Secondary 3 Chemistry Periodic Table Quiz

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]

END OF ANSWER KEY