Chemistry Olympiad — Free Diagnostic Mock

13-question diagnostic · real 2021 ASOE Section A · find your gaps
🎁 FREE Diagnostic

13 real 2021 ASOE questions to find exactly where your olympiad chemistry is strong — and where it's not

This is a free diagnostic for the Chemistry Olympiad course — Section A of the real 2021 ASOE paper. 13 multiple-choice questions from the real 2021 ASOE Section A — stoichiometry, periodicity, bonding, analysis and more.

It's an exam: answer all the questions first (you can change any answer freely), then submit the whole exam to see your report. The worked solutions and hints unlock only after you submit. Use the jump menu at the top to move around.

At the end you get a knowledge-point report: a bar for every topic you attempted, your weak spots flagged, and the exact lessons to revise.

  • ~15 minutes. No login needed. No marks deducted for a wrong answer — never leave one blank.
  • Self-paced, no time limit — but try to commit to an answer before you review, just like a real test.

Year 10–11 students preparing for the Australian Science Olympiad (Chemistry).

⭐ Q1ASOE 2021Difficulty ★★★Periodic › ionic radius
The Problem

Which of the following lists species in order of increasing ionic radius?

Your answer

❓ Which order is correct?

  • A Cs⁺, Rb⁺, Na⁺
  • B S²⁻, Cl⁻, K⁺
  • C O²⁻, Na⁺, Ba²⁺
  • D I⁻, Cl⁻, Br⁻
  • E Sr²⁺, Rb⁺, Br⁻
💡 Hint 1 — what's being tested

S²⁻, Cl⁻ and K⁺ are all isoelectronic with argon (18 electrons). More protons pull the same electron cloud in tighter, so radius falls as charge rises: S²⁻ > Cl⁻ > K⁺. The same rule works for any isoelectronic set: find the set, then order by proton count. The list must be printed smallest → largest to answer an "increasing" stem.

Spot the isoelectronic set, order by nuclear charge, and check which printed list is genuinely monotonic in radius.
💡 Hint 3 — set up the first step
  • S²⁻, Cl⁻, K⁺ all have 18 electrons — compare proton counts (16, 17, 19).
  • For same-charge ions down a group, radius grows downward — A and D run the wrong way.
  • C mixes ions that are not isoelectronic — its radii jump around.

Use the isoelectronic rule.

  1. S²⁻ (16 p), Cl⁻ (17 p), K⁺ (19 p) share 18 electrons. Radius decreases with more protons: S²⁻ > Cl⁻ > K⁺.
  2. E is the same rule on the krypton series: Sr²⁺ (38 p), Rb⁺ (37 p), Br⁻ (35 p) share 36 electrons, so radius grows as proton count falls — Sr²⁺ < Rb⁺ < Br⁻, printed smallest → largest. ✓
  3. B is the argon series printed largest → smallest (decreasing); A and D are not in increasing order; C is not an isoelectronic set.
✅ Answer: E — Sr²⁺, Rb⁺, Br⁻

Why this is the answer: the krypton-isoelectronic series Sr²⁺ < Rb⁺ < Br⁻ is the only list printed in increasing radius order. B is the argon series printed in decreasing order — the classic trap. The rest break the trend or charge consistency.

⭐ Q2ASOE 2021Difficulty ★★Solutions › solubility rules
The Problem

Which of the following pairs of compounds will form a precipitate when 0.1 mol L⁻¹ solutions of each are mixed?

Your answer

❓ Which pair precipitates?

  • A AgNO₃ and Ba(NO₃)₂
  • B K₂SO₄ and Cu(NO₃)₂
  • C Ca(NO₃)₂ and KBr
  • D NaOH and CuCl₂
  • E CuCl₂ and NH₄NO₃
💡 Hint 1 — what's being tested

Mixing two salts swaps the cations and anions. A precipitate forms only if one of the two new combinations is insoluble. Most hydroxides of transition metals are insoluble — Cu(OH)₂ is the precipitate here.

For each pair, write the two cross-products and ask whether either is insoluble.
💡 Hint 3 — set up the first step
  • All nitrates are soluble — any pair giving only nitrates stays in solution (A, C, E).
  • K₂SO₄ + Cu(NO₃)₂ → CuSO₄ (soluble) + KNO₃ — no precipitate.
  • Transition-metal hydroxides like Cu(OH)₂ are insoluble.

Cross-pair and test solubility.

  1. NaOH + CuCl₂ → 2 NaCl + Cu(OH)₂↓. Cu(OH)₂ is insoluble → precipitate.
  2. A, C, E give only soluble nitrates/halides — no solid.
  3. B gives soluble CuSO₄ and KNO₃ — no solid.
✅ Answer: D — NaOH and CuCl₂

Why this is the answer: only D produces an insoluble product, Cu(OH)₂. Every other pair recombines into fully soluble salts.

⭐ Q3ASOE 2021Difficulty ★★★Stoichiometry › gas volume
The Problem

A component of diesel fuel is the hydrocarbon C₁₂H₂₄, with density 0.790 g mL⁻¹. What volume of CO₂ (measured at 25 °C and 100 kPa) is produced from the complete combustion of 2.00 L of C₁₂H₂₄ in excess oxygen?

Your answer

❓ Volume of CO₂ produced?

  • A 2.79 L
  • B 3.53 L
  • C 1400 L
  • D 2790 L
  • E 3530 L
💡 Hint 1 — what's being tested

Volume × density gives the mass of fuel; ÷ molar mass gives moles. Each C₁₂H₂₄ burns to 12 CO₂. Multiply moles of CO₂ by the molar gas volume 24.79 L mol⁻¹ at 25 °C / 100 kPa.

2.00 L × 0.790 g mL⁻¹ → mass → moles fuel → ×12 → ×24.79 L mol⁻¹.
💡 Hint 3 — set up the first step
  • 2.00 L = 2000 mL; mass = 2000 × 0.790 = 1580 g.
  • M(C₁₂H₂₄) = 168.3 g mol⁻¹.
  • Twelve carbons → 12 CO₂; ignoring the density (using 2000 g instead of 1580 g) lands on 3530 L (E).

Mass → moles → ×12 → gas volume.

  1. Mass of fuel = 2000 mL × 0.790 g mL⁻¹ = 1580 g.
  2. Moles fuel = 1580 / 168.3 = 9.39 mol → CO₂ = ×12 = 112.7 mol.
  3. V(CO₂) = 112.7 × 24.79 = ≈ 2790 L.
✅ Answer: D — 2790 L

Why this is the answer: 1580 g of fuel is 9.39 mol, giving 112.7 mol of CO₂ at 24.79 L mol⁻¹ ≈ 2790 L. Ignoring the density gives 3530 L; using mL not L gives the tiny decoys. Dropping the ×12 gives 2390 L; using mL not L gives the tiny decoys.

⭐ Q4ASOE 2021Difficulty ★★Bonding › VSEPR shape
The Problem

Which of the following compounds has a trigonal pyramidal geometry?

Your answer

❓ Which is trigonal pyramidal?

  • A NCl₃
  • B CH₂Cl₂
  • C COCl₂
  • D CH₄
  • E BCl₃
💡 Hint 1 — what's being tested

Trigonal pyramidal needs 3 bonded atoms + 1 lone pair on the central atom (4 domains). NCl₃ fits: N has 3 N–Cl bonds and one lone pair.

Count domains and lone pairs on the central atom; trigonal pyramidal = 4 domains, 1 lone pair.
💡 Hint 3 — set up the first step
  • CH₄ and CH₂Cl₂ are tetrahedral (4 bonds, 0 lone pairs).
  • BCl₃ and COCl₂ are trigonal planar (3 domains, 0 lone pairs).
  • Only NCl₃ carries the lone pair that tilts it into a pyramid.

Match domains to shape.

  1. NCl₃: 3 bonds + 1 lone pair → 4 domains, 1 lone pair → trigonal pyramidal.
  2. CH₄, CH₂Cl₂: tetrahedral. COCl₂, BCl₃: trigonal planar.
✅ Answer: A — NCl₃

Why this is the answer: only NCl₃ has the 4-domain, 1-lone-pair arrangement that gives a pyramidal molecular shape (like NH₃).

⭐ Q5ASOE 2021Difficulty ★★★Stoichiometry › limiting reagent
The Problem

When the given amounts of each reagent are mixed together, which of the following will release the largest mass of CO₂?

Reactions: CuCO₃ + H₂SO₄ → CuSO₄ + H₂O + CO₂ (1 : 1), and CuCO₃ + 2 HCl → CuCl₂ + H₂O + CO₂ (1 : 2). Each mole of CuCO₃ that reacts releases one mole of CO₂.

Your answer

❓ Which mixture gives the most CO₂?

  • A 0.3 mol CuCO₃ and 0.1 mol H₂SO₄
  • B 0.2 mol CuCO₃ and 0.3 mol HCl
  • C 0.2 mol CuCO₃ and 0.2 mol H₂SO₄
  • D 0.3 mol CuCO₃ and 0.3 mol HCl
  • E 0.1 mol CuCO₃ and 0.3 mol H₂SO₄
  • F 0.1 mol CuCO₃ and 0.2 mol HCl
💡 Hint 1 — what's being tested

One CO₂ per CuCO₃ consumed. Find the limiting reagent in each mixture; H₂SO₄ reacts 1 : 1 with CuCO₃, HCl reacts 2 : 1 (½ mol CuCO₃ per mol HCl). The winner is the one that consumes the most CuCO₃.

For each, compute CuCO₃ that can react: min(CuCO₃, acid capacity). Acid capacity = mol H₂SO₄, or ½ × mol HCl.
💡 Hint 3 — set up the first step
  • HCl needs two moles per CuCO₃ — halve the HCl when converting to CuCO₃ capacity.
  • The CO₂ produced is the smaller of CuCO₃ available and acid capacity.
  • C: 0.2 H₂SO₄ can dissolve 0.2 CuCO₃ — all of it → 0.2 mol CO₂, the highest here.

Compute CuCO₃ reacted = CO₂ in each.

  1. A: H₂SO₄ caps at 0.1 → 0.1 mol CO₂.
  2. B: HCl caps at 0.3/2 = 0.15 → 0.15 mol CO₂.
  3. C: 0.2 H₂SO₄ vs 0.2 CuCO₃ → 0.2 mol CO₂.
  4. D: HCl caps at 0.3/2 = 0.15 → 0.15 mol CO₂.
  5. E: 0.1 CuCO₃ limits → 0.1 mol CO₂.
  6. F: HCl caps at 0.2/2 = 0.1 → 0.1 mol CO₂.
✅ Answer: C — 0.2 mol CuCO₃ and 0.2 mol H₂SO₄

Why this is the answer: C reacts a full 0.2 mol of CuCO₃ (0.2 mol CO₂), beating every other mixture once the 2 : 1 HCl ratio is applied.

⭐ Q6ASOE 2021Difficulty ★★Bonding › valence electron count
The Problem

What is the total number of valence electrons in the PO₂³⁻ ion?

Your answer

❓ Total valence electrons?

  • A 15
  • B 17
  • C 20
  • D 30
  • E 34
💡 Hint 1 — what's being tested

P contributes 5 valence electrons, each O contributes 6, and the 3− charge adds 3 more electrons.

5 (P) + 2 × 6 (O) + 3 (charge).
💡 Hint 3 — set up the first step
  • Two oxygens here, not four — read PO₂³⁻ carefully.
  • A negative charge adds electrons (+3), it does not subtract.

Add the contributions.

  1. P: 5; 2 O: 2 × 6 = 12; charge: +3.
  2. Total = 5 + 12 + 3 = 20.
✅ Answer: C — 20

Why this is the answer: 5 + 12 + 3 = 20. Counting all 34 electrons instead of only the valence electrons (E), or subtracting the 3− charge instead of adding it, are the usual slips.

⭐ Q7ASOE 2021Difficulty ★★Stoichiometry › percentage composition
The Problem

Which of the following molecules contains 29.67% sulfur by mass?

Your answer

❓ Which molecule is 29.67% S?

  • A SF₄
  • B SO₂Cl₂
  • C SOCl₂
  • D SF₆
  • E S₂F₁₀
💡 Hint 1 — what's being tested

Compute the molar mass of each candidate and check the sulfur fraction. SF₆: M = 32.07 + 6(19.00) = 146.07.

Target %S ≈ 29.67. Work %S = 32.07 / M for each candidate and find the match.
💡 Hint 3 — set up the first step
  • SF₆: M = 146.07; 32.07/146.07 = 21.95% — not the target.
  • S₂F₁₀ has the same S:F ratio per S as SF₅; %S = 64.14/254.07.
  • Compute each fraction rather than guessing from size.

Match the sulfur fraction.

  1. SF₆: M = 32.07 + 114.00 = 146.07; %S = 32.07/146.07 = 21.95% — does not match.
  2. Work every candidate: SF₄ M = 108.1 → 29.67% ✓; SO₂Cl₂ M = 135.0 → 23.76%; SOCl₂ M = 119.0 → 26.95%; SF₆ M = 146.1 → 21.95%; S₂F₁₀ M = 254.1 → 25.23%.
  3. Only SF₄ matches the printed 29.67% (option A).
✅ Answer: A — SF₄

Why this is the answer: 32.07 / 108.1 = 29.67% — only SF₄ reproduces the printed figure. SF₆ gives 21.95%, so the "biggest molecule" instinct is the trap here.

⭐ Q8ASOE 2021Difficulty ★★IMF › bonding type vs bp
The Problem

Which of the following lists substances in order of increasing boiling point?

Your answer

❓ Which order is correct?

  • A CO₂, PCl₃, CaO
  • B PCl₃, CaO, CO₂
  • C CaO, CO₂, PCl₃
  • D CaO, PCl₃, CO₂
  • E CO₂, CaO, PCl₃
  • F PCl₃, CO₂, CaO
💡 Hint 1 — what's being tested

CO₂ is a small non-polar molecule (sublimes very cold), PCl₃ is a larger polar molecule (liquid), CaO is an ionic lattice (very high bp). So boiling point rises CO₂ < PCl₃ < CaO.

Rank by bonding strength: small molecular < larger molecular < ionic.
💡 Hint 3 — set up the first step
  • CaO (ionic) must be highest — any list not ending in CaO is wrong.
  • Between molecules, the bigger/more polar PCl₃ boils higher than CO₂.

Order by interparticle force.

  1. CO₂: weak dispersion (lowest). PCl₃: dipole + larger dispersion (middle). CaO: ionic lattice (highest).
  2. Increasing: CO₂ < PCl₃ < CaO → option A.
✅ Answer: A — CO₂, PCl₃, CaO

Why this is the answer: the only list rising from the smallest molecular solid through the larger polar molecule to the ionic lattice.

⭐ Q9ASOE 2021Difficulty ★★★Periodic › ionisation energy
The Problem

First ionisation energy is defined as the energy required to remove one mole of electrons from one mole of gaseous atoms. Which of the following lists elements in order of increasing first ionisation energy?

Your answer

❓ Which order is correct?

  • A C, F, N, Li
  • B C, N, Li, F
  • C Li, C, N, F
  • D Li, N, F, C
  • E F, N, C, Li
  • F F, Li, N, C
💡 Hint 1 — what's being tested

Across period 2, first ionisation energy generally rises: Li < C < N < F (N is slightly high from its half-filled 2p, but stays below F). Li is lowest, F is highest.

Lowest IE = Li (left), highest = F (right). Order the period-2 set left to right.
💡 Hint 3 — set up the first step
  • Li must be first (lowest), F must be last (highest).
  • The N > O dip is not tested here; for Li, C, N, F the trend is clean.

Order by position across period 2.

  1. IE: Li (520) < C (1086) < N (1402) < F (1681) kJ mol⁻¹.
  2. Increasing: Li, C, N, F → option C.
✅ Answer: C — Li, C, N, F

Why this is the answer: the only list rising left-to-right from Li to F. Lists starting with C or F, or ending in Li/C, break the periodic trend.

⭐ Q10ASOE 2021Difficulty ★★Mole › particle count
The Problem

How many atoms are present in a 1.0 kg sample of C₂H₄O?

Your answer

❓ Number of atoms?

  • A 1.4 × 10²²
  • B 9.6 × 10²²
  • C 1.4 × 10²⁵
  • D 9.6 × 10²⁵
  • E 9.6 × 10²⁸
💡 Hint 1 — what's being tested

M(C₂H₄O) = 44.05 g mol⁻¹; each formula has 2 + 4 + 1 = 7 atoms.

1000 g ÷ 44.05 → moles → ×7 atoms × 6.022×10²³.
💡 Hint 3 — set up the first step
  • 1.0 kg = 1000 g.
  • Seven atoms per molecule — count C, H and O.

Mass → moles → atoms.

  1. Moles = 1000 / 44.05 = 22.70 mol.
  2. Atoms = 22.70 × 7 × 6.022×10²³ = 9.6 × 10²⁵.
✅ Answer: D — 9.6 × 10²⁵

Why this is the answer: 22.7 mol × 7 atoms × Nₐ ≈ 9.6 × 10²⁵. Forgetting the ×7 (molecules only) gives 1.4 × 10²⁵ (C).

⭐ Q11ASOE 2021Difficulty ★★Formulae › empirical vs molecular
The Problem

Which of the following is both an empirical formula and a molecular formula?

Your answer

❓ Which formula cannot be simplified?

  • A C₃F₆
  • B C₃F₈
  • C C₄F₆
  • D C₄F₈
  • E C₄F₁₀
💡 Hint 1 — what's being tested

A molecular formula is also empirical only when its subscripts have a greatest common divisor of 1. Check the GCD of the two subscripts in each option.

Find the GCD of the C and F subscripts; GCD = 1 means it is already empirical.
💡 Hint 3 — set up the first step
  • C₃F₆ → ÷3 = CF₂; C₄F₆ → ÷2 = C₂F₃; C₄F₈ → ÷4 = CF₂; C₄F₁₀ → ÷2 = C₂F₅.
  • C₃F₈: GCD(3,8) = 1 — cannot simplify.

Test each GCD.

  1. A 3,6 → GCD 3; C 4,6 → GCD 2; D 4,8 → GCD 4; E 4,10 → GCD 2 — all reducible.
  2. B 3,8 → GCD 1 → already empirical.
✅ Answer: B — C₃F₈

Why this is the answer: only C₃F₈ has subscripts (3 and 8) with no common factor, so its molecular formula is already the simplest whole-number ratio.

⭐ Q12ASOE 2021Difficulty ★★★Stoichiometry › find molar mass
The Problem

1.620 g of XF₆ can be produced from 1.000 g of element X. Which of the following could be element X?

Your answer

❓ Which element is X?

  • A W
  • B Se
  • C Mo
  • D Rh
  • E U
💡 Hint 1 — what's being tested

1.000 g of X gains (1.620 − 1.000) = 0.620 g of fluorine as 6 F atoms. The mole of X equals the mole of XF₆, so set up M(X) from the mass ratio: 1.000 g X holds the same moles as 0.620 g of 6 F.

Moles F = 0.620/19.00; moles X = moles F ÷ 6; M(X) = 1.000 ÷ moles X.
💡 Hint 3 — set up the first step
  • Mass of F added = 1.620 − 1.000 = 0.620 g.
  • Six fluorines per X, so divide moles of F by 6.
  • M(W) ≈ 184 g mol⁻¹.

Back out M(X).

  1. Moles F = 0.620/19.00 = 0.03263 mol; moles X = 0.03263/6 = 0.005438 mol.
  2. M(X) = 1.000 / 0.005438 = ≈ 184 g mol⁻¹.
  3. Tungsten is 183.8 g mol⁻¹ — element X = W. Check: 1.000 g W gives 1.000 + 6 × (1.000/183.8) × 19.00 = 1.620 g of WF₆ ✓.
✅ Answer: A — W

Why this is the answer: 0.620 g of fluorine is 0.0326 mol; a sixth of that is 0.00544 mol of X, so M(X) = 1.000/0.00544 ≈ 184 g mol⁻¹ — tungsten. Uranium (238) would give only 1.479 g of XF₆.

⭐ Q13ASOE 2021Difficulty ★★★Periodic › successive ionisation
The Problem

Of the following elements, which has the highest third ionisation energy?

Your answer

❓ Highest 3rd ionisation energy?

  • A Ar
  • B Si
  • C Mg
  • D Al
  • E Cl
💡 Hint 1 — what's being tested

The third electron removed from Mg comes from the closed Ne core (Mg²⁺ is already noble-gas). Breaking into that stable core costs an enormous amount of energy — far more than removing a third electron from the others, which still have valence electrons left.

Find which element must remove its 3rd electron from a noble-gas core — that one spikes.
💡 Hint 3 — set up the first step
  • Mg has only 2 valence electrons; the 3rd comes from the Ne core → huge IE₃.
  • Al, Si, Cl, Ar still have valence electrons available for the 3rd removal.

Spot the core break.

  1. Mg: [Ne]3s². Removing the 3rd electron means dipping into the [Ne] core → very high IE₃.
  2. Al ([Ne]3s²3p¹), Si, Cl, Ar all still have 3rd-shell electrons for the 3rd removal — much lower IE₃.
✅ Answer: C — Mg

Why this is the answer: Mg²⁺ is isoelectronic with neon, so the third ionisation breaks the closed core and costs by far the most energy.

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