S4 Chemistry Paper: Bonding, Structure And Properties (100 Marks)

S4 Chemistry Paper: Bonding, Structure and Properties (100 marks)

2026-11-11

S4 Chemistry – bonding, structure and properties

S4 Chemistry paper: 6 questions, 22 parts, 100 marks, 75 minutes. Covers ionic and covalent bonding, giant versus simple molecular structures, and metallic bonding.

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Paper details　S4 Chemistry practice paper on ionic and covalent bonding, giant and simple molecular structures, metallic bonding and formula mass calculations: 6 questions, 22 parts, 100 marks, 75 minutes, with full marking notes.

Part A Question 1: Ionic Bonding and Electron Transfer

Only main-group elements are involved. Write electron arrangements as layer counts separated by commas.

#	Question	Answer and marking notes
1	(a) Sodium has atomic number 11 and chlorine has atomic number 17. Write the electron arrangements of the sodium and chlorine atoms, and state how electrons are transferred when they form ions.	Answer: Sodium is 2,8,1 and chlorine is 2,8,7. Sodium loses one electron to form Na+ with arrangement 2,8; chlorine gains one electron to form Cl- with arrangement 2,8,8. Marking: two arrangements 1 each, direction and number 1, two ion arrangements 0.5 each.
2	(b) Magnesium has atomic number 12 and oxygen has atomic number 8. Write the electron arrangements of Mg2+ and O2-, and explain why the empirical formula of magnesium oxide is MgO.	Answer: Magnesium is 2,8,2 and loses two electrons to form Mg2+ with arrangement 2,8; oxygen is 2,6 and gains two electrons to form O2- with arrangement 2,8. Each magnesium ion carries two units of positive charge and each oxide ion two units of negative charge, so they pair one to one and the charges cancel, giving the empirical formula MgO. Marking: two arrangements 1 each, charge balance 1.5, formula 0.5.
3	(c) The aluminium ion is Al3+ and the oxide ion is O2-. Find the ratio of the two ions in the empirical formula of aluminium oxide, write the formula, and check the total charge.	Answer: The lowest common multiple of the two charges is 6, so 2 aluminium ions pair with 3 oxide ions, ratio 2:3, formula Al2O3. Check: total positive charge 2 times (+3) = +6 and total negative charge 3 times (-2) = -6, which sum to zero. Marking: LCM 1, ratio 1, formula 1, charge check 1.
4	(d) Magnesium oxide melts at about 2852 degrees Celsius while sodium chloride melts at about 801 degrees Celsius. Both are ionic compounds. Explain the difference in melting point.	Answer: The ions in magnesium oxide carry two units of charge while those in sodium chloride carry only one; higher ionic charge gives stronger ionic bonding and a larger lattice enthalpy, so the melting point is higher. The smaller ionic radii in magnesium oxide also shorten the ion separation and strengthen the attraction. Marking: charge comparison 1, bond strength and lattice enthalpy 1, radius factor 1.

Part B Question 2: Covalent Bonding and Electron Counting

Count valence electrons by group number: hydrogen 1, carbon 4, nitrogen 5, oxygen 6.

#	Question	Answer and marking notes
5	(a) In a water molecule oxygen forms one single bond with each hydrogen. Find the total number of valence electrons, the number of bonding electrons, and the number of lone-pair electrons.	Answer: Total valence electrons = 1 x 2 + 6 = 8; bonding electrons = 2 single bonds x 2 = 4; lone-pair electrons = 8 – 4 = 4, that is two lone pairs on oxygen. Marking: valence total 1, bonding 1, lone pairs and location 1.
6	(b) In a carbon dioxide molecule carbon forms one double bond with each oxygen. Find the total number of valence electrons, the bonding electrons, and the number of lone pairs.	Answer: Total valence electrons = 4 + 6 x 2 = 16; bonding electrons = 2 double bonds x 4 = 8; lone-pair electrons = 16 – 8 = 8, that is 4 pairs, two on each oxygen. Marking: valence total 1, double-bond counting 1.5, lone pairs 1.5.
7	(c) In a nitrogen molecule the two nitrogen atoms form a triple bond. Find the total number of valence electrons, the bonding electrons, and the lone-pair electrons.	Answer: Total valence electrons = 5 x 2 = 10; bonding electrons = 3 shared pairs x 2 = 6; lone-pair electrons = 10 – 6 = 4, one pair on each nitrogen. Marking: valence total 1, triple bond 1, lone pairs 1.
8	(d) Both carbon dioxide and water contain polar bonds. Explain why a carbon dioxide molecule has no net dipole while a water molecule has one.	Answer: Carbon dioxide is linear, so its two carbon-oxygen bond dipoles are equal and opposite and their vector sum is zero, giving no net dipole. Water is bent with a bond angle of about 104.5 degrees, so the two bond dipoles cannot cancel and their vector sum gives a net dipole with the oxygen end partially negative. Marking: linear shape and cancellation 1.5, bent shape 1, net dipole direction 1.5.

Part C Question 3: Giant Structures and Properties

When comparing properties, state the type of structure and the kind of attraction between the particles.

#	Question	Answer and marking notes
9	(a) Diamond and graphite are both made of carbon atoms, yet graphite conducts electricity and diamond does not, and graphite is softer. Explain both differences.	Answer: In diamond each carbon atom is covalently bonded to four others in a three-dimensional network, all valence electrons are used in bonding and none are free to move, so diamond does not conduct and the network is very hard. In graphite each carbon atom bonds to only three others, forming flat hexagonal layers, and the fourth valence electron is delocalised between the layers and can move, so graphite conducts; the layers are held only by weak intermolecular attraction and slide easily, which makes graphite soft. Marking: four bonds and no free electrons 1.5, delocalised electrons and conduction 1.5, layer sliding and hardness 1.
10	(b) Silicon dioxide melts at about 1710 degrees Celsius, while carbon dioxide sublimes at about -78 degrees Celsius. Their formulae look similar. Explain the large difference in melting point.	Answer: Silicon dioxide has a giant covalent structure in which every silicon atom is bonded to four oxygen atoms and every oxygen to two silicon atoms, so the whole crystal is one continuous covalent network; melting requires breaking many strong covalent bonds, hence the very high melting point. Carbon dioxide has a simple molecular structure: the molecules contain polar covalent bonds, but the molecules are held together only by weak intermolecular attraction, which is all that must be overcome on melting or subliming. Marking: silica structure and bond breaking 2, simple molecular structure 1, overcoming intermolecular attraction 1.
11	(c) In diamond each carbon atom is bonded to 4 carbon atoms; in graphite each carbon atom is bonded to 3 carbon atoms and every bond is shared by two atoms. Find the number of bonds per atom in graphite, and the ratio of bonds per atom in diamond to that in graphite.	Answer: Bonds per atom in graphite = 3 / 2 = 1.5; in diamond each atom has 4 bonds (bond to atom ratio 2:1, giving 4 bonds per atom). The ratio is 4 / 1.5 = 2.667. Marking: reason for dividing by two 1.5, graphite value 1, diamond value 0.5, ratio 1.

Part D Question 4: Metallic Bonding, Density and Alloys

In calculations show the formula, the substitution and the unit.

#	Question	Answer and marking notes
12	(a) Explain why a solid metal conducts electricity while a solid ionic compound does not.	Answer: In a metal the valence electrons are delocalised into a sea of free electrons that can move throughout the lattice, so they drift when a voltage is applied and the metal conducts. In a solid ionic compound the ions are held in fixed lattice positions by electrostatic attraction; although the ions are charged, there are no mobile charged particles, so it does not conduct. Marking: free electrons in metal 1.5, ions fixed in lattice 1.5.
13	(b) Explain why metals are malleable, that is they can be hammered into thin sheets.	Answer: Metallic bonding is non-directional, so the attraction between the metal ions and the free electrons is the same in all directions. When a force is applied, one layer of ions can slide over another into a new position and is still surrounded by free electrons, so the bonding is not broken and the metal deforms instead of shattering. Marking: non-directional bonding 1.5, layers still bonded after sliding 1.5.
14	(c) A cubic metal sample of side 2 cm has a mass of 54 g. Find its density, decide whether it is aluminium (density 2.7 g/cm3), and find the ratio of the two densities.	Answer: Volume = 2 cubed = 8 cm3; density = mass / volume = 54 / 8 = 6.75 g/cm3. The ratio is 6.75 / 2.7 = 2.5, so the sample is 2.5 times as dense as aluminium and is not aluminium. Marking: volume 1, density formula and value 1.5, unit 0.5, decision and ratio 1.
15	(d) Brass is an alloy of copper and zinc and is harder than pure copper. Explain this in terms of metallic structure.	Answer: In pure copper all atoms are the same size and form regular layers that slide over one another easily, so copper is soft. In brass the zinc atoms differ in size from the copper atoms and replace some of them, which disturbs the regular layers and prevents them from sliding past each other, so the alloy is harder. Marking: layer sliding in pure copper 1.5, different sized atoms blocking sliding 2, conclusion 0.5.

Part E Question 5: Formula Mass and Mole Calculations

Relative atomic masses: H 1, C 12, O 16, Al 27, S 32, Ca 40. Take the Avogadro constant as 6.02 x 10^23.

#	Question	Answer and marking notes
16	(a) Find the formula mass of calcium carbonate, CaCO3.	Answer: Formula mass = 40 + 12 + 16 x 3 = 40 + 12 + 48 = 100. Marking: correct atom counts 2, multiplication 1, total 1.
17	(b) Find the formula mass of aluminium sulphate, Al2(SO4)3, and state the total number of oxygen atoms in one formula unit.	Answer: One sulphate group has mass 32 + 16 x 4 = 96; three groups give 96 x 3 = 288; two aluminium atoms give 27 x 2 = 54; formula mass = 54 + 288 = 342. Number of oxygen atoms = 4 x 3 = 12. Marking: sulphate mass 2, times three 1, aluminium part 1, total and oxygen count 1.
18	(c) Take 20 g of calcium carbonate. Find the number of moles and the number of formula units, expressed in units of 10^23.	Answer: Moles = mass / formula mass = 20 / 100 = 0.2 mol; number of formula units = 0.2 x 6.02 x 10^23 = 1.204 x 10^23. Marking: mole formula 2, mole value 1, multiply by constant 1, value with unit 1.
19	(d) Take 0.25 mol of aluminium sulphate. Find its mass and the percentage by mass of aluminium, correct to one decimal place.	Answer: Mass = 0.25 x 342 = 85.5 g; mass of aluminium = 2 x 27 = 54, so the percentage = 54 / 342 x 100 = 15.8 per cent. Marking: mass calculation 2.5, aluminium mass 1.5, percentage expression 1, answer 1.

Part F Question 6: Integrated Application of Structure and Properties

Support each answer with the data given and name the type of structure you identify.

#	Question	Answer and marking notes
20	(a) Substance X is a white, odourless solid. Its aqueous solution conducts electricity, it melts at about 801 degrees Celsius, and solid X does not conduct. Identify the type of structure of X and explain each observation.	Answer: X is an ionic compound with a giant ionic lattice. The high melting point arises because melting must overcome many strong electrostatic attractions; the solid does not conduct because the ions are held in fixed lattice positions; in water the ions leave the lattice and move freely, so the solution conducts. Being white and odourless fits a typical salt, since ionic compounds do not evaporate as molecules. Marking: identification 2, melting point 2, conduction contrast 3, appearance 1.
21	(b) Substance A melts at about 1710 degrees Celsius, is insoluble in water, extremely hard, and does not conduct in either the solid or the liquid state. Substance B turns directly into a gas at about -78 degrees Celsius and does not conduct. Identify the structure type of A and of B, and explain the large difference in melting and boiling behaviour.	Answer: A has a giant covalent structure, for example silicon dioxide: every atom is joined to others by strong covalent bonds in a three-dimensional network, with no free ions or electrons, so it does not conduct, and melting requires breaking many covalent bonds, which explains the very high melting point and hardness. B has a simple molecular structure, for example carbon dioxide: the bonds within each molecule are covalent, but the molecules are held together only by weak intermolecular attraction, so only that weak attraction must be overcome and B sublimes at a low temperature. Marking: A structure and explanation 3, B structure and explanation 3, comparison of bond breaking versus intermolecular attraction 3.
22	(c) Two unlabelled bottles contain white solids: one is an ionic compound and the other a simple molecular compound. Using only a melting point test and a conductivity test, describe how to tell them apart, and explain the difference in conductivity between a solid ionic compound and its melt.	Answer: First test the melting point: an ionic compound normally melts at a high temperature of several hundred to over a thousand degrees Celsius, while a simple molecular compound melts at a low temperature and often melts or sublimes during gentle heating. Then test conductivity: both solids do not conduct; after heating each sample to melting, the molten ionic compound conducts while the molten simple molecular compound still does not. The reason is that in the solid the ions are fixed in the lattice by electrostatic attraction, and on melting the lattice breaks down so the ions move freely; a melted simple molecular compound consists of neutral molecules with no mobile charged particles. Marking: melting point test 2, conductivity test on the melt 3, explanation of solid versus molten conduction 3.

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