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MODULE • Chemistry FORM 4 49 UNIT 3 © Nilam Publication Sdn. Bhd. Write a balanced chemical equation for each of the following reactions: 1 Copper(II) carbonate Copper(II) oxide + Carbon dioxide CuCO3 CuO + CO2 2 Ammonia + Hydrogen chloride Ammonium chloride NH3 HCl + NH4Cl 3 Lead(II) nitrate + Potassium iodide Lead(II) iodide + Potassium nitrate Pb(NO3)2 + 2KI PbI2 + 2KNO3 4 Sulphuric acid + Sodium hydroxide Sodium sulphate + Water H2SO4 + 2NaOH Na2SO4 + 2H2O 5 Copper(II) oxide + Hydrochloric acid Copper(II) chloride + Water CuO + 2HCl CuCl2 + H2O 6 Sodium + Water Sodium hydroxide + Hydrogen 2Na + 2H2O 2NaOH + H2 7 Potassium oxide + Water Potassium hydroxide K2O + H2O 2KOH 8 Zinc oxide + Nitric Acid Zinc nitrate + Water ZnO + 2HNO3 Zn(NO3)2 + H2O 9 Lead(II) nitrate Lead(II) oxide + Nitrogen dioxide + Oxygen 2Pb(NO3)2 2PbO + 4NO2 + O2 10 Aluminium nitrate Aluminium oxide + Nitrogen dioxide + Oxygen 4Al(NO3)3 2Al2O3 + 12NO2 + 3O2 Example: The equation shows the reaction between zinc and hydrochloric acid. Zn + 2HCl ZnCl2 + H2 Calculate the mass of zinc required to react with excess hydrochloric acid to produce 6 dm3 of hydrogen gas at room conditions. [Relative atomic mass: Zn = 65, Cl = 35.5, 1 mole of gas occupies 24 dm3 at room conditions] Write a balanced equation Write the information from the question above the equation ? g Zn(s) excess 2HCl 6 dm3 + ➝ ZnCl2 + H2 Convert the given quantitity into moles by using the method shown in the following chart. Number of moles of H2 = 6 dm3 24 dm3 mol–1 = 0.25 mol Use the mole ratio of the substances involved to find the number of moles of the other substance. Remark: The coefficient of each formula shows the number of moles of reactants react and products formed. From the equation, 1 mol H2 : 1 mol Zn 0.25 mol H2 : 0.25 mol Zn Exercise Numerical Problems involving Chemical Equations LS 3.4.3 PL3 Quiz 03 U3 Chemistry F4(p27-52)csy2p.indd 49 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 50 UNIT 3 Mass (g) ÷ (RAM/RFM/RMM) g mol–1 × (RAM/RFM/RMM) g mol–1 No. of moles (n) Volume of gas (dm3 ) ÷ 24 dm3 mol–1/ 22.4 dm3 mol–1 × 24 dm3 mol–1/ 22.4 dm3 mol–1 Convert the mole into the quantity required in the question by using the method shown in the chart below. Mass of Zn = 0.25 mol × 65 g mol–1 = 16.25 g 1 The equation shows the reaction between potassium and oxygen. 4K + O2 2K2O Calculate the mass of potassium required to produce 23.5 g of potassium oxide. [Relative atomic mass: K = 39, O = 16] Number of moles of K2O = 23.5 g (2 × 39 + 16) g mol–1 = 23.5 94 = 0.25 mol From the equation, 2 mol K2O : 4 mol K 0.25 mol K2O : 0.5 mol K Mass of K = 0.5 mol × 39 g mol–1 = 19.5 g 2 8.0 g of copper(II) oxide powder is added to excess dilute nitric acid and heated. Calculate the mass of copper(II) nitrate produced. [Relative atomic mass: N = 14, O = 16, Cu = 64] CuO + 2HNO3 Cu(NO3)2 + H2O Number of moles of CuO = 8 g (64 + 16) g mol–1 = 0.1 mol From the equation, 1 mol CuO : 1 mol Cu(NO3)2 0.1 CuO : 0.1 mol Cu(NO3)2 Mass of Cu(NO3)22 = 0.1 mol × 188 g mol–1 = 18.8 g 3 1.3 g of zinc reacts with excess dilute sulphuric acid. The products are zinc sulphate and hydrogen. Calculate the volume of hidrogen gas released at STP. [Relative atomic mass: Zn = 65, 1 mol of gas occupies 22.4 dm3 mol–1 at STP] Zn + H2SO4 ZnSO4 + H2 Number of moles of Zn = 1.3 g 65 g mol–1 = 0.02 mol From the equation, 1 mol Zn : 1 mol H2 0.02 mol Zn : 0.02 mol H2 Volume of H2 = 0.02 mol × 22.4 dm3 mol–1 = 0.448 dm3 = 448 cm3 Subjective Questions PL3 PL3 PL3 HOTS ENRICHMENT EXERCISE 03 U3 Chemistry F4(p27-52)csy2p.indd 50 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 51 UNIT 3 © Nilam Publication Sdn. Bhd. 4 The equation shows the combustion of propane gas. C3H8 + 5O2 3CO2 + 4H2O 720 cm3 of propane gas (C3H8) at room conditions burns in excess oxygen. Calculate the mass of carbon dioxide formed. [Relative atomic mass: C = 12, O = 16, Molar volume of gas = 24 dm3 mol–1 at room conditions] C3H8 + 5O2 3CO2 + 4H2O Number of moles of C3H8 = 720 cm3 24 000 cm3 mol–1 = 0.03 mol From the equation, 1 mol C3H8 : 3 mol CO2 0.03 mol C3H8 : 0.09 mol CO2 Mass of CO2 = 0.09 mol × 44 g mol–1 = 3.96 g 5 The diagram below shows a car fitted with air bag which will inflate in an accident. The air bag contains solid sodium azide, NaN3 which will decompose rapidly to form sodium and nitrogen gas. The nitrogen gas formed fills the air bag. [Relative atomic mass: N = 14; H = 1; Na = 23 and 1 mol of gas occupies the volume of 24 dm3 at room temperature and pressure] (a) Construct equation for the decomposition of sodium azide. 2NaN3 → 2Na + 3N2 (b) In an accident, an air bag fills with 72 dm3 of nitrogen at room temperature and pressure. What is the mass of sodium azide needed to provide 72 dm3 of nitrogen? Number of moles of nitrogen = 72 dm3 24 dm3 mol–1 = 3 mol Number of moles of NaN3 = 2 mol Mass of NaN3 = 2 mol × [23 + 3(14)] g mol–1 = 130 g (c) Sodium azide, NaN3, reacts with dilute hydrochloric acid to produce sodium chloride and compound A. Compound A contains 2.33% of hydrogen and 97.7% of nitrogen by mass. (i) What is the empirical formula for compound A? Element H N Mass (g) 2.33 97.7 Number of mole 2.33 –—– 1 = 2.33 97.7 –—– 14 = 6.98 Simplest ratio 2.33 –—– 2.33 = 1 6.98 –—– 2.33 ≈ 3 Empirical formula: HN3 (ii) Construct the equation for the reaction between sodium azide and dilute hydrochloric acid. NaN3 + HCl → NaCl + HN3 PL3 PL3 HOTS 03 U3 Chemistry F4(p27-52)csy2p.indd 51 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 52 UNIT 3 1 The mass of one atom of element Y is two times more than an atom of oxygen. What is the relative atomic mass of element Y? [Relative atomic mass: O = 16] A 12 C 32 B 24 D 36 2 A bottle contains 3.01 × 1023 of gas particles. What is the number of moles of the gas in the bottle? [Avogadro constant = 6.02 × 1023 mol–1] A 0.5 mol C 3.0 mol B 1.0 mol D 6.0 mol 3 Which of the following gases contains 0.4 mol of atoms at room temperature and pressure? [Molar volume of gas = 24 dm3 mol–1 at room temperature and pressure] A 4.8 dm3 Ne C 4.8 dm3 CO2 B 4.8 dm3 O2 D 4.8 dm3 NH3 4 Abulb is filled with 1 800 cm3 of argon gas at room conditions. What is the number of argon atom in the bulb? [Molar volume of gas = 24 dm3 mol–1 at room conditions, Avogadro constant = 6.02 × 1023 mol–1] A 4.515 × 1022 C 8.03 × 1022 B 4.515 × 1023 D 8.03 × 1021 5 What is the number of oxygen atom in 0.1 mol of water? [Avogadro constant = 6.02 × 1023 mol–1] A 6.02 × 1022 C 6.02 × 1023 B 1.204 × 1022 D 1.204 × 1023 6 5 g of element X reacted with 8 g of element Y to form a compound with the formula XY2. What is the relative atomic mass of element X? [Relative atomic mass: Y = 80] A 25 C 50 B 40 D 100 7 Diagram below shows the standard representation for element of atom X and Y. 16 8 X 35 17 Y Elements X and Y react to form a compound. What is the relative formula mass of the compound? A 51 C 86 B 67 D 172 8 The diagram below shows the set-up of apparatus to determine the empirical formula of an oxide of metal X. Heat Metal X Which of the following is metal X? A Zinc C Tin B Lead D Copper 9 The equation shows a decomposition of magnesium nitrate when heated. 2Mg(NO3)2 2MgO + 4NO2 + O2 What is the number of oxygen molecules produced when 7.4 g magnesium nitrate decomposed when heated? [Relative formula mass of Mg(NO3)2 = 148; Avogadro constant = 6.02 × 1023 mol–1] A 1.505 × 1022 C 1.505 × 1023 B 3.010 × 1022 D 3.010 × 1023 10 The equation below shows the chemical equation of the combustion of ethanol in excess oxygen. C2H5OH + 3O2 2CO2 + 3H2O What is the volume of carbon dioxide gas released when 9.20 g ethanol burnt completely? [Relative atomic mass of H = 1, C = 12, O = 16, 1 mol of gas occupies 24 dm3 at room conditions] A 4.8 cm3 C 96.0 cm3 B 9.6 cm3 D 9 600 cm3 11 What is the percentage by mass of nitrogen content in urea, CO(NH2)2? [Relative atomic mass: C = 12, N = 14, H = 1 and O = 16] A 23.3% C 46.7% B 31.8% D 63.6% 12 1.72 g of an oxide of metal X contains 0.8 g oxygen. What is the empirical formula of the oxide? [Relative atomic mass : X = 46; O =16] A XO2 C X2O3 B X2O D X2O5 Objective Questions SPM PRACTICE PL2 PL3 PL3 PL2 PL2 PL3 PL3 PL4 PL4 PL4 PL3 PL4 Extra Questions 03 U3 Chemistry F4(p27-52)csy2p.indd 52 21/12/2022 3:04 PM

MODULE • Chemistry FORM 4 53 UNIT 4 © Nilam Publication Sdn. Bhd. PERIODIC TABLE Historical Development – Contribution of Scientists Group Period 3 (a) Metallic properties (shiny, conducts electricity, malleable, high tensile strength, high melting point & density) (b) Special characteristics: (i) Most elements form coloured compound. (ii) Most elements have more than one oxidation number. (iii) Many transition elements can form complex ion (iv) Many elements can act as a catalyst (a) Monoatomic and inert (b) Uses in daily life Physical properties & changes in physical properties down the group (a) Similar chemical properties. (React with H2O, O2 & Cl2) (b) Reactivity increases down the group. (a) Similar chemical properties. (React with H2O, NaOH & Fe) (b) Reactivity decreases down the group. Across Period 3 from left to right: (a) Change in atomic size (b) Change in electronegativity (c) Change in metallic properties (metal ➝ semi metal ➝ non metal) (d) Change in oxide properties (Basic oxide ➝ amphoteric oxide ➝ acidic oxide) Group 18 (Noble gases) Group 1 (Alkali metal) Group 17 (Halogen) Period Transition element Elements are arranged in order of increasing proton number Electron arrangement in an atom Number of shells occupied with electrons in an atom Located between Group 2 to Group 13 Atoms of elements have three shells occupied with electrons Atoms of elements have seven valence electrons Atoms of elements have one valence electron Atoms have achieved stable duplet/octet electron arrangement Number of valence electrons in an atom 4 UNIT Concept Map THE PERIODIC TABLE OF ELEMENTS Studied in Form 1, Unit 6: Periodic Table 04 U4 Chemistry F4(p53-76)csy2p.indd 53 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 54 UNIT 4 Scientists Discoveries Antoine Lavoisier • Substances were classified into 4 groups with similar chemical properties. J.W Dobereiner • Substances were arranged into groups of 3 elements with similar chemical properties. • Groups of element with similar chemical properties were called Triads. • Triad system was confined to some elements only. John Newlands • Elements were arranged in ascending atomic mass. • Law of Octaves because similar chemical properties were repeated at every eight element. • This system was inaccurate because there were some elements with wrong mass numbers. Lothar Meyer • The atomic volume = Mass of 1 mol (g) Density (g cm–3) • Plotted graph for the atomic volume against atomic mass. • Found that elements with similar chemical properties were positioned at equivalent places along the curve. Mendeleev • Elements were arranged in ascending order of increasing atomic mass. • Elements with similar chemical properties were in the same group. • Empty spaces were allocated for elements yet to be discovered. • Contributor to the formation of the modern Periodic table. Henry Moseley • Classified element based on concepts of proton number and arranged elements in order of increasing proton number. • Contributor to the formation of the modern Periodic Table. What is the Periodic Table? • It is an arrangement of the elements in the orders of increasing proton number. What is the advantages of arranging elements in the Periodic Table? Periodic Table enables: • Chemists to study, understand and remember the chemical and physical properties of all the elements and compounds in an orderly manner. • Properties of elements and their compounds to be predicted based on the position of elements in the Periodic Table. • Relationship between elements from different groups to be known. 4.1 THE DEVELOPMENT OF THE PERIODIC TABLE OF ELEMENTS CS 4.1 Advantages of Classifying the Elements in the Periodic Table LS 4.1.2 Contribution of Scientist to the Historical Development of the Periodic Table LS 4.1.1 PERFORMANCE LEVEL (PL) Mastered Not mastered PL1 PL2 Recall knowledge and basic skills about the Periodic Table of Elements. Understand and explain the Periodic Table of Elements. 04 U4 Chemistry F4(p53-76)csy2p.indd 54 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 55 UNIT 4 © Nilam Publication Sdn. Bhd. What is a Group? • The vertical column of elements in the Periodic Table arranged according to the number of valence electron in the outermost shell of an atom. How is the group number related to the number of valence electrons? • There are 18 vertical columns, called Group 1, Group 2, and Group 13 until Group 18. Number of valence electrons 1 2 3 4 5 6 7 8 (except Helium) Group 1 2 13 14 15 16 17 18 For atoms of elements with 3 to 8 valence electrons, the group number is: 10 + number of valence electrons. Certain groups have a special name. What are the names for these certain groups? Groups Special names 1 Alkali metals 2 Alkali-earth metals 3 – 12 Transition elements 17 Halogens 18 Noble gases Write the electron arrangement for atom of each element in the Periodic Table below. LS 4.2.1 Nucleon number Proton number Symbol of an element A ZX P E R I O D 1 18 1 H* 1 1 1 2 13 14 15 16 17 He 4 2 2 2 Li 7 3 2.1 Be 8 4 2.2 B11 5 2.3 C12 6 2.4 N14 7 2.5 O16 8 2.6 F 19 9 2.7 Ne 20 10 2.8 3 Na 23 11 2.8.1 Mg 24 12 2.8.2 3 4 5 6 7 8 9 10 11 12 Al 27 13 2.8.3 Si 28 14 2.8.4 P 31 15 2.8.5 S 32 16 2.8.6 Cl 35 17 2.8.7 Ar 40 18 2.8.8 4 K39 19 2.8.8.1 Ca 40 20 2.8.8.2 Br 80 35 GROUP TRANSITION METALS What is the basic principle of arranging the element in the Periodic Table? • Elements in the Periodic Table are arranged horizontally in increasing order of proton number . LS 4.2.2 State two main components of the Periodic Table: (i) Group (ii) Period LS 4.2.2 4.2 THE ARRANGEMENT IN THE PERIODIC TABLE OF ELEMENTS CS 4.2 Group 04 U4 Chemistry F4(p53-76)csy2p.indd 55 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 56 UNIT 4 How is the group number related to types of substance? Group 1, 2 and 13 Metals Group 3 – 12 Metals (Transition element) Group 14 – 18 Non-metal 1 Complete the table below. Elements Protons number Electrons arrangement Number of valence electrons Groups Number of shells Periods H 1 1 1 1 1 1 He 2 2 2 18 1 1 Li 3 2.1 1 1 2 2 Be 4 2.2 2 2 2 2 B 5 2.3 3 13 2 2 C 6 2.4 4 14 2 2 N 7 2.5 5 15 2 2 O 8 2.6 6 16 2 2 F 9 2.7 7 17 2 2 Ne 10 2.8 8 18 2 2 Na 11 2.8.1 1 1 3 3 Mg 12 2.8.2 2 2 3 3 Al 13 2.8.3 3 13 3 3 PL2 Short periods, # Period 3 will be studied in detail with respect to physical and chemical properties Long periods What is a Period? • The horizontal row of elements in the Periodic Table consists of the same number of shells occupied with electrons in an atom . How is the period number related to the number of shells? Number of shells 1 2 3 4 5 6 7 Periods 1 2 3 4 5 6 7 • Period 1 has 2 elements • Period 2 and 3 have 8 elements # • Period 4 and 5 have 18 elements • Period 6 has 32 elements • Period 7 has 23 elements LS 4.2.2 Period Exercise 04 U4 Chemistry F4(p53-76)csy2p.indd 56 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 57 UNIT 4 © Nilam Publication Sdn. Bhd. State the special name for Group 18 elements. • Noble gases Write the electron arrangement of the atoms of elements in Group 18. Elements Electrons arrangement Helium (He) 2 Neon (Ne) 2.8 Argon (Ar) 2.8.8 Krypton (Kr) 2.8.18.8 Xenon (Xe) – Radon (Rn) – 2 The diagram below shows the chemical symbols which represent elements X, Y and Z. X23 11 Z 39 Y 19 12 6 (a) Explain how to determine the position of element X in the Periodic Table. • The proton number of element X is 11 and the number of protons in atom X is 11 . • The number of electrons in atom X is 11 . • The electron arrangement of atom X is 2.8.1 . • Element X is located in Group 1 because atom X has one valence electron . • Element X is in Period 3 because atom X has three shells occupied with electrons . (b) (i) State the position of element Y in the Periodic Table. (ii) Explain how to determine the position of element Y in the Periodic Table. (i) Element Y is located in Group 14 and Period 2 . (ii) • The proton number of element Y is 6 and the number of proton in atom Y is 6 . • The electron arrangement of atom Y is 2.4 . • Element Y is located in Group 14 because atom Y has 4 valence electrons. • Element Y is in Period 2 because atom Y has 2 shells occupied/filled with electrons . (c) Which of the above elements show the similar chemical properties? Explain your answer. • Element X and element Z . • Electron arrangement of atom X is 2.8.1 and electron arrangement of atom Z is 2.8.8.1 . • Atoms X and Z have the same number of valence electron . PL3 4.3 ELEMENTS IN GROUP 18 CS 4.3 PERFORMANCE LEVEL (PL) Mastered Not mastered PL3 Apply knowledge about Periodic Table of Elements and its concept to explain the natural occurrences or phenomena and be able to carry out simple tasks. 04 U4 Chemistry F4(p53-76)csy2p.indd 57 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 58 UNIT 4 Group 18 are monoatomic gases. Explain what is meant by monoatomic. • These gases exist as a single uncombined atoms. Explain why the noble gases are monoatomic and chemically inert. • The atom has achieved duplet electron arrangement for helium and octet electron arrangement for others. • Noble gases do not react with other elements because the atom does not lose, gain or share electrons. State the uses of noble gases. LS 4.3.3 Noble gases Uses Helium To fill weather balloons and airship Neon To fill neon light (for advertisement board) Argon To fill electrical bulb Krypton To fill photographic flash lamp Radon To treat cancer State the physical properties and the changes going down Group 18. LS 4.3.2 1 All noble gases are insoluble in water and cannot conduct electricity in all conditions. 2 The melting point and boiling points are very low because atoms of noble gases are attracted by weak van der Waals forces, less energy is required to overcome these forces. 3 Going down Group 18: • The atomic size is increasing because the number of shells increases. • The density is low and increases gradually because the mass increases greatly compared to the volume. • The melting and boiling points increase because atomic size increases, causing the van der Waals forces to increase and more energy is required to overcome these forces. Explain why argon does not react with hot tungsten filament in term of electron arrangement. • Argon atom has achieved stable octet electron arrangement. • Argon atom does not need to gain , lose or share electrons with other elements. LS 4.3.1 LS 4.3.1 04 U4 Chemistry F4(p53-76)csy2p.indd 58 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 59 UNIT 4 © Nilam Publication Sdn. Bhd. State the special name for Group 1 elements. • Alkali metals List the elements in Group 1 of the Periodic Table and write the electron arrangements and number of shells of the atoms of elements. Elements Symbols Protons number Electrons arrangement Number of shells Lithium Li 3 2.1 2 Sodium Na 11 2.8.1 3 Potassium K 19 2.8.8.1 4 State the physical properties of Group 1 elements. • Grey solid with shiny surface. • Softer and the density is lower compared to other metals. • Lower melting and boiling points compared to other metals. Explain the changes in physical properties going down Group 1 elements. LS 4.4.1 • Atomic size increases because the number of shells increases. • Density increases because mass increases faster than the increase in radius. • Melting and boiling points decrease because when the atomic size increases, the metal bonds get weaker. Explain the similarities in chemical properties of the Group 1 elements. LS 4.4.4 Remark: 1 Proton is a positively charged sub atomic particle 2 Electron is negatively charged sub atomic particle • All atoms of elements in Group 1 have 1 valence electron and achieve a stable duplet/octet electron arrangement by releasing one electron to form +1 charged ions. Examples: (i) Lithium atom releases one electron to achieve stable duplet electron arrangement: Li Li+ + e– Electron arrangement: 2.1 Electron arrangement: 2 Number of protons = 3, total charge: +3 Number of electrons = 3, total charge: –3 Lithium atom is neutral . Number of protons = 3, total charge: +3 Number of electrons = 2, total charge: –2 Positively charges lithium ion, Li+ is formed. (ii) Sodium atom releases one electron to achieve stable octet electron arrangement: Na Na+ + e– Electron arrangement: 2.8.1 Electron arrangement: 2.8 Number of protons = 11, total charge: +11 Number of electrons = 11, total charge: –11 Sodium atom is neutral . Number of protons = 11, total charge: +11 Number of electrons = 10, total charge: –10 Positively charges sodium ion, Na+ is formed. LS 4.4.1 4.4 ELEMENTS IN GROUP 1 CS 4.4 PERFORMANCE LEVEL (PL) Mastered Not mastered PL4 Analyse knowledge about Periodic Table of Elements in the context of problem solving the natural occurrences or phenomena. 04 U4 Chemistry F4(p53-76)csy2p.indd 59 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 60 UNIT 4 Experiment for the Chemical Properties of Group 1 Elements: LS 4.4.2 (a) Metal Group 1 reacts with water to produce alkali and hydrogen gas. 2X + 2H2O 2XOH + H2 , X is the metal of Group 1 Water Lithium Procedures: (i) Pour water into a basin until half full. (ii) Cut a small piece of lithium using a knife and forcep. (iii) Dry the oil on the surface of the lithium with filter paper. (iv) Place the lithium slowly onto the water surface in a water trough. (v) When the reaction stop, test the solution produced with red litmus paper. (vi) Record the observation. (vii) Repeat steps (i) – (vi) using sodium and potassium to replace lithium one by one. Observations: Elements Observations Inferences Reactivity Li • Lithium moves slowly on the water surface and produces red flame. • The colourless solution formed turns red litmus paper to blue . • Lithium is the least reactive metal reacts with water to produce alkaline solution, lithium hydroxide: • Balanced chemical equation: 2Li + 2H2O 2LiOH + H2 Reactivity increases down Group 1 Na • Sodium moves quickly on the water surface and produces yellow flame. • The colourless solution formed turns red litmus paper to blue . • Sodium is reactive metal reacts with water to produce alkaline solution, sodium hydroxide. • Balanced chemical equation: 2Na + 2H2O 2NaOH + H2 K • Potassium moves very quickly on the water surface and produce purple flame. • The colourless solution formed turns red litmus paper to blue . • Potassium is the most reactive metal reacts with water to produce alkaline solution, potassium hydroxide. • Balanced chemical equation: 2K + 2H2O 2KOH + H2 • All elements in Group 1 have similar chemical properties because all atoms in Group 1 have one valence electron and achieve the stable duplet/octet electron arrangement by releasing its valence electron to form a positively charged ions. How are the Group 1 elements stored? Explain. LS 4.4.3 • The elements are stored under the paraffin oil. • To prevent them from reacting with water vapour and air. 04 U4 Chemistry F4(p53-76)csy2p.indd 60 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 61 UNIT 4 © Nilam Publication Sdn. Bhd. (b) Metal Group 1 reacts with oxygen to form metal oxide. The metal oxide dissolves in water to produce alkaline solution. 4X + O2 2X2O X2O + H2O 2XOH, X is a metal element of Group 1 (Li, Na and K) Procedures: (i) Cut a small piece of lithium using a knife and forcep. (ii) Dry the oil on the surface of the lithium with filter paper. (iii) Place the lithium in a combustion spoon and heat lithium until it start to burn. (iv) Put the burning lithium into a gas jar of oxygen. (v) When the reaction stop, add water to dissolve the compound formed. (vi) Add a few drops of universal indicator to the solution formed. (vii) Record the observation. (viii) Repeat steps (i) – (vii) using sodium and potassium to replace lithium one by one. Observations: Elements Observations Inferences Reactivity Li • Lithium burns slowly with a red flame to produce white solid . • The white solid dissolves in water to form colourless solution. • The solution turns green universal indicator to purple . • Lithium is the least reactive metal towards oxygen. • Lithium reacts with oxygen to produce lithium oxide. • Balanced chemical equation: 4Li + O2 2Li2O • Lithium oxide reacts with water to form alkaline solution, lithium hydroxide. • Balanced chemical equation: Li2O + H2O 2LiOH Reactivity increases down Group 1 Na • Sodium burns brightly with a yellow flame to produce white solid . • The white solid dissolves in water to form colourless solution. • The solution turns green universal indicator to purple . • Sodium is reactive metal towards oxygen. • Sodium reacts with oxygen to produce sodium oxide. • Balanced chemical equation: 4Na + O2 2Na2O • Sodium oxide reacts with water to form alkaline solution, sodium hydroxide. • Balanced chemical equation: Na2O + H2O 2NaOH K • Potassium burns very brightly with a purple flame to produce white solid . • The white solid dissolves in water to form colourless solution. • The solution turns green universal indicator to purple . • Potassium is the most reactive metal towards oxygen. • Potassium reacts with oxygen to produce potassium oxide. • Balanced chemical equation: 4K + O2 2K2O • Potassium oxide reacts with water to form alkaline solution, potassium hydroxide. • Balanced chemical equation: K2O + H2O 2KOH Combustion spoon Burning lithium Oxygen gas Gas jar 04 U4 Chemistry F4(p53-76)csy2p.indd 61 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 62 UNIT 4 (c) Metal Group 1 reacts with chlorine to produce metal chloride. 2X + Cl2 2XCl, X is a metal element of Group 1 (Li, Na and K) Combustion spoon Chlorine gas Gas jar Burning of Group 1 metal Procedures: (i) Cut a small piece of lithium using a knife and forcep. (ii) Dry the oil on the surface of the lithium with filter paper. (iii) Place the lithium in a combustion spoon and heat lithium until it start to burn. (iv) Put the burning lithium into a gas jar of chlorine. (v) When the reaction stop, add water to dissolve the compound formed. (vi) Add a few drops of universal indicator to the solution formed. (vii) Record the observation. (viii) Repeat steps (i) – (vii) using sodium and potassium to replace lithium one by one. Observations: Elements Observations Inferences Reactivity Li • Lithium burns slowly with a red flame to produce white solid. • Lithium is the least reactive metal towards chlorine. • Lithium reacts with chlorine to produce lithium chloride . Balanced chemical equation: 2Li + Cl2 2LiCl Reactivity increases down Group 1 Na • Sodium burns brightly with a yellow flame to produce white solid. • Sodium is reactive metal towards chlorine. • Sodium reacts with chlorine to produce sodium chloride . • Balanced chemical equation: 2Na + Cl2 2NaCl K • Potassium burns very brightly with a purple flame to produce white solid. • Potassium is the most reactive metal towards chlorine. • Potassium reacts with chlorine to produce potassium chloride . • Balanced chemical equation: 2K + Cl2 2KCl 04 U4 Chemistry F4(p53-76)csy2p.indd 62 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 63 UNIT 4 © Nilam Publication Sdn. Bhd. 3 Metal Group 1 reacts with oxygen or air to form metal oxide. The metal oxide reacts with water. 4X + O2 → 2X2O X2O + H2O → 2XOH (a) 4 Li + O2 → 2Li2O Li2O + H2O → 2LiOH (b) 4 Na + O2 → 2Na2O Na2O + H2O → 2NaOH (c) 4 K + O2 → 2K2O K2O + H2O → 2KOH Group 1 Metal Li, Na, K X Complete the following: LS 4.4.4 2 Metal Group 1 reacts with chlorine. 2X + Cl2 → 2XCl (a) 2 Li + Cl2 → 2LiCl (b) 2 Na + Cl2 → 2NaCl (c) 2 K + Cl2 → 2KCl 1 Metal Group 1 react with water. 2X + 2H2O → 2XOH + H2 (a) 2 Li + 2H2O → 2LiOH + H2 (b) 2 Na + 2H2O → 2NaOH + H2 (c) 2 K + 2H2O → 2KOH + H2 State the special name for Group 17 elements. • Halogens List the elements in Group 17 of the Periodic Table and write the electron arrangements and number of shells of the atoms of elements. Elements Symbols Protons number Electrons arrangement Number of shells Fluorine F2 9 2.7 2 Chlorine Cl2 17 2.8.7 3 Bromine Br2 35 2.8.18.7 4 Iodine I2 53 2.8.18.18.7 5 State the physical properties of Group 17 elements. LS 4.5.1 • Halogens cannot conduct heat and electricity in all states. 4.5 ELEMENTS IN GROUP 17 CS 4.5 04 U4 Chemistry F4(p53-76)csy2p.indd 63 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 64 UNIT 4 Explain the changes in physical properties going down Group 17 elements. (a) The melting and boiling points increase going down the group because: • The atomic size increases going down the Group 17 because of increasing in number of shell , the size of molecules get larger. • The inter molecular forces of attraction (van der Waals forces) between molecules become stronger. • More energy is needed to overcome the stronger attractive forces between molecules during melting or boiling. (b) Physical states change from gas (fluorine and chlorine) to liquid (bromine) and to solid (iodine) at room temperature due to increase in the strength of inter molecular forces from fluorine to iodine. (c) The density is low and increases. (d) The colour of the elements becomes darker : fluorine (light yellow), chlorine (greenish yellow), bromine (brown) and iodine (purplish black). Explain the similarities in chemical properties of the Group 17 elements. LS 4.5.2 Remark: 1 Proton is a positively charged sub atomic particle 2 Electron is negatively charged sub atomic particle (a) All atoms of elements in Group 17 have seven valence electrons and achieve a stable octet electron arrangement by accepting one electron to form negatively charged ions. Examples: (i) Fluorine atom receives one electron to achieve stable octet electron arrangement: Electron arrangement: 2.8 F– Electron arrangement: 2.7 F + e– Number of protons = 9, total charge: +9 Number of electrons = 9, total charge: –9 Fluorine atom is neutral . Number of protons = 9, total charge: +9 Number of electrons = 10, total charge: –10 Negatively charged fluoride ion, F– is formed. (ii) Chlorine atom receives one electron to achieve stable octet electron arrangement: Electron arrangement: 2.8.8 Cl– Electron arrangement: 2.8.7 Cl + e– Number of protons = 17, total charge: +17 Number of electrons = 17, total charge: –17 Chlorine atom is neutral . Number of protons = 17, total charge: +17 Number of electrons = 18, total charge: –18 Negatively charged chloride ion, Cl– is formed. (b) All elements in Group 17 have similar chemical properties because atoms in Group 17 have seven valence electrons and achieve the stable octet electron arrangement by receiving one electron to form a negatively charged ion. LS 4.5.1 F– 04 U4 Chemistry F4(p53-76)csy2p.indd 64 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 65 UNIT 4 © Nilam Publication Sdn. Bhd. Elements in Group 17 exist as diatomic molecules. Explain. • Group 17 elements exist as diatomic molecules. • Two atoms of element sharing one pair of valence electrons to achieve stable octet electron arrangement. Example: Two fluorine atoms share one pair of electrons to form one fluorine molecule: Fluorine atom Fluorine atom Fluorine molecule Share F F F F Chlorine, bromine and iodine exist as diatomic molecules. (Cl2, Br2 and I2) Experiments for the Chemical Properties of Group 17 Elements: LS 4.5.1 (a) Halogen reacts with water with different reactivity: X2 + H2O HX + HOX, X is halogen. (Cl2, Br2 and I2 ) Chlorine gas Bromine water Iodine crystals Chlorine or Bromine Klorin atau Bromin Haba Heat Haba Heat NaOH to absorb Chlorine / bromine NaOH untuk menyerap klorin / bromin Iron wool Wul Besi Iodine Iodin Fluorine, Chlorine Florin, Klorin air water Chlorine gas Water Procedures: • Chlorine gas is passed through water in a test tube. • The solution produced is tested with blue litmus paper. Water Bromine water Procedures: • A few drops of bromine water are added to water in a test tube. • The test tube is shaken. • The solution produced is tested with blue litmus paper. Iodine crystals Water Procedures: • Some iodine crystals are added to water in a test tube. • The test tube is shaken. • The solution produced is tested with blue litmus paper. Observations: • Chlorine dissolves rapidly in water to form light yellow solution: Cl2 + H2O HCl + HOCl • The solution changes blue litmus paper to red and quickly decolourises it. Observations: • Bromine dissolves slowly in water to form brown solution: Br2 + H2O HBr + HOBr • The solution changes blue litmus paper to red and slowly decolourises it. Observations: • Iodine dissolves very slow in water to form brown solution: I2 + H2O HI + HOI • The solution changes blue litmus paper to red . The litmus paper does not decolourise . Inference: • Chlorine, bromine and iodine react with water to form acidic solution. • Apart from the acidic solution, chloride and bromine also formed bleaching agent. • Solubility decreases from chlorine to iodine. 04 U4 Chemistry F4(p53-76)csy2p.indd 65 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 66 UNIT 4 (b) Halogens react with hot iron to form brown solid, iron(III) halide. Heat Heat Iodine Iron wool Chlorine or Bromine NaOH to absorb chlorine/bromine Iron wool Heat 2Fe + 3X2 2FeX3, X2 represents any halogen. (Cl2, Br2 or I2 ) Halogens Observations Chemical equations Chlorine • Iron wool burns very brightly and forms a brown solid when cooled. 2Fe + 3Cl2 2FeCl3 Bromine • Iron wool burns brightly and forms a brown solid when cooled. 2Fe + 3Br2 2FeBr3 Iodine • Iron wools glows slowly with a dull glow and forms a brown solid when cooled. 2Fe + 3I2 2FeI3 Experiment (a), (b) and (c) show that all halogens have similar chemical properties but their reactivity decreases going down the group: F2, Cl2, Br2 and l2 Reactvity decreases (c) Halogens react with sodium hydroxide solution X2 + 2NaOH NaX + NaOX + H2O, X2 is halogen. (Cl2, Br2 and I2 ) Complete the following: (i) Cl2 + 2NaOH NaCl + NaOCl + H2O (ii) Br2 + 2NaOH NaBr + NaOBr + H2O (iii) I2 + 2NaOH NaI + NaOI + H2O Reactivity decreases 1 There are seven periods known as period 1, 2, 3, 4, 5, 6, 7. 2 The number of period of an element represents the number of shells occupy with electrons in each atom of element. Elements Protons number Electrons arrangement Number of shells Period Li 3 2.1 2 2 Na 11 2.8.1 3 3 K 19 2.8.8.1 4 4 4.6 ELEMENTS IN PERIOD 3 CS 4.6 04 U4 Chemistry F4(p53-76)csy2p.indd 66 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 67 UNIT 4 © Nilam Publication Sdn. Bhd. List the elements of Period 3 and write the electron arrangement and number of shells of the atom of elements. Elements Protons number Electrons arrangement Number of shells Radius (nm) Na 11 2.8.1 3 0.191 Mg 12 2.8.2 3 0.160 Al 13 2.8.3 3 0.130 Si 14 2.8.4 3 0.118 P 15 2.8.5 3 0.110 S 16 2.8.6 3 0.102 Cl 17 2.8.7 3 0.099 Ar 18 2.8.8 3 0.095 Define electronegativity. LS 4.6.1 • The strength of an atom in a molecule to attract electron towards its nucleus. State the changes in properties of elements across Period 3 from left to right. LS 4.6.2 LS 4.6.3 (a) Physical state: • The physical state of elements in a period changes from solid to gas from left to right. • Metals on the left are solid while non-metals on the right are usually gases. (b) Changes in metallic properties and electrical conductivity: Element Na Mg Al Si P S Cl Ar Metallic properties Metal Semi metal or metaloid Non-metal Electrical conductivity Good conductors of electric. Weak conductor of electric but it increases with the increasing in temperature and the presence of boron or phosphorous. Uses: semiconductor Cannot conduct electricity (c) Changes in properties of oxide of elements Period 3: Na Mg Al Si P S Cl Basic oxide Amphoteric oxide Acidic oxide Basic oxide + Water Alkali Example: Na2O + H2O 2NaOH Basic oxide + Acid Salt + Water Example: MgO + 2HCl MgCl2 + H2O Amphoteric oxide + Acid Salt + Water Amphoteric oxide + Alkali Salt + Water Example: Al2O3 + 6HNO3 2Al(NO3)3 + 3H2O Al2O3 + 2NaOH 2NaAlO2 + H2O Acidic oxide + Water Acid Example: SO2 + H2O H2SO3 Acidic oxide + Alkali Salt + Water Example: SiO2 + 2NaOH Na2SiO3 + H2O 04 U4 Chemistry F4(p53-76)csy2p.indd 67 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 68 UNIT 4 Define basic oxide, amphoteric oxide and acidic oxide. • Basic oxide is metal oxide that can react with acid to form salt and water . • Acidic oxide is non-metal oxide that can react with alkali to form salt and water . • Amphoteric oxide is oxide that can react with both acid and alkali to form salt and water . Oxides elements in Period 3 consist of base oxides, acid oxides and amphoteric oxides. Conduct an experiment to investigate the properties of three different metal oxides. You are given with the following apparatus and materials. Test tube, spatula, glass rod, Bunsen burner, retort stand with clamp, 2.0 mol dm–3 of nitric acid, 2.0 mol dm–3 of sodium hydroxide, magnesium oxide powder, aluminium oxide powder, silicon(IV) oxide powder, 10 ml measuring cylinder and dropper 1 Carry out experiment by following the steps below. (i) Pour four spatula of magnesium oxide powder, MgO into two different test tube. (ii) Measure and add 5.0 cm3 of 2.0 mol dm–3 sodium hydroxide, NaOH into the first test tube. (iii) Measure and add 5.0 cm3 of 2.0 mol dm–3 nitric acid, HNO3 into the second test tube. (iv) Heat both test tube slowly and stir it by using glass rod as shown in Diagram 1. Nitric acid, HNO3 Magnesium oxide, MgO Sodium hydroxide, NaOH Heat Heat Diagram 1 (v) Observe the solubility of oxide in both solution and record your observation. (vi) Step (i) until (v) is repeat by replacing the magnesium oxide with aluminium oxide, Al2O3 and silicon(IV) oxide, SiO2. (vii) Record your observation in Table 1. Experiments to Investigate the Properties of the Oxide Elements Change Across Period 3: LS 4.6.2 SPM K3 S P O T S P M K 3 04 U4 Chemistry F4(p53-76)csy2p.indd 68 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 69 UNIT 4 © Nilam Publication Sdn. Bhd. Oxides Observations Reactions with nitric acid, HNO3 Reactions with sodium hydroxide, NaOH Magnesium oxide, MgO The white solid dissolves to form colourless solution. No change. The white solid does not dissolve. Aluminium oxide, Al2O3 The white solid dissolves to form colourless solution. The white solid dissolves to form colourless solution. Silikon(IV) oxide, SiO2 No change. The white solid does not dissolve. The white solid dissolves to form colourless solution. Table 1 [3 marks] (a) State the variables in this experiment (i) manipulated : Type of oxide of elements in Period 3 (ii) responding : Reaction of oxide with acid and alkali (iii) constant : Nitric acid and sodium hydroxide solution [3 marks] (b) Give the equation for the reaction of magnesium oxide, MgO and nitric acid, HNO3. MgO + 2 HNO3 → Mg(NO3)2 + H2O [2 marks] (c) State the hypothesis for this experiment. The oxide of elements across Period 3 change from basic oxide (magnesium oxide) to amphoteric oxide (aluminium oxide) and then to acidic oxide (silicone(IV) oxide). [2 marks] (d) State the operational definition for the acidic properties in this experiment. When aluminium oxide powder is added into sodium hydroxide solution, the powder dissolved. [2 marks] (e) Classify the following oxides into acidic oxide and basic oxide. • Sodium oxide • Carbon dioxide • Phosphorous pentoxide • Calcium oxide Acidic oxide Basic oxide Carbon dioxide Phosphorous pentoxide Sodium oxide Calcium oxide [3 marks] S P O T S P M K 3 04 U4 Chemistry F4(p53-76)csy2p.indd 69 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 70 UNIT 4 Explain Changes in Reactivity of Going Down Group 1 and 17 Elements as well as Changes in Size and Electronegativity Across Period 3 The atomic radius of the atoms decreases from sodium to chlorine Mg Al P S Na Si Cl • All the atoms of Period 3 elements have 3 shells occupied with electrons . Across Period 3 From Left to Right: • The proton number increases by one unit from sodium to chlorine. • Increasing in proton number causes the number of positive charge in the nucleus to increase . • The size of atom decreases across Period 3. • The strength of attraction from the proton in the nucleus to the electrons in the shells increases . • The electronegativity increases across Period 3 from sodium to chlorine. • Atoms of Group 1 metals achieve a stable duplet/octet electron arrangement by releasing one valence electron to form +1 charged ion. Going down Group 1, • The number of shells increases, the atomic size increases. • The strength of attraction from the proton in the nucleus towards the valence electron gets weaker . • The valence electron is easier be released by the Group 1 metal atom. • The reactivity of the elements increases. • Atoms of Group 17 have seven valence electrons and achieve a stable octet electron arrangement by accepting one electron to form –1 charged ion. Going down Group 17, • The number of shells increases, atomic size increases. • The strength of attraction from the proton in the nucleus towards electrons gets weaker . • The strength of a halogen atom to attract electron to the outermost shell decreases (electronegativity decreases). • The reactivity of the elements decreases. F Cl Br Reactivity decreases down Group 17 Reactivity increases down Group 1 Li NaK 04 U4 Chemistry F4(p53-76)csy2p.indd 70 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 71 UNIT 4 © Nilam Publication Sdn. Bhd. The Periodic Table Compare and explain the reactivity of elements X and Y. Element X Y Proton number 11 19 • Element Y is more reactive than element X. • Electron arrangement of X atom is 2.8.1 and Y atom is 2.8.8.1. • The number of shells occupied with electrons of atom Y is more than atom X. • The size of atom Y is larger than atom X. • Force of attraction between the nucleus towards the valence electron for atom Y is weaker than atom X. • Therefore, it is easier for atom Y to release the valance electron compared to X atom. Compare and explain the reactivity of elements X and Y. LS 4.5.4 Element X Y Proton number 9 17 • Element Y is less reactive than element X. • Electron arrangement of atom X is 2.7 and atom Y is 2.8.7. • The number shells occupied with electrons of atom Y is more than atom X. • The size of atom Y is larger than atom X. • Thus, the force of attraction the nucleus to attract one electrons on the outermost shells of atom Y is weaker than atom X. Compare and explain electronegativity of elements X and Y. LS 4.6.1 Element X Y Proton number 11 17 • Element Y is more electronegative than element X. • Electron arrangement of X atom is 2.8.1 and Y atom is 2.8.7 . • Atoms X and Y have same number shells occupied with electrons. • The number of protons in the nucleus of atom Y is more than atom X. • The attraction forces between the nucleus and the electrons in the shells of atom Y is stronger than atom X. • The size of atom Y is smaller than atom X. • The tendency to attract electrons of atom Y is stronger than atom X. To Compare Reactivity Down Group 1 and Group 17: To Compare Atomic Size / Radius and Electronegativity Across Period 3: (i) Compare number of shells in each atom. (ii) • Compare the strength of proton in the nucleus to attract valence electron (Group 1). • Compare the strength of proton in the nucleus to attract electron to the outermost shells (Group 17). (iii) • Compare tendency of the atom to release electron (Group 1). • Compare tendency of the atom to receive electron (Group 17). (iv) Compare the reactivity of the elements in the groups. (i) Compare number of shells in each atom. (ii) Compare number of proton in the nucleus. (iii) Compare the strength of attraction from the nucleus to the electrons in the shells . (iv) Compare the atomic size / Compare the electronegativity. 04 U4 Chemistry F4(p53-76)csy2p.indd 71 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 72 UNIT 4 State the position of the transition element in the Periodic Table • Situated between Groups 2 and 13 Examples: Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu and Zn. What are the metallic properties of transition element? (i) Shiny (ii) Conducts heat and electricity (iii) Malleable (iv) High tensile strength (v) High melting point and density What are the special characteristics of transition element? LS 4.7.2 • Most transition elements formed coloured compounds. Examples: (i) Iron(III) chloride is brown. (ii) Iron(II) chloride is green. (iii) Copper(II) sulphate is blue. • Most transition elements have more than one oxidation number in their compounds. Examples: Elements Compounds Oxidation number Copper Copper(I) chloride + 1 Copper(II) oxide + 2 Iron Iron(II) chloride + 2 Iron(III) chloride + 3 • Oxidation number of element in a compound will be studied in topic “redox”, Form 5. • Many of the transition elements are able to form complex ion: Elements Complex ions Formula Iron Hexacyanoferrate(II) Fe(CN)6 4- Copper Copper(II) tetramine Cu(NH4)4 2+ State the uses of transition elements in industry. LS 4.7.3 • Many of the transition elements can act as a catalyst in industries. • Catalyst is a substance that can change the rate of reaction. • A catalyst does not change chemically after a reaction. Examples: (i) Iron: Haber process in the manufacture of ammonia (ii) Vanadium(V) oxide: Contact process in the manufacture of sulphuric acid (iii) Platinum: Ostwald process in the manufacture of nitric acid LS 4.7.1 4.7 TRANSITION ELEMENTS CS 4.7 PERFORMANCE LEVEL (PL) Mastered Not mastered PL6 Invent by applying the knowledge about Periodic Table of Elements in the context of problem solving and decision-making or when carrying out an activity/task in new situations in a creatively and innovative; giving due considerations to the social/ economic/cultural values of the community. 04 U4 Chemistry F4(p53-76)csy2p.indd 72 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 73 UNIT 4 © Nilam Publication Sdn. Bhd. 1 The diagram below shows the electron arrangement for atoms P and Q. P P Q Q (a) Elements P and Q are placed in the same group in Periodic Table. State the group. Group 1 (b) How are elements P and Q kept in the laboratory? Give reason for your answer. In paraffin oil. To prevent them from reacting with oxygen or water vapour in the atmosphere. (c) (i) Write chemical equation for the reaction between element P with water. 2P + 2H2O 2POH + H2 (ii) What is the expected change of colour when a few drops of phenolphthalein are added into the aqueous solution of the product? Explain your answer. Colourless to purple / pink. The solution formed is alkaline. (iii) Between element P and element Q, which is more reactive in the reaction with water? Element Q is more reactive than P. (iv) Explain your answer in 1(c)(iii). • The size of atom Q is larger than atom P. • The valence electron of atom Q is further away from the nucleus compared to atom P. • The attractive forces between proton in the nucleus to the valence electron of atom Q is weaker than atom P. • It is easier for atom Q to release the valence electron compared to atom P. (d) Name one element that has the same chemical properties as P and Q. Potassium 2 The table below shows the number of neutron and relative atomic mass of eight elements represented as P, Q, R, S, T, U, V and W. Atoms P Q R S T U V W Number of neutron 12 12 14 14 16 16 18 22 Relative atomic mass 23 24 27 28 31 32 35 40 Number of proton 11 12 13 14 15 16 17 18 Electron arrangement 2.8.1 2.8.2 2.8.3 2.8.4 2.8.5 2.8.6 2.8.7 2.8.8 Subjective Questions PL2 PL1 PL3 PL3 PL4 PL4 PL2 ENRICHMENT EXERCISE Quiz 04 U4 Chemistry F4(p53-76)csy2p.indd 73 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 74 UNIT 4 (a) Complete the table by writing the number of proton and electron arrangement for the atom of each element. (b) (i) State the period of elements P – W in the Periodic Table. Explain your answer. Period 3 because P – W atoms have three shells occupied with electrons. (ii) State the proton number of another element that is in the same group as P. 3/19 (c) Write the standard representation for element Q. 24 12Q (d) Which element exists as monoatomic gas? W diatomic gas? T/ U/ V (e) (i) Which element can react vigorously with water to produce hydrogen gas? P (ii) Write the balanced equation for the reaction in 2(e)(i). 2P + 2H2O → 2POH + H2 (f) State the arrangement of elements T, U and V in the order of increasing atomic radius. Explain your answer. • V, U and T. • Atoms of T, U, and V have three shells occupied with electrons. • The proton number // positive charges in the nucleus increases from T to V. • The forces of attraction between proton in the nucleus towards the electrons in the shells increase from T to V. • The shells filled with electrons are pulled nearer to the nucleus from T to V. 3 The diagram below shows part of the Periodic Table of Elements. X, Y, Z, A, B, D, E, F and G do not represent the actual symbols. X Y Z A B D E F G (a) (i) State the position of element B in the Periodic Table. Period 3, Group 13 (ii) Explain your answer in 3(a)(i). Electron arrangement atom B is 2.8.3. Atom B has three valence electrons, element B is in Group 13. Atom B has three shells occupied with electrons, element B is in Period 3. (b) (i) Which element is monatomic gas? Element Y/Z (ii) Explain your answer in 3(b)(i). Atom Y has achieved stable duplet electron arrangement // has electron arrangement 2. OR Atom Z has achieved stable octet electron arrangement // has electron arrangement 2.8. PL1 PL2 PL2 PL2 PL3 PL2 PL2 PL3 PL2 PL2 PL2 PL2 PL2 04 U4 Chemistry F4(p53-76)csy2p.indd 74 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 75 UNIT 4 © Nilam Publication Sdn. Bhd. (c) Element X is hydrogen gas and element Y is helium gas. The diagram on the right shows a meteorological balloon filled with helium gas. (i) Explain why helium gas is used to fill the meteorological balloon. Helium gas is light and inert. (ii) Can hydrogen gas replace helium gas in the balloon? Give reason for your answer. Cannot. Hydrogen gas is flammable, it will explode with the presence of oxygen gas at high temperature. (d) Choose an element that: (i) exists in the form of molecule X / D / E (ii) forms acidic oxide D / E (iii) has atoms that have no neutron X (iv) is an alkali metal A / F (v) forms amphoteric oxide B (vi) has a proton number of 15 D (vii) is most electropositive F (viii) forms basic oxide A / F (ix) forms coloured compound G (e) Arrange Y, A, B, D and E according to the order of increasing atomic size. Y, E, D, B, A (f) (i) Write the electron arrangement for an atom of element: D: 2.8.5 E: 2.8.7 (ii) Compare electronegativity of elements D and E. Element E is more electronegative than element D. (iii) Explain your answer in 3(f)(ii). • Atoms E and D have the same number of shells occupied with electrons. • The number of proton in the nucleus of atom E is more than atom D. • The strength of proton in nucleus to attract electrons in the atom E is stronger than of atom D. HOTS PL2 PL5 PL2 PL2 PL2 PL4 PL4 Topical Exercise Gas helium 04 U4 Chemistry F4(p53-76)csy2p.indd 75 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 76 UNIT 4 1 Proton number of element P is 8. What is the position of this element in the Periodic Table of Elements? Group Period A 16 2 B 16 3 C 18 2 D 18 3 2 Potassium reacts with element Q from Group 17 in Periodic Table. Which of the following chemical equations is correct? A K + Q KQ B K+ + Q– KQ C 2K + Q2 2KQ D K + Q2 KQ2 3 The diagram below shows the standard representation for elements X, Y and Z. Y32 16 Z23 X 11 27 13 What type of oxides are formed by X, Y and Z? X oxide Y oxide Z oxide A Amphoteric Acidic Basic B Amphoteric Basic Acidic C Acidic Amphoteric Basic D Acidic Acidic Basic 4 The diagram below shows the position of elements X, Y and Z in the Periodic Table. X Y Z Which of the following statements is true? A All the elements can conduct electricity. B All the elements exist as gas at room temperature. C The boiling points of the elements increase from X → Y → Z. D The density of the elements decreases going down from X → Y → Z. 5 Which of the following elements can form acidic oxide? I Calcium II Sulphur III Potassium IV Nitrogen A I and II only B I and III only C II and IV only D III and IV only 6 The table below shows the properties of the oxide of elements X, Y and Z which are located in Period 3 of the Periodic Table. Element Property of the oxide formed X – Oxide of X reacts with nitric acid. – Oxide of X does not react with sodium hydroxide solution. Y – Oxide of Y reacts with sodium hydroxide solution. – Oxide of Y does not react with nitric acid. Z – Oxide of Z reacts with sodium hydroxide solution. – Oxide of Z reacts with nitric acid. What is the correct arrangement of elements X, Y and Z from left to right in Period 3 of the Periodic Table? A Z, X, Y C X, Y, Z B X, Z, Y D Y, Z, X 7 The following statements describe the characteristic of an element. • Used as a catalyst. • Forms coloured ions or compound. • Shows different oxidation number in its compound. Which of the following is the position of the element in the Periodic Table of Element? A B C D Objective Questions SPM PRACTICE PL2 PL3 PL2 PL3 PL4 PL4 PL4 04 U4 Chemistry F4(p53-76)csy2p.indd 76 21/12/2022 3:49 PM

MODULE • Chemistry FORM 4 77 UNIT 5 © Nilam Publication Sdn. Bhd. CHEMICAL BOND Positive ion Negative ion Simple molecule Giant molecular structure Ionic Bond Metallic Bond Covalent Bond Hydrogen Bond Ionic compound Transfer of electron from METAL ATOM to NON-METAL ATOM Sea of electrons from the valence electrons of METAL ATOM and positive METAL ION Electrostatic force between sea of electrons and metal ion Molecule of covalent compound Dative Bond Example: 2.8 2.8.8 To achieve stable duplet/octet electron arrangement Involves Involves Involves Electrons are contributed by both atoms Electrons are contributed by one atom only Between hydrogen atom and nitrogen/ oxygen/fluorine atom Hydrogen bond Dative bond Metal atom donates electron Non-metal atom receives electron Weak van der Waals forces between molecules Sodium chloride Ammonium ion Water Diamond Example: Example: Example: Example: Sharing one pair/two pairs/three pairs of electrons between NON-METAL ATOMS H δ+ H δ+ O δ– O δ– H δ+ H δ– H N H H H Strong covalent bond between atoms in the molecules Example: Carbon dioxide Concept Map 5 UNIT CHEMICAL BOND Strong electrostatic forces between positive and negative ions Simulation 05 U5 Chemistry F4(p77-97)csy2p.indd 77 21/12/2022 3:17 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 78 UNIT 5 What is compound? • A substance containing two or more types of elements chemically combined. Why Group 18 elements are inert gases? • The atoms have achieved duplet electron arrangement for helium and octet electron arrangement for others. What are chemical bonds? • Chemical bonds are formed when two or more atoms of elements bonded together. • There are two types of chemical bond, Ionic Bond and Covalent Bond. Why do certain atoms form chemicals bond with other atoms? • Atoms form chemical bonds to achieve a stable duplet or octet electron arrangement. What types of elements formed ionic bonds? • Ionic bond is formed between atoms of metal elements that release electrons to atoms of non-metal elements. How ionic bond is formed? • Atoms of elements that release electrons form positive ions and atoms that receive electrons form negative ions. • Ionic bond is usually formed between atoms from Groups 1, 2 and 13 (metal) with atoms from Groups 15, 16 and 17 (non-metal). Complete the following table: LS 5.2.1 Changes Na Na+ + e– Ca Ca2+ + 2e– O + 2e– O2– Cl + e– Cl– Electron arrangement 2.8.1 2.8 2.8.2 2.8 2.6 2.8 2.8.7 2.8.8 Total of positive charges (From number of proton) +11 +11 +12 +12 +8 +8 +17 +17 Total of negative charges (From number of electron) –11 –10 –12 –10 –8 –10 –17 –18 Total charges 0 +1 0 +2 0 –2 0 –1 Type of particles Sodium atom Sodium ion Calcium atom Calcium ion Oxygen atom Oxide ion Chlorine atom Chloride ion Define ionic bond. LS 5.2.1 The electrostatic force between the positive and negative ions forms ionic bond. LS 5.1.1 LS 5.1.1 LS 5.1.1 LS 5.2.1 5.1 BASIC OF COMPOUND FORMATION CS 5.1 5.2 IONIC BONDS CS 5.2 PERFORMANCE LEVEL (PL) Mastered Not mastered PL1 Recall knowledge and basic skills on chemical bonds. 05 U5 Chemistry F4(p77-97)csy2p.indd 78 21/12/2022 3:17 PM

MODULE • Chemistry FORM 4 79 UNIT 5 © Nilam Publication Sdn. Bhd. Example 1: Explain the formation of sodium chloride: Predict formula: Element Proton number Electron arrangement Na 11 2.8.1 Cl 17 2.8.7 Na Na+ + e– Cl + e– Cl– Na+ 1 Cl– 1 ⇒ NaCl Sodium atom, Na Sodium ion, Na+ Chlorine atom, Cl Chloride ion, Cl– Strong electrostatic force between ions Transfer Na Cl Na Cl (a) Electron arrangement of sodium atom is 2.8.1 . Sodium atom has one valence electron. Therefore, sodium atom is not stable . Sodium atom releases one electron to achieve a stable octet electron arrangement to form sodium ion , Na+ with electron arrangement 2.8 . (b) Electron arrangement of chlorine atom is 2.8.7 . Chlorine atom has seven valence electrons. Chlorine atom receives one electron to achieve stable octet electron arrangement to form chloride ion, Cl– with an octet arrangement of electron 2.8.8 . (c) Sodium ions , Na+ and chloride ions , Cl– ions are attracted with strong electrostatic force. The bond formed is called ionic bond. PERFORMANCE LEVEL (PL) Mastered Not mastered PL2 Understand and explain chemical bonds. Steps to Explain the Formation of Ionic Bond for Ionic Compound (c) • State the electrostatic force between the positive and negative ion formed. (b) • State the electron arrangement of non-metal atom (number of valence electron is 5/6/7). • State the number of electron received by the atom. • State the name of negative ion formed and the electron arrangement achieved (stable duplet/octet electron arrangement). Predict formula: Determine the formula of the ionic compound formed. Draw the electron arrangement for the compound formed. • Write the electron arrangement of metal atom and nonmetal atom. • By referring to the number of valence electron for each atom, determine the charge of ion formed from each atom. • Cross the coefficient of charge to get the formula of the ionic compound formed. • The number of positive ions and negative ions in the compound is based on the formula. • Draw all the electrons in the shells of positive ion and negative ion. • Write the charge of each ion. (a) • State the electron arrangement of metal atom (number of valence electron is 1/2/3). • State the number of electron released by the atom. • State the name of positive ion formed and the electron arrangement achieved (stable duplet/octet electron arrangement). Explanation AR 05 U5 Chemistry F4(p77-97)csy2p.indd 79 21/12/2022 3:17 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 80 UNIT 5 Example 2: Explain the formation of magnesium oxide: Element Proton number Electron arrangement Mg 12 2.8.2 O 8 2.6 Mg Mg2+ + 2e– O + 2e– O2– Mg2+ 2 O2– 2 ⇒ MgO Draw the electron arrangement of the compound formed. Magnesium atom, Mg Transfer Oxygen atom, O Mg O Magnesium ion, Mg2+ Oxide ion, O2– 2+ Mg O 2– (a) The electron arrangement of magnesium atom is 2.8.2 . Magnesium atom has 2 electrons in the outer shell. Therefore, magnesium atom is not stable . Magnesium atom releases 2 valence electrons to achieve a stable octet electron arrangement to form magnesium ion, Mg2+ with electron arrangement 2.8 . (b) The electron arrangement of oxygen atom is 2.6 . Oxygen atom is also unstable. Oxygen atom receives two electrons to achieve a stable octet electron arrangement to form oxide ion , O2– with electron arrangement 2.8 . (c) Strong electrostatic force is formed between magnesium ion , Mg2+ and oxide ion , O2– to form ionic bond. What types of elements formed covalent bonds? • Covalent bond is formed when similar or different non-metal atoms bond together. [Atoms from Groups 14, 15, 16 and 17] How covalent bond is formed? • This bond is formed when two or more similar or different atoms share their valence electrons to achieve stable duplet or octet electron arrangement to form a neutral molecule. • During the formation of covalent bond, each atom contributes same number of electrons for sharing. The number of electrons shared can be one pair, two pairs or three pairs. • The forces that exist between molecules are van der Waals forces that are weak. These forces become stronger when the molecule size increases. Define covalent bond. LS 5.3.1 Covalent bond is the bond formed by the shared of electrons between atoms. 5.3 COVALENT BONDS CS 5.3 05 U5 Chemistry F4(p77-97)csy2p.indd 80 21/12/2022 3:17 PM

MODULE • Chemistry FORM 4 81 UNIT 5 © Nilam Publication Sdn. Bhd. Example 1: Explain the formation of hydrogen molecule, H2 Share Covalent bond between atoms H H H H or: H + H H H or H – H Lewis Structure (a) Hydrogen atom has one electron in the first shell, with an electron arrangement of 1, needs one electron to achieve a stable duplet electron arrangement. (b) Two hydrogen atoms share a pair of electrons to form a hydrogen molecule. (c) Both hydrogen atoms achieve a stable duplet arrangement of electron. (d) The number of electron pairs shared is one pair. Single covalent bond is formed. Example 2: Explain the formation of oxygen molecule, O2 Share Oxygen atom, O Oxygen atom, O Oxygen molecule, O2 O O O O or: O O O O or O = O (a) Oxygen atom with an electron arrangement 2.6 needs two electrons to achieve a stable octet electron arrangement. (b) Two oxygen atoms share two pairs of electrons to achieve a stable octet arrangement of electron, form an oxygen molecule. Each oxygen atom achieves stable octet electron arrangement. (c) The number of electron pairs shared is 2 pairs. Double covalent bond is formed. Example 3: Explain the formation of nitrogen molecule, N2 Nitrogen atom, N Nitrogen atom, N Nitrogen molecule, N2 Share N N N N or: N N N N or (a) Nitrogen atom with an electron arrangement 2.5 needs 3 electrons to achieve stable octet electron arrangement. (b) Two nitrogen atoms share 3 pairs of electrons to achieve a stable octet electron arrangement, form a nitrogen molecule. Each nitrogen atom achieves stable octet electron arrangement. (c) The number of electron pairs shared is 3 pairs. Triple covalent bond is formed. LS 5.3.1 LS 5.3.1 AR 05 U5 Chemistry F4(p77-97)csy2p.indd 81 21/12/2022 3:17 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 82 UNIT 5 Steps to Explain the Formation of Covalent Bond for Covalent Compound (d) State the electron arrangement achieved by each atom in the molecule (stable duplet/octet electron arrangement). (e) State the type of covalent bond between atoms. Based on the drawing of the electron arrangement, state: (b) the number of pairs of electrons shared between atoms. (c) the number of electrons of each atom contribute for sharing in the molecule and type of covalent bond formed (single/double). (a) State the electron arrangement of each atom and the number of electrons needed to achieve stable duplet/octet electron arrangement. • Electrons are shared in pair. • The number of pairs of electrons shared is based on the number of electrons needed by each atom to achieve stable duplet/octet electron arrangement. • Write the electron arrangement of each atom. • By referring to the number of valence electrons for each atom, determine the number of electrons needed by each atom to achieve stable duplet/octet electron arrangement. • Cross the number of electrons needed to achieve stable duplet/octet electron arrangement. Draw the electron arrangement for the compound formed based on the formula. Predict formula: Determine the formula of the covalent compound formed. Explanations Example 4: Explain the formation of Hydrogen chloride, HCl Element Proton number Electron arrangement H 1 1 Cl 17 2.8.7 Predict formula: H Cl 1 electron needs needs 1 electron Cross the number of electrons each atom needs ⇒ HCl Hydrogen atom, H Chlorine atom, Cl Hydrogen chloride molecule, HCl H Cl H Cl Share or: H Cl H Cl or H – Cl (a) Hydrogen atom with an electron arrangement 1 needs one electron to achieve a stable duplet electron arrangement. Chlorine atom with an electron arrangement 2.8.7 needs one electron to achieve stable octet electron arrangement. (b) One chlorine atom shares one pair of electrons with one hydrogen atom to form hydrogen chloride molecule with the formula HCl . (c) One chlorine atom contributes one electron and one hydrogen atom contributes one electron for sharing. (d) Chlorine atom achieves stable octet electron arrangement and hydrogen atom achieves stable duplet electron arrangement. (e) One chlorine atom forms one single covalent bond with one hydrogen atom. 05 U5 Chemistry F4(p77-97)csy2p.indd 82 21/12/2022 3:17 PM

MODULE • Chemistry FORM 4 83 UNIT 5 © Nilam Publication Sdn. Bhd. Example 5: Explain the formation of carbon dioxide, CO2 Element Proton number Electron arrangement C 6 2.4 O 8 2.6 Predict formula: C O 4 electrons needs needs 2 electrons Cross the number of electrons each atom needs ⇒ CO2 Oxygen atom, O Oxygen atom, O Carbon dioxide molecule, CO2 Carbon atom, C Share Share O C O O C O or: O C O O C O or O = C = O (a) Carbon atom with an electron arrangement 2.4 needs four electrons to achieve a stable octet electron arrangement. Oxygen atom with an electron arrangement 2.6 needs two electrons to achieve stable octet electron arrangement. (b) One carbon atom shares four pairs of electrons with two oxygen atoms form carbon dioxide molecule with the formula CO2 . (c) One carbon atom contributes four electrons and each of the two oxygen atoms contributes two electrons for sharing to form double covalent bond. (d) Carbon atom achieves stable octet electron arrangement and oxygen atom achieves octet electron arrangement. (e) One carbon atom forms two double covalent bonds with two oxygen atoms. Comparing the Formation of Ionic and Covalent Bonds LS 5.3.2 Ionic Bond Covalent Bond State type of element involved. • Between metals (Groups 1, 2 and 13) and non-metals (Groups 15, 16 and 17). • Between non-metals and non-metals (Groups 14, 15, 16 and 17). How bonds are formed? • Electrons are released by metal atoms and received by non-metal atoms (electron transfer). • Pairs of electrons are shared by the same or different non-metals atoms. What are the type of particle produced? • Metal atom forms positive ion. • Non-metal atom forms negative ion. • Neutral molecule. How to predict the formulae? • Determine the coefficient of the charge of the ions and criss cross. • Determine the number of electrons needed to achieve stable duplet or octet electron arrangement and criss cross. AR 05 U5 Chemistry F4(p77-97)csy2p.indd 83 21/12/2022 3:17 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 84 UNIT 5 Define electronegativity. • Electronegativity is the strength of an atom in a molecule to attract electron towards its nucleus. • Examples of electronegative atoms are chlorine, oxygen and nitrogen. Remark: Electronegativity generally increases from left to right across a period and decreases down a periodic table group. What is hydrogen bond? LS 5.4.1 • Hydrogen bond is an attraction between hydrogen atoms with the electronegative atoms from other molecules when hydrogen atom is covalently bonded to a highly electronegative atoms such as oxygen, nitrogen or fluorine. Are all electronegative atoms bonded covalently with hydrogen causing hydrogen bonds to form between the molecules that are formed? Explain. • No. • The electronegative atom that is bonded to hydrogen atom has the following properties: (i) The atom should be electronegative in nature. (ii) The atom should possess at least one lone pair of unshared electron. (iii) The atom should be considerably small in size. • These conditions are fulfilled by only 3 atoms namely Nitrogen, Oxygen and Fluorine. Water is a polar molecule and hydrogen bonds are formed between the water molecules. Explain. LS 5.4.1 Polar Molecule and Hydrogen Bonds (a) When hydrogen atom covalently bonded with the electronegative oxygen atom, the pair electrons are more closely with the oxygen atom than with the hydrogen atom. (b) This leads to the formation of partially positive charge (δ+) on hydrogen atom, H and partially negative charge (δ–) on oxygen atom, O. (c) The partially positive charged hydrogen atom is then attracted by the other partially negative charged oxygen atom from other molecule is known as hydrogen bond. H δ+ Hydrogen bond H δ+ O δ– O δ– H δ+ H δ– Covalent bond Hydrogen bond δ+ δ+ δ+ δ+ δ– δ– O O O O H H H H H H H H O H H Hydrogen bond in hydrogen fluoride. LS 5.4.1 Fδ– Fδ– Fδ– Fδ– Hδ+ Hδ+ Hδ+ Hδ+ Covalent bond Hydrogen bond 5.4 HYDROGEN BOND CS 5.4 PERFORMANCE LEVEL (PL) Mastered Not mastered PL3 Apply knowledge and skills about chemical bonds and its concept to explain the natural occurrences or phenomena and be able to carry out simple tasks. 05 U5 Chemistry F4(p77-97)csy2p.indd 84 21/12/2022 3:17 PM

MODULE • Chemistry FORM 4 85 UNIT 5 © Nilam Publication Sdn. Bhd. Hydrogen bond in ammonia. Hδ+ Hδ+ Hδ+ Nδ– Nδ– Nδ– H H H H H H Covalent bond Hydrogen bond How do hydrogen bonds affect boiling points? LS 5.4.2 • The molecule held by the hydrogen bond has a higher boiling point than the molecule held by the Van der Waals force. Example 1: Substances Relative molecular mass Boiling point / ºC Ethanol (C2H5OH) H – C – C – O – H H H H H 46 +78 Propane (C3H8) H – C – C – C – H H H H H H H 44 –42 • Ethanol and propane have almost the same relative molecular mass and size. • The boiling point of ethanol is higher than propane. • Ethanol molecule has a hydrogen atom bonded to an oxygen atom as in a water molecule. • There are hydrogen bonds between ethanol molecules, more energy is needed to break the bonds before it boils. • Propane is a non-polar molecule. • Propane molecules are attracted by weak van der Waals force . • There are no hydrogen bonds between propane molecules. Example 2: Substances Relative molecular mass Boiling point / ºC Water (H2O) 18 100 Hydrogen sulfide (H2S) 34 –60 • The relative molecular mass of hydrogen sulfide is higher than water, but the boiling point of water is higher than hydrogen sulfide. • There are hydrogen bonds between water (H2O) molecules, more energy is needed to break the bonds before it boils. • There is no hydrogen bond between hydrogen sulfide (H2S) molecules due to the low electronegativity of sulphur atom. Quiz 05 U5 Chemistry F4(p77-97)csy2p.indd 85 21/12/2022 3:17 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 86 UNIT 5 Water is a polar solvent. Explain. Water: Polar Solvent • Water acts as a polar solvent because it can be attracted to either the positive or negative charge on a solute: (a) The partially positive hydrogen atom on water molecule attracts other partially negatively-charged regions of other molecules or negatively charged ionic compound. (b) The partially negative oxygen atom on water molecule attracts other partially positively-charged regions of other molecules or positively charged ionic compound. Cl Cl – – Na+ Na+ Water molecule Water molecule Remark: The explanation is related to water as a solvent of ionic compounds which will be studied in the physical properties of ionic compounds. How do hydrogen bonds affect solubility in water? • Molecules that can form hydrogen bond with water have a higher solubility in water. Examples: (a) • Ethanol (C2H5OH) is a polar molecule dissolves in water. • Ethanol molecules form hydrogen bonds with water molecules. H H – C – C – O – H H H H O O Hydrogen bond H | | H H | | H (b) • Other examples of polar molecules which form hydrogen bonds with water molecules are ammonia (NH3), hydrogen chloride (HCl), sugar (C6H12O6) and methanol (CH3OH). • The polarity of these molecules indicates that they will dissolve in water. 05 U5 Chemistry F4(p77-97)csy2p.indd 86 21/12/2022 3:17 PM

MODULE • Chemistry FORM 4 87 UNIT 5 © Nilam Publication Sdn. Bhd. Flipping the paper is easier when the fingertips are wet with water compared to dry fingertips. Explain. PL4 HOTS • The cellulose molecules that form the paper are polar molecules. • When fingertips are wet, there are hydrogen bonds between water molecules and cellulose molecules in paper, thus it is easier to flip paper. • When fingertips are dry, no hydrogen bond formed between water molecules and cellulose molecules in paper, thus it is difficult to flip the paper. Wet hair is sticky compared to dry hair. Explain. PL4 HOTS • The keratin molecules that form the outside layer of hair are the polar molecules. • When hair is wet, there are hydrogen bond between the water molecules and keratin molecules in the outer layer of hair. As a result, the hair becomes sticky. • When hair is dry, there is no hydrogen bonding formed between the water molecules and keratin molecules in the outer layer of hair. As a result, the hair is not sticky. 1 Cellulose and keratin are example of compounds that may form hydrogen bonding with other molecules. The diagram shows the structural formula of cellulose and keratin. Cellulose Keratin H | O H | O O | H O | H O O O H O H O H O O H O O HO H3C CH3 Cellulose is an organic compound mainly used to produce paper . Keratin is a strong natural protein and it is the main substance to form hair. PL6 Exercise PERFORMANCE LEVEL (PL) Mastered Not mastered PL4 Analyse knowledge on chemical bonds in the context of problem solving the natural occurrences or phenomena. 05 U5 Chemistry F4(p77-97)csy2p.indd 87 21/12/2022 3:17 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 88 UNIT 5 2 The diagram shows arrangement of water molecules in liquid water and ice. Hydrogen bond Ice (Solid) Hydrogen bonds are stable Water (Liquid) Hydrogen bonds constantly break and re-form Explain the arrangement of water molecules in water (liquid). • In water (liquid), water molecules are closely held together by hydrogen bonds and move randomly. Explain the arrangement of water molecules in ice (solid). • When water freezes, hydrogen bonds are stable, arranging the water molecules far apart from each other. • Hence, the volume of ice becomes greater than that of the water. What is the effect of increase in the volume of ice? • The effect of the increase in the volume of ice is that its density becomes lower than the density of water, thus ice becomes lighter than water. Why does ice float on water? • Ice floats because ice is less dense than water. PL5 What is dative bond? • Dative bond or coordinate bond is a type of covalent bond between two atoms in which two electrons are from one atom only. Explain the formation of dative bond in ammonium ion, NH4 + . (a) The reaction between ammonia and hydrogen chloride will produce ammonium chloride. NH3 + HCl NH4Cl (b) Ammonium ion, NH4 + , is formed by transferring a hydrogen ion (H+ ) from the hydrogen chloride to the lone pair of electrons on the ammonia molecule. Chloride ion, Cl– Ammonium ion, NH4 + Hydrogen chloride Ammonia Lone pair electrons Dative bond N Cl N Cl H H H H H H H H + + H+ NH3 + HCl → NH4 + + Cl– (c) Ammonium ion is positively charged because only the hydrogen ion, H+ is transferred from the chlorine atom to the nitrogen atom. (d) In ammonium ion, nitrogen and chlorine atoms have achieved stable octet electron arrangement while hydrogen atoms have achieved stable duplet electron arrangement. LS 5.5.1 5.5 DATIVE BOND CS 5.5 05 U5 Chemistry F4(p77-97)csy2p.indd 88 21/12/2022 3:17 PM

MODULE • Chemistry FORM 4 89 UNIT 5 © Nilam Publication Sdn. Bhd. Explain the formation of dative bond in hydroxonium ion, H3O+ . (a) Dissolving hydrogen chloride in water to make hydrochloric acid H2O + HCl H3O+ + Cl– (b) Hydroxonium ion, H3O+ is formed by transferring of a hydrogen ion (H+ ) from the hydrogen chloride to the lone pair of electrons on the water molecule. Chloride ion, Cl– Hydroxonium ion, H3O+ Hydrogen chloride Water Lone pair electrons Dative bond H O H Cl H O H Cl H H + + H+ Or H2O + HCl → H3O+ + Cl– (c) Hydroxonium ion, is positively charged because only the hydrogen ion, H+ is transferred from the hydrogen chloride to the lone pairs on the oxygen atom. (d) In hydroxonium ion, oxygen atom has achieved stable octet electron arrangement while hydrogen atoms have achieved stable duplet electron arrangement. LS 5.5.1 What is valence electron? • Valence electrons are the electrons in the outermost shell of an atom. What are delocalised electrons in metallic atom? • In metal, atoms are packed closely together in regular arrangement. • Metal atoms tend to lose their valence electrons and become positive ions. • The valence electrons from the metallic atom are free to move throughout the metal structure. • These mobile electrons are called delocalised electrons. What are sea of electrons? • Sea of electrons are the delocalised electrons that are free to move in the space between metal atoms. What is metallic bond? • It is the strong electrostatic force between the sea of electrons and the positive metal ions. Explain how are metallic bonds formed. LS 5.6.1 • When the valence electrons released by the metallic atom, the atom become positive metal ion. • Metallic bonds are formed from the strong electrostatic attraction between negatively charged sea of electrons and fixed, positively charged metal ions. Positive metal ion Sea of electron Valence electron • When an atom loses electrons, it become positive ion. • In metallic bonding, only the outer electrons are mobile. The positive metallic ions are immobile . 5.6 METALLIC BOND CS 5.6 05 U5 Chemistry F4(p77-97)csy2p.indd 89 21/12/2022 3:17 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 90 UNIT 5 State the physical properties of metal. Explain. Physical properties Explanations High melting and boiling points • A large amount of heat needed to overcome the strong electrostatic force between sea of electrons with the positive ions in metallic bonds. Good conductor of electricity • The delocalised electrons are able to move freely to conduct electricity. • The delocalised electrons can flow and carry the charge from the negative terminal to the positive terminal when the electric current is applied. LS 5.6.2 Comparing The Formation of Ionic, Covalent and Metallic Bond Ionic Bond Covalent Bond Metallic Bond State the type of element involved. • Between metals (Group 1, 2 and 13) and non-metals (Group 15, 16 and 17) • Between non-metals and non-metals (Group 14, 15, 16 and 17) • Between metal atoms How bonds are formed? • Electrons are released by metal atoms to form positive ions • Transfer of electron is to achieve stable octet electron arrangement • Electrons are received by non-metal atoms to form negative ions • Strong electrostatic force between positive and negative ions • Pairs of electrons are shared by same or different non -metal atoms. • Sharing of electron is to achieve stable octet electron arrangement. • Two different structures of covalent substances: (i) Simple molecules structure. (ii) Weak van der Waals forces between simple molecules (iii) Many atoms bonded to form a giant covalent structures • Metal atoms lose their valence electrons to form a sea of delocalized electrons • Strong electrostatic force between sea of electrons and metal ions • Many metal ions bonded together to form giant lattice structure Examples of electron arrangement in the particles + 2– + A E A Strong electrostatic forces between ions # Ionic bond is the strong electrostatic force of attraction between positively charged ion and negatively charged ion. Strong covalent bond between atoms in the molecules # Covalent bond is the shared pairs of electrons between atoms in a molecule. Strong electrostatic forces between sea of electrons and metal ions 05 U5 Chemistry F4(p77-97)csy2p.indd 90 21/12/2022 3:17 PM

MODULE • Chemistry FORM 4 91 UNIT 5 © Nilam Publication Sdn. Bhd. To compare electrical conductivity of ionic compound and covalent compound Manipulated variable: Lead(II) bromide and naphthalene // Ionic and covalent compounds Responding variable: Electrical conductivity/deflection of ammeter pointer Fixed variable: Carbon electrodes Hypothesis: Lead(II) bromide cannot conduct electricity in solid state but can conduct electricity in molten state. Naphthalene cannot conduct electricity in solid and molten states Materials: Lead(II) bromide, naphthalene Apparatus: Batteries, carbon electrodes, ammeter, Bunsen burner, connecting wires, crucible, tripod stand A Carbon electrode Lead(II) bromide Heat Procedures: 1 A crucible is filled with lead(II) bromide powder until it is half full. 2 Two carbon electrodes are dipped into lead(II) bromide and carbon electrodes are connected to batteries and ammeter using connecting wire. 3 The deflection on ammeter pointer is observed and recorded. 4 The lead(II) bromide powder is heated strongly until it melts. 5 The deflection on ammeter pointer is observed and recorded. 6 Steps 1 to 5 are repeated using naphthalene to replace lead(II) bromide. Observations: Compounds Deflection of ammeter pointer Solid Molten Lead(II) bromide ✗ ✓ Naphthalene ✗ ✗ Conclusion: Lead(II) bromide cannot conduct electricity in solid state but can conduct electricity in molten state. Naphthalene cannot conduct electricity in solid and molten states. To compare solubility of ionic compound and covalent compound in water and organic solvent Manipulated variable: Magnesium chloride and naphthalene // Ionic and covalent compounds Responding variable: Solubility of ionic and covalent compound in water and organic solvent Fixed variable: Water and cyclohexane // water and organic solvent Hypothesis: Magnesium chloride is soluble in water but insoluble in organic solvent. Naphthalene is insoluble in water but soluble in organic solvent Materials: Magnesium chloride, naphthalene, distilled water, cyclohexane Apparatus: Beaker, test tube, spatula Distilled water Magnesium chloride Naphthalene Procedures: 1 Half spatula of magnesium chloride and naphthalene powder are placed in two different test tubes. 2 About 5 cm3 distilled water is added into each test tube. 3 The test tubes are shaken. 4 All observations are recorded. 5 Steps 1 to 4 are repeated by replacing water with cyclohexane. Observations: Compounds Solubility Distilled water Cyclohexane Magnesium chloride Soluble Insoluble Naphthalene Insoluble Soluble Conclusion: Magnesium chloride is soluble in water but insoluble in organic solvent. Naphthalene is insoluble in water but soluble in organic solvent. To compare melting point of ionic compound and covalent compound Manipulated variable: Magnesium chloride and naphthalene// Ionic and covalent compounds Responding variable: Melting point Fixed variable: Amount of magnesium chloride and naphthalene Hypothesis: Magnesium chloride has a higher melting point than naphthalene Materials: Magnesium chloride, naphthalene, water Apparatus: Beaker, boiling tube, Bunsen burner, tripod stand Procedures: 1 Half spatula of magnesium chloride and naphthalene powder are placed in two different boiling tubes. 2 Both boiling tubes are heated in water until the water boils. 3 The changes in physical state are recorded. Observations: Compounds Observations Magnesium chloride No change. Naphthalene Melts easily. Liquid naphthalene evaporate Conclusion: Magnesium chloride has a higher melting point than naphthalene. Naphthalene Water Heat Magnesium chloride 5.7 IONIC AND COVALENT COMPOUND CS 5.7 Experiment to Study the Difference Between Ionic Compound and Covalent Compound LS 5.7.1 PERFORMANCE LEVEL (PL) Mastered Not mastered PL5 Evaluate knowledge on chemical bonds in the context of problem solving and decision-making to carry out a task. 05 U5 Chemistry F4(p77-97)csy2p.indd 91 21/12/2022 3:17 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 92 UNIT 5 Ionic compounds Covalent compounds Giant covalent compound Example of electron arrangement/ structure Sodium chloride, NaCl + + + + + _ _ _ _ + _ _ + + + _ _ + _ Strong electrostatic forces between positive and negative ions Carbon dioxide molecule, CO2 Strong covalent bond between atoms in the molecules Weak van der Waals forces between molecules Diamond Covalent bond between carbon atoms Carbon atoms Silicon dioxide, SiO2 State the type of forces between particles. Strong electrostatic force between ions. Weak van der Waals forces (intermolecular force) between molecule. Strong covalent bonds between atoms in the giant structure Compare and explain melting and boiling points. • High melting and boiling points because positive ions and negative ions are attracted by strong electrostatic force . • Large amount of energy is needed to overcome it. • Low melting and boiling points because of the weak “van der Waals” force between molecules. • Small amount of energy is needed to overcome it. • Very high melting point because large amount of energy is needed to break the strong covalent bonds between carbon atoms. Comparing Physical Properties of Ionic and Covalent Compounds 05 U5 Chemistry F4(p77-97)csy2p.indd 92 21/12/2022 3:17 PM

MODULE • Chemistry FORM 4 93 UNIT 5 © Nilam Publication Sdn. Bhd. Ionic compounds Covalent compounds Giant covalent compound Compare and explain electrical conductivity • Cannot conduct electricity when in solid form because the ions are not free to move . • Able to conduct electricity in molten or aqueous because the ions are free to move . • Cannot conduct electricity in all state. • Covalent compound is made up of neutral molecules . • No free moving ions in molten or aqueous state. • Diamond and silicone dioxide cannot conduct electricity at any state as there are no delocalized electrons. • Graphite can conduct electricity as there are delocalized electrons between hexagonal layers. Compare solubility in water and organic solvent • Most are soluble in water and insoluble in organic solvent* because the polarisation of water molecule. • Insoluble in water but soluble in organic solvents* (example: ether, alcohol, benzene, tetrachloromethane and propanone). * Organic solvents are covalent compounds that exist as liquid at room temperature • Insoluble in water and organic solvent. • Attractions between solvent molecules and carbon atoms is not strong enough to overcome the strong covalent bonds in the giant covalent structure. Example of ionic and covalent compounds Lead(II) bromide, PbBr2, Sodium chloride, NaCl Copper(II) sulphate, CuSO4 Naphthalene, C H8 10 Acetamide, CH CONH 3 2 Hexane, C H6 14 Diamond (carbon only) Graphite (carbon only) Silicone dioxide (silicone and oxygen) Uses in daily life • Potassium chloride as fertiliser • Common salt, Sodium chloride (NaCl) • Lime, Calcium oxide (CaO) • Magnesium hydroxide, Mg(OH)2 in antacid to reduce stomach acid • Baking powder, Sodium hydrogen carbonate (NaHCO3) • Turpentine is a solvent for paint • Carbon dioxide in fire extiguishers • Ethanol C H2 OH in perfumes 5 • Butane, C H4 10 in LPG gas for cooking Uses of graphite • Pencil • Electrode in batteries Uses of diamond • Jewellery • Cutting glass or drilling rocks 05 U5 Chemistry F4(p77-97)csy2p.indd 93 21/12/2022 3:17 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 94 UNIT 5 1 The table shows the melting points, boiling points and electrical conductivity of five substances, A to E. Substances Melting point / ºC Boiling point / ºC Electrical conductivity in solid state Electrical conductivity when dissolve in water A 850 2 100 Conducts Does not conduct B 3 550 4 830 Does not conduct Does not conduct C 0 100 Does not conduct Does not conduct D 789 1 447 Does not conduct Conducts E –98 –61 Does not conduct Does not conduct (a) Which substance are solids at room temperature? A, B, E (b) Which substance is an ionic compound? D (c) Which substance could be diamond? B (d) Which substance has delocalized electron? A 2 Table below shows information about two different substances. Substances P Q Structural formula C | C | C C | C | C C | C | C C C C | C | C C | C | C C | C | C C C H | C H H H Types of molecule Giant molecule Small molecule (a) Compare melting and boiling point of substances P and Q. Explain your answer. • Substance P has very high melting and boiling points. • Large amount of heat is needed to break the very strong covalents bond between carbon atom. • Substance Q has low melting and boiling points due to weak van der Waals force between molecules. • Small amount of energy is needed to overcome it. (b) Predict electrical conductivity of Substances P and Q. Explain your answer • Substances P and Q do not conduct electricity because both compounds are neutral molecules. • There are no free moving ions in both compounds. • Substance P also has no delocalized electrons because all the valence electrons are used for covalent bonds. 3 The diagram on the right shows the carbon dioxide fire extinguisher canister. Carbon dioxide is stored at high pressure in the liquid state in the fire extinguisher. (a) Carbon reacts with oxygen to produce carbon dioxide. [Given that proton number for carbon is 6 and proton number for oxygen is 8] (i) State the type of bond present in this compound and write the formula of the compound formed. Covalent bond and CO2 Subjective Questions PL2 Fire extinguisher canister contains liquid carbon dioxide ENRICHMENT EXERCISE Quiz Chemical Bond PL3 PL4 PL5 05 U5 Chemistry F4(p77-97)csy2p.indd 94 21/12/2022 3:17 PM

MODULE • Chemistry FORM 4 95 UNIT 5 © Nilam Publication Sdn. Bhd. (ii) Explain how a compound is formed between element carbon and oxygen based on their electron arrangement. • Carbon atom with electron arrangement 2.4 needs four electrons to achieve stable octet electron arrangement. • Oxygen atom with electron arrangement 2.6 needs two electrons to achieve stable octet electron arrangement. • One carbon atom share four pairs of electrons with two oxygen atoms to form a molecule with the formula CO2. • One carbon atom contributes four electrons and each of the two oxygen atoms contributes two electrons for sharing to form double covalent bond. • One carbon atom forms two double covalent bond with two oxygen atoms. • Carbon atom and oxygen atom achieve stable octet electron arrangement that is 2.8. (b) Draw the electron arrangement of carbon dioxide. O C O (c) Explain why carbon dioxide fire extinguishers are the only fire extinguisher recommended for fires involving electrical equipment. • Carbon dioxide is safe to be used on and around electrical equipment. • Carbon dioxide cannot conduct electricity. 4 The diagram below shows the electron arrangement of compound A. Compound A is formed from the reaction between element X and element Y. + X Y – (a) (i) Write the electron arrangement for atom of elements X and Y. X: 2.8.1 Y: 2.8.7 (ii) Compare the size of atoms of elements X and Y. Explain your answer. • Atom Y is smaller than atom X. • Atom X and atom Y have the same number of shells occupied with electrons. • The number of proton in the nucleus of atom Y is more than X. • The strength of nuclei attraction to the electrons in the shells of atom Y is stronger than atom X. (b) How are X ion and Y ion formed from their respective atoms? X ion: Atom X releases one electron Y ion: Atom Y receives one electron (c) (i) Write the formula for compound A and name type of bond in compound A. XY and ionic bond (ii) Write the chemical equation for the reaction between element X and element Y to form compound A. 2X + Y2 2XY PL2 PL3 PL2 PL2 PL3 PL4 PL3 HOTS PL3 05 U5 Chemistry F4(p77-97)csy2p.indd 95 21/12/2022 3:17 PM

MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 96 UNIT 5 (d) Y can react with carbon to form a compound. Draw the electron arrangement for the compound formed. [Given that proton number for carbon is 6] 5 The table below shows the melting point and electrical conductivity of substances V, W, X and Y. Substances Melting point (°C) Electrical conductivity Solid Molten V –7 Cannot conduct electricity Cannot conduct electricity W 80 Cannot conduct electricity Cannot conduct electricity X 808 Cannot conduct electricity Conduct electricity Y 1 080 Conduct electricity Conduct electricity (a) (i) Which of the substance is copper? Give reason for your answer. Y. It can conduct electricity in solid and molten state. (ii) Explain how copper conducts electricity in solid state. Copper atoms release electrons to form free moving delocalized electrons. When electric is applied, delocalized electrons flow and carry the charge from the negative terminal to the positive terminal (b) State the type of particles in substances V and W. Explain why substances V and W cannot conduct electricity in solid and molten state. Molecule. Substances V and W are made up of neutral molecules. No free moving ions in solid and molten state. (c) The boiling point of substance V is 59 °C. What is the physical state of substance V at room temperature? Liquid (d) Draw the arrangement of particle V at room temperature. (e) Explain why the melting and boiling points of substances V and W are low? Van der Waals forces between molecules are weak. Small amount of heat energy is required to overcome it. (f) State the type of particle in substance X. Explain why substance X cannot conduct electricity in solid but can conduct electricity in molten state. Ion. Ions are not freely moving // ions are in a fixed position in solid state. Ion can move freely in molten state. PL3 PL5 PL5 PL2 PL2 PL2 PL2 PL3 05 U5 Chemistry F4(p77-97)csy2p.indd 96 21/12/2022 3:17 PM

MODULE • Chemistry FORM 4 97 UNIT 5 © Nilam Publication Sdn. Bhd. 1 Which substance is an ionic compound? A Methane, CH4 B Carbon dioxide, CO2 C Propanol, C3H7OH D Copper(II) oxide, CuO 2 The diagram below shows the electron arrangement of a compound formed between atoms X and Y. Y Y Y X Y Which of the following statements is true about the compound? A It is an ionic compound. B The compound has high melting point. C The compound conducts electricity. D The compound is formed by sharing of electrons. 3 Which of the following is a property of zinc chloride? A Volatile B Has a low melting point C Insoluble in water D Conducts electricity in the molten state 4 The diagram below shows symbol of an element T. 24 12T What is the electron arrangement of ion formed by an atom of T? A 2.8 C 2.8.8 B 2.8.2 D 2.8.8.8 5 The table below shows the electron arrangements of atoms P, Q, R and S. Atom P Q R S Electron arrangement 2.4 2.8.1 2.8.2 2.8.7 Which pair of atoms forms a compound by transferring of electrons? A P and S C Q and S B P and R D Q and R 6 The table below shows the proton number of four elements P, Q, R and S. Element P Q R S Proton number 6 8 17 20 Which of the following pairs will form a compound with high melting and boiling points? A P and Q B Q and S C P and R D Q and R 7 The table below shows the proton number of elements X and Y. Element X Y Proton number 6 8 What type of bond and the chemical formula of the compound formed between atoms X and Y? Type of bond Chemical formula A Ion YX2 B Ion XY2 C Covalent XY2 D Covalent YX2 8 The diagram below shows the electron arrangement of ion X+ . X Which of the following is the position of element X in the Periodic Table? Group Period A 1 3 B 18 3 C 1 4 D 18 4 Objective Questions SPM PRACTICE PL2 PL2 PL3 PL2 PL3 PL3 PL4 PL3 05 U5 Chemistry F4(p77-97)csy2p.indd 97 21/12/2022 3:17 PM

© Nilam Publication Sdn. Bhd. 98 UNIT 6 MODULE • Chemistry FORM 4 Concept Map Meaning • Acid is a chemical substance which ionises in water to produce hydrogen ion. • Bases is a chemical substance that reacts with acid to produce salt and water only. • Alkali is a base that is soluble in water and ionises to hydroxide ion. Strength of acid and alkali • Strong acid and strong alkali • Weak acid and weak alkali Basicity of acid • Monoprotic acid • Diprotic acid ACID AND BASE Uses in daily life Preparation standard solution Concentration of hydrogen ion and hydroxide ion pH of solutions used in daily life Concentration of acid and alkali in mol dm–3 and g dm–3 Example pH value of acid and alkali • Titration acid-alkali • Application in daily life Chemical Properties Acid 1 Acid + Metal ➝ Salt + Hydrogen 2 Acid + Metal carbonate ➝ Salt + Water + Carbon dioxide 3 Acid + base ➝ Salt + Water (Neutralisation) Base 1 Base + Ammonium salt ➝ Salt + Water + Ammonia gas 2 Base + Acid ➝ Salt + Water (Neutralisation) 3 Base + Metal ion➝ Metal hydroxide Neutralisation Can be classified according to To compare and explain the chemical properties of acid/ alkali in water, without water or in other solvent Explain the role of water in the formation of hydrogen ions and hydroxide ions Based on ionisation of: • Strong and weak acid/ alkali • Monoprotic and diprotic acids pH = –log [H+ ] pH + pOH = 14 REMARK: • The order of content standard (CS) for Acid and Bases: (6.1) Role of water → (6.3) Strength of acid and alkali → (6.4) Chemical properties of acid and alkali → (6.5) Concentration of aqueous solution → (6.6) Standard solution → (6.2) pH value → (6.7) Neutralisation • CS 6.2 is placed after CS 6.6 to facilitate understanding of the concept because the pH value is related to the strength and concentration of acid/alkali that have been learned in CS 6.3, CS 6.5 and CS 6.6. 6 UNIT ACIDS, BASES AND SALTS 06 U6a Chemistry F4(p98-131)csy2p.indd 98 21/12/2022 3:50 PM