243 Chemistry Term 2 STPM Specimen Paper Term 2 9 Which of the following is true about Group 14 elements (C to Pb) in the Periodic Table? Antara berikut, manakah yang benar tentang sifat-sifat kumpulan 14 dalam Jadual Berkala? A Silicon can form two hydrides, GeH4 and Ge2 H6 . Silikon boleh membentuk dua hidrida, GeH4 dan Ge2 H6 . B The melting point of lead is lower than tin. Takat lebur plumbum lebih rendah daripada timah. C Lead(II) chloride can be prepared by heating lead with dilute hydrochloric acid. Plumbum(IV) klorida bolah disediakan dengan memanaskan plumbum dengan asid hidroklorik cair. D The tetrachlorides of carbon and silicon are covalent while the tetrachlorides of tin and lead are ionic. Tetraklorida bagi karbon dan silikon bersifat kovalen manakala tetraklorida stanum dan plumbum bersifat ion. 10 A silicate has the empirical formula of SiO3 2-. How many oxygen atoms are shared between each silicate unit? Suatu silikat mempunyai formula empirik SiO3 2-. Berapakah atom oksigen yang dikongsikan antara unit-unit silikat? A 1 C 4 B 2 D 5 11 Chlorine is flowed through cold dilute sodium hydroxide. Which statement is true about the reaction? Klorin dialirkan melalui natrium hidroksida cair yang sejuk. Pernyataan yang manakah adalah benar bagi tindak balas ini? A Chlorine reacts with OH- to produce O2 . Klorin bertindak balas dengan OH- untuk menghasilkan O2 . B Potassium chlorate(V) is one of the products. Salah satu hasil tindak balas ialah kalium klorat(VII). C Chlorine undergoes disproportionation. Klorin mengalami penyahkadaran. D Sodium reduces chlorine to chloride ion. Natrium menurunkan klorin kepada ion klorida. 12 Which of the following is not true about Group 17 elements and their compounds? Yang manakah antara berikut tidak benar tentang unsur-unsur Kumpulan 17 dan sebatiansebatiannya? A Silver bromide is used in black-andwhite photography instead of silver chloride because it is more sensitive to light. Argentum bromida dan bukan argentum klorida digunakan dalam fotografi hitamputih kerana ia lebih sentitif kepada cahaya. B The mole reacting ratio between chlorine and sodium hydroxide is the same irrespective of the concentration of the alkali or the temperature. Nisbah mol yang bertindak balas antara klorin dengan natrium hidroksida adalah tetap dan tidak bergantung kepada kepakatan natrium hidroksida atau suhu. C Astatine is a single element that is radioactive. Astatin merupakan unsur tunggal yang bersifat radioaktif. D HCl is a stronger acid than HClO4 . HCl adalah asid lebih kuat daripada HClO4 . 13 The chemical formula of a compound is K2 FeO4 . What is not true about the compound? Formula suatu sebatian ialah K2 FeO4 . Yang manakah tidak benar mengenai sebatian ini? A The oxidation state of iron is +6. Keadaan pengoksidaan besi adalah +6. B The IUPAC name is potassium tetraoxoiron(VI). Nama IUPAC sebatian ini ialah kalium tetraoksobesi(VI). C It is a powerful oxidising agent. Ia merupakan suatu agen pengoksidaan yang kuat. D It is brightly coloured. Ia berwarna cerah.
244 Chemistry Term 2 STPM Specimen Paper Term 2 14 Property X of the transition elements (Ti to Cu) is shown below. Sifat X unsur-unsur peralihan (Ti ke Cu) ditunjukkan di bawah. X Proton number Nombor proton Property X is most probably Sifat X kemungkinannya A atomic radius jejari atom B melting point takat lebur C maximum oxidation state keadaan pengoksidaan maksimum D density ketumpatan 15 Nickel forms a complex with EDTA. Which is not true about the complex? Nikel membentuk suatu kompleks dengan EDTA. Yang manakah tidak benar tentang kompleks ini? A The reaction mole ratio between Ni and EDTA is 1:1. Nisbah mol tindak balas antara nikel dan EDTA ialah 1:1. B The formula of the complex is [Ni(EDTA)]2+. Formula kompleks ialah [Ni(EDTA)]2+. C The complex is octahedral. Kompleks ini berbentuk octahedron. D The oxidation state of nickel in the complex is +2. Keadaan pengoksidaan nikel dalam kompleks ini ialah +2. Section B [15 marks] Bahagian B [15 markah] Answer all questions in this section. Jawab semua soalan dalam bahagian ini. 16 (a) The second ionisation energies (∆H°/kJ mol-1) of Period 3 elements (Na to Cl) are shown below: Tenaga pengionan kedua (∆H°/kJ mol-1) unsur-unsur Kala 3 (Na ke Cl) diberikan dalam jadual berikut: Element Unsur Na Mg Al Si P S Cl ∆H° 4560 1450 1820 1580 1900 2260 2300 (i) Write an equation to represent the second ionisation energy of chlorine. [1] Tuliskan suatu persamaan untuk mewakili tenaga pengionan kedua klorin. (ii) Plot a graph of second ionisation energy against the proton numbers of the Period 3 elements. [2] Plot suatu graf tenaga pengionan kedua melawan nombor proton unsur-unsur Kala 3. (iii) Explain the general trend of the graph. [2] Jelaskan trend umum graf. (iv) Explain any abnormally of the graph. [2] Jelaskan sebarang abnormali graf. (b) Sulphur dioxide, SO2 , is an atmospheric pollutant. Sulfur dioksida, SO2 , merupakan suatu pencemar udara. (i) Name two sources of SO2 in the atmosphere. [1] Namakan dua punca SO2 dalam udara. (ii) SO2 is catalytically oxidised to SO3 by atmospheric oxygen. The reaction is catalysed by nitrogen dioxide, NO2 . Write equations to illustrate the catalytic action of nitrogen dioxide. [1]
245 Chemistry Term 2 STPM Specimen Paper Term 2 SO2 dioksidakan secara bermangkin kepada SO3 oleh oksigen dalam atmosfera. Tindak balas ini dimangkinkan oleh nitrogen dioxida, NO2 . Tuliskan persamaan-persamaan untuk menunjukkan tindakan bermangkin NO2 . 17 The two allotropes of phosphorous (that exists as P4 molecules) are white phosphorous and red phosphorous. The standard enthalpy change of combustion of white phosphorous and red phosphorous are -2984 kJ mol-1 and -2967 kJ mol-1 respectively. Dua allotrop fosforus (yang wujud sebagai molekul P4 ) ialah fosforus putih dan fosforus merah. Perubahan entalpi pembakaran piawai bagi fosforus putih dan fosforus merah adalah -2984 kJ mol-1 dan -2967 kJ mol-1 masing-masing. (a) Define standard enthalpy change of combustion of phosphorous. [2] Takrifkan perubahan entalpi pembakaran fosforous. (b) Write an equation for the combustion of phosphorous. [1] Tuliskan persamaan untuk pembakaran fosforous. (c) Which allotrope is energetically more stable? [1] Allotrope yang manakah lebih stabil dari segi tenaga? (d) Calculate the enthalpy change for the following conversion: Hitung perubahan entalpi tindak balas berikut: Phosphorous(white) → Phosphorous(red) [2] Fosforus(putih) → Fosforus(merah) Section C [30 marks] Bahagian C [30 markah] Answer any two questions from this section. Jawab mana-mana dua soalan daripada bahagian ini. 18 (a) An aqueous solution of 0.15 mol dm-3 copper nitrate is electrolysed using platinum electrodes. Larutan kuprum nitrat 0.15 mol dm-3 dielektrolsiskan menggunakan elektrod-elektrod platinum. (i) Draw a diagram for the electrolysis cell and label the cathode and anode. [2] Lukis gambar rajah sel electrolisis and labelkan katod dan anod. (ii) Explain why copper rather than hydrogen is produced at the cathode. [1] Jelaskan mengapa kuprum dan bukan hidrogen dihasilkan pada katod. (iii) As electrolysis progresses, the electrolyte becomes acidic. Explain why. [1] Semasa elektrolisis, larutan menjadi berasid. Jelaskan mengapa. (iv) When a current of 0.30 A flows through the electrolyte for 25 minutes, 1.48 g of copper is deposited at the cathode. Calculate a value for the Avogadro’s constant. [3] [Ar of Cu = 63.5, Charge on one electron = 1.60 × 10-19 C] Apabila arus 0.30 A dialirkan melalui elektrolit selama 25 minit, 1.48 g kuprum dienapkan pada katod. Hitung suatu nilai untuk Pemalar Avogadro. [Ar Cu = 63.5, Cas pada satu elektron = 1.60 × 10-19 C] (b) An electrochemical cell is shown below: Suatu sel elektrokimia ditunjukkan di bawah: Zn(s) Zn2+(aq) Cu2+(aq) Cu(s) Zn(p) Zn2+(ak) Cu2+(ak) Cu(p) Given/Diberi: Cu2+(aq) + 2e– Cu(s) E° = +0.34 V Zn2+(aq) + 2e– Zn(s) E° = -0.76 V (i) Calculate the standard e.m.f. of the cell at 25°C. [2] Hitung d.g.e. piawai sel pada 25°C. (ii) Calculate the equilibrium constant of the following reaction: Hitung pemalar kesimbangan tindak balas berikut:
246 Chemistry Term 2 STPM Specimen Paper Term 2 Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) [2] Zn(p) + Cu2+(ak) Zn2+(ak) + Cu(p) (iii) What is the e.m.f. of the cell when [Zn2+] = 1.10 M, and [Cu2+] = 0.55 M? [2] Berapakah d.g.e. sel apabila [Zn2+] = 1.10 M dan [Cu2+] = 0.55 M? (iv) How would you expect the equilibrium constant to change under the new set of conditions? Explain your answer. [2] Bagaimanakah pemalar keseimbangan dijangka berubah pada keadaan yang baru itu? Jelaskan jawapan anda. 19 Explain the following. Jelaskan yang berikut. (a) Some liquid lead(IV) chloride is placed in a flask and gently warmed. A gas is evolved and a white solid is left. When the gas is bubbled through aqueous potassium iodide, purple fumes are produced. [3] Sedikit cecair plumbum(IV) klorida diletakkan di dalam suatu kelalang dan dipanaskan. Suatu gas dibebaskan dan suatu pepejal tertinggal. Apabila gas itu dialirkan melalui kalium iodide berair, wap ungu dihasilkan. (b) Carbon dioxide is stable towards heat but lead dioxide decomposes upon heating. [3] Karbon dioksida stabil terhadap haba manakala plumbum dioksida terurai apabila dipanaskan. (c) All the dioxides of the elements in Group 14 are high melting point solids except carbon dioxide. [3] Semua dioksida unsur-unsur Kumpulan 14 merupakan pepejal dengan takat lebur tinggi kecuali karbon dioksida. (d) When magnesium nitrate is heated over a Bunsen flame, a brown gas is evolved. But no brown fume is evolved with barium nitrate. [3] Apabila magnesium nitrat dipanaskan dengan penunu Bunsen, suatu gas perang dibebaskan. Tetapi tiada gas perang diperhatikan dengan barium nitrat. (e) Hydrogen chloride has a lower boiling point than hydrogen iodide, but hydrogen chloride is more stable towards heat than hydrogen iodide. [3] Takat didih hidrogen klorida lebih rendah daripada hidrogen iodida, tetapi hidrogen klorida lebih stabil terhadap haba. 20 (a) (i) Using manganese as an example, illustrate three characteristic properties of transition elements. [3] Dengan menggunakan mangan sebagai contoh, perihalkan tiga sifat cirian unsur-unsur peralihan. (ii) Describe the observations when an aqueous solution of potassium manganate(VII) is added slowly to an acidified solution of hydrogen peroxide until potassium manganate(VII) is in excess. Write an ionic equation for the reaction taking place. [4] Jelaskan pemerhatian apabila larutan kalium manganat(VII) dicampur secara perlahan-lahan kepada larutan berasid hidrogen peroksida sehingga kalium manganat(VII) adalah berlebihan. Tuliskan suatu persamaan ion bagi tindak balas yang berlaku. (b) Comment on the fact that it is possible to isolate three different forms of chromium(III) chloride hexahydrate, CrCl3 .6H2 O. How would you differentiate between them chemically? [5] Komen tentang fakta bahawa adalah mungkin untuk mengasingkan tiga bentuk berlainan kromium(III) klorida heksahidrat, CrCl3 .6H2 O. Bagaimanakah anda dapat membezakannya secara kimia? (c) Explain why scandium (21Sc) and zinc (30Zn) are not transition elements but copper (29Cu) is. [3] Jelaskan mengapa skandium (21Sc) dan zink (30Zn) masing-masing bukan unsur peralihan tetapi kuprum (29Cu) adalah suatu unsur peralihan.
247 Periodic Table of Elements 1HHydrogen 1 1HHydrogen 1 Proton number 1234567 Period 3Li Lithium 7 4 Be Beryllium 9 11 Na Sodium 23 12 Mg Magnesium 24 19KPotassium 39 20 Ca Calcium 40 37 Rb Rubidium 85.5 38 Sr Strontium 88 55 Cs Caesium 133 56 Ba Barium 137 57 – 71 Lanthanoids 89 – 103 Actinoids 87 Fr Francium 88 Ra Radium 105 Db Dubnium 106 Sg Seaborgium 107 Bh Bohrium 108 Hs Hassium 109 Mt Meitnerium 110 Ds Darmstadtium 111 Rg Roentgenium 112 Cn Copernicium 113 Nh Nihonium 114 Fl Flerovium 115 Mc Moscovium 116 Lv Livermorium 21 Sc Scandium 4539YYttrium 89 22 Ti Titanium 4840 Zr Zirconium 9172 Hf Hafnium 178.5 23VVanadium 5141 Nb Niobium 9373 Ta Tantalum 181 24 Cr Chromium 5242 Mo Molybdenum 9674 WTungsten 184 25 Mn Manganese 5543 Tc Technetium 9875 Re Rhenium 186 26 FeIron 5644 Ru Ruthenium 101 76 Os Osmium 190 27 Co Cobalt 5945 Rh Rhodium 103 77 Ir Iridium 192 28 Ni Nickel 5946 Pd Palladium 106 78 Pt Platinum 195 29 Cu Copper 6447 Ag Silver 108 79 AuGold 197 30 ZnZinc 6548 Cd Cadmium 112 80 Hg Mercury 201 31 Ga Gallium 7049 InIndium 115 81 Tl Thallium 204 32 Ge Germanium 7350 SnTin 119 82 PbLead 207 33 As Arsenic 7551 Sb Antimony 122 83 Bi Bismuth 209 34 Se Selenium 7952 Te Tellurium 128 84 Po Polonium 209 35 Br Bromine 8053IIodine 127 85 At Astatine 210 117 Ts Tennessine 118 Og Organesson 36 Kr Krypton 8454 Xe Xenon 131 86 Rn Radon 222 5BBoron 1113 Al Aluminium 27 6CCarbon 1214 Si Silicon 28 7NNitrogen 1415P Phosphoros 31 8OOxygen 1616SSulphur 32 9FFluorine 1917 Cl Chlorine 35.5 2 He Helium 410 Ne Neon 2018 Ar Argon 40 58 La Lanthanum 139 90 Ac Actinium 59 Ce Cerium 140 91 Th Thorium 232 60 Pr Praseodymium 141 92 Pa Protactinum 231 61 Nd Neodymium 144 93UUranium 238 62 Pm Promethium 94 Np Nepturium 63 SmSamarium 150 95 Pu Plutonium 64 Eu Europium 152 96 Am Americium 65 Gd Gadolinium 157 97 CmCurium 66 Tb Terbium 159 98 Bk Berkelium 67 Dy Dysprosium 162 99 Cf Californium 68 Ho Holmium 165 100 Es Elinsteinium 69 Er Erbium 167 101 FmFermium 70 TmThulium 169 102 Md Mendelevium 71 Yb Ytterbium 173 103 No Nobelium 71 Lu Lutetium 175 103 Lr Lawrencium 104 Rf Rutherfordium Symbol of element Name of element Relative atomic mass Transition elements Lantanide series Actinide series I1 II2 3 4 5 6 7 8 9 10 Group 11 12 III 13 IV14 V15 VI 16 VII 17 VIII 18 Metals Key: Semi-metals Non-metals
248 Chemistry Term 2 STPM Appendix Summary of Key Quantities and Units Quantity Common symbol Unit mass m kg, g length l m time t s electric current I A amount of substance n mol temperature T °C, K temperature change q, ∆T °C, K volume V m3 , dm3 , cm3 density ρ kg m–3, g dm–3, g cm–3 pressure p Pa, atm frequency v Hz wavelength l m, mm, nm speed of light c m s–1 Planck constant h J s electronic charge e C standard electrode potential E° V standard reduction potential E° V standard cell potential E° cell V cell potential Ecell V electromotive force E V gas constant R J K–1 mol–1 half-life t—1 2 s atomic mass unit – a.m.u. relative atomic mass Ar – relative isotopic mass Ar –
249 Chemistry Term 2 STPM Appendix Quantity Common symbol Unit relative molecular mass Mr – molar mass M g mol–1 Molar volume of gas Vm dm3 mol–1 nucleon number A – proton number Z – neutron number N – Avogadro constant L mol–1 Faraday constant F C mol–1 specific heat capacity c J g–1 °C–1 heat capacity C J °C–1 heat change q J, kJ enthalpy change of reaction ∆H J, kJ standard enthalpy change of reaction ∆H° J mol–1, kJ mol–1 ionisation energy I kJ mol–1 lattice energy ∆H° lattice kJ mol–1 bond energy – kJ mol–1 electron affinity – kJ mol–1 rate constant k as appropriate equilibrium constant K, Kp , Kc as appropriate acid dissociation constant Ka mol dm–3 base dissociation constant Kb mol dm–3 mole fraction x – concentration c mol dm–3 solubility product Ksp as appropriate ionic product of water Kw mol2 dm–6
250 Chemistry Term 2 STPM Appendix Chemical Constants Standard Electrode Potentials and Standard Redox Potentials (at 298 K) Speed of light in vacuum c = 3.0 × 108 m s–1 Planck constant h = 6.63 × 10–34 J s = 3.99 × 10–13 kJ mol–1 s Molar gas constant R = 8.31 J K–1 mol–1 Faraday constant F = 9.65 × 104 C mol–1 Rydberg constant RH = 1.097 × 107 m–1 Avogadro constant L, NA = 6.02 × 1023 mol–1 Mass of proton mp = 1.67 × 10–27 kg Mass of neutron mn = 1.67 × 10–27 kg Mass of electron me = 9.11 × 10–31 kg Electronic charge e = –1.60 × 10–19 C Ionic product of water Kw = 1.0 × 10–14 mol2 dm–6 (at 298 K) Molar volume of gas Vm = 22.4 dm3 (at s.t.p.) = 24.0 dm3 (at room conditions) Specific heat capacity of water c = 4.18 J g–1 K–1 p = 3.142 ln x = 2.303 log10 x Electrode reaction Standard potential E°/V Ag+(aq) + e– Ag(s) +0.80 AgCl(s) + e– Ag(s) + Cl– (aq) +0.22 [Ag(NH3 )2 ]+(aq) + e– Ag(s) + 2NH3 (aq) +0.37 Al3+(aq) + 3e– Al(s) –1.66 Au3+(aq) + 3e– Au(s) +1.50 Ba2+(aq) + 2e– Ba(s) –2.90 Be2+(aq) + 2e– Be(s) –1.85 Br2 (aq) + 2e– 2Br– (aq) +1.07 Ca2+(aq) + 2e– Ca(s) –2.76 Cd2+(aq) + 2e– Cd(s) –0.40
251 Chemistry Term 2 STPM Appendix Electrode reaction Standard potential E°/V Ce4+(aq) + e– Ce3+(aq) +1.44 Cl2 (g) + 2e– 2Cl– (aq) +1.36 ClO– (aq) + H2 O(l) + 2e– Cl– (aq) + 2OH– (aq) +0.90 ClO2 – (aq) + H2 O(l) + 2e– ClO– (aq) + 2OH– (aq) +0.59 ClO3 – (aq) + H2 O(l) + 2e– ClO2 – (aq) + 2OH– (aq) +0.35 ClO4 – (aq) + H2 O(l) + 2e– ClO3 – (aq) + 2OH– (aq) +0.17 Co2+(aq) + 2e– Co(s) –0.28 Co3+(aq) + e– Co2+(aq) +1.82 [Co(NH3 )6 ]2+(aq) + 2e– Co(s) + 6NH3 (aq) –0.43 Cr2+(aq) + 2e– Cr(s) –0.91 Cr2 O7 2–(aq) + 14H+(aq) + 6e– 2Cr3+(aq) + 7H2 O(l) +1.33 Cr3+(aq) + 3e– Cr(s) –0.74 Cr3+(aq) + e– Cr2+(aq) –0.41 Cu+(aq) + e– Cu(s) +0.52 Cu2+(aq) + 2e– Cu(s) +0.34 Cu2+(aq) + e– Cu+(aq) +0.15 [Cu(NH3 )4 ]2+(aq) + 2e– Cu(s) + 4NH3 (aq) –0.05 F2 (g) + 2e– 2F– (aq) +2.87 Fe2+(aq) + 2e– Fe(s) –0.41 Fe3+(aq) + 3e– Fe(s) –0.04 Fe3+(aq) + e– Fe2+(aq) +0.77 [Fe(CN)6 ]3– + e– [Fe(CN)6 ]4–(aq) +0.36 FeO4 2–(aq) + 8H+ + 3e– Fe3+(aq) + 4H2 O(l) +2.20 H+(aq) + e– — 1 2 H2 (g) 0.00 H2 (g) + 2e– 2H– (aq) –2.25
252 Chemistry Term 2 STPM Appendix Electrode reaction Standard potential E°/V H2 O(l) + e– —1 2 H2 (g) + OH– (aq) –0.83 H2 O2 (aq) + 2H+(aq) + 2e– 2H2 O(l) +1.77 Hg+(aq) + e– Hg(l) +0.80 Hg2+(aq) + 2e– Hg(l) +0.85 Hg2+(aq) + e– Hg+(aq) +0.90 HNO2 (aq) + H+(aq) + e– NO(g) + H2 O(l) +0.99 HOCl(aq) + H+(aq) + e– — 1 2 Cl2 (g) + H2 O(l) +1.64 I2 (s) + 2e– 2I– (aq) +0.54 IO– (aq) + H2 O(l) + 2e– I2 (aq) + 2OH– (aq) +0.49 IO3 – (aq) + 6H+(aq) + 5e– —1 2 I2 (aq) + 3H2 O(l) +1.19 K+(aq) + e– K(s) –2.92 Li+(aq) + e– Li(s) –3.04 Mg2+(aq) + 2e– Mg(s) –2.38 Mn2+(aq) + 2e– Mn(s) –1.18 Mn3+(aq) + e– Mn2+(aq) +1.51 MnO2 (s) + 4H+(aq) + 2e– Mn2+(aq) + 2H2 O(l) +1.23 MnO4 – (aq) + 4H+(aq) + 3e– MnO2 (s) + 2H2 O(l) +1.67 MnO4 – (aq) + 8H+(aq) + 5e– Mn2+(aq) + 4H2 O(l) +1.49 MnO4 – (aq) + e– MnO4 2–(aq) +0.56 Na+(aq) + e– Na(s) –2.71 Ni2+(aq) + 2e– Ni(s) –0.25 [Ni(NH3 )6 ]2+(aq) + 2e– Ni(s) + 6NH3 (aq) –0.51 NO3 – (aq) + 10H+(aq) + 8e– NH4 +(aq) + 3H2 O(l) +0.87 NO3 – (aq) + 2H+(aq) + 2e– NO2 (g) + H2 O(l) +0.81 NO3 – (aq) + 3H+(aq) + 2e– HNO2 (aq) + H2 O(l) +0.94
253 Chemistry Term 2 STPM Appendix Electrode reaction Standard potential E°/V NO3 – (aq) + 4H+(aq) + 3e– NO(g) + 2H2 O(l) +0.96 NO3 – (aq) + 5H+(aq) + 4e– — 1 2 N2 (g) + — 5 2 H2 O(l) +1.11 NO3 – (aq) + 6H+(aq) + 5e– —1 2 N2 (g) + 3H2 O(l) +1.24 O2 (g) + 2H+(aq) + 2e– H2 O2 (l) +0.68 O2 (g) + 2H2 O(l) + 4e– 4OH– (aq) +0.40 O2 (g) + 4H+(aq) + 4e– 2H2 O(l) +1.23 O3 (g) + 2H+(aq) + 2e– O2 (g) + H2 O(l) +2.07 Pb2+(aq) + 2e– Pb(s) –0.13 Pb4+(aq) + 2e– Pb2+(aq) +1.69 PbO2 (s) + 4H+(aq) + 2e– Pb2+(aq) + 2H2 O(l) +1.47 S(s) + 2e– S2–(aq) –0.51 S(s) + 2H+(aq) + 2e– H2 S(g) +0.14 S2 O8 2–(aq) + 2e– 2SO4 2–(aq) +2.01 S4 O6 2–(aq) + 2e– 2S2 O3 2–(aq) +0.09 Sc3+(aq) + 3e– Sc(s) –2.08 Sn2+(aq) + 2e– Sn(s) –0.14 Sn4+(aq) + 2e– Sn2+(aq) +0.15 SO4 2–(aq) + 4H+(aq) + 2e– H2 SO3 (g) + H2 O(l) +0.17 Ti2+(aq) + 2e– Ti(s) –1.63 Ti3+(aq) + e– Ti2+(aq) –0.37 V2+(aq) + 2e– V(s) –1.20 V3+(aq) + e– V2+(aq) –0.26 VO2+(aq) + 2H+(aq) + e– V3+(aq) + H2 O(l) +0.34 VO2 +(aq) + 2H+(aq) + e– VO2+(aq) + H2 O(l) +1.00 VO3 – (aq) + 4H+(aq) + e– VO2+(aq) + 2H2 O(l) +1.00 Zn2+(aq) + 2e– Zn(s) –0.76
254 Chemistry Term 2 STPM Appendix Ionisation Energies (kJ mol–1) Z Name 1st 2nd 3rd 4th 5th 6th 7th 1 H 1310 2 He 2370 5250 3 Li 519 7300 11 800 4 Be 900 1760 14 800 21 000 5 B 799 2420 3660 25 000 32 828 6 C 1090 2350 4610 6220 37 830 42 270 7 N 1400 2860 4590 7480 9440 53 270 64 360 8 O 1310 3390 5320 7450 10 990 13 320 71 330 9 F 1680 3370 6040 8410 11 020 15 160 17 870 10 Ne 2080 3950 6150 9290 12 180 15 240 20 000 11 Na 494 4560 6940 9540 13 350 16 610 21 110 12 Mg 736 1450 7740 10 500 13 630 17 800 21 700 13 Al 577 1820 2740 11 600 14 830 18 380 23 290 14 Si 786 1580 3230 4360 16 090 15 P 1060 1900 2920 4960 3270 21 270 16 S 1000 2260 3390 4540 7010 8500 27 110 17 Cl 1260 2300 3850 5150 6540 9360 11 020 18 Ar 1520 2660 3950 5770 7240 8780 12 000 19 K 418 3070 4600 5860 20 Ca 590 1150 4940 6480 21 Sc 632 1240 2390 7110 8840 10 720 22 Ti 661 1310 2720 4170 9570 11 520 23 V 648 1370 2870 4600 6290 12 360 24 Cr 653 1590 2990 4770 6690 8740 25 Mn 716 1510 3250 5190 6990 9200 26 Fe 762 1560 2960 5400 7240 9600 27 Co 757 1640 3230 5100 7670 9840 28 Ni 736 1750 3390 5400 7280 10 400 29 Cu 745 1960 3550 5690 7710 9940 30 Zn 908 1730 3828 5980 7970 10 400 31 Ga 577 1980 2960 6190 32 Ge 762 1540 3300 4390 9020 35 Br 140 2080 3460 4850 5760 8550 9940 38 Sr 548 1060 4120 5440 50 Sn 707 1410 2940 3930 6980 53 I 1010 18 470 2040 4030 56 Ba 502 966 3390 82 Pb 716 1450 3080 4080 6640
255 Chemistry Term 2 STPM Appendix Electronegativity (Pauling’s Scale) 19 K 0.8 37 Rb 0.8 55 Cs 0.7 87 Fr 0.7 20 Ca 1.0 38 Sr 1.0 56 Ba 0.9 1 H 2.1 3 Li 1.0 11 Na 0.9 4 Be 1.5 12 Mg 1.2 21 Sc 1.3 39 Y 1.2 57 La 1.1 22 Ti 1.5 40 Zr 1.4 72 Hf 1.3 23 V 1.6 41 Nb 1.6 73 Ta 1.5 24 Cr 1.6 42 Mo 1.8 74 W 1.7 25 Mn 1.5 43 Tc 1.9 75 Re 1.9 26 Fe 1.8 44 Ru 2.2 76 Os 2.2 27 Co 1.9 45 Rh 2.2 77 Ir 2.2 28 Ni 1.9 46 Pd 2.2 78 Pt 2.2 29 Cu 1.9 47 Ag 1.9 79 Au 2.4 30 Zn 1.6 48 Cd 1.7 80 Hg 1.9 31 Ga 1.6 49 In 1.7 81 Tl 1.8 32 Ge 1.8 50 Sn 1.8 82 Pb 1.9 83 Bi 1.9 34 Se 2.4 52 Te 2.1 84 Po 2.0 35 Br 2.8 33 As 2.0 51 Sb 1.9 53 I 2.5 85 At 2.2 36 Kr 54 Xe 86 Rn 5 B 2.0 13 Al 1.5 6 C 2.5 14 Si 1.8 15 P 2.1 8 O 3.5 16 S 2.5 7 N 3.0 9 F 4.0 17 Cl 3.0 2 He 10 Ne 18 Ar 88 Ra 0.9
256 Alkaline earth metal Logam alkali bumi The Group 2 elements Alkaline metal Logam alkali The Group 1 elements Anode Anod The electrode where oxidation occurs Anodisation Penganodan An electrolytic passivation process used to increase the thickness of the natural oxide layer on the surface of metal Catalyst Mangkin A substance that changes the rate of a chemical reaction but remains chemically unchanged at the end of it Cathode Katod The electrode where reduction occurs Charge density Ketumpatan cas The charge over size ratio of an ion Complex ion Ion kompleks A central metal ion bonded to a group of atoms, molecules or ions by coordinate (dative) bonds Corrosion Kakisan The gradual destruction of material, usually metal, by a chemical reaction with its environment Covalent radius Jejari kovalen Half the distance between the centres of two adjacent atoms that are held by a covalent bond Decomposition potential Keupayaan penguraian The minimum voltage that is required before electrolysis can occur Diagonal relationship Hubungan penjuru A relationship within the periodic table by which certain elements in the second period have a close chemical similarity to their diagonal neighbours in the next group of the third period. For example, Li and Mg; Be and Al. Effective nuclear charge Cas nukleus berkesan The net positive charge experienced by an electron in a multi-electron atom Electrochemical cell Sel elektrokimia A device used for generating an electromotive force (voltage) and current from spontaneous redox reactions Electrolysis Elektrolisis Decomposition of an electrolyte by electrical current Electrolytic cell Sel elektrolisis A device that uses electricity to cause a nonspontaneous redox reaction to occur Electron affinity Afiniti elektron The energy released when an electron is added to a gaseous atom per mole of the atom under standard conditions Electronegativity Keelektronegatifan The relative tendency of an atom to attract electron in a covalent bond towards itself Endothermic reaction Tindak balas endotermik A reaction where there is a net gain in energy Energy Tenaga The ability to do work
257 Chemistry Term 2 STPM Glossary Enthalpy Entalpi A measure of the total energy of a thermodynamic system Enthalpy change of reaction Perubahan entalpi tindak balas The amount of energy absorbed or released in a chemical reaction for the number of moles of reactants as indicated by the chemical equation Exothermic reaction Tindak balas eksotermik A reaction where there is a net loss in energy Faraday’s first law of electrolysis Hukum elektrolisis pertama Faraday The amount of substance produced during electrolysis is directly proportional to the amount of electrical charge that flows through the electrolyte Faraday’s second law of electrolysis Hukum elektrolisis kedua Faraday The amount of substances produced during electrolysis by the same amount of electrical charge is inversely proportional to the charge of the ions First ionisation energy Tenaga pengionan pertama The minimum energy required to remove the most loosely held electron from a gaseous atom, per mole of the atom, under standard conditions Fuel cell Sel bahan api An electrochemical cell that converts the chemical energy from a continuous supply of reactants into electrical energy Galvanisation Penggalvanian The process of applying a protective zinc coating to steel or iron, in order to prevent rusting Group Kumpulan The vertical column in the Periodic Table Half-cell Setengah sel A structure consisting of an electrode in contact with its aqueous ion Halogen Halogen The Group 17 elements Heat capacity Muatan haba The amount of heat energy required to raise the temperature of a certain mass of a substance through 1 K Hess’ law Hukum Hess The total enthalpy change of a chemical reaction does not depend on how the reaction is completed Heterogeneous catalyst Mangkin heterogen A catalyst whose physical state is different from those of the reactants Homogeneous catalyst Mangkin homogen A catalyst whose physical state is the same as the reactants Inert pair effect Kesan pasangan lengai The increasing stability of oxidation states that there are 2 units less than the group valency for the heavier elements of Group 14, 15 and 16 Isoelectronic Isoelektronik Atoms, ions or molecules having the same number of electrons (and same electronic configuration) Lattice energy Tenaga kekisi The energy released when one mole of an ionic solid is formed from its gaseous ions under standard conditions Ligand Ligan Electron pair donors. They are Lewis bases. Metallic radius Jejari logam Half the distance between the centres of two adjacent atoms in a giant metallic structure
258 Chemistry Term 2 STPM Glossary Overpotential Keupayaan lampau The additional electric potential that has to be applied in an electrolysis cell in addition to the theoretical potential required for electrolysis to occur Oxidation Pengoksidaan A process of electron loss Oxidising agent Agen pengoksidaan An electron acceptor Period Kala The horizontal row in the Periodic Table Photochromatic lens Kanta fotokromik Lens that darkens when exposed to bright sunlight but turns clear again away from sunlight Redox reaction Tindak balas redoks A reaction involving the transfer of electrons from one species to another Reducing agent Agen penurunan An electron donor Reduction Penurunan A process of electron gain Screening/shielding effect Kesan pengskrinan The decrease in attraction between an electron and the nucleus in any atom with more than one electron shell Semiconductor Semikonduktor A substance that does not conduct electricity at room conditions but becomes conducting at elevated temperatures or when doped with other elements Standard electrode potential Keupayaan elektrod piawai The electrode potential between an electrode and its aqueous ions of 1.0 mol dm–3, with respect to that of the standard hydrogen electrode under standard conditions Standard enthalpy change of atomisation Perubahan entalpi pengatoman piawai The energy absorbed when one mole of free gaseous atoms is formed from an element under standard conditions Standard enthalpy change of combustion Perubahan entalpi pembakaran piawai The energy released when one mole of a substance is burned in excess oxygen under standard conditions Standard enthalpy change of formation Perubahan entalpi pembentukan piawai The enthalpy change when one mole of a substance is formed from its constituent elements under standard conditions Standard enthalpy change of hydration Perubahan entalpi penghidratan piawai The energy released when one mole of gaseous ions dissolved in water to form hydrated ions of infinite dilution under standard conditions Standard enthalpy change of neutralisation Perubahan entalpi peneutralan piawai The energy released when one mole of water is produced from the reaction between an acid and a base under standard conditions Standard enthalpy change of solution Perubahan entalpi keterlarutan piawai The enthalpy change when one mole of a substance dissolves in water to form aqueous solution of infinite dilution under standard conditions Standard ionisation energy Tenaga pengionan piawai The minimum energy required to remove the most loosely held electron from a gaseous atom or gaseous ion, per mole of the atom or ion under standard conditions Transition element Unsur peralihan An element that can form at least one stable ion with an incompletely filled d orbitals
259 Chapter 7 Chemical Energetics Quick Check 7.1 1 (a) Ag+ (aq) + Cl– (aq) → AgCl(s) ∆H° = –66 kJ mol–1 (b) +1.32 kJ 2 (a) +354 kJ (b) –885 kJ 3 (a) –8.55 kJ (b) 29.12 °C 4 (a) –100 kJ mol–1 (b) 45.02 kJ 5 (a) –150.3 kJ (b) –58.94 kJ (c) 970.1 g 6 (a) 831.82 kJ (b) –2868.34 kJ Quick Check 7.2 1 (a) Ca(s) + C(s) + —3 2 O2(g) → CaCO3(s) (b) Cu(s) + S(s) + 5H2(g) + —9 2 O2(g) → CuSO4•5H2O(s) (c) 2C(s) + 2H2(g) + O2(g) → CH3COOH(l) (d) —1 2 N2(g) + O2(g) → NO2(g) (e) —1 2 H2(g) + —1 2 I2(s) → HI(g) (f) C(s) + 2Cl2(g) → CCl4(l) (g) Cd(s) + 2C(s) + N2(g) → Cd(CN)2(s) Quick Check 7.3 1 –876.0 kJ 2 +127 kJ mol–1 Quick Check 7.4 1 (a) Na(s) + —1 4 O2(g) → —1 2 Na2O(s) (b) H2C2O4(s) + —1 2 O2(g) → 2CO2(g) + H2O(l) (c) SO2(g) + —1 2 O2(g) → SO3(g) (d) NH3(g) + —3 4 O2(g) → —1 2 N2(g) + —3 2 H2O(l) 2 –720 kJ mol–1 3 91.65 °C 4 (a) 23.76 kJ (b) 570.24 kJ mol–1 5 (a) C2H5OH + 3O2 → 2CO2 + 3H2O (b) –1337.6 kJ mol–1 (c) 2.91 × 104 kJ Quick Check 7.5 1 29.1 °C 2 (a) CH3COOH + NaOH → CH3COONa + H2O (b) 1421.2 kJ (c) –56.8 kJ mol–1 Quick Check 7.6 1 –110 kJ mol–1 2 –110 kJ mol–1 3 (a) C(Diamond) C(Graphite) –394 kJ CO2 –396 kJ (b) –2 kJ mol–1 (c) Graphite 4 –848 kJ 5 +45.0 kJ mol–1 6 (a) 2C(s) + —1 2 O2(g) + 3H2(g) → C2H5OH(l) (b) –279 kJ mol–1 7 –1260 kJ mol–1 8 –334.6 kJ 9 (a) C2H5OH(l) + —1 2 O2(g) → CH3CHO(l) + H2O(l) (b) –200.2 kJ mol–1 10 –466 kJ mol–1 Quick Check 7.7 1 (a) Na+ (g) + Br– (g) → NaBr(s) (b) Ag+ (g) + Cl– (g) → AgCl(s) (c) 2Na+ (g) + O2–(g) → Na2O(s) (d) Ca2+(g) + 2F– (g) → CaF2(s) (e) 2Al3+(g) + 3O2–(g) → Al2O3(s) Quick Check 7.8 1 (a) MgO (b) MgSO4 (c) Al2O3 (d) LiF (e) Na2O Quick Check 7.9 1 –2050.6 kJ mol–1 2 –494 kJ mol–1 3 –325 kJ mol–1 Quick Check 7.10 1 (a) –18 kJ mol–1 (b) Soluble 2 –86.7 kJ mol–1 3 12.8 °C 4 (a) +81 kJ mol–1 (b) Insoluble STPM Practice 7 Objective Questions 1 B 2 B 3 B 4 A 5 A 6 D 7 B 8 C 9 B 10 C 11 D 12 A 13 B 14 C 15 C 16 C 17 C 18 A 19 B 20 D 21 B 22 B 23 B 24 A 25 C 26 C 27 D 28 A 29 B 30 C 31 C 32 A 33 B 34 A 35 D 36 B 37 C 38 A 39 B Structured and Essay Questions 1 (a) Ag+ (g) + Br– (g) → Ag+ Br– (s) (b) (i) (+285) + (+112) + (+731) + (–325) + ∆H = –100 ∆H = –903 kJ mol–1 (ii) Silver bromide has significant covalent character. (iii) Less exothermic. The size of I– is larger than Br– . (c) Less. The size of I– is larger than that of Br– . The electron is less strongly held by the nucleus. ANSWERS
260 Chemistry Term 2 STPM Answers 2 (a) Born-Haber cycle (b) ∆H1 = Enthalpy of atomisation of Mg ∆H2 = (First + second) ionisation energy of Mg ∆H3 = Enthalpy of atomisation of oxygen ∆H4 = (First + second) electron affinity of oxygen ∆H5 = Lattice energy of MgO ∆H6 = Enthalpy of formation of MgO (c) (+150) + (736 + 1450) + (+248) + (+702) + ∆H5 = –603 ∆H5 = –3889 kJ mol–1 (d) Less. The size of Ca2+ is larger than Mg2+. 3 (a) The standard enthalpy change of formation of a compound is the heat change when one mole of the compound is formed from its constituent elements under standard conditions. (b) (i) 1 ––2N2(g) + 1 ––2O2(g) → NO(g) ∆H = + 90 kJ mol–1 (ii) A lot of energy is required to break the strong triple bond in the N N molecules. (c) 2Al(s) + 6HCl(aq) → Al2Cl6(aq) + 3H2(g) –1007 3H2(g) + 3Cl2(g) → 6HCl(g) 3(–185) 6HCl(g) → 6HCl(aq) 6(–73) Al2Cl6(aq) → Al2Cl6(s) +646 —————————————————————– 2Al(s) + 3Cl2(g) → Al2Cl6(s) ∆H = –1354 kJ mol–1 4 (a) The heat energy released when one mole of a substance is completely burned in excess oxygen under standard conditions. (b) (i) C2H6(g) + —7 2 O2(g) → 2CO2(g) + 3H2O(l) (ii) q = mc∆T = 40.5 × 4.2 × (85 – 25) = 10 206 kJ No. of moles of C2H6 required= ——100 80 × ——— 10 206 1540 = 8.28 mol Volume required = 8.28 × 24 dm3 = 198.72 dm3 (c) (i) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) H2(g) + —1 2 O2(g) → H2O(l) (ii) For CH4: ∆H = – —– 1 16 × 890 kJ = –55.63 kJ g–1 For H2: ∆H = —1 2 × (–286) = –143 kJ g–1 (iii) Weight by weight, hydrogen is a better fuel than methane. 5 (a) (i) –2 (ii) N2H4(l) + O2(g) → N2(g) + 2H2O(g) (iii) Q = 3.5 —–32 × (–540) = –59.1 kJ Temperature change = —–– 59.1 6.25 = 9.46 °C (b) (i) N2(g) + 2H2(g) → N2H4(l) (ii) N2H4(l) + O2(g) → N2(g) + 2H2O(g) ∆H = –540 kJ ∆Hf : a 0 0 2(–286 + 44) 2(–286 + 44) – a = –540 a = +56 kJ mol–1 6 (a) Lattice energy is the energy released when one mole of an ionic solid is formed from its constituent gaseous ions under standard conditions. (b) There is close agreement between the theoretical (assuming that the bond is 100% ionic) and experimental values for NaCl. On the other hand, there is a large difference between the theoretical and experimental values for AgCl. This is because the difference in electronegativity between Na and Cl is large. Hence, the bond in NaCl is 100% ionic. However, the difference in electronegativity between Ag and Cl is small leading to a significant amount of covalent character in AgCl. The presence of covalent character in AgCl causes the discrepancy in the theoretical and experimental values. 7 (a) Heat change when 1 mole of a substance is burned completely in oxygen under standard conditions. (b) (i) C8H18 + —– 25 2 O2 → 8CO2 + 9H2O (ii) C8H18 + —– 25 2 O2 → 8CO2 + 9H2O ∆Hf °: –250 0 8(–394) 9(–286) ∆H = [8(–394) + 9(–286)] – [–250] = –5476 kJ mol–1 8 (a) (i) Na2CO3 + H2SO4 → Na2SO4 + H2O + CO2 (ii) Q = 30 × 4.18 × 5.2 = 652.08 J (released) ∆H = – 652.08 × ——–– 1 0.020 = –32 604 J mol–1 (b) (i) NaHCO3 + —1 2 H2SO4 → —1 2 Na2SO4 + H2O + CO2 (ii) Q = 30 × 4.18 × 3.9 = 489.06 J (absorbed) ∆H = +489.06 × ——–– 1 0.020 = + 24 453 J mol–1 (c) (i) 2NaHCO3 → Na2CO3 + CO2 + H2O (ii) 2NaHCO3 ⎯→ Na2CO3 + CO2 + H2O Na2SO4 + 2CO2 + 2H2O 2(+24 453) –32 604 x By Hess’ Law: x – 32 604 = 2(24 453) x = +81 510 J = +81.51 kJ 9 (a) Heat changes when one mole of ethane is formed from carbon and hydrogen under standard conditions. 2C(s) + 3H2(g) → C2H6(g) (b) 2C(s) + 3H2(g) ⎯→ C2H6(g) 2CO2 + 3H2O 2(–394) + 3(–286) –1560 x By Hess’ Law: x – 1560 = –1646 x = –86 kJ mol–1 10 (a) (i) C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) (ii) Heat absorbed by water = mc∆T = 200 × 4.18 × (75 – 25) = 41 800 J = 41.8 kJ Heat released by ethanol = ——100 68 × 41.8 kJ = 61.48 kJ ∆H° = –—– 46 2 × 61.48 kJ mol–1 = –1414 kJ mol–1
261 Chemistry Term 2 STPM Answers (b) C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ∆Hf : a 0 2(–394) 3(–286) 2(–394) + 3(–286) – a = –1414 a = –232 kJ mol–1 11 (a) 2AgClO3 + Cl2 → 2AgCl + 2ClO2 + O2 ∆Hf : 2(–25) 0 2(–125.8) 2a 0 2(–125.8) + 2a – 2(–25) = 0 a = +100.8 kJ mol–1 (b) Chlorine dioxide is energetically unstable compared to its elements, chlorine and oxygen. Hence, we would expect chlorine dioxide to decompose when heated. 2ClO2(g) ⎯→ Cl2(g) + 2O2(g) ∆ 12 N2O4 2NO2 ∆Hf : +9.7 2a 2a – 9.7 = 57.0 a = +33.4 kJ mol–1 13 (a) The total heat change of a chemical reaction does not depend on how the reaction is completed. (b) Set I: Zn(s) + S(s) → ZnS(s) –167 ZnS(s) + 2O2(g) → ZnSO4(s) –795 —————————————————————– Zn(s) + S(s) + 2O2(g) → ZnSO2(s) ∆H° = –962 kJ Set II: Zn(s) + —1 2 O2(g) → ZnO(s) –355 S(s) + O2(g) → SO2(g) –293 SO2(g) + —1 2 O2(g) → SO3(g) –92 ZnO(s) + SO3(g) → ZnSO4(s) –222 —————————————————————– Zn(s) + S(s) + 2O2(g) → ZnSO4(s) ∆H° = –962 kJ mol–1 The total heat change for Set I is the same as that of Set II. Hence, Hess’ law is verified. 14 (a) CO(NH2)2 + H2O → CO2 + 2NH3 ∆Hf : –320 –286 –414 2(–81) ∆H = –414 + 2(–81) – (–320 – 286) = +30 kJ mol–1 (b) (i) 4CH3NHNH2 + N2O4 → 4CO2 + 9N2 + 12H2O ∆Hf : 4(56.2) –20 4(–394) 0 12(–286) ∆H = 4(–394) + 12(–286) –4(56.2) + 20 = –5212.8 kJ (ii) A lot of energy is released during the reaction. A large amount (volume) of gaseous products is formed. 15 (a) Q = 50 × 4.18 × (35 – 27) = 1672 J (b) CuSO4(s) → CuSO4(aq) ∆H = –159.6 —–– 4 × 1672 × 10–3 kJ mol–1 = – 66.71 kJ mol–1 (c) Q = 50 × 4.18 × (27 – 25.7) = 271.7 J (d) CuSO4•5H2O(s) → CuSO4(aq) ∆H = + —–— 249.6 4 × 271.7 × 10–3 = +16.95 kJ mol–1 (e) CuSO4(s) + 5H2O(l) ⎯→ CuSO4•5H2O(s) CuSO4(aq) –66.71 +16.95 a a + 16.95 = –66.71 a = –83.66 kJ mol–1 16 (a) C(s) + 1 2 O2(g) → CO(g) ∆H = -99 kJ molC(s) + O2(g) → CO2(g) ∆H = -294 kJ mol-1 (b) (i) Carbon monoxide: (Linear) :C#O: Carbon dioxide: (Linear) O"C"O (ii) Two carbon-oxygen double bond is stronger than one carbon-oxygen triple bond. (c) 2CO + O2 ⎯→ 2CO2 2C + 2O2 –99 × 2 –394 x Based on Hess’ law: x + (-99 × 2) = -394 x = -196 kJ 17 (a) K+ (g) + H– (g) → K+ H– (s) (b), (c) K+ (g) K(g) +418 1 2 +90 +78 ∆Ho lattice K+ H– (s) K(s) H– (g) –78 +218 H2(g) H(g) Based on Hess’ law: 90 + 418 + 218 – 78 + ∆H = +78 ∆Hlattice = –570 kJ mol–1 (d) More exothermic. The Na+ ion is smaller than that of K+ . 18 (a) Sr(s) + Cl2(g) → SrCl2(s) (b) Sr2+(g) Sr+ (g) +1064 –829 X CaCl2(s) 2Cl– (g) 2(–349) +242 Cl2(g) Sr(g) 2Cl(g) Sr(s) +550 +165 By Hess law: 165 + 550 + 1064 + 242 – 2(349) + x = –829 x = –2152 kJ mol–1 (c) (i) (I) SrCl is stable with respect to its elements. SrCl3 is unstable with respect to its elements. (II) SrCl is less stable compared to SrCl2. SrCl3 is unstable compared to SrCl3. (ii) More exothermic. This is due to the high charge density of the smaller Sr3+ ion.
262 Chemistry Term 2 STPM Answers 19 (a) Na+(g) + e– + Cl(g) Na+(g) + Cl– (g) Na(g) + Cl(g) 1 Na(g) + —Cl2 (g) 2 1 Na(s) + —Cl2 (g) 2 Na+Cl– (s) (b) The enthalpy of formation of sodium chloride is exothermic. It is stable under standard conditions. 20 (a) Hess’ law states that the enthalpy change of a chemical reaction is independent of its pathway. (b) Ag+(g) + e– + I(g) Ag+(g) + I – (g) Ag(g) + I(g) 1 Ag(g) + —I2 (s) 2 1 Ag(s) + —I 2 (s) 2 Ag+I– (s) 21 (a) As the size of the cations increases from Li+ to Na+ and K+ , the charge density decreases. As a result, the bonds form between the ions and water molecules get weaker, and the process becomes less exothermic. (b) The standard enthalpy of atomisation of hydrogen fluoride refers to the following process: H–F(g) → H(g) + F(g) ∆H = +562 kJ mol–1 Hence, the H–F bond energy is +562 kJ mol–1. Energy required to break one H–F bond = + 562 ————–— 6.02 × 1023 kJ = 9.3 × 10–22 kJ Chapter 8 Electrochemistry Quick Check 8.1 (a) (b) (c) (d) 1 R.A. Mg Sn2+ Sn2+ I – O.A. Fe2+ CI2 Fe3+ S2O8 2– [R.A.: Reducing agent; O.A.: Oxidising agent] Quick Check 8.2 1 +6 6 +2.5 2 +6 7 +2 3 +7 8 +1 4 +4 9 +6 5 +5 10 +2 Quick Check 8.3 Element being reduced Element being oxidised Oxiding agent Reducing agent (a) Cr N Cr2O7 –2 NO2 – (b) Mn O MnO4 – H2O2 (c) N I HNO3 I2 (d) S I H2SO4 HI (e) Co O Co3+ H2O (f) I O IO3 – H2O2 Quick Check 8.4 1 Yes 6 Yes 2 Yes 7 Yes 3 No 8 Yes 4 No 9 No 5 Yes 10 Yes Quick Check 8.5 1 (a) +7 (e) +2 (b) +4 (f) +1 (c) +5 (g) +6 (d) +3 2 (a) 2MnO4 – + 5C2O4 2– + 16H+ → 2Mn2+ + 10CO2 + 8H2O (b) VO3 – + 6H+ + 2Fe2+ → V3+ + 2Fe3+ + 3H2O (c) 2S2O3 2– + I2 → S4O6 2– + 2I– (d) Cr2O7 2– + 3H2O2 + 8H+ → 2Cr3+ + 7H2O + 3O2 (e) N2H5 + + 2Cl2 → N2 + 5H+ + 4Cl– (f) 2IO3 – + 5Co + 12H+ → 5Co2+ + I2 + 6H2O (g) 2MnO2 + 3H2O2 → 2MnO4 – + 2H+ + 2H2O (h) 4MnO4 – + 2H2O + 3Pb → 3PbO2 + 4MnO2 + 4OH– (i) Cl2 + IO3 – + H2O → IO4 – + 2H+ + 2Cl– Quick Check 8.6 1 (a) V Pt [Mn2+] = 1 M H2 (g, 1 atm) H+ (aq, 1 M) Pt [H+ ] = 1 M [MnO– 4] = 1 M
263 Chemistry Term 2 STPM Answers (b) V AgCl(s) H2 (g, 1 atm) H+ (aq, 1 M) Pt Cl– (aq,1 M) Ag (c) V Pt H2 (g, 1 atm) [H+ ] = 1.0 M Pt [H+ ] = 1 M [H2 O2 ] = 1 M (d) V Hg(l) Cl– (aq, 1 M) H2 (g, 1 atm) [H+ ] = 1.0 M Pt Hg2 Cl2 (s) Pt 2 Potassium would react with water. 2K+ (s) + 2H2O(l) → 2KOH(aq) + H2(g) 3 So that only a minimum amount of current is drawn out from the cell. This is to ensure that the change in the concentration of the ions is negligible. Quick Check 8.7 1 (a) Mg | Mg2+ || Zn2+ | Zn (b) Mg → Mg2+ + 2e– Zn2+ + 2e– → Zn (c) Mg + Zn2+ → Mg2+ + Zn (d) 1.62 V 2 (a) Pt | Sn2+, Sn4+ || Cl2, Cl– | Pt (b) Sn2+ → Sn4+ + 2e– Cl2 + 2e– → 2Cl– (c) Cl2 + Sn2+ → 2Cl– + Sn4+ (d) 1.21 V 3 (a) Pt | Br– , Br2 || MnO4 – , H+ , Mn2+ | Pt (b) 2Br– → Br2 + 2e– MnO4 – + 8H+ + 5e– → Mn2+ + 4H2O (c) 2MnO4 – + 16H+ + 10Br– → 5Br2 + 2Mn2+ + 8H2O (d) 0.45 V 4 (a) Pt | S2O3 2–, S4O6 2– || I2, I– | Pt (b) 2S2O3 2– → S4O6 2– + 2e– I2 + 2e– → 2I– (c) 2S2O3 2– + I2 → S4O6 2– + 2I– (d) 0.45 V 5 (a) Cr | Cr3+ || Fe2+ | Fe (b) Cr → Cr3+ + 3e– Fe2+ + 2e– → Fe (c) 2Cr + 3Fe2+ → 2Cr3+ + 3Fe (d) 0.30 V Quick Check 8.8 1 Oxidising agent: Fe3+ Cr2O7 2– MnO4 – H2O2 F2 Reducing agent: F– H2O Mn2+ Cr3+ Fe2+ 2 (a) Cr3+ I2 Fe3+ Ag+ Cl2 (b) Cl– Ag Fe2+ I– Cr Quick Check 8.9 1 2 3 4 5 6 E°cell /V –0.03 +0.75 +0.41 –0.13 +0.34 +0.37 Feasibility No Yes Yes No Yes Yes Quick Check 8.10 1 No 4 No 2 Yes 5 Yes 3 Yes 6 Yes Quick Check 8.11 1 (a) Ecell = – 0.13 V (b) Use concentrated HCl with heat. 2 (a) Species MnO4 – MnO2 MnO4 2– Oxidation state +7 +4 +6 (b) No. MnO4 – is being reduced, while MnO2 is being oxidised. (c) Use concentrated KOH with heat. Quick Check 8.12 1 (a) [Mn(H2O)6] 2+ (b) O2 + 4H+ + 4e– 2H2O E° = +1.23 V Mn3+ + e– Mn2+ E° = +1.51 V Mn3+ will oxidise water to oxygen. 4Mn3+ + 2H2O → 2Mn2+ + O2 + 4H+ E° = +0.28 V 2 (a) Mn2+ and Co2+ (b) Ti3+, V3+, Cr3+, Fe3+ Quick Check 8.13 1 (a) Increases (b) Decreases 2 (a) Decreases (b) Increases (c) Increases 3 (a) Increases (b) Decreases (c) No change Quick Check 8.14 1 (a) Zn | Zn2+ || Fe3+, Fe2+ | Pt (b) 1.53 V (c) Zn + 2Fe3+ → Zn2+ + 2Fe2+ (d) (i) Increases (iii) Increases (ii) Increases (iv) Decreases 2 (a) Fe3+ + e– Fe2+ +0.77 V Ag+ + e– Ag +0.80 V (b) From Fe3+/Fe2+ half-cell to Ag+ /Ag half-cell (c) Cathode: Ag+ /Ag Anode: Fe3+/Fe2+ (d) 0.030 V (e) Increase the concentration of Ag+ . Decrease the concentration of Fe3+. Quick Check 8.15 1 (a) +0.356 V (c) +0.752 V (b) –0.177 V (d) +1.326 V 2 22.7 3 0.0202 mol dm–3
264 Chemistry Term 2 STPM Answers 4 (a) (b) [Zn2+] E/V log [Zn2+] E/V 5 (a) (b) [Cl – ] E/V log [Cl – ] E/V Quick Check 8.16 1 (a) –0.0295 V (b) –0.245 V (c) +1.44 V Quick Check 8.17 1 2.52 × 10–3 mol dm–3 2 (a) 0.665 V (b) 0.0347 V (c) 0.0314 V 3 6.91 4 (a) (b) log [Zn2+] E.m.f log [Cu2+] E.m.f Quick Check 8.18 1 (a) 1.08 × 1022 (d) 1.87 × 106 (b) 7.32 × 1051 (e) 1.08 × 1062 (c) 1.17 × 104 2 2.18 Quick Check 8.19 1 (a) So that the solution is saturated. (b) 0.504 V (c) 1.0 × 10–5 mol dm–3 (d) 1.0 × 10–10 mol2 dm–6 Quick Check 8.20 1 (a) 1.48 V (b) 0.617 V 2 pH = 1.36 Quick Check 8.21 1 H2 and O2 3 H2 and I2 2 H2 and O2 4 H2 and O2 Quick Check 8.22 1 The difference in E° values between the two competing species is too large. F2 + 2e– 2F– +2.87 V O2 + 4H+ + 4e– 2H2O +1.23 V Furthermore, if F2 is produced it will oxidise water to oxygen. 2F2 + 2H2O → 4HF + O2 2 Consider the two E° values: Cl2 + 2e– 2Cl– +1.36 V O2 + 4H+ + 4e– 2H2O +1.23 V Water will be oxidised in preference to Cl– . However, as more water is oxidised, the concentration of Cl– ions increases causing its E° value to decrease, and hence will eventually be discharged. Quick Check 8.23 1 The oxygen liberated at the anode reacts with graphite to produce oxides of carbon. 2H2O → O2 + 4H+ + 4e– C + O2 → CO2 2C + O2 → 2CO 2 Lead, oxygen and oxides of carbon 3 The copper anode dissolves while copper gets deposited on the cathode. Quick Check 8.24 1 (a) 0.68 × 4 g (c) 2 × 0.68 g (b) 0.68 g (d) 0.68 g 2 66.3 cm3 3 +3 4 63 hours, 40 minutes and 48.7 seconds 13.3 dm3 Quick Check 8.25 1 0.448 g 2 +3 STPM Practice 8 Objective Questions 1 B 2 B 3 C 4 B 5 B 6 B 7 A 8 B 9 D 10 D 11 C 12 C 13 D 14 C 15 C 16 B 17 B 18 C 19 A 20 A 21 B 22 D 23 B 24 B 25 D 26 A 27 C 28 B 29 C 30 D 31 B 32 B 33 B 34 C 35 A 36 B 37 B 38 B 39 C 40 C 41 C 42 A 43 B 44 C 45 A 46 A 47 D 48 C 49 D 50 A 51 C 52 B 53 C 54 A 55 A 56 B 57 B 58 C 59 B 60 B 61 D 62 D 63 C Structured and Essay Questions 1 (a) It is the e.m.f. of a cell consisting of the electrode in contact with its aqueous ions of concentration 1.0 mol dm–3, and the standard hydrogen electrode at 298 K and 101 kPa.
265 Chemistry Term 2 STPM Answers (b) V Zn [Zn2+] (aq) = 1.0 M H2 (1 atm) [H+ ] (aq) = 1.0 M Salt bridge Pt (c) Calcium will react with water. Ca(s) + 2H2O(l) → Ca(OH)2(aq) + H2(g) 2 (a) H2 (1 atm) Pt H+ (aq, 1 M) (b) (i) The cell reaction is: H2(g) + 2H+ (1.0 M) → 2H+ (1 × 10–5 M) + H2(g) Using the Nernst equation: E = E° – 0.059 2 log (1 × 10–5) 2 12 = 0 – (–0.295) = 0.295 V (Note: This is an example of a concentration cell.) (ii) The cell reaction is: Zn(s) + 2H+ (aq, 1 M) → Zn2+(aq, 1 M) + H2(g, 0.04 atm) Using the Nernst equation: E = E° – 0.059 2 log 0.04 = 0.76 – (– 0.041) = 0.801 V 3 (a) A device that converts chemical energy to electrical energy (b) (i) V M M 2+(aq) H2 (g, 1 atm) [H+ ] (aq) = 1.0 M Pt Electron (ii) H2(g) + M2+(aq) → M(s) + 2H+ (aq) (iii) From platinum to M (iv) At the anode (Hydrogen half-cell), the concentration of H+ will increase. At the cathode (M2+/M half-cell), the concentration of M2+ will decrease. (v) E.m.f. = E°(M2+/M) – E°(hydrogen half-cell) 0.56 = E°(M2+/M) – 0 ∴ E°(M2+/M) = +0.56 V (vi) E.m.f. = E°(M2+/M) – E°(hydrogen half-cell) 0.62 = 0.56 – E(H+ /H2) E(H+ /H2) = –0.06 V Using Nernst equation: E = E° + 0.059 2 log[H+ ] 2 –0.06 = 0 – 0.059 pH ∴ pH = 1.02 4 (a) The activation energy of the reaction is too high. The particles, at standard conditions, do not have enough energy to overcome the activation energy. (b) Oxidising agents are electron acceptors. [Fe(H2O)6] 3+ + e– [Fe(H2O)6] 2+ E° = +0.77 V [Fe(CN)6] 3– + e– [Fe(CN)6] 4– E° = +0.36 V It is more difficult for an anion, such as [Fe(CN)6] 3–, to accept electron because of mutual repulsion. As a result, the oxidising strength is less than that of [Fe(H2O)6] 3+ which do not experience such repulsion. (c) Potassium will react with water. 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g) (d) In acidified conditions, the purple MnO4 – ion is reduced to the almost colourless Mn2+ ion. But in aqueous solution, the purple MnO4 – ion is reduced to the brown solid, manganese(IV) oxide, MnO2. 5 (a) A fuel cell is a special kind of electrochemical cell which converts the chemical energy from a continuous supply of reactants into electrical energy. (b) (i), (ii) V H2 SO4 (aq) Pt C2 H5 OH(aq) + H2 SO4 (aq) Pt O2 (g) Porous partition (Anode) (Cathode) (iii) C2H5OH → CH3CHO + 2H+ + 2e– O2 + 4H+ + 4e– → 2H2O (iv) 2C2H5OH + O2 → 2CH3CHO + 2H2O (v) E.m.f. = (+1.23) – (+0.20) = 1.03 V 6 (a) V Platinum (Cathode) Platinum (Anode) Salt bridge Fe3+(aq;I.0 M) Fe2+(aq;I.0 M) Sn4+(aq;I.0 M) Sn2+(aq;I.0 M) (b) Cathode: Fe3+ + e– ⎯→ Fe2+ Anode: Sn2+ ⎯→ Sn4+ + 2e– Overall cell reaction: 2Fe3+ + Sn2+ → Sn4+ + 2Fe2+ E.m.f. = E°(Cathode) – E°(Anode) = 0.77 – 0.15 = +0.62 V (c) Using: E° = 0.059 z log KC 0.62 = 0.059 2 log KC KC = 1.04 × 1021
266 Chemistry Term 2 STPM Answers (d) The cell reaction: 2Fe3+ + Sn2+ ⎯→ 2Fe2+ + Sn4+ On the addition of a little oxidising agent to the Fe3+/ Fe2+ half-cell causes the Fe3+ concentration to increase. The above reaction will shift to the right-hand side causing the e.m.f. to increase. 7 (a) + – Na2 SO4 (aq) (b) At the cathode: 2H2O(l) + 2e– → 2OH– (aq) + H2(g) At the anode: 2H2O(l) → O2(g) + 4H+ (aq) + 4e– (c) The net reaction is the decomposition of water to its elements, hydrogen and oxygen which are in the mole ratio of 2 : 1. (d) Cathode: pH increases (OH– is formed) Anode: pH decreases (H+ is formed) (e) No. of moles of H2 = 35.6 22 400 = 1.59 × 10–3 mol Quantity charge used = 2(1.59 × 10–3) × 96 500 C = 306.9 C Using, Q = It 306.9 = I × (30 × 60) \ I = 0.17 A 8 When zinc is immersed in aqueous zinc sulphate, the zinc atom from the zinc plate will dissolve to form zinc(II) ions. Zn(s) → Zn2+(aq) + 2e– At the same time, the zinc(II) ion from solution will pick up electrons from the zinc plate and get reduced to zinc atoms. Zn2+(aq) + 2e– → Zn(s) At equilibrium: Zn(s) Zn2+(aq) + 2e– This sets up a potential difference between the zinc plate and its aqueous ions. This is called electrode potential. 9 (a) Pt(s) | Hg(l), Hg2Cl2(s), Cl– (aq) || Fe3+(aq), Fe2+(aq) |Pt(s) (b) X (c) E.m.f. = 0.77 – 0.24 = 0.53 V (d) 2Hg(l) + 2Fe3+(aq) + 2Cl– (aq) → Hg2Cl2(s) + 2Fe2+(aq) (e) (i) Volume of Z 0.53 E.m.f (ii) The Nernst equation is given by: E = 0.53 – 0.059 2 log [Fe2+] 2 [Fe3+] 2 [Cl– ] 2 Z will oxidise Fe2+ to Fe3+. This causes the ratio [Fe2+] 2 [Fe3+] 2 [Cl– ] 2 to decrease. As a result, the e.m.f. increases. (iii) The Nernst equation for the Fe3+/Fe2+ half-cell is E = +0.77 – 0.059 log [Fe2+] [Fe2+] = +0.77 – 0.059 log 1 1 × 1010 = 1.36 V E.m.f. = 1.36 – 0.24 = 1.12 V 10 (a) The electrolytic extractioin of aluminium from purified bauxite is carried out using the following electrolytic cell. Graphite Electrolyte Outlet Steel container Graphite Liquid aluminium + – The electrolyte is a molten mixture of bauxite and cryolite (Na3AlF6) maintain at 900°C. The cathode and anode are graphite. Al2O3(s) ⎯→ 2Al3+(l) + 3O2-(l) At the cathode: Al3+(l) + 3e– ⎯→ Al(l) The molten aluminium sinks to the bottom of the cell and is drained off from time to time and fresh bauxite is added at the top. At the anode: 2O2-(l) ⎯→ O2(g) + 4e– The cell operates at a low voltage of 5 – 6 volts but a large current of about 100,000 A. The heat generated by the high current helps to keep the cell at 950°C (so that the electrolyte does not solidify). At the temperature of the cell, the carbon anode burns in the oxygen liberated to produce carbon monoxide and carbon dioxide: C(s) + 1 2 O2(g) ⎯→ CO(g) C(s) + O2(g) ⎯→ CO2(g) Hence, constant replacement is required. (b) The extraction of aluminium is expensive due to the high current involved. The CO, CO2, F2 and fluorine compounds (from the cryolite) released contribute to atmospheric pollution. Vegetation has to be cut down and the land cleared to build the extraction plant and roads for the transportation of the aluminium. Recycling of aluminium uses only about 5% of the energy used during extraction. Recycling also cuts down environmental pollution from discarded aluminium objects. (c) Kidney and liver demage 11 (a) Faraday’s First Law of electrolysis states that the amount of substance produced during electrolysis is directly proportional to the amount of electrical charge that flows through the electrolyte. Faraday’s Second Law states that the amount of substances produced during electrolysis by the same amount of electricity is inversely proportional to the charge on the ions.
267 Chemistry Term 2 STPM Answers (b) + Copper cathode (–) Cu2 SO4 (aq) Copper anode (+) – A piece of copper plate is cleaned and weighed. It is used as the cathode. An electrolytic cell is set up as shown above. The circuit is closed and the rheostat adjusted to give a steady flow of current through the cell for a time interval of t seconds. The cathode is removed, and rinsed with water, then with propanone, dried and weighed. Magnitude of current = I amp Duration of electrolysis = t seconds Increase in mass of cathode = m g Sample calculation: Cu2+(aq) + 2e– → Cu(s) Quantity of charge = It C It C = m g of Cu ∴ 63.5 g of Cu = 63.5 m × It C 63.5 m × It C = charge on 2 moles of electrons. Charge on 1 mole of electrons = 1 2 × 63.5 m × It C No. of electrons in one mole of electrons = 1 2 × 63.5 m × It × 1 e (Where e = charge on one electron) ∴ Avogadro’s constant = 1 2 × 63.5 m × It × 1 e × mol–1 12 (a) X = Anode (+) Y = Cathode (–) (b) Aqueous copper(II) sulphate (c) Electrode X: Cu(s) → Cu2+(aq) + 2e– Electrode Y: Cu2+(aq) + 2e– → Cu(s) (d) X diminishes in size. Y increases in size. (e) Zinc will dissolve to form Zn2+(aq) ions. Zn(s) → Zn2+(aq) + 2e– Silver will collect at the base of the anode as anode sludge. 13 (a) Zn(s) + 2H+ (aq) → Zn2+(aq) + H2(g) (b) E.m.f. = E°(H+ /H2) – E°(Zn2+/Zn) 0.76 V = 0 – E°(Zn2+/Zn) \ E°(Zn2+/Zn) = –0.76 V (c) (i) The piece of zinc diminishes in size. A brown solid is formed. The blue colour of aqueous copper(II) sulphate slowly fades. (ii) Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) E° = +0.34 – (–0.76) V = +1.10 V 14 (a) Brine (saturated sodium chloride) (b) (i) 2Cl– (aq) → Cl2(g) + 2e– (ii) 2H2O(l) + 2e– → H2(g) + 2OH– (aq) (c) To prevent the products of electrolysis from mixing (d) Aqueous sodium hydroxide (e) (i) Cl2(g) + 2NaOH(aq) → NaCl(aq) + NaOCl(aq) + H2O(l) (ii) 3Cl2(g) + 6NaOH(aq) → 5NaCl(aq) + NaClO3(aq) + 3H2O(l) (f) Domestic bleach 15 (a) 1s 2 2s 2 2p6 3s 2 3p6 3d10 (b) White. This is because the 3d sub-shell is fully filled. (c) The brown solid is copper metal. The blue solution contains the Cu2+ aqueous ions. Cu2+(aq) + e– Cu+ (aq) E° = +0.15 V Cu+ (aq) + e– Cu(s) E° = +0.52 V The E° for the reaction: 2Cu+ (aq) → Cu2+(aq) + Cu(s) E° = 0.52 – 0.15 V = +0.37 V A positive E° value indicates that the reaction is feasible under standard conditions. (d) Disproportionation 16 (a) (i) Cathode (positive terminal) is the silver/silver chromate half-cell. (ii) 2Li(s) + Ag2CrO4(s) → 2Li+ (aq) + 2Ag(s) + CrO4 2_(aq) (iii) E.m.f. = +0.45 – (–3.05) V = 3.50 V (b) (i) In the fuel cell, the reactants are supplied from an external source. Whereas the reactants in an ‘ordinary’ electrochemical cell are contained in the cell. (ii) V –ve +ve O2 H2O H2 H2O e– NaOH(aq) Porous graphite coated with nickel At the positive terminal, oxygen gas diffuses through the porous nickel partition and gets reduced to OH– . O2(g) + 2H2O(l) + 4e– → 4OH– (aq) E° = +0.40 V At the negative terminal, hydrogen gas diffuses through the porous nickel partition and reacts with OH– ions to produce H2O. H2(g) + 2OH– (aq) → 2H2O(l) + 2e– The overall cell reaction is: 2H2(g) + O2(g) → 2H2O(l) E°cell = 0.83 + 0.40 = 1.23 V (iii) Advantage: The only by-product is water. Disadvantage: Hydrogen gas is highly explosive. 17 (a) Species: Mg Explanation: It has the most negative E° value. (b) Cell diagram: Fe(s) | Fe3+(aq) || Fe3+(aq), Fe2+(aq) | Pt(s) Calculation: E.m.f. = (+0.77) – (–0.04) V = 0.81 V Cathode: Platinum (c) – When iron is in contact with a metal that is less electronegative than iron. – The presence of dissolved electrolytes. – At higher temperature.
268 Chemistry Term 2 STPM Answers Chapter 9 Periodic Table: Periodicity Quick Check 9.1 1 (a) 19 – 18 = 1 (d) 13 – 10 = 3 (b) 3 – 2 = 1 (e) 82 – 78 = 4 (c) 35 – 28 = 7 Quick Check 9.2 1 The ionic radius will be smaller than Al3+. Both Si4+ and Al3+ are isoelectronic. However, Si has more proton than Al. 2 Be2+. Both Li+ and Be2+ are isoelectronic, but the nuclear charge of Be2+ is larger. Quick Check 9.3 1 Species with lower proton electron ratio will have a larger size. (a) Mg : proton electron = 12 12 = 1 Mg2+ : proton electron = 12 10 = 1.2 \ Mg . Mg2+ (b) Br : proton electron = 35 35 = 1 Br– : proton electron = 35 36 = 0.97 \ Br– . Br (c) Si4+ : proton electron = 14 10 = 1.4 Si4– : proton electron = 14 18 = 0.78 \ Si4– . Si4+ Quick Check 9.4 1 (a) N3– (b) • N x x• x x • x (c) N3– + 3H2O ⎯→ 3OH– + NH3 2 P P P P Quick Check 9.5 1 (a) The acid has three displaceable hydrogen. (b) H3PO4 + NaOH ⎯→ NaH2PO4 + H2O NaH2PO4 + NaOH ⎯→ Na2HPO4 + H2O Na2HPO4 + NaOH ⎯→ Na3PO4 + H2O 2 (a) – O— O– (b) O O x x x x x x x •• •• • • • (c) Na2O2 + 2H2O ⎯→ 2NaOH + H2O2 3 Sulphuric(VI) acid Sulphuric(VI) acid has two electron-withdrawing oxygen atoms, while sulphuric(IV) acid has only one. HO OH O O S O S HO OH : 4 High charge density of Al3+ polarises the O2– ion. STPM Practice 9 Objective Questions 1 D 2 B 3 A 4 A 5 D 6 C 7 A 8 B 9 B 10 B 11 B 12 B 13 C 14 A 15 A 16 B 17 B 18 C 19 C 20 C 21 A 22 D 23 A 24 D 25 A 26 D 27 B 28 C 29 C Structured and Essay Questions 1 (a) The heat change when one mole of a substance is changed from the liquid phase to the vapour phase at the boiling point of the substance under standard conditions. (b) Na and Al: Giant structure with strong metallic bond. The metallic bond in aluminium is stronger than that in sodium because aluminium has three valence electrons while sodium has only one. Si: Giant structure with strong covalent bonds holding the atoms together. S and Cl: Simple molecular structure with weak van der Waals force between the molecules. Sulphur exists as S8 molecules while chlorine exists as Cl2 molecules. The van der Waals force between S8 molecules is stronger because of its larger size and larger number of electrons. 2 (a) The first ionisation energy is the energy required to remove one electron from every gaseous atom in one mole of the atoms under standard conditions. Cl(g) ⎯→ Cl+ (g) + e– (b) X X X X X X X X Na Al Mg Si P S Cl Ar (i) Going across the period, the atomic radius decreases but the nuclear charge increases. (ii) Magnesium has an s 2 configuration and phosphorous has a p3 configuration. These configurations are relatively more stable. As a result, more energy is required. (c) (i) The second ionisation energy is the energy required to remove one electron from every unipositive gaseous ion in one mole of the ions under standard conditions. Na+ (g) ⎯→ Na2+(g) + e– (ii) The second electron removed is from an inner 2p sub-shell which is completely filled.
269 Chemistry Term 2 STPM Answers 3 (a) • x • x O S O O x• x• •x •x (b) Na2O and Al2O3 are ionic. The ionic bond in Al2O3 is stronger due to its higher charge. SO3: Simple molecule with weak van der Waals forces between molecules. (c) Na2O + H2O ⎯→ 2NaOH SO3 + H2O ⎯→ H2SO4 4 (a) There are mobile electrons in the metallic bonds of aluminium. Chlorine exists as discrete Cl2 molecules. There are no mobile electrons present. (b) (i) 4Al(s) + 3O2(g) ⎯→ 2Al2O3(s) 2Al(s) + N2(g) ⎯→ 2AlN(s) (ii) AlN(s) + 3H2O(l) ⎯→ Al(OH)3(s) + NH3(aq) (c) When heated with aqueous sodium hydroxide or dilute sulphuric acid, it dissolves to form colourless solutions. Al2O3 + 2NaOH(aq) + 3H2O(l) ⎯→ 2Na[Al(OH)4](aq) Al2O3 + 3H2SO4(aq) ⎯→ Al2(SO4)3(aq) + 3H2O(l) (d) Aluminium reacts with chlorine to produce aluminium chloride, a covalent compound. 2Al(s) + 3Cl3(g) ⎯→ Al2Cl6(s) In the solid or molten state, there are no ions present to conduct electricity. However, when dissolves in water, it dissociates to form Al3+ and Cl- aqueous ions which is responsible for electrical conduction. Al2Cl6(s) + aq ⎯→ 2Al3+(aq) + 6Cl- (aq) (e) Chlorine dissolves in water to produce a mixture of HCl and HOCl. Cl2(g) + H2O(l) ⎯→ HCl(aq) + HOCl(aq) When exposed to strong sunlight, the HOCl decomposes to HCl with the liberation of oxygen that rekindles a glowing splinter. 2HOCl(aq) ⎯→ 2HCl(aq) + O2(g) 5 (a) (ii) MgO AI2 O3 SiO2 P4 O10 SO3 (ii) MgO and Al2O3 are ionic oxides with strong ionic bonds, while P2O3 and SO3 are covalent molecular oxides with weak van der Waals forces between molecules. This is why the melting points of MgO and Al2O3 are higher than those of P2O3 and SO3. On the other hand, SiO2 exists in the form of giant covalent structure with strong covalent bonds holding the atoms together. Hence, even though it is a covalent oxide, its melting point is higher than expected. The melting point of Al2O3 is lower than that of MgO. This is due to the presence of covalent nature in the ionic bond due to the high polarising power of the small Al3+ ion. This weakens the strength of the ionic bond in Al2O3. P2O3 has more electrons than SO3 making the van der Waals force in P2O3 stronger. This accounts for the higher melting point of P2O3 compared to SO3. (b) Basic oxide: MgO(s) + H2SO4(aq) ⎯→ MgSO4(aq) + H2O Amphoteric oxide: Al2O3(s) + 3H2SO4(aq) ⎯→ Al2(SO4)3(aq) + 3H2O(l) Al2O3(s) + 2NaOH(aq) + 3H2O(l) ⎯→ 2Na[Al(OH)4](aq) 6 (a) Na, Mg and Al have high melting point due to the strong metallic bond in the solid lattice. Si has exceptionally high melting point because of its giant covalent structure. P, S, Cl and Ar are all simple molecules with weak van der Waals forces between the molecules. Hence, they have relatively lower melting point. (b) Electronegativity increases from Na to Cl as the size of the atoms decreases. (c) Na, Mg and Al are good conductors due to the presence of delocallised electrons in the solid lattice. Silicon, a metalloid, is a semiconductor. P, S, Cl and Ar do not have delocallised electrons. Hence, they are non-conductors. 7 (a) Aluminium. There are more delocalised electrons in the solid lattice of aluminium compared to magnesium. (b) (i) The large Cl– ion is greatly polarised by small and highly charged Al3+ ion. [OR: The small lattice energy of AlCl3 cannot compensate for the high energy involved in the formation of the Al3+ ion.] (ii) Al2Cl6 Cl Al Cl Cl Cl Cl Cl Al Cl Al Cl Cl Cl Cl Cl Al (iii) The charge density of Al3+ is higher than that of Mg2+. 8 (a) Sulphur (b) The valance shell configuration of W (sulphur) is 3s 2 3p4 . The first electron removed is from a p-orbital that is occupied by two electrons. These two electrons experience mutual electrostatic repulsion making it easier to be removed than expected. 9 (a) Proton number Ionic radius 0 Na+ Mg2+ Al3+ P3– S2– Cl– (b) The electronic configuration of the cations and anions are 2.8 and 2.8.8 respectively. The anions with one extra shell of electrons are larger than the cations.
270 Chemistry Term 2 STPM Answers The cations have the same number of electrons but the nuclear charge increases from Na+ to Al3+. This leads to a decrease in size from Na+ to Al3+. The anions have the same number of electrons but the nuclear charge increases from P3- to Cl- , causing the ionic size to decrease in the same order. 10 (a) The metallic bond becomes stronger because the number of electrons per atom used to form the bond increases from 1 to 3 from Na to Al. (b) Silicon has a giant covalent structure with strong covalent bonds holding the atoms together. (c) The size and the number of electrons in sulphur (S8) is larger than those in phosphorous (P4). As a result, the van der Waals force in sulphur is stronger. 11 (a) Na2O, MgO and Al2O3 are ionic oxides and exist in the form of giant ionic structure with strong ionic bonds holding the ions together. On the other hand, P4O10 and SO3 are covalent oxides, with strong covalent bonds holding the atoms together in the molecule, but the intermolecular force is the weak van der Waals force. As a result, the melting point of the ionic oxides are higher than those of the covalent oxides. However, the melting point of Al3O3 is lower than expected. This is because the bond in Al2O3 is not purely ionic but with a significant amount of covalent character due to the polarisation of the O2– ion by the highly charged and small size Al3+ ion. This reduces the strength of the ionic bond causing the melting point to be lower. The charge density of Mg2+ is higher than that of Na+ due to its smaller size and higher charge. As result, the ionic bond in MgO is stronger than that in Na2O resulting in the higher melting point of MgO. The van der Waal’s force in P4O10 is stronger than SO3 due to the its larger size and greater number of electrons. This accounts for the higher melting point of P4O10. Molecule P4O10 SO3 Total number of electrons 140 40 Silicon dioxide, a covalent oxide, exists in the form of giant covalent structure with strong covalent bonds holding the atoms together in a three dimensional array. = Silicon atom = Oxygen atom A lot of energy is needed to break the covalent bonds. As a result, its melting point is higher than expected. (b) Oxide Na2O MgO Al2O3 SiO2 P4O10 SO3 Natura of oxide Basic Ampoteric Acidic Basic oxides: Na2O(s) + 2HCl(aq) → 2NaCl(aq) + H2O(l) MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l) Amphotheric oxide: Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(l) Al2O3(s) + 3H2O(l) + 2NaOH(aq) → 2NaAl(OH)4(aq) Acidic oxide: P4O10(s) + 12NaOH(aq) → 4Na3PO4 + 6H2O(l) SO3(g) + 2NaOH(aq) → Na2SO3(aq) + H2O(l) Chapter 10 Group 2 Quick Check 10.1 1 (a) Mg2+(aq) + 2e– Mg(s) –1.85 V Ba2+(aq) + 2e– Ba(s) –2.90 V (b) Barium has lower ionisation energy than magnesium. 2 (a) TiCl4 + 2Mg ⎯⎯→ 2MgCl2 + Ti 189.9 g 24.3 g 47.9 g Mass of Mg required = 1000 24.3 × 189.9 kg = 7814.8 kg (b) Theoretical yield = 1000 189.9 × 47.9 kg = 252.2 kg Actual yield = 252.2 × 65 100 kg = 163.9 kg Quick Check 10.2 1 NaOH is a strong base. The heat of neutralisation will be very exothermic and will damage the stomach tissues. 2 Barium hydroxide reacts with carbon dioxide from the air to form insoluble barium carbonate. Ba(OH)2(aq) + CO2(g) → BaCO3(s) + H2O(l) 3 Being basic, they are added to neutralise soils that are acidic. STPM Practice 10 Objective Questions 1 A 2 D 3 C 4 B 5 B 6 C 7 D 8 C 9 C 10 D 11 A 12 C 13 B 14 A 15 A 16 D 17 A 18 B 19 C 20 C 21 B 22 A Structured and Essay Questions 1 (a) Beryllium reacts slowly with steam. Magnesium reacts slowly with hot water but vigorously with steam. Calcium reacts rapidly with hot water but slowly with cold water. Strontium and barium react vigorously with cold water.
271 Chemistry Term 2 STPM Answers The reactivity of the Group 2 elements with water increases down the group. The reaction with water is a redox reaction where water is reduced to hydrogen. M(s) + 2H2O ⎯→ M(OH)2 + H2 Going down the group, the reducing power of the Group 2 elements increases (as the ionisation energy decreases) thereby causing the reactivity to increase. (b) Solubility of the Group 2 sulphates depends on the difference between the lattice energy and the hydration energy of the sulphates. M2+(g) + SO4 2-(aq) –Lattice energy Hydration energy MSO4(s) Heat of solution M2+(aq) + SO4 2-(aq) By Hess law: Heat of solution = Hydration energy – lattice energy Between MgSO4 and BaSO4 there is little change in the lattice energy. However, there is a large decrease in the hydration energy due to the larger size of Ba2+. This causes the heat of solution to become less exothermic (or more endothermic). As a result, the solubility of MgSO4 is higher than that of BaSO4. 2 (a) The Group 2 carbonates decompose according to the equation: MCO3(s) ⎯→ MO(s) + CO2(g) The decomposition is caused by the polarisation of the large CO3 2- ion by the small and highly charged M2+ ions. Going from Be to Ba, The size of the cations increases causing the polarising power to decrease. As a result, the carbonates become more stable and need higher temperature to decompose. (b) The CO3 2- ion is larger than the OH- ion. As a result, CO3 2- is more easily polarised by Ca2+ compared to the smaller OH- . 3 (a) Going down Group 2, the size of the cations increases causing a decrease in the charge density and the polarising power of the cations. As a result, the thermal stability of the carbonates increases. (b) Magnesium hydroxide. The OH– ion is smaller than the CO3 2– ion and hence is more difficult to be polarised by the cations, M2+. 4 (a) BeCl2(s) + 2H2O(l) ⎯→ Be(OH)2(aq) + 2HCl(g) (b) (i) van der Waals forces (ii) Be2Cl4 (iii) Cl—Be Be—Cl Cl Cl 5 (a) Be(s) + Cl2(g) → BeCl2(s) (b) Cl—Be Be—Cl Cl—Be—Cl Cl Dimer (Trigonal planar) Monomer (Linear) Cl (c) (i) A white solid is obtained. White fumes liberated. (ii) BeCl2(s) + 2H2O(l) → Be(OH)2(s) + 2HCl(g) (d) Due to its small size and low screening effect, the energy required to form the Be2+ ion is high. The Be2+ ion with its high charge density is able to polarise any anion associated with it to give a significant amount of covalent character. 6 (a) (i) 2Mg(NO3)2(s) → 2MgO(s) + O2(g) + 4NO2(g) The white residue is magnesium oxide. The brown gas is nitrogen dioxide. (ii) The decomposition of the Group 2 nitrates is due to the polarisation of the NO3 – ion by the small and highly charged cations. The greater the polarisation of the NO3 – ion, the less stable the nitrate. The Ba2+ ion is larger than the Mg2+ ion. As a result, the charge density as well as the polarising power of Ba2+ is lower than Mg2+. Hence, it is more difficult to decompose Ba(NO3)2 than Mg(NO3)2. Thus, the decomposition temperature of Ba(NO3)2 is higher than 200 °C. (b) When heated using bunsen flame, magnesium carbonate undergoes thermal decomposition to form magnesium oxide with the release of carbon dioxide gas. MgCO3(s) → MgO(s) + CO2(g) The carbon dioxide released reacts with limewater to form insoluble magnesium carbonate that causes the limewater to turn chalky. Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l) The decomposition of the carbonate is due to the polarisation of the large CO3 2– ion by the small and highly charged Mg2+ ion. The polarisation of the CO3 2– ion results in the weakening of the carbon-oxygen bond in CO3 2–. The stability of the carbonate thus depends on the polarising power of the M2+ ion. The Ba2+ ion is very large compared to the Mg2+ ion. As a result, the polarising power of Ba2+ is lower than Mg2+. As a result, relatively, BaCO3 is more stable to heat than MgCO3 and needs higher temperature for it to decompose. However, the temperature of the bunsen flame is not high enough to cause the decomposition of BaCO3, although the bunsen flame temperature is high enough to cause decompostion of MgCO3. Chapter 11 Group 14 Quick Check 11.1 1 Due to the large size of the iodine atom, the C—I is extremely weak and will decompose spontaneously. 2 Lead(IV) chloride decomposes on heating. PbCl4(l) ⎯→ PbCl2(s) + Cl2 3 PbCl4(l) + 2H2O ⎯→ PbO2 + 4HCl Lead dioxide STPM Practice 11 Objective Questions 1 B 2 C 3 C 4 B 5 A 6 C 7 C 8 A 9 D 10 D 11 B 12 A 13 B 14 C 15 B 16 D 17 C 18 A 19 B 20 C 21 C 22 D 23 D 24 D 25 A 26 D 27 C 28 D 29 B 30 C 31 B 32 B 33 A 34 C 35 D 36 B
272 Chemistry Term 2 STPM Answers Structured and Essay Questions 1 (a) The tetrachlorides decompose according to the equation: SnCl4(l) ⎯→ SnCl2(s) + Cl2(g) PbCl4(l) ⎯→ PbCl2(s) + Cl2(g) The ease of decomposition depends on the strength of the element⎯chlorine bond. Lead is lower than tin in Group 14 and the lead atom is larger than the tin atom. The Pb⎯Cl bond is longer and weaker than the Sn⎯Cl bond. Thus, PbCl4 is thermally less stable than SnCl4. (b) Silicon and diamond have giant covalent structure with covalent bonds holding the atoms together. However, silicon is larger than carbon, and the covalent bond in diamond is stronger and needs more energy to break. This accounts for the higher melting point of diamond. (c) (i) Pb2+ + 2H2O ⎯→ PbO2 + 4H+ + 2e– OCl- + 2H+ + 2e– ⎯→ Cl- + H2O Pb2+ + H2O + OCl- ⎯→ PbO2 + Cl- + 2H+ .……..(I) Add 2OH- to both sides of equation (I): Pb2+ + H2O + OCl- + 2OH- ⎯→ PbO2 + Cl- + 2H+ + 2OH- Overall equation: Pb2+ + OCl- + 2OH- ⎯→ PbO2 + Cl- + H2O (ii) From the above equation: PbO2 : OCl- = 1 : 1 No. of mol of OCl- = 50.0 1000 × 0.11 mol = 5.50 × 10-3 mol Mass of PbO2 produced = (5.50 × 10-3)(207 + 32) g = 1.31 g 2 (a) (i) CO2: Acidic SiO2: Acidic GeO2: Amphoteric SnO2: Amphoteric PbO2: Amphoteric (ii) SiO2 + 2NaOH → Na2SiO3(aq) + H2O(l) PbO2(s) + 2NaOH(aq) + 2H2O(l) → Na2Pb(OH)6(aq) PbO2(s) + 4HCl(aq) → PbCl2(s) + Cl2(g) + 2H2O(l) (b) (i) Manganate(VII) ion, MnO4 – (ii) 5PbO2(s) + 2Mn2+(aq) + 4H+ (aq) → 2MnO4 – (aq) + 5Pb2+(aq) + 2H2O(l) (c) (i) SiCl4(l) + 2H2O(l) → SiO2(s) + 4HCl(aq) (ii) Making glass 3 The conversion of CO2 to CO is endothermic and energetically unfavourable. Hence, the +4 oxidation state of carbon is more stable compared to the +2. The conversion of PbO2 to PbO is exothermic and energetically favourable. Hence, for lead, the +2 oxidation state is more stable than the +4. 4 (a) (i) Cl Si Cl Cl Cl Bond angle = 109.5° (ii) SiCl4(l) → Si(g) + 4Cl(g) (iii) Let the standard enthalpy of formation of SiCl4 = x kJ mol–1 Si(g) + 4Cl(g) Si(s) + 2Cl2(g) ⎯⎯→ SiCl4(l) +368 2(+242) +1492 x Based on Hess’s law: x + 1492 = 368 + 2(242) x = –640 kJ mol–1 (iv) The Si–Cl bond energy refers to the energy required to break the Si–Cl bond to form free atoms. All species involved are in the gaseous state: SiCl4(g) → Si(g) + 4Cl(g) ∆H = 4 (Si–Cl bond energy) However, the atomisation of SiCl4 is the conversion of one mole of SiCl4, in the liquid state (the physical state of SiCl4 under standard conditions) to free gaseous atoms. 5 (a) The acid/base property of the monoxides are given in the table below: Monoxide CO SiO GeO SnO PbO Nature Neutral Amphoteric Reaction with acid: GeO(s) + 2HCl(aq) → GeCl2(aq) + H2O(l) SnO(s) + 2HCl(aq) → SnCl2(aq) + H2O(l) PbO(s) + 2HCl(aq) → PbCl2(aq) + H2O(l) Reaction with base: GeO(s) + 2NaOH(aq) + H2O(l) → Na2Ge(OH)4(aq) SnO(s) + 2NaOH(aq) + H2O(l) → Na2Sn(OH)4(aq) PbO(s) + 2NaOH(aq) + H2O(l) → Na2Pb(OH)4(aq) The acid/base property of the dioxides are given in the table below: Dioxide CO2 SiO2 GeO2 SnO2 PbO2 Nature Acidic Amphoteric The acidic oxides: CO2(g) + 2NaOH(aq) → Na2CO3(aq) + H2O(l) SiO2(s) + 2NaOH(aq) → Na2SiO3(aq) + H2O(l) The amphoteric oxides: Reaction with acid: GeO2(s) + 4HCl → GeCl4(aq) + 2H2O(l) SnO2(s) + 4HCl → SnCl4(aq) + 2H2O(l) PbO2(s) + 4HCl → PbCl4(aq) + 2H2O(l) Reaction with base: GeO2(s) + 2NaOH(aq) + 2H2O(l) → Na2Ge(OH)6(aq) SnO2(s) + 2NaOH(aq) + 2H2O(l) → Na2Sn(OH)6(aq) PbO2(s) + 2NaOH(aq) + 2H2O(l) → Na2Pb(OH)6(aq) (b) (i) Permanganate ion or manganate(VII) ion, MnO4 – (ii) Oxidising agent (iii) 2Mn2+(aq) + 4H+ (aq) + 5PbO2(s) ⎯→ 2MnO4 – (aq) + 5Pb2+(aq) + 2H2O(l) Chapter 12 Group 17 Quick Check 12.1 Chlorine water contains chloric(I) acid, HOCl. When exposed to strong sunlight, it decomposes to hydrochloric acid and oxygen. 2HOCl(aq) ⎯→ 2HCl(aq) + O2(g)
273 Chemistry Term 2 STPM Answers Quick Check 12.2 1 The oxidising strength of the halogen decreases with increasing proton number due to the increase in the size of the atom and decreasing electron affinity. 2 Sodium thiosulphate reduces iodine to iodide. 2S2O3 2– + I2 ⎯→ S4O6 2– + 2I– Orange Colourless Quick Check 12.3 3ClO– ⎯→ 2Cl– + ClO3 – (Disproportionation) Quick Check 12.4 The lens contains microscopic silver halide crystals embedded in the glass lattice. When exposed to sunlight, the lens turns dark due to the deposition of metallic silver that blocks off some of the visible light. For example, AgCl(s) ⎯→ Ag(s) + Cl(g) When the lens is removed from sunlight, the reverse reaction occurs and the lens turns clear. Ag(s) + Cl(g) ⎯→ AgCl(s) STPM Practice 12 Objective Questions 1 C 2 C 3 A 4 C 5 A 6 B 7 A 8 B 9 A 10 B 11 B 12 D 13 D 14 A 15 C 16 B Structured and Essay Questions 1 (a) Chlorine reacts explosively with hydrogen when exposed to bright sunlight to form hydrogen chloride. Cl2(g) + H2(g) ⎯→ 2HCl(g) 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) Iodine vapour reacts reversibly with hydrogen at 450 °C in the presence of nickel or platinum catalyst to produce hydrogen iodide. The reaction is reversible. H2(g) + I2(g) 2HI(g) Going from chlorine to iodine, the size of the atom increases causing the oxidising power of the halogens to decrease. As a result, the reactivity decreases. (b) (i) Bromide ion (ii) AgBr(s) + 2NH3(aq) → [Ag(NH3)2] + (aq) + Br– (aq) (c) (i) Iodine cannot form hydrogen bonds with water molecules. (ii) Iodine forms a water soluble compiles with aqueous I– ions. I2 + I– I3 – 2 (a) The halogens exist as simple diatomic molecules with van der Waals forces between the molecules. Going from fluorine to iodine, the size of the molecules as well as the total number of electrons in the molecules increases. This causes the van der Waals force, as well as the melting point and boiling point to increase down the group. Fluorine and chlorine are gases, bromine is a volatile liquid and iodine is a solid. (b) (i) The equation for the reaction is: X2(s,l,g) + H2(g) ⎯→ 2HX(g) In the reaction, H2 is oxidised to H+ . Going from F2 to I2, the oxidising power of the halogens decreases (due to the increase in the size of the atoms) down the group causing the reactivity to decrease. F2 and Cl2 react vigorously with H2 in the presence of strong sunlight. Br2 reacts with H2 at 450°C in the presence of nickel catalyst. I2 reacts reversibly with H2 at 450°C in the presence of nickel catalyst. (ii) The variation of the boiling point of the hydrogen halides is shown below: Boiling point HF HCI HBr HI As the size and the total number of electrons in the molecules increase from HCl to HI, the intermolecular van der Waals force becomes stronger. This accounts for the increase in the boiling point from HCl to HI. However, HF can also form the stronger intermolecular hydrogen bonding. Thus, its boiling point is higher than expected. (c) (i) Indicator: Starch solution. End-point: From blue to colourless. (ii) Equations of reactions: Ca(OCl)2 ⎯→ Ca2 + + 2ClO- ClO- + 2I- + 2H+ ⎯→ Cl- + I2 + H2O I2 + 2S2O3 2- ⎯→ 2I- + S4O6 2- More reaction ratio Ca(OCl2) : S2O3 2- = 1 : 4 No. of moles of S2O3 2- = 0.10 × 18.50 1000 = 1.85 × 10-3 mol No. of moles of Ca(OCl)2 = 1 4 (1.85 × 10-3) mol = 4.63 × 10-4 mol Mass of Ca(OCl)2= (4.63 × 10-4) × 143.1 g = 6.62 × 10-2 g % Purity = 6.62 × 10-2 0.50 × 100% = 13.2% 3 (a) Silver bromide is more sensitive to light than silver chloride. (b) Concentrated sulphuric acid will oxidise the HI formed to iodine. KI(s) + H2SO4(aq) ⎯⎯→ KHSO4(aq) + HI(g) ∆ 2HI(g) + H2SO4(aq) ⎯⎯→ I2(g) + SO2(g) + 2H2O(aq) ∆ (c) Chlorine reacts with sodium hydroxide to form sodium chloride and sodium chlorate(l), which are non-volatile. Cl2 + 2NaOH ⎯→ NaCl + NaOCl + H2O (d) The size of HI is larger than HCl. Hence, the intermolecular van der Waals force in HI is stronger. Due to the larger size of the iodine atom, the H⎯I bond is weaker and easier to dissociate. 4 (a) Aqueous sodium chlorate(I) decomposes to sodium chloride and oxygen gas. 2NaOCl(aq) ⎯→ 2NaCl(aq) + O2(g)
274 Chemistry Term 2 STPM Answers (b) Chlorine gas is liberated. MnO2(s) + 4HCl(aq) → Mn2+(aq) + 2H2O(l) +2Cl– (aq) + Cl2(g) 5 (a) H2(g) + Cl2(g) ⎯→ 2HCl(g) (b) Mr of HX Boiling point 0 HF HC HI HBr Going from HCl to HI, the size of the HX molecules as well as the total number of electrons in the molecule increases. This causes the strength of the van der Waals forces to increase from HCl to HI. This accounts for the increase in the boiling point from HCl to HI. HF, although with the smallest size and the least number of electrons, has an exceptionally high boiling point. This is due to the presence of hydrogen bonding between the HF molecules. (c) The thermal stability decreases from HF to HI. 2H—X(g) ⎯→ H2(g) + X2(g) The strength of the H—X bond decreases from HF to HI due to the increase in the size of the halogen atom, X. 6 (a) Chlorine, a non-polar molecule, cannot form hydrogen bond with water molecules. Cl2(g) + H2O(l) ⎯→ HCl(aq) + HOCl(aq) (b) Disproportionation. Cl2 is reduced to chloride and oxidised to chlorate(I). (c) (i) Oxygen (ii) 2HOCl(aq) ⎯→ 2HCl(aq) + O2(g) (d) Sodium chlorate(I): Domestic bleach Carbon tetrachloride: Non-polar solvent 7 (a) Cl2 + 2e– 2Cl– E° = +1.36 V Br2 + 2e– 2Br– E° = +1.07 V I2 + 2e– 2I– E° = +0.54 V Oxidising agents are electron-acceptors. The oxidising strength decreases in the order: Cl2 . Br2 . I2 as indicated by their respective standard electrode potentials. [Note: The more positive the E°, the stronger the oxidising power.] Going from chlorine to iodine, the size of the atoms increases. The attraction for electrons and hence the oxidising power decreases. (b) NaOH + spent brine Diaphragm Chlorine Hydrogen Brine + – Steel cathode Titanium anode Anode: 2Cl– (aq) ⎯→ Cl2(g) + 2e– Cathode: 2H2O(l) + 2e– ⎯→ H2(g) + 2OH– (aq) (c) (i) Cl2(g) + 2NaOH(aq) ⎯→ NaCl(aq) + NaOCl(aq) + H2O(l) (ii) 3Cl2(g) + 6NaOH(aq) ⎯→ 5NaCl(aq) + NaClO3(aq) + 3H2O(l) 8 (a) When solid sodium chloride is heated with concentrated sulphuric acid, a white fume of hydrogen chloride is liberated. NaCl(s) + H2SO4(aq) → NaHSO4(aq) + HCl(g) When solid potassium bromide is heated with concentrated sulphuric acid, initially a white fume of hydrogen bromide is liberated. After sometimes, a brown fume of bromine is liberated. KBr(s) + H2SO4(aq) → KHSO4(aq) + HBr(g) 2HBr(g) + H2SO4(aq) → Br2(g) + SO2(g) + 2H2O(g) When solid potassium iodide is heated with concentrated sulphuric acid, initially a white fume of hydrogen iodide is liberated. After sometimes, a purple fume of iodine is liberated. KI(s) + H2SO4(aq) → KHSO4(aq) + HI(g) 2HI(g) + H2SO4(aq) → I2(g) + SO2(g) + 2H2O(g) This is because, concentrated sulphuric acid is a stronger oxidising agent than bromine and iodine. As a result, it is able to oxidise the Br– and I– ions to Br2 and I2 respectively. However, concentrated sulphuric acid is a weaker oxidising agent than chlorine. As a result, it is not able to oxidise the Cl– ions to Cl2. (b) (i) NaOH at room temperature. 2NaOH(aq) + Cl2(g) → NaCl(aq) + NaOCl(aq) + H2O(l) (ii) NaOH at 70 °C. 6NaOH(aq) + 3Cl2(g) → 5NaCl(aq) + NaClO3(aq) + 3H2O(l) (iii) Sodium chlorate(I): As domestic bleach Sodium chlorate(V): As a weed killer (c) (i) Water is a polar molecule, whereas the halogens are non-polar. The halogens cannot form hydrogen bonds with water molecules. (ii) Iodine reacts with the iodide ions to form a soluble complex. I2(s) + I– (aq) → I3 – (aq) Chapter 13 Transition Elements Quick Check 13.1 1 The electronic configuration of manganese is [Ar]3d 5 4s 2 . There is greater repulsion between the completely halffilled 3d sub-shell and the fully filled 4s sub-shell. This causes the electron cloud to drift further away from the nucleus. 2 All the transition elements can make use of the inner 3d electrons as well as the outer 4s electrons to form metallic bonds. The metallic bonds are strong and need a lot of energy to break. 3 The electronic configurations of calcium and nickel are: 20Ca: [Ar]4s 2 28Ni: [Ar]3d 8 4s 2
275 Chemistry Term 2 STPM Answers The size of nickel is smaller than calcium. The nuclear charge of nickel is higher than calcium. As a result, the attraction between the nucleus and the valence electrons in nickel is stronger. This leads to higher first ionisation energy for nickel. Nickel can make use of the 3d as well as the 4s electrons in forming the metallic bonds. Calcium can make use of the 4s electrons only to form the metallic bonds. As a result, there are more delocalised electrons in nickel compared to calcium. 4 The third ionisation energy of iron involves the following process: Fe2+(g) ⎯→ Fe3+(g) + e– The electronic configuration of the iron(II) ion is: Fe2+: [Ar]3d 6 In the formation of Fe3+, the electron removed is from a d orbital that is occupied by two electrons. These two electrons experience mutual electrostatic repulsion. 3d Fe2+ 3d Fe3+ This repulsion makes the electron easier to be removed than expected. STPM Practice 13 Objective Questions 1 C 2 A 3 B 4 D 5 D 6 B 7 A 8 B 9 A 10 B 11 C 12 D 13 B 14 B 15 D 16 C 17 B 18 A 19 C 20 D 21 D Structured and Essay Questions 1 (a) Transition element is an element that can form at least one stable ion with an incompletely filled 3d orbitals. (b) Scandium forms only the Sc3+ ion where the 3d orbitals are empty. Zinc forms only the Zn2+ ion where the 3d orbitals are fully filled. Copper forms Cu2+ with d 9 configuration. (c) (i) Cr: [Ar] 3d5 4s 1 (ii) According to Aufbau rule, the configuration of chromium should be [Ar] 3d4 4s 2 . (d) (i) Orange (ii) Cr2O7 2– + 14H+ + 6e– ⎯→ 2Cr3+ + 7H2O (iii) Cr2O7 2– + 14H+ + 6Fe2+ → 2Cr3+ + 7H2O + 6Fe3+ 25.0(0.10) (0.025)(V) = 6 V = 16.70 cm3 2 (a) Transition elements have incompletely filled inner shells. Ca: [Ar]4s 2 Fe: [Ar]3d6 4s 2 (b) Variable oxidation states. Examples: Fe3+, Fe2+ Form complexes. Example: [Cu(NH3)4] 2+, [Fe(CN)6] 4– Form coloured compounds. Examples: K2Cr2O7 (orange), KMnO4 (purple) Catalytic properties. Examples: Fe3+(aq) in the oxidation of iodide ions by peroxodisulphate ions. Nickel in hydrogenation of alkenes. 3 (a) High melting point and high density (b) (i) Cu: [Ar]3d104s 1 (ii) Cu+ : [Ar]3d10 (iii) Cu2+: [Ar]3d 9 (c) (i) The 3d orbitals in Cu(I) are fully filled. No d-d transition can occur. The 3d orbitals in Cu(II) are partially filled. d-d transition can take place. (ii) Copper(I) oxide, reddish brown solid (d) (i) Copper (ii) [Cu(H2O)6] 2+ (iii) Cu2SO4(aq) ⎯→ CuSO4(aq) + Cu(s) (Disproportionation) 4 (a) Co: [Ar] 3d7 4s 2 Maximum oxidation state = +5 (b) (i) [Co(H2O)6] 2+ (ii) Formation of [CoCl4] 2– which is blue. [Co(H2O)6] 2+ + 4Cl– [CoCl4] 2– + 6H2O ∆H = +ve Cooling the solution shifts the equilibrium to the left, and the pink [Co(H2O)6] 2+ is regenerated. (c) Co3+ + e– Co2+ E° = +1.82 V O2 + 4H+ + 4e– 2H2O E° = +1.23 V Co3+ oxidises H2O to O2. 4Co3+ + 2H2O ⎯→ 4Co2+ + O2 + 4H+ E° = 1.82 – 1.23 = +0.59 V 5 (a) Hexaaquacopper(I) (b) Colourless. The 3d sub-shell is fully filled. No d-d transition can occur. (c) +3 (d) The colour changes from blue to yellow. Ligand displacement occurs: [Cu(H2O)6] 2+ + 4Cl- ⎯→ CuCl4 2- + 6H2O Blue Yellow 6 (a) An ion consisting of a central metal cation bonded to a group of ligands via coordinate bonds. (b) (i) Trisethanedioatechromate(III) (ii) Triamminetrichloronickelate(II) (c) (i)– OOC—CH2 – OOC—CH2 N—CH2—CH2—N CH2COO– CH2COO– (ii) [Cu(NH3)4] 2+, square planar (iii) [Cu(EDTA)]2–, coordination number = 6 7 (a) Mn3+ + e– Mn2+ E° = +1.51 V The large positive E° value indicates that Mn2+ is more stable than Mn3+ in standard conditions. The electronic configuration of Mn is [Ar] 3d5 4s 2 The electronic configuration of Mn2+ and Mn3+ are as follows: Mn2+ : [Ar] 3d5 Mn3+: [Ar] 3d4 The 3d5 configuration (3d sub-shell is half-filled) is more stable than the 3d4 configuration (3d configuration is partially filled). (b) The electronic configurations of Ca and Cr are: Ca: [Ar] 4s2 Cr: [Ar] 3d5 4s 1
276 Chemistry Term 2 STPM Answers The chromium atom is smaller than calcium. On top of that, chromium can use both the 3d and 4s electrons to form metallic bond, whereas, calcium can make use of the 4s electrons only to form metallic bond. As a result, the melting point of chromium would be higher than that of calcium. Chromium is smaller than calcium, but the effective nuclear charge of chromium is higher than calcium. As a result, the first ionisation energy of chromium would be higher than that of calcium. (c) An aqueous solution of iron(III) chloride is acidic due to the hydrolysis of the Fe3+(aq) ion. FeCl3(s) + 6H2O(l) ⎯→ [Fe(H2O)6] 3+ (aq) + 3Cl- (aq) [Fe(H2O)6] 3+ + H2O(l) ⎯→ [Fe(H2O)5OH]2+(aq) + H3O+ (aq) The added CO3 2- ion reacts with the H3O+ to liberate carbon dioxide. 2H3O+ (aq) + CO3 2-(aq) ⎯→ CO2(g) + 3H2O(l) The brown precipitate is iron(III) hydroxide. (d) (i) [Ni(EDTA)]2- (ii) Shape: Octahedral O ' C CH2 O & & & Ni2+ & & & O O ' C!O CH2 N CH2 !!CH2 N CH2 CH2 C ' O C"O O 2– 8 (a) An element that can form at least one stable ion with an incompletely filled d sub-shell (b) (i) [Fe(H2O)6] 3+ (ii) [Fe(H2O)6] 3+(aq) + SCN– (aq) ⎯→ [Fe(H2O)5 (SCN)]2+(aq) + H2O(l) (iii) Ligand displacement or substitution (c) (i) [Fe(CN)6] 3– (ii) NC NC CN 3– CN Fe CN CN (iii) [Fe(H2O)6] 3+ + e– [Fe(H2O)6] 2+(aq) E° = +0.77 V [Fe(CN)6] 3–(aq) + e– [Fe(CN)6] 4–(aq) E° = +0.36 V Oxidising agents are electron-acceptors. From the E° values, [Fe(CN)6] 3– is a weaker oxidising agent than [Fe(H2O)6] 3+. This is because it is more difficult for a negative ion to accept an electron compared to a positive ion. 9 Ni : [Ar]3d8 4s 2 Ni2+ : [Ar]3d8 The complex between nickel(II) and ethylenediamine has the formula of [Ni(H2NCH2CH2NH2)3] 2+ Ethylenediamine is a bidentate ligand. Each ethylenediamine molecule can form two coordinate bonds with the Ni(II) ions. 2+ Cu NH2 CH2 CH2 CH2 CH2 NH2 NH2 CH2 NH2 CH2 H2N H2N Ni(II) ions make use of the empty 4s, 4p and two of the 4d orbitals to accept the six lone pairs of electrons from three ethylenediamine molecules. 4s 4p 4d Orbitals used in the complex formation 10 (a) (i) It is an element that can form at least one stable ion with an incompletely filled d sub- shell. (ii) Ni: 1s 2 2s 2 2p6 3s 2 3p6 3d 8 4s 2 (iii) Catalytic property. (iv) Nickel. Nickel can make use of electrons from both the 3d and 4s orbitals to form metallic bond. Whereas, calcium (1s 2 2s 2 2p6 3s 2 3p6 4s 2 ) can only make use of the 4s electrons to form metallic bond. (v) Trisethane-1,2-diamminecopper(II). (b) (i) Ethylenediamminetetraethanoate ion. (ii) O C C – :O N : N : – :O O:– O:– CH2 CH2 CH2 CH2 CH2 CH2 O O C C O STPM Model Paper Term 2 Section A 1 B 2 C 3 D 4 B 5 B 6 D 7 C 8 C 9 A 10 B 11 C 12 D 13 B 14 D 15 B SECTION B 16 (a) (i) Cl+ (g) ⎯→ Cl2 + (g) (ii) 4000 3000 2000 1000 Na Mg AI Si P S CI
277 Chemistry Term 2 STPM Answers (iii) From Mg (2.8.2.) to Cl (2.8.7.), the second electrons removed are from the same third shell. While the atomic radius decreases, the effective nuclear charge increases causing the ionisation energy to increase. The second ionisation energy of Al is higher than expected because the electron removed is from a stable 3s 2 configuration. (iv) The second ioniastion energy for sodium is extraordinary high because the second electron is from an inner second fully-filled shell that is closer to the nucleus. Na+ (g) ⎯→ Na2+(g) + e– 2.8 2.7 (b) (i) Contact process for the manufacture of sulphuric acid Volcanic action (ii) SO2(g) + NO2(g) ⎯→ SO3(g) + NO(g) 2NO(g) + O2(g) ⎯→ 2NO2(g) 17 (a) The enthalpy change when one mole of a substance is completely burned in excess oxygen under standard conditions. (b) P4(s) + 5O2(g) ⎯→ P4O10(s) (c) Red phosphorous (d) Using the energy cycle: x P4(white) P4(red) -2984 -2967 P4O10(s) Applying Hess’s law: x + (-2967) = -2984 x = -17 kJ mol-1 18 (a) (i) Anode Cathode Cu(NO3 )2 (aq) Platinum Platinum (ii) The two possible reactions occur at the cathode are the reduction of Cu2+ and H2O. Cu2+(aq) + 2e- ⎯→ Cu(s) E° = +0.34 V 2H2O(l) + 4e- ⎯→ 2OH- (aq) + H2(g) E° = -0.83 V Since the E° value for the first reaction is more positive, it will occur in preference to the second reaction. [Alternative explanation: The reaction occurring at the cathode is reduction. Hence, species which are stronger oxidising agents (with more positive E° value) will get reduced first]. (iii) At the anode, oxidation occurs. The reaction at the anode is the oxidation of water according to the equation: 2H2O(l) ⎯→ O2(g) + 4H+ (aq) The production of H+ makes the solution acidic. [NOTE: The NO3 – ion cannot be further oxidised because nitrogen is already at its maximum oxidation state of +5] (iv) Equation for the cathode reaction: Cu2+(aq) + 2e- ⎯→ Cu(s) Quantity of charge = It = 3.00 × 25 × 60 C = 4500 C 1.48 g of Cu requires 4500 C 1.0 mol of Cu (63.5) g would require = 63.5 1.48 × 4500 C = 1.93 × 105 C Charge carried by 2 mol of electrons = 1.93 × 105 C Charge on one mole of electron = 1 2 (1.93 × 105 ) C = 9.65 × 104 C No. of electrons in 1 mol of electrons = 9.65 × 104 1.60 × 10-19 = 6.03 × 1023 Avogadro’s constant = 6.03 × 1023 mol-1 (b) (i) E.m.f. = E°(cathode) – E°(Anode) = +0.34 – (-0.76) V = +1.10 V (ii) The Nernst’s equation for the cell is: E = 1.10 – 0.059 2 log [Zn2+] [Cu2+] At equilibrium, E = 0.00, and [Zn2+] [Cu2+] = Kc 1.10 = 0.059 2 log Kc Solving, Kc = 1.94 × 1037 (iii) Using the Nernst’s equation: E = 1.10 – 0.059 2 log [Zn2+] [Cu2+] = 1.10 – 0.059 2 log 1.10 0.55 = 1.091 V (iv) The value of Kc remains the same. This is because equilibrium constant is only affected by a change in the temperature. 19 (a) When heated, PbCl4 decomposes to PbCl2 (a white ionic solid) and chlorine gas. PbCl4(l) → PbCl2(s) + Cl2(g) D When the gas is bubbled into aqueous potassium iodide, chlorine (a stronger oxidising agent than iodine) oxidises iodide ion to iodine (purple fume). Cl2(g) + 2I- (aq) ⎯→ 2Cl- (aq) + I2(g) (b) Due to the inert-pair effect, lead(II) compounds are more stable than lead(IV) compounds. However, the inert-pair effect in carbon is negligible and carbon(IV) compounds are more stable than carbon(II) compounds. Thus, when heated, lead(IV) oxide decomposes to the more stable lead(II) oxide. 2PbO2(s) ⎯→ 2PbO(s) + O2(g) On the other hand, carbon dioxide does not decompose to carbon monoxide.
278 Chemistry Term 2 STPM Answers (c) Silicon dioxide, SiO2, has a giant covalent structure with strong covalent bonds holding the atoms together. A lot of energy is required to break these covalent bonds. As a result, silicon dioxide is a solid. Germanium dioxide, tin dioxide and lead dioxide have giant ionic structures with some covalent character . A lot of energy is required to break the ionic bonds. They are all solids. On the other hand, carbon dioxide, CO2, is a simple non-polar molecule with strong covalent bonds holding their atoms together in the molecule, but the intermolecular force is the weak van der Waals force which can be broken easily. As a result, carbon dioxide is a gas. (d) When heated, the Group 2 nitrates decompose to form respective oxides with the liberation of nitrogen dioxide (brown gas) and oxygen. For example, 2Mg(NO3)2(s) ⎯→ 2MgO(s) + 4NO2(g) + O2(g) 2Ba(NO3)2(s) ⎯→ 2BaO(s) + 4NO2(g) + O2(g) The relative stability of the nitrates towards heat depends on the polarising power of the cation, M2+. – O N O O M2+ Bond is weaken due to polarisation of electron cloud by cation As barium is below magnesium in Group 2, Ba2+ is larger and hence weaker polarising power. As a result, higher temperature is needed to decompose Ba(NO3)2 compared to Mg(NO3)2. The temperature of the Bunsen flame is high enough to cause Mg(NO3)2 to decompose but is not enough to cause Ba(NO3)2 to decompose. (e) Hydrogen chloride and hydrogen iodide exist as simple diatomic molecules with weak van der Waals forces between the molecules. Due to the larger size of the HI molecule as well as greater number of electrons, the van der Waals force in HI is stronger than in HCl. This causes the boiling point of HCl to be lower than HI. Thermal decomposition of the hydrogen halides follows the equation: 2H⎯Cl(g) ⎯→ H2(g) + Cl2(g) 2H⎯I(g) ⎯→ H2(g) + I2(g) The relative stability depends on the strength of the hydrogen-halogen covalent bond. As iodine is below chlorine in Group 17, it is larger than chlorine. Thus, the H⎯Cl bond is shorter and stronger than the H⎯I bond making HCl more stable to heat. 20 (a)(i) They exhibit variable oxidation states. For example, MnSO4 (+2), MnO2 (+4), MnO4 2- (+6) and MnO4 – (+7). They exhibit catalytic properties. For example, MnO2 is used as a catalyst for the thermal decomposition of potassium chlorate(VII). KClO4(s) ⎯→ KCl(s) + 2O2(g) They form coloured compounds. For example, MnO4 2- (green) and MnO4 – (purple). (ii) When KMnO4 is added slowly to acidified H2O2, the purple colour of KMnO4 is decolourised and bubbles of gas are liberated. However, the solution turns pink when the KMnO4 is in excess. 2MnO4- + 6H+ + 5H2O2⎯→ 2Mn2+ + 5O2 + 8H2O In the reaction, the MnO4 – ion oxidises H2O2 to oxygen and the MnO4 – is converted to the colourless Mn2+ ion. (b) This is because CrCl3.6H2O can exist in three structural isomeric forms depending on how chlorine ligands are bonded covalently to the Cr3+ ion. The three possible isomers are: [Cr(H2O)6] 3+3Cl- (I) No Cl bonded to Cr3+ [Cr(H2O)5Cl]2+2Cl- H2O (II) 1 Cl bonded to Cr3+ [Cr(H2O)4Cl2 ] + Cl- 2H2O (III) 2 Cl bonded to Cr3+ To differentiate them, we first prepare aqueous solutions of the three salts with the same concentration (the three isomers have the same molar mass). Excess aqueous silver nitrate is then added to precipitate the Cl- as silver chloride, AgCl. The Cl that is covalently bonded to the Cr3+ in the complex ion would not be precipitated. [Cr(H2O)6] 3+3Cl- + aq ⎯→ [Cr(H2O)6] 3+(aq) + 3Cl- (aq) [Cr(H2O)5Cl]2+2Cl- H2O ⎯→ [Cr(H2O)5Cl]2+(aq) + 2Cl- (aq) + H2O(l) [Cr(H2O)4Cl2] + Cl- 2H2O ⎯→ [Cr(H2O)4Cl2] + (aq) + Cl- (aq) + 2H2O(l) Then, Ag+ (aq) + Cl- (aq) ⎯→ AgCl(s) The precipitates are filtered and dried. They are then weighed. The mass of the precipitate from the three isomers (I), (II) and (III) are in the ratio of 3 : 2 : 1 respectively. (c) A transition element is defined as an element that can form at least one stable ion with an incomplete 3d subshell. The electronic configurations of Sc, Zn and Cu are shown below: Sc: [Ar] 3d1 4s 2 Zn: [Ar] 3d10 4s 2 Cu: [Ar] 3d10 4s 2 Scandium forms only one stable ion, Sc3+, where the 3d sub-shell is empty. Sc3+: [Ar] 3d° Zinc forms only one stable ion, Zn2+, where the 3d subshell is completely filled. Zn2+: [Ar] 3d10 As a result, they are not transition elements. On the other hand, Cu forms two stable ions, Cu+ and Cu2+. Cu+ : [Ar] 3d10 Cu2+: [Ar] 3d9 The Cu2+ ion has an incomplete 3d sub-shell. Thus, copper is a transition element.
279 Chemistry Term 1 STPM Index Index A Absorption 228 Active electrode 87 Adsorption 228 Alloy, tin 184 Aluminium nitride 132 Aluminium, recycle 94 Aluminium/air fuel cell 81 Amphiboles 180 Anode 57 Anodisation 95 Anticlockwise rule 62 Asbestos 180 Atomic radius 110 B Beryllium, anomalous behaviour of 156 Bidentate ligand 232 Black-and-white photography 201 Bond energy 13 Born-Haber cycle 21 C Catalyst, transition elements 226 Cathode 57 Cathodic protection 93 Cell diagram 57 Ceramics 183 Chain silicate 179 Clay 181 Complex 229 Complex, geometry of 233 Corrosion of iron 91 Covalent radius 110 D Daniell cell 39 d-d transition, complex 219 Diagonal relationship 157 Diaphragm cell 95 Disproportionation 44 Dynamic equilibrium 73 E Effective nuclear charge 111 Effluent, treatment of 96 Electric car 81 Electrochemical cell 57 Electrode potential 51 Electrolysis 82 Electron affinity 17 Electronegativity 122 Electroplating 96 Endothermic reaction 2 Enthalpy change of reaction 2 Enthalpy of dilution 10 Equilibrium constant 73 Exothermic reaction 2 Extraction of aluminium 93 F Faraday’s first law of electrolysis 89 Faraday’s second law of electrolysis 90 First ionisation energy 16 Framework silicate 181 Fuel cell 79 G Galvanisation 92 Giant structure silicate 181 Glass 182 Group (Periodic Table) 109 H Half-equation 39 Halogen 190 Hess’ law 19 Heterogeneous catalysis 228 Hexadentate ligand 233 Homogeneous catalysis 227 Hydrogen fuel cell 80 Hydroquinone 202 I Ionic radius 115 L Lattice energy 21 Ligand 231 M Metallic radius 110 Mica 181 Molar enthalpy of vaporisation 118 Monodentate ligand 231 N Negative electrode potential 51 Nernst equation 69 Nuclear charge 110 O Overpotential 86 Oxidation 39 Oxidation number 41 Oxidation state 41 Oxidising agent 40 P Period (Periodic Table) 109 Photochromic lens 202 Positive electrode potential 52 Pyroxenes 179 R Redox reaction 39 Reducing agent 40 Reduction 39 Rusting of iron 91 S Sacrificial anode 93 Salt bridge 39 Screening effect 110 Second ionisation energy 16 Sheet silicate 180 Silicate 178 Silicone 177 Solubility product 74 Standard electrode potential 50 Standard electrode potential series 56 Standard enthalpy change of atomisation 12 Standard enthalpy change of formation 5 Standard enthalpy change of hydration 17 Standard enthalpy change of neutralisation 9 Standard enthalpy change of reaction 3 Standard enthalpy change of solution 18 Standard enthalpy of combustion 8 Standard hydrogen electrode 53 Standard ionisation energy 16 Standard reduction potential 52 T Talcum 181 Thermochemical equation 3 Transition element, definition 209 Z Zeolite 182
280 Chemistry Term 1 STPM Index Notes