Pra-U STPM Chemistry Penggal 2 2019 CB039349b

Chemistry Term 2 STPM Chapter 12 Group 17 193 12 12.2 Reactions of Selected Group 17 Elements Bond Energy 1 Bond energy is defined as the energy required to break a particular covalent bond per mole of the bond. X⎯X(g) ⎯→ 2X(g) ΔH° = Bond energy 2 The bond energy of the halogens is given in the table below: Bond F⎯F Cl⎯Cl Br⎯Br I⎯I Bond energy/kJ mol–1 +158 +242 +193 +151 3 Generally, the bond energy decreases down the group. This is as expected because the size of the atom increases from fluorine to iodine. As a result, the covalent bond gets longer and weaker. 10 20 30 Proton number 40 50 Bond energy/kJ mol–1 200 180 160 140 260 240 220 60 F–F Cl–Cl Br–Br I–I 4 However, the bond energy of fluorine (F⎯F) is lower than expected even though it has the smallest atomic size. F⎯F(g) ⎯→ 2F(g) ΔH° = +158 kJ mol–1 5 The low bond energy of F2 is due to the closeness of the atoms in the molecules which resulted in considerable repulsion between the non-bonding electrons and thus weakening the bond. • • • • • • • F F Repulsion Repulsion fifi fi fi fi fi fi Exam Tips Bond energy – All the species involved are in the gaseous state.

Chemistry Term 2 STPM Chapter 12 Group 17 194 12 As Oxidising Agents 1 The halogens are the most reactive group of elements in the Periodic Table due to their low bond dissociation energy and high electronegativity. 2 In most reactions, the halogens act as electron acceptor in order to achieve octet configurations. X(g) + e– X– (aq) 3 They are all powerful oxidising agents as indicated by their large positive standard electrode potentials. F2(g) + 2e– 2F– (aq) E° = +2.87 V Cl2(g) + 2e– 2Cl– (aq) E° = +1.36 V Br2(aq) + 2e– 2Br– (aq) E° = +1.09 V I2(s) + 2e– 2I– (aq) E° = +0.54 V 4 As can be seen from the above E° values, the oxidising strength of the halogens decreases down the group. 5 The atomic size increases down the group causing a decrease in the tendency to accept electrons. As a result, the oxidising strength decreases. 6 Chlorine can oxidise bromide and iodide to bromine and iodine. Cl2(g) + 2Br– (aq) ⎯→ 2Cl– (aq) + Br2(aq) Cl2(g) + 2I– (aq) ⎯→ 2Cl– (aq) + I2(aq) 7 Bromine can only oxidise iodide to iodine. Br2(aq) + 2I– (aq) ⎯→ 2Br– (aq) + I2(aq) 8 On the other hand, iodine is unable to oxidise chloride or bromide to chlorine and iodine respectively. Example 12.2 Iodine reacts with aqueous sodium thiosulphate, Na2S2O3 to sodium tetrathionate, Na2S4O6. On the other hand, chlorine reacts with sodium thiosulphate to form sodium sulphate, Na2SO4. Explain the different behaviours of the two halogens. Solution The average oxidation state of sulphur in: S2O3 2– is +2 S4O6 2– is +2.5 SO4 2– is +6 Iodine, a weak oxidising agent, can oxidise sulphur from +2 to +2.5 whereas, chlorine, a stronger oxidising agent, is able to oxidise sulphur from +2 to +6. 2015/P2/Q20(b) 2011/P2/Q7(b)

Chemistry Term 2 STPM Chapter 12 Group 17 195 12 Example 12.3 When fluorine gas is passed through water at room conditions, a vigorous reaction occurs and a gas that supports combustion is liberated. Use appropriate data from the Data Booklet to explain the observations. Solution Considering the following standard electrode potentials: F2(g) + 2e– 2F– (aq) E° = +2.87 V O2(g) + 4H+(aq) + 4e– 2H2O(l) E° = +1.23 V From the E° values, fluorine is a stronger oxidising agent than oxygen. When fluorine is passed through water, it oxidises water to oxygen (a gas that supports combustion). 2F2(g) + 2H2O(l) ⎯→ 4HF(aq) + O2(g) E° = +1.64 V Reaction with Hydrogen 1 Under suitable conditions, halogens react with hydrogen to form hydrogen halides. X2(g) + H2(g) ⎯→ 2HX(g) 2 The above is a redox reaction where hydrogen is oxidised from the oxidation state of ‘0’ to ‘+1’. 3 As the oxidising strength of the halogens decreases down the group, the reactivity towards hydrogen decreases from chlorine to iodine. 4 Chlorine reacts explosively with hydrogen when exposed to bright sunlight to form hydrogen chloride. Cl2(g) + H2(g) ⎯→ 2HCl(g) The reaction is slower in the dark. 5 Bromine vapour reacts with hydrogen when heated to about 500 °C in the presence of a catalyst to form hydrogen bromide. Br2(g) + H2(g) ⎯→ 2HBr(g) 6 Iodine vapour reacts reversibly with hydrogen at 450 °C in the presence of nickel or platinum catalyst to produce hydrogen iodide. I2(g) + H2(g) 2HI(g) Example 12.4 How would you expect fluorine to react with hydrogen? Solution Due to its high oxidising power, fluorine will react explosively with hydrogen at room temperature and in the dark to form hydrogen fluoride. H2(g) + F2(g) ⎯→ 2HF(g) The E° value for F2 is more positive than that for O2. 2013/P2/Q20(a) Element Reaction with H2 F2 Explosive in the dark Cl2 Explosive in bright sunlight Br2 • Heat • Catalyst I2 • Heat • Catalyst • Reversible 2015/P2/Q12 2017/P2/Q19(a)

Chemistry Term 2 STPM Chapter 12 Group 17 196 12 Hydrogen Halides Boiling Point 1 The hydrogen halides exist as simple HX molecules. 2 They are all colourless gases which fume in moist air. 3 Going from HCl to HI, the size of the molecule and the total number of electrons in the molecule increase. This causes the intermolecular van der Waals forces to get stronger as reflected in their boiling points. Hydrogen halide HF HCl HBr HI Boiling point/°C 20 –85 –67 –35 10 20 30 Molar mass/g 40 50 Boiling point/°C –40 –60 –80 –100 20 0 –20 60 HF HCl HBr HI Example 12.5 Despite its small size, the boiling point of hydrogen fluoride is exceptionally high (20 °C). Explain why is it so. Solution The high boiling point of hydrogen fluoride is due to the presence of strong intermolecular hydrogen bonds between the molecules. HF HF Hydrogen Bond H X • fi 2014/P2/Q13 2018/P2/Q17

Chemistry Term 2 STPM Chapter 12 Group 17 197 12 Thermal Stability 1 All the hydrogen halides decompose to their elements on heating. 2HX(g) ⎯→ H2(g) + X2(g) 2 During decomposition, the covalent bond between hydrogen and the halogen atom breaks. 2H⎯⎯X(g) ⎯→H2(g) + X2(g) Δ 3 Therefore, the thermal stability of the hydrogen halides depends on the strength of the H⎯X bond. Bond H⎯F H⎯Cl H⎯Br H⎯I Bond energy/kJ mol–1 +562 +430 +370 +300 4 Going down the group, the H⎯X bond gets longer and weaker. As a result, thermal stability decreases down the group as shown by the degree of dissociation at various temperatures. Temperature/°C Degree of dissociation / % HCl HBr Hl 600 – 0.04 20 1000 0.01 0.40 25 2000 0.40 4.20 30 5 For example, when a heated platinum wire is introduced into a jar of hydrogen iodide gas, a purple fume is observed. The purple fume is iodine vapour formed by the decomposition of hydrogen iodide. 2HI(g) ⎯→ H2(g) + I2(g) Δ 2009/P1/Q25 Quick Check 12.2 1 The reactivity of the halogens with hydrogen decreases with increasing proton number of the halogen. Explain. 2 When aqueous sodium thiosulphate is added to a solution of iodine in aqueous potassium iodide, the solution changes from orange to colourless. Use appropriate equations to explain the reaction involved. Reaction of Chlorine with Sodium Hydroxide 1 When chlorine gas is bubbled through dilute sodium hydroxide, sodium chloride and sodium chlorate(I) are produced. Cl2(g) + 2NaOH(aq) ⎯→ NaCl(aq) + NaOCl(aq) + H2O(l) Oxidation state: 0 –1 +1 2 This is an example of disproportionation where chlorine is reduced to chloride and at the same time is oxidised to chlorate(I). 3 Sodium chlorate is used as a domestic bleach. 4 When chlorine is bubbled through concentrated sodium hydroxide at 70 °C, sodium chlorate(V) is formed instead. Info Chem Hydrogen fluoride is stable to heat. Info Chem When NaOCl is warmed, it disproportionates to sodium chloride and sodium chlorate(V): 3NaOCl ⎯→∆ NaClO3 + 2NaCl 2011/P2/Q7(c) 2012/P2/Q3(a) 2013/P2/Q14 2016/P2/Q13

Chemistry Term 2 STPM Chapter 12 Group 17 198 12 3Cl2(g) + 6NaOH ⎯→ 5NaCl(aq) + NaClO3(aq) + 3H2O(l) Oxidation state: 0 Δ –1 +5 5 This is another example of disproportionation. Chlorine is reduced to chloride and at the same time is oxidised to chlorate(V). 6 Sodium chlorate(V) is used in explosive and as a weed-killer. Example 12.4 Draw the Lewis diagram for the chlorate(I) and chlorate(V) ion. Predict the shape of the chlorate(V) ion. Solution • • • • • • Cl O– Cl O– O O • • fifi fifi fi fi fi fi fi fi fi fi The central chlorine atom in chlorate(V) is surrounded by 3 bond pairs (2 single bonds and 1 double bond) and 1 lone pair. It is pyramidal. Example 12.7 When potassium chlorate(V) is heated to a temperature just above its melting point (370 °C), it is converted into a mixture of potassium chloride and potassium chlorate(VII). However, when potassium chlorate is heated together with manganese dioxide, it decomposes at temperature below its melting point to oxygen and potassium chloride. (a) Write equations to represent the decomposition of potassium chlorate(V) under the two different conditions. (b) Which of the two reactions is disproportionation? Explain your answer. Solution (a) 4KClO3(s) ⎯→ KCl(s) + 3KClO4 Δ 2KClO3(s) ⎯→ 2KCl(s) + 3O2(g) Δ (b) The first reaction is disproportionation because chlorine is reduced to chloride and at the same time is oxidised to chlorate(VII). Cl O– O RO R RR !!

Chemistry Term 2 STPM Chapter 12 Group 17 199 12 Quick Check 12.3 When an aqueous solution containing the chlorate(I) ion is heated, it decomposes to the chlorate(V) and chloride ions. Write a balanced equation for the decomposition and name the type of reaction taking place. 12.3 Reactions of Selected Halide Ions Reactions of Halide Ions with Silver Nitrate 1 When aqueous silver nitrate is added to an aqueous solution containing halide ions, a coloured precipitate is formed. Ag+(aq) + X– (aq) ⎯→ AgX(s) 2 The colour of the precipitate depends on the nature of the halide ions: Ag+(aq) + Cl– (aq) ⎯→ AgCl(s) White Ag+(aq) + Br– (aq) ⎯→ AgBr(s) Cream Ag+(aq) + I– (aq) ⎯→ AgI(s) Yellow 3 Silver chloride dissolves in both dilute and concentrated ammonia to produce a colourless solution by forming the water-soluble complex ion, [Ag(NH3)2]+. AgCl(s) + 2NH3(aq) ⎯→ [Ag(NH3)2]+(aq) + Cl– (aq) Diamminesilver 4 Silver bromide is insoluble in dilute ammonia but dissolves in concentrated ammonia to form a colourless solution. AgBr(s) + 2NH3(aq) ⎯→ [Ag(NH3)2]+(aq) + Br– (aq) 5 Silver iodide is insoluble in ammonia, dilute or concentrated. 6 Summary: Silver halide AgCl AgBr AgI Colour White Cream Yellow Solubility in dilute ammonia Soluble Insoluble Insoluble Solubility in concentrated ammonia Soluble Soluble Insoluble Reactions of Halides with Concentrated Sulphuric Acid 1 When a solid halide is heated with concentrated sulphuric acid, hydrogen halide is formed. X– (s) + H2SO4(aq) ⎯→ HX(g) + HSO4 – (aq) Δ 2007/P1/Q25 2007/P1/Q25 2008/P1/Q47 2010/P1/Q25 2015/P2/Q13 2010/P2/Q2(a) 2018/P2/Q13 2013/P2/Q20(c) 2017/P2/Q19(b)

Chemistry Term 2 STPM Chapter 12 Group 17 200 12 2 For example, when solid sodium chloride is heated with concentrated sulphuric acid, white fumes of hydrogen chloride is liberated. NaCl(s) + H2SO4(aq) ⎯→ NaHSO4(aq) + HCl(g) Δ 3 When solid potassium bromide is heated with concentrated sulphuric acid, initially a white fume of hydrogen bromide is liberated. KBr(s) + H2SO4(aq) ⎯→ KHSO4(aq) + HBr(g) Δ 4 However, concentrated sulphuric acid is also an oxidising agent. On prolong heating, the hydrogen bromide released is further oxidised to bromine, a reddish brown fume. 2HBr(g) + H2SO4(aq) ⎯→ Br2(g) + SO2(g) + 2H2O(g) Δ 5 Similarly, heating solid potassium iodide with concentrated sulphuric acid produces hydrogen iodide (white fumes) and iodine (violet fumes). KI(s) + H2SO4(aq) ⎯→ KHSO4(aq) + HI(g) Δ 2HI(g) + H2SO4(aq) ⎯→ I2(g) + SO2(g) + 2H2O(g) Δ Solid halide NaCl KBr KI Reaction with concentrated H2SO4 White fumes White fumes + reddish brown fumes White fumes + violet fumes 6 As a result, concentrated sulphuric acid is not a suitable agent for the preparation of hydrogen bromide and hydrogen iodide. Instead, concentrated phosphoric acid, H3P4, which is non-oxidising is used instead. KBr(s) + H3PO4(aq) ⎯→ HBr(g) + KH2PO4(aq) Δ KI(s) + H3PO4(aq) ⎯→ HI(g) + KH2PO4(aq) Δ Example 12.8 Explain why concentrated sulphuric acid is unable to oxidise hydrogen chloride to chlorine. Solution Chlorine is a stronger oxidising agent than sulphuric acid. Thus, the following reaction cannot take place. 2HCl(g) + H2SO4(aq) ⎯→ Cl2(g) + SO2(g) + 2H2O(g) Δ Because a weak oxidising agent cannot displace a stronger oxidising agent.

Chemistry Term 2 STPM Chapter 12 Group 17 201 12 12.4 Industrial Applications of Halogens and Their Compounds Uses of Chlorine 1 Chlorine is used in the purification of water (e.g. in swimming pools and in piped water). 2 Chlorine is used as a bleaching agent in the paper industry. 3 Sodium chlorate(I) is used as domestic bleach and as an antiseptic. 4 Chlorinated organic compounds are useful solvents. For example, CCl4 (carbon tetrachloride) and CHCl3 (chloroform). 5 Trichloroethane is used as a dry-cleaning agent. 6 Freons such as CCl2F2 and CFCl3 are used as cleaning agents, refrigerants and as propellant in aerosol cans. 7 Dichlorodiphenyltrichloroethane, DDT is a powerful insecticide. 8 Potassium chlorate(V) is used as a weed-killer as well as in explosives. 9 Chloroethene, CH2==CHCl is a monomer for PVC, a plastic-like substance used mainly to make pipes and as substitute for leather. Uses of Bromine 1 Bromine is used in the manufacture of dyes and drugs. 2 1,2-dibromoethane, BrCH2CH2Br is used as a petrol additive to remove lead compounds from internal combustion engines. 3 Silver bromide is used in the manufacture of photographic films (AgBr is more sensitive to light than AgCl). Uses of Iodine 1 Iodine is used to make dyes and in colour photography. 2 Iodoform (triiodomethane) is an antiseptic. 3 Silver iodide is used in cloud seeding, where it provides the ‘nucleus’ for the condensation of water vapour in the cloud. 4 An aqueous solution of iodine in ethanol is used as an antiseptic. Black-and-white Photography 1 Silver halides are sensitive to light. When exposed to light, they turn dark as the silver halides decompose to form metallic silver and the halogen. AgCl(s) ⎯→ Ag(s) + 1 – 2Cl2(g) AgBr(s) ⎯→ Ag(s) + 1 – 2 Br2(l) AgI(s) ⎯→ Ag(s) + 1 – 2 I2(s) Info Chem The structure of DDT: Cl CH Cl CCl3 Info Chem Radioactive iodine (131I) is used to treat goitre (hyperthyroidism) Chlorine dissolves in water to produce chloric(I) acid. Cl2 + H2O → HOCl + HCl Chloric(I) acid has the ability to kill germs and bacteria. INFO Uses of Halogens 2017/P2/Q19(a) 2018/P2/Q14

Chemistry Term 2 STPM Chapter 12 Group 17 202 12 SUMMARY SUMMARY 2 The photosensitive nature of the silver halides forms the basis of black-and-white photography. 3 In black-and-white photography, silver bromide is preferred because it is more sensitive to light. 4 Black-and-white photographic film is a clear cellulose strip coated with grains of silver bromide. 5 When light strikes the film, the silver bromide that is exposed to the light gets activated. sunlight AgBr(s) ⎯⎯⎯→ AgBr*(s) 6 The exposed film is then treated with aqueous hydroquinone, C6H2O2 (a reducing agent), which reduces the activated silver bromide to silver. 2AgBr* (s) + C6H6O2(aq) ⎯→ 2Ag(s) + 2HBr(aq) + C6H4O2(aq) Hydroquinone Quinone 7 The film is then immersed in aqueous sodium thiosulphate (called ‘hypo’) where unactivated silver bromide is removed from the film. AgBr(s) + 2S2O3 2–(aq) ⎯→ Ag(S2O3)2 3–(aq) + Br– (aq) 8 Hence, the area on the film that was struck by light would appear black (due to the deposition of metallic silver). The film now is called a negative. 9 When light passes through the negative onto a photosensitive paper, the dark area would appear white, while the clear area would appear black. A print of the photograph is obtained. 2017/P2/Q13 OH OH Hydroquinone O O Quinone + 2H+ + 2e– Quick Check 12.4 Photochromatic lens turns dark when exposed to bright sunlight, but turns clear again when indoor. The working of the lens follows the same principle as black-and-white photography. Explain the action of the lens. 1 The Group 17 elements are called the halogens. 2 The halogens exist as diatomic molecules under room conditions. 3 Going down Group 17, • the atomic radius increases • the melting point and boiling point increases • the electronegativity decreases • electron affinity decreases(less exothermic) • the bond length increases and the bond energy decreases • the oxidising power decreases 4 The reaction of the halogens towards hydrogen get less vigorous down the group. 5 The hydrogen halides are colourless gas that fume in moist air. INFO Black-and-White Photography

Chemistry Term 2 STPM Chapter 12 Group 17 203 12 1 An aqueous solution Y has the following properties. • Decomposes in the presence of sunlight to evolve a gas that supports combustion • Reacts with aqueous silver nitrate to form a precipitate that is insoluble in ammonia Y could be a mixture of A HCl and HBr B HBr and HOBr C HI and HOCl D NH4Cl and NH3 2 A halide ion, X– , reacts with aqueous silver nitrate to produce a coloured precipitate which is insoluble in dilute ammonia, but dissolves in concentrated ammonia. What is the colour of the precipitate? A Greenish yellow C Pale yellow B Purple D Reddish brown 3 Fluorine reacts explosively with hydrogen in bright sunlight. Iodine reacts reversibly with hydrogen at 450°C and in the presence of platinum. What factor explains this observation? A Fluorine is a more powerful oxidising agent than iodine. B Fluorine is less stable than iodine. C HF is more stable than HI. D Fluorine has higher bond energy. 4 The compound obtained when chlorine is bubbled through hot, concentrated sodium hydroxide can be used A as a bleach B to make photographic papers C as a weed-killer D to manufacture hydrochloric acid 6 Going down the group, • the boiling point of the hydrogen halides increases due to the increase in the van der Waals force • thermal stability decreases • the bond energy decreases 7 Hot concentrated sulphuric acid reacts with solid halide salts as follows: • With chlorides to form hydrogen chloride • With bromides to form hydrogen bromide and bromine (reddish brown fumes) • With iodide to form hydrogen iodide and iodine (violet fumes) 8 All aqueous halides form precipitates with aqueous silver nitrate. • With chloride to form AgCl (white) that is soluble in both dilute and concentrated ammonia • With bromide to form AgBr (cream) that is insoluble in dilute ammonia but dissolves in concentrated ammonia • With iodide to form AgI (yellow) that is insoluble in either dilute or concentrated ammonia 9 Chlorine reacts with cold, dilute aqueous sodium hydroxide to form sodium chlorate(I) which is used as a bleach. 10 Chlorine reacts with hot, concentrated aqueous sodium hydroxide to form sodium chlorate(V) which is used as a weed-killer and also in explosives. Objective Questions STPM PRACTICE 12

Chemistry Term 2 STPM Chapter 12 Group 17 204 12 5 Which of the following is not the use of the halogens and their compounds? A Production of TNT B Manufacture of insecticide C Sterilising drinking water D Refrigerants 6 Astatine is the last member of the Group 17 elements. Which of the following statements is not true about astatine and its compounds? A Astatine has the highest melting point among the Group 17 elements. B Astatine is the most electronegative elements among the Group 17 elements. C Silver astatide is insoluble in concentrated ammonia. D Astatine is produced when solid potassium astatide is warmed with concentrated sulphuric acid. 7 All of the following are redox products of the reaction between potassium iodide and concentrated sulphuric acid except A HSO4 – B SO2 C I2 D I3 – 8 Chorine is bubbled into an aqueous solution of sodium hydroxide at 75°C. Which statement is true about the reaction? A Chlorine acts as an oxidising agent. B The reaction is disproportionation. C Potassium chlorate(VII) is one of the products. D Sodium reduces chlorine to chloride ion. 9 Which of the following is true when descending Group 17 (from Cl to I)? A Volatility decreases B Oxidising power increases C The bond enthalpy increases D Reactivity towards hydrogen increases 10 Silver bromide is widely used in black-and-white photography because A it is the cheapest B it is more sensitive to light C it is chemically more inert D it is a stronger reducing agent 11 Which statement about the halogens (Cl2, Br2 and I2) is true? A The oxidising power increases in the order: Cl2 < Br2 < I2. B The reactivity of the elements with hydrogen increases in the order: I2 < Br2 < Cl2. C All the halogens, except Cl2, can exhibit variable oxidation states in their compounds. D All the halogens can oxidise sodium thiosulphate (Na2S2O3) to sodium sulphate (Na2SO4). 12 Chlorine reacts with aqueous sodium hydroxide according to the equation: 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O Which statement about the reaction is not true? A Chlorine undergoes disproportionation. B The reaction requires heat. C The oxidation state of chlorine changes from 0 to –1 and +5. D Chlorine acts as an oxidising agent. 13 When solid potassium iodide is heated with concentrated sulphuric acid, the gaseous product(s) include(s) I HI II SO2 III I2 A I only B II only C I and III D I, II and III 14 Astatine, At, is situated below iodine in the Periodic Table. Which statement(s) about astatine is/are correct? I Hydroastatic acid is a strong acid. II Astatine reacts vigorously with hydrogen in the dark. III Astatine is a weaker oxidising agent than iodine. IV Silver astatide is soluble in concentrated ammonia. A I and III B II and IV C I, II and IV D I, II, III and IV

Chemistry Term 2 STPM Chapter 12 Group 17 205 12 15 Which of the following statement about the Group 17 elements and their compounds is not correct? A The boiling point of the hydrides increases in the order: HCl , HBr , HI , HF B The reducing power of the halide ions increases C The elements show a constant oxidation state of –1 in their compounds D The H–Cl bond is more polar than the H–I bond 16 Which of the following compound is coloured? A ScCl3 B CuO C ZnSO4 D TiCl4 Structured and Essay Questions 1 (a) Describe the reactions of the halogens (Cl to I) with hydrogen. Offer an explanation for the trend of reactivity of the halogens towards hydrogen. (b) An aqueous solution of NaX gives a pale yellow precipitate with aqueous silver nitrate. The precipitate is insoluble in dilute ammonia but dissolves in concentrated ammonia. (i) Identify X. (ii) Write an equation to show the reaction between the pale yellow precipitate and ammonia. (c) Iodine is sparingly soluble in water but dissolves in aqueous potassium iodide. (i) Explain the low solubility of iodine in water. (ii) Explain why iodine dissolves in aqueous KI. Illustrate your answer with a balanced equation. 2 (a) Explain the variation in the physical states of halogens (F to I). (b) Halogens react with hydrogen under suitable conditions to produce hydrogen halides. (i) Describe and explain the reactivity of halogens with hydrogen. (ii) Explain the trend of the boiling points of hydrogen halides. (c) In acidic solution, chlorate(I) ion oxidises iodide to iodine. 0.50 g of a sample bleaching powder, Ca(OCl)2, is added to excess acidified potassium iodide solution. The iodine released requires 18.50 cm3 of 0.10 mol dm-3 sodium thiosulpahte, Na2S2O3 for complete reaction. (i) Name a suitable indicator and state the colour change at the end-point. (ii) Calculate the percentage purity of the sample. 3 Explain the following observations. (a) Silver bromide rather than silver chloride is used to make black-and-white photographic films. (b) Concentrated sulphuric acid is not a suitable reagent to produce hydrogen iodide from solid potassium iodide. (c) The smell of chlorine in water can be removed by the addition of aqueous sodium hydroxide. (d) The boiling point of hydrogen iodide is higher than that of hydrogen chloride, but hydrogen iodide is thermally less stable than hydrogen chloride. 4 Explain the following observations. (a) On standing, an aqueous solution of sodium chlorate(I) gives off a gas which supports combustion. (b) When a mixture containing manganese(IV) oxide and concentrated hydrochloric acid is warmed, a gas that bleaches moist blue litmus paper is evolved.

Chemistry Term 2 STPM Chapter 12 Group 17 206 12 5 The halogens (from F to I) react with hydrogen, under suitable conditions, to produce the corresponding hydrogen halides. (a) Write a balanced equation for the reaction between hydrogen and chlorine to produce hydrogen chloride. (b) Sketch a graph to show the variation of the boiling points of the hydrogen halides with the molar mass of the halides. Explain the shape of the curve. (c) Discuss the thermal stability of the hydrogen halides. 6 Chlorine gas is sparingly soluble in water to produce ‘chlorine water’, which contains hydrochloric acid and hypochlorous acid, HOCl. (a) Explain why chlorine is only sparingly soluble in water. Write a balanced equation for the reaction taking place. (b) Name and explain the type of reaction occurring in (a). (c) When ‘chlorine water’ is exposed to bright sunlight, a colourless gas is given off. (i) Name the colourless gas. (ii) Write an equation for its production. (d) State two compounds of chlorine and the use of each of the compounds. 7 (a) Use appropriate data from the Data Booklet, explain the relative strength of the halogens (from Cl2 to I2) as oxidising agents. (b) Draw a labelled diagram of the diaphragm cell in the manufacture of chlorine from brine. Write equations for the reactions taking place at the anode and cathode. (c) Chlorine reacts with aqueous sodium hydroxide, under different conditions, to give different products. Write the equation for the reaction of chlorine with aqueous sodium hydroxide under the following conditions: (i) Room temperature (ii) At 70 °C 8 (a) How do the chloride, bromide and iodide ions differ in their reactions with hot, concentrated sulphuric acid? Explain any differences in the reactions. (b) Chlorine reacts with aqueous sodium hydroxide to produce either sodium chlorate(I) or sodium chlorate(V), depending on the conditions of the reaction. (i) State the condition for the production of sodium chlorate(I), and write an equation for the reaction. (ii) State the condition for the production of sodium chlorate(V), and write an equation for the reaction. (iii) State one use each for sodium chlorate(I) and sodium chlorate(V). (c) Chlorine, bromine and iodine are sparingly soluble in water. However, iodine dissolves completely in aqueous potassium iodide. (i) Explain the low solubility of the halogen in water. (ii) Why is iodine completely soluble in aqueous potassium iodide? Write an equation for any reaction taking place.

CHAPTER 13 TRANSITION ELEMENTS Concept Map Transition Elements Physical Properties • Atomic radius • Melting/Boiling points • Density • Ionisation energy Nomenclature and Bonding of Complexes • Naming of complexes • Bonding in complexes • Classification of ligands • Geometry of complexes Uses of Transition Elements and Their Compounds Formation of Complex Ions and Ligand Exchange Chemical Properties • Formation of coloured compounds • Variable oxidation states • Standard reduction potentials and the relative stability of aqueous ions • Redox properties of various oxidation states • Homogeneous catalysis • Heterogeneous catalysis Learning earning Outcomes Students should be able to: Physical properties of first row transition elements • define a transition element in terms of incomplete d orbitals in at least one of its ions; • describe the similarities in physical properties such as atomic radius, ionic radius and first ionisation energy; • explain the variation in successive ionisation energies; • contrast qualitatively the melting point, density, atomic radius, ionic radius, first ionisation energy and conductivity of the first row transition elements with those of calcium as a typical s-block element. Chemical properties of first row transition elements • explain variable oxidation states in terms of the energies of 3d and 4s orbitals; • explain the colours of transition metal ions in terms of a partially filled 3d orbitals; • state the principal oxidation numbers of these elements in their common cations, oxides and oxo ions; • explain qualitatively the relative stabilities of these oxidation states; • explain the uses of standard reduction potentials in predicting the relative stabilities of aqueous ions; • explain the terms complex ion and ligand; • explain the formation of complex ions and the colour changes by exchange of ligands. (Examples of ligands: water, ammonia, cyanide ions, thiocyanate ions, ethanedioate ions, ethylenediaminetetraethanoate, halide ions; examples of complex ions: [Fe(CN)6] 4–, [Fe(CN)6] 3–, [Fe(H2O)5(SCN)]2+); • explain the use of first row transition elements in homogeneous catalysis, as exemplifed by Fe2+ or Fe3+ in the reaction between I – and S2O8 2–; • explain the use of first row transition elements in heterogeneous catalysis, as exemplifed by Ni and Pt in the hydrogenation of alkenes. Nomenclature and bonding of complexes • name complexes using International Union of Pure and Applied Chemistry (IUPAC) nomenclature; • discuss coordinate bond formation between ligands and the central metal atom/ion, and state the types of ligands, i.e. monodentate, bidentate and hexadentate. Uses of first row transition elements and their compounds • describe the use of chromium (in stainless steel), cobalt, manganese, titanium (in alloys) and TiO2 (in paints).

Chemistry Term 2 STPM Chapter 13 Transition Elements 208 13 13.1 Physical Properties of First Row Transition Elements 1 The d-block elements are a block of elements slotted in between the electropositive elements of the s-block and those of the electronegative elements of the p-block. The first row of d-block elements consists of ten elements from scandium to zinc. d-block elements Sc Ti V Cr Mn Fe Co Ni Cu Zn 2 The electronic configurations of the d-block elements are shown in the table below: Element Symbol Proton number Electronic configuration Scandium Sc 21 [Ar]3d1 4s2 2.8.9.2 Titanium Ti 22 [Ar]3d2 4s2 2.8.10.2 Vanadium V 23 [Ar]3d3 4s2 2.8.11.2 Chromium Cr 24 [Ar]3d5 4s1 2.8.13.1 Manganese Mn 25 [Ar]3d5 4s2 2.8.13.2 Iron Fe 26 [Ar]3d6 4s2 2.8.14.2 Cobalt Co 27 [Ar]3d7 4s2 2.8.15.2 Nickel Ni 28 [Ar]3d8 4s2 2.8.16.2 Copper Cu 29 [Ar]3d104s1 2.8.18.1 Zinc Zn 30 [Ar]3d104s2 2.8.18.2 3 The electronic configuration of the d-block elements differs from those of the main group elements (such as the Period 3 elements) in that they have incompletely filled inner sub-shell (3d), except for copper and zinc. 4 All the elements have two electrons in the outermost sub-shell except for chromium and copper, which have one electron respectively. This is because a 3d5 4s 1 configuration is more stable than 3d4 4s 2 configuration for chromium. Similarly, for copper, the 3d 10 4s 1 configuration is more stable than the 3d9 4s 2 configuration. 3d 4s Cr: 3d 4s Cu: A fully filled 3d orbitals and a half-filled 3d orbitals have extra stability due to the uniform distribution of electron density in the five 3d orbitals. 2017/P2/Q14

Chemistry Term 2 STPM Chapter 13 Transition Elements 209 13 5 Previously, a transition element is defined as one with an incompletely filled d sub-shells. 6 This definition would include scandium ([Ar]3d1 4s 2 ) but exclude copper ([Ar]3d104s 1 ) and zinc ([Ar]3d104s 2 ). 7 However, scandium and zinc are not transition elements whereas copper is. 8 In view of that, a transition element is now defined as an element that can form at least one stable ion with an incompletely filled d sub-shell. 9 With this new definition, scandium and zinc are not transition elements because scandium forms only one cation, Sc3+, where there are no electrons in the d sub-shell, whereas zinc forms only Zn2+ where the d sub-shells are fully occupied. Sc: [Ar] 3d1 4s 2 Sc3+: [Ar] 3d0 Zn: [Ar] 3d10 4s 2 Zn2+: [Ar] 3d10 10 On the other hand, copper forms two cations, Cu+ and Cu2+, with Cu2+ having an incompletely filled d sub-shells. Cu+ : [Ar] 3d10 Cu2+: [Ar] 3d9 11 Thus, the first row transition element consists of only eight elements from titanium to copper. Atomic Radius 1 The atomic radii of the transition elements are shown in the table below. Element No. of protons No. of inner electrons Radius/nm Titanium 22 20 0.147 Vanadium 23 21 0.135 Chromium 24 23 0.129 Manganese 25 23 0.137 Iron 26 24 0.126 Cobalt 27 25 0.125 Nickel 28 26 0.125 Copper 29 28 0.128 x x x x x x x x Radius/nm 0.160 0.150 0.140 0.130 0.120 22 23 24 25 26 27 28 29 Proton number Exam Tips Exam Tips The atomic radius of Mn is larger than expected due to the greater repulsion between a half-filled 3d orbitals and a fully-filled 4s orbitals. 2008/P2/Q8(a) Info Chem Scandium and zinc do not have the properties of transition elements.

Chemistry Term 2 STPM Chapter 13 Transition Elements 210 13 2 Going from titanium to copper, the number of protons and electrons increases. However, each additional electron is added to an inner 3d sub-shell (except for manganese). 3 The increase in the screening effect (caused by the increasing number of inner electrons) to a certain extent balanced out the increase in the nuclear charge. As a result, the effective nuclear charge increases only slightly causing little change in the atomic radius of the atoms. Melting and Boiling Points 1 The table and graph below show the variation of melting and boiling points of the transition elements. Element Melting point/ °C Boiling point/ °C Titanium 1680 3260 Vanadium 1900 3400 Chromium 1890 2480 Manganese 1240 2100 Iron 1540 3000 Cobalt 1500 2900 Nickel 1450 2730 Copper 1080 2600 4000 3000 2000 1000 0 Ti V Cr Mn Fe Co Ni Cu Boiling point Melting point Temperature/o C 2 The melting points of the transition elements are all above 1000 °C, while the boiling points are in excess of 2000 °C. This indicates the presence of strong metallic bonds. 3 Due to the small difference in energy between the 3d and 4s sub-shells, transition elements can make use of electrons from both these sub-shells to participate in metallic bond formation. 4 The melting and boiling points of chromium and manganese are lower than expected because electrons in the 3d5 configuration are less available to take part in the formation of metallic bond. 2015/P2/Q14

Chemistry Term 2 STPM Chapter 13 Transition Elements 211 13 Density 1 The table below shows the variation of the density of the transition elements. Element Density/g cm–3 Titanium 4.43 Vanadium 6.10 Chromium 7.19 Manganese 7.21 Iron 7.90 Cobalt 8.70 Nickel 8.90 Copper 8.92 2 Due to the small size of the atoms and the strong metallic bonds, which cause the atoms to be tightly packed, the transition elements have high densities (compared to aluminium = 2.70 g cm–3). 3 Going across the d-block, the relative atomic mass of the elements increases while the atomic size does not change much. This causes a gradual increase in the mass volume ratio and thus the density. X X X X X X X X 10 9 8 7 6 5 4 3 22 23 24 25 26 27 28 29 30 Proton number Density/g cm–3 First Ionisation Energy 1 The table below shows the variation of the first ionisation energy (1st I.E.) of the first row transition elements. Element Proton number Electronic configuration 1st I.E./kJ mol–1 Titanium 22 [Ar]3d 2 4s2 661 Vanadium 23 [Ar]3d 3 4s2 648 Chromium 24 [Ar]3d 5 4s1 653 Manganese 25 [Ar]3d 5 4s2 716 Iron 26 [Ar]3d 6 4s2 762 Cobalt 27 [Ar]3d 7 4s2 757 Nickel 28 [Ar]3d 8 4s2 736 Copper 29 [Ar]3d 10 4s1 745

Chemistry Term 2 STPM Chapter 13 Transition Elements 212 13 2 When compared to the Period 3 elements, there is little change in the first ionisation energy of the first row transition elements. 3 This is because the first electron removed from each atom is from the outer 4s orbital which is shielded from the nucleus by increasing screening effect as electrons are added to the inner 3d sub-shell. This to a certain extent cancels out the increase in the nuclear charge. Successive Ionisation Energies 1 The variation of the first, second, third and fourth ionisation energies of the transition elements are shown in the table and the graph below. Element 1st I.E./kJ mol–1 2nd I.E./kJ mol–1 3rd I.E./kJ mol–1 4th I.E./kJ mol–1 Ti 661 1310 2720 4170 V 648 1370 2870 4600 Cr 653 1590 2990 4770 Mn 716 1510 3250 5190 Fe 762 1560 2960 5400 Co 757 1640 3230 5100 Ni 736 1750 3390 5400 Cu 745 1960 3550 5690 2 The first and second ionisation energies show little change across the period. This is because the electrons are removed from the outer 4s orbitals which are shielded from the nucleus by increasing screening effect. 3 The second ionisation energies are higher than the corresponding first ionisation energies. This is because, it is more difficult to remove an electron from a cation compared to a neutral atom. Element M M+ Ionisation energy/kJ mol–1 0 1000 2000 3000 4000 5000 6000 7000 Ti V Cr Mn Fe Co Ni Cu M+ M2+ M2+ M3+ M3+ M4+

Chemistry Term 2 STPM Chapter 13 Transition Elements 213 13 Ti(g) ⎯→ Ti+ (g) + e– ΔH = +661 kJ mol–1 Ti+ (g) ⎯→ Ti2+(g) + e– ΔH = +1310 kJ mol–1 4 The second ionisation energies of chromium and copper are higher than expected. This is because, the second electron removed from both these two elements is from an inner 3d sub-shell, which experience greater pull by the nucleus. Cr(g) ⎯→ Cr+ (g) + e– [Ar]3d5 4s 1 [Ar]3d5 Cr+ (g) ⎯→ Cr2+(g) + e– [Ar]3d5 [Ar]3d4 Cu(g) ⎯→ Cu+ (g) + e– [Ar]3d104s 1 [Ar]3d10 Cu+ (g) ⎯→ Cu2+(g) + e– [Ar]3d10 [Ar]3d9 5 The third ionisation energy of manganese is higher than expected. This is because the third electron is removed from a stable configuration of 3d5 . Extra energy is required. Mn2+ Mn3+ 6 On the other hand, the third ionisation energy of iron is lower than expected. This is because removal of one electron from Fe2+ produces a more stable Fe3+ ion with a configuration of 3d 5 . Fe2+ Fe3+ 7 The fourth ionisation energy of iron is higher than expected. This is because the fourth electron is removed from a stable 3d 5 configuration. Fe3+ Fe4+ Comparison between the Physical Properties of the Transition Elements with that of Calcium 1 Calcium (20Ca) is a typical s-block element. Like the transition elements, calcium is a metal. 2 In this section, we shall compare qualitatively the atomic radius, ionic radius, density, melting point, conductivity and first ionisation energy of the first row transition elements with those of calcium. Exam Tips Exam Tips The 3d5 and 3d10 configurations have extra stability. More energy is required to remove an electron from these two configurations. 2009/P2/Q1(b)(ii) Exam Tips Exam Tips The two electrons occupying the same orbital in Fe2+ experience electrostatic repulsion making it easier to be removed. 2017/P2/Q20(d)

Chemistry Term 2 STPM Chapter 13 Transition Elements 214 13 Atomic Radius 1 The table below lists the atomic radius of titanium and calcium. Element No. of protons Electronic configuration Atomic radius/nm Calcium 20 2.8.8.2 or [Ne]4s2 0.197 Titanium 22 2.8.10.2 or [Ne]3d2 4s2 0.147 2 Both calcium and titanium have four electronic shells that are filled with electrons. Both have two electrons in their outermost shells. 3 Going from calcium to titanium, the number of protons as well as the number of inner electrons increases. Thus, we would expect little change in the effective nuclear charge and also in the atomic radius. 4 However, this is not the case. This is because the additional electrons for titanium are added to the inner 3d sub-shell. The screening effect of d electrons is not as effective as those of the s and p electrons. 5 The increase in the screening effect is less than the increase in the nuclear charge. As a result, going from calcium to titanium, there is an increase in the effective nuclear charge and the attraction between the nucleus and the electron clouds gets stronger resulting in a decrease in the atomic radius. Ionic Radius Ion No. of protons Electronic configuration Atomic radius/nm Ca2+ 20 2.8.8. or [Ne] 0.099 Ti2+ 22 2.8.10. or [Ne]3d2 0.090 1 Both the ions have three electronic shells that are filled with electrons. 2 Going from Ca2+ to Ti2+, the increase in the screening effect is less than the increase in the nuclear charge. This is because d electrons are poor screeners. 3 As a result, the effective nuclear charge increases causing a decrease in the ionic radius from Ca2+ to Ti2+. Density Element Atomic radius/nm Molar mass/g mol–1 Density/g cm–3 Calcium 0.197 40.1 1.55 Titanium 0.147 47.9 4.43 1 The size of the titanium atom is smaller than that of calcium. However, the molar mass of titanium is higher than that of calcium. 2 As a result, the mass volume ratio (the density) is higher for titanium compared to calcium.

Chemistry Term 2 STPM Chapter 13 Transition Elements 215 13 Melting Point Element Atomic radius/nm Melting point/°C Calcium 0.197 839 Titanium 0.147 1680 1 Both calcium and titanium exist as giant metallic structures with strong metallic bonds holding the atoms together in the solid lattice. 2 Due to the small difference in energy between the 3d and 4s sub-shells; titanium can make use of a maximum of four electrons (3d 2 4s 2 ) to form the metallic bonds. On the other hand, calcium can make use of the 4s 2 electrons only to form the metallic bonds. 3 Furthermore, the atomic size of titanium is smaller than that of calcium. As a result, the metallic bond in titanium is stronger than that in calcium. This results in the higher melting point of titanium. Conductivity 1 Titanium is a better electrical conductor than calcium. 2 Titanium can make use of the 3d as well as the 4s electrons to form metallic bonds. Whereas calcium can make use of its 4s electrons only to form metallic bonds. 3 As a result, titanium has more delocalised electrons than calcium, which makes it a better conductor. First Ionisation Energy Element Proton number Atomic size/nm 1st I.E./kJ mol–1 Calcium 20 0.197 590 Titanium 22 0.147 661 1 Going from calcium to titanium, the atomic size decreases but the nuclear charge increases. 2 As a result, the valence electrons in titanium are more strongly held by the nucleus compared to calcium. 3 More energy is required to remove the valence electron in titanium compared to calcium. Exam Tips Exam Tips Small atoms can be more closely packed than larger ones. Quick Check 13.1 1 Explain why the atomic radius of manganese is larger than expected. 2 Explain in terms of structure and bonding, why the melting points of the first row transition elements are all above 1000 °C. 3 Both calcium and nickel have two electrons in their outermost shell. However, the first ionisation energy and the electrical conductivity of nickel are higher than that of calcium. Explain. 4 Explain in terms of electronic configuration why the third ionisation energy of iron is lower than expected. 2012/P2/Q7(a) 2014/P2/Q17(a)(iii)

Chemistry Term 2 STPM Chapter 13 Transition Elements 216 13 13.2 Chemical Properties of First Row Transition Elements Transition elements have the following chemical properties: (a) They exhibit variable oxidation states in their compounds. (b) They form complexes. (c) Their compounds are mostly coloured. (d) They exhibit catalytic properties. Variable Oxidation States 1 The relative energies of the 3d and 4s sub-shells are shown below. Energy 3d 3p 4s 2 Due to the very small difference in energy between the 3d and the 4s sub-shells, transition elements can lose variable number of electrons from these two sub-shells to form ions of comparable stability. As a result, they show variable oxidation states in their compounds. 3 The table below lists the oxidation states of the transition elements. The more common oxidation states are enclosed in circles. Ti V Cr Mn Fe Co Ni Cu +1 +2 +2 +2 +2 +2 +2 +2 +2 +3 +3 +3 +3 +3 +3 +3 +4 +4 +4 +4 +4 +4 +4 +5 +5 +5 +5 +5 +6 +6 +6 +7 4 The maximum oxidation state of the elements increases from +4 for titanium to +7 for manganese. Thereafter, it decreases to +2 for copper. 5 The maximum oxidation states refer to the maximum number of electrons available (3d and 4s) to take part in chemical reactions. 6 This increases from 4 for titanium (3d 2 4s 2 ) to 7 for manganese (3d 5 4s 2 ). The oxidation state of sodium ferrate(VI), Na2FeO4 (a red solid) is +6. 2014/P2/Q17(a)(i)(ii) 2015/P2/Q17

Chemistry Term 2 STPM Chapter 13 Transition Elements 217 13 3d 4s Ti: Maximum electrons (4s + unpaired 3d) 2 + 2 = 4 3d 4s Mn: 2 + 5 = 7 7 However, after manganese, the pairing of the 3d electrons occurs. The paired electrons in the 3d sub-shell somehow are reluctant to take part in chemical reactions. This causes a gradual decrease in the maximum oxidation state from iron to copper. Ti: Maximum electrons (4s + unpaired 3d) 2 + 2 = 4 V: 3 + 2 = 5 Cr: 5 + 1 = 6 Mn: 5 + 2 = 7 Fe: 4 + 2 = 6 Co: 3 + 2 = 5 Ni: 2 + 2 = 4 Cu: 0 + 1 = 1 Formation of Coloured Compounds 1 A substance appears blue because when sunlight (or white light) strikes the object and due to the molecular structure of the substance, light in the red and orange regions are absorbed, while blue light (and maybe some green) is unaffected. The reflected light gives the colour of the substance. 2 However, if none of the visible light components is absorbed, the object will appear white (or colourless). 3 On the other hand, if all the visible light components are absorbed, the object will appear black. 2013/P2/Q15 2014/P2/Q14 2017/P2/Q20(a) 2018/P2/Q20(a)

Chemistry Term 2 STPM Chapter 13 Transition Elements 218 13 Total absorption Black White Colour Total reflection Partial absorption White light 4 Most complexes of transition elements are coloured because they absorb parts of the visible light. 5 For example, the absorption spectrum of [Cu(NH3)4] 2+(aq) ion is shown below: Relative absorbance Violet Indigo Blue Green Yellow Orange Red 6 The spectrum shows that the violet, yellow, orange and red region of the visible spectrum are absorbed, while light in the indigo, blue and green regions are reflected. Hence, a aqueous solution of the ion appears blue. 7 The ability to absorb visible light is due to the splitting of the five 3d orbitals in the presence of ligands. 8 In a free gaseous ion, the five 3d orbitals are degenerate. That is they have the same energy. However, in the presence of ligands, for example in an octahedral complex, the five 3d orbitals are split into two groups with different energy. Energy Free gaseous ion Octahedral complex ion ΔE Exam Tips Exam Tips • The colour of transition metal ions is the consequence of partially filled d-orbitals. • Transition metal ions with an empty or fully filled d-orbitals are colourless. 2014/P2/Q14

Chemistry Term 2 STPM Chapter 13 Transition Elements 219 13 9 Electrons in the lower lying 3d orbitals can absorb certain component (wavelength) of the visible light and get promoted to the 3d orbitals of higher energy (this is known as d-d transition). The wavelengths that are not absorbed give the colour of the complex. 10 The wavelength absorbed by a complex depends on: (a) the nature of the central metal cation (b) the oxidation state of the central metal cation (c) the nature of the ligands 11 For example: [Fe(H2O)6] 2+ is green while [Fe(H2O)6] 3+ is yellow. [Cu(H2O)6] 2+ is light blue while [Cu(NH3)4] 2+ is dark blue. [Fe(H2O)6] 3+ is yellow while [Fe(H2O)5SCN]2+ is red. [CoCl4] 2– is blue while [Co(H2O)6] 2+ is pink. 12 The table below lists the colours of some common complexes of the transition elements. Element Complex Oxidation state Colour Titanium [Ti(H2O)6] 3+ +3 violet Vanadium [V(H2O)6] 2+ [V(H2O)6] 3+ [V(H2O)5O]2+ [V(H2O)4O2] + +2 +3 +4 +5 violet green blue yellow Chromium [Cr(H2O)6] 2+ [Cr(H2O)6] 3+ [Cr(OH)6] 3– [Cr(H2O)5Cl]2+ [Cr(H2O)4Cl2] + Cr2O7 2– CrO4 2– +2 +3 +3 +3 +3 +6 +6 blue green bright green light green dark green orange yellow Manganese [Mn(H2O)6] 2+ MnO4 2– MnO4 – +2 +6 +7 light pink green purple Iron [Fe(H2O)6] 2+ [Fe(CN)6] 7– [Fe(H2O)6] 3+ [Fe(CN)6] 3– [Fe(H2O)5NO]2+ [FeO4] 2– +2 +2 +3 +3 +3 +6 green yellowish green yellow yellowish green brown red Cobalt [Co(H2O)6] 2+ [Co(NH3)6] 2+ [CoCl4] 2– [Co(H2O)6] 3+ [Co(NO2)6] 3– +2 +2 +2 +3 +3 pink yellow blue brown yellow Nickel [Ni(H2O)6] 2+ [Ni(NH3)6] 2+ [Ni(CN)4] 2– [Ni(DMG)2] [Ni(edta)]2– +2 +2 +2 +2 +2 light green light blue green red (solid) blue Copper [Cu(H2O)6] 2+ [Cu(NH3)4] 2+ [CuCl4] 2– [Cu(edta)]2– +2 +2 +2 +2 blue dark blue yellow light blue [Mn(H2O)6] 3+ is purple-red

Chemistry Term 2 STPM Chapter 13 Transition Elements 220 13 Example 13.1 Explain why most titanium(IV) compounds are colourless. Solution The electronic configuration of titanium is [Ar]3d2 4s 2 and the electronic configuration of titanium(IV) is [Ar]3d0 . There are no electrons in the 3d orbitals. As a result, no d-d transition can occur. This accounts for the lack of colour of titanium(IV) compounds. Example 13.2 Explain why an aqueous solution of nickel(II) sulphate is green in colour. Solution Nickel sulphate dissolves in water to form the hydrated nickel(II) complex. NiSO4(s) + 6H2O(l) ⎯→ [Ni(H2O)6] 2+(aq) + SO4 2–(aq) The electronic configuration of Ni2+ is [Ar]3d 8 . In the hexaaquacomplex, the five 3d orbitals are split into two groups. When exposed to sunlight, electrons in the lower energy group absorb certain wavelengths of the light and get promoted to the 3d orbitals of higher energy. The wavelength that is not absorbed (which corresponds to green light) gives the colour of the complex. White light ⎯→ [Ni(H2O)6] 2+(aq) ⎯→ Green light One possible d-d transition is shown below: White light ⎯⎯⎯⎯⎯→ ⎯⎯⎯⎯⎯→ Green [Ni(H2O)6] 2+ * [Ni(H2O)6] 2+ The Principle Oxidation Numbers of the Elements in their Common Oxides, Cations and Ions 1 The table below lists the stable oxides of the transition elements. Ti V Cr Mn Fe Co Ni Cu +1 Cu2O +2 MnO FeO CoO NiO CuO +3 Ti2O3 V2O3 Cr2O3 Mn2O3 Fe2O3 Co2O3 +4 TiO2 MnO2 NiO2 +5 V2O5 +6 CrO3 K2FeO4 +7 Mn2O7 Exam Tips The 3d orbitals of Ti4+ are empty. No d-d transition is possible. Exam Tips Exam Tips The 3d orbitals of Ni2+ are partially filled. As a result, d-d transition can occur. 2017/P2/Q20(b)

Chemistry Term 2 STPM Chapter 13 Transition Elements 221 13 2 The table below lists the common aqueous cations and oxo-ions of the transition elements. Ti V Cr Mn Fe Co Ni Cu +1 Cu+ +2 Ti2+ V2+ Cr2+ Mn2+ Fe2+ Co2+ Ni2+ Cu2+ +3 Ti3+ V3+ Cr3+ Fe3+ Co3+ +4 TiO2+ VO2+ +5 VO2 + VO3 – +6 CrO4 2– Cr2O7 2– MnO4 2– FeO4 2– +7 MnO4 – 3 The +2 and +3 oxidation states of the elements exist as simple aqueous cations such as [Ti(H2O)6] 2+ and [Fe(H2O)6] 3+. 4 However, higher oxidation states do not exist as simple aqueous cations but exist in the form of oxo-cations or oxo-anions. For example, VO2 + , MnO4 – , Cr2O7 2– and FeO4 2–. 5 This is because the high charge density of the cations of higher oxidation states tends to polarise the water molecules that are bonded to it and causes the simple aqueous ions to decompose to the oxo-cations or oxo-anions. 6 Take vanadium(V) as an example. The simple V(H2O)6 5+ cation would have the following structure. H2O H2O OH2 OH2 H2O H2O V5+ 7 However, due to the very high charge density of the V5+ ion, the water molecules that are coordinated to it are greatly polarised. This polarisation weakens the O⎯H bonds in the H2O molecules, which then dissociates, with the elimination of H+ ions to form the [V(H2O)4O2] + cation, or written simply as VO2 + . OH2 O H H O OH2 H2O H2O V5+ H H OH2 O2– O2– OH2 + 4H+ H2O H2O V5+ All bonds between the water molecules and V5+ are coordinate bonds.

Chemistry Term 2 STPM Chapter 13 Transition Elements 222 13 The Relative Stability of the Oxidation States 1 The most common oxidation states of the transition elements are +2 and +3. 2 For titanium, vanadium, chromium and iron, the +3 oxidation state is more stable than the +2 oxidation state. 3 On the other hand, for manganese, cobalt, nickel and copper, the +2 oxidation state is more stable than the +3 oxidation state. Element Stable oxidation state Ti +3 V +3 Cr +3 Mn +2 Fe +3 Co +2 Ni +2 Cu +2 4 This is because, as we go across the d block elements, it gets more and more difficult to remove a third electron from the element concern due to the decrease in the size of the atom and the increase in the nuclear charge. 5 The table below lists the third ionisation energy (3rd I.E. in kJ mol–1) for the transition elements. M 2+(g) ⎯→ M 3+(g) + e– Element 3rd I.E. Ti 2720 V 2870 Cr 2990 Mn 3250 Fe 2960 Co 3230 Ni 3390 Cu 3550 As a result, the +3 cations get progressively more difficult to form. 6 Oxidation states of +4 and above are relatively less stable because the energies involved are too high. For example, the total energy required to form the Cr6+ ion is 25 433 kJ mol–1. Standard Reduction Potentials and the Relative Stability of Aqueous Ions 1 The most common oxidation states of the transition elements are +2 and +3 states which correspond to the following processes: M ⎯→ M 2+ + 2e– M ⎯→ M 3+ + 3e– The 3rd ionisation energy of iron is lower than expected. This is because removal of one electron from Fe2+ produces a more stable Fe3+ with a d5 configuration Fe2+ ⎯⎯→ Fe3+ [Ar]3d6 [Ar]3d5 2017/P2/Q20(c)

Chemistry Term 2 STPM Chapter 13 Transition Elements 223 13 2 The relative stability of the +2 and +3 states can be predicted by looking at the standard electrode potentials of the M 3+/M 2+ system: System E°/V Cr3+ + e– Cr2+ –0.41 Ti3+ + e– Ti2+ –0.37 V3+ + e– V2+ –0.24 Fe3+ + e– Fe2+ +0.77 Mn3+ + e– Mn2+ +1.51 Co3+ + e– Co2+ +1.81 Compare to the oxygen system below: O2(g) + 4H+ (aq) + 4e– 2H2O(l) E° = +1.23 V (All ions are aqueous ions.) 3 The half-cells above are represented by the graph below: x x x x x x M2+ is more stable. M3+ is more stable. +1.23 E°/V 2.0 1.0 0.0 –1.0 Ti V Cr Mn Fe Co 4 For half-cells where the E° values are less than +1.23 V, oxygen from the air (or dissolved oxygen) will oxidise the +2 state to the +3 state. 4Ti2+(aq) + O2(g) + 4H+ (aq) ⎯→ 4Ti3+(aq) + 2H2O(l) E° = +1.60 V 4V2+(aq) + O2(g) + 4H+ (aq) ⎯→ 4V3+(aq) + 2H2O(l) E° = +1.47 V 4Cr2+(aq) + O2(g) + 4H+ (aq) ⎯→ 4Cr3+(aq) + 2H2O(l) E° = +1.64 V 4Fe2+(aq) + O2(g) + 4H+ (aq) ⎯→ 4Fe3+(aq) + 2H2O(l) E° = +0.46 V Hence, for Ti, V, Cr and Fe, the +3 state are relatively more stable than the +2 state. 5 For half-cells where the E° is more positive than +1.23 V, the +3 aqueous ion will oxidise water to oxygen and itself reduced to +2. 4Mn3+(aq) + 2H2O(l) ⎯→ 4Mn2+(aq) + O2(g) + 4H+ (aq) E° = +0.28 V 4Co3+(aq) + 2H2O(l) ⎯→ 4Co2+(aq) + O2(g) + 4H+ (aq) E° = +0.58 V Thus, for Mn and Co, +2 is more stable than +3. +2 more stable +3 more stable Mn, Co Ti, V, Cr, Fe

Chemistry Term 2 STPM Chapter 13 Transition Elements 224 13 Redox Properties of the Various Oxidation States 1 Most lower oxidation states of the transition elements are reducing as shown by their negative standard electrode potentials. System E°/V Cr3+ + e– Cr2+ –0.41 Ti3+ + e– Ti2+ –0.37 V3+ + e– V2+ –0.24 2 For example, Cr2+ can reduce Fe3+ to Fe2+. Fe3+(aq) + Cr2+(aq) ⎯→ Fe2+(aq) + Cr3+(aq) 3 On the other hand, the higher oxidation states are strongly oxidising. System E°/V MnO4 2– + 4H+ + 2e– MnO2 + 2H2O +2.26 FeO4 2– + 8H+ + 3e– Fe3+ + 4H2O +2.20 MnO4 – + 8H+ + 5e– Mn2+ + 4H2O +1.52 Cr2O7 2– + 14H+ + 6e– 2Cr3+ + 7H2O +1.33 VO2 + + 2H+ + e– VO2+ + H2O +1.00 VO2+ + 2H+ + e– V3+ + H2O +0.34 4 For example, in acidic solution, potassium manganate(VII) will oxidise chloride ions to chlorine. MnO4 – + 8H+ + 10Cl– ⎯→ 5Cl2 + Mn2+ + 4H2O In acidic solution, potassium dichromate will oxidise iron(II) to iron(III). Cr2O7 2– + 14H+ + 6Fe2+ ⎯→ 2Cr3+ + 7H2O + 6Fe3+ Example 13.3 Potassium ferrate(VI) is a red solid. When added to water, effervescence occurs and a gas that supports combustion is released. Explain the observation. Solution Consider the two following half-cells: FeO4 2–(aq) + 8H+ (aq) + 3e– Fe3+(aq) + 4H2O(l) E° = +2.20 V O2(g) + 4H+ (aq) + 4e– 2H2O(l) E° = +1.23 V The ferrate(VI) ion is a very powerful oxidising agent. When added to water, it oxidises water to oxygen according to the equation: 4FeO2– + 20H+ ⎯→ 4Fe3+ + 3O2 + 10H2O E° = +0.97 V Exam Tips The more negative the E° value, the stronger the reducing agent. Exam Tips Exam Tips The more positive the E° value, the stronger the oxidising agent.

Chemistry Term 2 STPM Chapter 13 Transition Elements 225 13 Example 13.4 25.0 cm3 of an aqueous solution containing iron(II) ions requires 23.50 cm3 0.020 M potassium manganate(VII) in the presence of excess dilute sulphuric acid for complete reaction. Calculate the concentration of iron(II) in g dm–3. Solution The equation of reaction is: MnO4 – + 8H+ + 5Fe2+ ⎯→ 5Fe3+ + Mn2+ + 4H2O (MV)Fe2+ = 5 (MV)MnO4 – M × 25.0 = 5 0.020 × 23.5 M = 0.094 mol dm–3 = 0.094 × 55.8 g dm–3 = 5.25 g dm–3 Formation of Complex Ions and Ligand Exchange 1 A complex ion is defined as an ion consisting of a central metal cation bonded to a group of atoms/molecules/ions (known as ligand) by coordinate bonds. 2 An example of complex ion is the [Cu(NH3)4] 2+ ion formed when excess ammonia is added to aqueous copper(II) sulphate. 3 At first, a blue precipitate of copper hydroxide is formed: Cu2+(aq) + 2NH3(aq) + 2H2O(l) ⎯→ Cu(OH)2(s) + 2NH4 + (aq) On addition of excess ammonia, the blue precipitate dissolves to form a dark blue solution containing the [Cu(NH3)4] 2+ complex ion. [Cu(H2O)6] 2+(aq) + 4NH3(aq) [Cu(NH3)4] 2+(aq) + 6H2O(l) 4 The reaction is that of a ligand exchange process. The H2O ligand is substituted by the NH3 ligand. 5 Another example is: [Fe(H2O)6] 2+ + SCN– (aq) [Fe(H2O)5(SCN)]2+(aq) + H2O(l) Yellow Red 6 The bonding in the [Cu(NH3)4] 2+ complex ion is shown below: H3N NH3 H3N NH3 Cu2+ 2012/P1/Q26 2016/P2/Q14; Q17(c)(d) A complex ion: L     L  L   M  L n+ (L = Ligand) INFO Formation of Complex Ions

Chemistry Term 2 STPM Chapter 13 Transition Elements 226 13 7 In the formation of the complex, the ammonia molecules donate their lone pair electrons to the empty orbitals in the Cu2+ ion to form coordinate bonds. Thus, the metal cation acts as Lewis acid and the ammonia molecule acts as Lewis base. 8 Other examples of complex ions are [Fe(CN)6] 3– and [Fe(CN)6] 4–. Catalytic Properties 1 A catalyst is a substance that increases the rate of a chemical reaction but remains chemically unchanged at the end of the reaction. 2 A catalyst takes part in the chemical reaction but is not consumed by the reaction. 3 The energy profile of a reaction with and without catalyst is shown below: Activation energy without catalyst Activation energy with catalyst Products Progress of reaction Reactants Energy 4 The Boltzmann distribution curve of the reaction is shown below: Number of particles with energy greater than the lowered activation energy Number of particles with energy greater than the activation energy Particles which do not have enough energy to react Energy Activation energy in the absence of catalyst Activation energy in the presence of catalyst Number of particles 5 There are generally two types of catalytic actions: homogeneous catalysis and heterogeneous catalysis. Exam Tips Exam Tips Catalyst provides an alternative pathway, with lower activation energy, for the reaction to occur. L     L  L   M  L Lewis acid Lewis base

Chemistry Term 2 STPM Chapter 13 Transition Elements 227 13 Homogeneous Catalysis 1 In homogeneous catalysis, the physical state of the catalyst is the same as those of the reactants. 2 Examples of homogeneous catalysis are: Reaction Catalyst 2SO2(g) + O2(g) 2SO3(g) NO2(g) 2KClO3(s) ⎯→ 2KCl(s) + 3O2(g) MnO2(s) 2I – (aq) + S2O8 2–(aq) ⎯→ I2(aq) + 2SO4 2–(aq) Fe3+(aq) 3 Homogeneous catalysis is usually explained by the intermediate product theory. 4 Using the oxidation of iodide ions by the peroxodisulphate ions as an example. S2O8 2–(aq) + 2I– (aq) ⎯→ 2SO4 2–(aq) + I2(aq) (a) The activation energy for the uncatalysed reaction is high because it involves the direct collision of negative charge ions, which tends to repel one another. (b) However, in the presence of Fe3+(aq) as catalyst, the reaction proceeds via the following pathway: 2Fe3+(aq) + 2I– (aq) ⎯→ 2Fe2+(aq) + I2(aq) 2Fe2+(aq) + S2O8 2–(aq) ⎯→ 2Fe3+(aq) + 2SO4 2–(aq) (c) The above pathway has a lower activation energy because it involves the collision of opposite charged particles. (d) During the reaction, Fe3+ is temporarily converted to Fe2+. However, it is regenerated at the end of the reaction. Catalysed reaction Uncatalysed reaction Energy Progress of reaction 5 Transition elements and their compounds are good homogeneous catalysts because they exhibit variable oxidation states. INFO Catalysis

Chemistry Term 2 STPM Chapter 13 Transition Elements 228 13 Heterogeneous Catalysis 1 In heterogeneous catalysis, the physical state of the catalyst is different from those of the reactants. 2 Examples of heterogeneous catalysis are: Reaction Catalyst N2(g) + 3H2(g) 2NH3(g) Fe(s) 2SO2(g) + O2(g) 2SO3(g) V2O5(s) H2(g) + I2(g) 2HI(g) Ni(s) CH2 = CH2(g) + H2(g) ⎯→ CH3CH2(g) Ni(s) CH3CH2OH(g) ⎯→ CH2=CH2(g) + H2O(g) Al2O3(s) 3 Heterogeneous catalysis is usually explained by the adsorption theory. One should distinguish between adsorption and absorption. 4 Adsorption occurs on the surface of the substance, while absorption occurs in the bulk of the substance. Adsorption Absorption 5 An example of heterogeneous catalysis is the hydrogenation of ethene. CH2=CH2(g) + H2(g) ⎯→ CH3⎯CH3(g) The catalyst used is solid nickel although platinum is sometimes used. 6 The catalysis process is as follows: (a) Nickel atoms on the surface of the nickel metal make use of their empty orbitals to form temporary bonds with the ethene and hydrogen molecules. Ni Ni Ni Ni Ni Ni CH2 CH2 H H CH2 CH2 Ni Nickel surface (b) Adsorption brings the ethene and hydrogen molecules closer to one another (thereby increasing the concentration) and holds the molecules in the correct orientation for new bonds to be formed in the product. Ni Ni Ni Ni Ni Ni CH2 CH2 H H CH2 CH2 Ni Nickel surface New bond forming (c) After the new bonds are formed, the ethane molecules leave the surface of nickel so that other ethene and hydrogen molecules can be adsorbed. 2014/P2/Q15 2018/P2/Q15 VIDEO Heterogenous Catalysis

Chemistry Term 2 STPM Chapter 13 Transition Elements 229 13 13.3 Nomenclature and Bonding of Complexes Naming of Complexes 1 The ligands are named first, in alphabetical orders (the prefix of di, tri and tetra do not alter this order) followed by the name of the metal ion, all in one word with no space between them. 2 The oxidation number of the metal cation is written in Roman numerals enclosed in bracket after the name of the metal cation. 3 For neutral or cationic complexes, the normal names of the metals are used. 4 For anionic complexes, the metal is named as its salt (except for EDTA complex). However, if the metal has a Latin name, the Latin name is used for naming of its salt. 5 The names of some metals (including non-transition elements) in anionic complexes are given below: Metal Name Titanium Titanate Vanadium Vanadate Chromium Chromate Manganese Manganate Iron Ferrate Cobalt Cobaltate Nickel Nickelate Copper Cuperate Zinc Zincate Aluminium Aluminate Tin Stanate Lead Plumbate Beryllium Beryllate 6 The prefix of bis, tris and tetrakis are used to replace di, tri and tetra if a ligand itself already has the later prefix. 7 For example, [Ni(C2O4)3] 4– is named as tris-ethanedioatenickelate(II) and not tri-ethanedioatenickelate(II). 8 However, the complex, [Cr(NH3)4Cl2] + is named as tetraamminedichlorochromium(III) and not tetrakisamminebischlorochromium(III). 2014/P2/Q17(a)(iv) 2016/P2/Q17(a)(b) 2015/P2/Q15 2017/P2/Q15;Q20(a) INFO Nomenclature of Complex Ions

Chemistry Term 2 STPM Chapter 13 Transition Elements 230 13 Example 13.5 Name the following complexes. (a) [V(H2O)6] 3+ (b) [Fe(H2O)5SCN]2+ (c) [CuCl2] – (d) Ni(CO)4 (e) [V(H2O)4O2] + (f) Ni(H2NCH2CH2NH2)2Cl2 (g) [Ni(EDTA)]2– (h) [Fe(CN)6] 4– (i) [Co(C2O4)3] 3– Solution (a) Hexaaquavanadium(III) (b) Pentaaquathiocyanatoiron(III) (c) Dichlorocuperate(I) (d) Tetracarbonylnickel (Note: This is one of the few complexes where the central metal atom is neutral.) (e) Tetraaquadioxovanadium(V) (f) Bis-ethane-1,2-diaminedichloronickel(II) (g) (EDTA)nickel (h) Hexacyanoferrate(II) (i) Trisethanedioatecobaltate(III) or Trioxalatecobaltate(III) Bonding in Complexes 1 During the formation of complexes, the central metallic cation makes use of its empty orbitals (in its valence shell) to accept lone pair electrons from the ligand to form coordinate bonds. 2 An example is the formation of the hexacyanoferrate(II) complex ion, [Fe(CN)6] 4–. Fe2+(aq) + 6CN– (aq) [Fe(CN)6] 4– 3 The electronic configuration of Fe2+ is [Ar]3d6 . 3d 4s 4p Fe2+ 4 The first step involves the pairing of the six 3d electrons in the Fe2+ ion. 3d 4s 4p Fe2+ 2018/P2/Q20(b)

Chemistry Term 2 STPM Chapter 13 Transition Elements 231 13 5 This results in six empty orbitals which are used to accept six lone pair electrons from six CN– ions to form six coordinate bonds. 3d 4s 4p [Fe(CN)6] 4– 6 :CN– (Note: The Fe2+ ion first undergoes d2 sp3 hybridisation before forming the six coordinate bonds. However, hybridisation involving d orbitals is beyond the scope of the STPM syllabus.) Example 13.6 Nickel hydroxide dissolves in excess of aqueous ammonia to form a light blue complex according to the equation: [Ni(H2O)6] 2+(aq) + 6NH3(aq) [Ni(NH3)6] 2+(aq) + 6H2O(l) (a) In view of the above reaction, comment on the strength of H2O and NH3 as ligands. (b) Draw the structure of the light blue complex showing clearly the nature of the metal-ligand bonds. Solution (a) Ammonia is a stronger ligand than water because it can displace water ligand in the reaction. (b) NH3 NH3 NH3 NH3 H3N H3N Ni2+ Classification of Ligands 1 Ligands have lone pair electrons that can be donated to a metal cation to form coordinate bonds. 2 Ligands can be classified according to the number of lone pairs per ligand molecule that can be donated to the central metal cation. Monodentate Ligand 1 A monodentate ligand is a species that has one donor atom and forms only one coordinate bond per unit ligand with the central metal cation. 2010/P1/Q27 2012/P1/Q27 2015/P2/Q15

Chemistry Term 2 STPM Chapter 13 Transition Elements 232 13 2 The table below lists some examples of monodentate ligands and their names. Ligand Name H2O aqua NH3 ammine F – fluoro Cl – chloro OH – hydroxo O2– oxo CN– cyano CO carbonyl SCN– thiocyanate 3 Examples of complexes containing monodentate ligands are: [V(H2O)6] 2+; [Fe(H2O)5SCN]2+; [Cr(NH3)4Cl2] + ; Ni(CO)4; [V(H2O)5O]2+ Bidentate Ligand 1 Bidentate ligands are species that have two donor atoms and form two coordinate bonds per unit ligand with the central metal cation. Examples are (the donor atoms are in bold face): (a) Ethane-1,2-diamine, H2NCH2CH2NH2 • • • • H2N⎯CH2⎯CH2⎯NH2 (b) Ethanedioate ion or oxalate ion, C2O4 2– – O⎯C⎯C⎯O– • • ⏐⏐ ⏐⏐ • • O O The oxalate ligand is sometimes represented by ‘ox’. 2 Bidentate ligands can form ring structures with the central metal cation. An example is [Ni(C2O4)3] 2–: O O O O O O O O C C C C C C O O O Ni2+ O 3 These ligands are known as chelating agents and the complexes are called chelate compounds or chelates. The ethane-1,2-diamine ligand is sometimes represented by ‘en’. The complex has three five-membered rings. 2014/P2/Q17(b)

Chemistry Term 2 STPM Chapter 13 Transition Elements 233 13 Hexadentate Ligand 1 One of the most important hexadentate ligand is EDTA or ethylenediaminetetraacetate (or ethylenediaminetetraethanoate or bis(1,1-dicarboxymethyl)-1,2-diaminoethane). 2 It is used in its anionic form, EDTA4–, with the following structure: N⎯CH2⎯CH2⎯N – O⎯C⎯CH2 CH2⎯C⎯O – O O O O – O⎯C⎯CH2 CH2⎯C⎯O – 3 There are all together six donor atoms in the EDTA structure. 4 EDTA forms very stable complex ions with most metal cations especially the divalent cations. For example, [Ni(EDTA)]2–. O O O O N O CH2 C O C C O O C N Ni2+ CH2 CH2 CH2 CH2 CH2 Geometry of Complexes 1 Complex ions are molecular ions with covalent bonds (coordinate bonds) holding the individual atoms together in the ion. 2 The general geometry depends on the number of coordinate bonds surrounding the central metal cation. 3 The number of coordinate bonds surrounding the central atom is called the coordination number of the complex. 4 Here we shall look at complexes with coordination numbers 2, 4 and 6. Coordination Number 2 1 Complexes with coordination number 2 are linear. L → M ← L 2 Examples are: [CuCl2] – : Cl → Cu+ ← Cl [Ag(NH3)2] + : H3N → Ag+ ← NH3 Exam Tips Exam Tips EDTA chelation theraphy is used to treat heavy metal poisoning by mercury, chromium, lead or arsenic through our diet and daily lives. The heavy metals accumulate in our body and cause adverse side effects. EDTA works by binding with the metal to form a stable complex and removes the metal from the tissues that it has accumulated and allows the body to dispose of the complex as waste. Info Chem EDTA complexes have five five-membered rings. 2015/P2/Q15

Chemistry Term 2 STPM Chapter 13 Transition Elements 234 13 Coordination Number 4 1 Complexes with coordination number 4 are either tetrahedral or square planar. L L L L M L L L L M 2 The different geometries are due to the different type of hybridisation undergone by the central atom. Generally, tetrahedral involves sp3 hybridisation, while the square planar involves dsp2 hybridisation. 3 Examples of complexes with coordination number 4 are: [Cu(NH3)4] 2+ Square planar [CuCl4] 2– Tetrahedral Ni(CO)4 Tetrahedral Coordination Number 6 1 Complexes with coordination number 6 are octahedral. L L L L L L M 2 There are many examples of octahedral complexes. For example: [Co(H2O)6] 2+ [Cr(NH3)3Cl3] H2O H2O OH2 OH2 H2O H2O Co2+ NH3 NH3 Cl Cl H3N Cl Cr3+ 2008/P2/Q8(a)

Chemistry Term 2 STPM Chapter 13 Transition Elements 235 13 [Co(H2NCH2CH2NH2)3] 3+ NH2 CH2 CH2 NH2 NH2 NH2 CH2 CH2 CH2 CH2 H2N H2N Co3+ [Ni(EDTA)]2– O O O O N O CH2 C O C C O O C N Ni2+ CH2 CH2 CH2 CH2 CH2 Example 13.7 Nickel forms a neutral complex with carbon monoxide with the formula of Ni(CO)4. (a) State the oxidation state of nickel in the complex. (b) Draw the structure of the complex showing all the bonds in the complex. Solution (a) Zero (b) O C Ni C O C O C O Exam Tips Exam Tips All EDTA complexes are octahedral.

Chemistry Term 2 STPM Chapter 13 Transition Elements 236 13 Example 13.8 Chromium forms a complex with the oxalate ion with the formula [Cr(C2O4)3] 3–. (a) What is the oxidation state of chromium in the complex. (b) Predict the shape of the complex. Explain your answer. Solution (a) +3 (b) Octahedral. The oxalate ion, C2O4 2– is a bidentate ligand. Hence, with three oxalate ions, there are all together six coordinate bonds surrounding the chromium ion. 13.4 Uses of First Row Transition Elements and Their Compounds Uses of Chromium 1 When chromium (and a little nickel and carbon) is added to iron, stainless steel is produced. Chromium increases the tensile strength of steel and also increases the steel’s resistant to corrosion by forming an impervious chromium oxide layer that protects the iron. 2 Alloy of chromium with vanadium and tungsten is used in high speed cutting tools. 3 Alloys of nickel and chromium (which contains about 20% chromium) are used to make electrical heating elements. Uses of Titanium 1 Titanium and alloys of titanium have the same tensile strength as steel. However, it has two advantages over steel that it is lighter and more resistant to corrosion. 2 Titanium(IV) oxide, TiO2, which is a white solid is used as a white pigment in paints. It is non-toxic (unlike BaSO4) and does not blacken on exposed to air containing hydrogen sulphide gas. Uses of Manganese and Cobalt 1 The main use of cobalt and manganese is in the manufacture of alloys. 2 Alloy of manganese and steel is used to make high speed cutting tools. 3 Alloy of cobalt and samarium (Sm) is used to make permanent magnets. 2016/P2/Q15

Chemistry Term 2 STPM Chapter 13 Transition Elements 237 13 SUMMARY SUMMARY 1 A transition element is an element that forms at least one stable ion which an incompletely filled 3d sub-shell. 2 All transition elements are hard metals with high melting/boiling point and high density. 3 The characteristic properties of the transition elements are: • They exhibit variable oxidation states in their compounds. • Ability to form complexes. • Their compounds are mostly coloured. • The elements or their compounds have catalytic properties. 4 Complex ion is an ion consisting of a central metal cation bonded to a group of ligand via coordinate bonds. 5 A ligand is a Lewis base that can donate its lone pair electrons to a metal cation to form coordinate bonds. 6 Ligand can be classified as monodentate or polydentate. 7 The geometry of complex ions depends on the coordination number, that is the number of coordinate bonds (not the number of ligand) surrounding the central metal cation. 8 d-d transition is responsible for the colour of the complexes. 1 Which is the correct definition of a transition element? A An element with incompletely filled d sub-shells. B An element that exhibits variable oxidation state. C An element that forms ions with incompletely filled d sub-shells. D An element that forms coloured compounds. 2 Which of the following complexes is colourless? A [CuCl2] – C [Ni(NH3)6] 2+ B [CuCl4] 2– D [Fe(CN)6] 4– 3 The electronic configuration of M is [Ar] 3d 10 4s 2 . Which statement is not true about M? A It is a d– block element. B It is denser than copper. C M2+(aq) ion is colourless. D It is a Period 4 element. 4 All except one of the following statements about the complex [Fe(CN)6] 4– is incorrect. A Iron exhibits a valency of +2. B It is brightly coloured. C It is octahedral. D It is insoluble in water. 5 The chemical formula of a compound is K3[Fe(C2O4)3]. What is true about the compound? A Its IUPAC name is potassium triethanedioateiron(III). B It is planar. C Its ligand is hexadentate. D It is brightly coloured. 6 Chromium reacts with a ligand L in the mole ratio of 1 : 3 to form a complex. Ligand L is most likely to be A CN– C EDTA4– B C2O4 2– D NH3 7 Which of the following transition metal cations has five unpaired electrons? A Fe3+ C Ti3+ B V5+ D Cr3+ 8 Most of the transition elements or their compounds can act as homogeneous catalyst because A they have low ionisation energy B they can alter their oxidation states C they have good absorption characteristics D they can form stable complexes Objective Questions STPM PRACTICE 13

Chemistry Term 2 STPM Chapter 13 Transition Elements 238 13 9 T h e I U PAC n a m e o f a c o m p o u n d i s pentaamminechlorocobalt(II) sulphate. What is its chemical formula? A [CoCl(NH3)5]2SO4 C Co[Cl(NH3)5]2SO4 B [CoCl(NH3)5]SO4 D Co2Cl[(NH3)5SO4 10 Which of the following best explains the action of transition element as heterogeneous catalyst? A It increases the rate of collision. B It provides a suitable surface for the adsorption of the reactant molecules. C It exhibits variable oxidation states. D It can form complexes with the reactant species. 11 In which of the following compounds does the transition element have the highest oxidation state? A NH4VO3 C K2FeO4 B K2CuCl4 D [Cu(NH3)4(H2O)2] 2+ 12 Which of the following elements exhibits only one oxidation state in its compounds? A Vanadium C Copper B Cobalt D Zinc 13 Most copper(I) salts are white, whereas copper(II) salts are coloured. This is because A copper(I) compounds are less stable than copper(II) compounds B all the 3d orbitals in copper(I) are fully filled, whereas those of copper(II) are partially filled C copper(I) salts absorb all the wavelengths of the visible spectrum D copper(II) salts have higher lattice energy than copper(I) salts 14 Which of the following is not correct about the uses of transition elements and their compounds? A Alloys of cobalt are used to make permanent magnets. B Chromium oxide is used as a white pigment in paints. C Titanium is used to build aircraft jet engines. D Iron(II) sulphate is used as a catalyst in the reaction between sodium peroxodisulphate and potassium iodide. 15 Which of the following reagents when added to aqueous iron(III) chloride produces a blood-red solution? A KCN B H2NCH2CH2NH2 C EDTA D NH4SCN 16 During the titration between potassium manganate(VII) and aqueous iron(II) salts, a brown precipitate is formed if insufficient dilute sulphuric acid is used. What is the nature of the brown precipitate? A Fe2O3 C MnO2 B MnO4 2– D Mn(OH)2 17 Zinc, 30Zn has the lowest melting point amongst the d-block elements (Sc to Zn). Which statement can best explain this observation? A The first ionisation energy of Zn is the lowest. B Zn cannot use 3d electrons to form metallic bond. C The screening effect is the highest. D Zinc has a close-packed structure. 18 Which of the following explains why transition elements formed coloured ions? A Incompletely filled 3d subshell B Ability to form complex ions C Ability to act as catalysts D Variable oxidation states 19 Which statement is true of the transition elements? A All compounds formed by transition elements are coloured. B All their atoms have incompletely filled 3d sub-shell in their ground states. C They exhibit variable oxidation states. D All their ions have incompletely filled 3d sub-shell. 20 An element W has the electronic configuration of [Ar]3d 6 4s 2 . Which statement(s) is/are true of W? I It exhibits oxidation states of +2, +3 and +6. II It has 4 unpaired electrons in the ground state. III It forms an oxoanion with formula of WO4 2–. IV It forms coloured oxides. A I and III C II, III and IV B II and IV D I, II, III and IV 21 Nickel is used in the hydrogenation of ethene. CH2=CH2 + H2 → C2H6 What is the function of nickel in the reaction? A To decrease the enthalpy of the reaction. B To form a more reactive complex species with ethene molecules. C To increrase the rate of collision between ethene and hydrogen molecules. D To decrease the activation energy.

Chemistry Term 2 STPM Chapter 13 Transition Elements 239 13 Structured and Essay Questions 1 (a) Define ‘transition element’. (b) Explain why scandium and zinc are not considered as transition elements whereas copper is one. (c) (i) Write the electronic configuration of a chromium atom. (ii) Explain why this configuration violates the Aufbau rule. (d) In acidic solution, the dichromate(VI) ion is a powerful oxidising agent. (i) What is the colour of the dichromate(VI) ion? (ii) Write an ion/electron equation to illustrate the oxidising property of the dichromate(VI) ion. (iii) Calculate the volume of 0.025 M aqueous potassium dichromate that is required to react completely with 25.0 cm3 of 0.10 M iron(II) sulphate. 2 (a) How does the electronic configuration of a transition metal differs from the elements in the main groups of the Periodic Table? You may use iron and calcium as examples. (b) Give four characteristic properties of transition elements or their compounds. Illustrate your answer by reference to suitable examples. 3 (a) Give two physical properties of copper metal which show it to be typical of a transition element. (b) Give the electronic configuration of the (i) copper atom, (ii) copper(I) ion, and (iii) copper(II) ion. (c) (i) Explain why copper(II) compounds are usually blue, but those of copper(I) is white. (ii) Name a compound of copper(I) which is coloured and state the colour of the compound. (d) When copper(I) sulphate is added to water, a reddish-brown solid in a blue solution is formed. (i) What is the reddish-brown solid? (ii) What ion causes the blue colour of the solution? (iii) Write a balanced equation for the reaction. State the type of reaction involved. 4 (a) Give the electronic configuration of the cobalt atom and state the maximum oxidation state of cobalt. (b) Aqueous cobalt(II) chloride is pink. When concentrated hydrochloric acid is added to cobalt(II) chloride, the solution turns blue. When the blue solution is cooled in ice, the colour slowly changes to pink again. (i) Write the formula of the species responsible for the pink colour of cobalt(II) chloride. (ii) Using balanced equation(s), explain the observations. (c) When cobalt(III) chloride is added to water, effervescence occurs and a pink solution is formed. By quoting suitable data from the Data Booklet, explain the observation. 5 Copper (29Cu) forms the following complexes: [Cu(H2O)6] + [Cu(H2O)6] 2+ [Cu(EDTA)]- (a) Give the IUPAC name of [Cu(H2O)6] + . (b) State the colour of [Cu(H2O)6] + . Explain your answer. (c) What is the oxidation state of copper in [Cu(EDTA)]– ? (d) Explain the colour change when concentrated hydrochloric acid is added to [Cu(H2O)6] 2+. 6 (a) Define complex ion. (b) Name the following complex ions. (i) [Cr(C2O4)3] 3– (ii) [Ni(NH3)3Cl3] –

Chemistry Term 2 STPM Chapter 13 Transition Elements 240 13 (c) When excess ammonia is added to aqueous copper(II) sulphate, a dark blue solution is formed. However, the intensity of the blue colour fades on addition of aqueous EDTA. (i) Draw the structure of EDTA. (ii) Give the formula of the complex ions responsible for the dark blue solution and state its shape. (iii) Write the formula of the copper-EDTA complex. What is the coordination number of the complex? 7 (a) Discuss the relative stability of Mn2+ and Mn3+ ions. [E° of Mn3+/Mn2+ half-cell is +1.51 V] (b) Compare the melting point and first ionisation energy between chromium and calcium. (c) When aqueous sodium carbonate is added to iron(III) chloride solution, effervescence occurs and a reddish brown precipitate is formed. Explain these observations. (d) Nickel forms a complex with EDTA. (i) Give the IUPAC name of EDTA. (ii) Draw the structural formula of the complex and state its shape. 8 (a) What do you understand by the term transition element? (b) An aqueous solution of iron(III) chloride is yellowish. When a few drops of potassium thiocynate, KSCN, is added to aqueous iron(III) chloride, a red solution is obtained. (i) Give the formula of the ion responsible for the yellow colour of iron(III) chloride. (ii) Write an ionic equation for the production of the red solution. (iii) Name the type of reaction taking place. (c) Iron(III) reacts with excess of CN– ions to form hexacyanoferrate(III) ions. (i) Give the formula of the complex ion. (ii) Draw the shape of the complex ion. (iii) Compare the oxidising power of the iron(III) aqueous ion and the hexacyanoferrate(III) ion. Explain your answer. 9 Ethylenediamine forms a stable complex ion with nickel(II) ions. Explain the formation of the complex ion. 10 (a) Nickel (Proton number = 28) is a transition element. (i) Define transition element. (ii) Write the electronic configuration of the nickel atom. (iii) State one characteristic property of nickel other than the ability to form complex ions and coloured ions. (iv) Between nickel and calcium, which metal is expected to have the higher melting point? Explain your answer. (v) A complex of nickel is shown below. Give the IUPAC name of the complex. CH2 NH2 Cu CH2 NH2 CH2 CH2 NH2 NH2 2+ H2N CH2 H2N CH2 (b) The EDTA ion is an example of a hexadentate ligand. (i) Give the IUPAC name of the EDTA ion. (ii) Draw the structure of EDTA ion.

241 STPM Model Paper (962/2) Section A [15 marks] Bahagian A [15 markah] Answer all questions in this section. Jawab semua soalan dalam bahagian ini. 1 The enthalpy changes involved in the dissolution of an ionic solid is shown below: Perubahan entalpi yang terlibat apabila suatu pepejal ion larut dalam air ditunjukkan di bawah: Gaseous ions Ion bergas ∆H1               ∆H2 Ionic solid ∆H3 Aqueous solution Pepejal ion ⎯→ Larutan berair Which of the enthalpy changes is/are always negative? Yang manakah antara berikut sentiasa bernilai negatif? A ∆H1 C ∆H1 and/dan ∆H2 B ∆H2 D ∆H2 and/dan ∆H3 2 The standard enthalpy of combustion of aluminium is -835 kJ mol-1. Which of the following deduction is not correct? Perubahan entalpi pembakaran piawai aluminium adalah -835 kJ mol-1. Deduksi yang manakah tidak benar? A 150.3 kJ is liberated when 4.86 g of aluminium burns in excess oxygen. Apabila 4.86 g aluminium dibakar dalam oksigen berlebihan, 150.3 kJ dibebaskan. B -58.94 kJ of heat is involved in the formation of 3.60 g of aluminium oxide. -58.94 kJ terlibat dalam pembentukan 3.60 g aluminium oksida. C 242.5 g of aluminium has to be burned completely in excess oxygen to generate 3.00 × 104 kJ of heat. 242.5 g aluminium perlu dibakar dengan lengkap untuk menghasilkan 3.00 x 104 kJ haba. D The enthalpy change for the following reaction is -3340 kJ. 4Al(s) + 3O2 (g) → 2Al2 O3 (s) Perubahan entalpi bagi tindak balas berikut ialah -3340 kJ. 4Al(p) + 3O2 (g) → 2Al2 O3 (p) 3 The standard reduction potentials of two half-cells are given below: Keupayaan penurunan piawai bagi dua setengahsel diberikan berikut: Mn3+(aq) + e– Mn2+(aq) E° = + 1.51 V Ti3+(aq) + e– Ti2+(aq) E° = -0.37 V Which of the following is not correct? Yang manakah antara berikut tidak benar? A Under standard conditions, the most stable oxidation of Mn is +2 and for Ti is +3. Di bawah keadaan piawai, keadaan pengoksidaan yang paling stabil bagi Mn ialah +2 dan bagi Ti ialah +3. B In an electrochemical cell, the Mn3+/ Mn2+ is the cathode. Dalam suatu sel elektrokimia, Mn3+/Mn2+ merupakan katod. C The e.m.f. of the cell is +1.88 V. D.g.e. sel ialah +1.88 V. D The cell notation is: Notasi sel adalah: Pt(s)  Ti3+(aq); Ti2+(aq)  Mn3+(aq); Mn2+(aq)  Pt(s) Pt(p)  Ti3+(ak); Ti2+(ak)  Mn3+(ak); Mn2+(ak)  Pt(p) 4 Which of the following statements is not correct? Pernyataan yang manakah antara berikut adalah tidak benar? A The negative terminal of an electrochemical cell is called the anode. Kutub negatif suatu sel elektrokimia disebut anod.

242 Chemistry Term 2 STPM Specimen Paper Term 2 B The main purpose of anodisation is to increase the conductivity of aluminium. Tujuan utama menganodisasi adalah untuk meningkatkan kekonduksian aluminium. C In an electrolysis cell, the cathode is negatively charged. Dalam suatu sel elektrolisis, katod adalah bercas negatif. D The voltage used in the electrolytic extraction of aluminium from bauxite is about 4 V. Voltan yang digunakan dalam pengekstrakan aluminium dari bauksit adalah lebih kurang 4 V. 5 The melting point of magnesium oxide is higher than that of aluminium oxide because Takat lebur magnesium oksida lebih tinggi daripada aluminium oksida. Ini adalah kerana A magnesium oxide has a giant ionic structure while aluminium oxide has a simple ionic structure. magnesium oksida mempunyai struktur ion gergasi manakala aluminium oksida berstruktur ion ringkas. B Al3+ has a higher charge density than Mg2+. ketumpatan cas Al3+ lebih tinggi daripada Mg2+. C magnesium is more electronegative than aluminium. magnesium lebih elektronegatif daripada aluminium. D the interionic distance in MgO is shorter than in Al2 O3 . jarak antara ion dalam MgO lebih pendek daripada dalam Al2 O3 . 6 Which is the best statement that explains why sodium oxide is a basic oxide? Yang manakah antara berikut paling sesuai untuk menerangkan mengapa natrium oksida merupakan oksida berbes? A It reacts with acids to form salt and water only. Ia bertindak balas dengan asid untuk manghasilkan garam dan air sahaja. B Sodium is a metal. Natrium adalah suatu logam. C Oxygen is more electronegative than sodium. Oksigen lebih elektronegatif daripada natrium. D It contains O2- ions. Ia menagandungi ion-ion O2-. 7 Which of the following is not of the properties of Group 2 elements and their compounds? Yang manakah antara berikut bukan sifat unsurunsur kumpulan 2 dan sebatiannya? A Only strontium and barium react vigorously with water at room temperature. Hanya strontium dan barium bertindak balas dengan cergas dengan air pada suhu bilik. B Their peroxides dissolve in water to produce hydrogen peroxide. Peroksida masing-masing larut dalam air untuk menghasilkan hidrogen peroksida. C Strontium sulphate is more soluble than calcium sulphate. Strontium sulfat lebih larut berbanding dengan kalsium sulfat. D Most beryllium compounds are covalent because of its high ionisation energy and strong polarising power. Kebanyakan sebatian-sebatian berillium adalah kovalen disebabkan oleh tenaga pengionan yang tinggi dan kuasa pengutuban yang kuat. 8 The melting point of beryllium chloride is 405°C. This is because Takat lebur berillium klorida adalah 405°C. Ini disebabkan A beryllium chloride contains Be2+ and Clions. berillium klorida mengandungi ion-ion Be2+ dan Cl- . B it exists as simple BeCl2 molecules. ia wujud dalam bentuk molekul BeCl2 ringkas. C in the solid state, BeCl2 molecules are polymerised to form long chains. dalam keadaan pepejal, molekul-molekul BeCl2 berpolimeran untuk membentuk rantai-rantai panjang. D it exists as dimer Be2 Cl4 in the solid state. dalam keadaan pepejal, ia wujud dalm bentuk dimer, Be2 Cl4 .