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HKDSE Chemistry Recap: Microscopic World I | Tia (2026)

HKDSE Chemistry — Microscopic World I

Subject: Chemistry

Tia (2026)

Issue date: Wednesday, 14 October 2026

Total: 60 marks  |  Suggested time: 80–90 minutes  |  Topic: atoms, the Periodic Table, chemical bonding and structures

Instructions

  1. Question 1 consists of 10 multiple-choice questions. Each has one correct answer. Write your answers in the answer boxes at the end of Question 1.
  2. In Question 2, state whether each statement is True or False. If a statement is false, rewrite it correctly.
  3. Questions 3 to 6 are structured questions. Show your working clearly and give numerical answers with appropriate units where needed.
  4. The letters used for elements and substances in the questions are not the actual symbols, unless stated otherwise.
  5. A Periodic Table may be used. The atomic numbers of some elements are given below.

Paper structure

Question Type Marks Score
1 Multiple choice (10 questions) 10
2 True or false (10 statements) 10
3 Comprehensive structured question 1 10
4 Comprehensive structured question 2 10
5 Comprehensive structured question 3 10
6 Comprehensive structured question 4 10
Total 60
Data — atomic numbers: H = 1, Li = 3, C = 6, N = 7, O = 8, F = 9, Na = 11, Mg = 12, Al = 13, Si = 14, P = 15, S = 16, Cl = 17, Ar = 18, K = 19, Ca = 20.

Total score: / 60

01

Section A — Microscopic World I

Multiple Choice

Each question has one correct answer. Choose the best option. (1 mark each)

  1. An ion X2− has 18 electrons and 18 neutrons. Which of the following is correct for an atom of X?
    1. 18 protons; mass number 36; Group 0
    2. 16 protons; mass number 34; Group II
    3. 16 protons; mass number 34; Group VI
    4. 20 protons; mass number 38; Group II
  2. Element Q has two isotopes, 69Q and 71Q. The relative atomic mass of Q is 69.8. What is the percentage abundance of 71Q?
    1. 20%
    2. 40%
    3. 60%
    4. 80%
  3. Element X has atomic number 13 and element Y has atomic number 8. Which of the following statements about the compound formed between X and Y is correct?
    1. Its formula is XY and it is a covalent compound.
    2. Its formula is X3Y2 and it contains X2+ ions and Y3− ions.
    3. Its formula is X2Y3 and it consists of simple molecules.
    4. Its formula is X2Y3 and it has a giant ionic structure.
  4. Rubidium (Rb) is in Group I, directly below potassium in the Periodic Table. Which of the following predictions about rubidium is INCORRECT?
    1. Rubidium atoms lose their outermost electron less readily than potassium atoms do.
    2. Rubidium reacts with water more vigorously than potassium does.
    3. The formula of rubidium oxide is Rb2O.
    4. Rubidium has a lower melting point than potassium.
  5. Astatine (At) is the Group VII element below iodine in the Periodic Table. Which of the following predictions about astatine are correct?
    • (1) Astatine is a dark-coloured solid at room temperature and pressure.
    • (2) Aqueous astatine displaces iodine from potassium iodide solution.
    • (3) Chlorine water displaces astatine from sodium astatide solution.
    1. (1) and (2) only
    2. (1) and (3) only
    3. (2) and (3) only
    4. (1), (2) and (3)
  6. Which of the following shows the molecules in order of increasing number of lone pairs of electrons per molecule?
    1. HCl < N2 < CO2
    2. N2 < CO2 < HCl
    3. N2 < HCl < CO2
    4. CO2 < HCl < N2
  7. Which of the following best explains why a metal is malleable but an ionic crystal is brittle?
    1. In a metal, layers of cations can slide over one another while still being held by the delocalised electrons. In an ionic crystal, sliding a layer brings ions of the same charge next to each other, and the repulsion breaks the crystal.
    2. Metallic bonds are much stronger than ionic bonds, so a metal does not break when it is hammered.
    3. A metal contains mobile ions that rearrange when it is hammered, whereas an ionic crystal contains only fixed electrons.
    4. The bonds between atoms in a metal are directional and can reform in a new direction, but ionic bonds cannot.
  8. Which of the following properties of the elements increase from sodium to chlorine across Period 3?
    • (1) Nuclear charge
    • (2) Atomic radius
    • (3) Number of outermost-shell electrons
    • (4) Metallic character
    1. (1) and (2) only
    2. (2) and (4) only
    3. (3) and (4) only
    4. (1) and (3) only
  9. Silicon dioxide melts at about 1710 °C, but solid carbon dioxide changes directly to gas at about −78 °C. Which of the following statements best explains this difference?
    1. Silicon atoms are larger than carbon atoms, so silicon dioxide molecules have much stronger intermolecular forces.
    2. Silicon dioxide has a giant covalent structure, so many strong covalent bonds have to be broken to melt it, whereas only weak intermolecular forces between CO2 molecules have to be overcome.
    3. Silicon dioxide is an ionic compound, whereas carbon dioxide is a covalent compound.
    4. The covalent bonds in silicon dioxide are stronger than the covalent bonds in carbon dioxide.
  10. Fig. 1 shows part of the Periodic Table. The letters are not the actual symbols of the elements. Which of the following statements are correct?
    Group I II III IV V VI VII 0 Period 2 Period 3 Period 4 R X Z Q W Y D E
    Fig. 1 — Part of the Periodic Table (letters are not the actual symbols)
    • (1) R and X combine to form a covalent compound.
    • (2) E reacts with water more vigorously than Z does.
    • (3) Q and Y combine to form an ionic compound of formula QY2.
    1. (1) and (2) only
    2. (1) and (3) only
    3. (2) and (3) only
    4. (1), (2) and (3)
1
2
3
4
5
6
7
8
9
10
10 marks
02

Section B — Microscopic World I

True or False

For each statement, state whether it is True or False. If it is false, rewrite the statement correctly. (1 mark each)

  1. Isotopes of an element have the same chemical properties because their atoms have the same electronic arrangement, but they differ in physical properties such as mass and density.
  2. The relative atomic mass of an element is always a whole number, because it is equal to the mass number of the most abundant isotope of the element.
  3. Going down Group I from lithium to potassium, the melting point decreases and the reactivity towards water increases, because the atoms become larger and the outermost electron is held less strongly by the nucleus.
  4. Going down Group VII from chlorine to iodine, the melting point and boiling point increase because the covalent bond within each diatomic molecule becomes stronger.
  5. Noble gases exist as single atoms because each atom has a stable (duplet or octet) outermost shell and has little tendency to gain, lose or share electrons.
  6. Transition metals generally have lower densities and lower melting points than Group I metals, and their compounds are usually colourless.
  7. Solid sodium chloride consists of NaCl molecules, each containing one Na+ ion and one Cl− ion joined by an ionic bond.
  8. Graphite conducts electricity but diamond does not, because in graphite each carbon atom uses only three of its four outermost-shell electrons for covalent bonding and the fourth electron is delocalised within the layer, whereas in diamond all four electrons are used in bonding.
  9. Silicon dioxide has the formula SiO2 because it consists of discrete molecules, each containing one silicon atom and two oxygen atoms, held together by weak intermolecular forces.
  10. A metal is malleable because layers of metal cations can slide over one another without breaking the metallic bond, as the delocalised electrons are not fixed to particular cations.
10 marks
03

Section C — Comprehensive Structured Question 1

Atomic Structure, Isotopes and Relative Atomic Mass

The table below shows the numbers of subatomic particles in five species P, Q, R, S and T. Each species is either a neutral atom or a monatomic ion. The letters are not the actual symbols of the elements. R is an ion of chlorine, S is an atom of argon and T is an ion of potassium.

Species Number of protons Number of neutrons Number of electrons
P 17 18 17
Q 17 20 17
R 17 20 18
S 18 22 18
T 19 20 18

(a)

  1. Which two of the neutral atoms in the table are isotopes of each other? Explain your answer. (2 marks)
  2. Explain why isotopes of an element have the same chemical properties. (1 mark)

(b) R, S and T have the same electronic arrangement.

  1. Write the electronic arrangement of R, S and T. (1 mark)
  2. Write the nuclide symbol of R, in the form AZXn± (showing the mass number, atomic number, symbol and charge). (1 mark)

(c) The relative atomic masses of argon and potassium are 39.9 and 39.1 respectively. Explain why argon is placed before potassium in the Periodic Table although argon has the greater relative atomic mass. (1 mark)

(d) Natural chlorine consists of only two isotopes, the atoms P and Q, with relative isotopic masses of 34.97 and 36.97 respectively. The relative atomic mass of chlorine is 35.45.

  1. Calculate the percentage abundance of P in natural chlorine. (3 marks)
  2. A sample of chlorine gas, Cl2, contains molecules made up of atoms P and Q. Using the mass numbers of P and Q, state the relative molecular masses of all the different types of Cl2 molecule in the sample. (1 mark)
10 marks
04

Section C — Comprehensive Structured Question 2

The Periodic Table: Group I Metals and the Halogens

Part 1 — Group I. A small piece of potassium is dropped onto water in a trough. A few drops of universal indicator have been added to the water, and the indicator is initially green.

(a)

  1. State the colour change of the indicator and one other observation. (2 marks)
  2. Write the chemical equation for the reaction, with state symbols. (1 mark)
  3. When a piece of sodium of similar size is used, the reaction is less vigorous. By considering the atomic structures of sodium and potassium, explain why. (2 marks)

Part 2 — Group VII. The halogens X2, Y2 and Z2 are chlorine, bromine and iodine, but not necessarily in this order. Aqueous solutions of X2, Y2 and Z2 were added separately to aqueous solutions of the sodium halides NaX, NaY and NaZ. The results are shown below.

Halogen solution added NaX(aq) NaY(aq) NaZ(aq)
X2(aq) — Reaction occurs Reaction occurs
Y2(aq) No reaction — Reaction occurs
Z2(aq) No reaction No reaction —

(b)

  1. Arrange X2, Y2 and Z2 in order of decreasing reactivity, and hence identify X, Y and Z. (2 marks)
  2. Write the ionic equation for the reaction between Y2(aq) and NaZ(aq), using the actual formulae of the halogens and halide ions. (1 mark)
  3. By considering atomic structure, explain why the reactivity of the halogens changes in this way down the group. (2 marks)
10 marks
05

Section C — Comprehensive Structured Question 3

Ionic and Covalent Bonding

The table below shows the atomic numbers of six elements. The letters are not the actual symbols of the elements.

Element P Q R S T U
Atomic number 6 7 8 11 12 17

(a)

  1. Write the electronic arrangements of atoms of T and U. (1 mark)
  2. T reacts with U to form a compound. Deduce the formula of the compound (using the letters T and U) and state the type of chemical bond in it. (1 mark)

(b) Draw the electron diagram of the compound formed between T and U. Show the electrons in the outermost shell of each ion only. Use a cross (×) for an electron originally from an atom of T, and a dot (•) for an electron originally from an atom of U. (2 marks)

Space for electron diagram

(c) Hydrogen cyanide, HCN, is a covalent compound of hydrogen, P and Q. In a molecule of HCN, the P atom is bonded to both the H atom and the Q atom. Draw the electron diagram of an HCN molecule, showing all the outermost-shell electrons of each atom. Each of P and Q has eight electrons in its outermost shell in the molecule. (2 marks)

Space for electron diagram

(d) The compound formed between T and U does not conduct electricity in the solid state, but it conducts electricity in the molten state. Explain. (2 marks)

(e) The compound formed between T and R has a melting point of 2852 °C, which is much higher than that of the compound formed between S and U (801 °C). Explain. (2 marks)

10 marks
06

Section C — Comprehensive Structured Question 4

Structures and Melting Points of the Period 3 Elements

Fig. 2 shows the melting points of the eight elements in Period 3. Phosphorus is white phosphorus, which exists as P4 molecules, and sulphur exists as S8 molecules.

1500 1000 500 0 −200 98 650 660 1410 44 115 −101 −189 Na Mg Al Si P S Cl Ar Melting point / °C Element in Period 3
Fig. 2 — Melting points of the Period 3 elements

(a)

  1. Describe the structure of solid sodium. (1 mark)
  2. Explain why the melting point of magnesium is much higher than that of sodium. (2 marks)

(b) Silicon has the highest melting point among the Period 3 elements. Describe the structure of silicon and explain why its melting point is so high. (2 marks)

(c) Phosphorus (P4), sulphur (S8) and chlorine (Cl2) are all simple molecular substances. Explain why sulphur has the highest melting point and chlorine the lowest among the three. Your answer should state what is overcome when each of them melts. (3 marks)

(d) Argon has the lowest melting point in Period 3. State why argon exists as single atoms, and explain why its melting point is lower than that of chlorine. (2 marks)

10 marks

Question 1 — Multiple Choice

1. C
2. B
3. D
4. A
5. B
6. C
7. A
8. D
9. B
10. C
  1. C — 16 protons; mass number 34; Group VI. X2− has gained 2 electrons, so the neutral atom has 18 − 2 = 16 electrons and therefore 16 protons. Mass number = 16 + 18 = 34. The electronic arrangement of the atom is 2,8,6, so X is in Group VI. (A and D come from using 18 or 20 electrons for the atom. B has the correct numbers but confuses the charge of the ion with the group number.)
  2. B — 40%. Let the abundance of 69Q be x (as a fraction): 69x + 71(1 − x) = 69.8, so 71 − 2x = 69.8 and x = 0.60. The abundance of 71Q is 1 − 0.60 = 0.40 = 40%. (60% is the abundance of 69Q.)
  3. D — X2Y3, giant ionic structure. X (atomic number 13) has the electronic arrangement 2,8,3 and loses 3 electrons to form X3+. Y (atomic number 8) has the arrangement 2,6 and gains 2 electrons to form Y2−. For electrical neutrality, 2 X3+ combine with 3 Y2−, giving X2Y3 (aluminium oxide-type), a giant ionic lattice. A metal and a non-metal combine by electron transfer, not by sharing.
  4. A — This prediction is incorrect. The outermost electron of Rb is in a shell farther from the nucleus and is more shielded by the inner shells, so it is lost more readily than that of K. B, C and D are all correct: reactivity increases, Rb is in Group I so its oxide is Rb2O, and the melting point of Group I metals decreases down the group.
  5. B — (1) and (3) only. (1) is correct: the halogens get darker and change from gas to liquid to solid down the group. (2) is wrong: reactivity decreases down the group, so astatine cannot displace iodine. (3) is correct: chlorine is more reactive than astatine, so it displaces astatine from its salt.
  6. C — N2 < HCl < CO2. N≡N: each N has one lone pair, so 2 lone pairs per molecule. H–Cl: Cl has 3 lone pairs and H has none, so 3 lone pairs. O=C=O: each O has 2 lone pairs, so 4 lone pairs. The order of increasing number is 2 < 3 < 4.
  7. A. In a metal, the delocalised electrons are shared by all the cations and keep holding them together even when a layer of cations slides. In an ionic crystal, a slip of half a unit brings like-charged ions face to face; the strong repulsion cracks the crystal. (B is not an explanation; C is false because metals contain cations and delocalised electrons, not mobile ions; D is false because metallic bonding is non-directional.)
  8. D — (1) and (3) only. Across Period 3, the nuclear charge increases from +11 to +17 and the number of outermost-shell electrons increases from 1 to 7. The electrons are added to the same shell, so the attraction of the nucleus increases and the atomic radius decreases. The metallic character also decreases.
  9. B. SiO2 has a giant covalent structure, so melting it needs the breaking of a large number of strong covalent bonds throughout the lattice. CO2 is a simple molecular substance, so only weak intermolecular forces between the molecules are overcome. (D is the most tempting wrong answer: it is the structure, not the strength of an individual bond, that explains the large difference. A is wrong because SiO2 has no discrete molecules. C is wrong because SiO2 is not ionic.)
  10. C — (2) and (3) only. (1) is wrong: R (Group I metal) and X (Group VII non-metal) combine by electron transfer to form an ionic compound RX. (2) is correct: E is below Z in Group I, so it is more reactive. (3) is correct: Q is in Group II and forms Q2+, Y is in Group VII and forms Y−, so the ionic compound is QY2.

Question 2 — True or False

  1. True. Isotopes have the same number of protons and electrons but different numbers of neutrons, so their masses (and hence densities) differ.
  2. False. The relative atomic mass is the weighted average of the relative isotopic masses, based on the natural abundances of the isotopes. It is not necessarily a whole number (for example, chlorine has a relative atomic mass of about 35.5).
  3. True.
  4. False. The melting point and boiling point increase because the intermolecular forces (van der Waals’ forces) between the molecules become stronger as the molecules become larger. The covalent bonds within the molecules are not broken on melting (and the X–X bond actually gets weaker down the group).
  5. True.
  6. False. Transition metals generally have higher densities and higher melting points than Group I metals, and their compounds are usually coloured.
  7. False. Sodium chloride has a giant ionic structure (lattice) with no discrete molecules. Each ion is attracted by several oppositely charged ions around it. NaCl shows only the simplest ratio of ions, 1 : 1.
  8. True.
  9. False. Silicon dioxide has a giant covalent structure. Each Si atom is bonded to four O atoms and each O atom to two Si atoms, so the ratio Si : O is 1 : 2. There are no discrete molecules, and all the atoms are joined by strong covalent bonds.
  10. True.

Question 3 — Atomic Structure, Isotopes and Relative Atomic Mass

(a) Isotopes

(i) P and Q (1)
They are both neutral atoms with the same number of protons (17) but different numbers of neutrons (18 and 20), i.e. different mass numbers (35 and 37). (1)
(ii) Isotopes have the same number of electrons and the same electronic arrangement (2,8,7). Chemical properties depend on the electronic arrangement, so they react in the same way. (1)
Marking note: R is an ion of the same element (17 protons) but it is not a neutral atom, so do not award the first mark for “P, Q and R”.

(b) Species with the same electronic arrangement

(i) All three have 18 electrons: 2,8,8 (1)
(ii) R has 17 protons and 20 neutrons, so its mass number is 37. It has one more electron than proton, so its charge is 1−: 3717Cl− (1)

(c) Order of argon and potassium

Elements are arranged in the Periodic Table in order of increasing atomic number, not relative atomic mass. Argon (18) has a smaller atomic number than potassium (19). (1)

(d) Relative atomic mass of chlorine

(i) Let the percentage abundance of P be x. Then Q is (100 − x)%.
34.97x + 36.97(100 − x)100 = 35.45 (1)
3697 − 2.00x = 3545 (1)
x = 76% (P); Q is 24%. (1)
(ii) 35Cl–35Cl: 70; 35Cl–37Cl: 72; 37Cl–37Cl: 74. The relative molecular masses are 70, 72 and 74. (1)
Check: 0.76 × 34.97 + 0.24 × 36.97 = 26.58 + 8.87 = 35.45.
Common errors: (1) including the ion R as an “isotope” of P and Q; (2) dividing by 2 (a simple mean) instead of using abundances; (3) giving only two relative molecular masses (70 and 74) for Cl2.

Question 4 — The Periodic Table: Group I Metals and the Halogens

(a) Potassium and water

(i) The indicator changes from green to purple (blue/violet) (1)
One other observation, e.g. the potassium floats and moves quickly on the water / melts into a silvery ball / burns with a lilac flame / fizzing (hissing) occurs (1)
(ii) 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g) (1)
(iii) A potassium atom has one more electron shell than a sodium atom, so its outermost electron is farther from the nucleus (and more shielded by the inner electrons). (1)
The nucleus attracts this electron more weakly, so potassium atoms lose their outermost electron more easily and potassium is more reactive. (1)

(b) Halogens

(i) X2 displaces both Y− and Z−, Y2 displaces only Z− and Z2 displaces neither. So the order of decreasing reactivity is X2 > Y2 > Z2 (1)
Reactivity of the halogens decreases down the group, so X = chlorine, Y = bromine, Z = iodine (1)
(ii) Br2(aq) + 2I−(aq) → 2Br−(aq) + I2(aq) (1)
(iii) Down the group the atomic size increases (more electron shells), so the incoming electron enters a shell farther from the nucleus and more shielded. (1)
The attraction of the nucleus for the incoming electron is weaker, so the atoms gain an electron less easily and the reactivity decreases. (1)
Common errors: (1) applying the Group I explanation (“loses electrons more easily”) to the halogens; (2) writing the equation for Y2 + NaZ as a full formula equation with an unbalanced ionic charge; (3) not using the table to deduce the order of reactivity before naming the halogens.

Question 5 — Ionic and Covalent Bonding

(a) Electronic arrangements and formula

(i) T: 2,8,2; U: 2,8,7 (1)
(ii) T loses 2 electrons to form T2+; U gains 1 electron to form U−. The formula is TU2 and the bond is an ionic bond. (1)

(b) Electron diagram of TU2

[ T with 8 × in its outermost shell ]2+   and two   [ U with 7 • and 1 × in its outermost shell ]−
Correct ions (one T2+ and two U−), each in square brackets with the correct charge (1)
Eight electrons in the outermost shell of each ion, with the correct dots and crosses (each U− has seven • and one ×) (1)
Marking note: T2+ has the arrangement 2,8, so its outermost shell contains 8 electrons, all shown as ×. Accept T2+ drawn with no outermost-shell electrons, i.e. [T]2+, only if the candidate clearly shows it as an ion that has lost its two outer electrons.

(c) Electron diagram of HCN

H–C≡N, with one lone pair on N
One shared pair of electrons between H and C, and three shared pairs between C and N (a triple bond) (1)
One lone pair on N; C and N each have 8 electrons in the outermost shell and H has 2 (1)
Check: Total outer electrons = 1 (H) + 4 (C) + 5 (N) = 10, which is 4 bond pairs (8 electrons) + 1 lone pair (2 electrons). C has 2 + 6 = 8 electrons around it; N has 6 + 2 = 8.

(d) Conduction of TU2

In the solid state, the T2+ and U− ions are held in fixed positions in the giant ionic lattice by strong electrostatic attraction, so they are not free to move and cannot carry charge. (1)
In the molten state, the lattice breaks down and the ions become mobile, so they can carry charge. (1)

(e) Melting points of TR and SU

TR consists of T2+ and R2− ions (R: 2,6 gains 2 electrons), whereas SU consists of S+ and U− ions. Both have giant ionic structures. (1)
The ions in TR carry higher charges (and are smaller), so the electrostatic attraction (ionic bonds) between the ions is much stronger and much more energy is needed to break down the lattice. (1)
Common errors: (1) drawing a shared pair in the compound between T and U; (2) drawing only a double bond in HCN, or omitting the lone pair on N; (3) saying that “electrons” (not ions) carry the charge in the molten ionic compound; (4) saying that “molecules” of TR or SU are separated on melting.

Question 6 — Structures and Melting Points of the Period 3 Elements

(a) Sodium and magnesium

(i) A giant metallic structure: a lattice of Na+ ions surrounded by a ‘sea’ of delocalised electrons, held together by metallic bonds (electrostatic attraction between the cations and the delocalised electrons). (1)
(ii) Each Mg atom contributes two outermost electrons to the sea of delocalised electrons (Na contributes one), and Mg2+ has a higher charge and is smaller than Na+. (1)
The metallic bond in magnesium is therefore much stronger, so more energy is needed to overcome it. (1)

(b) Silicon

Silicon has a giant covalent structure: each Si atom is joined to four other Si atoms by covalent bonds in a three-dimensional network. (1)
A very large amount of energy is needed to break the large number of strong covalent bonds, so the melting point is very high. (1)

(c) P4, S8 and Cl2

On melting, only the weak intermolecular forces (van der Waals’ forces) between the molecules are overcome; the covalent bonds within the molecules are not broken. (1)
S8 molecules are the largest and Cl2 molecules are the smallest of the three (P4 is in between), so the van der Waals’ forces are strongest between S8 molecules and weakest between Cl2 molecules. (1)
More energy is therefore needed to melt sulphur, and the least is needed to melt chlorine: S8 > P4 > Cl2 in melting point. (1)

(d) Argon

An argon atom already has a stable (full) outermost shell (2,8,8), so it does not gain, lose or share electrons and exists as single atoms (monatomic). (1)
Argon atoms are smaller than Cl2 molecules, so the van der Waals’ forces between argon atoms are weaker and the melting point is lower. (1)
Common errors: (1) saying that covalent bonds are broken when sulphur or chlorine melts; (2) explaining the melting point of magnesium only by “more electrons” without mentioning the stronger metallic bond; (3) describing silicon as having “molecules”.
Mark weighting: Section A (10) + Section B (10) + Question 3 (10) + Question 4 (10) + Question 5 (10) + Question 6 (10) = 60 marks.
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