References only, not question text. Open Official paper ↗ where Cambridge publishes the paper,
or see the official Cambridge past-papers page; for other papers, ask your
school. “Paper 42” means Paper 4, variant 2.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Ionisation energies · Core · Paper 1 style · 1 mark
(a)Which equation represents the first ionisation energy of chlorine?[1]
Show answer
(a) A Cl(g) → Cl⁺(g) + e⁻
First ionisation energy is for one mole of gaseous atoms each losing one electron to form one mole of gaseous 1+ ions. B starts from the molecule (it includes atomisation); C is electron affinity; D removes an electron from an anion.
Q2Ionisation energies · Core · Paper 2 style · 5 marks
(a)Define the term first ionisation energy.[2]
(b)Write an equation, with state symbols, for the second ionisation energy of magnesium.[1]
(c)Explain why the second ionisation energy of magnesium is greater than its first ionisation energy.[2]
Show answer
(a) The energy needed to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous 1+ ions.
M1 energy to remove one electron from each atom in 1 mole of atomsM2 atoms and ions are gaseous (1+ ions formed)
(b) Mg⁺(g) → Mg²⁺(g) + e⁻
B1 correct species and state symbols
(c) The electron is removed from a positive ion; there is the same nuclear charge but fewer electrons, so less repulsion and the remaining electrons are held more tightly / the outer electron is closer to the nucleus.
M1 electron removed from a positive ion / greater attraction (effective nuclear charge per electron greater)M2 same number of protons but fewer electrons so less shielding/repulsion; radius of ion smaller
Q3Ionisation energies · Core · Paper 2 style · 5 marks
(a)State three factors that affect the size of the first ionisation energy of an atom.[3]
(b)State whether sodium or potassium has the lower first ionisation energy and explain your answer.[2]
Show answer
(a) Nuclear charge; distance of the outer electron from the nucleus (atomic radius); shielding by inner electrons. (Spin-pair repulsion also accepted.)
B1 each for any three of: nuclear charge; atomic radius/distance; shielding; spin-pair repulsion / sub-shell
(b) Potassium; its outer electron is in the 4th shell, further from the nucleus with more shielding, so it is less strongly attracted.
B1 potassiumB1 outer electron further from nucleus / more shielding (outweighs larger nuclear charge)
Q4Ionisation energies · Standard · Paper 2 style · 6 marks
The first ionisation energies of the Period 3 elements, in kJ/mol, are:Na 496, Mg 738, Al 578, Si 787, P 1012, S 1000, Cl 1251, Ar 1521.
(a)Explain why there is a general increase in first ionisation energy from sodium to argon.[2]
(b)Explain why the first ionisation energy of aluminium is lower than that of magnesium.[2]
(c)Explain why the first ionisation energy of sulfur is lower than that of phosphorus.[2]
Show answer
(a) Nuclear charge increases across the period while shielding stays roughly the same (electrons added to the same shell), so the outer electron is held more strongly / atomic radius decreases.
M1 nuclear charge (number of protons) increasesM2 similar shielding / same shell, so greater attraction for outer electrons
(b) In Al the electron is removed from a 3p orbital, which is higher in energy (further from the nucleus) and is shielded by the 3s electrons, so less energy is needed.
M1 Al outer electron in 3p (Mg in 3s)M2 3p is higher in energy / further from nucleus / extra shielding by 3s²
(c) In S the electron is removed from a 3p orbital containing a pair of electrons; repulsion between the paired electrons makes it easier to remove. P has three unpaired 3p electrons.
M1 S has a paired electron in one 3p orbital (3p⁴); P is 3p³ with no pairsM2 spin-pair repulsion makes the paired electron easier to remove
Q5Ionisation energies · Standard · Paper 2 style · 5 marks
The first six successive ionisation energies of element Y, in kJ/mol, are:578, 1817, 2745, 11 577, 14 842, 18 379.
(a)Deduce the group of the Periodic Table that element Y belongs to. Explain your answer.[2]
(b)Y is in Period 3. Identify Y and write the electronic configuration of the ion Y³⁺.[2]
(c)Explain why each successive ionisation energy of Y is larger than the one before.[1]
Show answer
(a) Group 13; the large jump is between the 3rd and 4th ionisation energies, so Y has 3 outer-shell electrons.
M1 big jump between IE3 (2745) and IE4 (11 577)M2 three electrons in outer shell, so Group 13
(b) Aluminium; Al³⁺ is 1s² 2s² 2p⁶
B1 aluminiumB1 1s² 2s² 2p⁶
(c) Each electron is removed from an increasingly positive ion, so there is less repulsion and greater attraction to the nucleus.
B1 removing electron from more positive ion / same nuclear charge holding fewer electrons
Q6Ionisation energies · Standard · Paper 1 style · 1 mark
(a)The first six successive ionisation energies of an element, in kJ/mol, are:787, 1577, 3232, 4356, 16 091, 19 805In which group of the Periodic Table is the element?[1]
Show answer
(a) C Group 14
The large jump comes between the 4th (4356) and 5th (16 091) ionisation energies, so the element has 4 outer electrons: Group 14 (silicon). Group 15 would show the jump after the 5th value.
Q7Ionisation energies · Standard · Paper 2 style · 5 marks
(a)Write the electronic configurations of nitrogen and oxygen atoms using orbital notation (e.g. 1s² …).[1]
(b)The first ionisation energy of nitrogen is 1402 kJ/mol and that of oxygen is 1314 kJ/mol. Explain why oxygen's value is lower although it has more protons.[2]
(c)Calculate the energy, in kJ, needed to convert 0.500 g of gaseous nitrogen atoms into N⁺ ions. Give your answer to 3 significant figures. [Ar: N, 14.0][2]
Show answer
(a) N 1s² 2s² 2p³; O 1s² 2s² 2p⁴
B1 both correct
(b) Oxygen's electron is removed from a 2p orbital that holds two electrons; the repulsion between the paired electrons makes it easier to remove. In nitrogen each 2p orbital holds one electron.
M1 O has a pair of electrons in one 2p orbitalM2 spin-pair repulsion outweighs the extra nuclear charge
(a)Calculate the total energy needed to convert one mole of gaseous atoms into one mole of gaseous 2+ ions for (i) magnesium and (ii) calcium.[2]
(b)Explain the difference between your two answers.[3]
(c)Suggest how this difference relates to the reactions of magnesium and calcium with cold water.[1]
Show answer
(a) (i) 2189 kJ/mol (ii) 1735 kJ/mol
B1 738 + 1451 = 2189B1 590 + 1145 = 1735
(b) Calcium's outer electrons are in the 4th shell, further from the nucleus and shielded by more inner shells. Although Ca has more protons, the extra distance and shielding outweigh this, so attraction is weaker and less energy is needed.
M1 Ca outer electrons in a higher shell / larger atomic radiusM2 more shielding by inner electronsM3 increase in nuclear charge is outweighed, so weaker attraction
(c) Calcium reacts much more readily with cold water than magnesium because it loses its two outer electrons more easily.
B1 Ca more reactive (reacts readily with cold water; Mg very slowly) because ions form more easily
Q9Ionisation energies · Challenge · Paper 2 style · 6 marks
A student plots log₁₀(ionisation energy) against the number of the electron removed for all 19 electrons of a potassium atom.
(a)Between which successive ionisation energies would you expect the three large jumps to occur? Explain.[3]
(b)Explain why, between the jumps, the ionisation energies rise steadily rather than staying constant.[2]
(c)State the sub-shell from which the 2nd electron is removed.[1]
Show answer
(a) Between the 1st and 2nd, between the 9th and 10th, and between the 17th and 18th, because potassium is 2,8,8,1: each jump marks the start of removing electrons from a shell closer to the nucleus.
B1 1st→2ndB1 9th→10th and 17th→18thB1 each jump = removing from a new inner shell, closer to nucleus / less shielded
(b) Each electron removed leaves an ion with more positive charge relative to the remaining electrons, so less repulsion and the ion shrinks; each next electron is attracted more strongly.
M1 increasing positive charge on ion / fewer electrons for same nuclear chargeM2 less repulsion / smaller radius so stronger attraction
(c) 3p
B1 3p
Q10Ionisation energies · Challenge · Paper 1 style · 1 mark
(a)Which Period 2 element has a first ionisation energy greater than that of both of the elements next to it in the period?[1]
Show answer
(a) C nitrogen
N (1402 kJ/mol) is above C (1086) and O (1314): N has a half-filled 2p sub-shell with no paired electrons, while O has one paired 2p electron that is easier to remove. Boron is below beryllium; carbon is below nitrogen.
Q11Particles, isotopes & ions · Core · Paper 1 style · 1 mark
(a)Which species has more electrons than protons and more neutrons than protons?[1]
Show answer
(a) C ³⁷Cl⁻
³⁷Cl⁻: 17 protons, 20 neutrons, 18 electrons. ¹⁶O²⁻ has more electrons but equal protons and neutrons (8 and 8); ³⁹K⁺ and ²⁷Al³⁺ have fewer electrons than protons.
Q12Particles, isotopes & ions · Core · Paper 2 style · 5 marks
(a)Give the numbers of protons, neutrons and electrons in each of: ⁵⁶Fe³⁺, ³²S²⁻ and ⁶⁵Cu⁺.[3]
(b)State the relative mass and relative charge of a neutron and of an electron.[2]
Show answer
(a)⁵⁶Fe³⁺: 26 p, 30 n, 23 e³²S²⁻: 16 p, 16 n, 18 e⁶⁵Cu⁺: 29 p, 36 n, 28 e
B1 for each correct species (all three numbers right)
(b) Neutron: mass 1, charge 0. Electron: mass 1/1836 (about 0.0005), charge −1.
B1 neutron 1 and 0B1 electron 1/1836 and −1
Q13Particles, isotopes & ions · Standard · Paper 2 style · 6 marks
A beam of particles is passed through an electric field between two parallel charged plates in a vacuum.
(a)Describe what happens to beams of protons, neutrons and electrons as they pass between the plates.[3]
(b)Explain why electrons are deflected through a much larger angle than protons.[1]
(c)A beam of ¹H⁺ ions is deflected through 12.0°. Assuming the angle is proportional to charge/mass, calculate the angle of deflection for a beam of ⁴He²⁺ ions under the same conditions.[2]
Show answer
(a) Protons are deflected towards the negative plate; electrons are deflected towards the positive plate, through a much larger angle; neutrons pass straight through undeflected.
B1 protons towards negative plateB1 electrons towards positive plate, larger angleB1 neutrons not deflected
(b) Same size of charge but an electron's mass is about 1/1836 of a proton's, so its charge/mass ratio is much bigger.
B1 deflection depends on charge/mass; electron much lighter
(c) 6.0°
M1 charge/mass for ⁴He²⁺ = 2/4 = 0.5 of that of ¹H⁺A1 12.0 × 0.5 = 6.0°
Q14Particles, isotopes & ions · Standard · Paper 2 style · 4 marks
A sample of boron used in borosilicate glass at a factory in Lahore contains 19.9% ¹⁰B (isotopic mass 10.013) and 80.1% ¹¹B (isotopic mass 11.009) by number of atoms.
(a)Define the term isotopes.[1]
(b)Explain why ¹⁰B and ¹¹B have the same chemical properties.[1]
(c)Calculate the relative atomic mass of boron in this sample. Give your answer to 4 significant figures.[2]
Show answer
(a) Atoms of the same element with the same number of protons but different numbers of neutrons.
B1
(b) They have the same number of electrons / the same electronic configuration (chemistry depends on electrons).
B1
(c) 10.81
M1 (10.013 × 19.9 + 11.009 × 80.1)/100A1 10.81
Q15Particles, isotopes & ions · Standard · Paper 2 style · 4 marks
(a)An ion X²⁻ has 18 electrons and a nucleon number of 34. Identify X and give the number of neutrons in the ion.[2]
(b)An ion Z³⁺ has the same electronic configuration as a neon atom and contains 14 neutrons. Write the full symbol of the ion, showing nucleon number and proton number.[2]
Show answer
(a) Sulfur (³⁴S²⁻); 18 neutrons
B1 sulfur (16 protons)B1 34 − 16 = 18 neutrons
(b) ²⁷₁₃Al³⁺
M1 13 protons (10 + 3) so aluminiumA1 nucleon number 27
Q16Particles, isotopes & ions · Standard · Paper 2 style · 3 marks
(a)Arrange the beams ¹H⁺, ³He²⁺, ⁴He²⁺ and ⁷Li⁺ in order of increasing angle of deflection in an electric field. Show your reasoning.[2]
(b)State the direction in which all four beams are deflected and why.[1]
Show answer
(a) ⁷Li⁺ < ⁴He²⁺ < ³He²⁺ < ¹H⁺
M1 charge/mass: 0.143, 0.5, 0.667, 1A1 correct order
(b) Towards the negative plate, because they are all positively charged.
B1
Q17Particles, isotopes & ions · Challenge · Paper 2 style · 5 marks
Gallium, used in LEDs, has two stable isotopes, ⁶⁹Ga and ⁷¹Ga. Its relative atomic mass is 69.72.
(a)Calculate the percentage abundance of ⁶⁹Ga. Assume the isotopic masses are 69.0 and 71.0.[2]
(b)Give the numbers of protons, neutrons and electrons in the ion ⁷¹Ga³⁺.[1]
(c)Gallium's first ionisation energy is lower than that of zinc, the element before it. Suggest why, in terms of electronic configuration.[2]
Show answer
(a) 64.0%
M1 69x + 71(100 − x) = 6972A1 x = 64.0
(b) 31 protons, 40 neutrons, 28 electrons
B1 all three
(c) Ga's outer electron is a single 4p electron, higher in energy and shielded by the full 4s² (and 3d¹⁰); Zn loses an electron from a full 4s sub-shell.
M1 Ga electron removed from 4p (Zn from 4s)M2 4p higher energy / further / shielded by 4s
Q18Particles, isotopes & ions · Challenge · Paper 2 style · 5 marks
Natural chlorine contains 75% ³⁵Cl and 25% ³⁷Cl by number of atoms. A sample of chlorine gas, Cl₂, is analysed in a mass spectrometer.
(a)Give the m/z values of the three peaks caused by Cl₂⁺ ions.[1]
(b)Calculate the ratio of the heights of these three peaks. Show your working.[3]
(c)State the total number of neutrons in one ³⁷Cl–³⁷Cl molecule.[1]
Q20Electronic configuration · Standard · Paper 1 style · 1 mark
(a)What is the electronic configuration of a Cr³⁺ ion?[1]
Show answer
(a) A [Ar] 3d³
Cr is [Ar] 3d⁵ 4s¹. On forming ions the 4s electron is lost first, then two 3d electrons: [Ar] 3d³. Option B wrongly removes 3d electrons before 4s.
Q21Electronic configuration · Standard · Paper 2 style · 5 marks
(a)Write the electronic configurations of chromium and copper atoms in the form [Ar] ….[2]
(b)Write the electronic configuration of the Cu²⁺ ion and state the number of unpaired electrons it has.[2]
(c)Explain why, when a transition element forms a positive ion, the 4s electrons are removed before the 3d electrons.[1]
Show answer
(a) Cr [Ar] 3d⁵ 4s¹; Cu [Ar] 3d¹⁰ 4s¹
B1 each
(b) [Ar] 3d⁹; one unpaired electron
B1 [Ar] 3d⁹B1 one
(c) Once the 3d orbitals are occupied, the 4s electrons are higher in energy (further out) than the 3d electrons, so they are lost first.
B1 4s is higher in energy / outermost once 3d filled
Q22Electronic configuration · Challenge · Paper 2 style · 5 marks
(a)An ion Z³⁺ has the configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵. Identify element Z.[1]
(b)An ion W²⁺ has the configuration [Ar] 3d¹⁰. Identify W and explain why W²⁺ is colourless in solution.[2]
(c)State the block of the Periodic Table to which selenium belongs and write the configuration of the Se²⁻ ion.[2]
Show answer
(a) Iron (Fe)
B1 23 electrons + 3 = 26 protons
(b) Zinc; the 3d sub-shell is full so no d–d electron transitions are possible.
B1 zincB1 full 3d / no d–d transitions
(c) p-block; [Ar] 3d¹⁰ 4s² 4p⁶ (same as krypton)
B1 p-blockB1 configuration
Q23Electronic configuration · Challenge · Paper 1 style · 1 mark
(a)Which pair of ions has the same number of unpaired electrons?[1]
Show answer
(a) B Mn²⁺ and Fe³⁺
Mn²⁺ and Fe³⁺ are both [Ar] 3d⁵ with five unpaired electrons. Fe²⁺ 3d⁶ has 4 but Co²⁺ 3d⁷ has 3; Cu⁺ 3d¹⁰ has 0, Cu²⁺ 3d⁹ has 1; Ni²⁺ 3d⁸ has 2, Ti³⁺ 3d¹ has 1.
Q24Atomic & ionic radius · Core · Paper 2 style · 4 marks
(a)Explain why atomic radius decreases from sodium to chlorine across Period 3.[2]
(b)Explain why atomic radius increases down Group 1 from lithium to caesium.[2]
Show answer
(a) Nuclear charge increases while electrons are added to the same shell with similar shielding, so outer electrons are pulled closer.
M1 increasing nuclear charge / more protonsM2 same shell / similar shielding, so stronger attraction
(b) Each element has an extra occupied shell, so outer electrons are further from the nucleus and more shielded.
M1 more shellsM2 more shielding (outweighs increased nuclear charge)
Q25Atomic & ionic radius · Standard · Paper 2 style · 4 marks
(a)State what is meant by an isoelectronic series and give the electronic configuration shared by all six ions.[2]
(b)Explain the trend in ionic radius shown in the table.[2]
Show answer
(a) Species with the same number of electrons (same configuration); 1s² 2s² 2p⁶
B1 same number of electronsB1 1s² 2s² 2p⁶
(b) Proton number increases from N (7) to Al (13) while the number of electrons stays at 10, so the electrons are attracted more strongly and pulled closer.
M1 same number of electrons / same shieldingM2 increasing nuclear charge pulls electrons closer
Q26Atomic & ionic radius · Standard · Paper 2 style · 3 marks
(a)Place the species S²⁻, Cl⁻, Ar and K⁺ in order of decreasing radius.[1]
(b)Explain why S²⁻ is the largest of these species.[2]
Show answer
(a) S²⁻ > Cl⁻ > Ar > K⁺
B1
(b) All have 18 electrons, but S²⁻ has the fewest protons (16), so the attraction of its nucleus for the electrons is weakest.
M1 same number of electrons (18)M2 fewest protons so weakest attraction
Q27Atomic & ionic radius · Challenge · Paper 2 style · 5 marks
Radii (nm): Na atom 0.186, Na⁺ 0.102, Cl atom 0.099, Cl⁻ 0.181
(a)Calculate the percentage change in radius when (i) a sodium atom forms Na⁺ and (ii) a chlorine atom forms Cl⁻. Give each to 3 significant figures.[2]
(b)Explain why Na⁺ is much smaller than a Na atom.[2]
(c)Explain why Cl⁻ is much larger than a Cl atom.[1]
(b) Na⁺ has lost its only 3s electron, so it has one fewer shell; the remaining 10 electrons are attracted by 11 protons.
M1 loses the outer (third) shellM2 more protons than electrons so electrons pulled in more strongly / less repulsion
(c) The extra electron increases repulsion between electrons in the same shell, while the nuclear charge is unchanged, so the electron cloud expands.
B1 extra electron, more repulsion, same nuclear charge
Q28Orbitals & sub-shells · Core · Paper 2 style · 4 marks
(a)Describe the shapes of an s orbital and a p orbital.[2]
(b)State the number of orbitals in a 3d sub-shell.[1]
(c)State the maximum number of electrons that the third shell (n = 3) can hold.[1]
Show answer
(a) s: spherical. p: dumbbell (two lobes on opposite sides of the nucleus).
B1 each
(b) 5
B1
(c) 18
B1 (3s 2 + 3p 6 + 3d 10)
Q29Orbitals & sub-shells · Standard · Paper 2 style · 5 marks
(a)Explain what is meant by Hund's rule and use it to describe how the three 2p electrons are arranged in a nitrogen atom.[2]
(b)State the number of unpaired electrons in atoms of phosphorus, sulfur and chlorine.[2]
(c)Explain why the 4s sub-shell fills before the 3d sub-shell in potassium and calcium.[1]
Show answer
(a) Electrons occupy orbitals of equal energy singly, with parallel spins, before pairing; nitrogen has one electron in each of 2px, 2py and 2pz.
M1 orbitals of same energy filled singly before pairingM2 one electron in each 2p orbital, parallel spins
(b) P 3, S 2, Cl 1
B2 all three; B1 two correct
(c) In these atoms the 4s sub-shell is lower in energy than the 3d sub-shell.
B1
Q30Orbitals & sub-shells · Challenge · Paper 1 style · 1 mark
(a)How many orbitals are fully occupied (contain two electrons) in a ground-state sulfur atom?[1]
Show answer
(a) B 7
S is 1s² 2s² 2p⁶ 3s² 3p⁴: 1s, 2s, three 2p and 3s are full (6) and the 3p⁴ has one full orbital and two half-filled, giving 7. Answer A forgets the paired 3p orbital.
Similar questions
Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.
References only: find each question in its paper (official link where available, otherwise through your school).
