Questions :
CHM 001: General Chemistry
1(a)(i) First ionization energy is the energy required to remove one mole of electrons from one mole of gaseous atoms in their ground state to form one mole of gaseous 1+ ions.
(ii) Al(g) → Al⁺(g) + e⁻
(iii) P=799, Q=1090, R=1400, T=1310 correspond to B, C, N, O. The trend reverses from R(N) to T(O) because nitrogen has a stable, half-filled 2p³ configuration, making it harder to remove an electron than expected; oxygen's paired 2p electron experiences extra electron–electron repulsion, making it easier to remove — so its first IE is lower than nitrogen's despite having one more proton.
1(b)(i) The species at m/e 65 is ⁶⁵Cu⁺ (the copper-65 isotope ion)
(ii) The two peaks arise because copper occurs naturally as two isotopes, ⁶³Cu and ⁶⁵Cu, with different masses and relative abundances.
(iii) Relative atomic mass = (63×69.10 + 65×30.90)/100 = (4353.3 + 2008.5)/100 = 63.62
1(c)(i) Sodium metal conducts electricity in the solid state; sodium chloride conducts only when molten or in aqueous solution.
(ii) Na conducts because it has mobile, delocalized electrons throughout the metallic lattice. Solid NaCl does not conduct because its ions are fixed in a rigid lattice; only when molten or dissolved do the ions become free to move and carry charge.
CHM 001 Q2
2(a)(i) Silicon has a giant covalent (macromolecular) structure with strong covalent bonds extending throughout the crystal, requiring a large amount of energy to break — hence high melting point. Its valence electrons are localized in these covalent bonds (not free/delocalized), so it has low conductivity.
(ii) A dative (coordinate) covalent bond forms: H₃N→BH₃
(iii) BH₃ is electron-deficient (boron has only 6 electrons) with an empty p-orbital, so it readily accepts the lone pair from NH₃. AlCl₃ exists predominantly as the dimer Al₂Cl₆, in which aluminium's empty orbital is already satisfied by bridging chlorine lone pairs, making it far less available to accept a lone pair from NH₃.
2(b) mol K = 5.85/39 = 0.15; mol O = 2.40/16 = 0.15 → ratio 1:1
(i) Empirical formula: KO
(ii) Empirical mass = 39+16 = 55; 110/55 = 2 → Molecular formula: K₂O₂
(iii) 2K(s) + O₂(g) → K₂O₂(s)
2(c) Two limitations of Bohr's model:
- It could not explain the spectra of multi-electron atoms (only worked for hydrogen-like species).
- It could not explain the fine splitting of spectral lines in magnetic/electric fields (Zeeman/Stark effects), and it contradicts the Heisenberg uncertainty principle by assigning electrons fixed orbits with definite position and momentum.
CHM 002: Physical Chemistry
3(a)(i) Hess's Law: the total enthalpy change of a reaction is independent of the route taken, provided the initial and final states are the same.
(ii) C₂H₄ + H₂ → C₂H₆
Bonds broken: 1×C=C (612) + 1×H–H (436) = 1048 kJ
Bonds formed: 1×C–C (348) + 2×C–H (2×412=824) = 1172 kJ
ΔH = 1048 − 1172 = −124 kJ/mol
3(b)(i) Ksp(BaCl₂) = [Ba²⁺][Cl⁻]²
(ii) For precipitation, K = 1/Ksp = 1/(2.00×10⁻³¹) = 5.0×10³⁰
ΔG° = −RT ln K = −(8.314)(298)ln(5.0×10³⁰) = −(8.314)(298)(70.7)
ΔG° ≈ −175.2 kJ/mol (large negative value → precipitation is strongly favoured)
3(c) 2CO(g) + O₂(g) → 2CO₂(g); CO = 350 cm³, O₂ = 200 cm³
O₂ required for all CO = 175 cm³ < 200 cm³ available → CO is limiting
CO₂ formed = 350 cm³; O₂ used = 175 cm³; O₂ remaining = 200−175 = 25 cm³
(i) After reaction: CO₂ = 350 cm³, O₂ = 25 cm³, CO = 0 cm³
(ii) Alkaline pyrogallol absorbs O₂ only, so the volume left = 350 cm³ (CO₂)
CHM 003: Inorganic Chemistry
5(a)(i) Reactions with water:
- Be: does not react with water
- Mg: reacts very slowly with cold water; more readily with steam — Mg(s) + 2H₂O(g) → Mg(OH)₂(aq) + H₂(g)
- Ca: Ca(s) + 2H₂O(l) → Ca(OH)₂(aq) + H₂(g)
- Sr: Sr(s) + 2H₂O(l) → Sr(OH)₂(aq) + H₂(g)
- Ba: Ba(s) + 2H₂O(l) → Ba(OH)₂(aq) + H₂(g)
(ii) Reactivity with water increases down the group (Be < Mg < Ca < Sr < Ba)
5(b)(i) Transition metals form complexes because: (1) their ions are small with high charge density; (2) they have empty/available d-orbitals that can accept lone pairs from ligands to form dative bonds.
(ii) They are coloured because of d–d electron transitions: ligands split the degenerate d-orbitals into different energy levels, and electrons absorb specific wavelengths of visible light to jump between them — the colour seen is complementary to the light absorbed.
5(c)(i) Acid rain is precipitation with an abnormally low pH (below ~5.6), caused by atmospheric SO₂ and NOₓ dissolving in rainwater to form sulfuric and nitric acids.
(ii)
- Trees: leaches essential nutrients (Ca²⁺, Mg²⁺) from soil and releases toxic Al³⁺ which damages roots, weakening trees and increasing susceptibility to disease.
- Fish: lowers the pH of lakes/rivers, damaging gills and disrupting osmoregulation; also mobilizes toxic aluminium ions that are lethal to fish and fish eggs.
- Building materials: reacts with carbonate stone (limestone/marble): CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂, causing corrosion and erosion; also corrodes exposed metal.
CHM 003 Q6
6(a) A transition metal is one that forms at least one stable ion with a partially filled d-subshell.
(b) Atomic radius decreases only slightly (remains nearly constant) from Cr→Mn→Fe because the added electrons enter inner d-orbitals, which shield the increasing nuclear charge poorly; the increased effective nuclear charge pulls the outer electrons in, roughly balancing the effect of added electrons.
(c) Any three: variable oxidation states; formation of coloured compounds; formation of complex ions; catalytic activity; paramagnetism.
(d)(i) Greenhouse effect: the trapping of infrared (heat) radiation from the Earth's surface by gases in the atmosphere, causing the planet's surface and lower atmosphere to warm.
(ii) Burning of fossil fuels; deforestation; industrial processes (e.g., cement manufacture); agriculture/livestock farming (methane emissions).
CHM 004: Organic Chemistry
7. Two types: structural isomerism and geometric isomerism.
(a)(i) Two structural isomers of C₂H₄Br₂:
- 1,1-dibromoethane: CH₃–CHBr₂
- 1,2-dibromoethane: CH₂Br–CH₂Br
(ii) cis- and trans-1,2-dibromoethene (C₂H₂Br₂): both Br atoms on the same side of the C=C double bond (cis) versus on opposite sides (trans).
(b) Cyclopentane does not decolourise aqueous bromine, since — unlike the pentene isomers of C₅H₁₀ — it is saturated and contains no C=C double bond.
8(a) (bond-enthalpy calculation) — done above in 3(a)(ii): ΔH = −124 kJ/mol
(b)(i) Ksp expression and precipitation ΔG — done above in 3(b).
(c) Compound A = C₃H₇OH (propan-1-ol):
- B (with H⁺, 180 °C — acid-catalysed dehydration, elimination): CH₃–CH=CH₂, prop-1-ene
- C (with KMnO₄ — oxidation of a primary alcohol): CH₃CH₂COOH, propanoic acid
(i) Structures and names as above (B = prop-1-ene, C = propanoic acid)
(ii) Reaction between A and C is esterification (condensation reaction), catalysed by acid, forming propyl propanoate and water.
CHM 004 Q8
8(a)(i) Two alcohols that give pent-2-ene on dehydration:
- Pentan-2-ol: CH₃–CH(OH)–CH₂–CH₂–CH₃
- Pentan-3-ol: CH₃–CH₂–CH(OH)–CH₂–CH₃
(ii) Pentan-2-ol exhibits optical isomerism (C2 bears four different groups: OH, H, CH₃, CH₂CH₂CH₃). Its two optical isomers are the non-superimposable mirror images (R)-pentan-2-ol and (S)-pentan-2-ol (drawn with wedge/dash bonds showing the mirrored arrangement about C2). Pentan-3-ol is not chiral, since C3 carries two identical ethyl groups.
(b) Natural rubber's monomer is isoprene (2-methylbuta-1,3-diene):
CH₂=C(CH₃)–CH=CH₂
CHM 001: General Chemistry
1(a)(i) First ionization energy is the energy required to remove one mole of electrons from one mole of gaseous atoms in their ground state to form one mole of gaseous 1+ ions.
(ii) Al(g) → Al⁺(g) + e⁻
(iii) P=799, Q=1090, R=1400, T=1310 correspond to B, C, N, O. The trend reverses from R(N) to T(O) because nitrogen has a stable, half-filled 2p³ configuration, making it harder to remove an electron than expected; oxygen's paired 2p electron experiences extra electron–electron repulsion, making it easier to remove — so its first IE is lower than nitrogen's despite having one more proton.
1(b)(i) The species at m/e 65 is ⁶⁵Cu⁺ (the copper-65 isotope ion)
(ii) The two peaks arise because copper occurs naturally as two isotopes, ⁶³Cu and ⁶⁵Cu, with different masses and relative abundances.
(iii) Relative atomic mass = (63×69.10 + 65×30.90)/100 = (4353.3 + 2008.5)/100 = 63.62
1(c)(i) Sodium metal conducts electricity in the solid state; sodium chloride conducts only when molten or in aqueous solution.
(ii) Na conducts because it has mobile, delocalized electrons throughout the metallic lattice. Solid NaCl does not conduct because its ions are fixed in a rigid lattice; only when molten or dissolved do the ions become free to move and carry charge.
CHM 001 Q2
2(a)(i) Silicon has a giant covalent (macromolecular) structure with strong covalent bonds extending throughout the crystal, requiring a large amount of energy to break — hence high melting point. Its valence electrons are localized in these covalent bonds (not free/delocalized), so it has low conductivity.
(ii) A dative (coordinate) covalent bond forms: H₃N→BH₃
(iii) BH₃ is electron-deficient (boron has only 6 electrons) with an empty p-orbital, so it readily accepts the lone pair from NH₃. AlCl₃ exists predominantly as the dimer Al₂Cl₆, in which aluminium's empty orbital is already satisfied by bridging chlorine lone pairs, making it far less available to accept a lone pair from NH₃.
2(b) mol K = 5.85/39 = 0.15; mol O = 2.40/16 = 0.15 → ratio 1:1
(i) Empirical formula: KO
(ii) Empirical mass = 39+16 = 55; 110/55 = 2 → Molecular formula: K₂O₂
(iii) 2K(s) + O₂(g) → K₂O₂(s)
2(c) Two limitations of Bohr's model:
- It could not explain the spectra of multi-electron atoms (only worked for hydrogen-like species).
- It could not explain the fine splitting of spectral lines in magnetic/electric fields (Zeeman/Stark effects), and it contradicts the Heisenberg uncertainty principle by assigning electrons fixed orbits with definite position and momentum.
CHM 002: Physical Chemistry
3(a)(i) Hess's Law: the total enthalpy change of a reaction is independent of the route taken, provided the initial and final states are the same.
(ii) C₂H₄ + H₂ → C₂H₆
Bonds broken: 1×C=C (612) + 1×H–H (436) = 1048 kJ
Bonds formed: 1×C–C (348) + 2×C–H (2×412=824) = 1172 kJ
ΔH = 1048 − 1172 = −124 kJ/mol
3(b)(i) Ksp(BaCl₂) = [Ba²⁺][Cl⁻]²
(ii) For precipitation, K = 1/Ksp = 1/(2.00×10⁻³¹) = 5.0×10³⁰
ΔG° = −RT ln K = −(8.314)(298)ln(5.0×10³⁰) = −(8.314)(298)(70.7)
ΔG° ≈ −175.2 kJ/mol (large negative → precipitation strongly favoured)
3(c) 2CO(g) + O₂(g) → 2CO₂(g); CO = 350 cm³, O₂ = 200 cm³
O₂ required for all CO = 175 cm³ < 200 cm³ available → CO is limiting
CO₂ formed = 350 cm³; O₂ used = 175 cm³; O₂ remaining = 25 cm³
(i) After reaction: CO₂ = 350 cm³, O₂ = 25 cm³, CO = 0 cm³
(ii) Alkaline pyrogallol absorbs O₂ only, so volume left = 350 cm³ (CO₂)
CHM 003: Inorganic Chemistry
5(a)(i) Reactions with water:
- Be: does not react
- Mg: reacts very slowly with cold water, more readily with steam — Mg(s) + 2H₂O(g) → Mg(OH)₂(aq) + H₂(g)
- Ca: Ca(s) + 2H₂O(l) → Ca(OH)₂(aq) + H₂(g)
- Sr: Sr(s) + 2H₂O(l) → Sr(OH)₂(aq) + H₂(g)
- Ba: Ba(s) + 2H₂O(l) → Ba(OH)₂(aq) + H₂(g)
(ii) Reactivity increases down the group (Be < Mg < Ca < Sr < Ba)
5(b)(i) Transition metals form complexes because: (1) their ions are small with high charge density; (2) they have empty/available d-orbitals that accept lone pairs from ligands to form dative bonds.
(ii) Coloured because of d–d electron transitions: ligands split degenerate d-orbitals into different energy levels; electrons absorb specific wavelengths of visible light to jump between them, and the colour seen is complementary to the light absorbed.
5(c)(i) Acid rain is precipitation with abnormally low pH (below ~5.6), caused by atmospheric SO₂ and NOₓ dissolving in rainwater to form sulfuric and nitric acids.
(ii)
- Trees: leaches nutrients (Ca²⁺, Mg²⁺) from soil and releases toxic Al³⁺, damaging roots and weakening trees.
- Fish: lowers pH of lakes/rivers, damaging gills and disrupting osmoregulation; mobilizes toxic aluminium ions lethal to fish and eggs.
- Building materials: reacts with carbonate stone: CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂, causing corrosion/erosion; also corrodes exposed metal.
CHM 003 Q6
6(a) A transition metal forms at least one stable ion with a partially filled d-subshell.
(b) Atomic radius stays nearly constant from Cr→Mn→Fe because added electrons enter inner d-orbitals, which shield the increasing nuclear charge poorly; the rising effective nuclear charge pulls outer electrons in, roughly balancing the effect of the added electrons.
(c) Any three: variable oxidation states; coloured compounds; formation of complex ions; catalytic activity; paramagnetism.
(d)(i) Greenhouse effect: trapping of infrared (heat) radiation from Earth's surface by atmospheric gases, warming the surface and lower atmosphere.
(ii) Burning fossil fuels; deforestation; industrial processes (e.g., cement manufacture); agriculture/livestock farming (methane).
CHM 004: Organic Chemistry
7. Two types: structural isomerism and geometric isomerism.
(a)(i) Two structural isomers of C₂H₄Br₂:
- 1,1-dibromoethane: CH₃–CHBr₂
- 1,2-dibromoethane: CH₂Br–CH₂Br
(ii) cis- and trans-1,2-dibromoethene: both Br atoms on the same side of the C=C (cis) versus opposite sides (trans).
(b) Cyclopentane does not decolourise aqueous bromine — unlike the pentene isomers of C₅H₁₀, it is saturated with no C=C double bond.
8(a) Bond-enthalpy calculation (same as 3a-ii): ΔH = −124 kJ/mol
(b) Ksp expression and precipitation ΔG — as in 3(b) above.
(c) Compound A = C₃H₇OH (propan-1-ol):
- B (H⁺, 180°C — acid-catalysed dehydration): CH₃–CH=CH₂, prop-1-ene
- C (KMnO₄ — oxidation of primary alcohol): CH₃CH₂COOH, propanoic acid
(ii) Reaction between A and C is esterification (condensation), acid-catalysed, forming propyl propanoate and water.
CHM 004 Q8
8(a)(i) Two alcohols giving pent-2-ene on dehydration:
- Pentan-2-ol: CH₃–CH(OH)–CH₂–CH₂–CH₃
- Pentan-3-ol: CH₃–CH₂–CH(OH)–CH₂–CH₃
(ii) Pentan-2-ol exhibits optical isomerism (C2 bears four different groups: OH, H, CH₃, CH₂CH₂CH₃), giving two enantiomers (R)-pentan-2-ol and (S)-pentan-2-ol (mirror images, non-superimposable). Pentan-3-ol is not chiral (C3 has two identical ethyl groups).
(b) Natural rubber's monomer is isoprene (2-methylbuta-1,3-diene): CH₂=C(CH₃)–CH=CH₂