S4 Chemistry Practice Paper: Chemical Bonding And Structure (12 Questions, 100 Marks, 75 Minutes, With Answers)

S4 Chemistry Practice Paper: Chemical Bonding and Structure (12 Questions, 100 Marks, 75 Minutes, with Answers)

2026-11-09

HKDSE Chemistry: Chemical Bonding and Structure

12 questions, 100 marks, 75 minutes on ionic, covalent and metallic bonding, mole calculations, empirical and molecular formulae, bond angles and conductivity, with answers.

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Instructions: This paper has three sections, A, B and C. Answer all questions. Total marks: 100. Time allowed: 75 minutes.

Show all working. Answers without working earn no method marks. Write down a formula before substituting numbers.

Give numerical answers to three significant figures unless stated otherwise, and state the correct units.

Spend about 35 minutes on Section A, then use the remaining time for the mole calculations in Section B and the structure and design questions in Section C.

Section A: Types of Chemical Bond

Formation, examples and electron transfer in ionic, covalent and metallic bonds

#	Question	Answer and marking notes
1	Describe the formation of ionic, covalent and metallic bonds, giving one compound or substance as an example of each.	(a) An ionic bond forms when a metal atom loses electrons to become a cation and a non-metal atom gains electrons to become an anion, and the ions are held by electrostatic attraction (2 marks). (b) Example: sodium chloride (1 mark). (c) A covalent bond forms when two atoms share one or more pairs of electrons (2 marks). (d) Example: water or methane (1 mark). (e) A metallic bond is the electrostatic attraction between metal cations and the delocalised electrons (2 marks).
2	Magnesium burns in oxygen to form magnesium oxide (Mg = 24, O = 16): (a) write the balanced equation; (b) find the relative formula mass of magnesium oxide; (c) find the mass of magnesium oxide formed when 4.8 g of magnesium burns completely; (d) find the number of moles of oxygen needed.	(a) 2Mg + O2 gives 2MgO (2 marks). (b) Relative formula mass = 24 + 16 = 40 (1 mark). (c) Moles of magnesium = 4.8 divided by 24 = 0.2 mol (1 mark); by the equation the moles of magnesium oxide are also 0.2 mol (1 mark); mass = 0.2 x 40 = 8.0 g (2 marks). (d) Moles of oxygen = 0.2 divided by 2 = 0.1 mol (2 marks).
3	Ammonium nitrate (NH4NO3) is a common nitrogen fertiliser (N = 14, H = 1, O = 16): (a) find its relative formula mass; (b) find the percentage by mass of nitrogen; (c) find the mass of nitrogen in 80 g of the fertiliser.	(a) Relative formula mass = 14 x 2 + 1 x 4 + 16 x 3 = 80 (2 marks). (b) Total mass of nitrogen = 28 (1 mark); percentage = 28 divided by 80 x 100% = 35% (3 marks). (c) 80 g x 35% = 28 g (2 marks).
4	An organic compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass, and has a relative molecular mass of 60 (C = 12, H = 1, O = 16): (a) find the mole ratio of the elements; (b) find the empirical formula; (c) find the molecular formula.	(a) C : H : O = 40.0 divided by 12, 6.7 divided by 1, 53.3 divided by 16, which gives 3.33 : 6.7 : 3.33 (2 marks). (b) Simplifying to whole numbers gives 1 : 2 : 1 (2 marks). (c) The empirical formula is CH2O (1 mark); its mass is 30 (1 mark); 60 divided by 30 = 2 (1 mark), so the molecular formula is C2H4O2 (1 mark).

Section B: Calculations and Structure

Relative formula mass, moles and limiting reagents, empirical and molecular formulae, shapes and bond angles

#	Question	Answer and marking notes
5	Compare the structures of diamond, graphite and copper, and explain why graphite conducts electricity while diamond does not.	(a) In diamond each carbon atom is joined to four other carbon atoms by covalent bonds, forming a three-dimensional network (2 marks). (b) In graphite each carbon atom is joined to three others, forming hexagonal layers, with only weak forces between the layers (2 marks). (c) Copper consists of regularly arranged metal cations with outer electrons delocalised into a sea of electrons (2 marks). (d) Each carbon atom in graphite has one delocalised electron that is free to move within the layer, so graphite conducts (2 marks). (e) In diamond all valence electrons are used in covalent bonds, so there are no free electrons and it does not conduct (1 mark).
6	State the shape and bond angle of the methane, ammonia and water molecules, and explain why the bond angles in ammonia and water are smaller than in methane.	(a) Methane is tetrahedral with a bond angle of 109.5 degrees (2 marks). (b) Ammonia is trigonal pyramidal with a bond angle of 107 degrees (2 marks). (c) Water is angular (V shaped) with a bond angle of 104.5 degrees (2 marks). (d) The nitrogen atom in ammonia has one lone pair and the oxygen atom in water has two lone pairs; lone pair repulsion is greater than bonding pair repulsion, so the bond angles are reduced (2 marks).
7	Sodium chloride melts at 801 degrees but does not conduct electricity when solid; when molten or dissolved in water it does conduct. Explain these three observations.	(a) Sodium chloride has a giant ionic structure in which each ion is surrounded by several oppositely charged ions, giving strong electrostatic attraction (2 marks). (b) Melting requires many ionic bonds to be broken, which needs much energy, so the melting point is high (2 marks). (c) In the solid the ions are held in fixed positions in the lattice and cannot move, so it does not conduct (2 marks). (d) When molten or dissolved, the ions are free to move and act as charge carriers, so it conducts (2 marks).
8	Use electronegativity differences to explain why hydrogen chloride is a polar covalent compound while sodium chloride is an ionic compound.	(a) Chlorine is more electronegative than hydrogen, but the difference is only about 1.0, so the atoms share an electron pair that is pulled towards chlorine, giving a polar covalent bond (3 marks). (b) The electronegativity difference between chlorine and sodium is about 2.1, which is large, so an electron is effectively transferred from sodium to chlorine and ions are formed (3 marks). (c) Compounds with an electronegativity difference greater than about 1.7 tend to be ionic (2 marks).

Section C: Structure and Properties, Experiments

Giant structures and conductivity, flame tests and precipitation tests, experimental design

#	Question	Answer and marking notes
9	Describe how flame tests are used to identify sodium, potassium and calcium ions, and how silver nitrate solution is used to test for chloride ions, including the observations.	(a) Dip a platinum or nichrome wire into the sample and hold it in a colourless flame (1 mark). (b) Sodium gives a golden yellow flame (1 mark). (c) Potassium gives a lilac flame, seen through blue glass to screen out the yellow of sodium (1 mark). (d) Calcium gives a brick red flame (1 mark). (e) Acidify the sample solution with dilute nitric acid, then add silver nitrate solution (2 marks). (f) A white precipitate shows that chloride ions are present (2 marks), and the precipitate is silver chloride (1 mark).
10	Hydrogen reacts with oxygen to form water (H2 = 2, O2 = 32): 6.0 g of hydrogen is mixed with 24 g of oxygen and ignited. (a) Find the number of moles of each gas; (b) identify the limiting reagent; (c) find the mass of water formed.	(a) Moles of hydrogen = 6.0 divided by 2 = 3.0 mol (1 mark); moles of oxygen = 24 divided by 32 = 0.75 mol (1 mark). (b) From 2H2 + O2 gives 2H2O, 3.0 mol of hydrogen needs 1.5 mol of oxygen, but only 0.75 mol is present, so oxygen is the limiting reagent (2 marks). (c) Water formed = 0.75 x 2 = 1.5 mol (2 marks); mass = 1.5 x 18 = 27 g (2 marks).
11	Explain why graphite is used as a lubricant and as an electrode, while diamond is used in cutting tools.	(a) The forces between the layers of graphite are weak, so the layers slide easily and graphite acts as a lubricant (2 marks). (b) Graphite has delocalised electrons that are free to move, so it can be used as an electrode (2 marks). (c) Diamond is a three-dimensional covalent network in which the covalent bonds are very strong and fixed in direction (2 marks). (d) Diamond is extremely hard, so it is used for cutting and drilling tools (2 marks).
12	Design an experiment to compare the electrical conductivity of sodium chloride solution and sucrose solution, stating four essential points of the design.	(a) Prepare the two solutions at the same concentration, for example 0.1 mol dm-3 (2 marks). (b) Use the same pair of electrodes and keep the electrode separation and depth of immersion the same (2 marks). (c) Connect the same power supply and an ammeter (or lamp) in series and measure the current at the same voltage (3 marks). (d) Use distilled water as a control, and rinse and dry the electrodes after each test (2 marks).

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