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HKDSE Chemistry Recap: Relative Atomic Mass, Electronic Arrangement, Periods and Groups | Vanilla (2026)

HKDSE Chemistry — Relative Atomic Mass, Electronic Arrangement, Periods and Groups

Subject: Chemistry

Vanilla (2026)

Issue date: Friday, 9 October 2026

Total: 70 marks  |  Suggested time: 90–100 minutes  |  Atomic numbers: H 1, He 2, Li 3, Be 4, B 5, C 6, N 7, O 8, F 9, Ne 10, Na 11, Mg 12, Al 13, Si 14, P 15, S 16, Cl 17, Ar 18, K 19, Ca 20. Relative isotopic masses are taken to be equal to the mass numbers unless stated otherwise. Give numerical answers to 3 significant figures unless stated otherwise. In Questions 4 to 6, the letters used do not represent the actual symbols of the elements.

01

Section A — Atomic Mass, Electronic Arrangement, Periods and Groups

True or False

For each statement, state whether it is True or False. If it is false, rewrite the statement correctly. (1 mark each: the mark for a false statement is awarded only if the rewritten statement is also correct.)

  1. The relative isotopic mass of carbon-12 is exactly 12, so the relative atomic mass of carbon is also exactly 12.
  2. The relative atomic mass of chlorine is 35.5. This means that every chlorine atom has a mass equal to 35.5 times 112 of the mass of a carbon-12 atom.
  3. An element X has two isotopes, with relative isotopic masses of 63 and 65. The relative atomic mass of X is 63.6. The isotope with a relative isotopic mass of 65 is the more abundant isotope.
  4. Natural boron consists of 10B and 11B only, and its relative atomic mass is 10.8. A sample of boron which contains a higher percentage of 10B than natural boron has a relative atomic mass lower than 10.8, but its chemical properties are the same as those of natural boron.
  5. A sample of magnesium contains 79% of 24Mg, 10% of 25Mg and 11% of 26Mg by number of atoms. The relative atomic mass of magnesium calculated from these data is 24.3 (to 3 significant figures).
  6. Hydrogen has two isotopes, 1H and 2H, and chlorine has two isotopes, 35Cl and 37Cl. A natural sample of hydrogen chloride therefore contains molecules of four different relative molecular masses.
  7. A potassium atom (atomic number 19) has the electronic arrangement 2,8,9, because the third shell is completely filled before electrons enter the fourth shell.
  8. A particle with the electronic arrangement 2,8,8 may be an argon atom, a chloride ion or a calcium ion.
  9. A Ca2+ ion has the same electronic arrangement as an argon atom, so Ca2+ ions and argon atoms have the same chemical properties.
  10. Across Period 3 from sodium to chlorine, the number of occupied electron shells stays the same, but the atomic size decreases because the nuclear charge increases.
  11. Across Period 3 from sodium to sulphur, the oxides change from basic (Na2O), through amphoteric (Al2O3), to acidic (SO2).
  12. Fluorine and chlorine are in the same period of the Periodic Table because the atoms of both elements have seven electrons in the outermost shell.
  13. When a small piece of potassium is added to water, it melts, moves rapidly on the water surface and the hydrogen given off burns with a lilac flame. The solution formed turns blue litmus paper red.
  14. The reactivity of the Group I metals increases down the group, because the outermost electron of the atom is farther from the nucleus and more shielded by the inner electron shells, and so is lost more easily.
  15. Magnesium, like calcium, reacts steadily with cold water to give hydrogen and the metal hydroxide.
  16. Although potassium and calcium are in the same period, calcium reacts less vigorously with water than potassium. This is because the outermost electrons of a calcium atom are attracted more strongly by a greater nuclear charge, and each calcium atom has to lose two electrons instead of one.
  17. Iodine solution can displace bromine from potassium bromide solution, because iodine is more reactive than bromine.
  18. Down Group VII from chlorine to iodine, the colour of the halogen becomes darker and its physical state at room temperature changes from gas, through liquid, to solid.
  19. Every element in Group 0 has eight electrons in the outermost shell of its atoms.
  20. The Group 0 elements are unreactive and exist as monatomic gases because their atoms have fully filled outermost shells. The boiling points of the Group 0 elements increase down the group.
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2
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19
20
20 marks
02

Section B — Comprehensive Structured Question 1

Relative Isotopic Mass and Relative Atomic Mass

Naturally occurring neon consists of three isotopes. Fig. 1 shows the relative abundance (percentage by number of atoms) of each isotope. The relative isotopic mass of each isotope is equal to its mass number.

0 20 40 60 80 100 Relative abundance (%) 90.5 0.3 9.2 20Ne 21Ne 22Ne Isotope
Fig. 1 — Relative abundances of the isotopes of neon (the bar for 21Ne is too thin to see on this scale)

(a) State the meaning of the term isotopes. (1 mark)

(b)

  1. Calculate the relative atomic mass of neon, correct to 3 significant figures. (2 marks)
  2. Explain why the relative atomic mass of neon is not a whole number although the relative isotopic mass of each of its isotopes is a whole number. (1 mark)

(c) Element Y exists as three isotopes, 24Y, 25Y and 26Y. The relative atomic mass of Y is 24.30 and the relative abundance of 25Y is 10.0%. Calculate the percentage abundances of 24Y and 26Y. (3 marks)

(d) Natural boron consists of 20.0% of 10B and 80.0% of 11B by number of atoms. Pure 10B atoms are added to 200 atoms of natural boron, so that the relative atomic mass of the mixture becomes 10.5. Calculate the number of 10B atoms added. (3 marks)

10 marks
03

Section B — Comprehensive Structured Question 2

Electronic Arrangement of Atoms and Ions

The electronic arrangement of an atom or ion shows how its electrons are distributed in the electron shells. Use the atomic numbers given on page 1.

(a) Write the electronic arrangement of each of the following. (3 marks)

  1. a magnesium atom
  2. a chloride ion, Cl−
  3. an aluminium ion, Al3+

(b) Complete Table 1. The letter X is not the actual symbol of an element. (3 marks)

Table 1

Particle Atomic number Electronic arrangement of the particle Period of the element Group of the element
atom 15
atom 2,8,8,2
ion X+ 2,8,8

(c) An ion W2− has the same electronic arrangement as an argon atom. Deduce the atomic number of W, and state the period and the group of the Periodic Table in which W is found. (2 marks)

(d) Fig. 2 shows the electron diagrams of two particles, Q and R. The number in each nucleus is the number of protons. Each dot represents an electron. Deduce the formula, including the charge, of each particle. You are given that Q and R are an ion of one element and an ion of another element, from the first 20 elements. (2 marks)

11+ Particle Q 17+ Particle R
Fig. 2 — Electron diagrams of particles Q and R
10 marks
04

Section B — Comprehensive Structured Question 3

Periods of the Periodic Table: Trends Across Period 3

Fig. 3 shows part of the Periodic Table. Each of the letters A to M represents one of the first 20 elements. The letters are not the actual symbols of the elements.

Group I II III IV V VI VII 0 Period 2 Period 3 A B C D E F G H J L M
Fig. 3 — Part of the Periodic Table (only some elements are shown; the letters are not the actual symbols)

(a) Elements E, F, G, H, J, L and M are all in Period 3.

  1. Explain, in terms of electronic arrangement, why these elements are placed in the same period. (1 mark)
  2. Write the electronic arrangement of an atom of G. (1 mark)

(b) The atomic size of L is smaller than that of E. Explain why. (2 marks)

(c) The elements E, G and J each form an oxide.

  1. State whether the oxide of each of E, G and J is acidic, basic or amphoteric. (2 marks)
  2. Write a chemical equation, with state symbols, for the reaction of the oxide of E with water. (1 mark)

(d) E reacts violently with cold water, whereas F reacts only very slowly with cold water. By considering the structure of their atoms, explain why E is more reactive than F. (2 marks)

(e) D and M are in the same group, but they are in different periods. Explain this, in terms of the electronic arrangement of their atoms. (1 mark)

10 marks
05

Section B — Comprehensive Structured Question 4

Groups of the Periodic Table: Group I and Group II Metals

Lithium, sodium and potassium are Group I metals (alkali metals). Magnesium and calcium are Group II metals.

(a) Group I metals are stored under paraffin oil. Explain why. (1 mark)

(b) A small piece of sodium is added to a trough of water.

  1. State two observations that can be made. (2 marks)
  2. Write a chemical equation, with state symbols, for the reaction. (1 mark)

(c) Potassium reacts with water more vigorously than sodium does. By considering the structure of their atoms, explain why. (2 marks)

(d) Table 2 shows what is observed when small pieces of three metals are added separately to cold water.

Table 2

Metal Observation with cold water
Potassium Reacts violently; the metal melts and the gas given off burns with a lilac flame.
Calcium Reacts steadily; a stream of bubbles is formed and the mixture turns milky.
Magnesium Reacts extremely slowly; only a few tiny bubbles appear after several minutes.
  1. Write a chemical equation, with state symbols, for the reaction of calcium with water. (1 mark)
  2. Explain why calcium is more reactive than magnesium. (1 mark)
  3. Calcium and potassium are in the same period, but calcium is less reactive than potassium. Explain why. (2 marks)
10 marks
06

Section B — Comprehensive Structured Question 5

Groups of the Periodic Table: Group VII and Group 0 Elements

Chlorine, bromine and iodine are halogens (Group VII). Helium, neon and argon are noble gases (Group 0).

(a)

  1. Complete Table 3 by stating the physical state and the colour of each halogen at room temperature and pressure. (2 marks)
  2. Astatine lies below iodine in Group VII. Predict its physical state and colour at room temperature and pressure. (1 mark)

Table 3

Halogen Physical state Colour
Chlorine
Bromine
Iodine

(b) A student carries out five displacement experiments. Table 4 shows the results.

Table 4

Experiment Halogen solution added Halide solution Observation
1 chlorine water potassium bromide solution turns orange
2 chlorine water potassium iodide solution turns brown
3 bromine water potassium iodide solution turns brown
4 bromine water potassium chloride no visible change
5 iodine solution potassium bromide no visible change
  1. Write an ionic equation, with state symbols, for the reaction in Experiment 1. (1 mark)
  2. Using the results in Table 4, arrange chlorine, bromine and iodine in order of decreasing reactivity. (1 mark)
  3. By considering the structure of the halogen atoms, explain why the reactivity changes in this way down Group VII. (2 marks)

(c) Table 5 gives some data about three Group 0 elements.

Table 5

Element Atomic number Electronic arrangement Boiling point / °C
Helium 2 2 −269
Neon 10 2,8 −246
Argon 18 2,8,8 −186
  1. Using Table 5, explain why the Group 0 elements are chemically unreactive and exist as monatomic gases. (1 mark)
  2. Krypton has atomic number 36. Predict whether its boiling point is higher or lower than that of argon. (1 mark)
  3. State one use of helium or argon, and the property of the gas that makes it suitable for this use. (1 mark)
10 marks

Question 1 — True or False

1. F
2. F
3. F
4. T
5. T
6. T
7. F
8. T
9. F
10. T
11. T
12. F
13. F
14. T
15. F
16. T
17. F
18. T
19. F
20. T
  1. False. The relative isotopic mass of carbon-12 is exactly 12 (by definition), but the relative atomic mass of carbon is about 12.0 (12.01), because natural carbon also contains a very small proportion of carbon-13. The relative atomic mass is the weighted average over all the isotopes.
  2. False. No individual chlorine atom has this mass. Each atom is either 35Cl or 37Cl. The value 35.5 is the weighted average mass of the atoms in natural chlorine (35Cl : 37Cl = 3 : 1 by number of atoms), compared with 112 of the mass of a carbon-12 atom.
  3. False. The relative atomic mass 63.6 is closer to 63, so the isotope with relative isotopic mass 63 is more abundant. If x% is 65X: 63(100 − x) + 65x = 6360, so x = 30. The abundances are 70% (63X) and 30% (65X).
  4. True. Natural boron: 10(0.200) + 11(0.800) = 10.8. A sample richer in the lighter isotope 10B has a lower average mass. The isotopes have the same electronic arrangement, hence the same chemical properties.
  5. True. (24 × 79 + 25 × 10 + 26 × 11) ÷ 100 = (1896 + 250 + 286) ÷ 100 = 24.32 = 24.3 (3 s.f.).
  6. True. The possible molecules are 1H35Cl (36), 2H35Cl (37), 1H37Cl (38) and 2H37Cl (39), giving four different relative molecular masses.
  7. False. The electronic arrangement of a potassium atom is 2,8,8,1. For the first 20 elements, the third shell is stable with 8 electrons, so the 19th electron goes into the fourth shell.
  8. True. An Ar atom (18 electrons), a Cl− ion (17 + 1 = 18 electrons) and a Ca2+ ion (20 − 2 = 18 electrons) all have the electronic arrangement 2,8,8.
  9. False. Ca2+ and Ar have the same electronic arrangement but different chemical properties. They have different numbers of protons (20 and 18), and Ca2+ is a charged ion which takes part in reactions (e.g. it forms a white precipitate of calcium carbonate with carbonate ions), whereas argon is an unreactive monatomic gas.
  10. True. Na, Mg, Al, Si, P, S, Cl and Ar all have three occupied shells. From Na to Cl the nuclear charge increases, so the outermost electrons are pulled in more strongly and the atomic size decreases.
  11. True. Na2O is basic (it forms NaOH with water), Al2O3 is amphoteric (it reacts with both acids and alkalis) and SO2 is acidic (it forms an acidic solution with water).
  12. False. Fluorine (2,7) and chlorine (2,8,7) are in the same group (Group VII) because they have the same number of outermost-shell electrons. They are in different periods (Period 2 and Period 3) because they have different numbers of occupied electron shells.
  13. False. The solution formed (potassium hydroxide) is alkaline, so it turns red litmus paper blue. (Blue litmus paper stays blue.)
  14. True. Down Group I the atoms have more electron shells, so the outermost electron is farther from the nucleus and more shielded. The attraction is weaker, so the electron is lost more easily.
  15. False. Magnesium reacts only extremely slowly with cold water. (It reacts rapidly with steam to give magnesium oxide and hydrogen.) Calcium reacts steadily with cold water.
  16. True. Both are in Period 4. A Ca atom has one more proton and is smaller, so the outermost electrons are held more strongly. A Ca atom must also lose two electrons to form Ca2+ while a K atom loses only one electron to form K+.
  17. False. Reactivity decreases down Group VII, so iodine is less reactive than bromine. Iodine cannot displace bromine from potassium bromide (no reaction); bromine can displace iodine from potassium iodide.
  18. True. Chlorine is a pale yellowish-green gas, bromine is a reddish-brown liquid and iodine is a dark grey (purple-black) solid.
  19. False. Helium has only 2 electrons in its outermost (first) shell, which is already full. The other noble gases (Ne, Ar, etc.) have 8 electrons in the outermost shell.
  20. True. The outermost shells are fully filled, so the atoms are stable and do not gain, lose or share electrons, and exist as separate atoms. Boiling points rise down the group (He −269 °C, Ne −246 °C, Ar −186 °C).

Question 2 — Relative Isotopic Mass and Relative Atomic Mass

(a) Isotopes

Isotopes are atoms of the same element (same atomic number / same number of protons) with different numbers of neutrons (different mass numbers). (1)

(b) Neon

(i) Relative atomic mass = 20 × 90.5 + 21 × 0.3 + 22 × 9.2100 (1)
= 2018.7100 = 20.187 = 20.2 (1)
(ii) The relative atomic mass is the weighted average of the relative isotopic masses, taking the relative abundances into account. No single neon atom has a mass of 20.2. (1)
Common errors: (1) taking a simple average, (20 + 21 + 22) ÷ 3 = 21.0; (2) forgetting to divide by 100; (3) stating that the RAM is the mass of the most abundant isotope.

(c) Element Y

Let the percentage abundance of 24Y be a and that of 26Y be b.
a + b + 10.0 = 100, so a + b = 90.0 (1)
24a + 25(10.0) + 26b100 = 24.30, so 24a + 26b = 2180 (1)
Substituting a = 90.0 − b: 2160 + 2b = 2180, so b = 10.0 and a = 80.0
24Y = 80.0%; 26Y = 10.0% (1)
Check: (24 × 80.0 + 25 × 10.0 + 26 × 10.0) ÷ 100 = (1920 + 250 + 260) ÷ 100 = 24.30. ✓
Alternative method: Compare with 24: the excess mass is 0.30 per atom on average, so (10.0 × 1 + b × 2) ÷ 100 = 0.30, which gives b = 10.0.

(d) Boron

In 200 atoms of natural boron: 10B = 0.200 × 200 = 40 atoms; 11B = 160 atoms (1)
Let n atoms of 10B be added:
10(40 + n) + 11(160)200 + n = 10.5 (1)
400 + 10n + 1760 = 2100 + 10.5n, so 60 = 0.5n
n = 120 (1)
Check: The relative atomic mass 10.5 lies exactly midway between 10 and 11, so the mixture must contain equal numbers of 10B and 11B atoms. With 160 atoms of 11B, there must be 160 atoms of 10B, so 160 − 40 = 120 atoms must be added. ✓

Question 3 — Electronic Arrangement of Atoms and Ions

(a) Electronic arrangements

(i) Mg atom (12 electrons): 2,8,2 (1)
(ii) Cl− (17 + 1 = 18 electrons): 2,8,8 (1)
(iii) Al3+ (13 − 3 = 10 electrons): 2,8 (1)

(b) Table 1

Particle Atomic number Electronic arrangement of the particle Period of the element Group of the element
atom 15 2,8,5 3 V
atom 20 2,8,8,2 4 II
ion X+ 19 2,8,8 4 I
Marking: 1 mark for each row, awarded only if all the blanks in that row are correct. In row 3, X+ has 18 electrons, so the atom X has 19 electrons (2,8,8,1). Its atom has four occupied shells, so X is in Period 4 (not Period 3) and Group I, since it has one outermost-shell electron.

(c) Ion W2−

An argon atom has 18 electrons, so W2− has 18 electrons and the W atom has 16 electrons.
Atomic number of W = 16 (1)
The W atom is 2,8,6: three occupied shells and six outermost-shell electrons, so W is in Period 3, Group VI (1)

(d) Particles Q and R

Q: 11 protons and 2 + 8 = 10 electrons, so the charge is 11 − 10 = +1. Q is Na+ (1)
R: 17 protons and 2 + 8 + 8 = 18 electrons, so the charge is 17 − 18 = −1. R is Cl− (1)
Common errors: (1) writing the electronic arrangement of a cation by removing electrons from the wrong shell, or writing the atom’s arrangement for an ion; (2) giving the period of X+ from the number of shells of the ion (3) instead of the atom (4); (3) writing the third shell of K and Ca as 9 and 10 electrons.

Question 4 — Periods of the Periodic Table: Trends Across Period 3

Identification of the elements (from Fig. 3): E = Na (2,8,1); F = Mg (2,8,2); G = Al (2,8,3); H = Si (2,8,4); J = S (2,8,6); L = Cl (2,8,7); M = Ar (2,8,8); A = Li; B = N; C = F; D = Ne.

(a) Period 3

(i) The atoms of these elements all have the same number of occupied electron shells (three). (1)
(ii) G has 13 electrons: 2,8,3 (1)

(b) Atomic size of L and E

Atoms of E (11 protons) and L (17 protons) have the same number of occupied shells (three), but the nuclear charge of L is greater. (1)
The attraction between the nucleus and the outermost-shell electrons is therefore stronger in L, which pulls the electron shells closer to the nucleus. L has a smaller atomic size. (1)

(c) Oxides

(i) Oxide of E (Na2O): basic; oxide of G (Al2O3): amphoteric; oxide of J (SO2): acidic
(2 marks for all three correct; 1 mark for two correct)
(ii) Na2O(s) + H2O(l) → 2NaOH(aq) (1)

(d) Reactivity of E and F

An E atom has fewer protons (11 against 12) and the same number of shells as an F atom, so the outermost electron of E is held less strongly by the nucleus. (1)
An E atom has to lose only one electron to form E+, while an F atom has to lose two electrons to form F2+. E loses electrons more easily and is more reactive. (1)

(e) D and M

D (2,8) and M (2,8,8) both have a fully filled outermost shell of eight electrons, so they are in the same group (Group 0). D has two occupied shells (Period 2) while M has three occupied shells (Period 3). (1)
Common errors: (1) saying that atomic size decreases across a period because the number of shells decreases (it stays the same); (2) saying that the elements in the same period have the same number of outermost-shell electrons; (3) not stating that the nuclear charge increases while the number of occupied shells stays constant.

Question 5 — Groups of the Periodic Table: Group I and Group II Metals

(a) Storage

Group I metals react readily with oxygen and water vapour (moisture) in the air. The paraffin oil keeps these out and prevents the reaction. (1)

(b) Sodium and water

(i) Any two of the following (1 mark each):
• sodium floats on the water
• sodium melts into a silvery ball
• sodium moves rapidly on the water surface
• a hissing / fizzing sound and bubbles of a colourless gas
• sodium becomes smaller and finally disappears
(ii) 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) (1)

(c) Potassium compared with sodium

A potassium atom has more occupied electron shells (four against three), so it has a larger atomic size and its outermost electron is farther from the nucleus and more shielded by the inner electrons. The attraction between the nucleus and the outermost electron is weaker. (1)
The outermost electron is lost more easily, so potassium is more reactive. (1)

(d) Group II metals

(i) Ca(s) + 2H2O(l) → Ca(OH)2(aq) + H2(g) (1) (accept Ca(OH)2(s), as it is only sparingly soluble)
(ii) A calcium atom has more shells (four against three) and a larger atomic size than a magnesium atom, so its outermost electrons are farther from the nucleus and held less strongly. They are lost more easily. (1)
(iii) Calcium has a greater nuclear charge than potassium (20 against 19) with the same number of shells, so the outermost electrons of Ca are attracted more strongly by the nucleus. (1)
A Ca atom must lose two electrons but a K atom only one electron, so more energy is needed for Ca to react. (1)
Common errors: (1) writing the product of Ca with water as CaO or Ca(OH) instead of Ca(OH)2; (2) explaining the reactivity trend by “more electrons” without referring to the distance from the nucleus and the shielding; (3) forgetting to compare the number of electrons lost when comparing a Group II metal with a Group I metal.

Question 6 — Groups of the Periodic Table: Group VII and Group 0 Elements

(a) Physical states and colours

Halogen Physical state Colour
Chlorine gas pale yellowish-green
Bromine liquid reddish-brown
Iodine solid dark grey (purple-black)
(i) 2 marks if all six entries are correct; 1 mark if four or five are correct.
(ii) Astatine is a solid and is black (a very dark colour, darker than iodine). (1)

(b) Displacement reactions

(i) Cl2(aq) + 2Br−(aq) → 2Cl−(aq) + Br2(aq) (1)
(ii) chlorine > bromine > iodine (1)
(Chlorine displaces both bromine and iodine, bromine displaces iodine but not chlorine, and iodine displaces neither.)
(iii) Down the group, the halogen atoms have more occupied shells and a larger atomic size. The outermost shell is farther from the nucleus and more shielded by the inner electrons, so the attraction of the nucleus for an incoming electron is weaker. (1)
The atom gains an electron less readily to form X−, so the reactivity decreases down the group. (1)

(c) Group 0

(i) The atoms have fully filled outermost shells (2 for He; 8 for Ne and Ar), which are stable. They do not gain, lose or share electrons, so they do not react and exist as separate (monatomic) atoms. (1)
(ii) Higher. The boiling points increase down the group (−269, −246, −186 °C). (1)
(iii) Any one of the following (1):
• helium for filling airships / weather balloons: it is less dense than air and does not burn
• argon for filling filament lamps: it is unreactive, so the hot tungsten filament does not burn away / oxidise
Common errors: (1) giving “ability to lose an electron” as the reason for the halogen trend (a halogen atom gains an electron); (2) writing Cl2 + 2Br2 or leaving the spectator K+ ions in the ionic equation; (3) saying that all the noble gases have eight outermost electrons.
Mark weighting: Section A (20) + Question 2 (10) + Question 3 (10) + Question 4 (10) + Question 5 (10) + Question 6 (10) = 70 marks.
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