Secondary 3 Chemistry Atomic Structure Bonding Quiz

Secondary 3 Chemistry Quiz - Atomic Structure Bonding

Answer Key

Section A: Multiple Choice

1. B [2]
Reasoning: An atom with 12 protons has 12 electrons. The electronic configuration fills up as 2, 8, 2 (first shell holds 2, second holds 8, remaining 2 go to third shell).
Common mistake: Students may choose A (2, 8, 1) if they miscount, or C (2, 8, 8) if they confuse it with a different element.

2. C [2]
Reasoning: An electron has a negative charge (−1) and a relative mass of approximately 1/1840 (negligible compared to protons and neutrons).
Common mistake: Students may select proton or neutron, confusing charge or mass.

3. C [2]
Reasoning: NaCl is an ionic compound with a giant ionic lattice. The strong electrostatic forces of attraction between oppositely charged Na⁺ and Cl⁻ ions require a large amount of energy to overcome, resulting in a high melting point.
Common mistake: Students may choose B (giant covalent) if they confuse ionic and covalent structures.

4. B [2]
Reasoning: Ionic bonds form between metals and non-metals. Sodium (Na) is a metal and chlorine (Cl) is a non-metal. The other options involve only non-metals, which form covalent bonds.
Common mistake: Students may choose C (HCl) thinking it is ionic, but HCl is a covalent molecule.

5. D [2]
Reasoning: Isotopes are atoms of the same element (same proton number) with different numbers of neutrons, and therefore different mass numbers.
Common mistake: Students may choose A or C, confusing proton number with mass number.

Section B: Short Answer

6. Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons. [2]
Marking: 1 mark for "same element/same proton number"; 1 mark for "different number of neutrons."
Common mistake: Students may say "same mass number" instead of "same proton number."

7.
(a) 17 [1]
Reasoning: In a neutral atom, number of electrons = number of protons = 17.

(b) 18 [1]
Reasoning: Number of neutrons = mass number − proton number = 35 − 17 = 18.

8. [3 — 1 mark per correct cell]

9.
(a) Ionic bonding [1]
(b) Covalent bonding [1]
(c) Metallic bonding [1]
Common mistake: Students may say "intermolecular forces" for water instead of covalent bonding (the question asks about bonding within the molecule).

10. [3]
In solid NaCl, the ions (Na⁺ and Cl⁻) are held in fixed positions in the giant ionic lattice and are not free to move, so they cannot carry charge and the solid does not conduct electricity. [1]
When NaCl is melted (molten), the ions are free to move [1] and can carry electrical charge through the liquid, so molten NaCl conducts electricity. [1]
Common mistake: Students may say "electrons carry the charge" — in ionic compounds, it is ions that carry the charge, not electrons.

11. [3]
Methane (CH₄): Carbon has 4 outer electrons; each hydrogen has 1 outer electron. Carbon shares one pair of electrons with each of the four hydrogen atoms, forming four single covalent bonds.

The dot-and-cross diagram should show:

• C at the centre with 4 electrons (e.g., as dots ×)
• 4 H atoms around C, each with 1 electron (e.g., as crosses •)
• Each C–H bond shown as a shared pair (one dot + one cross)

Marking: 1 mark for correct number of bonds (4); 1 mark for showing shared pairs correctly; 1 mark for showing outer shell electrons only (no inner shells).

12.
(a) Group 6 [1]
Reasoning: The number of outer electrons = 6, which corresponds to Group 6.

(b) Period 3 [1]
Reasoning: The number of electron shells = 3 (2, 8, 6), so it is in Period 3.

(c) Non-metal [1] because it has 6 outer electrons and tends to gain 2 electrons to achieve a stable noble gas configuration [1].
Common mistake: Students may say "metal" if they confuse electron configuration trends.

13. [4]
Potassium (K) has electronic configuration 2, 8, 8, 1. It loses 1 electron to form K⁺ with configuration 2, 8, 8. [1]
Fluorine (F) has electronic configuration 2, 7. It gains 1 electron to form F⁻ with configuration 2, 8. [1]
The transfer of an electron from potassium to fluorine results in the formation of oppositely charged ions, K⁺ and F⁻. [1]
The strong electrostatic force of attraction between the oppositely charged ions forms the ionic bond. [1]
Common mistake: Students may forget to state the electronic configurations of the ions formed, or may not mention electrostatic attraction.

Section C: Structured Response

14.
(a) R and S are isotopes [1] because they have the same proton number (17) but different mass numbers (35 and 37) [1].
Common mistake: Students may choose P and Q because they are close in proton number, but they are different elements.

(b) P forms a positive ion with configuration 2, 8 [1]. P has 11 electrons (configuration 2, 8, 1); it loses 1 electron to form P⁺ with configuration 2, 8 [1].
Common mistake: Students may choose Q (which loses 2 electrons to form Q²⁺ with configuration 2, 8) — this is also correct. Accept Q as well with valid reasoning.

(c) PR [1]
Reasoning: P (proton number 11) is in Group 1 and forms P⁺. R (proton number 17) is in Group 7 and forms R⁻. The charges balance in a 1:1 ratio, so the formula is PR [2].
Common mistake: Students may write P₂R or PR₂ if they confuse the charges.

15.
(a) Giant ionic lattice [1]
Reasoning: The alternating + and − ions in a regular arrangement indicate a giant ionic lattice.

(b) The structure has a high melting point [1] because there are strong electrostatic forces of attraction between the oppositely charged ions in all directions in the lattice, requiring a large amount of energy to overcome [1].

(c) It is brittle / hard / soluble in water [1]. When a force is applied, ions of like charge come next to each other and repel, causing the lattice to shatter (for brittle) [1].
Accept any valid property with correct explanation.

16.
(a) In diamond, each carbon atom is covalently bonded to four other carbon atoms in a tetrahedral arrangement, forming a giant covalent (macromolecular) structure [2].
Marking: 1 mark for "each carbon bonded to 4 others"; 1 mark for "giant covalent structure."

(b) In graphite, each carbon atom is bonded to three others in layers, leaving one delocalised electron per carbon atom [1]. These delocalised electrons are free to move between the layers and can carry electrical charge, so graphite conducts electricity [1]. In diamond, all four outer electrons of each carbon atom are involved in covalent bonds, so there are no free electrons or ions to carry charge, and diamond does not conduct electricity [1].

(c) Graphite is used as a lubricant / in pencil leads [1].
Reasoning: The layers in graphite can slide over each other because the forces between layers are weak.

17.
(a) Substance A has an ionic structure [1]. It has a high melting point (801 °C), does not conduct electricity as a solid, but does conduct when molten — these are characteristic properties of ionic compounds [1].

(b) Substance B has a simple molecular structure [1]. It has a very low melting point (−7 °C) and does not conduct electricity in any state, indicating weak intermolecular forces and no free ions or electrons [1].

(c) Substance D has a metallic structure [1]. It conducts electricity in both solid and molten states due to the presence of delocalised electrons, and has a moderately high melting point (660 °C) [1].

18.
(a) 2, 8, 3 [1]
Reasoning: 13 electrons fill as 2, 8, 3.

(b) [3]
Al₂O₃ — The dot-and-cross diagram should show:

• Two Al³⁺ ions (each with lost 3 outer electrons, shown as empty outer shell or 2, 8 configuration)
• Three O²⁻ ions (each with gained 2 electrons, shown as 2, 8 configuration with 8 outer electrons)
• The diagram should show electron transfer: each Al loses 3 electrons (total 6), each O gains 2 electrons (total 6)

Marking: 1 mark for correct ions; 1 mark for correct electron transfer; 1 mark for correct formula Al₂O₃.

(c) Ionic bonding [1]. It has a high melting point because of the strong electrostatic forces of attraction between the Al³⁺ and O²⁻ ions in the giant ionic lattice [1].

19.
(a)
J: Giant ionic structure [1]
K: Giant metallic structure [1]
L: Simple molecular structure [1]
Reasoning: J has a high melting point and conducts electricity only when dissolved (ions free in solution). K conducts in all states (delocalised electrons). L has a very low melting point and does not conduct (weak intermolecular forces, no ions).

(b) Solid J does not conduct electricity because the ions are held in fixed positions in the ionic lattice and cannot move to carry charge [1]. When J is dissolved in water, the ions are free to move and can carry electrical charge, so aqueous J conducts electricity [1].

20.
(a) Element M belongs to Group I [1]. There is a very large jump between the 1st and 2nd ionisation energies (520 → 7300 kJ/mol), indicating that the first electron is easily removed (outer shell) but the second electron is removed from a stable inner shell. This means M has 1 outer electron, so it is in Group I [1].

(b) Element N belongs to Group II [1]. There is a large jump between the 2nd and 3rd ionisation energies (1760 → 14900 kJ/mol), indicating that the first two electrons are removed from the outer shell, but the third electron is removed from a stable inner shell. This means N has 2 outer electrons, so it is in Group II [1].

(c) MCl [1]
Reasoning: M is in Group I (forms M⁺) and chlorine forms Cl⁻. The formula is MCl.

End of Answer Key