Pra-U STPM Chemistry Penggal 2 2019 CB039349b

Chemistry Term 2 STPM Chapter 8 Electrochemistry 93 8 (b) This is because iron (E° = –0.44 V) is more reactive than tin (–0.14 V). Use of Sacrificial Anode 1 This method make use of the principle that if iron is in contact with a more electropositive element (i.e. one with a more negative E° value than –0.44 V), the metal will get oxidised instead of iron. 2 One example is galvanised iron described above. Here, zinc acts as a sacrificial anode. This method is sometimes called cathodic protection where iron is prevented from rusting by making it the cathode of an electrochemical cell. 3 Another example of sacrificial anode is magnesium. Blocks of magnesium are bolted to the hull of ships to prevent the iron structure from rusting. 4 Blocks of magnesium are also connected to underground pipes for water or oil to prevent the pipes from rusting. The magnesium blocks are replaced from time to time. 5 Note that iron will rust faster under the following conditions: (a) Presence of dissolved salt As rusting is an electrolytic process, anything that increases the conductivity of the electrolyte (water in this case) would increase the rate of rusting. That is why iron objects in the vicinity of the sea would rust faster. (b) At elevated temperature This is a kinetic factor. Increasing temperature will increase the rate of reaction. In this case, the rate of rusting. (c) Total surface area of iron Rusting is a heterogeneous reaction involving iron (solid), oxygen (gas) and water (liquid). Hence, the larger the surface area of iron, the faster will be the rusting process. That is why a piece of iron with pitted surface will rust faster than one with a smooth surface. (d) When iron is in contact with less reactive metal, a short-circuit cell is formed in the presence of an electrolyte. Iron now forms the anode of the cell and will undergo oxidation. Extraction of Aluminium 1 Aluminium is extracted electrolytically from its ore, bauxite, which is hydrated aluminium oxide, Al2O3•2H2O. 2 Due to the high melting point of aluminium oxide (2070 °C), it is impractical to electrolyse molten aluminium oxide directly. 3 Instead, purified bauxite is dissolved in molten cryolite, Na3AlF6, at 900 °C. A little sodium fluoride is added to further lower the melting point of aluminium oxide. Molten cryolite, 900 °C Al2O3(s) ⎯⎯⎯⎯⎯⎯⎯⎯→ 2Al3+(l) + 3O2–(l) Sacrifical anode and cathodic protection Mg2+ + 2e– Hull of ship Mg block Ground surface Magnesium Iron/steel pipe Mg2+ + 2e– Conditions the accelerate rusting The function of cryolite is to lower the melting point of aluminium oxide. 2015/P2/Q18(b) 2017/P2/Q5 2018/P2/Q16

Chemistry Term 2 STPM Chapter 8 Electrochemistry 94 8 4 The molten electrolyte is electrolysed using graphite electrodes. The steel tank is lined with graphite, which acts as cathode. Graphite Electrolyte Outlet Steel container Graphite Liquid aluminium + – 5 At the cathode, aluminium(III) ion is reduced to aluminium (melting point = 660 °C) Al3+(l) + 3e– ⎯→ Al(l) The liquid aluminium (with density higher than the electrolyte) sinks to the bottom of the cell and can either be tapped out or siphoned off into moulds, where it solidifies into blocks. 6 At the anode, oxygen gas is produced. 2O2–(l) ⎯→ O2(g) + 4e– The oxygen liberated reacts with the graphite anode to form oxides of carbon. C(s) + O2(g) ⎯→ CO2(g) 2C(s) + O2(g) ⎯→ 2CO(g) As a result, the anode has to be replaced from time to time. 7 A high current of about 40 000 A is used to generate the heat necessary to keep the electrolyte in the molten state. 8 A low voltage of about 4 V is used to prevent the F– ions from being discharged. This is because fluorine gas and fluorine compounds (such as CF4) are poisonous. 2F– (l) ⎯→ F2(g) + 2e– C(s) + 2F2(g) ⎯→ CF4(g) 9 However, electrolytic extraction of aluminium is expensive because of the large amount of electricity used. It requires about 297 kJ of electrical energy to produce 1 mole of aluminium from bauxite. 10 Hence, recycling of aluminium is important to reduce the cost of producing aluminium. 11 In recycling aluminium, the aluminium objects (such as cans for drinks) are first cleaned and then heated to 660 °C where aluminium melts. The molten aluminium is then drained into moulds for it to solidify. 12 The energy required to produce 1 mole of aluminium through recycling is about 26 kJ, which amounts to only 8.8% of the energy required for the extraction process. 13 The advantages of recycling aluminium are: (a) It is cost saving. (b) It cuts down environmental pollution from discarded aluminium containers. The anode burns away as CO2 and CO. A high current is required to maintain the temperature at 900 °C. Expensive way of producing aluminium Advantages of recycling The combustion of graphite provides the heat to keep the electrolyte at 900°C VIDEO Extraction of Aluminium

Chemistry Term 2 STPM Chapter 8 Electrochemistry 95 8 (c) It reduces toxic waste that is produced during the purification of bauxite. (d) It conserves environment by reducing the number of aluminium mines. Anodisation 1 Aluminium (though a reactive metal, E° = –1.66 V) is resistant to corrosion due to the presence of a thin protective oxide layer (about 10–8 m in thickness) on its surface. This impervious oxide layer shields aluminium from moisture and air. 2 Sometimes, it is desirable to increase the thickness of this oxide layer (to  10–5 m) to further increase its resistance to corrosion (for example in cooking utensils). This is done electrolytically by a process called anodisation. 3 The electrolytic cell on the right is used for the purpose of anodisation of aluminium. 4 The electrolyte is either dilute sulphuric acid or dilute chromic acid. The aluminium object is made as the anode, while the cathode can be either graphite or platinum. 5 At the anode, water is oxidised to oxygen: 2H2O(l) ⎯→ O2(g) + 4H+ + 4e– The oxygen liberated oxidises aluminium to aluminium oxide. 4Al(s) + 3O2(g) ⎯→ 2Al2O3(s) 6 At the cathode, water is reduced to hydrogen gas. 2H2O(l) + 2e– ⎯→ H2(g) + 2OH– (aq) 7 The overall reaction is: 2Al(s) + O2(g) + H2O(l) ⎯→ Al2O3(s) + H2(g) 8 If dyes are added to the electrolyte, the aluminium object will be coloured when the dyes are adsorbed onto the oxide layer. The Diaphragm Cell (Manufacture of Chlorine) 1 A diagram of the diaphragm cell is shown below. Steel Cl2 + – H2 NaOH(aq) + NaCl(aq) Asbestos diaphragm Brine NaCl(aq) Titanium Brine enters the cell through the anode compartment. 2 At the titanium anode, chloride ions are oxidised to chlorine gas. 2Cl– (aq) ⎯→ Cl2(g) + 2e– This leaves an excess of Na+ in the anode compartment. Aluminium is resistant to corrosion. Increasing the thickness of the oxide layer Al C/Pt H2 SO4 (aq) Sulphuric acid or chromic acid is used as electrolytic. Anodised aluminium is usually used to make cooking utensils. Brine is concentrated sodium chloride solution. 2013/P2/Q7 INFO Manufacture of Chlorine 2011/P1/Q18 2009/P2/Q2(b) 2016/P2/Q5 2013/P2/Q20(b) 2018/P2/Q6

Chemistry Term 2 STPM Chapter 8 Electrochemistry 96 8 3 At the steel cathode, hydrogen is liberated. 2H2O(l) + 2e– ⎯→ H2(g) + 2OH– (aq) This leaves an excess of OH– in the cathode compartment. 4 The level of the electrode in the anode compartment is deliberately higher than the cathode compartment to force the electrolyte through the asbestos diaphragm, carrying with it the excess Na+ ions, where it combines with OH– to form sodium hydroxide. 5 The higher level in the anode compartment is also to prevent the migration of OH– into the anode compartment where it might be discharged as oxygen gas. 6 Sodium hydroxide and spent brine is drained out from the cell. This solution contains about 10% by mass of NaOH and 15% by mass of sodium chloride. 7 The solution is evaporated to 1 3 its original volume where the less soluble NaCl will precipitate out, leaving an aqueous solution containing 50% by mass of NaOH and 1% by mass of NaCl. Treatment of Effluent 1 Industrial effluents (discharges) containing inorganic and organic wastes are often discharged into streams, rivers or seas. 2 These effluents sometimes contain heavy metal cation such as Cr3+, Cd2+ and Ni2+, which is harmful to human when ingested. 3 One way to remove these heavy metal cations is by way of electrolysis. Cr3+(aq) + 3e– Cr(s) E° = –0.25 V Cd2+(aq) + 2e– Cd(s) E° = –0.40 V Ni2+(aq) + 2e– Ni(s) E° = –0.91 V By controlled electrolysis, these metal cations can be reduced to their elemental state and removed from the water. Electroplating 1 The metal to be electroplated is made the cathode in an electrolytic cell. The anode is the electroplating metal and the electrolyte contains aqueous ions of the electroplating metal ions. 2 The following example is the electroplating of chromium onto iron to prevent the rusting of iron. 3 At the anode, chromium dissolves to form Cr3+(aq) ions. Cr(s) ⎯→ Cr3+(aq) + 3e– At the cathode, chromium is deposited onto the surface of the iron object. Cr3+(aq) + 3e– ⎯→ Cr(s) 4 The overall reaction is the transfer of chromium from the anode to the cathode. No mercury is used in this process. Exam Tips Exam Tips The diaphragm also prevents the chlorine produced to react with NaOH to produce sodium chlorate(I). Cl2 + 2NaOH → NaCl + NaOCl + H2O Sodium chlorate(I) is used as a domestic bleach. 2015/P2/Q18(b) Iron object Cr3+(aq) Cr(s) Cr Pt Industrial waste water The electrolytic cell for the removal of Cr3+(aq) ions.

Chemistry Term 2 STPM Chapter 8 Electrochemistry 97 8 SUMMARY SUMMARY 5 Electroplating can also be carried out on plastic objects. The plastic object is first coated with a graphite paste to make it conducting before the electroplating process begins. Plastic object coated with graphite Cr3+(aq) Cr(s) 1 Redox reactions are reactions that involve transfer of electrons from one species to another. 2 Oxidation is a process of electron loss. Reduction is a process of electron gain. 3 Oxidising agents are electron acceptors. Reducing agents are electron donors. 4 In redox reactions, the oxidising agents undergo reduction, while reducing agents undergo oxidation. 5 Anode is the electrode where oxidation takes place. Cathode is the electrode where reduction takes place. 6 The positive terminal of an electrochemical cell is the cathode of the cell. The negative terminal of an electrochemical cell is the anode of the cell. 7 The standard electrode potential is the potential difference between a metal in contact with its aqueous ions (of 1.0 M) relative to that of the standard hydrogen electrode at 298 K and 101 kPa. 8 The standard electrode potential can also be defined as the e.m.f. of a cell consisting of the standard hydrogen electrode and the metal in contact with its aqueous ions of 1.0 M at 298 K and 101 kPa. 9 In the standard electrode potential series, oxidising agents are found on the left-hand side of the half-equations while reducing agents are found on the right-hand side of the half-equations. 10 The more positive the value of E°, the more powerful is the oxidising agent. The more negative the value of E°, the more powerful is the reducing agent. 11 In an electrochemical cell, the half-cell with more positive Eo value forms the positive terminal (cathode) of the cell, while the half-cell with more negative Eo value forms the negative terminal (anode) of the cell. 12 The e.m.f. of an electrochemical cell = E°(Cathode) – E°(Anode). 13 The short-hand notation (or cell diagram) of an electrochemical cell: Electrons Electrode(s) | Electrolyte(aq) || Electrolyte(aq) | Electrode(s) Anode/negative Cathode/positive terminal terminal 14 A fuel cell is an electrochemical cell that converts the chemical energy of a continuous supply of reactants (or fuels) into electrical energy. 15 The factors affecting the values of electrode potentials are (a) concentration of the aqueous ions, (b) pressure of gases involved, (c) pH, (d) temperature, (e) complex formation. 2014/P2/Q7

Chemistry Term 2 STPM Chapter 8 Electrochemistry 98 8 16 The Nernst equation is given by: 0.059 [Reduced form] E = E° – ——– log ——————— z [Oxidised form] 0.059 [Oxidised form] or E = E° + ——– log ——————— z [Reduced form] 0.059 [Product ions] or Ecell = E°cell – ——– log —————— z [Reactant ions] 0.059 [Reactant ions] or Ecell = E°cell + ——– log —————— z [Product ions] 17 The equilibrium constant, Kc, can be calculated using the following formula: 0.059 E°cell = ——– log Kc 2 18 Rusting of iron is an electrochemical process. Water and oxygen must be present for iron to rust. 19 Rusting of iron can be prevented or slowed down by the following methods: (a) using a protective layer, (b) using a sacrificial anode, (c) formation of alloy. 20 Iron will rust faster: (a) in the presence of dissolved salts, (b) at elevated temperature, (c) when the total surface area of iron is increased, (d) when iron is in contact with a less reactive metal. 21 Electrolysis is the decomposition of an electrolyte using electricity. 22 In an electrolysis cell, oxidation occurs at the anode while reduction occurs at the cathode. 23 Factors affecting the products of electrolysis are: (a) the E° values, (b) the concentration of the aqueous ions, (c) the type of electrodes used. 24 Faraday’s first law of electrolysis states that the amount of substance produced during electrolysis is directly proportional to the quantity of charge (Q) that flows through the electrolyte, Q = It. 25 One Faraday (F) is the quantity of charge carried by one mole of electrons. F = Le = 96 500 C 26 Faraday’s second law of electrolysis states that the amount of substances produced by the same quantity of electrical charge is inversely proportional to the charge on the ions. 27 Electrolysis is used for (a) extraction of aluminium and chlorine, (b) anodisation of aluminium, (c) electroplating, (d) effluence treatment. STPM PRACTICE 8 Objective Questions 2 The graph below shows the standard electrode potentials, E°, of four elements P, Q, R and S. P Q R S E°/ V Based on the above graph, it can be concluded that, from P to S, 1 Consider the following three electrochemical cells: Zn/Zn2+ || Cu2+/Cu E°cell = 1.10 V Zn/Zn2+ || H+/H2/Pt E°cell = 0.76 V Cu/Cu2+ ||Fe3+/Fe2+/Pt E°cell = 0.43 V What is the E° value for the Fe3+/Fe2+ half-cell? A –0.44 V B +0.77 V C +1.19 V D +0.34 V

Chemistry Term 2 STPM Chapter 8 Electrochemistry 99 8 A the electrical conductivity increases. B the strength of the elements as oxidising agent increases. C the strength of the elements as reducing agents increases. D the physical state of the elements changes from liquid to solid. 3 An electrochemical cell is constructed using the following two half-cells. Cu2+(aq) + 2e– Cu(s) E° = +0.34 V Sn4+(aq) + 2e– Sn2+(aq) E° = +0.15 V Which of the following statements is not correct of the cell? A The e.m.f. of the cell is 0.19 V. B The Sn4+/Sn2+ half-cell forms the anode of the cell. C Electrons flow from the Cu2+/Cu half-cell to the Sn4+/Sn2+ half-cell. D The e.m.f. of the cell increases with increasing Cu2+ concentration. 4 The standard reduction potentials of two half-cells are given below: Fe3+(aq) + e– Fe2+(aq) Eo = +0.77 V Cr3+(aq) + e– Cr2+(aq) Eo = -0.41 V Which statement is true about the above data? A Cr3+(aq) is more stable than Cr2+(aq). B In an electrochemical cell, the Cr3+/Cr2+ halfcell is the cathode. C Fe3+(aq) is a stronger oxidising agent than Cr3+. D The e.m.f. of the cell constructed from the above two cells is 1.18 V. 5 Chlorine dioxide which is used to bleach paper pulp can be prepared by the following reaction: 2NaClO3(aq) + SO2(g) → 2ClO2(g) + Na2SO4(s) Which of the following statements regarding the reaction is not correct? A The SO2 acts as a reducing agent. B The ClO3 – ion undergoes disproportionation. C The oxidation state of sulphur increases by 2 units. D The ClO2 molecule has a bent structure. 6 The E° for several half-cells is given below: X2++ 2e– X –0.23 V Y3++ 3e– Y +0.57 V Z + e– Z– –1.05 V Which of the following species is the strongest reducing agent? A X C X2+ B Z– D Z 7 An electrolytic cell is prepared by using a zinc electrode immersed in aqueous zinc sulphate and a piece of platinum immersed in potassium hexacyanoferate(III). Zn2+ + 2e– Zn –.076 V [Fe(CN)6]3– + e– [Fe(CN)6]4– +0.36 V Which of the following statements is not true regarding the above cell? A The e.m.f. of the cell is 0.40 V. B The zinc electrode diminished in size. C Electrons flow from the zinc electrode to the platinum electrode. D The platinum electrode is the cathode. 8 Iron will rust faster except A at high temperature B when iron is in contact with zinc C in sea water D when iron is in contact with tin 9 The rusting of iron results in the formation of hydrated iron(III) oxide. Which is not true about the process? A It is an electrochemical process. B Oxygen and water must be present for rusting to occur. C The presence of dissolved salts accelerates the rusting process. D It occurs at the cathode. 10 Which of the following is not a redox reaction? A Rusting of iron B Anodisation of aluminium C Decomposition of hydrogen peroxide D Interconversion between chromate and dichromate ions 11 Given the following standard reduction potentials: Cu2+(aq) + 2e– Cu(s) +0.34 V Zn2+(aq) + 2e– Zn(s) -0.76 V What is the equilibrium constant of the following reaction at 25°C and 101 kPa? Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) A 1.1 × 1021 B 3.5 × 1022 C 1.9 × 1037 D Insufficient data

Chemistry Term 2 STPM Chapter 8 Electrochemistry 100 8 12 Battery for electric cars should have all of the properties listed below except A it must be cheap B it is able to supply high current density C it must be big D it must be able to store charge for long period of time 13 The standard reduction potentials of two half-cells are shown below: X: I2(aq), I– (aq)  Pt(s) +0.54 V Y: Cr2O7 2–(aq), Cr3+(aq), H+(aq)  Pt(s) +1.33 V Which statement is true about the electrochemical cell constructed from the two half-cells? A Platinum electrode in cell X is the cathode. B The e.m.f. of the cell is 1.87 V. C The colour of the solution changes from purple to colourless in half-cell Y. D Electrons flow from half-cell Y to half-cell X. 14 The reaction occurring in the Daniell cell is given below: Zn(s) + Cu2+(aq) ⎯→ Zn2+(aq) + Cu(s) What happens when the e.m.f. of the cell drops to zero? A All the zinc has been used up. B The electrode potentials of both the half-cells become zero. C The electrode potentials of both the half-cells have the same value. D All the Cu2+ ions have been reduced to Cu. 15 Which of the following statements best describes an electrolyte? A Electrolytes conduct electricity. B Electrolytes are ionic compounds. C Electrolytes are decomposed by the passage of an electric current. D Electrolytes are liquids at room conditions. 16 Which statement is not true about the diaphragm cell in the production of chlorine? A Titanium is used as the anode. B One of the by-products is sodium chlorate(V). C Hydrogen gas is produced at the cathode. D Asbestos is used as the diaphragm. 17 Copper(I) aqueous ions disproportionate according to the equation 2Cu+(aq) ⎯→ Cu(s) + Cu2+(aq) What conclusion can be drawn from the above information? A Copper(I) compound cannot exist. B The e.m.f. of the reaction is positive. C Copper(I) compounds are colourless. D The following reaction can also take place: 3Fe2+(aq) ⎯→ 2Fe3+(aq) + Fe(s) 18 Consider the following electrochemical cell: Pt(s) | X2+(aq), X4+(aq) || YO3 – (aq), Y– | Pt(s) When a current is drawn from the cell, which of the following species is reduced? A X2+ C YO3 – B X4+ D Y– 19 Which of the following half-equations is balanced? A 2H2O + 2e– ⎯→ H2 + 2OH– B Cl2 + e– ⎯→ 2Cl– C Cl2 + Ag ⎯→ AgCl + Cl– D S2O8 2– ⎯→ 2SO4 2– + 2e– 20 In the following equation, 2MnO4 – + 16H+ + 10Cl– → 2Mn2+ + 5Cl2+ 8H2O Which element is being reduced? A Mn C Cl B H D O 21 Which of the following is not an oxidation reaction? A S2O3 2– ⎯→ S4O6 2– C SO4 2– ⎯→ S2O8 2– B VO2 + ⎯→ VO2+ D H2O2 ⎯→ O2 22 The Nernst equation is given as follows: 0.059 [Reduced species] E = E° – ——– lg ——————–— n [Oxidised species] The e.m.f. of the cell: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) is 1.10 V under standard conditions. What would be the e.m.f. of the cell if the concentration of Cu2+(aq) is increased to 2.0 M and the concentration of Zn2+(aq) is maintained at 1.0 mol dm–3? A 0.34 C 1.09 B 0.76 D 1.11 23 In electrolysis of brine using the diaphragm cell, the mole ratio of Cl2 : NaOH : H2 produced is A 1 : 1 : 1 C 2 : 1 : 2 B 1 : 2 : 1 D 1 : 2 : 3 24 What is the most probable effect on the electrode potential of the following half-cell when the pressure of the gas is increased at constant temperature? Cl2(g) + 2e– 2Cl– (aq) A Unchanged C Decrease B Increase D Increase then decrease

Chemistry Term 2 STPM Chapter 8 Electrochemistry 101 8 25 Dilute sulphuric acid is electrolysed using copper as the anode and platinum as the cathode. What happens at the anode and cathode? Anode Cathode A Copper deposited. Copper dissolves. B Copper deposited. Hydrogen is produced. C Oxygen is produced. Hydrogen is produced. D Copper dissolve. Copper deposited. 26 The reaction of potassium manganate(VII) with concentrated hydrochloric acid is shown below: 2MnO4 – + 10HCl + 6H+ → 2Mn2+ + 8H2O + 5Cl2 Which of the following statements regarding the reaction is not correct? A HCl is reduced to Cl2. B The change in oxidation state of manganese is 5 units. C HCl is present as impurity in the Cl2 collected. D K+ is a spectator ion. 27 Aluminium is produced via the electrolysis of purified bauxite. Which statement is not true about the process? A A low voltage is used to prevent the discharge of F– ions. B The electrode used is graphite. C The electrolyte consists of molten bauxite. D Cryolite is added to lower the melting point of aluminium oxide. 28 Consider the following redox equation: aHNO3 + bI2 ⎯→ cNO2 + dHIO3 + eH2O What are the values of a, b, c, d and e in a balanced equation? a b c d e A 1 10 1 2 3 B 10 1 10 2 4 C 8 3 8 3 12 D 3 8 3 8 12 29 Disproportionation occurs when A a single substance undergoes decomposition B there is no net transfer of electrons during a reaction C the oxidation state of an atom is both raised and lowered in a reaction D two substances react to form a single compound 30 What quantity of electric charge is required to deposit all the silver metal onto a suitable cathode from 150 cm3 of 0.25 M silver nitrate solution? A 96 500 C C 24 125 C B 48 250 C D 3620 C 31 A cell diagram is shown below: Zn(s)  Zn2+(1.10 M)  Cu2+(0.55 M)  Cu(s) Given the standard e.m.f. of the cell is 1.10 V, what is the e.m.f. of the cell at 25°C? A +0.920 V C +0.980V B +1.09 V D +1.10 V 32 What is the standard e.m.f. of an electrochemical cell consisting of the two following half-cells? Cu2+(aq) + 2e– Cu(s) E°= +0.34 V VO2+(aq) + 2H+ + e– V3+ + H2O E°= +0.34 V A –0.68 V C +0.34 V B 0.00 V D +0.68 V 33 Blocks of magnesium are welded to the hull of ships to protect the steel from rusting. In this set-up, A the steel hull acts as the anode B the magnesium loses electrons C the magnesium acts as the cathode D the steel hull loses electrons 34 The reaction taking place in a vanadium battery is: V5+(aq) + V2+(aq) ⎯→ V3+(aq) + V4+(aq) During discharging, V2+ is oxidised to V3+. What is the reaction at the cathode when the cell is discharging? A V2+ ⎯→ V3+ + e– B V4+ ⎯→ V2+ + 2e– C V5+ + e– ⎯→ V4+ D V5+ + 2e– ⎯→ V3+ 35 The reaction in the zinc/mercury cell is: Zn(s) + HgO(s) + H2O(l) → Zn(OH)2(s) + Hg(l) During the discharging process, A zinc acts as anode and is oxidised B zinc acts as cathode and is reduced C electrons are transferred from the mercury to zinc D mercury acts as cathode and is oxidised 36 At the anode of an electrochemical cell, A oxidation takes place and its polarity is positive B oxidation takes place and its polarity is negative C reduction takes place and its polarity is positive D reduction takes place and its polarity is negative 37 Water can be converted to its elements by electrolysis using inert electrodes: 2H2O(l) ⎯→ 2H2(g) + O2(g)

Chemistry Term 2 STPM Chapter 8 Electrochemistry 102 8 What happens at the anode during the process? A Oxygen is produced and the pH around the electrode increases. B Oxygen is produced and the pH around the electrode decreases. C Oxygen is produced and the pH around the electrode remains constant. D Hydrogen is produced and the pH around the electrode increases. 38 The cell notation of an electrochemical cell is shown below: Zn(s)  Zn2+(0.10 M)  Zn2+(1.0 M)  Zn(s) Given: Zn2+(0.10 M) + 2e– Zn(s) Eo = -0.79 V Zn2+(1.0 M) + 2e– Zn(s) Eo = -0.76 V Which statement is not true about the electrochemical cell? A Electrons flow from the left to the right of the half-cell. B The concentration of Zn2+ on the right of halfcell increases. C The e.m.f. of the cell is 0.030 V. D The left of half-cell is the anode. 39 The standard electrode potentials, E°, of two halfcells are given below: O2(g) | H+(aq)||H2O2(aq) | Pt(s) +0.68 V Cr2O7 2–(aq) | H+(aq)||Cr3+(aq) | Pt(s) +1.15 V Based on the above information, which of the following is true? A O2 is a stronger oxidising agent than Cr2O7 2–. B The e.m.f. of a cell constructed by combining the above two half-cells is 1.83 V. C The colour of acidified Cr2O7 2– changes from orange to green when H2O2 is added to it. D H2O2 is a stronger oxidising agent than Cr3+. 40 The reaction taking place at the anode of a methane fuel cell is: CH4(g) + 2H2O(l) ⎯→ CO2(g) + 8H+ + 8e– What is the volume of methane (measured at s.t.p.) required to generate a current of 0.85 A for 3 hours? A 0.074 cm3 C 266.4 cm3 B 4.44 cm3 D 17.0 dm3 41 The standard e.m.f. of the Daniell cell is 1.10 V. What change can be made to increase the e.m.f. of the cell? Cu2+(aq) + Zn(s) ⎯→ Cu(s) + Zn2+(aq) A Increase the surface area of the copper electrode. B Adding a reagent that forms a complex with Cu2+(aq) ions. C Adding more solid CuSO4 to the copper halfcell. D Increase the concentration of Zn2+(aq) ions. 42 A current is passed through two cells connected in series. One contains XCl and the second contains YCl2. What is the ratio of the mass of X deposited to the mass of Y deposited at the cathode? [Relative atomic mass of X is twice the relative atomic mass of Y.] A 1 : 1 B 1 : 2 C 1 : 3 D 4 : 1 43 When 5 moles of electrons are passed through a solution containing the [AuCl4]– ion, what is the maximum mass of gold deposited at the anode? [Au = 197] A 118.2 g B 328.3 g C 591.0 g D 985.0 g 44 The standard e.m.f. of the Daniell cell is 1.10 V. If the concentrations of Cu2+ and Zn2+ were reduced to 0.25 M respectively, what is the value of the e.m.f. of the cell? A 0.00 V C 1.10 V B 0.96 V D 1.24 V 45 The standard electrode potentials for the Cu2+/Cu electrode and the Fe3+/Fe2+ electrode are +0.34 V and +0.77 V respectively. From this, it can deduced that A copper will reduce Fe3+ to Fe2+. B copper will oxidise Fe2+ to Fe3+. C copper is the cathode of the cell formed by combining the two half-cells. D copper and iron are both transition elements. 46 Which of the following is not true during the electrolysis of aqueous potassium iodide? A Potassium is deposited at the cathode. B The region around the anode turns reddish brown. C Bubbles of gas are formed at the cathode. D The final solution contains potassium hydroxide.

Chemistry Term 2 STPM Chapter 8 Electrochemistry 103 8 47 The standard electrode potential of copper is +0.34 V. Which of the following is true from this data? A Copper can exhibit two different oxidation states. B Copper compounds are coloured. C Copper dissolves readily in concentrated alkali. D Copper will form the cathode when combined with the hydrogen electrode to form a cell. 48 Chlorine is produced in industry using the diaphragm cell. Which statement is true about the cell? A Chlorine is produced at the cathode. B The diaphragm is a porous platinum. C The anode is titanium. D Cryolite is added to increase the electrical conductivity of the electrolyte. 49 The standard reduction potentials of several halfcell reactions are given below: Fe3+ + e–   Fe2+ +0.77 V (I) Ag+ + e–   Ag +0.80 V (II) I2 + 2e–   2I– +0.54 V (III) Which statement deduced from the above data is true? A Ag is the strongest oxidising agent. B When aqueous silver nitrate is added to aqueous potassium iodide, silver and iodine are produced. C The e.m.f. of the electrochemical cell formed by the combination of (I) and (II) is –0.03 V. D When aqueous iron(III) chloride is added to aqueous potassium iodide, the solution turns dark brown. 50 The standard electrode potentials for two half-cells are given below: Fe3+(aq) + e–   Fe2+(aq) +0.77 V Sn4+(aq) + 2e–   Sn2+(aq) +0.15 V The two half-cells are combined to form an electrochemical cell. Which of the following is the correct cell notation? A Pt(s)|Sn2+(aq), Sn4+(aq)||Fe3+(aq), Fe2+(aq)|Pt(s) B Pt(s)|Fe3+(aq), Fe2+(aq)||Sn4+(aq), Sn2+(aq)|Pt(s) C Pt(s)|Sn4+(aq), Sn2+(aq)||Fe2+(aq), Fe3+(aq)|Pt(s) D Pt(s)|Fe2+(aq), Fe3+(aq)||Sn2+(aq), Sn4+(aq)|Pt(s) 51 Consider the electrochemical cell below: Cu(s)|Cu2+(aq)||Fe3+(aq), Fe2+(aq)|Pt(s) What would happen to the e.m.f. of the cell, if a reducing agent is added to the Fe3+/Fe2+ half-cell? A The e.m.f. remains constant. B The e.m.f. increases. C The e.m.f. decreases. D The e.m.f. increases then decreases. 52 Which of the following statements is correct for the anodisation of aluminium? A The electrolyte is aqueous aluminium sulphate. B Aluminium forms the anode of the cell. C The purpose of the anodisation is to purify the aluminium. D Oxidation occurs at the cathode. 53 The cell diagram of an electrochemical cell is shown below: Pt(s)|Sn2+(aq), Sn4+(aq)||Cu2+(aq)|Cu(s) Which of the following statements is not true? A The mass of copper increases with time. B The e.m.f. of the cell increases when the concentration of Cu2+ increases. C Electron flows from the copper electrode to the platinum electrode. D There is no net transfer of electron when the e.m.f. of the cell is zero. 54 When a quantity of electric current passes through an aqueous solution of silver nitrate, 2.16 g of silver is deposited at the cathode. If the same quantity of electric current is passed through an aqueous solution of copper(II) nitrate, what is the maximum mass of copper that will be deposited at the cathode? A 0.64 C 1.28 B 1.08 D 2.16 55 An aqueous solution of potassium iodide is electrolysed using silver electrodes. Which of the following is true of the electrolysis? A The solution around the cathode turns brown. B Silver is deposited at the cathode. C Bubbles are formed at both the electrodes. D A yellow precipitate is formed in the solution. 56 A current of 1.00 A is passed through a 0.25 mol dm–3 aqueous solution of copper(II) sulphate for 25.0 minutes. What is the maximum volume of gas (measured at s.t.p.) collected at the anode? A 22 cm3 C 93 cm3 B 87 cm3 D 372 cm3 57 The standard electrode potentials of two half-cells are shown below: Zn2+(aq) + 2e– Zn(s) –0.76 V Sn4+(aq) + 2e– Sn2+(aq) +0.15 V

Chemistry Term 2 STPM Chapter 8 Electrochemistry 104 8 Which statement(s) is/are true of an electrochemical cell constructed by combining the above two cells? I The e.m.f. of the cell is –0.61 V. II Zn2+(aq) is reduced to Zn. III The Sn4+/Sn2+ half-cell forms the cathode of the cell. A I only C II and III B III only D I, II and III 58 The standard electrode potentials of two half-cells are given below: AgCl(s) + e– Ag(s) + Cl– (aq) +0.22 V Cu2+(aq) + 2e– Cu(s) +0.34 V What is the standard electrode potential for the following half-cell? Cu2+(aq) + 2Ag(s) + 2Cl– (aq) Cu(s) + 2AgCl(s) A –0.12 V C +0.12 V B –0.56 V D +0.56 V 59 A diaphragm cell is used to manufacture chlorine. Which of the following statement is not true of the cell? A Chlorine is produced at the anode. B The diaphragm is made of platinum. C Sodium hydroxide is one of the by-product. D The electrolyte used is brine. 60 The half-cell equation and the standard reduction potential for zinc half-cell and silver half-cell are shown below: Ag+(aq) + e  Ag(s) E° = +0.80 V Sn4+(aq) + 2e  Sn2+(aq) E° = +0.15 V What is the e.m.f. of the cell constructed from the two half-cells? A (0.80) – 2(0.15) V C (0.80) + (0.15) V B (0.80) – (0.15) V D (–0.80) + 2(0.15) V 61 The standard reduction potential of the following half-cell is +1.36 V. Cl2(g) + 2e  2Cl– (aq) What is the reduction potential of the following half-cell? —1 2 Cl2(g, 5 atm) + e  Cl– (aq, 1.0 M) A –0.68 V C +1.36 V B +0.68 V D +1.38 V 62 An electrochemical cell is shown below: Salt bridge Pt V + – H2 at 1 atm H+ Fe (1.0 M) 3+(1.0 M) Fe2+(1.0 M) The standard reduction ppotential for the Fe3+/ Fe2+ half-cell is +0.77 V. Which of the following statement is true of the above cell? A Electrons flow from the Fe3+/Fe2+ half-cell to the standard hydrogen electrode. B The e.m.f. of the cell is -0.77 V C The cell diagram is Pt(s), H2(g) | H+(aq) || Fe2+(aq), Fe3+(aq) | Pt(s) D The standard hydrogen electrode forms the anode of the cell. 63 A graphite paste is first spread on the surface of plastic materials before the plastic material is electroplated. Why is graphite used in the process? A It is the most stable allotrope of carbon. B Graphite has a very high melting point. C Graphite is a conductor. D Graphite forms strong bonds with the plastic materials. Structured and Essay Questions 1 (a) Define standard electrode potential. (b) Draw the set-up of the apparatus you would use to measure the standard electrode potential of zinc. (c) Explain why the standard electrode potential of calcium cannot be measured directly by the method described in (b). 2 (a) Draw a labelled diagram of the standard hydrogen electrode. (b) Calculate the e.m.f. of the following cell. (i) Pt(s)/H2(g, 1 atm) | H+(pH = 5.0) || H+(pH = 0) | H2(g, 1 atm)/Pt(s) (ii) Zn(s) | Zn2+(aq, 1 M) || H+(aq, 1 M) | H2(g, 0.04 atm) / Pt(s) 3 (a) What is an electrochemical cell?

Chemistry Term 2 STPM Chapter 8 Electrochemistry 105 8 (b) The standard e.m.f. of the following cell is 0.56 V. Pt(s) | H2(g, 1 atm) | HCl(aq, 1.0 M) || M2+(aq) | M(s) (i) Draw a labelled diagram of the cell. (ii) Write a balanced equation for the overall cell reaction. (iii) Indicate the direction of the flow of electrons through the external circuit. (iv) Explain how the concentration of the aqueous solutions at the anode and cathode would change after the cell has been discharged for some time. (v) What is the standard electrode potential for the M2+/M half-cell? (vi) When the acid in the above cell is replaced by dilute sulphuric acid, the e.m.f. of the cell is 0.62 V at 298 K and 101 kPa. Calculate the concentration and the pH of the sulphuric acid. 4 Explain the following observations. (a) Hydrogen cannot reduce Cu2+(aq) under standard conditions even though the e.m.f. of the reaction is positive. (b) The [Fe(CN)6]3– ion is a weaker oxidising agent than [Fe(H2O)6]3+ ion. (c) The standard electrode potential of potassium cannot be directly measured. (d) When potassium manganate(VII) is added to acidified iron(II) sulphate, the purple colour of potassium manganate is decolourised. When potassium manganate(VII) is added to aqueous iron(II) sulphate, the purple colour of potassium manganate is also decolourised, but a brown precipitate is formed. 5 (a) What do you understand by the term fuel cell? (b) Ethanol, C2H5OH can be oxidised electrolytically to ethanal, CH3CHO, by oxygen. In the reaction, oxygen is reduced to water. You are provided with an aqueous solution of ethanol in dilute sulphuric acid, oxygen gas, platinum electrodes, a porous partition and aqueous sulphuric acid. (i) Sketch a diagram of the fuel cell. (ii) Label the anode and cathode. (iii) Construct two half-equations to represent the reactions. (iv) Write an overall equation for the cell reaction. (v) Given that the standard electrode potential for the anode is +0.20 V, calculate the standard e.m.f. of the cell. 6 An electrochemical cell is set up using the following two half-cells: Sn4+(aq) + 2e– Sn2+(aq) Eo = +0.15 V Fe3 +(aq) + e– Fe2+(aq) Eo = +0.77 V (a) Draw a labelled diagram of the cell and label the anode and cathode of the cell and the direction of flow of electrons. (b) Write a balanced equation for the cell reaction and calculate the Eo cell of the cell. (c) Calculate the equilibrium constant, Kc, of the cell reaction at 25°C. (d) What is the effect of the Eo cell when a few drops of an oxidising agent is added to the Fe3+/Fe2+ half-cell. Explain your answer. 7 Aqueous sodium sulphate is electrolysed using platinum electrodes. The volumes of the gas collected at the cathode and anode are in the ratio of 2 : 1. (a) Draw a diagram of the electrolysis cell. (b) Write the balanced equations for the reactions taking place at the anode and cathode. (c) Explain why the two volumes are in the ratio 2 : 1. V Fe3+ (aq) Hg2 Cl2 (s) Pt wire Fe2+ (aq) Cl– (aq) Pt Hg(l) X Y

Chemistry Term 2 STPM Chapter 8 Electrochemistry 106 8 (d) What are the acidity of the solutions surrounding the cathode and anode? Explain your answer. (e) When a current flows through the electrolysis cell for 30 minutes, 35.6 cm3 of gas (measured at s.t.p.) is collected at the cathode. Calculate the magnitude of the current. 8 Explain how equilibrium is achieved when a zinc plate is partially immersed in aqueous zinc sulphate. 9 The diagram shows an electrochemical cell consisting of the calomel electrode, X and the Fe3+/Fe2+ half-cell, Y. The half-cell reactions are: Hg2Cl2(s) + 2e– 2Hg(l) + 2Cl– (aq) E° = +0.24 V Fe3+(aq) + e– Fe2+(aq) E° = +0.77 V (a) Draw a simplified diagram for the cell. (b) Name the cell (X or Y ) that acts as the anode. (c) Calculate the e.m.f. of the cell under standard conditions. (d) Write a balanced equation for the overall reaction taking place as current is drawn from the cell. (e) An oxidising agent, Z, is added gradually to cell Y. (i) Sketch the graph of e.m.f. against volume of Z added. (ii) Explain the shape of your graph. [Fe3+] (iii) Calculate the e.m.f. of the cell if the ——– = 1.0  1010. [Fe2+] 10 (a) Describe the industrial manufacture of aluminium from purified bauxite, and list any interesting features of the process. (b) Explain why recycling of aluminium is preferred to the electrolytic extraction of aluminium. (c) A sample of industrial effluent contains Cr3+ ion that is detriment to health. (i) State one adverse effect of Cr3+ on health. 11 (a) State Faraday’s first and second law of electrolysis. (b) Describe, with the aid of a diagram, how you would determine the value of the Avogadro’s constant by an electrolysis experiment. You are provided with: Aqueous copper(II) sulphate, 1.0 mol dm–3, two copper plates, propanone, connecting wires, ammeter, rheostat, stopwatch, weighing machine. 12 The diagram shows the electrolytic cell used to purify copper. (a) Give the polarity of electrode X and Y. (b) Name a substance that is suitable to be used as the electrolyte. (c) Write equations to represent the reactions taking place at electrode X and Y. (d) State any changes in the physical appearance of electrodes X and Y. (e) The piece of impure copper contains zinc and silver. Explain what happens to the impurities. 13 The cell diagram for an electrochemical cell with a standard e.m.f. of +0.76 V is shown below: Zn(s)|Zn2+(aq)||H+(aq), H2(g)|Pt(s) (a) Write the overall cell reaction. (b) Calculate the standard electrode potential of the following half-cell reaction. Zn2+(aq) + 2e–   Zn(s) (c) Given that the standard electrode potential of the following half-cell: Cu2+(aq) + 2e–   Cu(s) is +0.34 V. Pure copper Impure copper Electrolyte X Y

Chemistry Term 2 STPM Chapter 8 Electrochemistry 107 8 (i) Explain what happens when a piece of zinc is added to an aqueous solution of copper(II) sulphate. (ii) Write a balanced equation for the reaction in (c)(i). 14 The diagram below shows a diaphragm cell used in the manufacture of chlorine. Compound X + spent electrolyte Electrolyte + – Diaphragm (a) Name the electrolyte. (b) Write an ionic equation for the reaction taking place at the (i) anode (ii) cathode (c) What is the purpose of the asbestos diaphragm? (d) Name the by-product X. (e) Write a balanced equation for the reaction between chlorine and substance X at (i) room temperature, and (ii) 70 °C. (f) Give one commercial use of the product formed in (e)(i). 15 When solid copper(I) chloride is added to water, a brown solid and a blue solution is obtained. (a) Write the electronic configuration of the copper(I) ion. (b) What is the colour of copper(I) chloride? Explain your answer. (c) Use the appropriate standard electrode potentials to explain the reaction taking place. (d) What is the name given to the above reaction? 16 (a) An electrochemical cell is constructed using the following two half-cells. Ag2CrO4(s) + 2e– 2Ag(s) + CrO4 2–(aq) E° = +0.45 V Li+(aq) + e– Li(s) E° = –3.05 V (i) Name the half-cell that will form the cathode of the cell. (ii) Write an equation for the cell reaction. (iii) Calculate the e.m.f. of the cell under standard conditions. (b) This question is about the hydrogen/oxygen fuel cell. (i) What is the difference between a fuel cell and an 'ordinary' electrochemical cell? (ii) Draw a labelled diagram of the cell and explain its action. (iii) Name one advantage and one disadvantage of the fuel cell. 17 (a) The standard reduction potentials for three half-cells at 298 K are given below: Half-cell reaction E°/V F2(g) + 2e ↔ 2F– (aq) +2.87 MnO4 – (aq) + 8H+(aq) + 5e ↔ Mn2+(aq) + 4H2O(l) +1.19 Mg2+(aq) + 2e ↔ Mg(s) –2.38 Which species is the most powerful reducing agent? Explain your answer. (b) An electrochemical cell was set up using the following half-cells: Fe3+(aq) + e ↔ Fe2+(aq) E° = +0.77 V Fe3+(aq) + 3e ↔ Fe(s) E° = –0.04 V Write the cell diagram and calculate the standard e.m.f. of the cell. Name the cathode of the cell. (c) Corrosion of iron is an electrochemical process where iron is oxidised to rust, Fe2O3 .xH2O. State three conditions under which iron will rust at a faster rate.

CHAPTER PERIODIC TABLE: 9 PERIODICITY Concept Map Learning earning Outcomes The Periodic Table and Periodicity Periodicity of Physical Properties • Atomic radius • Ionic radius • Melting/Boiling points and molar enthalpy of vaporisation • Electrical conductivity • Electronegativity • Ionisation energy Chemical Properties of Period 3 Elements • Reaction with oxygen • Physical properties of the Period 3 oxides • Acid/Base properties of oxides • Reactions of elements with water Students should be able to: Physical properties of elements of Period 2 and Period 3 • interpret and explain the trend and gradation of atomic radius, melting point, boiling point, enthalpy change of vaporisation and electrical conductivity in terms of structure and bonding; • explain the factors influencing ionisation energies; • explain the trend in ionisation energies across Period 2 and Period 3 and down a group; • predict the electronic configuration and position of unknown elements in the Periodic Table from successive values of ionisation energies. Reactions of Period 3 elements with oxygen and water • describe the reactions of Period 3 elements with oxygen and water; • interpret the ability of elements to act as oxidising and reducing agents. Acidic and basic properties of oxides and hydrolysis of oxides • explain the acidic and basic properties of the oxides of Period 3 elements; • describe the reactions of the oxides of Period 3 elements with water; • describe the classification of the oxides of Period 3 elements as basic, amphoteric or acidic based on their reactions with water, acid and alkali; • describe the use ofsulphur dioxide in food preservation.

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 109 9 9.1 Physical Properties of Elements of Period 2 and Period 3 1 In the Periodic Table, the elements are arranged in order of increasing proton numbers. 2 The modern Periodic table has many forms. In 1984, the International Union of Pure and Applied Chemistry (IUPAC) officially adopted the form shown below. 3 The elements in the Periodic Table are arranged into seven horizontal rows (called Period), eighteen vertical columns (called Groups) and four blocks (s, p, d and f). 4 The Group numbers run from 1 to 18 and the Period numbers run from 1 to 7. 5 Groups 1, 2, 13, 14, 15, 16, 17 and 18 are also represented by the Roman numerals of I, II, III, IV, V, VI, VII and VIII respectively. 7 periods and 18 groups 1 H 3 Li 4 Be 11 Na 12 Mg 19 K 20 Ca 37 Rb 38 Sr 55 Cs 56 Ba 87 Fr 88 Ra 21 Sc 22 Ti 39 Y 40 Zr 57 La 72 Hf 89 Ac 104 Rf 23 V 41 Nb 73 Ta 90 Th 58 Ce 91 Pa 59 Pr 92 U 60 Nd 93 Np 61 Pm 94 Pu 62 Sm 95 Am 63 Eu 96 Cm 64 Gd 97 Bk 65 Tb 98 Cf 66 Dy 99 Es 67 Ho 100 Fm 68 Er 101 Md 69 Tm 102 No 70 Yb 103 Lr 71 Lu 105 Db 24 Cr 42 Mo 74 W [106] 25 Mn 43 Tc 75 Re [107] 26 Fe 44 Ru 76 Os [108] 27 Co 45 Rh 77 Ir [109] 28 Ni 46 Pd 78 Pt 29 Cu 47 Ag 79 Au 30 Zn 48 Cd 80 Hg 31 Ga 49 In 81 Tl 32 Ge 50 Sn 82 Pb 33 As 51 Sb 83 Bi 34 Se 52 Te 84 Po 35 Br 53 I 85 At 36 Kr 54 Xe 86 Rn 3 4 5 6 7 8 9 10 11 12 (III) 13 (IV) 14 (V) 15 (VI) 16 (VII) 17 (VIII) 18 5 B 6 C 7 N 8 O 9 F 2 He 10 Ne 13 Al 14 Si 15 P 16 S 17 Cl 18 Ar (II) 2 (I) 1 s-Block p-Block d-Block f-Block Period 1 2 3 4 5 6 7 6 We shall look at how the physical properties of the elements changes with the proton numbers of the elements. 7 In this section, we shall study the periodic trend of atomic radius, ionic radius, melting point, boiling point, enthalpy change of vaporisation, electric conductivity, electronegativity. and ionisation energy.

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 110 9 Atomic Radius 1 Strictly, it is not possible to measure the absolute radius of an atom because the electron cloud theoretically extends to infinity. 2 So, we take the distance between the centres of adjacent atoms in a particular structure and halved it to give the atomic radius. 3 There are two main types of atomic radius that we will be dealing with: (a) Metallic radius The metallic radius is defined as half the distance between the centres of two adjacent atoms in the giant metallic structure. d Atomic radius = —d 2 (b) Covalent radius This is defined as half the distance between the centres of two adjacent atoms that are held together by a covalent bond. d Atomic radius = —d 2 4 The atomic radius of an atom (whether metallic or covalent) is affected by two factors: The nuclear charge and the screening effect. 5 The nuclear charge refers to the actual number of protons in the nucleus of an atom. The more protons an atom has, the stronger the attraction between the nucleus and the electron cloud. This will cause the atomic radius to decrease. 6 The screening effect arises from two sources: (a) Repulsion between electron shells As electron shells are negatively charged, they tend to repel one another. As a result, the more shells that are filled with electrons, the greater the repulsion. + Mutual repulsion Electron shell Repulsion between electron shells Info Chem The more electronic shells an atom has, the larger its size.

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 111 9 (b) The shielding of outer electrons from the nucleus. As electron shells are negatively charged, the inner shell electrons can shield the outer electrons from the ‘actual pull’ of the nucleus, much like a wall helps to shield us from direct sunlight. 7 Hence, the larger the screening effect, the weaker the attraction between the nucleus and the electron clouds. The electron clouds can drift further away from the nucleus causing the atomic radius to increase. 8 The combining effect of (a) the nuclear charge, and (b) the screening effect, gives rise to what is known as the effective nuclear charge. This is the net attractive force between the nucleus and the electrons concerned. Effective nuclear charge  (no. of protons) – (no. of inner electrons) 9 For example, the approximate effective nuclear charge felt by the outermost electrons in a magnesium atom is calculated as follows: Effective nuclear charge = (12) – (10) = +2 The shielding effect 12+ These two electrons feel the pull of 2 protons only Quick Check 9.1 Calculate the effective nuclear charge on the outermost electron(s) of the following atoms. (a) K (d) Al (b) Li (e) Pb (c) Br Variation of Atomic Radius Across a Period 1 The Period 3 elements consist of eight elements from sodium to argon. 2 The electronic configuration as well as the approximate effective nuclear charge of the atoms are given below: Element Na Mg Al Si P S Cl Ar Proton number 11 12 13 14 15 16 17 18 Electronic configuration 2.8.1 2.8.2 2.8.3 2.8.4 2.8.5 2.8.6 2.8.7 2.8.8 Effective nuclear charge 1 2 3 4 5 6 7 8 Atomic radius/ nm 0.191 0.160 0.143 0.117 0.110 0.104 0.099 0.094 2007/P1/Q5 INFO Periodicity

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 112 9 3 The graph below shows the variation of the atomic radius across Period 3. 9 10 11 12 13 14 15 16 17 18 19 11 12 13 14 15 16 17 18 Proton number Radius (× 10–2)/nm Na Mg Al Si P S Cl Ar 4 Going from sodium to argon, the number of protons and the number of electrons increase by one progressively. 5 But the additional electrons are added to the same electronic shell (the third shell). Hence, there is not much change in the screening effect (as the number of inner electrons remains the same), but there is a corresponding increase in the nuclear charge. 6 As such, the effective nuclear charge increases from +1 for sodium to +8 for argon. As a result, the outer electrons are being pulled closer to the nucleus causing the atomic radius to decrease across the period. 7 The same trend is also seen in Period 2 elements. Element Proton number Electronic configuration Effective nuclear charge Atomic radius/nm Li 3 2.1 1 0.157 Be 4 2.2 2 0.125 B 5 2.3 3 0.090 C 6 2.4 4 0.077 N 7 2.5 5 0.075 O 8 2.6 6 0.073 F 9 2.7 7 0.071 Ne 10 2.8 8 0.065 Atomic radius decreases across a period. Info Chem Effective nuclear charge increases, but the screening effect remains the same.

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 113 9 6 7 8 9 10 11 12 13 14 15 3 4 5 6 7 8 9 10 Proton number Radius (× 10–2)/nm Li Be B C N O F Ne Variation of Atomic Radius Down a Group 1 Group 2 of the Periodic Table consists of five elements from beryllium to barium (Radium is excluded as it is radioactive). 2 The table and graph below show the variation of the atomic radius of the elements. Element Be Mg Ca Sr Ba Proton number 4 12 20 38 56 Electronic configuration 2.2 2.8.2 2.8.8.2 2.8.18.8.2 2.8.18.18.8.2 Effective nuclear charge 2 2 2 2 2 Atomic radius/ nm 0.112 0.160 0.197 0.215 0.222 3 Going down the group: (a) the number of protons in the nucleus increases, (b) the number of electrons increases, (c) the number of electronic shells that is filled with electrons increases causing an increase in the screening effect, (d) as a result, the effective nuclear charge remains almost constant. Info Chem Effective nuclear remains almost constant, but the number of shells increases.

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 114 9 4 However, each subsequent atom has one extra electronic shell compared to the preceding ones. As a result, the attraction between the nucleus and electron clouds decreases, causing the atomic radius to increase. 12 11 13 14 15 16 17 18 19 20 21 22 10 20 30 40 50 60 Proton number Radius (× 10–2)/nm Ba Sr Ca Mg Be 5 The same trend is also observed for the other groups, such as the halogens (Group 17). Element Proton number Electronic configuration Effective nuclear charge Atomic radius/nm Fluorine 9 2.7 7 0.071 Chlorine 17 2.8.7 7 0.099 Bromine 35 2.8.18.7 7 0.114 Iodine 53 2.8.18.18.7 7 0.183 Atomic radius increases down a group.

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 115 9 8 10 12 14 16 18 20 22 Br Cl I 0 10 20 30 40 50 60 Proton number Radius (× 10–2)/nm 6 As a generalisation: Atomic radius decreases Atomic radius increases Ionic Radius Similar to that of the atomic radius, the ionic radius of an ion also depends on the nuclear charge and screening effect. Variation of Ionic Radius Across a Period 1 The table and graph below show the variation of the ionic radius with proton numbers for Period 3 elements. Ion Na+ Mg2+ Al3+ P3– S2– Cl– Ionic radius/nm 0.095 0.065 0.050 0.212 0.184 0.181 Proton number 11 12 13 15 16 17 No. of electrons 10 10 10 18 18 18 Electronic configuration 2.8 2.8 2.8 2.8.8 2.8.8 2.8.8 2 Sodium, magnesium and aluminium are metals. They form cations of Na+, Mg2+ and Al3+ respectively. 2007/P1/Q41, 2008/P2/Q7(a)

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 116 9 3 Phosphorous, sulphur and chlorine are non-metals. They form anions of P3–, S2– and Cl– respectively. 4 Silicon, on the other hand, does not form simple ions of either Si4+ or Si4–. 5 The anions are larger than the cations. This is because all the anions (with configuration of 2.8.8) have one extra electron shell compared to the cations (with configuration 2.8). 6 For the series of anions, the ionic radius decreases with increasing proton number. This is because all the anions are isoelectronic (having the same number of electrons), but the number of protons increases from P3–, to S2– and Cl– . This increases the attraction between the nucleus and the electrons. 7 The same trend is shown by the cations. The cations are isoelectronic, but the number of protons increases with the proton number of the elements. 2 4 6 8 10 12 14 16 18 20 P3– Na+ Mg2+ Al3+ S2– Cl– 11 12 13 14 15 16 17 18 P roton number Radius (× 10–2)/nm Anions are larger than the cations.

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 117 9 Quick Check 9.2 1 If silicon were to form the Si4+ ion, what would the ionic radius be? Explain your answer. 2 Which of the two ions, Li+ or Be2+, is expected to have a smaller ionic radius? Explain your reason. Variation of Ionic Radius Down a Group 1 The ionic radius of the Group 2 elements is shown below: Ion Be2+ Mg2+ Ca2+ Sr2+ Ba2+ Proton number 4 12 20 38 56 Electronic configuration 2 2.8 2.8.8 2.8.18.8 2.8.18.18.8 Ionic radius/nm 0.031 0.065 0.099 0.113 0.135 4 3 5 6 7 8 9 10 11 12 13 10 20 30 40 50 60 Proton number Radius (× 10–2)/nm Ba2+ Sr2+ Ca2+ Mg2+ Be2+ 2 The variation of the ionic radius follows the same trend as the atomic radius. Ionic radius increases down a group.

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 118 9 3 Each subsequent ion has one electronic shell more than the preceding one. 4 The cations are smaller than their corresponding neutral atoms. This is because the cations have two electrons less than their neutral atoms, but the number of protons is the same in both species. 5 The same trend is also shown by the halogens (Group 17). Ion F– Cl– Br – I – Proton number 9 17 35 53 Electronic configuration 2.8 2.8.8 2.8.18.8 2.8.18.18.8 Ionic radius/nm 0.133 0.181 0.196 0.216 10 20 30 Proton number 40 50 Radius (fi 10–2)/nm 22 20 18 16 14 12 10 F– Cl– Br– I – Quick Check 9.3 1 Which is larger? Explain your answer. (a) Mg or Mg2+ (b) Br or Br– (c) Si4+ or Si4– Melting/Boiling Point and Molar Enthalpy of Vaporisation (ΔH°V) 1 The melting point, boiling point and molar enthalpy of vaporisation of an element are a measure of the strength of the forces that holds the atoms/molecules of the elements together. 2 Molar enthalpy of vaporisation is the amount of energy required to convert one mole of a substance from the liquid state to vapour at the boiling point of the substance. Definition of ΔH°v Exam Tips Exam Tips Generally, species with higher proton ———–––– electron ratio will have a smaller size.

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 119 9 3 The stronger the ‘intermolecular force’, the higher the melting/ boiling point and molar enthalpy of vaporisation, as more energy is required to break or weaken the ‘intermolecular force’. 4 The table below lists the melting/boiling point and molar enthalpy of vaporisation of Period 3 elements. Element Na Mg Al Si P S Cl Ar M.p./°C 98 650 660 1423 44 120 –101 –189 B.p./°C 890 1120 2450 2680 280 445 –34 –186 ΔHv/kJ mol–1 89 129 294 377 12.4 9.7 10.2 6.5 Na Mg Al Si P S Cl Ar Temperature/°C Boiling point 2750 2500 2250 2000 1750 1500 1250 1000 750 500 250 0 –250 Melting point 11 12 13 Proton number 14 P S Cl Ar 15 Molar enthalpy of vaporisation/kJ mol–1 300 250 200 150 100 50 0 400 350 16 17 18 Na Mg Al Si Exam Tips Exam Tips The boiling point and the melting point of argon are almost similar. This indicates that the van der Waals forces in liquid argon and solid argon are of almost the same strength. 2011/P2/Q1(b) 2007/P1/Q5

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 120 9 5 The nature of Period 3 elements is given in the table below: Element Na Mg Al Si P S Cl Ar Structure Giant metallic Giant covalent Simple molecule Intermolecular force Metallic Covalent van der Waals Property Metal Metalloid Non-metal The elements change from giant metallic structures to simple molecular structures. 6 The melting/boiling point and molar enthalpy of vaporisation of the non-metals (P, S, Cl, Ar) are relatively lower than those of the metals (Na, Mg, Al). 7 This is because the ‘intermolecular forces’ between atoms of the metals are the strong metallic bonds, but the ‘intermolecular forces’ between the non-metallic elements are the weak van der Waals forces. 8 For the metals, the strength of the metallic bond increases from sodium to aluminium as the number of valence electrons donated to form the bonds increases from 1 for sodium to 3 for aluminium. This is reflected in the increase of the melting/boiling points and molar enthalpy of vaporisation from sodium to aluminium. 9 For the non-metals, phosphorous exists in the form of P4 molecules, sulphur as S8 molecules, chlorine as Cl2 molecules and argon is monoatomic. P4 molecule S8 molecule Cl2 molecule Ar 10 The number of electrons in the non-metallic ‘molecules’ is given below: Species P4 S8 Cl2 Ar No. of electrons 60 128 34 18 As a result, the strength of the van der Waals force as well as the melting/boiling point and molar enthalpy of vaporisation increase in the order: Ar < Cl2 < P4 < S8 11 The number of electrons in the S8 molecule is so large that the van der Waals force is even stronger than the metallic bond in sodium. 12 Silicon, a metalloid has exceptionally high melting/boiling point and molar enthalpy of vaporisation. This is because silicon has a giant covalent structure with strong covalent bonds holding the atoms together in a three dimension array. 13 Melting, boiling or vaporisation of silicon involves the complete breaking of the strong covalent bonds. Exam Tips Exam Tips The larger the number of electrons in a molecule, the stronger the van der Waals force. Info Chem The strength of the metallic bond: Na < Mg < Al

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 121 9 14 The same trend is also observed in Period 2 elements from lithium to neon. Element Li Be B C N O F Ne M.p./°C 181 1278 2300 3652 –210 –218 –220 –248 B.p./°C 1342 2961 2550 4827 –196 –183 –188 –246 ΔHv/kJ mol–1 135 295 539 717 2.79 3.41 3.27 1.77 Structure Giant metallic Giant covalent Simple molecule (a) Lithium and beryllium are metals. Their melting/boiling points and molar enthalpy of vaporisation are relatively higher than those of the non-metals: nitrogen, oxygen, fluorine and neon. (b) Boron and carbon have exceptionally high melting/boiling points and molar enthalpy of vaporisation because of their giant covalent structures. Electrical Conductivity 1 The graph below shows the variation of the electrical conductivity (μΩ–1) of the Period 3 elements. 11 12 13 14 15 16 17 Elements 18 Electrical conductivity/µΩ–1 Na Mg Al Si P S Cl Ar 2 Lithium and beryllium (of Period 2), and sodium, magnesium and aluminium (of Period 3) are metals and are good electrical conductors because of the presence of delocalised electrons in their metallic bonds. 3 Boron, carbon (diamond), nitrogen, oxygen, fluorine and neon (of Period 2), and phosphorous, sulphur, chlorine and argon (of Period 3) all exists as simple molecules. They do not have delocalised electrons in their molecular structure. As a result, they are all non-conductors. 4 Silicon is a metalloid which has partial metallic character. Thus, its conductivity is lower than the metals but is higher than the nonmetals. It is a semiconductor. 2016/P2/Q20(a) 2017/P2/Q6 2010/P2/Q6(a) Info Chem Pure silicon is not a conductor at room conditions. It starts conducting when heated to above 700 K or when it is doped with other elements such as boron or phosphorous.

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 122 9 Example 9.1 Explain why the electrical conductivity of the elements increases in the order Na < Mg < Al. Solution In the solid state, the sodium, magnesium and aluminium atoms are held together by metallic bonds. The conductivity of the metallic elements depends on the number of delocalised electrons that make up the metallic bonds. The number of electrons per atom that are contributed to form the metallic bonds increases from one for sodium, to two for magnesium and three for aluminium. This results in the increasing conductivity from sodium to aluminium Electronegativity 1 Electronegativity is a measure of the ability of an atom to attract the bonding electrons in a covalent bond to which the atom is bonded. It is a measure of the ‘electron pulling power’ of an atom. 2 Take the example of the hydrogen chloride molecule that is formed when the hydrogen atom and the chlorine atom share one electron each to form a covalent bond: H • Cl H Cl 3 The diagram above shows that the electron density in the ‘molecular orbital’ is evenly distributed. 4 However, this is not the case. Experiments show that chlorine exerts a stronger ‘pull’ for the electron cloud compared to hydrogen, causing a distortion. 5 The end of the molecule closer to chlorine now acquires a partial negative charge, while the end closer to hydrogen acquires a partial positive charge. The covalent bond is said to undergo polarisation. 6 In this case, chlorine has a higher electronegativity than hydrogen. 7 The electronegativity of an atom depends on the atomic radius and the nuclear charge. The smaller the atomic radius and/or the higher the nuclear charge, the higher the electronegativity. Definition of electronegativity Polarisation of the H – Cl bond H Cl δ+ δ– HCl is a polar molecule.

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 123 9 Variation of Electronegativity Across a Period 1 The electronegativity (according to the Pauli’s scale) for Period 3 elements is given below. Element Na Mg Al Si P S Cl Electronegativity 0.9 1.2 1.5 1.8 2.1 2.5 3.0 2 Going from sodium to chlorine, the atomic radius decreases while the effective nuclear charge increases. 3 As a result, the ‘electron-pulling power’ increases resulting in an increase in the electronegativity of the elements. 4 Similar trend is also shown by Period 2 elements. Period 2: Element Li Be B C N O F Electronegativity 1.0 1.5 2.0 2.5 3.0 3.5 4.0 11 12 13 Proton number 14 15 Electronegativity 4 3 2 1 0 16 17 3 4 5 6 7 8 9 Na Mg Al Si P S Cl Li Be B C N O F Period 3 Period 2 Period 3 Period 2 Variation of Electronegativity Down a Group 1 The electronegativity of the halogens (Group 17) is given in the table below. Element F Cl Br I Electronegativity 4.0 3.0 2.8 2.5 Electronegativity increases across a period. Info Chem The electronegativity of the noble gases (Group 18) is zero.

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 124 9 2 3 4 Br I F Cl 10 20 30 40 50 60 Proton number Electronegativity 2 Going down the group, the atomic radius increases while the effective nuclear charge remains almost constant. 3 This causes the ‘electron-pulling effect’ to decrease as reflected in their decreasing electronegativity. 4 Summary: Electronegativity increases Electronegativity decreases Ionisation Energy and Factors Affecting Ionisation Energy 1 Ionisation energy is the minimum energy required to remove the most loosely held electron from a gaseous atom or gaseous ion per mole of the atom/ion under standard conditions. 2 The first ionisation energy is the minimum energy required to remove the most loosely bound electron from a gaseous atom to form a unipositive ion, per mole of the atom under standard conditions. M(g) → M+(g) + e– For example, Na(g) → Na+(g) + e– ΔH = +494 kJ mol–1 Cl(g) → Cl+(g) + e– ΔH = +1260 kJ mol–1 Electronegativity decreases down a group. Fluorine is the most electronegative element. Definition of ionisation energy Definition of 1st ionisation energy

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 125 9 3 The second ionisation energy is the minimum energy required to remove the most loosely held electron from a gaseous unipositive ion to form a gaseous dipositive ion per mole of the ion under standard conditions. M+(g) → M2+(g) + e– For example, Na+(g) → Na2+(g) + e– ΔH = +4560 kJ mol–1 Cl+(g) → Cl2+(g) + e– ΔH = +2300 kJ mol–1 4 Higher ionisation energies are defined in the same manner. For example, the third and fourth ionisation energy of sodium is: Na2+(g) → Na3+(g) + e– ΔH = +6940 kJ mol–1 Na3+(g) → Na4+(g) + e– ΔH = +9540 kJ mol–1 5 The magnitude of the ionisation energy depends on the size of the atom, the effective nuclear charge and the type of electrons involved. 6 The smaller the size of the atom and/or the higher the effective nuclear charge, the more strongly the electrons are bound by the nucleus, the higher the ionisation energy. + + Stronger attractive force Weaker attractive force 7 In a particular principle shell, the ionisation energy increases in the order d < p < s. 8 More energy than expected is required to remove an electron from these 'stable' configurations: s 2 , p3 , p 6 , d 5 and d10. Trend of the First Ionisation Energy Across a Period 1 The table and graph below show the trend of the first ionisation energy of the Period 3 elements with proton numbers. Element Na Mg Al Si P S Cl Ar Proton number 11 12 13 14 15 16 17 18 Atomic radius/ nm 0.186 0.160 0.143 0.117 0.110 0.104 0.099 0.094 1st I.E./ kJ mol–1 496 738 578 798 1012 1000 1251 1520 Definition of 2nd ionisation energy Info Chem The higher the energy of an electron, the lower the ionisation energy. Exam Tips Exam Tips Energy d p s 2008/P2/Q2(b) 2015/P2/Q7 2013/P2/Q8 2016/P2/Q6 2007/P1/Q5

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 126 9 500 600 700 800 900 1000 1100 1200 1300 1400 1500 I.E. / kJ mol–1 11 12 13 14 15 16 17 18 Proton number Na Mg Ar Al Si S P Cl 2 The first ionisation energy generally increases across the period. This is due to a decrease in the atomic radius and an increase in the nuclear charge, the electrons are being held more strongly by the nucleus. 3 However, the increase is not a smooth one. There is a reversal of trend between magnesium and aluminium, and between phosphorous and sulphur. 4 The electronic configurations of magnesium and aluminium are: 3s 3p Mg:[Ne] 3s 2 Al: [Ne] 3s 2 3p1 Energy 738 kJ mol–1 10 inner electrons 12 inner electrons n = ∞ n = ∞ 578 kJ mol–1 [Ne] Magnesium [Ne] Aluminium The first electron removed from aluminium is from a 3p sub-shell which is of higher energy than the 3s sub-shell in magnesium. There is a general increase in the 1st I.E. across a period. Info Chem The outermost electron in aluminium is from a 3p orbital of higher energy. Hence, less energy is required to ‘promote’ it to n = ∞.

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 127 9 5 Furthermore, this electron in the 3p orbital also enjoys extra shielding by the two inner 3s electrons. This makes the electron easier to remove than expected. 6 The first electron removed from magnesium is from a fully filled 3s orbital. This arrangement has extra stability and requires more energy than expected. 7 The electronic configurations of phosphorous and sulphur are: P: [Ne] 3s 2 3p3 S: [Ne] 3s 2 3p4 3s 3p P: [Ne] 3s 2 3p3 S: [Ne] 3s 2 3p4 Energy 1012 kJ mol–1 1000 kJ mol–1 [Ne] Phosphorous [Ne] Sulphur Mutual repulsion n = ∞ n = ∞ 8 The first electron removed from sulphur is from a 3p sub-shell which is occupied by two electrons. The two electrons in the same orbital experience mutual repulsion, making the electron easier to remove. 9 The first electron removed from phosphorous is from a set of halffilled 3p sub-shell which has extra stability, thus requiring more energy than expected. 10 The same trend is also shown by the Period 2 elements. Element Li Be B C N O F Ne Proton number 3 4 5 6 7 8 9 10 Atomic radius/ nm 0.152 0.111 0.088 0.077 0.070 0.066 0.064 0.062 1st I.E./ kJ mol–1 520 900 801 1086 1402 1314 1681 2081 Info Chem Repulsion between the two electrons makes it easier to be removed. Extra stability of the 3p3 configuration in phosphorus. INFO First Ionisation Energy Across a Period

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 128 9 400 600 800 1000 1200 1400 1600 1800 2000 2200 I.E. / kJ mol–1 3 4 5 6 7 8 9 10 Proton number Ne F O N C B Be Li 11 The lower than expected values for boron and oxygen is because of the configuration of [He] 2s 2 2p1 and [He] 2s 2 2p4 respectively. Trend of the First Ionisation Energy Down a Group 1 The first ionisation energies of Group 2 elements are shown below: Element Be Mg Ca Sr Ba Proton number 4 12 20 38 56 Atomic radius/nm 0.11 0.16 0.20 0.21 0.23 1st Ionisation energy/kJ mol–1 900 740 590 550 500 0 500 400 700 800 900 600 Mg Be Ba Sr Ca 10 20 30 40 50 60 Proton number I.E. / kJ mol–1 2011/P1/Q4 Ionisation energy decreases down a group. Exam Tips The higher than expected values for Be and N is due to the stable 2s2 and 2s2 2p3 configurations respectively.

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 129 9 2 Going down Group 2, the atomic radius increases but the effective nuclear charge remains almost constant. 3 The attraction between the nucleus and electrons gets weaker causing a steady decrease in the first ionisation energies. 4 The same trend is also shown by the halogens (Group 17). Element F CI Br I Proton number 9 17 35 53 Atomic radius/nm 0.133 0.099 0.114 0.183 1st Ionisation energy/kJ mol–1 1680 1260 1140 1010 10 20 30 Proton number 40 50 First I.E./kJ mol–1 1500 1400 1300 1200 1100 1000 0 1700 1600 60 F Cl Br I The Electronic Configuration and Position of Elements from Successive Ionisation Energies 1 We can predict the valence-shell electronic configuration of an element by analysing the successive ionisation energy graph of the element and also from its position in the Periodic Table. 2 The successive ionisation energy graph of carbon, 6C is shown below. 1 2 3 Order of electron removed 4 5 Ionisation energy 0 6

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 130 9 3 The graph shows that the carbon atom has six electrons. There is a large increase in the ionisation energy between electron 4 and electron 5. This indicates that the six electrons in carbon are arranged in two different shells. 4 Thus, the electronic configuration of carbon is 2.4 or 1s 2 2s 2 2p2 . It is in Group 14 of the Periodic Table. 5 We can also predict the valence shell configuration of an element if we know the position of that element in the Periodic Table. 6 The group number of the element gives the number of electrons in its valence shell using the following formula: For elements in Group 1 and Group 2: No. of valence electrons = group number For elements in Group 13 to 18: No. of valence electrons = group number – 10 Summary: Group number 1 2 13 14 15 16 17 18 No. of valence electrons 1 2 3 4 5 6 7 8 7 The period number of the element gives the principle quantum number (n) of the valence-shell. 8 For example, Antimony (51Sb) is in Period 5 and Group 15 of the Periodic Table. (a) It has five valence electrons with the configuration of s 2 p3 . (b) The valence electrons are in the 5th shell. (c) The valence-shell configuration of antimony is 5s 2 5p3 . 9 This method of predicting the valence-shell electronic configurations is not applicable for the d block elements and the transition elements. This is because they have incomplete inner shells. Example 9.2 Predict the valence-shell configurations of the following elements: (a) 9F (b) 19K (c) 52Te (d) 31Ga Solution Element 9F 19K 52Te 31Ga Group number 17 1 16 13 No. of valence electrons 7 1 6 3 Period number 2 4 5 4 Valence-shell configuration 2s2 2p5 4s1 5s2 5p4 4s2 4p1

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 131 9 9.2 Reactions of Period 3 Elements with Oxygen and Water 1 The Period 3 elements consist of metals (sodium, magnesium and aluminium), metalloid (silicon), non-metals (phosphorous, sulphur and chlorine). 2 Argon is inert under normal conditions. Reactions of the Elements with Oxygen 1 In the reaction with oxygen to form oxides, the oxygen is reduced to the O2– ions. 2 All the Period 3 elements react with oxygen under suitable conditions to form their respective oxides, except chlorine. 3 Chlorine is too powerful as an oxidising agent to reduce oxygen. 4 The table below lists the oxides of the elements. Element Na Mg Al Si P S Cl Formula of oxide Na2O MgO Al2O3 SiO2 P4O6 P4O10 SO2 SO3 Cl2O Cl2O7 Oxidation state of element +1 +2 +3 +4 +3 +5 +4 +6 +1 +7 Physical state Solid Liquid Solid Gas Gas Gas Liquid 5 Sodium burns in air with a bright flame to produce sodium oxide, a white solid. 4Na(s) + O2(g) ⎯→ 2Na2O(s) Sodium is very little attacked by dry air. However, when exposed to moist air at room temperature, it reacts to form the oxide which dissolves in water from the air to form caustic soda (concentrated sodium hydroxide) Na2O(s) + H2O(l) ⎯→ 2NaOH(aq) 6 Magnesium burns in air, when heated, with a brilliant white flame to form magnesium oxide, a white solid. 2Mg(s) + O2(g) ⎯→ 2MgO(s) However, the oxide is always contaminated with a little nitride. 3Mg(s) + N2(g) ⎯→ Mg3N2(s) The nitride liberates ammonia gas when heated with water. Mg3N2(s) + 6H2O(l) ⎯→ 3Mg(OH)2(s) + 2NH3(g) Info Chem Chlorine does not react with oxygen under any circumstances. Formation of magnesium nitride Release of ammonia gas Info Chem Sodium also forms sodium peroxide with oxygen: 2Na + O2 → Na2O2 Sodium peroxide dissolves in water to produce hydrogen peroxide: Na2O2 + 2H2O → 2NaOH + H2O2 2012/P1/Q20 2016/P2/Q7; Q19 2013/P2/Q9; Q10 2017/P2/Q7; Q18 2014/P2/Q19(a)(i) 2018/P2/Q8 2010/P1/Q20

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 132 9 7 When exposed to air, aluminium acquires a very thin layer of oxide on its surface and this protects it from further attack. It has little change when heated in air to about 700 °C. However, at around 1000 °C, it catches fire and forms aluminium oxide, a white solid and a little nitride. 4Al(s) + 3O2(g) ⎯→ 2Al2O3(s) 2Al(s) + N2(g) ⎯→ 2AlN(s) When heated with water, the nitride undergoes hydrolysis to liberate ammonia gas. AlN(s) + 3H2O(l) ⎯→ Al(OH)3(s) + NH3(g) 8 Silicon burns slowly in oxygen, at red heat, to form silicon dioxide, a white solid. Si(s) + O2(g) ⎯→ SiO2(g) Sand is an impure form of silicon dioxide. 9 Phosphorous (which exists as P4 molecules) exists in two principal allotropic forms: white phosphorous and red phosphorous. Red phosphorus ignites in air at about 300 °C to form phosphorous(V) oxide, a white solid. P4(s) + 5O2(g) ⎯→ P4O10(s) White phosphorous catches fire at about 40 °C to produce the same oxide. In limited amount of air/oxygen, a colourless liquid phosphorous(III) oxide is formed. P4(s) + 3O2(g) ⎯→ P4O6(l) 10 When heated in air, sulphur melts and catches fire, burning with a blue flame, to produce white fumes of sulphur dioxide. S(s) + O2(g) ⎯→ SO2(g) In the presence of excess air and vanadium(V) oxide as catalyst, sulphur dioxide is further oxidised to sulphur trioxide at about 500 °C. 2SO2(g) + O2(g) 2SO3(g) 11 Oxygen has no effect on chlorine. The two main oxides of chlorine is chlorine(I) oxide, Cl2O, a yellowish gas and chlorine(VII) oxide, Cl2O7, a colourless liquid. Formation of aluminium nitride The nitride contains the N3– ion. Red phosphorous: White phosphorous: Info Chem White phosphorus is used in chemical warfare as an incendiary weapon which can set any combustibles on fire and causes serious burns or death. Exam Tips Exam Tips Al2O3 is insoluble in water due to the strong ionic bonds.

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 133 9 Example 9.3 How would you expect the reactivity of the Period 3 elements (Na to Cl) towards oxygen to change with increasing proton number of the elements? Solution The Period 3 elements react with oxygen to form their respective oxides. In the reaction, oxygen is reduced to the oxide, O2–. O2 + 4e– → 2O2– Going across Period 3, the reducing power of the elements decreases as the size of the atoms decreases and the ionisation energy increases. As a result, reactivity towards oxygen decreases. Example 9.4 The melting points of magnesium oxide and aluminium oxide are given in the table below: Compound MgO Al2O3 Melting point/oC 2852 2072 Comment on their values. Solution Both MgO and Al2O3 are ionic compounds with strong ionic bonds. This accounts for their high melting points. However, due to the high charge density of the Al3+ ion (due to its smaller size and higher charge), it is able to polarised the O2– anion to a larger extent than Mg2+. As a result, there is significant covalent character in Al2O3. This reduces the strength of the ionic bond resulting in its lower melting point compared to MgO. Quick Check 9.4 1 Magnesium nitride is an ionic compound. (a) Write the formula for the anion in magnesium nitride. (b) Draw the Lewis structure of the anion. (c) Write a balanced ionic equation for the reaction of the anion with water. 2 At room conditions, phosphorous exits in the form of P4 molecules. Draw a diagram to show the shape of a P4 molecule. Reactions of the Elements with Water 1 When a freshly cut piece of sodium is added to cold water, it catches fire and a violently exothermic reaction occurs to form sodium hydroxide with the release of hydrogen gas. 2Na(s) + 2H2O(l) ⎯→ 2NaOH(aq) + H2(g) Na reacts with cold water. Info Chem Reducing agents are electron donors: M → Mn+ + ne– Exam Tips Exam Tips Aluminium oxide has higher degree of covalent character than magnesium oxide. INFO Reactions of Period 3 Elements

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 134 9 2 Magnesium does not react with cold water. It reacts very slowly with hot water, but rapidly with steam to produce magnesium hydroxide and hydrogen gas. Mg(s) + 2H2O(g) ⎯→ Mg(OH)2(s) + H2(g) 3 Aluminium has no reaction with water because of the presence of a protective oxide layer on its surface. However, if the oxide layer is removed (by mechanical or chemical means), then aluminium reacts exothermically with cold water to produce aluminium oxide and hydrogen gas. 2Al(s) + 3H2O(l) ⎯→ Al2O3(s) + 3H2(g) 4 Silicon reacts slowly with steam at red heat to form silicon dioxide. Si(s) + 2H2O(g) ⎯→ SiO2(s) + 2H2(g) Δ 5 Phosphorous and sulphur do not react with water under any conditions. 6 Chlorine is sparingly soluble in water to form ‘chlorine water’, which is a mixture of hydrochloric acid and chloric(I) acid. Cl2(aq) + H2O(l) HCl(aq) + HClO(aq) This is an example of disproportionation, where chlorine is simultaneously oxidised and reduced. In strong sunlight, chloric(I) acid undergoes photochemical decomposition to liberate oxygen gas. Sunlight 2HClO(aq) ⎯⎯⎯→ 2HCl(aq) + O2(g) 7 The reactions of the Period 3 elements with water is summarised in the table below. Element Reaction with water Na Vigorous with cold water to form sodium hydroxide. Hydrogen is liberated. Mg Vigorous with steam to form magnesium hydroxide. Hydrogen is liberated. Al Vigorous if protective oxide layer is removed to form aluminium oxide and hydrogen gas. Si Slow with steam at red heat. P No reaction. S No reaction. Cl Slight reaction to form acids. Ability of the Elements to Act as Oxidising and Reducing Agents 1 Oxidising agents are electron-acceptors. For example, Cl2 + 2e– → 2Cl– Mg reacts with steam. Presence of a protective oxide layer Si reacts with steam at red heat. P and S do not react with water. Cl H ! O :: ! Info Chem Sodium chlorate(I), NaOCl, is used as a domestic bleach.

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 135 9 2 Reducing agents are electron-donors. For example, Mg → Mg2+ + 2e– 3 Due to their larger size and lower ionisation energies, the elements in Group 1, Group 2 and Group 13 can lose electrons readily and thus are reducing agents. Na → Na+ + e– Mg → Mg2+ + 2e– Al → Al3+ + 3e– 4 For example: (a) Magnesium is used in the extraction of titanium by reducing titanium(IV) chloride. 2Mg + TiCl4 → 2MgCl2 + Ti (b) Aluminium is used in the extraction of chromium by reducing chromium oxide. 2Al + Cr2O3 → Al2O3 + 2Cr 5 Due to their small size and higher ionisation energies, elements in Group 14 to Group 17 do not lose electrons easily. As a result, they are not reducing agents. 6 Due to their small size and higher nuclear charge, phosphorous, sulphur and chlorine readily accepts electrons. They are oxidising agents. P + 3e– → P3– S + 2e– → S2– Cl2 + 2e– → 2Cl– 7 For example, chlorine can oxidise iron(II) to iron(III). Cl2 + 2Fe2+ → 2Cl– + 2Fe3+ Physical Properties of the Period 3 Oxides 1 The table below lists some of the properties of the Period 3 oxides. Oxide Na2O MgO Al2O3 SiO2 P4O6 P4O10 SO2 SO3 Cl2O Cl2O7 Nature of bond Ionic Covalent Structure Giant ionic Giant covalent Simple covalent Intermolecular force Ionic Covalent van der Waals force 2 Oxides of sodium, magnesium and aluminium are ionic. Strong ionic bonds hold the particles together. They are all solid with high melting/boiling points. 3 Silicon dioxide, though a covalent compound has high melting point because it has a giant covalent structure. Strong covalent bonds hold the atoms together in a three dimensional array. A lot of energy is required to break the strong covalent bonds. 2007/P2/Q8(a) 2016/P2/Q20(c) silicon oxygen

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 136 9 4 The oxides of phosphorous, sulphur and chlorine are simple molecules with weak intermolecular van der Waals force. They are mostly liquids, gases or are solid with low melting points, as melting only involves the weakening of the weak van der Waals forces. 5 The total number of electrons in the molecules of the covalent oxides is shown below: Oxide P4O10 P4O6 SO3 SO2 Cl2O7 Cl2O No. of electrons 140 108 40 32 90 42 The strength of the van der Waals force increases in the order: SO2 < SO3 < Cl2O < Cl2O7 < P4O6 < P4O10 The melting/boiling point also increases in the same order. 6 The diagrams below show the structures of the various oxides. x x S SO2 P molecule 4O6 molecule P P P P P4O10 molecule P P P P CI 2O molecule x x x x CI CI CI 2O7 molecule CI CI SO3 molecule S O O O O O O O O O O O O O O O O O O O O O O O O O O O O O 9.3 Acidic and Basic Properties of Oxides and Hydrolysis of Oxides Oxide Na2O MgO Al2O3 SiO2 P4O6 P4O10 SO2 SO3 Cl2O Cl2O7 Acid/base nature Basic Amphoteric Acid 1 The ionic oxides (Na2O, MgO) are all basic oxides, while the covalent oxides (SiO2, P4O6, P4O10, SO2, SO3, Cl2O and Cl2O7) are all acidic. 2 Aluminium oxide is an ionic compound but with a significant amount of covalent character (≈ 60% ionic and 40% covalent). As a result, aluminium oxide is amphoteric. 3 Sodium oxide dissolves readily in water to form sodium hydroxide, a strong base. Na2O(s) + H2O(l) → 2NaOH(aq) Al2O3 is amphoteric. 2014/P2/Q8, Q19(a)(ii) Exam Tips Exam Tips Metallic oxides contain the O2- ions that are Bronsted − Lowry base: 2O2− + 2H2O → 4OH−

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 137 9 4 Magnesium oxide is insoluble in water. However, it dissolves slowly in hot dilute acids. MgO(s) + H2SO4(aq) → MgSO4(aq) + H2O(l) Δ 5 Aluminium oxide dissolves slowly in hot, dilute mineral acids and also in hot, concentrated alkalis. Al2O3(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2O(l) Δ Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2NaAl(OH)4(aq) Δ Sodium aluminate 6 The aluminate is easily decomposed by dilute acid or carbon dioxide to aluminium hydroxide, a white solid. NaAl(OH)4(aq) + HCl(aq) → Al(OH)3(s) + H2O(l) + NaCl(aq) 2NaAl(OH)4(aq) + CO2(g) → 2Al(OH)3(s) + Na2CO3(aq) + H2O(l) 7 Due to its giant covalent structure, silicon dioxide is insoluble in water. It dissolves slowly in hot caustic alkali to form silicates. SiO2(s) + 2NaOH(aq) → Na2SiO3(aq) Δ 8 Oxides of phosphorous, sulphur and chlorine dissolve in water to form acidic solutions. P4O6(l) + 6H2O(l) → 4H3PO3(aq) Phosphoric(III) acid or phosphorous acid P4O10(s) + 6H2O(l) → 4H3PO4(aq) Phosphoric(V) acid or phosphoric acid SO2(g) + H2O(l) → H2SO3(aq) Sulphuric(IV) acid or sulphurous acid SO3(g) + H2O(l) → H2SO4(aq) Sulphuric(VI) acid or sulphuric acid Cl2O(g) + H2O(l) → 2HClO(aq) Chloric(I) acid Cl2O7(l) + H2O(l) → 2HClO4(aq) Chloric(VII) acid 9 Oxides of phosphorous, sulphur and chlorine react with alkalis to form salts. P4O6(l) + 12NaOH(aq) → 4Na3PO3(aq) + 6H2O(l) Sodium phosphate(III) P4O10(s) + 12NaOH(aq) → 4Na3PO4(aq) + 6H2O(l) Sodium phosphate(V) OH OH OH HO ! Al !3+ OH HO OH ! P : OH HO OH ! P O R HO OH !! O S R: HO OH !! O O R R S OH O ! R Cl O O R R Exam Tips Exam Tips Covalent oxides do not contain O2− ions.

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 138 9 SUMMARY SUMMARY SO2(g) + 2NaOH(aq) → Na2SO3(aq) + H2O(l) Sodium sulphate(IV) or sodium sulphite SO3(g) + 2NaOH(aq) → Na2SO4(aq) + H2O(l) Sodium sulphate(VI) or sodium sulphate Cl2O(g) + 2NaOH(aq) → 2NaOCl(aq) + H2O Sodium chlorate(I) Cl2O7(l) + 2NaOH(aq) → 2NaClO4(aq) + H2O(l) Sodium chlorate(VII) Use of Sulphur Dioxide in Food Preservation 1 Sulphur dioxide is widely used in the food and drinks industries for its properties as a preservative and antioxidant (reducing agent). 2 Due to its mild reducing and acidic properties, it can prevent the growth of bacteria which causes the food to turn rancid. Quick Check 9.5 1 Phosphoric acid is a weak triprotic acid. It reacts with aqueous sodium hydroxide to form three different types of salt depending on the relative amount of base present. (a) What do you understand by the term ‘triprotic acid’? (b) Write equations to represent the formation of the three salts when phosphoric acid reacts with aqueous sodium hydroxide. 2 Under suitable conditions, sodium can react with oxygen to form sodium peroxide, Na2O2. (a) Give the formula of the peroxide ion. (b) Draw a Lewis structure for the peroxide ion. (c) Sodium peroxide dissolves in water to form sodium hydroxide and hydrogen peroxide. Construct a balanced equation for the reaction. 3 Which is the stronger acid, sulphuric(IV) acid or sulphuric(VII) acid. Explain your answer. 4 Explain the origin of the covalent character in aluminium oxide. 1 The Modern Periodic Table consists of 18 Groups and 7 Periods. 2 Going across a period: • the atomic radius decreases • the ionisation energy generally increases • the electronegativity increases 3 Going down a group: • the atomic radius increases • the ionisation energy decreases • the electronegativity decreases 4 The Period 3 is a mixture of metals and non-metals. 5 All the elements react with oxygen except chlorine. 6 The oxides change from ionic (basic) to covalent (acidic). 7 Aluminium oxide is the only amphoteric oxide in Period 3. 8 All elements, except phosphorous and sulphur, react with water.

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 139 9 Objective Questions 1 Going across the Period 3 elements (Na to Cl) in the Periodic Table, A the atomic radius increases B the first ionisation decreases C the electronegativity decreases D the maximum oxidation state increases 2 Which of the following oxides is amphoteric? A Sodium oxide C Silicon dioxide B Aluminium oxide D Chlorine(VII) oxide 3 Which of the following elements forms an ionic hydride? A Magnesium C Sulphur B Aluminium D Chlorine 4 The first ionisation energy of nitrogen is higher than that of oxygen. Which statement best explains the difference? A Nitrogen has a more stable electronic configuration. B Nitrogen is in Group 15 of the Periodic Table while oxygen is in Group 16. C Nitrogen is more stable due to the presence of NN triple bond. D Nitrogen is more electronegative than oxygen. 5 The second ionisation energy of sodium is the highest among the Period 3 elements (Na to Cl) because A sodium is a very electropositive element B sodium has the highest effective nuclear charge C it is very difficult for sodium to form the Na2+ ion D the second electron removed is from an inner filled shell 6 Arrange the following anions in the order of increasing ionic radius. P3–, S2–, Cl– A P3–, S2–, Cl– C Cl– , S2–, P3– B P3–, Cl– , S2– D S2–, P3–, Cl– 7 Which oxide when dissolved in water would produce a solution with the highest pH? A Na2O C SiO2 B MgO D SO3 8 Which of the following statements regarding the oxides of Period 3 is not correct? A SiO2 and P4O10 are solid at room conditions. B Al2O3 and MgO are amphoteric oxides. C SiO exists only in the gaseous state. D Chlorine(I) oxide is prepared by passing dry clorine over mercury(II) oxide at 400 °C. 9 Going across Period 3, the reducing power of the elements (Na to Cl) decreases. This is because A the electronegativity of the elements increases. B the ionisation energy of the elements increases. C the oxidation state of the elements increases. D the polarising power of the ions increases. 10 Which of the following oxides is amphoteric? A Na2O2 B Al2O3 C MgO D Cl2O7 11 Which of the following is not true regarding silicon dioxide? A It is a white solid. B It conducts electricity in the molten state. C It dissolves in hot, concentrated sodium hydroxide. D It has a giant structure. 12 Which of the following statements is not true regarding the Period 3 elements (Na to Ar)? A Chlorine forms the most number of oxides. B Chlorine is the only gaseous element in Period 3. C Elements with giant structures include sodium, magnesium aluminium and silicon. D Chlorine is the only element that does not react with oxygen. 13 Which of the following compounds would produce an acidic solution when added to water? I SO3 II P4O6 III Al2O3 A I only C I and II B III only D I and III STPM PRACTICE 9

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 140 9 14 Which of the following statements is correct with regards to the Period 3 elements? A Sodium oxide is the only oxide that dissolves in water to give a strong alkaline solution. B Generally, the reactivity of the elements towards oxygen increases. C Chlorine is the only gaseous element. D Silicon dioxide has the highest melting point. 15 Which statement(s) is/are true regarding the oxides of Na2O, MgO and Al2O3? I Na2O is the only oxide that is soluble in water. II The melting point increases in the order of Na2O < MgO < Al2O3. III MgO and Al2O3 are amphoteric. A I only C I and II B II only D II and III 16 Which is true about the properties of the metallic elements of Period 3 (sodium to aluminium)? A Semiconductors B Act as reducing agents C Form basic oxides D React vigorously with water at room temperature 17 Boron is the first element in Group 13 of the Periodic Table. Which of the following statements about boron and its compounds is not true? A An aqueous solution of boron chloride is acidic. B Boron oxide, B2O3, is basic. C It is a poor electrical conductor. D Boron is added to silicon to increase the conductivity of silicon. 18 Which of the following groups contains a basic, an acidic and an amphoteric oxide? A MgO, Al2O3, Na2O B Cl2O7, SO3, P4O10 C Al2O3, MgO, Cl2O D SiO2, SO2, Na2O 19 Which of the following ions has the smallest ionic radius? A F– C Al3+ B Na+ D P3– 20 Bismuth is in the same group as phosphorous in the Periodic Table. Which of the following sets of oxidation states is bismuth likely to show in its oxides? A +5 only C +3 and +5 B +2 and +3 D +4 and +6 21 X is an element in Period 3 (sodium to chlorine) in the Periodic Table. It burns in oxygen to produce an oxide that is insoluble in water. X has no reaction with water at room conditions. Which of the following is true? A X has a giant structure. B The oxide of X has the molecular formula of X2O7. C X is an oxidising agent. D X is a metal. 22 Which of the following best explains why the ionic size decreases in the order of P3– > S2– > Cl– ? A An increase in the total number of protons and electrons B A decrease in the total number of electrons and an increase in the number of protons C A decrease in the total number of electrons and in the number of protons D The total number of electrons remains the same and an increase in the total number of protons 23 Which element has the highest second ionisation energy? A Na B Al C S D Cl 24 Which statement explains why the first ionisation energy of aluminium is lower than that of magnesium? A Aluminium is more reactive than magnesium. B The effective nuclear charge of aluminium is lower than that of magnesium. C The size of aluminium is larger than that of magnesium. D The screening effect in aluminium is larger than that in magnesium. 25 Which statement is correct with regards to the Period 3 elements (Na → Cl) as the proton number increases? A The reactivity of the elements with oxygen generally decreases. B The reducing power of the elements increases. C The oxides become more basic. D The reactivity of the elements with water increases.

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 141 9 26 Which property(ies) of the elements in Period 2 and Period 3 of the Periodic Table change(s) periodically? I First ionisation energy II Boiling point III Electronegativity A III only B I and II C II and III D I, II and III 27 X, Y and Z are elements in Period 3 of the Periodic Table. The physical and chemical properties of the oxides of the elements are shown below: Oxides of element X Y Z Physical state Gas Liquid Solid Solubility in water Soluble Soluble Insoluble Acidity Acidic Acidic Basic What could elements X, Y and Z be? X Y Z A Magnesium Chlorine Sodium B Sulphur Phosphorous Magnesium C Chlorine Sodium Sulphur D Silicon Sulphur Aluminium 28 An oxide X has a melting point of 1600 °C. It dissolves in hot, aqueous sodium hydroxide but not in water or acid. Oxide X could be A MgO C SiO2 B Al2O3 D P4O10 29 Which of the following oxides is soluble in water? A Al2O3 B SiO2 C P4O10 D Fe3O4 Structured and Essay Questions 1 The standard enthalpy change of vaporisation for some of the elements in Period 3 of the Periodic Table is given below: Element ΔHv/kJ mol–1 Na 100 Al 280 Si 300 S 63 Cl 10 (a) Define standard enthalpy of vaporisation. (b) Explain in terms of structure and bonding the trend of the standard enthalpy change of vaporisation of the elements. 2 (a) Define first ionisation energy. Illustrate your answer with reference to chlorine. (b) Sketch a graph of the variation of the first ionisation energy with proton number for the Period 3 elements (Na to Ar) and suggest explanation for the following: (i) the general increases from Na to Ar (ii) the discontinuities between magnesium and aluminium, and between phosphorous and sulphur (c) (i) Define second ionisation energy. Illustrate your answer with reference to sodium. (ii) Among the Period 3 elements, sodium has the highest second ionisation energy. Explain why is it so.

Chemistry Term 2 STPM Chapter 9 Periodic Table: Periodicity 142 9 3 The melting points of sodium oxide, aluminium oxide and sulphur trioxide are 1275 °C, 2070 °C and 17 °C respectively. (a) Draw the Lewis diagram for the sulphur trioxide molecule. (b) Explain their melting points with reference to their structures and bondings. (c) State how the oxides react with water. Write equations where appropriate. 4 Aluminium and chlorine are elements from Period 3 of the Periodic Table. (a) Explain why aluminium is an electric conductor while chlorine is a non-conductor. (b) When aluminium burns in air, aluminium oxide and aluminium nitride are produced. (i) Write the balanced equations for the formation of the oxide and nitride. (ii) When aluminium nitride is heated with water, a gas that turns red litmus blue is produced. Write the balanced equation for the hydrolysis of aluminium nitride. (c) Describe with the help of equations, what happens when aluminium oxide is heated with aqueous sodium hydroxide and dilute sulphuric acid. (d) Aluminium reacts with chlorine to form a white solid that conducts electricity when dissolved in water but not in the solid or molten state. Explain this observation in terms of bonding. (e) Chlorine dissolves sparingly in water to produce an acidic solution. When this solution is exposed to strong sunlight, a gas that rekindles a glowing splinter is released. Explain this observations. 5 Period 2 elements form the following oxides: MgO, Al2O3, SiO2, P2O3 and SO3 (a) (i) Sketch a graph to show the variation of the melting points of these oxides. (ii) Explain the trend of the melting points of the oxides in terms of structure and bonding. (b) Describe with the aid of balanced equations, the acid/base characteristics of the oxides. 6 Discuss the trend of the following properties of the Period 3 elements (Na to Ar). (a) Melting point (b) Electronegativity (c) Electrical conductivity 7 Aluminium and magnesium are metals in Period 3 of the Periodic Table. Both elements conduct electricity and react with chlorine to form chlorides. (a) Which is a better electric conductor, magnesium or aluminium? Explain your answer. (b) Aluminium chloride is a covalent compound whereas magnesium chloride is an ionic compound. (i) Explain the covalent nature of aluminium chloride. (ii) At room conditions, aluminium chloride exists as a dimer. Write the formula for the dimer and draw its structure.