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IB Chemistry practice questions.

1,461 exam-style questions with answers and explanations, organised by the official syllabus. Try the samples below, then practise the full bank free.

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The successive ionisation energies (kJ/mol) of an element are: 738, 1451, 7733, 10540. Which group does the element belong to?

  • Group 1
  • Group 2
  • Group 13
  • Group 14

Answer: Group 2

There is a large jump between the 2nd (1451) and 3rd (7733) ionisation energies. This means two electrons are removed easily before reaching a stable inner (noble-gas) shell, so the element has 2 valence electrons and is in Group 2 (this is magnesium).

According to VSEPR theory, what is the molecular geometry and bond angle of ammonia (NH3)?

  • Trigonal planar, with bond angles of exactly 120 degrees
  • Trigonal pyramidal, about 107 degrees
  • Tetrahedral, 109.5 degrees
  • Bent, about 104.5 degrees

Answer: Trigonal pyramidal, about 107 degrees

NH3 has 4 electron domains around N (3 bonding pairs + 1 lone pair). The electron geometry is tetrahedral, but the shape describing atom positions is trigonal pyramidal. The lone pair repels more strongly, compressing the H-N-H angle from 109.5 to about 107 degrees.

Which species has the smallest radius: Na+, Mg2+, F-, or O2- (all isoelectronic with neon)?

  • Na+
  • Mg2+
  • F-
  • O2-

Answer: Mg2+

All four are isoelectronic (10 electrons). The radius decreases as nuclear charge increases because more protons pull the same electron cloud inward. Mg2+ has 12 protons, the most of the four, so it has the smallest radius.

Heating 100 g of water (c = 4.18 J/g/K) raises its temperature by 20 K. How much heat energy is absorbed? Use Q = mcΔT.

  • 8.36 kJ
  • 83.6 kJ
  • 0.836 kJ
  • 836 kJ

Answer: 8.36 kJ

Q = mcΔT = 100 × 4.18 × 20 = 8360 J = 8.36 kJ (divide by 1000 to convert J to kJ). Mass is in grams to match the units of c; a slip in the J-to-kJ conversion gives the other options.

A compound contains 40.0% C, 6.7% H and 53.3% O by mass. (Ar: C = 12.0, H = 1.0, O = 16.0.) What is its empirical formula?

  • CH2O
  • C2H4O2
  • CHO
  • C2H6O

Answer: CH2O

Divide each percentage by its Ar: C = 40.0/12.0 = 3.33, H = 6.7/1.0 = 6.7, O = 53.3/16.0 = 3.33. Divide by the smallest (3.33): C = 1, H = 2, O = 1. The empirical formula is CH2O.

A weak acid has Ka = 1.0 x 10^-5. What is its pKa?

  • 5
  • -5
  • 9
  • 1.0 x 10^-5

Answer: 5

pKa = -log(Ka) = -log(1.0 x 10^-5) = 5. A smaller Ka (larger pKa) means a weaker acid.

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