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Secondary 3 Chemistry Periodic Table Quiz
Free Sec 3 Chemistry Periodic Table quiz with questions, answers, and O Level-style practice for Singapore students preparing for school assessments.
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Questions
Secondary 3 Chemistry Quiz - Periodic Table
Name: _________________________ Class: _________________________
Date: _________________________ Score: _______ / 60
Duration: 40 minutes
Total Marks: 60
Instructions:
- Answer ALL questions.
- Write your answers in the spaces provided.
- For questions involving calculations, show all working clearly.
- Marks are awarded for correct chemical equations, reasoning, and working.
Section A: Multiple Choice (Questions 1–5)
Choose the best answer. Each question carries 2 marks.
1. Which statement correctly describes a periodic trend in the Periodic Table? <image_placeholder> id: Q1-fig1 type: diagram linked_question: Q1 description: A simplified Periodic Table outline showing three periods (Periods 2–4) and Groups I, II, VII, 0, with arrows indicating trends labels: Periods 2, 3, 4; Groups I, II, VII, 0; arrows labeled "increasing atomic radius", "increasing electronegativity", "increasing ionization energy" values: None must_show: Direction of each trend arrow, group and period labels clearly visible, simplified block format without element symbols </image_placeholder>
A. Atomic radius decreases down a group because more electron shells are added. B. Electronegativity increases across a period from left to right. C. Ionization energy decreases across a period from left to right. D. Group 0 elements have the lowest ionization energies in their respective periods.
Answer: _________________________ [2]
2. Element X is in Period 3, Group III. Which statement about X is correct?
A. X forms an ion with a 3– charge. B. X has 3 electron shells and 3 valence electrons. C. X is a non-metal with a giant covalent structure. D. X is more reactive than sodium in the same period.
Answer: _________________________ [2]
3. The table below shows properties of four elements P, Q, R, and S.
| Property | P | Q | R | S |
|---|---|---|---|---|
| Melting point / °C | 98 | 1535 | -7 | 114 |
| Electrical conductivity | Good | Good | Poor | Poor |
| Density / g cm⁻³ | 0.97 | 7.87 | 1.58 | 2.07 |
| Reaction with water | Vigorous | No reaction | Violent | No reaction |
Which element is most likely to be a transition metal?
A. P B. Q C. R D. S
Answer: _________________________ [2]
4. Which pair of elements would form a compound with the greatest ionic character?
A. Lithium and fluorine B. Sodium and chlorine C. Potassium and bromine D. Magnesium and oxygen
Answer: _________________________ [2]
5. In which group of the Periodic Table do elements exist as simple diatomic molecules at room temperature and pressure?
A. Group I B. Group II C. Group VII D. Group 0
Answer: _________________________ [2]
Section B: Short Answer (Questions 6–12)
Answer in the spaces provided. Marks for each question are shown in brackets.
6. (a) Define the term ionization energy. [1]
(b) Explain why potassium has a lower first ionization energy than sodium. [2]
(c) Write the equation for the second ionization energy of magnesium, including state symbols. [2]
Total: [5]
7. The Periodic Table below shows part of Period 3.
<image_placeholder> id: Q7-fig1 type: diagram linked_question: Q7 description: Extract of Period 3 of the Periodic Table showing elements Na, Mg, Al, Si, P, S, Cl, Ar in boxes with their symbols labels: Na (Z=11), Mg (Z=12), Al (Z=13), Si (Z=14), P (Z=15), S (Z=16), Cl (Z=17), Ar (Z=18) values: Atomic numbers 11–18 must_show: All 8 element symbols in order, atomic numbers below symbols, period labeled "Period 3" to left </image_placeholder>
(a) Identify the element with the highest melting point in Period 3. Explain your answer in terms of structure and bonding. [3]
(b) Explain why the electrical conductivity of silicon is lower than that of aluminium. [2]
Total: [5]
8. Francium (Fr) is the element below caesium in Group I of the Periodic Table.
(a) Predict two physical properties of francium. [2]
(b) Write a balanced equation for the reaction of francium with water. Include state symbols. [2]
(c) Explain why francium is more reactive than caesium. [2]
Total: [6]
9. Astatine (At) is the element below iodine in Group VII.
(a) Predict the physical state of astatine at room temperature. Explain your reasoning. [2]
(b) Write the formula of astatide ion. [1]
(c) Predict whether astatine will displace bromide ions from potassium bromide solution. Explain your answer. [2]
Total: [5]
10. The graph below shows the first ionization energies of the elements in Period 2.
<image_placeholder> id: Q10-fig1 type: graph linked_question: Q10 description: Line graph of first ionization energy (y-axis, kJ/mol) against atomic number for Period 2 elements (x-axis, Z=3 to 10) labels: x-axis: Atomic number (3 to 10), with tick marks at Li(3), Be(4), B(5), C(6), N(7), O(8), F(9), Ne(10) y-axis: First ionization energy / kJ mol⁻¹, approximate scale 500–2500 values: Li ~520, Be ~900, B ~801, C ~1086, N ~1402, O ~1314, F ~1681, Ne ~2081 (general trend with dip at B and O) must_show: All data points clearly marked, line connecting them showing general upward trend with characteristic dips at B and O, axes labeled with units, element symbols at relevant points </image_placeholder>
(a) Explain the general increase in first ionization energy across Period 2 from lithium to fluorine. [2]
(b) Explain why boron has a lower first ionization energy than beryllium, despite the general trend. [2]
(c) Explain why neon has the highest first ionization energy in Period 2. [2]
Total: [6]
11. The table below shows some properties of elements in Group II.
| Element | Atomic radius / nm | Ionic radius of M²⁺ / nm | First ionization energy / kJ mol⁻¹ |
|---|---|---|---|
| Be | 0.112 | 0.045 | 900 |
| Mg | 0.160 | 0.072 | 738 |
| Ca | 0.197 | 0.100 | 590 |
| Sr | 0.215 | 0.118 | 550 |
| Ba | 0.222 | 0.135 | 503 |
(a) Explain why the atomic radius increases down Group II. [2]
(b) Explain why the ionic radius of M²⁺ is always smaller than the atomic radius of the corresponding element. [2]
(c) Explain why the first ionization energy decreases down Group II. [2]
Total: [6]
12. Chlorine, bromine, and iodine are halogens in Group VII.
(a) Describe and explain the trend in boiling point down Group VII. [3]
(b) When chlorine gas is bubbled through sodium bromide solution, the solution turns orange-brown.
(i) Write the ionic equation for this reaction. [2]
(ii) Explain why this reaction occurs with reference to oxidizing ability. [2]
Total: [7]
Section C: Structured Response (Questions 13–20)
Answer in the spaces provided. Show all working and reasoning.
13. Element T has the following properties:
- Melting point: 1414 °C
- Boiling point: 3265 °C
- Electrical conductivity: Moderate (semiconductor)
- Forms an oxide with formula TO₂
- In Period 3
(a) Identify element T. [1]
(b) Draw a diagram to show the bonding in the oxide TO₂. Show outer shell electrons only. [2]
(c) Explain why TO₂ has a high melting point. [2]
(d) Explain why pure TO₂ does not conduct electricity, but conducts when heated or doped. [3]
Total: [8]
14. The table below shows some information about elements W, X, Y, and Z.
| Element | Proton number | Relative atomic mass | Density / g cm⁻³ |
|---|---|---|---|
| W | 19 | 39.1 | 0.86 |
| X | 20 | 40.1 | 1.55 |
| Y | 35 | 79.9 | 3.12 |
| Z | 53 | 126.9 | 4.93 |
(a) Identify which elements belong to the same group in the Periodic Table. Explain your reasoning. [2]
(b) Element W reacts with element Y to form a compound.
(i) Predict the formula of this compound. [1]
(ii) Draw a dot-and-cross diagram to show the bonding in this compound. Show outer shell electrons only. [2]
(iii) Explain whether the compound conducts electricity when molten or in aqueous solution. [2]
Total: [7]
15. Consider the elements sodium (Na), magnesium (Mg), aluminium (Al), and silicon (Si) in Period 3.
(a) Compare the structure and bonding of sodium, magnesium, and aluminium. Explain how this affects their electrical conductivity. [4]
(b) Explain why aluminium has a higher melting point than magnesium. [3]
Total: [7]
16. The diagram below shows a section of the Periodic Table with some elements labeled A–F.
<image_placeholder> id: Q16-fig1 type: diagram linked_question: Q16 description: Section of Periodic Table showing 8 elements in two periods (Periods 2 and 3) with labels A-F assigned to specific positions labels: Period 2: A (Group I), B (Group II), C (Group VI) Period 3: D (Group I), E (Group II), F (Group VII), plus unlabelled elements in Groups III-V and 0 values: None specifically, positions define relationships must_show: Grid layout with period numbers (2, 3) and group numbers (I, II, III, IV, V, VI, VII, 0) clearly indicated; letters A-F placed in correct group/period positions with surrounding elements shown as empty boxes or generic symbols </image_placeholder>
(a) Which element is the strongest reducing agent? Explain your answer. [2]
(b) Write the formula of the compound formed between B and F. [1]
(c) Compare the atomic radius of C and F. Explain your answer. [2]
(d) Predict whether the oxide of A is acidic, basic, or amphoteric. Write an equation to support your answer. [2]
Total: [7]
17. Gallium (Ga) is in Group III, Period 4 of the Periodic Table.
(a) Write the electron configuration of a gallium atom. [1]
(b) Predict the charge of the gallium ion. Explain your answer. [2]
(c) Aluminium is in the same group as gallium. Compare the reactivity of aluminium and gallium with dilute hydrochloric acid. Explain your answer. [3]
Total: [6]
18. The table below shows the melting points and boiling points of the noble gases.
| Noble gas | Melting point / K | Boiling point / K |
|---|---|---|
| He | 1 | 4 |
| Ne | 25 | 27 |
| Ar | 84 | 87 |
| Kr | 116 | 120 |
| Xe | 161 | 165 |
(a) Explain why noble gases have low melting and boiling points. [2]
(b) Explain the trend in boiling point from helium to xenon. [2]
(c) Helium is used in balloons rather than hydrogen, despite hydrogen being less dense. Suggest a reason related to the properties of the elements in the Periodic Table. [1]
Total: [5]
19. The graph below shows the atomic radii of the elements in Period 3.
<image_placeholder> id: Q19-fig1 type: graph linked_question: Q19 description: Bar chart or line graph showing atomic radius decreasing across Period 3 from Na to Ar labels: x-axis: Elements Na, Mg, Al, Si, P, S, Cl, Ar; y-axis: Atomic radius / nm, scale approximately 0.1 to 0.2 nm values: Na ~0.186, Mg ~0.160, Al ~0.143, Si ~0.117, P ~0.110, S ~0.104, Cl ~0.099, Ar ~0.071 nm (decreasing trend) must_show: Continuous decrease across period, clear element labels, numerical values or clear trend, Na has largest radius, Ar has smallest, y-axis labeled with units </image_placeholder>
(a) Explain why atomic radius decreases across Period 3 from sodium to argon. [3]
(b) The ionic radius of Na⁺ is 0.102 nm, while the atomic radius of Na is 0.186 nm. Explain this difference. [2]
(c) The ionic radius of Cl⁻ is 0.181 nm, while the atomic radius of Cl is 0.099 nm. Explain this difference. [2]
Total: [7]
20. Scandium (Sc) is a transition metal with proton number 21. It forms a stable ion Sc³⁺.
(a) Write the electron configuration of: (i) A scandium atom: _________________________ [1] (ii) A Sc³⁺ ion: _________________________ [1]
(b) Explain why Sc³⁺ is more stable than Sc²⁺. [2]
(c) Transition metals show different properties from Group I and Group II metals. Give two properties of scandium (or transition metals in general) that are different from calcium in Group II. [2]
(d) Scandium oxide has the formula Sc₂O₃. Calculate the percentage by mass of oxygen in scandium oxide. [2]
Total: [8]
END OF QUIZ
Answers
Secondary 3 Chemistry Quiz Answers - Periodic Table
Total Marks: 60
Section A: Multiple Choice
1. Answer: B [2 marks]
- Correct: Electronegativity increases across a period from left to right due to increasing nuclear charge with the same shielding, so electrons are attracted more strongly.
- Why A is wrong: Atomic radius increases down a group (more electron shells), not decreases. This statement has the trend direction backwards for the justification.
- Why C is wrong: Ionization energy increases across a period (general trend), not decreases.
- Why D is wrong: Group 0 elements have the highest ionization energies in their periods due to stable full outer shell; they are very unreactive.
2. Answer: B [2 marks]
- Correct: Period 3 means 3 electron shells; Group III means 3 valence electrons.
- Why A is wrong: Group III metals form 3+ cations (lose 3 electrons), not 3– anions.
- Why C is wrong: Group III elements are metals (aluminium), not non-metals with giant covalent structures.
- Why D is wrong: Aluminium is less reactive than sodium; reactivity of metals decreases across a period.
3. Answer: B [2 marks]
- Correct: Q has high melting point, good conductivity, high density, and does not react with water — characteristic properties of a transition metal (like iron).
- Why A is wrong: P is too light and reacts vigorously with water — alkali metal.
- Why C is wrong: R has very low melting point (-7°C) and density — not a metal.
- Why D is wrong: S is non-conductive and doesn't react with water — non-metal (like iodine).
4. Answer: A [2 marks]
- Correct: Greatest ionic character comes from the largest electronegativity difference. Li (0.98) to F (3.98) gives difference ≈ 3.0; this is the largest gap among options using Pauling scale values. Also, smaller ions mean stronger ionic bonding character.
- Working: Electronegativity differences: Li-F ≈ 3.0, Na-Cl ≈ 2.2, K-Br ≈ 2.0, Mg-O ≈ 2.3. Lithium fluoride has the greatest difference.
5. Answer: C [2 marks]
- Correct: Group VII elements (halogens: F₂, Cl₂, Br₂, I₂) exist as simple diatomic molecules with covalent bonding between two atoms.
- Why A is wrong: Group I metals have giant metallic structures.
- Why B is wrong: Group II metals have giant metallic structures.
- Why D is wrong: Group 0 elements exist as monatomic gases (single atoms).
Section B: Short Answer
6. (a) Ionization energy is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous positive ions. [1]
(b) Potassium has a lower first ionization energy than sodium because:
- Potassium is below sodium in Group I, so it has one more electron shell (4 shells vs 3 shells) [1]
- The outer electron in potassium is further from the nucleus and experiences more shielding, so the attraction between the nucleus and the valence electron is weaker [1]
(c) Mg⁺(g) → Mg²⁺(g) + e⁻ [2]
- Correct equation with Mg⁺ on left and Mg²⁺ + e⁻ on right [1]
- Correct state symbols (all gaseous) [1]
Total: [5]
7. (a) Silicon (Si) has the highest melting point in Period 3. [1]
- Silicon has a giant covalent structure where each silicon atom is covalently bonded to four other silicon atoms in a tetrahedral network [1]
- Many strong covalent bonds must be broken to melt silicon, requiring a large amount of energy [1]
(b) Silicon's electrical conductivity is lower than aluminium because:
- Silicon is a semiconductor with a giant covalent structure; it has no free electrons at room temperature (electrons are held in covalent bonds) [1]
- Aluminium is a metal with a giant metallic structure containing a 'sea' of delocalized electrons that are free to move and carry charge [1]
Total: [5]
8. (a) Two physical properties of francium: Any two from:
- Soft, silvery metal [1]
- Low melting point (lower than caesium, ~28°C) [1]
- Low density (higher than caesium due to relativistic effects but still low, ~2 g cm⁻³) [1]
- Good conductor of heat and electricity [1]
(Accept any 2 reasonable predictions based on Group I trends)
(b) 2Fr(s) + 2H₂O(l) → 2FrOH(aq) + H₂(g) [2]
- Correct formulae and balanced [1]
- Correct state symbols [1]
(c) Francium is more reactive than caesium because:
- Francium has one more electron shell than caesium (6 vs 5) [1]
- The outer electron is further from the nucleus with more shielding, so it is lost more easily; weaker attraction between nucleus and valence electron [1]
Total: [6]
9. (a) Astatine is a solid at room temperature. [1]
- Down Group VII, melting/boiling points increase due to stronger intermolecular forces (more electrons, larger electron cloud, stronger instantaneous dipole-induced dipole/London dispersion forces) [1]
(b) At⁻ [1]
(c) No, astatine will not displace bromide ions. [1]
- Reactivity of halogens decreases down the group; astatine is below iodine so is the weakest oxidizing agent/weakest halogen. It cannot oxidize bromide ions to bromine because it is less reactive/less easily reduced than bromine. [1]
Total: [5]
10. (a) General increase across Period 2 because:
- Nuclear charge increases (proton number increases from 3 to 9) [1]
- Electrons are added to the same shell, so shielding remains relatively constant; the increased nuclear pull makes it harder to remove an electron [1]
(b) Boron has a lower first ionization energy than beryllium because:
- Beryllium has a stable full 2s subshell: 1s² 2s² [1]
- Boron's electron configuration is 1s² 2s² 2p¹; the 2p electron is slightly higher in energy and is shielded by the 2s electrons, so it is easier to remove [1]
(c) Neon has the highest first ionization energy because:
- It has the highest nuclear charge in Period 2 (10 protons) [1]
- It has a stable full outer shell (1s² 2s² 2p⁶) — removing an electron would disrupt this stable noble gas configuration, requiring the most energy [1]
Total: [6]
11. (a) Atomic radius increases down Group II because:
- Each successive element has one more occupied electron shell [1]
- The outer electrons are further from the nucleus despite increasing nuclear charge, so the atomic radius increases [1]
(b) Ionic radius of M²⁺ is smaller than atomic radius because:
- The atom loses two electrons to form the 2+ ion, so the outermost shell is removed [1]
- There are fewer electron-electron repulsions, and the remaining electrons are pulled closer to the nucleus by the same nuclear charge [1]
(c) First ionization energy decreases down Group II because:
- The outer electron is further from the nucleus as you go down the group (more electron shells) [1]
- There is more shielding from inner shells, so the nuclear attraction on the valence electron is weaker, making it easier to remove [1]
Total: [6]
12. (a) Boiling point increases down Group VII because:
- The halogens exist as simple diatomic molecules with weak intermolecular forces (instantaneous dipole-induced dipole forces/London dispersion forces) [1]
- Down the group, the number of electrons increases, making the electron cloud more polarizable [1]
- Stronger intermolecular forces require more energy to overcome, so boiling point increases [1]
(b) (i) Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq) [2]
- Correct formulae and products [1]
- Balanced with state symbols [1]
(ii) Chlorine is a stronger oxidizing agent than bromine (or chlorine has a greater tendency to gain electrons/be reduced) [1]
- Chlorine can oxidize bromide ions to bromine by accepting electrons from them; chlorine is more reactive/has a higher standard electrode potential [1]
Total: [7]
Section C: Structured Response
13. (a) Silicon (Si) [1]
(b) Dot-and-cross diagram of SiO₂:
- Central silicon atom with 4 bonding pairs (2 double bonds or 4 single bonds to two oxygen atoms)
- Each oxygen has 2 lone pairs
- All atoms achieve noble gas configuration [2] (Students should show Si with 4 outer electrons, each O with 6, forming two double bonds or showing 4 single bonds in network — either representation acceptable)
(c) TO₂ (SiO₂) has a high melting point because:
- It has a giant covalent structure with strong covalent bonds between silicon and oxygen atoms [1]
- These bonds extend throughout the entire structure in a network; many strong bonds must be broken to melt it, requiring much energy [1]
(d) Pure TO₂ does not conduct electricity because:
- All electrons are involved in covalent bonds; there are no free electrons or mobile ions to carry charge [1]
When heated or doped:
- At higher temperatures, some electrons gain enough energy to break free from bonds and move [1]
- When doped with impurities (e.g., phosphorus or boron), extra electrons or 'holes' are introduced that can move and carry charge, creating an n-type or p-type semiconductor [1]
Total: [8]
14. (a) Elements W and D belong to Group I; elements Y and Z belong to Group VII. [2]
- W (19, K) and D (35 — wait, recheck: W is Z=19, D would need comparison. Actually Y (35) and Z (53) are both halogens (Group VII) with similar outer electron configuration (7 valence electrons), showing pattern of increasing mass and density. W (19) is potassium, X (20) is calcium.
- Correct grouping: Y and Z are in Group VII (both have 7 valence electrons, similar properties — non-metals, diatomic, similar density trends) [1]
- W and X are in Period 4 (consecutive proton numbers), not same group. Actually W (19) and D is not given properly. Looking at pattern: W=K (Group I), X=Ca (Group II), Y=Br (Group VII), Z=I (Group VII).
- Y and Z are in Group VII — both have 7 valence electrons; show trend of increasing density and atomic mass down the group [1]
(b) (i) KY or KBr [1] — Potassium bromide
(ii) Dot-and-cross diagram:
- Potassium loses 1 electron to K⁺
- Bromine gains 1 electron to Br⁻
- Ionic structure with K⁺ and Br⁻ ions shown with correct charges and electron transfer indicated [2]
(iii) The compound conducts electricity when molten or in aqueous solution because:
- In solid state: ions are fixed in lattice, cannot move [1]
- When molten or dissolved: ionic bonds broken, ions are free to move and carry electric charge [1]
Total: [7]
15. (a) Sodium, magnesium, and aluminium all have giant metallic structures [1]
- Positive metal ions arranged in a lattice with delocalized electrons forming a 'sea' [1]
- Increasing number of delocalized electrons per atom: Na (1), Mg (2), Al (3) [1]
- This causes increasing metallic bond strength and therefore increasing electrical conductivity from Na to Mg to Al; more mobile charge carriers per unit volume [1]
(b) Aluminium has a higher melting point than magnesium because:
- Both have giant metallic structures [1]
- Al³⁺ ions have a higher charge than Mg²⁺ ions, and the ionic radius of Al³⁺ is smaller than Mg²⁺ [1]
- This results in stronger electrostatic attraction between the positive ions and the delocalized electrons, requiring more energy to break the metallic bonds [1]
Total: [7]
16. (a) Element A (or more precisely, D if in Period 3) — actually A is in Period 2 Group I, D is Period 3 Group I. Element D is the strongest reducing agent. [1]
- It is lowest in Group I shown (Period 3 vs Period 2), with the most electron shells, weakest nuclear attraction on valence electron, so loses electron most readily [1]
(b) BF₂ — B is Group II (Mg), F is Group VII (Cl), so MgCl₂ [1]
(c) F has a smaller atomic radius than C. [1]
- F (Group VII, Period 3) has more protons/greater nuclear charge than C (Group VI, Period 2), and F has its electrons in a higher shell. Actually: C is in Period 2 Group VI (oxygen family), F is in Period 3 Group VII. Need to compare: C=O/S? Actually C is Group VI Period 2. F is Group VII Period 3.
- F is in Period 3 so has an extra electron shell compared to C in Period 2, so F would be LARGER. Wait — let me re-read: C is in Period 2 Group VI, F is in Period 3 Group VII.
- Correction: C (Period 2) has smaller radius than F (Period 3). The increase down a group (extra shell) dominates over the increase across a period. So C has smaller radius than F. [1]
- C is in Period 2 with 2 shells; F is in Period 3 with 3 shells. The additional electron shell in F more than compensates for the increased nuclear charge. [1]
(d) The oxide of A is basic. [1]
- A is a Group I metal (alkali metal). Example equation: Na₂O + H₂O → 2NaOH or 4Na + O₂ → 2Na₂O [1]
Total: [7]
17. (a) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p¹ or [Ar] 3d¹⁰ 4s² 4p¹ [1]
(b) Ga³⁺ [1]
- Gallium has 3 valence electrons (Group III); it tends to lose these 3 electrons to achieve a stable noble gas configuration (like argon) [1]
(c) Aluminium is more reactive than gallium (with dilute HCl). OR Gallium is slightly more reactive due to d-block contraction — actually Group III trend is complex.
- Standard expected answer: Aluminium appears less reactive due to protective oxide layer; if oxide removed, reactivity should increase down group (opposite to observed for Al).
- More precise: Aluminium has a protective oxide layer that makes it appear unreactive; once this is removed or in certain conditions, gallium may show different behavior. For Sec 3 level: Gallium is more reactive than aluminium as the outer electrons are further from nucleus with more shielding [2]
- However, aluminium's apparent unreactivity is due to its protective oxide coating [1]
Note: This is a challenging question. Accept either trend with valid explanation. Mark for reasoning quality.
Total: [6]
18. (a) Noble gases have low melting and boiling points because:
- They exist as individual atoms with weak intermolecular forces (instantaneous dipole-induced dipole forces/London dispersion forces) between them [1]
- Very little energy is needed to overcome these weak forces [1]
(b) Boiling point increases from He to Xe because:
- The number of electrons increases down the group, making the electron cloud larger and more polarizable [1]
- Stronger instantaneous dipole-induced dipole forces result, requiring more energy to boil [1]
(c) Helium is unreactive/noble gas/inert, while hydrogen is flammable/explosive [1]
- Helium will not burn or react, making it safer for balloons despite being slightly less lifting capacity.
Total: [5]
19. (a) Atomic radius decreases across Period 3 because:
- Nuclear charge increases (number of protons increases from 11 to 18) [1]
- Electrons are added to the same principal quantum shell/clectron shell, so shielding is roughly constant [1]
- The increased nuclear attraction pulls the electron cloud closer to the nucleus, making the atom smaller [1]
(b) Na⁺ is smaller than Na because:
- Na loses its outermost (3rd) electron shell to form Na⁺ [1]
- The remaining 2 shells experience greater effective nuclear charge per electron, pulling remaining electrons closer; also fewer electron-electron repulsions [1]
(c) Cl⁻ is larger than Cl because:
- Cl gains an electron to form Cl⁻, increasing electron-electron repulsion [1]
- The extra electron reduces the effective nuclear charge per electron; the electron cloud expands [1]
Total: [7]
20. (a) (i) Scandium atom: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹ 4s² or [Ar] 3d¹ 4s² [1] (ii) Sc³⁺ ion: 1s² 2s² 2p⁶ 3s² 3p⁶ or [Ar] [1]
(b) Sc³⁺ is more stable than Sc²⁺ because:
- Sc³⁺ has the electron configuration of argon, a noble gas with a full and stable outer shell [1]
- The 4s² and 3d¹ electrons are all removed, leaving a stable octet in the outer shell (actually full n=2 and n=3 shells, argon configuration) [1]
(c) Two properties different from calcium:
- Scandium forms ions with variable oxidation states / Sc³⁺ is common but other states possible; Ca only forms Ca²⁺ [1]
- Scandium and its compounds are often colored / act as catalysts; calcium compounds are typically white and not catalytic [1]
- Scandium has higher melting point / higher density than calcium [1]
- Scandium forms complexes with ligands more readily [1] (Any 2 valid comparisons)
(d) Percentage by mass of oxygen in Sc₂O₃:
- Relative formula mass of Sc₂O₃ = 2(45) + 3(16) = 90 + 48 = 138 [1]
- Mass of oxygen = 48
- Percentage of oxygen = (48 / 138) × 100% = 34.8% (accept 34.78% or 35%) [1]
Total: [8]