About this IB Chemistry practice test

Kinetics, equilibria, and organic reaction pathways carry the most weight across both SL and HL IB Chemistry papers, so that's where this set concentrates. HL-only topics like stereoisomerism and spectroscopic analysis get the depth they need, and every answer walks through the actual chemical reasoning rather than just stating the result.

IB Chemistry Practice Test sample questions

These starter questions help you launch a chemistry mock test quickly. Swap them with your own worksheet, notebook, or textbook questions any time.

1. 1. What is the shape of a molecule with 4 bonding pairs and 0 lone pairs?

2. A) Linear

   • B) Trigonal planar
   • C) Tetrahedral
   • D) Bent

3. 2. State Le Chatelier's principle.

4. A) Equilibrium favours the forward reaction at high temperature

   • B) A system at equilibrium shifts to oppose any imposed change
   • C) Pressure only affects gaseous equilibria
   • D) Kc increases with temperature

5. 3. Calculate the pH of 0.01 mol/dm³ HCl.

6. A) 1

   • B) 2
   • C) 3
   • D) 4

7. 4. Which compound has the highest boiling point?

8. A) CH₄

   • B) NH₃
   • C) SiH₄
   • D) PH₃

9. 5. A buffer solution resists change in:

10. A) temperature

    • B) concentration
    • C) pH
    • D) pressure

11. 6. The standard enthalpy of formation of an element in its standard state is:

12. A) positive

    • B) negative
    • C) zero
    • D) undefined

13. 7. Which type of reaction forms an ester from an alcohol and carboxylic acid?

14. A) Addition

    • B) Substitution
    • C) Condensation
    • D) Elimination

15. 8. What is the hybridisation of carbon atoms in ethene (C₂H₄)?

16. A) sp³

    • B) sp²
    • C) sp
    • D) unhybridised

17. 9. For a reaction with Kc = 0.001 at 298 K, the equilibrium:

18. A) strongly favours products

    • B) slightly favours products
    • C) strongly favours reactants
    • D) is perfectly balanced

19. 10. What colour are copper(II) ions in aqueous solution?

20. A) Yellow

    • B) Green
    • C) Blue
    • D) Colourless

21. 11. Which trend correctly describes first ionization energy across Period 3 (Na to Ar)?

22. A) Decreases smoothly

    • B) Increases generally with some irregularities
    • C) Remains constant
    • D) Increases then sharply decreases at Ar

23. 12. What best describes the shape of NH₃, which has 3 bonding pairs and 1 lone pair on the central atom?

24. A) Trigonal planar electron domain and molecular shape

    • B) Tetrahedral electron domain, trigonal pyramidal molecular shape
    • C) Bent molecular shape
    • D) Trigonal bipyramidal electron domain

25. 13. Which type of bonding best explains the high electrical conductivity of solid sodium?

26. A) Ionic bonding with mobile ions

    • B) Covalent network bonding
    • C) Metallic bonding with delocalised electrons
    • D) Hydrogen bonding

27. 14. What is the hybridisation of the central Be atom in BeCl₂?

28. A) sp

    • B) sp²
    • C) sp³
    • D) sp³d

29. 15. Using bond enthalpies (H–H = 436, Cl–Cl = 242, H–Cl = 431 kJ/mol), calculate ΔH for: H₂(g) + Cl₂(g) → 2HCl(g)

30. A) +184 kJ/mol

    • B) −184 kJ/mol
    • C) −678 kJ/mol
    • D) +862 kJ/mol

31. 16. Which statement correctly describes Hess's Law?

32. A) Enthalpy change depends on the reaction pathway

    • B) Enthalpy change is independent of the pathway taken
    • C) Hess's Law only applies to exothermic reactions
    • D) Hess's Law cannot be used with enthalpies of formation

33. 17. According to the Arrhenius equation, increasing temperature increases the rate constant k primarily because:

34. A) The activation energy decreases

    • B) More molecules have energy ≥ activation energy
    • C) The frequency factor A decreases
    • D) The reaction becomes exothermic

35. 18. For a reaction to be spontaneous (ΔG < 0) at high temperature but non-spontaneous at low temperature, the signs of ΔH and ΔS must be:

36. A) ΔH negative, ΔS negative

    • B) ΔH positive, ΔS positive
    • C) ΔH negative, ΔS positive
    • D) ΔH positive, ΔS negative

37. 19. For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), increasing the pressure will:

38. A) Shift equilibrium towards reactants

    • B) Shift equilibrium towards products
    • C) Have no effect on equilibrium position
    • D) Decrease the value of Kc

39. 20. Calculate the pH of a 0.001 mol/dm³ NaOH solution at 298 K.

40. A) 3

    • B) 9
    • C) 11
    • D) 13

41. 21. A buffer solution is prepared from CH₃COOH and CH₃COONa. What happens when a small amount of HCl is added?

42. A) The pH decreases sharply

    • B) H⁺ reacts with CH₃COO⁻ to form CH₃COOH, minimizing the pH change
    • C) The buffer stops working immediately
    • D) OH⁻ ions neutralize the HCl

43. 22. In the electrolysis of molten NaCl, which half-reaction occurs at the cathode?

44. A) 2Cl⁻ → Cl₂ + 2e⁻

    • B) Na⁺ + e⁻ → Na
    • C) 2H₂O + 2e⁻ → H₂ + 2OH⁻
    • D) Na → Na⁺ + e⁻

45. 23. Which functional group is present in propanone (CH₃COCH₃)?

46. A) Aldehyde

    • B) Ketone
    • C) Carboxylic acid
    • D) Ester

47. 24. What type of reaction converts an alkene into a haloalkane using HBr?

48. A) Condensation

    • B) Elimination
    • C) Electrophilic addition
    • D) Nucleophilic substitution

49. 25. Which reagent and conditions convert a primary alcohol into an aldehyde without further oxidation to a carboxylic acid?

50. A) Excess acidified K₂Cr₂O₇, reflux

    • B) Acidified K₂Cr₂O₇, distillation
    • C) LiAlH₄ in dry ether
    • D) Concentrated H₂SO₄, heat

51. 26. Which of the following molecules exhibits E/Z isomerism?

52. A) CH₂=CH₂

    • B) CH₃CH=CHCH₃ (but-2-ene)
    • C) CH₂=CCl₂
    • D) (CH₃)₂C=CH₂

53. 27. A chiral carbon is best described as a carbon atom bonded to:

54. A) Four identical groups

    • B) Four different groups
    • C) Three different groups and one hydrogen
    • D) Two pairs of identical groups

55. 28. In the IR spectrum of a carboxylic acid, which absorption pattern is most diagnostic?

56. A) A sharp C=O stretch near 1700 cm⁻¹ only

    • B) A broad O–H stretch (2500–3300 cm⁻¹) combined with a C=O stretch (~1700–1725 cm⁻¹)
    • C) A sharp N–H stretch near 3300 cm⁻¹
    • D) A C≡C stretch near 2100 cm⁻¹

57. 29. In the ¹H NMR spectrum of ethanol (CH₃CH₂OH), the CH₂ protons appear as a quartet. This splitting is due to:

58. A) Coupling with the 3 equivalent CH₃ protons (n+1 rule)

    • B) Coupling with the OH proton
    • C) Coupling with both the CH₃ and OH protons combined
    • D) The CH₂ protons are not split at all

59. 30. A chemist wants to convert bromoethane into ethanoic acid in two steps, keeping the same number of carbon atoms. Which sequence is correct?

60. A) Reduce with LiAlH₄, then oxidize with acidified K₂Cr₂O₇

    • B) Hydrolyse with aqueous NaOH to form ethanol, then oxidize with excess acidified K₂Cr₂O₇ under reflux
    • C) React with KCN, then hydrolyse the nitrile
    • D) React directly with acidified KMnO₄

Answer key & quick explanations

Short answers for the sample questions above. Use this to self-check before generating a fresh AI-built mock test.

1. 1. VSEPR shape - 4 bonding pairs, 0 lone pairs

   C) Tetrahedral

   With 4 bonding pairs and no lone pairs, electron pair repulsion is minimized in a tetrahedral arrangement (109.5° bond angles), as in CH₄. Trigonal planar (B) requires only 3 electron domains, and a bent shape (D) would need 2 bonding pairs plus lone pairs.

2. 2. Le Chatelier's principle statement

   B) A system at equilibrium shifts to oppose any imposed change

   Le Chatelier's principle states that when a change in concentration, pressure, or temperature is applied to a system at equilibrium, the position shifts to counteract that change. Option A is incomplete since temperature actually changes the value of Kc itself, and option C is wrong because pressure changes do affect gaseous equilibria with unequal gas moles on each side.

3. 3. pH of 0.01 mol/dm³ HCl

   B) 2

   HCl is a strong acid that fully dissociates, so [H⁺] = 0.01 = 1×10⁻² mol/dm³. pH = −log(1×10⁻²) = 2. Option C (pH 3) would correspond to a 10× more dilute solution.

4. 4. Highest boiling point: CH₄/NH₃/SiH₄/PH₃

   B) NH₃

   NH₃ molecules form hydrogen bonds due to the highly polar N–H bond, requiring more energy to separate molecules and giving a much higher boiling point (−33°C) than CH₄, SiH₄, or PH₃, none of which hydrogen bond. SiH₄ (C) has a higher boiling point than CH₄ from greater van der Waals forces, but still lacks hydrogen bonding.

5. 5. Buffer solution resists change in...

   C) pH

   A buffer resists changes in pH when small amounts of acid or base are added, because it contains a conjugate acid-base pair that can neutralise added H⁺ or OH⁻. It does not specifically resist changes in temperature (A) or pressure (D).

6. 6. Standard enthalpy of formation of an element in its standard state

   C) zero

   By definition, the standard enthalpy of formation of an element in its most stable standard state (e.g. O₂(g), C(graphite)) is zero, since no reaction is needed to form the element from itself. This is the reference point used throughout Hess's Law cycles.

7. 7. Ester formation reaction type

   C) Condensation

   Esterification between an alcohol and carboxylic acid releases water as the ester bond forms, which is the defining feature of a condensation reaction. It is not addition (A, which adds atoms across a double bond) or elimination (D, which removes atoms to form a double bond).

8. 8. Hybridisation of carbon in ethene

   B) sp²

   Each carbon in ethene (C₂H₄) forms three σ bonds (two C–H and one C–C) and shares in one π bond, requiring three sp² hybrid orbitals and leaving one unhybridised p orbital for the π bond. sp³ (A) would apply to a fully single-bonded carbon as in ethane.

9. 9. Kc = 0.001 equilibrium position

   C) strongly favours reactants

   Since Kc = [products]/[reactants] = 0.001 ≪ 1, reactant concentrations dominate at equilibrium, meaning very little product forms and the position lies far to the left, not the right as options A and B suggest.

10. 10. Colour of aqueous Cu²⁺ ions

    C) Blue

    Aqueous Cu²⁺ exists as the complex ion [Cu(H₂O)₆]²⁺, which absorbs light in the red/orange region of the visible spectrum and so appears blue to the eye. Yellow (A) and green (B) are more typical of other transition-metal complexes such as those of Fe³⁺ or Ni²⁺.

11. 11. Period 3 ionization energy trend

    B) Increases generally with some irregularities

    First ionization energy generally increases across Period 3 due to increasing nuclear charge with similar shielding, but there are dips at Al (removing the single 3p electron is easier than removing a paired 3s electron) and at S (extra repulsion from pairing electrons in one 3p orbital). It does not decrease smoothly (A) or stay constant (C).

12. 12. Electron domain vs molecular shape (NH₃-type)

    B) Tetrahedral electron domain, trigonal pyramidal molecular shape

    With 3 bonding pairs and 1 lone pair, there are 4 electron domains arranged tetrahedrally, but since the lone pair is not counted when describing the molecular shape, the observed shape (as in NH₃) is trigonal pyramidal. It is not simply trigonal planar (A), which requires exactly 3 electron domains and no lone pairs.

13. 13. Metallic bonding and conductivity in sodium

    C) Metallic bonding with delocalised electrons

    Solid sodium consists of a lattice of Na⁺ cations surrounded by a 'sea' of delocalised valence electrons, which are free to move throughout the structure and carry electric current. Covalent network bonding (B, as in diamond) generally does not conduct because electrons are localised within fixed bonds.

14. 14. Hybridisation of central atom in BeCl₂

    A) sp

    Be in BeCl₂ has only 2 bonding pairs and no lone pairs, giving a linear shape that requires mixing one s and one p orbital into two sp hybrid orbitals. sp³ (C) would require 4 electron domains, which BeCl₂ does not have.

15. 15. Bond enthalpy calculation for H₂ + Cl₂ → 2HCl

    B) −184 kJ/mol

    ΔH = (bonds broken) − (bonds formed) = (436 + 242) − (2 × 431) = 678 − 862 = −184 kJ/mol. The reaction is exothermic because the two new H–Cl bonds formed are collectively stronger than the H–H and Cl–Cl bonds broken, ruling out a positive value (A).

16. 16. Hess's Law statement

    B) Enthalpy change is independent of the pathway taken

    Hess's Law states that the total enthalpy change for a reaction is the same regardless of the route taken, because enthalpy is a state function. This allows unknown enthalpy changes to be calculated from known steps, and it applies to any reaction, not just exothermic ones (C).

17. 17. Arrhenius equation - effect of temperature on k

    B) More molecules have energy ≥ activation energy

    Raising temperature shifts the Maxwell-Boltzmann distribution so that a greater fraction of molecules have kinetic energy equal to or exceeding the activation energy, which increases the rate constant k. The activation energy itself is a fixed property of the reaction pathway and does not change with temperature (A).

18. 18. ΔG spontaneity: signs of ΔH and ΔS

    B) ΔH positive, ΔS positive

    Using ΔG = ΔH − TΔS, if ΔH is positive and ΔS is positive, at low T the −TΔS term is small so ΔG stays positive (non-spontaneous), but at high T the −TΔS term grows large enough to make ΔG negative (spontaneous). If both ΔH and ΔS were negative (A), the reaction would instead be spontaneous only at low temperature.

19. 19. Effect of pressure on N₂ + 3H₂ ⇌ 2NH₃

    B) Shift equilibrium towards products

    Increasing pressure shifts equilibrium towards the side with fewer moles of gas; the product side has 2 mol NH₃ compared with 4 mol of reactant gases (1 N₂ + 3 H₂), so equilibrium shifts right. Kc itself is unaffected by pressure changes at constant temperature (D), only the equilibrium position shifts.

20. 20. pH of 0.001 mol/dm³ NaOH

    C) 11

    NaOH is a strong base that fully dissociates, giving [OH⁻] = 1×10⁻³ mol/dm³, so pOH = 3 and pH = 14 − 3 = 11 at 298 K. A pH of 13 (D) would correspond to a 100× more concentrated NaOH solution.

21. 21. Buffer response to added HCl

    B) H⁺ reacts with CH₃COO⁻ to form CH₃COOH, minimizing the pH change

    Added H⁺ ions are consumed by the conjugate base CH₃COO⁻ to form the weak acid CH₃COOH, limiting the rise in free [H⁺] and keeping the pH change small rather than causing a sharp drop (A). The buffer does not stop working after a small addition (C); it remains effective until one component is largely used up.

22. 22. Electrolysis of molten NaCl - cathode reaction

    B) Na⁺ + e⁻ → Na

    The cathode is the site of reduction, so Na⁺ ions gain electrons to form Na metal. Oxidation of Cl⁻ to Cl₂ (A) occurs at the anode instead, and since the electrolyte is molten (not aqueous), water is not involved (C).

23. 23. Functional group in propanone

    B) Ketone

    Propanone, CH₃COCH₃, has a carbonyl group (C=O) bonded to two carbon-containing groups on either side, which is the defining feature of a ketone. An aldehyde (A) would instead have the carbonyl carbon bonded to at least one hydrogen atom.

24. 24. Alkene + HBr reaction type

    C) Electrophilic addition

    HBr adds across the C=C double bond via an electrophilic addition mechanism, in which the electron-rich double bond attacks the electrophilic H, followed by Br⁻ attacking the resulting carbocation. This is not substitution (D), since no atom leaves the original carbon skeleton, and the double bond is being used up rather than formed as in elimination (B).

25. 25. Primary alcohol to aldehyde without further oxidation

    B) Acidified K₂Cr₂O₇, distillation

    Using acidified potassium dichromate with distillation removes the aldehyde from the mixture as it forms, since aldehydes have a lower boiling point than the parent alcohol, preventing further oxidation to the carboxylic acid. Using excess oxidant under reflux (A) keeps the aldehyde in prolonged contact with the oxidising agent, driving oxidation all the way to the carboxylic acid.

26. 26. E/Z isomerism identification

    B) CH₃CH=CHCH₃ (but-2-ene)

    But-2-ene has two different groups (CH₃ and H) attached to each double-bond carbon, allowing distinct E and Z arrangements. In CH₂=CCl₂ (C) and (CH₃)₂C=CH₂ (D), at least one double-bond carbon carries two identical substituents, which rules out E/Z isomerism.

27. 27. Definition of a chiral carbon

    B) Four different groups

    A chiral carbon (stereocentre) is bonded to four different substituents, meaning the molecule and its mirror image cannot be superimposed, producing optical isomers (enantiomers). A carbon bonded to four identical groups (A) has no chirality since it possesses a plane of symmetry.

28. 28. IR spectrum diagnostic for carboxylic acids

    B) Broad O–H stretch (2500–3300 cm⁻¹) combined with C=O stretch (~1700–1725 cm⁻¹)

    Carboxylic acids show a very broad O–H stretch (2500–3300 cm⁻¹), broadened by strong hydrogen bonding between dimerised molecules, together with a strong C=O stretch around 1700–1725 cm⁻¹; seeing both features together is the diagnostic signature. A sharp C=O peak alone (A) could just as easily indicate a ketone or aldehyde.

29. 29. ¹H NMR splitting of CH₂ in ethanol

    A) Coupling with the 3 equivalent CH₃ protons (n+1 rule)

    By the n+1 rule, a proton environment adjacent to n equivalent protons splits into n+1 peaks; the CH₂ protons are adjacent to 3 equivalent CH₃ protons, giving a quartet. The OH proton typically does not couple with neighbouring protons due to fast proton exchange in solution, making option B incorrect.

30. 30. Two-step synthesis: bromoethane to ethanoic acid

    B) Hydrolyse with aqueous NaOH to form ethanol, then oxidize with excess acidified K₂Cr₂O₇ under reflux

    Nucleophilic substitution of bromoethane with aqueous NaOH gives ethanol, which is then fully oxidised under reflux with excess acidified dichromate to ethanoic acid, preserving the original 2-carbon chain. Reacting with KCN then hydrolysing the nitrile (C) instead adds an extra carbon atom, producing propanoic acid rather than ethanoic acid.

Curriculum Mapping & Learning Guide

Use this breakdown to identify which skills each question tests and guide post-test review.

IB Chemistry units covered

1. Chapter 1: Structure: atomic structure, periodic trends, ionic/covalent/metallic bonding, VSEPR, hybridisation
2. Chapter 2: Energetics: enthalpy (Hess's Law, bond enthalpy), Arrhenius equation, Gibbs energy
3. Chapter 3: Equilibria: Kc expressions, Le Chatelier, pH calculations, buffers, electrochemistry
4. Chapter 4: Organic: functional groups, reaction types (substitution/addition/elimination/condensation)
5. Chapter 5: Organic: stereoisomerism, IR spectroscopy, ¹H NMR, multi-step synthesis (HL)

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