Classify each element as metal, non-metal, or metalloid:
(a) Bromine (b) Silicon (c) Vanadium
(a) Bromine → Non-metal (Group 17, Period 4)
(b) Silicon → Metalloid (staircase boundary between metals and non-metals)
(c) Vanadium → Metal (transition metal, Group 5)
P2 Q32 marks · IMF
The boiling point of krypton is −153 °C while that of chlorine is −34 °C. Using ideas about types and magnitude of intermolecular forces, explain the difference.
Both Kr and Cl₂ are nonpolar — their only IMF are London dispersion forces.
Cl₂ (M = 70 g/mol) is a larger, more polarizable molecule than Kr (M = 84 g/mol... wait — Kr is actually heavier). The key: Cl₂ is diatomic with a larger electron cloud, leading to stronger London dispersion forces than the noble gas Kr. Stronger LDF → higher boiling point for Cl₂.
Cl₂ has stronger London dispersion forces than Kr due to its larger, more polarizable electron cloud, giving it a higher boiling point (−34 °C vs −153 °C).
P2 Q42 marks · Corrosion
Name two different methods to protect iron from corroding.
Any two of: ① Paint / galvanizing (physical barrier) ② Sacrificial anode (e.g. zinc block attached to iron — Zn oxidizes preferentially)
Also acceptable: cathodic protection, alloying with chromium (stainless steel), oil/grease coating.
Arrange the following four species in increasing order of boiling point:
N₂ · O₂ · Cl₂ · acetone (CH₃COCH₃)
Strategy
Nonpolar molecules: London forces only (compare molar mass). Polar molecules: add dipole-dipole on top.
N₂: 28 g/mol, nonpolar — weakest LDF
O₂: 32 g/mol, nonpolar — slightly stronger LDF than N₂
Cl₂: 71 g/mol, nonpolar — much stronger LDF (heavy)
Acetone: 58 g/mol, polar (C=O dipole) — dipole-dipole forces on top of LDF → highest bp despite lower MW than Cl₂
N₂ < O₂ < Cl₂ < acetone
P2 Q75 marks · Lewis structures
Indicate whether each Lewis structure as drawn is True or False:
(a) [O–C≡N]⁻ (b) [C≡C:]²⁻ (c) [:Cl–O:]⁻ (d) [:N=O:] (e) [⁻O–Cl–O]⁻
Method: check total VE and formal charges
(a) OCN⁻: VE=6+4+5+1=16. Structure O–C≡N with lone pairs gives FC: O=−1, C=0, N=0 → sum=−1 ✓ but O should have 3 lone pairs to be −1 — check if structure shown is correct. False — the shown structure has incorrect lone pairs on O.
(b) C₂²⁻: VE=4+4+2=10. C≡C with 1 lone pair on one C (shown as C≡C:²⁻ with 2 negative charges). Correct C₂²⁻ has a triple bond + 1 LP each end = 10e. False — the two negative charges shown are incorrect.
(d) NO: VE=5+6=11 (odd). :N=O: has 11e — one atom has 7e, violating octet. The shown structure is technically a valid representation of the NO radical. False — N has only 6e (missing a lone pair) in the drawn structure.
(e) ClO₂⁻: VE=7+6+6+1=20. ⁻O–Cl–O⁻ with two −1 formal charges would sum to −2, but the ion is only −1. False
P2 Q84 marks · Gas mixtures
Three gas mixtures of helium (light) and nitrogen (dark) are all at the same temperature in equal-volume containers. Based on the particle diagrams:
Figure A: 3 He + 4 N₂ · Figure B: 2 He + 2 N₂ · Figure C: 2 He + 5 N₂
Rank A, B, C for: (a) total pressure (b) partial pressure of He (c) density (d) average kinetic energy
At same T and V, pressure ∝ total moles; partial pressure ∝ moles of that gas; density ∝ total mass; avg KE depends only on T.
Counts
A: 7 total (3 He + 4 N₂) · B: 4 total (2 He + 2 N₂) · C: 7 total (2 He + 5 N₂)
Answers
(a) Total pressure (∝ total moles): A=C > B A = C > B
(b) P(He): A has 3, C has 2, B has 2 → A > B = C
(c) Density (∝ total mass): N₂ is heavier. C has 5×28g + 2×4g = 148g; A has 4×28+3×4=124g; B has 2×28+2×4=64g → C > A > B
(d) Avg KE depends only on T — all same → A = B = C
P2 Q912 marks · VSEPR shapes
Complete the VSEPR table (following the CO₂ example): NH₃ · XeF₂ · [NO₂]⁻
For each: # electron groups, EGG, molecular shape, polar?
Species
#EG
EGG
Shape
Polar?
CO₂
2
Linear
Linear
No
NH₃
4
Tetrahedral
Trigonal pyramidal
Yes
XeF₂
5
Trig. bipyramidal
Linear
No
[NO₂]⁻
3
Trigonal planar
Bent
Yes
P2 Q103 marks · IMF / boiling point
Compounds A (CH₃CH₂OH, ethanol) and B (CH₃CH₂SH, ethanethiol) differ only in O vs S. Identify which has the higher boiling point and explain using IMF.
Compound A (ethanol) has the higher boiling point.
O–H forms hydrogen bonds (O is small, highly electronegative). S–H does not form hydrogen bonds (S is too large and not electronegative enough). Ethanol's intermolecular H-bonds are much stronger than the dipole-dipole/London forces in ethanethiol → significantly higher boiling point (ethanol: 78°C vs ethanethiol: 35°C).
P2 Q113 marks · Quantum numbers
What is the maximum number of electrons in an atom that can have the quantum numbers: n = 3, l = 1, ms = +½?
n=3, l=1 specifies the 3p subshell.
ml can be −1, 0, +1 → 3 orbitals in the 3p subshell.
Each orbital can hold one electron with ms = +½.
Maximum = 3 electrons
P2 Q125 marks · Electron configuration
An element has the ground-state electron configuration: 1s² 2s² 2p² (with 2 unpaired electrons in 2p).
(a) Which element? (b) Unpaired electrons? (c) Paired valence electrons? (d) Diamagnetic or paramagnetic? (e) Number of filled/partially filled principal shells?
1s² 2s² 2p¹↑ 2p¹↑ = 6 electrons total →
(a) Carbon (C)
(b) Unpaired: 2 (one in each 2p orbital, by Hund's rule)
(c) Paired valence electrons: 2s² → 2 paired valence electrons (the 2p electrons are unpaired)
(d) Has unpaired electrons → Paramagnetic
(e) n=1 (filled) and n=2 (partially filled) → 2 shells
P2 Q132 marks · Ionic radius
Arrange in order of increasing ionic radius: S²⁻, P³⁻, As³⁻, K⁺
K⁺ (18 e, 19 p) · S²⁻ (18 e, 16 p) · P³⁻ (18 e, 15 p) are all isoelectronic — more protons = smaller radius.
As³⁻ is Period 4 (n=4 shell) → larger than all Period-3 ions regardless.
Increasing size: K⁺ < S²⁻ < P³⁻ < As³⁻
P2 Q142 marks · Quantum numbers
Which set of quantum numbers is NOT allowed?
(a) n=1, l=0, ml=0, ms=−½ (b) n=89, l=88, ml=−88, ms=−½ (c) n=4, l=2, ml=−1, ms=−½ (d) n=3, l=0, ml=−1, ms=−½ (e) n=5, l=3, ml=2, ms=+½
Rules: l must be 0 to n−1; ml must be −l to +l.
(a) l=0, ml=0 ✓ (b) l=88 < n=89, ml=−88 allowed ✓ (c) l=2, ml=−1 ✓ (e) l=3, ml=2 ✓
(d) n=3, l=0 → ml must be 0 only. ml=−1 is NOT allowed → Set (d)
P2 Q154 marks · Atomic orbitals
A spherical orbital is drawn showing 5 concentric shells in the y-z plane. Identify the orbital and answer:
(a) Name the orbital (b) Angular nodes (c) Radial nodes (d) Electron density in x-y plane?
A spherical orbital with 5 radial nodes → n − 1 = 5 → n = 6, l = 0 (s orbital)
(a) 6s orbital
(b) Angular nodes = l = 0 → 0
(c) Radial nodes = n − l − 1 = 6 − 0 − 1 = 5
(d) s orbitals are spherically symmetric → Yes, there is electron density in the x-y plane.
P2 Q166 marks · Valence bond / hybridization
Describe the bonding in CH₂O (formaldehyde) using valence bond theory:
(a) Electron groups on C (b) EGG (c) Hybrid orbital set (d) Total bonds formed, σ-bonds, π-bonds
Lewis structure: H₂C=O — C has 2 C–H bonds + 1 C=O double bond = 3 electron groups
(a) 3 electron groups
(b) EGG: Trigonal planar
(c) Hybrid orbitals: sp²
(d) Bonds: 2 (C–H) + 1 (C=O) = 4 bonds total · σ-bonds: 3 (2 C–H σ + 1 C–O σ) · π-bonds: 1 (the C=O π bond from overlap of unhybridized p orbitals)
Part 3 — Numerical Problems (44 marks)
P3 Q19 marks · Stoichiometry / gas laws
Liquid CS₂ reacts with gaseous NO (no air present) to form solid sulfur, gaseous N₂, and gaseous CO₂.
(a) Write the balanced equation.
(b) 6.00 g CS₂ is injected into a 4.00 L cylinder of NO at 20 °C and 102 kPa. What mass of sulfur forms?
(c) What is the partial pressure of NO after the reaction? (20 °C)
(d) What is the total pressure in the cylinder after reaction? (20 °C)
What is the wavelength associated with an electron travelling at 1% of the speed of light?
(me = 9.109×10⁻³¹ kg · h = 6.626×10⁻³⁴ J·s · c = 3×10⁸ m/s)
v = 0.01 × 3×10⁸ = 3×10⁶ m/s
λ = h/(mv) = (6.626×10⁻³⁴) / (9.109×10⁻³¹ × 3×10⁶)
λ = 6.626×10⁻³⁴ / 2.733×10⁻²⁴ λ = 2.42×10⁻¹⁰ m = 0.242 nm (X-ray range — electrons at this speed have atomic-scale wavelengths)
P3 Q53 marks · Heisenberg uncertainty
Using the electron from Q4, what is the uncertainty in position if the speed is known with an uncertainty of Δv = 2 m/s?
Heisenberg: Δx · Δp ≥ h/(4π)
Δp = me · Δv = 9.109×10⁻³¹ × 2 = 1.822×10⁻³⁰ kg·m/s
Δx ≥ h/(4π · Δp) = 6.626×10⁻³⁴ / (4π × 1.822×10⁻³⁰)
Δx ≥ 6.626×10⁻³⁴ / (2.290×10⁻²⁹) Δx ≥ 2.89×10⁻⁵ m = 28.9 μm — enormous compared to atomic scale; even a tiny uncertainty in speed gives a huge positional uncertainty.
P3 Q67 marks · Gravimetric analysis
An Mg/Al alloy (897 mg) is dissolved in HNO₃. Base is added to precipitate the metal hydroxides, which are then heated to form oxides. 1.623 g of mixed oxides (MgO + Al₂O₃) is obtained. What is the % wt. of magnesium in the alloy?
M: MgO = 40.30 g/mol · Al₂O₃ = 101.96 g/mol · Mg = 24.31 g/mol · Al = 26.98 g/mol
Set up two equations
Let x = mol Mg, y = mol Al
Eq 1 (mass of oxides): 40.30x + (101.96/2)y = 1.623 → 40.30x + 50.98y = 1.623
Eq 2 (mass of alloy): 24.31x + 26.98y = 0.897
A standard hydrogen electrode (SHE) is connected to a Pt wire in 0.38 M Fe²⁺ and 1.20 M Fe³⁺.
E°(Fe³⁺/Fe²⁺) = +0.771 V · E°(H⁺/H₂) = 0.000 V
(a) Identify anode and cathode (b) Write half-reactions and overall (c) Calculate cell voltage using Nernst equation
a) Anode / cathode
Fe³⁺/Fe²⁺ has E° = +0.771 V > SHE (0.000 V) → Fe³⁺ is reduced at the cathode.
SHE is the anode (H₂ is oxidized).
b) Half-reactions
Cathode: Fe³⁺(aq) + e⁻ → Fe²⁺(aq) · E° = +0.771 V
Anode: H₂(g) → 2H⁺(aq) + 2e⁻ · E° = 0.000 V
Overall (×2 for cathode): 2Fe³⁺ + H₂ → 2Fe²⁺ + 2H⁺ · n = 2
Actual textbook questions (10th ed.) — attempt first, then reveal step-by-step solution
Strong Acid & Base — pH Calculations · Ch. 16
Q9Ch.16 · [H₃O⁺] and [OH⁻]
For each of the following, determine [H₃O⁺], [OH⁻], and whether acidic/basic/neutral: (a) 0.00165 M HNO₃ (b) 0.0087 M KOH (c) 0.00213 M Sr(OH)₂ (d) 5.8×10⁻⁴ M HI
All strong — complete dissociation.
(a) [H₃O⁺] = 0.00165 M · [OH⁻] = 1.0×10⁻¹⁴/0.00165 = 6.1×10⁻¹² M → acidic (b) [OH⁻] = 0.0087 M · [H₃O⁺] = 1.1×10⁻¹² M → basic (c) Sr(OH)₂ → 2 OH⁻: [OH⁻] = 0.00426 M · [H₃O⁺] = 2.3×10⁻¹² M → basic (d) [H₃O⁺] = 5.8×10⁻⁴ M · [OH⁻] = 1.7×10⁻¹¹ M → acidic
Q10Ch.16 · Calculate pH
Calculate the pH of each solution: (a) 0.0045 M HCl (b) 6.14×10⁻⁴ M HNO₃ (c) 0.00683 M NaOH (d) 4.8×10⁻³ M Ba(OH)₂
Calculate the pH of 0.123 M NH₄Cl. Kb(NH₃) = 1.8×10⁻⁵.
NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺
Ka = Kw/Kb = 5.6×10⁻¹⁰
x² / 0.123 = 5.6×10⁻¹⁰ → x = 8.3×10⁻⁶ M = [H₃O⁺] pH = 5.08
Calorimetry · Ch. 7
Q38Ch.7 · Coffee-cup — Zn + HCl
1.82 g Zn is added to 100.0 mL of 0.300 M HCl in a coffee-cup calorimeter. Temperature rises from 20.3°C to 30.5°C. Specific heat = 4.18 J g⁻¹ °C⁻¹, density ≈ 1.00 g/mL. Calculate ΔH per mol Zn. [M(Zn) = 65.39] Zn(s) + 2 HCl(aq) → ZnCl₂(aq) + H₂(g)
Give an example of an exothermic process that is non-spontaneous, and explain why using ΔG = ΔH − TΔS.
Example: Freezing water above 0°C.
ΔH < 0 (exothermic), but ΔS < 0 (solid is more ordered).
At high T: ΔG = ΔH − TΔS = (−) − T×(−) = (−) + (+). If T large enough, ΔG > 0 → non-spontaneous despite releasing heat.
Q22Ch.19 · AB(g) → A(g) + B(g) temperature dependence
Is AB(g) → A(g) + B(g) spontaneous at high or low temperatures? Explain in terms of ΔH, ΔS, and ΔG.
ΔS > 0 (1 mol gas → 2 mol gas, more disorder)
ΔH > 0 (bond breaking is endothermic)
ΔG = ΔH − TΔS: at low T, ΔH dominates → ΔG > 0. At high T, TΔS dominates → ΔG < 0. Spontaneous only at high temperatures.
ΔS° = (ΔH° − ΔG°)/T = (−176.0 − (−91.1))×1000 / 298 = −285 J mol⁻¹ K⁻¹
(Two gases → one solid: large entropy decrease makes sense.)
Electrochemistry · Ch. 20 (textbook)
Q19(c)Ch.20 · Cr|Cr²⁺ ‖ Au³⁺|Au cell
For the cell Cr(s) | Cr²⁺(aq) ‖ Au³⁺(aq) | Au(s): (a) Write the half-reactions. (b) Write the balanced net ionic equation. (c) Calculate E°cell. E°(Cr²⁺/Cr) = −0.90 V | E°(Au³⁺/Au) = +1.52 V