element that forms one or more stable ions wi...

24.1 What is a transition element? The transition elements are found in the d block of the Periodic Table, between Groups 2 and 13. However, not all d-block elements are classified as transition elements. IMPORTANT A transition element is a d-block element that forms one or more stable ions with an incomplete d sub- shell. We do not define scandium (Sc) and zinc (Zn) as transition elements. Scandium, with the electronic configuration [Ar] 3d1 4s2, forms only one ion, Sc3+. This 3+ ion has no electrons in its 3d sub-shell: the electronic configuration of Sc3+ is just its argon core, [Ar]. Zinc, with the electronic configuration [Ar] 3d10 4s2, forms only one ion, Zn2+. This 2+ ion has a complete 3d sub-shell: the electronic configuration of Zn2+ is [Ar] 3d10. So neither scandium nor zinc forms an ion with an incomplete d sub-shell. In this chapter we will be looking at the transition elements in the first row of the d block. These are the metals titanium (Ti) through to copper (Cu), according to the definition above. Electronic configurations Atoms of transition elements Table 24.1 shows the electronic configurations of the atoms in the first row of the transition elements. In atoms of the transition elements, the 4s sub-shell is normally filled and the rest of the electrons occupy orbitals in the 3d sub-shell. However, chromium and copper atoms are the exceptions. Chromium atoms have just one electron in the 4s sub-shell. The remaining five electrons are arranged in the 3d sub-shell so that each of its five orbitals is occupied by one electron. Copper atoms also have just one electron in the 4s sub-shell. The remaining ten electrons are arranged in the 3d sub-shell so that each orbital is filled by two electrons. Element Electronic configuration titanium (Ti) 1s2 2s2 2p6 3s2 3p6 3d2 4s2 vanadium (V) 1s2 2s2 2p6 3s2 3p6 3d3 4s2 chromium (Cr) 1s2 2s2 2p6 3s2 3p6 3d5 4s1 manganese (Mn) 1s2 2s2 2p6 3s2 3p6 3d5 4s2 iron (Fe) 1s2 2s2 2p6 3s2 3p6 3d6 4s2 cobalt (Co) 1s2 2s2 2p6 3s2 3p6 3d7 4s2 nickel (Ni) 1s2 2s2 2p6 3s2 3p6 3d8 4s2 copper (Cu) 1s2 2s2 2p6 3s2 3p6 3d10 4s1 Table 24.1: Electronic configurations of the first row of transition elements. Ions of transition elements The transition elements are all metals. In common with all metals, their atoms tend to lose electrons so they form positively charged ions. However, each transition metal can form more than one positive ion. For example, the common ions of copper are Cu+ and Cu2+. We say that the transition metals have variable oxidation states. The resulting ions are often different colours. Look at the ions of vanadium in their different oxidation states (Figure 24.2). Figure 24.2: Vanadium and its oxidation states: a a solution containing VO2+ ions, b a solution containing VO2+ ions, c a solution containing V3+ ions, d a solution containing V2+ ions. Table 24.2 shows the most common oxidation states of the first row of the transition elements. Element Most common oxidation states titanium (Ti) +3, +4 vanadium (V) +2, +3, +4, +5 chromium (Cr) +3, +6 manganese (Mn) +2, +4, +6, +7 iron (Fe) +2, +3 cobalt (Co) +2, +3 nickel (Ni) +2 copper (Cu) +1, +2 Table 24.2: Common oxidation states of the transition elements. It is the similarity in the energy of the 3d and 4s atomic orbitals that make these variable oxidation numbers possible. Because there are variable oxidation states, the names of compounds containing transition elements must have their oxidation number included, e.g. manganese(IV) oxide, MnO2, and cobalt(II) chloride, CoCl2. IMPORTANT When transition elements form ions, their atoms lose electrons from the 4s sub-shell first, followed by 3d electrons. Notice the partially filled d sub-shells in the following examples of ions (see definition of transition element earlier in this section): V atom = 1s2 2s2 2p6 3s2 3p6 3d3 4s2 → V3+ ion = 1s2 2s2 2p6 3s2 3p6 3d2 4s0 Fe atom = 1s2 2s2 2p6 3s2 3p6 3d6 4s2 → Fe3+ ion = 1s2 2s2 2p6 3s2 3p6 3d5 4s0 Cu atom = 1s2 2s2 2p6 3s2 3p6 3d10 4s1 → Cu2+ ion = 1s2 2s2 2p6 3s2 3p6 3d9 4s0 The most common oxidation state is +2, usually formed when the atom of a transition element loses its two 4s electrons. The maximum oxidation number of the transition elements at the start of the row involves all the 4s and 3d electrons in the atoms. For example, vanadium’s maximum oxidation state is +5, involving its two 4s electrons and its three 3d electrons. At the end of the row, from iron onwards, the +2 oxidation state becomes most common as 3d electrons become increasingly harder to remove as the nuclear charge increases across the period. The higher oxidation states of the transition elements are found in complex ions or in compounds formed with oxygen or fluorine. Common examples are the chromate(VI) ion, CrO42−, and the manganate(VII) ion, MnO4−. Question 1 a State the electronic configurations of each of the following atoms and ions: i Ti ii Cr iii Co iv Fe3+ v Ni2+ vi Cu+ b Explain why scandium (which forms only one ion, Sc3+) and zinc (which forms only one ion, Zn2+) are not called transition elements. c Explain why the maximum oxidation state of manganese is +7. d Look back at the different oxidation states of vanadium shown in Figure 24.2. State the oxidation state of the vanadium in each test-tube a–d. e Zirconium (Zr) is in the second row of transition elements beneath titanium in the Periodic Table. Its electronic configuration is [Kr] 4d2 5s2, where [Kr] represents the electronic configuration of krypton, the noble gas with atomic number 36. i Predict the maximum stable oxidation state of zirconium and explain your answer. ii Give the formula of the oxide of zirconium, assuming zirconium exhibits the oxidation state given in part e i. Swap your answers to Question 1 with another learner and find any differences. Discuss the differences and explain your reasoning to each other before agreeing on a final answer. Now mark your answers together against the correct ones provided. 24.2 Physical and chemical properties of the transition elements The transition elements commonly have physical properties that are typical of most metals: they have high melting points they have high densities they are hard and rigid, and so are useful as construction materials they are good conductors of electricity and heat they have variable oxidation states they behave as catalysts they form complex ions (see Section 24.1) they form coloured ions (see Section 24.1). Redox reactions We have seen how the transition elements can exist in various oxidation states. When a compound of a transition element is treated with a suitable reagent, the oxidation state of the transition element can change in the reaction. Whenever a reaction involves reactants that change their oxidation states, the reaction is a redox reaction. Redox reactions involve the transfer of electrons. Remember that a species (atom, molecule or ion) is reduced when its oxidation state is reduced. The oxidation state changes to a lower value. An oxidation state is lowered when a species gains electrons, and, when it is being reduced, it acts as an oxidising agent. Example 1 In the half-equation: Fe3+ has been reduced to Fe2+ by gaining one electron. In the equation above, Fe3+ is acting as an oxidising agent, an electron acceptor. For this reaction to happen, another half-equation is needed in which the reactant loses one or more electrons. The full reaction needs a reducing agent, an electron donor. In Chapter 20 we saw how we can use standard electrode potential values, E⦵, to predict whether or not such reactions could possibly take place. Another half-equation we could consider combining with the reduction of Fe3+ half-equation is: Here the MnO4− ion is also acting as an oxidising agent, an electron acceptor. Both half-equations are written below as they appear in tables of data showing standard electrode potentials: Fe3+(aq) + e− → Fe2+(aq) E⦵ = +0.77 V MnO4−(aq) + 8H+(aq) + 5e− → Mn2+(aq) + 4H2O(l) E⦵ = +1.52 V So the question is: Can Fe3+ ions oxidise Mn2+ to MnO4− ions, or can MnO4− ions in acid solution oxidise Fe2+ ions to Fe3+ ions? In other words, which is the more powerful oxidising agent, Fe3+ ions or MnO4− ions? The magnitude of the positive values provides a measure of the tendency of the half-equations to proceed to the right-hand side. The values show us that MnO4−(aq) is more likely to accept electrons and proceed in the forward direction, changing to Mn2+(aq), than Fe3+(aq) is to accept electrons and change to Fe2+(aq). MnO4−(aq) is a more powerful oxidising agent than Fe3+(aq). Therefore, MnO4−(aq) ions are capable of oxidising Fe2+(aq) ions to form Fe3+(aq). So the top half-equation proceeds in the reverse direction. We can now combine the two half-cells to get the overall reaction. Note that the sign of the Fe3+ / Fe2+ half- cell has changed by reversing its direction. The Fe2+ / Fe3+ equation also has to be multiplied by 5 so that the electrons on either side of the equation cancel out when we add the half-equations together. This does not affect the value of E⦵. Combining the two half-equations: ⦵ ⦵ ⦵ The relatively large positive value of E⦵ (+0.75 V) tells us that the reaction is likely to proceed in the forward direction as written. PRACTICAL ACTIVITY 24.1 Redox titration We can use the reaction above to calculate the amount of iron (Fe2+ ions) in a sample, such as an iron tablet, by carrying out a titration. A known volume (e.g. 25 cm3) of an unknown concentration of Fe2+(aq) is placed in a conical flask. A solution of a known concentration of potassium manganate(VII) is put in a burette. The potassium manganate(VII) solution is titrated against the solution containing Fe2+(aq) in the conical flask. During the reaction of MnO4−(aq) with Fe2+(aq) in the flask, the purple colour of the manganate(VII) ions is removed. The end-point is reached when the Fe2+(aq) ions have all reacted and the first permanent purple colour appears in the conical flask. This is when the MnO4−(aq) ions become in excess in the reaction mixture (Figure 24.3).  Figure 24.3: Manganate(VII) ions can be used to determine the percentage of Fe2+ in an iron tablet. You can achieve a more accurate result for the mass of Fe2+ in a solution by using dichromate(VI) ions, Cr2O72−(aq), to oxidise it in a titration. This is because compounds such as potassium dichromate(VI) can be prepared to a higher degree of purity than potassium manganate(VII). In a titration with Fe2+(aq) and dichromate(VI), we need an indicator of the end-point that will be oxidised as soon as the Fe2+(aq) has all reacted. The half-equation and value for E⦵ for the use of dichromate as an oxidising agent is: Cr2O72−(aq) + 14H+(aq) + 6e− → 2Cr3+(aq) + 7H2O(l) E⦵ = +1.33 V WORKED EXAMPLE 1 0.420 g of iron ore were dissolved in acid, so that all of the iron present in the original ore was then present as Fe2+(aq). The solution obtained was titrated against 0.0400 mol dm−3 KMnO4(aq). The titre was 23.50 cm3. a Calculate the number of moles of MnO4− in the titre. Solution to part a Use the equation: n=V×c where n = number of moles, V = volume of solution in dm3 and c = concentration. b Calculate the number of moles of Fe2+ in the solution. Solution to part b The equation for the reaction in the titration is: 5Fe2+ + MnO4− + 8H+ → 5Fe3+ + Mn2+ + 4H2O(l) c Calculate the mass of iron in the solution (Ar of iron is 55.8). Solution to part c moles of Fe = moles of Fe2+ = 0.004 70 mol d Calculate the percentage mass of iron in the 0.420 g of iron ore. Solution to part d Example 2 Another example of a redox titration is the reaction of manganate(VII) ions with ethanedioate ions, C2O42−. The ethanedioate ion is derived from ethanedioic acid, formerly known as oxalic acid. The ethanedioate ions can be oxidised by potassium manganate(VII) in an acidic solution. A similar titration to the one above can be carried out to find the unknown concentration of ethanedioate ions (toxic) in a solution or to standardise potassium manganate(VII) solution (when the sodium ethanedioate solution needs to be warmed before titrating). The half-equation for the oxidation of ethanedioate ions is: As we have just seen, the reduction of manganate(VII) ions to manganese(II) is: Combining the two half-equations to get the full redox equation we need to multiply the ethanedioate ions half-equation by 5, and multiply the manganate(VII) half-equation by 2. This will ensure that the 10 electrons cancel out on either side of the full equation: Combining the two half-equations we get the full redox equation: Note from the redox equation that, in a titration calculation, the manganate(VII) ion reacts with the ethanedioate ion in the ratio 2 : 5. This reaction is an example of autocatalysis, as the Mn2+(aq) ions formed act as a catalyst for reaction. The transition elements and their ions can act as catalysts because they can vary their oxidation state in the course of a reaction, accepting or losing electrons, and then reverting back to their original state. They can also accept pairs of electrons into vacant d orbitals forming dative bonds with ligands. Example 3 In this final redox example, we can consider the reaction of copper(II) ions with iodide ions in aqueous solution. We can add the two half-equations together if we multiply the Cu2+ / Cu+ half-equation by 2 in order to cancel out the electrons: If we add excess iodide ions to a solution of Cu2+(aq) ions, we get iodine forming in solution and a precipitate of copper(I) iodide: 2Cu2+(aq) + 4I−(aq) → 2CuI(s) + I2(aq) equation 1 We can use this reaction to find the concentration of copper(II) ions in a solution. The formation of iodine in the reaction mixture means that we can use starch indicator to judge an accurate end-point in a titration experiment. With excess iodide ions added to the copper(II) solution, we can add sodium thiosulfate solution of a known concentration from a burette to the reaction mixture in a flask. The aqueous iodine, I2(aq), formed in the reaction mixture reacts with the thiosulfate ions added: I2(aq) + 2S2O32−(aq) → 2I−(aq) + S4O62−(aq) equation 2 The brownish colour of the solution gets lighter as the iodine is used up. Now we can add the starch solution, which turns blue / black with the remaining I2(aq) in the flask, to help us get a sharp end-point. Then titrate slowly until the blue / black colour disappears when all the iodine has reacted, noting the reading on the burette. So, knowing the number of moles of thiosulfate ions added in the titration, from equation 2 we can deduce that half that number of moles of I2(aq) must have been formed in equation 1. So, the number of moles of Cu2+(aq) that we started with in equation 1 must be twice that of the I2(aq), i.e. the same number of moles as the thiosulfate ions added in the titration. From this we can calculate the concentration of the Cu2+(aq) ions in the initial copper(II) solution (i.e. the number of moles in 1 dm3 or 1000 cm3).  WORKED EXAMPLE 2 A 25.00 cm3 of copper(II) sulfate solution of unknown concentration reacted with excess potassium iodide solution. The reaction mixture was then titrated against 0.0850 mol dm−3 sodium thiosulfate solution, using a starch indicator. The titre was 15.40 cm3 when the blue / black colour disappeared. Calculate the concentration of the copper(II) sulfate solution. Solution First, calculate the number of moles of thiosulfate ions, S2O32−(aq), in the titre. Use the equation: n=V×c where n = number of moles, V = volume of solution in dm3 and c = concentration. n= × 0.0850 mol of thiosulfate ions Working backwards from the two equations (as explained above this worked example): 2Cu2+(aq) + 4I−(aq) → 2CuI(s) + I2(aq) I2(aq) + 2S2O32−(aq) → 2I−(aq) + S4O62−(aq) we can deduce that the number of moles of copper(II) ions in the original solution is the same as the number of moles of thiosulfate ions in the titre i.e. × 0.0850 mol Therefore, we can calculate the concentration (c) of the 25.00 cm3 of copper(II) sulfate solution by re-arranging the equation n = V × c to Question 2 a Write two half-equations for the reactions that take place when Fe2+(aq) is oxidised by dichromate(VI) ions. b Combine the two half-equations and write the equation for the oxidation of Fe2+(aq) by dichromate(VI) ions. c Work out the E⦵ value of the cell made when the two half-cells in part a are connected and the reaction in part b takes place. Discuss and explain what this value predicts about the likelihood of Fe2+(aq) being oxidised by dichromate(VI) ions. d How many moles of Fe2+(aq) can 1 mole of dichromate(VI) oxidise? e In a titration, 25.0 cm3 of a solution containing Fe2+(aq) ions was completely oxidised by 15.30 cm3 of 0.001 00 mol dm−3 potassium dichromate(VI) solution. i Calculate the number of moles of potassium dichromate(VI) in 15.30 cm3 of 0.001 00 mol dm−3 solution? ii Calculate how many moles of Fe2+ were present in the 25.0 cm3 of solution. iii Calculate the concentration of the Fe2+(aq) in the flask at the start of the titration. REFLECTION Working as a pair, one person tells the other how to carry out a titration and process the results. Then swap and let the other person explain how you can use the results of the titration to work out the unknown concentration of a solution. Discuss any ways you can think of that would improve both explanations. At this point, reflect with your partner on your understanding of: a the electronic configurations of the atoms and ions of transition elements b the use of standard electrode potentials to predict the feasibility of reactions c redox reactions, including adding up half-equations d titration calculations. How much did you know about these topics before starting this chapter? Has this improved? Do you need to increase your knowledge further? 24.3 Ligands and complex formation In the section above on redox reactions, we learned about the oxidation of Fe2+(aq) ions. When these ions are in solution the Fe2+ ion is surrounded by six water molecules. Each water molecule bonds to the central Fe2+ ion by forming a dative (co-ordinate) bond from the oxygen atom into vacant orbitals on the Fe2+ ion (Figure 24.4). The water molecules are called ligands and the resulting ion is called a complex ion. Its formula is written as [Fe(H2O)6]2+. The shape of a complex with six ligands is octahedral. Figure 24.4: [Fe(H2O)6]2+; the complex ion formed between an Fe2+ ion and six water molecules. It is called a hexaaquairon(II) ion. All ligands can donate an electron pair to a central transition metal ion. The co-ordination number of a complex is the number of co-ordinate (dative) bonds to the central metal ion. Some ligands can form two co-ordinate (dative) bonds from each ion or molecule to the transition metal ion. These are called bidentate ligands, as shown in Figure 24.6. Most ligands, such as water and ammonia, form just one co-ordinate (dative) bond and are called monodentate ligands. Figure 24.5: The complex ion formed between a transition metal ion and a larger ligand can only fit four, not six, ligands around the central ion. These are arranged in either a square planar shape (as in a [Ni(CN)4]2−) or a tetrahedral shape (as in b [CoCl4]2−). Figure 24.6: [Co(en)3]2+ is an example of a complex ion containing the bidentate ligand NH2CH2CH2NH2 (abbreviated to ‘en’). Another common bidentate ligand is the ethanedioate ion, C2O42− (abbreviated to ‘ox’) e.g. in [Fe(ox)3]3. Name of ligand Formula Example of Co- Shape of complex ordination complex number water H 2O [Fe(H2O)6]2+ 6 octahedral (see Figure 24.4) ammonia NH3 [Co(NH3)6]3+ 6 octahedral chloride ion Cl− [CoCl4]2− 4 tetrahedral (see Figure 24.5b) cyanide ion CN− [Ni(CN)4]2− 4 square planar (see Figure 24.5a) hydroxide ion OH− [Cr(OH)6]3− 6 octahedral thiocyanate ion SCN− [FeSCN]2+ or 6 octahedral [Fe(SCN) (H2O)5]2+ ethanedioate ion (abbreviated to ‘ox’ – oxalate – −OOC─COO− [Mn(ox)3]3− 6 octahedral in the formulae of complexes) 1,2-diaminoethane (abbreviated to ‘en’ in the NH2CH2CH2NH2 [Co(en)3]3+ 6 octahedral (see formulae of complexes) Figure 24.6) Table 24.3: Some common ligands and their complexes. A few transition metal ions (e.g. copper(I), silver(I), gold(I)) form linear complexes with ligands. The co- ordination number in these complexes is 2 (Figure 24.7). Figure 24.7: The diamminesilver(I) cation has a linear structure. Table 24.3 shows some common ligands. Note that the charge on a complex is simply the sum of the charges on the central metal ion and on each ligand in the complex. Some complexes will carry no charge, e.g. Cu(OH)2(H2O)4. EDTA4− ions can act as ligands. A single EDTA4− ion can form six co-ordinate bonds to a central transition metal ion to form an octahedral complex. It is an example of a polydentate ligand called a hexadentate ligand. Question 3 a State the oxidation number of the transition metal in each of the following complexes: i [Co(NH3)6]3+ ii [Ni(CN)4]2− iii [Cr(OH)6]3− iv [Co(en)3]3+ v Cu(OH)2(H2O)4 b Give the formula of a complex formed between Ni2+ and EDTA4−.  c Which ligands in Table 24.3 are bidentate? Stereoisomerism in transition metal complexes In Chapter 14 we learned about two types of stereoisomerism: geometric and optical isomerism. The presence of a double bond in 1,2-dibromoethene means that two geometrical isomers (cis/trans isomers) are possible. Geometric isomerism is also possible in transition metal complexes, where no double bond exists. In this case, the term ‘geometric isomerism’ refers to complexes with the same molecular formula but different geometrical arrangements of their atoms. Examples are any square planar complexes with the general formula [M(A)2(B)2], where M is the transition element, and A and B are its surrounding ligands. Cis- and trans-platin are geometric isomers (Figure 24.8). In cis-platin, the chlorine atoms are next to each other in the square complex but in trans-platin, they are diagonally opposite. Figure 24.8: The geometrical isomers, cis-platin and trans-platin. Cis-platin has been used as an anti-cancer drug. It acts by binding to sections of the DNA in cancer cells, preventing cell division (see the ‘Fighting cancer’ section at the start of Chapter 24). Trans-platin does not have the same beneficial medical effects. The properties of geometrical isomers can be different. Differences in electronegativity of the atoms in ligands forming the dative bonds to the central transition metal ion can result in polar and non-polar complexes. For example, the cis-isomer will have two identiical groups on one side of any square planar complex. So any difference in electronegativity between the two pairs of isomers will cause an imbalance of charge, making a polar complex. The atoms with the higher electronegativity have a stronger pull on the electrons in the dative bonds and will carry a partial negative charge. However, the trans-isomer will have identical ligands directly opposite each other at the corners of the square planar complex. So the pull on the electrons in the dative bonds to the central metal ion are perfectly balanced. This means that the charge is balanced and results in a non-polar complex. Octadedral complexes, with the general formula [M(A)4(B)2], can also display geometric isomerism. An example is the cobalt(II) complex ion, [Co(NH3)4(H2O)2]2+ (see Figure 24.9). Figure 24.9: The cis- and trans- isomers of [Co(NH3)4(H2O)2]2+ Again, the cis-isomer is a slightly polar complex, whereas the trans-isomer is non-polar. The imbalance of charge in cis-isomer results from the asymmetric shape of the cis-[Co(NH3)4(H2O)2]2+ isomer and the difference in electronegativity between the oxygen and nitrogen atoms bonded to the central cobalt(II). The side of the complex with the water ligands will be partially negative as oxygen is more electronegative than the nitrogen in the ammonia ligands. The differences in electronegativity still apply in the trans- [Co(NH3)4(H2O)2]2+ isomer but its symmetical arrangement ensures the charge in the complex is spread evenly around the complex. Stereoisomerism is also commonly shown by octahedral (six co-ordinate) complexes associated with bidentate ligands. An example is the complex containing nickel as the transition metal and 1,2- diaminoethane (NH2CH2CH2NH2) as the bidentate ligand (Figure 24.10). The two isomers are stereoisomers because the two different molecules are mirror images of each other and cannot be superimposed. They are optical isomers, differing only in their ability to rotate the plane of polarised light in opposite directions. Figure 24.10: The two non-superimposable optical isomers of [Ni(NH2CH2CH2NH2)3]2+: a the full structure, b a simplified structure with ‘en’ representing a molecule of 1,2-diaminoethane. The complex ion consisting of nickel(II) bonded to two bidentate 1,2-diaminoethane (en) ligands and two monodentate ligands, such as water or chloride ions, can form both geometric (cis/trans) isomers and optical isomers. Look at the cis- and trans-isomers of [Ni(H2NCH2CH2NH2)2(H2O)2]2+ in Figure 24.11, simplified to [Ni(en)2(H2O)2]2+. Figure 24.11: The geometric, cis- and trans- isomers of [Ni(H2NCH2CH2NH2)2(H2O)2]2+ Of the two cis- and trans- isomers of [Ni(en)2(H2O)2]2+ only the cis- isomer is optically active, with its two non-superimposable mirror images. The symmetrical nature of the trans-isomers of [Ni(en)2(H2O)2]2+ means that its mirror images can be superimposed so they do not display optical isomerism. Look at the optical isomers of [Ni(en)2(H2O)2]2+ in Figure 24.12. Figure 24.12: The non-superimposable, optical isomers of cis-[Ni(en)2(H2O)2]2+ Question 4 a Cobalt forms a complex with the simplified structure: i Give the co-ordination number of this complex. ii Draw the stereoisomer of this complex. iii Explain why this is a stereoisomer. b i Draw the two geometrical isomers of [Ni(CN)2(Cl)2]2−. Label the cis-isomer and the trans- isomer. ii Deduce the overall polarity of the trans-isomer of [Ni(CN)2(Cl)2]2−. Discuss your answers to Question 4, and your reasoning, with a partner. Substitution of ligands Copper complexes The ligands in a complex can be exchanged, wholly or partially, for other ligands. This is a type of substitution reaction. It happens if the new complex formed is more stable than the original complex. The complexes of copper(II) ions can be used to show ligand substitution reactions, also called ligand exchange reactions. Whenever we write Cu2+(aq) we are really referring to the complex ion [Cu(H2O)6]2+. This ion gives a solution of copper sulfate its blue colour. On adding sodium hydroxide solution, we see a light blue precipitate forming. Two water ligands are replaced by two hydroxide ligands in the reaction: If you now add concentrated ammonia solution, the pale blue precipitate dissolves and we get a deep blue solution: The first reaction can also be achieved by adding concentrated ammonia solution to copper sulfate solution drop by drop or by adding a dilute solution of ammonia. The pale blue precipitate formed will then dissolve and form the deep blue solution when excess ammonia is added. The structure of [Cu(NH3)4(H2O)2]2+(aq) is shown in Figure 24.13. Figure 24.13: The octagonal structure of [Cu(NH3)4(H2O)2]2+(aq) Water ligands in [Cu(H2O)6]2+ can also be exchanged for chloride ligands if we add concentrated hydrochloric acid drop by drop. A yellow solution forms, containing the complex ion [CuCl4]2− : The mixture of blue and yellow solutions in the reaction mixture gives it a greenish colour (Figure 24.14). Figure 24.14: The equations for the changes are: [Cu(H2O)6]2+ + 4Cl− ⇌ [CuCl4]2− + 6H2O [Cu(H2O)6]2+ + 4NH3 ⇌ [Cu(NH3)4(H2O)2]2+ + 4H2O Cobalt complexes Aqueous cobalt(II) compounds also form complex ions. Whenever we write Co2+(aq), we are referring to the complex ion [Co(H2O)6]2+. This ion gives an aqueous solution of cobalt(II) sulfate its pink colour. On adding sodium hydroxide solution, we see a blue precipitate of cobalt(II) hydroxide forming, which turns red when warmed if the alkali is in excess. Water ligands in [Co(H2O)6]2+ can also be exchanged for ammonia ligands if we add concentrated aqueous ammonia drop by drop. The brown cobalt(II) complex ion is oxidised by oxygen in the air to [Co(NH3)6]3+(aq), a cobalt(III) complex ion. Adding concentrated hydrochloric acid drop by drop to an aqueous solution of cobalt(II) ions results in the formation of a blue solution containing the tetrahedral complex [CoCl4]2−(aq). Aqueous cobalt(II) ions usually form tetrahedral complexes with monodentate anionic ligands such as Cl−, SCN− and OH−. Question 5 a Blue cobalt chloride paper gets its blue colour from [CoCl4]2− ions. Give the oxidation number of the cobalt in this complex. b Blue cobalt chloride paper is used to test for water. If water is present, the paper turns pink as a complex forms between the cobalt ion and six water ligands (Figure 24.15). Write an equation to show the ligand substitution reaction that takes place in a positive test for water. c Which equation correctly describes the reaction of hydrated copper(II) ions with concentrated hydrochloric acid? A [Cu(H2O)6]2+(aq) + 4Cl−(aq) → [CuCl4]2−(aq) + 6H2O(l) B [Cu(H2O)6]2+(aq) + 6Cl−(aq) → [CuCl6]4−(aq) + 6H2O(l) C [Cu(H2O)6]2+(aq) + 2Cl−(aq) → [CuCl2]2−(aq) + 6H2O(l) D [Cu(H2O)6]2+(aq) + 4Cl−(aq) → [Cu(H2O)2Cl4]2−(aq) + 4H2O(l) Discuss with a partner the thought processes you go through to deduce the oxidation number of the central transition element in a complex (Question 5a). Working together, list the sequence of steps that another learner could follow. Discuss the option you chose in Question 5c, explaining why you think the three other options are incorrect. Figure 24.15: A positive test for the presence of water using anhydrous cobalt chloride paper. Stability constants Aqueous solutions of transition element ions are hydrated. They are complex ions with water molecules acting as ligands, forming dative (coordinate) bonds to the central metal ion. Different ligands form complexes with different stabilities. For example, when we add concentrated aqueous ammonia to an aqueous solution of copper(II) sulfate, the ammonia ligands displace water ligands in a stepwise process. [Cu(H2O)6]2+(aq) + NH3(aq) ⇌ [Cu(NH3)(H2O)5]2+(aq) + H2O(l) [Cu(NH3)(H2O)5]2+(aq) + NH3(aq) ⇌ [Cu(NH3)2(H2O)4]2+(aq) + H2O(l) As we increase the concentration of ammonia, this process continues until four of the water molecules are replaced by ammonia molecules. The solution formed is a deep blue colour. The overall ligand substitution reaction is: We can think of this exchange of ligands in terms of competing equilibria of the forward and backward reactions. The position of equilibrium lies in the direction of the more stable complex. In this case, the complex with ammonia as a ligand is more stable than the complex with just water as a ligand. If we dilute the complex with water, the position of equilibrium shifts to the left and a complex with more water molecules as ligands forms. The stability of the complex is expressed in terms of the equilibrium constants for ligand displacement. This is called the stability constant. Usually an overall stability constant, Kstab, is given rather than the stepwise constants. The method for writing equilibrium expressions for stability constants is similar to the one we used for writing Kc (see Section 21.1). So for the equilibrium: [Cu(H2O)6]2+ + 4Cl−(aq) ⇌ [CuCl4]2−(aq) + 6H2O(l) the expression for the stability constant is: Note: water does not appear in the equilibrium expression because it is in such a large excess that its concentration is regarded as being constant the units for the stability constant are worked out in the same way as for the units of Kc (see Section 8.3). For example, in the above case: Stability constants are often given on a log10 scale. When expressed on a log10 scale, they have no units (see Table 24.4). Stability constants can be used to compare the stability of any two ligands. The values quoted usually give the stability of the complex relative to the aqueous ion where the ligand is water. IMPORTANT The higher the value of the stability constant, the more stable the complex. Table 24.4 gives some values of stability constants for various copper(II) complexes relative to their aqueous ions. Ligand log10 Kstab chloride, Cl− 5.6 ammonia, NH3 13.1 2-hydroxybenzoate 16.9  1,2-dihydroxybenzene 25.0 Table 24.4: The stability constants of some complexes of copper. From Table 24.4 you can see that, in general, complexes with bidentate ions (2-hydroxybenzoate and 1,2- dihydroxybenzene) have higher stability constants than those with monodentate ligands. We can use the values of the stability constants to predict the effect of adding different ligands to complex ions. For example, the addition of excess ammonia to the complex [CuCl4]2−(aq) should result in the formation of a dark blue solution of the ammonia complex. That is because the stability constant of the ammonia complex is higher than that of the chloride complex. The position of equilibrium is shifted to the right in the direction of the more stable complex. Addition of excess 1,2-dihydroxybenzene to the dark blue ammonia complex results in the formation of a green complex with 1,2-dihydroxybenzene. This is because the stability constant with the 1,2- dihydroxybenzene is much higher than that for ammonia. You can see from Table 24.4 that the log of Kstab for 1,2-dihydroxybenzene is 25.0, so the actual value of Kstab is 10 raised to the power 25, compared to the actual value for ammonia of 10 raised to the power 13.1. This shows that the copper complex with 1.2- dihydroxybenzene is much more stable than the complex ion formed with ammonia. These values of Kstab predict that there will be hardly any of the copper complex with ammonia left in the equilibrium mixture shown in the equation below: WORKED EXAMPLE 3 When concentrated hydrochloric acid is added to copper(II) sulfate solution, the aqueous solution formed contains [CuCl4]2− and [Cu(H2O)4]2+ complex ions. a Write the expression for the stability constant for [CuCl4]2− in an aqueous solution. Solution First of all write the chemical equation for the ligand exchange equilibrium mixture formed in the reaction: [Cu(H2O)4]2+(aq) + 4Cl−(aq) ⇌ [CuCl4]2−(aq) + 4H2O Then write the expression for Kstab. (Remember that products go on the top line and reactants on the bottom line in equilibrium expressions, and that H2O is not included) b Work out the units of Kstab for the expression in part a. Solution Substitute the unit of concentration (mol dm−3) into your equilibrium expression from part a in this question. Then cancel out the units. c Using Table 24.4, explain the proportions of the two copper(II) complex ions present in the equilibrium mixture. Solution  As the value of log10Kstab for the chloride ion is given in the table as 5.6, this shows that the actual value of Kstab is 10 raised to the power of 5.6 i.e. a very large number. This shows that the concentration of products in the equilibrium expression, in this case the [CuCl4]2− ions, is much greater than the concentration of the reactant [Cu(H2O)6]2+ ions. d When concentrated ammonia solution is added to copper(II) chloride, explain what you would see happen, using Table 24.4. Solution As the Kstab value for the copper(II) complex with ammonia is a much larger value than that of the complex with chloride ions, the ammonia molecules will displace the chloride ions from the [CuCl4]2− complex, forming a more stable complex of [Cu(NH3)4(H2O)2]2+. This will turn the (greenish) yellow solution of the chloride complex a deep blue colour. e The exchange of ligands occurs step-by-step as successive ligands are replaced. For example the second step in the exchange of water ligands for chloride ion ligands is shown by this equation: [Cu(H2O)5Cl]+(aq) + Cl–(aq) ⇌ Cu(H2O)4Cl2(aq) + H2O(l) It was found for this reaction that when a 0.15 mol dm−3 solution of the complex ion [(Cu(H2O)5Cl]+ was mixed with 0.15 mol dm−3 hydrochloric acid, the equilibrium concentration of the Cu(H2O)4Cl2(aq) complex was 0.10 mol dm−3. i Write an expression for the stability constant for this ligand exchange reaction. Solution ii Calculate the value for Kstab for the reaction giving the units. Solution Work out the concentration of each of the ions at equilibrium: [Cu(H2O)5Cl]+(aq) Cl−(aq) Cu(H2O)4Cl2(aq) At start 0.15 mol dm−3 0.15 mol dm−3 0 At equilibrium (0.15 − 0.10) mol dm−3 (0.15 − 0.10) mol dm−3 0.10 mol dm-3 Then substitute the concentrations at equilibrium into the expression for Kstab in e i: Question 6 a Write expressions for the stability constants for the following reactions: i [PtCl4]2−(aq) + 2NH3(aq) ⇌ PtCl2(NH3)2(aq) + 2Cl−(aq) ii [Cr(H2O)6]3+(aq) + 2Cl−(aq) ⇌ [Cr(H2O)4Cl2]+(aq) + 2H2O(l) iii [Ni(H2O)6]2+(aq) + 4NH3(aq) ⇌ [Ni(NH3)4(H2O)2]2+(aq) + 4H2O(l) b An iron(III) ion, Fe3+, in aqueous solution has six water molecules bonded to it as ligands. i Draw the structure of this ion. ii When thiocyanate ions, SCN−, are added to an aqueous solution of iron(III) ions, the solution turns red and one water molecule is replaced by a thiocyanate ion. Use the concept of stability constants to explain why the reaction occurs. iii Deduce the formula of the ion forming the red solution.  iv The stability constant for aqueous Fe3+ ions with SCN− as a ligand is 891 dm3 mol−1. The stability constant for aqueous Fe3+ ions with fluoride ions, F−, as a ligand is 2 × 105 dm3 mol −1. A solution containing fluoride ions is added to the red solution. Would you expect to observe any changes? Explain your answer. Now share and discuss your answers to Question 6 with another learner. Agree on a final set of answers before checking them against the answers provided. The colour of complexes You will have now seen the striking colours of complexes containing transition metal ions. But how do these colours arise? White light is made up of all the colours of the visible spectrum. When a solution containing a transition metal ion in a complex appears coloured, part of the visible spectrum is absorbed by the solution. The colour we see is called the complementary colour, made up of light with frequencies not absorbed. For example, copper(II) ions absorb light from the red end of the spectrum, so the complementary colour seen is a pale blue (called cyan). Here is a list of complementary pairs of colours: red cyan yellow blue green magenta You should be aware that mixing light of different frequencies does not produce the same colour as mixing different colours of paint. However, that still doesn’t explain why part of the spectrum is absorbed by transition metal ions. To answer this question, we must look in more detail at the d orbitals in the ions (Figure 24.16). Figure 24.16: The degenerate d orbitals in a transition metal atom. The five d orbitals in an isolated transition metal atom or ion are described as degenerate orbitals, meaning they are all at the same energy level. In the presence of ligands, a transition metal ion is not isolated. The dative (or co-ordinate) bonding from the ligands causes the five d orbitals in the transition metal ion to split into two sets. The two sets of orbitals are described as non-degenerate orbitals as they are at slightly different energy levels (see Figure 24.17). In a complex with six ligands, the ligands are arranged in an octahedral shape around the central metal ion. The lone pairs donated by the ligands into the transition metal ion repel electrons in the two and orbitals shown in Figure 24.17 more than those in the other three d orbitals. This happens because these two d orbitals line up with the dative (co-ordinate) bonds in the complex’s octahedral shape. As the electrons in the and orbitals are closer to the bonding electrons in the octahedral arrangement, the repulsion between electrons increases. Therefore, the d orbitals are split, with these two d orbitals at a slightly higher energy level than the dyz, dxz and dxy orbitals. Look at the left-hand side of Figure 24.17. Figure 24.17: The splitting of the 3d orbitals in the octahedral [Cu(H2O)6]2+ complex ion. A Cu2+ ion has an electronic configuration of [Ar] 3d9. Figure 24.17 shows how the nine d electrons are distributed between the non-degenerate orbitals formed in a complex with ligands. The difference in the energy between the non-degenerate d orbitals is labelled ΔE. ΔE corresponds to part of the visible spectrum of light. So, when light shines on the solution containing the [Cu(H2O)6]2+ complex, an electron absorbs this amount of energy. It uses this energy to jump into the higher of the two non-degenerate energy levels. In copper complexes, the rest of the visible spectrum that passes through the solution makes it appear blue in colour. The exact energy difference (ΔE) between the non-degenerate d orbitals in a transition metal ion is affected by many factors. One of these factors is the identity of the ligands that surround the transition metal ion. As you have seen, a solution containing [Cu(H2O)6]2+ is a light blue, whereas a solution containing [Cu(NH3)4(H2O)2]2+ is a very deep shade of blue. The colour change arises because the presence of the ammonia ligands causes the d orbitals to split by a different amount of energy. This means that the size of ΔE changes and this results in a slightly different amount of energy being absorbed by electrons jumping up to the higher orbitals. Therefore, a different colour is absorbed from visible light, so a different colour is seen. In tetrahedral complexes, such as [CoCl4]2−, the splitting of the d orbitals is different (Figure 24.18). The bonding pairs of electrons from four ligands line up with the dyz, dxz and dxy orbitals of the transition metal ion. Unlike the octahedral arrangement, the and orbitals in a tetrahedral complex lie between the metal–ligand bonds. So there is less repulsion between electrons in or orbitals and the lone pairs of bonding electrons donated by the ligands. Therefore, when the d orbitals split in this case, the or are at a lower, more stable energy level than the dyz, dxz and dxy orbitals. Figure 24.18: The splitting of the 3d orbitals in the tetrahedral [CoCl4]2− complex ion. Question