Secondary 3 Chemistry Atomic Structure Bonding Quiz

Secondary 3 Chemistry Quiz - Atomic Structure Bonding (Answer Key)

Total Marks: 40

Section A: Multiple Choice Questions (10 marks)

1. D — F⁻ has 9 protons and 10 electrons (2.8), same as Ne (2.8). Na⁺ and Mg²⁺ also have 10 electrons, but F⁻ is the only anion listed. All three are isoelectronic with Ne. [1]

2. A — Electronic configuration 2.8.5 means 3 electron shells (Period 3) and 5 valence electrons (Group V). [1]

3. B — Mg atom: 2.8.2. Mg²⁺ loses 2 electrons from outer shell → 2.8. [1]

4. B — Y has 7 valence electrons (Group VII), forms Y⁻. Ca (Group II) forms Ca²⁺. Formula: CaY₂. [1]

5. C — NH₄Cl: NH₄⁺ has covalent bonds (N-H, one dative) and ionic bond between NH₄⁺ and Cl⁻. [1]

6. C — Substance R (3550 °C) matches diamond (giant covalent). P and S are low (simple molecular). Q (801 °C) is ionic (NaCl). [1]

7. B — Isotopes: same protons, different neutrons. Same electronic configuration → similar chemical properties. [1]

8. A — 12 protons = Mg. 10 electrons = 2+ charge. Mg²⁺. [1]

9. C — O (Group VI) and S (Group VI) both have 6 valence electrons. [1]

10. C — Diamond: giant covalent, each C bonded to 4 others tetrahedrally → very hard, very high melting point. [1]

Section B: Short Answer Questions (15 marks)

11. [3 marks — 1 mark per fully correct row]

Marking notes:

• ²³Na: protons = 11, neutrons = 23-11 = 12, electrons = 11 (neutral), config = 2.8.1
• ²⁴Mg²⁺: protons = 12, neutrons = 24-12 = 12, electrons = 12-2 = 10, config = 2.8
• ¹⁶O²⁻: protons = 8, neutrons = 16-8 = 8, electrons = 8+2 = 10, config = 2.8

12. (a) Ionic bonding / Electrovalent bonding [1]

(b) X (Group I) loses 1 valence electron to form X⁺ with stable noble gas configuration. Y (Group VII) gains 1 electron to form Y⁻ with stable noble gas configuration. The electrostatic attraction between oppositely charged ions forms the ionic bond. [2]

• 1 mark: X loses electron, Y gains electron
• 1 mark: Both achieve noble gas configuration / electrostatic attraction

(c) Solid state: Does not conduct electricity (ions fixed in lattice). Aqueous solution: Conducts electricity (ions mobile and free to move). [2]

• 1 mark: Correct conductivity for both states
• 1 mark: Correct explanation (mobile ions in solution, fixed in solid)

13. [2 marks]

   O = C = O
  (4)   (4)  (4)

Dot-and-cross diagram showing:

• Carbon: 4 valence electrons (●)
• Each oxygen: 6 valence electrons (×)
• Two double bonds (C=O), each with 2 shared pairs
• Each atom achieves octet (8 electrons in outer shell)

Marking:

• 1 mark: Correct sharing (two double bonds, 4 shared pairs total)
• 1 mark: Correct outer shell electrons (C: 8, each O: 8)

14. (a) Silicon dioxide has a giant covalent structure. Each silicon atom is covalently bonded to four oxygen atoms in a tetrahedral arrangement. Each oxygen atom is bonded to two silicon atoms. The network extends in three dimensions with strong Si-O covalent bonds throughout. [2]

• 1 mark: Giant covalent / tetrahedral / each Si bonded to 4 O
• 1 mark: 3D network / strong covalent bonds throughout

(b) Very high melting point because all atoms are held by strong covalent bonds in a giant 3D network. A large amount of energy is energy is needed to overcome these strong covalent bonds throughout the structure. [2]

• 1 mark: Strong covalent bonds / giant structure
• 1 mark: Large energy needed to break bonds

15. [4 marks — 1 mark each]

• Substance A: Giant ionic structure (high MP, conducts when molten/aq, soluble) — e.g., NaCl
• Substance B: Simple molecular structure (low MP, no conductivity, molecular) — e.g., ethanol
• Substance C: Giant covalent structure (very high MP, no conductivity, insoluble) — e.g., SiO₂/diamond
• Substance D: Giant metallic structure (high MP, conducts in solid and molten, insoluble) — e.g., Fe, Cu

Marking notes: Accept "ionic lattice", "molecular covalent", "macromolecular", "metallic lattice" etc.

Section C: Structured Questions (15 marks)

16. (a) Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons (different nucleon numbers). [1]

(b) Relative atomic mass = (35 × 75 + 37 × 25) / 100 = (2625 + 925) / 100 = 3550 / 100 = 35.5 [2]

• 1 mark: Correct formula/working
• 1 mark: Correct answer (35.5) to 1 d.p.

(c) Chemical properties depend on electronic configuration. Isotopes have the same number of protons and electrons, hence same electronic configuration. [1]

17. (a) 2Mg(s) + O₂(g) → 2MgO(s) [1]

• 1 mark: Correct formulae, balancing, and state symbols

(b) [3 marks]

Mg: 2.8.2        O: 2.6
Mg loses 2e⁻ → Mg²⁺ (2.8)
O gains 2e⁻ → O²⁻ (2.8)

Dot-and-cross:
  ××     ●●
×  Mg  ×  →  [Mg]²⁺  [O]²⁻
  ××     ●●

Description: Mg (●) transfers 2 electrons to O (×). Mg²⁺ has 2.8 config, O²⁻ has 2.8 config. Ionic lattice forms.

Marking:

• 1 mark: Correct electron transfer (Mg loses 2, O gains 2)
• 1 mark: Correct resulting ions with noble gas configurations
• 1 mark: Charges shown correctly (Mg²⁺, O²⁻)

(c) MgO has a giant ionic lattice structure with strong electrostatic forces of attraction between Mg²⁺ and O²⁻ ions. High charge density (2+ and 2-) leads to very strong ionic bonds. Large amount of energy needed to overcome these forces. [2]

• 1 mark: Giant ionic lattice / strong electrostatic forces
• 1 mark: High charge (2+/2-) → strong attraction / high energy needed

18. (a) Similarity: Both have covalent bonds between carbon atoms. Difference: Diamond has 3D tetrahedral network (each C bonded to 4 others); graphite has layered structure (each C bonded to 3 others in hexagonal layers with delocalised electrons). [2]

• 1 mark: Valid similarity (covalent C-C bonds)
• 1 mark: Valid difference (3D vs layered, 4 vs 3 bonds, delocalised electrons)

(b) Graphite: Each carbon bonded to 3 others, leaving 1 delocalised electron per carbon. These delocalised electrons are mobile between layers and can carry charge. Diamond: Each carbon bonded to 4 others, all 4 electrons used in covalent bonds, no free electrons/mobile charge carriers. [2]

• 1 mark: Graphite has delocalised/mobile electrons
• 1 mark: Diamond has no free electrons / all electrons in bonds

(c) Weak van der Waals forces between layers allow layers to slide over each other easily. [1]

19. (a) [3 marks]

    H
    |
H → N → H
    |
    H

Ammonium ion NH₄⁺:

• N (5 valence electrons) shares 3 electrons with 3 H atoms (3 normal covalent bonds)
• 4th H⁺ (no electrons) accepts a lone pair from N → dative covalent bond (arrow from N to H)
• N has formal charge +1, overall ion charge +1
• All 4 N-H bonds equivalent

Marking:

• 1 mark: 3 normal covalent bonds + 1 dative bond shown with arrow
• 1 mark: Correct electron distribution (N: 8 electrons, each H: 2 electrons)
• 1 mark: Correct charge (+1 on ion)

(b) Solid state: Ions (NH₄⁺ and Cl⁻) fixed in lattice, cannot move. Aqueous: Ions dissociate and are mobile, free to move and carry charge. [2]

• 1 mark: Solid — ions fixed
• 1 mark: Aqueous — ions mobile/dissociated

20. (a) [3 marks — 1 each]

• X: Giant metallic structure (shiny, high MP, conducts solid/molten, insoluble)
• Y: Giant ionic structure (crystalline, high MP, conducts molten/aq only, soluble)
• Z: Simple molecular structure (gas at room temp, very low MP, no conductivity, slightly soluble)

(b) Mobile ions (NH₄⁺ and Cl⁻ / cations and anions) [1]

(c) In metals, atoms are arranged in layers of positive ions in a sea of delocalised electrons. When force is applied, layers of ions can slide over each other without breaking the metallic bonds because the delocalised electrons hold the structure together non-directionally. [2]

• 1 mark: Layers of cations / sea of delocalised electrons
• 1 mark: Layers slide / non-directional bonding / bonds not broken

End of Answer Key