Form 4 Chemistry Practice Paper: Atomic Structure, Periodic Table And Bonding

Form 4 Chemistry Practice Paper: Atomic Structure, Periodic Table and Bonding

2026-11-05

Form 4 Chemistry – 12 questions – 100 marks – 75 minutes

Form 4 Chemistry paper: 12 questions, 100 marks, 75 minutes covering atomic structure, isotopes, ionic and covalent bonding, electron configuration and periodic table trends.

Download　Question paper PDF　|　Answer key PDF

Paper details　Form 4 Chemistry practice paper with 12 structured questions on atoms, isotopes, bonding and the periodic table; includes full worked answers and two printable PDFs.

Section A: Atomic Structure and Isotopes (Q1-Q3)

Answer all questions; show the expression, substitution and unit for every calculation.

#	Question	Answer and marking notes
1	An atom X has atomic number (Z) 17 and mass number (A) 35. (a) State the number of protons, neutrons and electrons. (b) Write the electron configuration of the atom.	(a) Protons = 17, neutrons = 35 – 17 = 18, electrons = 17. (b) Electron configuration: 2, 8, 7. Marking notes: protons (2 marks), neutrons (2 marks), electrons (2 marks), electron configuration (2 marks).
2	Chlorine has two isotopes, 35Cl and 37Cl, with abundances of 75% and 25%. Calculate the relative atomic mass of chlorine.	Relative atomic mass = 0.75 x 35 + 0.25 x 37 = 26.25 + 9.25 = 35.5. Marking notes: correct expression (3 marks), correct substitution of abundances (3 marks), answer with sensible significant figures (2 marks).
3	Magnesium (Z = 12) reacts with oxygen (Z = 8) to form magnesium oxide. (a) Write the electron configuration of the magnesium ion and the oxide ion. (b) State which noble gas has the same number of electrons as these ions.	(a) A magnesium atom loses 2 electrons to form the magnesium ion, configuration 2, 8 (10 electrons). An oxygen atom gains 2 electrons to form the oxide ion, configuration 2, 8 (10 electrons). (b) Both ions have 10 electrons, the same as neon (Ne, Z = 10). Marking notes: magnesium ion configuration (3 marks), oxide ion configuration (3 marks), naming neon (3 marks).
4	Explain how ionic bonding is formed in sodium chloride (NaCl), and explain why sodium chloride has a high melting point.	A sodium atom (2, 8, 1) transfers its single outermost electron to a chlorine atom (2, 8, 7), forming Na+ and Cl- ions. The strong electrostatic attraction between oppositely charged ions is the ionic bond, and it holds the ions in a giant ionic lattice. Melting requires breaking many strong electrostatic attractions, so a large amount of energy is needed and the melting point is high. Marking notes: electron transfer (3 marks), ions and ionic bond (2 marks), lattice and melting point (3 marks).

Section B: Bonding and Formulae (Q4-Q8)

Relative atomic masses: H = 1, C = 12, N = 14, O = 16, S = 32, Ca = 40; Avogadro constant = 6.02 x 10^23 per mol.

#	Question	Answer and marking notes
5	State the number of shared electron pairs in a water molecule (H2O), an ammonia molecule (NH3) and a methane molecule (CH4), and give the total.	Water has 2 shared pairs (two O-H single bonds); ammonia has 3 shared pairs; methane has 4 shared pairs. Total = 2 + 3 + 4 = 9 pairs. Marking notes: water (2 marks), ammonia (2 marks), methane (2 marks), total (2 marks).
6	An element Y has atomic number 16. (a) Write the electron configuration of a Y atom. (b) State the group and period of Y in the periodic table. (c) State the number of outermost shell electrons in a Y atom.	(a) Electron configuration: 2, 8, 6. (b) There are 6 outermost electrons, so Y is in Group VI (Group 16); it has 3 occupied shells, so it is in Period 3. (c) The number of outermost shell electrons is 6. Marking notes: electron configuration (3 marks), group and period (3 marks), outermost electrons (3 marks).
7	Write the chemical formulae of aluminium oxide, calcium chloride and sodium sulphate.	Aluminium oxide: the lowest common multiple of the charges of Al3+ and O2- is 6, so the formula is Al2O3. Calcium chloride: Ca2+ combines with Cl- in the ratio 1:2, giving CaCl2. Sodium sulphate: Na+ combines with SO4 2- in the ratio 2:1, giving Na2SO4. Marking notes: aluminium oxide (3 marks), calcium chloride (2 marks), sodium sulphate (3 marks).
8	Calculate the relative molecular masses of H2SO4 and Ca(OH)2. (H = 1, O = 16, S = 32, Ca = 40)	H2SO4: 2 x 1 + 32 + 4 x 16 = 2 + 32 + 64 = 98. Ca(OH)2: 40 + 2 x (16 + 1) = 40 + 34 = 74. Marking notes: H2SO4 working (3 marks), H2SO4 answer (1 mark), Ca(OH)2 working (3 marks), Ca(OH)2 answer (1 mark), no unit stated (1 mark).

Section C: Mole Calculations and Structure (Q9-Q12)

Answer explanation questions in full sentences; give numerical answers to sensible significant figures.

#	Question	Answer and marking notes
9	Find the mass of 0.25 mol of carbon dioxide (CO2) and the number of molecules it contains. (C = 12, O = 16, Avogadro constant = 6.02 x 10^23 per mol)	Molar mass = 12 + 2 x 16 = 44 g/mol. Mass = 0.25 x 44 = 11.0 g. Number of molecules = 0.25 x 6.02 x 10^23 = 1.505e+23, that is 1.505 x 10^23 molecules. Marking notes: molar mass (3 marks), mass with unit (3 marks), number of molecules (2 marks).
10	Calculate the percentage by mass of nitrogen in ammonium nitrate (NH4NO3). (N = 14, H = 1, O = 16)	Molar mass = 2 x 14 + 4 x 1 + 3 x 16 = 28 + 4 + 48 = 80 g/mol. Percentage by mass of nitrogen = 28 / 80 x 100% = 35.0%. Marking notes: molar mass (5 marks), percentage expression and answer (4 marks).
11	An ion X2+ contains 18 electrons. (a) Find the atomic number of element X. (b) Write the electron configuration of a X atom.	(a) A X atom has 18 + 2 = 20 electrons, so the atomic number is 20 (calcium). (b) Electron configuration: 2, 8, 8, 2. Marking notes: deducing the number of electrons (4 marks), electron configuration (4 marks).
12	Explain why graphite conducts electricity but diamond does not.	Graphite and diamond are both giant covalent structures of carbon, but their bonding differs. In graphite each carbon atom bonds to only 3 other carbon atoms, leaving one valence electron free to move between the layers (a delocalised electron), so graphite conducts electricity. In diamond each carbon atom bonds to 4 carbon atoms, so all 4 valence electrons are fixed in covalent bonds and there are no free electrons; diamond therefore does not conduct. Marking notes: delocalised electrons in graphite (4 marks), all electrons bonded in diamond (4 marks).

Find a tutor — free Tutor Registration →

A TutorZone practice paper; original questions and passages, not publisher material.