🏆 Module 1 Quiz

❓ Multiple Choice — All 25

1. C — Pure water = compound (H₂O, fixed composition). Salt water = NaCl + H₂O, mixed at molecular level = homogeneous mixture.

2. B — Filter first (sand is an undissolved solid). Then evaporate water / crystallise to recover salt. Distillation could separate water from salt solution but crystallisation is more appropriate for solid recovery.

3. D — Chromatography: differential migration based on relative affinity for stationary vs mobile phase (solubility-based distribution). Not BP, not particle size, not density of phases.

4. A — BaSO₄ is essentially insoluble (Ksp ≈ 1×10⁻¹⁰) — precipitates quantitatively. Filter, dry, weigh → calculate SO₄²⁻ mass. Classic gravimetric analysis.

5. C — Ethanol (BP 78°C) and water (BP 100°C) are miscible liquids separated by fractional distillation exploiting the 22°C BP difference. Filtration doesn't apply (both liquids). Crystallisation doesn't separate two miscible liquids efficiently.

6. B — Homogeneous mixture: uniform at molecular scale, variable composition, separable by physical means. Pure substance: fixed composition, constant properties (fixed MP/BP), cannot be separated by physical means without chemical change.

7. D — Multiple spots on chromatogram → mixture. Blue (highest Rf = most soluble in mobile phase) + yellow (medium Rf) + green (low Rf, possibly incompletely separated). The green ink appears green because it's a blend of blue and yellow pigments. Each spot = a different component.

8. B — CH₄: covalent molecular, only dispersion forces → lowest MP (−182°C). H₂O: covalent molecular with H-bonds → MP 0°C. NaCl: ionic lattice → MP 801°C. Diamond: covalent network → MP >3500°C.

9. C — Copper: metal with delocalised electrons → conducts as solid AND in any state. NaCl: insulates as solid (ions fixed), conducts when molten/dissolved. Glucose: non-electrolyte, non-conductor in any state. Diamond: no free electrons, insulator.

10. A — NH₃: N is electronegative (3.0); N–H···N H-bonds form between molecules. PH₃: P is less electronegative (2.2); P–H bond not polar enough for H-bonding → only dispersion forces. H-bonds much stronger → higher BP for NH₃ despite lower molar mass.

11. D — –O–CO– is the ester linkage (–COO–): –O– from the diol, –CO– from the diacid. Both the –O– and –CO– appear in the repeat unit. This is a condensation polymerisation (ester linkage formed with H₂O released). Not amide (no N). Not addition (no C=C and byproduct present).

12. B — Both Cl₂ and Br₂ are non-polar → only dispersion forces. Br₂ has 70 electrons vs Cl₂'s 34 → larger, more polarisable electron cloud → stronger dispersion forces → higher BP (58°C for Br₂ vs −34°C for Cl₂) → Br₂ is liquid at room temperature (~20°C).

13. C — Solvent extraction: I₂ non-polar → hexane. KBr ionic → water. Two immiscible layers form; I₂ in hexane layer (upper/lower depending on density; hexane less dense → floats). Separate by separating funnel.

14. A — Metallic: electron sea allows lattice planes to slide (cations remain surrounded by electrons at all times). Ionic: slip shifts like charges into contact → repulsion → brittle fracture. Not about bond strength (ionic bonds are often stronger than metallic).

15. D — S–S cross-links = covalent bonds between chains = thermosetting. Cannot melt — covalent bonds don't break on heating. The repeat unit shows addition polymer (C–C backbone, –CH₂–CH(OH)–) but the S–S inter-chain links make it thermoset. This resembles vulcanised rubber.

16. B — All hydrocarbons are non-polar → only dispersion forces. Longer chain = more electrons = larger surface area = stronger dispersion forces = higher BP. Longer chains condense lower in the distillation tower where it's hotter; shorter chains (petrol fractions) exit near the top where it's cooler.

17. C — Most through with little deflection → mostly empty space. The rare back-scatter → nucleus is tiny and dense. The plum pudding model (diffuse positive charge) would give uniform slight deflection, not mostly straight-through with rare large deflections.

18. A — Ar = (69 × 0.601) + (71 × 0.399) = 41.469 + 28.329 = 69.798 ≈ 69.8. Pulled toward the more abundant isotope (⁶⁹Ga at 60%).

19. D — Cl (Z=17): 1s²2s²2p⁶3s²3p⁵. Cl⁻ gains 1 electron → 3p⁵+1 = 3p⁶ → 1s²2s²2p⁶3s²3p⁶ = [Ar] (18 electrons). A is the neutral Cl. B is incorrect (3p⁴). C is [Ar]+4s¹ = K⁺... no.

20. B — Down Group 2: each element has one more electron shell. New shells are further from the nucleus; more inner electrons provide more shielding → Z_eff increases slowly → atomic radius increases. A describes across-period trend (wrong direction). C is wrong (2 valence electrons throughout but radius still changes).

21. C — P: [Ne]3s²3p³ — three 3p orbitals each singly occupied. S: [Ne]3s²3p⁴ — one 3p orbital doubly occupied. The paired electrons in S repel each other → lower energy required to remove one → IE₁(S) < IE₁(P) despite higher Z. This is the Group 15/16 anomaly.

22. A — Large jump between IE₂ (1451) and IE₃ (7733): ratio ≈5.3. Two valence electrons → Group 2. IE₁(Mg)=738, IE₂(Mg)=1451 kJ/mol — exact match → Magnesium. Na would show jump after IE₁. Al would show jump after IE₃.

23. D — H–F: |Δχ|=1.8, at/above the ~1.7 ionic/polar threshold — very strongly polar or ionic. C–F: |Δχ|=1.4 → polar covalent. In both, F is δ− (most electronegative). H–F can donate H-bonds (H attached to F); C–F cannot (H is not on the F atom). Entirely correct statement in D.

24. B — Covalent network: very high MP (must break covalent bonds throughout entire 3D lattice), insoluble (no IMFs to displace, covalent bonds intact), does not conduct (no free electrons or ions), extremely hard. A = molecular solid. C = ionic compound. D = metal.

25. C — Going down Group 14: atomic radius increases, IE decreases, electronegativity decreases → tendency to lose electrons increases → metallic character increases. C has high IE and EN → forms covalent bonds exclusively. Si/Ge: intermediate (semiconductors/metalloids). Sn/Pb: metallic character dominates. All still have 4 valence electrons (A is wrong). Electronegativity DECREASES down the group (B is wrong). Number of shells INCREASES down the group (D is wrong).

📝 Short Answer Model Answers

Q26 (3 marks): (a) Fractional distillation (1 mark). Hexane (BP 69°C) and ethanol (BP 78°C) are miscible liquids with different boiling points. When the mixture is heated, the more volatile component (hexane, lower BP) vaporises preferentially and rises up the fractionating column, condenses, and is collected first. As temperature increases, ethanol vapour rises and is collected separately. The two components are separated based on their different vapour pressures (different BPs) (1 mark). (b) Filtration separates solids from liquids based on particle size (solids are trapped by the filter paper). Both hexane and ethanol are liquids — neither is a solid — so there are no particles to filter. Both liquids would pass through the filter paper without separation (1 mark).

Q27 (4 marks): Oxygen (Z=8): config [He]2s²2p⁴, χ=3.4. Sulfur (Z=16): config [Ne]3s²3p⁴, χ=2.6. The O–H bond: |Δχ| = 3.4−2.2 = 1.2 → strongly polar (1 mark). Oxygen's high electronegativity (Period 2, high Z_eff, small atom) means the O–H bond creates a large δ+ on H and δ− on O. Crucially, O is electronegative enough (χ ≥ 3.0 needed) and small enough for strong O–H···O hydrogen bonds to form between water molecules (1 mark). The S–H bond: |Δχ| = 2.6−2.2 = 0.4 → only weakly polar. Sulfur (Period 3, larger atom, lower χ) is not electronegative enough for H-bonding. H₂S molecules are held together by dipole-dipole forces and dispersion forces only (weaker than H-bonds) (1 mark). H-bonds (in H₂O) require significantly more energy to overcome than the dipole-dipole/dispersion forces (in H₂S) → water has a much higher boiling point (100°C) despite lower molar mass (18 vs 34 g/mol). The higher molar mass of H₂S gives it stronger dispersion forces but these still do not compensate for the absence of H-bonding (1 mark).

Q28 (4 marks): (a) Large jump between IE₃ (3660 kJ/mol) and IE₄ (25,020 kJ/mol): ratio = 25020/3660 ≈ 6.8 (much larger than ratios of ~2–3 between IE₁, IE₂, IE₃). This indicates 3 valence electrons are removed with steadily increasing energy; the 4th electron comes from an inner shell → W has 3 valence electrons → Group 13 (1 mark). (b) Possible configuration: [Ne]3s²3p¹ (aluminium, Z=13) or [Ar]3d¹⁰4s²4p¹ (gallium, Z=31) — both are Group 13 with 3 valence electrons. The given IE₁=800 kJ/mol matches Al (IE₁=577 kJ/mol) reasonably well if adjusted; or Ga (IE₁=579 kJ/mol). Either accepted with justification (1 mark). (c) Property 1: W is a solid metal at room temperature with moderate/high melting point — justified because Group 13 metals (Al: MP 660°C, Ga: MP 30°C) have metallic bonding (delocalised valence electrons from 3s²3p¹ or 4s²4p¹ form the electron sea that holds metal cations together) (1 mark). Property 2: W conducts electricity as a solid — justified because the delocalised electrons from the 3 valence electrons (ns²np¹) are mobile and can carry charge through the solid under an applied potential difference (1 mark).

Q29 (6 marks — extended response): (a) Carbon (Z=6) has electron configuration [He]2s²2p² with 4 valence electrons in the n=2 shell. It is a non-metal in Group 14 with moderate-to-high electronegativity (χ=2.6) and forms 4 covalent bonds to reach an octet (1 mark). (b) Diamond: each C atom forms 4 covalent bonds in a tetrahedral arrangement (sp³ hybridisation), bonding to 4 neighbouring C atoms. All 4 valence electrons of each C are used in C–C sigma bonds. This creates an infinite 3D covalent network extending throughout the crystal with no molecular boundaries (1 mark). Graphite: each C atom forms 3 covalent bonds to 3 neighbours within a planar hexagonal layer (sp² hybridisation), using 3 of the 4 valence electrons. The remaining 1 electron per C is not used in sigma bonding and instead becomes part of a delocalised π-electron system spread across the entire layer. Adjacent layers are held together only by weak dispersion forces (1 mark). The key structural difference: diamond uses all 4 valence electrons in localised bonds; graphite uses only 3 in localised bonds and leaves 1 per C atom delocalised across the layer. (c) Graphite in-plane conductivity: the delocalised π-electrons (one per C atom across the entire layer) are mobile — they can move freely parallel to the layer under an applied electric field, just as conduction electrons move in metals. These electrons act as charge carriers → graphite conducts electricity along the layers (1 mark). Perpendicular to layers: the π-electrons are confined within each layer; between layers there are only weak dispersion forces and no electron pathway → graphite does not conduct perpendicular to the layers (anisotropic conductivity) (1 mark). Diamond: all 4 valence electrons of each C are localised in C–C sigma bonds throughout the 3D lattice. There are no delocalised electrons — no mobile charge carriers exist. No electric field can move bonding electrons through the rigid covalent network without breaking bonds (which requires enormous energy) → diamond is an electrical insulator. The same element (C) produces opposite electrical behaviour purely because of the structural arrangement of its bonds (1 mark).

📝 New Short Answer Model Answers (Q30–40)

Q30 (3 marks): Equipment: pre-weighed evaporating dish, analytical balance, hotplate or oven, desiccator. Step 1: Weigh the clean, dry evaporating dish on an analytical balance (record m₁) (1 mark). Step 2: Transfer 250 mL of river water sample into the dish. Heat on a hotplate to evaporate all the water — continue until the residue appears dry and no further mass change occurs. Cool in a desiccator (prevents moisture reabsorption from air). Step 3: Weigh the dish + dry residue (record m₂) (1 mark). Calculation: mass of dissolved solids = m₂ − m₁ (grams). Concentration of dissolved solids = (m₂ − m₁) / 0.250 L, expressed in g/L. If m₁ = 52.340 g and m₂ = 52.518 g, then dissolved solids = 0.178 g, concentration = 0.712 g/L (1 mark).

Q31 (3 marks): (a) Amino acid C has the greatest affinity for the stationary phase — it has the lowest Rf value (0.21), meaning it travelled the shortest distance relative to the solvent front. Low Rf indicates the substance spends more time adsorbed to the stationary phase (paper) and less time dissolved in the mobile phase (solvent), i.e. it is more attracted to the stationary phase than to the solvent (1 mark). (b) Amino acids A (Rf=0.82) and B (Rf=0.45) would be hardest to separate, as they have the largest Rf difference (0.37) — no wait, the hardest to separate are those with the most SIMILAR Rf values. Comparing: A vs B = 0.37 difference; A vs C = 0.61; B vs C = 0.24. B and C are hardest to separate (smallest Rf difference of 0.24) (1 mark). (c) Rf = distance spot / distance solvent front. 0.45 = d / 12.0 cm. d = 0.45 × 12.0 = 5.4 cm (1 mark).

Q32 (4 marks): P: Metallic bonding — conducts electricity as a solid (delocalised electrons carry charge through the solid lattice) AND has a very high MP (1540°C, consistent with strong metallic bonding in iron or similar transition metal). Both properties are hallmarks of metals (1 mark). Q: Covalent molecular — very low MP (80°C, only weak IMFs between molecules need to be overcome) and non-conducting in all states (no ions or free electrons), and soluble in non-polar hexane (like dissolves like — non-polar molecules dissolve in non-polar solvents) (1 mark). R: Covalent network solid — extremely high MP (3550°C, must break covalent bonds throughout 3D lattice — consistent with diamond or SiO₂) and non-conducting in all states (no free electrons, no ions even when melted) (1 mark). S: Ionic compound — non-conducting as solid (ions locked in fixed lattice positions) but conducts when molten (ions become free to move as charge carriers) and dissolves in water giving a conducting solution (ions released into solution) (1 mark).

Q33 (3 marks): Both molecules contain an oxygen atom, but the key difference is the type and accessibility of H-bonding (1 mark). Propan-1-ol (CH₃CH₂CH₂OH) has a free –OH group: the O–H bond has |Δχ| = 3.4−2.2 = 1.2, making H sufficiently δ+ to donate H-bonds to water molecules, and O has lone pairs to accept H-bonds from water. The –OH group is hydrophilic; the short 3-carbon chain is not large enough to disrupt the water H-bond network significantly → fully miscible (1 mark). Diethyl ether (CH₃CH₂–O–CH₂CH₃) has no O–H bond — it cannot donate H-bonds to water. Its oxygen atom has lone pairs and can weakly accept H-bonds from water, but this one interaction is insufficient to compensate for the disruption caused by the two large, non-polar ethyl groups (–CH₂CH₃) on either side of the O. These hydrophobic chains disrupt the water H-bond network without providing compensating interactions → limited miscibility (1 mark).

Q34 (4 marks): (a) Condensation polymerisation — both monomers are bifunctional: ethylene glycol has two –OH groups (one at each end) and terephthalic acid has two –COOH groups (one at each end). Both functional groups react at both ends of each monomer, building a long polymer chain. A byproduct (H₂O) is released at each condensation step (1 mark). (b) Ester linkage (–COO– or –O–CO–) is formed when the –OH of ethylene glycol reacts with the –COOH of terephthalic acid. The small molecule released is water (H₂O), one molecule per ester bond formed (1 mark). (c) PET is thermoplastic — it is a linear chain polymer. The chains are held to each other by intermolecular forces (dipole-dipole interactions from the C=O of the ester group, and dispersion forces along the chain backbone). These IMFs can be overcome by heating → the polymer softens and can be remoulded. There are no covalent cross-links between chains (which would make it thermosetting) (1 mark). Additional: PET can be recycled by melting (consistent with thermoplastic) — used in bottle-to-bottle recycling (1 mark — if included).

Q35 (3 marks): (a) Fractions: m/z=54 → 0.058; m/z=56 → 0.917; m/z=58 → 0.022. Check: 0.058+0.917+0.022 = 0.997 ≈ 1.000 ✓ (rounding from tabulated data). Ar = (54×0.058) + (56×0.917) + (58×0.022) = 3.132 + 51.352 + 1.276 = 55.760 ≈ 55.8 (1 mark). (b) Ar ≈ 55.8 → Iron (Fe, Z=26, literature Ar=55.85). The three isotopes ⁵⁴Fe, ⁵⁶Fe, ⁵⁸Fe match the three peaks (1 mark). (c) Most abundant isotope: ⁵⁶Fe (91.7%). Neutrons = A − Z = 56 − 26 = 30 neutrons (1 mark).

Q36 (4 marks): (a) Overall trend: IE₁ generally increases from Na to Si: 496 → 738 → 577 → 786 kJ/mol. Moving left to right across Period 3, atomic number Z increases from 11 (Na) to 14 (Si). All four elements have valence electrons in the n=3 shell. The inner electron shielding (from 1s²2s²2p⁶ — same for all four) stays constant. As Z increases, Z_eff increases by approximately 1 per element → the nucleus attracts valence electrons more strongly → more energy is required to remove a valence electron → IE₁ generally increases (1 mark). (b) Anomaly: Al (IE₁=577 kJ/mol) < Mg (IE₁=738 kJ/mol) despite Al having Z=13 vs Mg's Z=12 (1 mark). Explanation: Mg has configuration [Ne]3s². The electron removed is from the 3s subshell. Al has configuration [Ne]3s²3p¹. The electron removed is from the 3p¹ subshell, which is at higher energy than 3s (1 mark). The 3p orbital is further from the nucleus on average, less able to penetrate close to the nucleus than 3s, and is partially shielded by the 3s² electrons below it. Despite Al having one more proton, the 3p¹ electron is more loosely held than Mg's 3s² electrons — less energy is needed to remove it. This subshell energy difference (3s → 3p) overrides the effect of the extra proton (1 mark).

Q37 (3 marks): (a) Ca (Z=20): full config 1s²2s²2p⁶3s²3p⁶4s². Ca²⁺: remove both 4s electrons → 1s²2s²2p⁶3s²3p⁶ (18 electrons). Same as Argon (Ar, Z=18) (1 mark). (b) N (Z=7): full config 1s²2s²2p³. N³⁻: gain 3 electrons into 2p → 1s²2s²2p⁶ (10 electrons). Same as Neon (Ne, Z=10) (1 mark). (c) Fe (Z=26): [Ar]3d⁶4s². Fe³⁺: lose 4s² first, then one 3d → [Ar]3d⁵ = 1s²2s²2p⁶3s²3p⁶3d⁵ (23 electrons). No noble gas has 23 electrons — noble gases: He(2), Ne(10), Ar(18), Kr(36). Fe³⁺ has a partially filled 3d subshell; noble gases always have completely filled subshells. Transition metal ions with partially filled d orbitals cannot match any noble gas configuration (1 mark).

Q38 (5 marks): (a) Same reaction type — Na and K are both in Group 1 with valence configuration ns¹ (Na: [Ne]3s¹; K: [Ar]4s¹). Both have exactly one valence electron in an s subshell (1 mark). Both react by donating this one electron to water: the metal atom is oxidised (loses 1e⁻ to form M⁺), and water is reduced (H₂O accepts the electron, releases H₂ and forms OH⁻). The ionic equation is identical: 2M(s) + 2H₂O(l) → 2M⁺(aq) + 2OH⁻(aq) + H₂(g). Same products (MOH and H₂) because both form 1+ ions — the number of valence electrons determines what ions form and therefore what products result (1 mark). (b) K (Period 4, Z=19) has a larger atomic radius than Na (Period 3, Z=11) because K has 4 occupied electron shells vs Na's 3 (1 mark). Going from Na to K down Group 1: the additional electron shell means the 4s¹ valence electron in K is further from the nucleus and shielded by more inner electrons (1s²2s²2p⁶3s²3p⁶ vs Na's 1s²2s²2p⁶). The effective nuclear charge experienced by the valence electron is lower in K → weaker nuclear attraction on the 4s¹ electron (1 mark). This means K has a lower first ionisation energy (IE₁: K=419 kJ/mol vs Na=496 kJ/mol) — the valence electron is more readily donated to water → the reaction proceeds faster, releases more heat per unit time, H₂ is produced faster → the heat ignites the H₂ spontaneously in K's case (1 mark).

Q39 (4 marks): (a) Pure substance — evidence: (1) Sharp, fixed melting point at 156°C: pure substances have a fixed, precise melting point whereas mixtures melt over a temperature range. (2) Single spot on TLC: only one component is present — a mixture would produce multiple spots of different Rf values (1 mark). (b) Covalent molecular solid: MP=156°C is consistent with overcoming weak-to-moderate IMFs between discrete molecules (too low for ionic: >600°C; too low for covalent network: >1000°C). Non-conducting in solid and liquid state: no free ions (rules out ionic) and no delocalised electrons (rules out metallic). Dissolves in hot water but not hexane: the molecule has polar character (consistent with covalent molecular substance with polar groups like –OH or –COOH that can interact with water via H-bonds); insoluble in hexane confirms it is not purely non-polar (1 mark for each correctly argued property — 2 marks for bonding classification with evidence). (c) Citric acid (C₆H₈O₇, MP=153–156°C) is the best match: MP exactly 156°C, polar molecule with three –COOH and one –OH group → dissolves well in hot water via H-bonding, insoluble in hexane (too polar for non-polar solvent), single pure compound. Other acceptable answers: succinic acid (MP 185°C — close), benzoic acid (MP 122°C, though less water-soluble) (1 mark for reasonable suggestion with justification).

Q40 (6 marks): Electron configurations: S (Z=16): [Ne]3s²3p⁴. Se (Z=34): [Ar]3d¹⁰4s²4p⁴. Te (Z=52): [Kr]4d¹⁰5s²5p⁴. All three central atoms have 6 valence electrons (Group 16), and their electronegativity values: S=2.6, Se=2.6, Te=2.1 — none are electronegative enough to form H-bonds with the H atoms bonded to them (1 mark). IMFs in H₂S, H₂Se, H₂Te: all are polar covalent molecules (dipole moment due to bent geometry and electronegativity difference), so they have dipole-dipole forces plus dispersion forces. No H-bonding (central atoms not electronegative enough) (1 mark). Trend H₂S→H₂Se→H₂Te: BP increases from −60°C to −41°C to −2°C. Down Group 16: atomic radius increases, electron cloud of central atom becomes larger and more polarisable (more electrons, higher shells) → stronger dispersion forces between molecules → more energy required to separate molecules in the liquid → higher BP. Dispersion forces dominate and increase with molar mass (H₂S=34, H₂Se=81, H₂Te=130 g/mol) (1 mark). H₂O anomaly: O (Z=8) config [He]2s²2p⁴, χ=3.4. The O–H bond: |Δχ| = 3.4−2.2 = 1.2 → strongly polar. O is both small (Period 2) and highly electronegative — these conditions allow strong O–H···O hydrogen bonds to form between water molecules (1 mark). H-bonds in water are significantly stronger than dipole-dipole or dispersion forces — they require much more energy (~20 kJ/mol per H-bond) to overcome (1 mark). Extrapolating the H₂S→H₂Se→H₂Te trend to H₂O would predict a BP of approximately −80°C. The actual BP (100°C) is ~180°C higher than predicted — entirely due to H-bonding. Without this anomaly, water would be a gas at room temperature, making liquid water (and therefore life as we know it) impossible on Earth (1 mark).