IB DP Chemistry Structure 3.1: The periodic table: classification of elements – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This set tests Structure 3.1, The periodic table: classification of elements, based on the IB Chemistry guide for the Diploma Programme (February 2023 edition, first examined in 2025). Questions 1 to 7 cover understandings 3.1.1 to 3.1.6, which apply at SL and HL. Questions 8 to 11 test the additional higher level understandings 3.1.7 to 3.1.10, so each carries an HL only label. It applies from 2025 onwards.

First read the study guide or the revision notes. See also the course hub, the checklist and the free diagnostics.

Questions

1. Element X is in period 3 and group 14 of the periodic table.

(a) State the block that X belongs to. [1] (b) Deduce the full electron configuration of X. [1] (c) Classify X as a metal, a metalloid or a non-metal. [1]

2. An atom has the condensed electron configuration [Ar] 3d³ 4s².

(a) Identify the element, and state its period and group. [2] (b) State which of these classifications applies to it: alkali metal, halogen, transition element, noble gas. [1]

3. Explain each observation.

(a) Atomic radius decreases from sodium to chlorine across period 3. [2] (b) A magnesium ion, Mg²⁺, is smaller than a magnesium atom. [2] (c) Electronegativity decreases from fluorine to iodine. [2]

4. A small piece of rubidium is dropped into a trough of water.

(a) Give a balanced equation for what happens, including state symbols. [2] (b) Explain why rubidium reacts more vigorously than potassium. [3] (c) Universal indicator is added to the water afterwards. State and explain what it shows. [1]

5. Aqueous bromine is added to two test tubes, one holding potassium iodide solution and one holding potassium chloride solution.

(a) Describe the observation in each of the two tubes. [2] (b) Write the ionic equation for the change in the iodide tube. [1] (c) Explain, in terms of atomic structure, the difference between the two tubes. [2]

6. Oxides and the environment.

(a) Write equations for the reactions with water of caesium oxide and of sulfur dioxide. [2] (b) Aluminium oxide reacts with both hydrochloric acid and sodium hydroxide solution. State the term for this behaviour. [1] (c) Explain, using equations, how a rise in atmospheric carbon dioxide lowers the pH of seawater and affects organisms that build carbonate shells. [4] (d) State how the acid–base character of the oxides changes across period 3, and link it to metallic character. [1]

7. Deduce the oxidation state of the named atom in each species.

(a) Hydrogen in MgH₂ [1] (b) Oxygen in MgO₂ (magnesium peroxide) [1] (c) Vanadium in VO₃⁻ [1] (d) Chromium in Cr₂O₇²⁻ [1] (e) Explain why the oxidation state of oxygen in ozone, O₃, is zero. [1] (f) Give names, using Roman numerals for the oxidation state, for KNO₂ and Na₂SO₃. [2]

8. HL only. The first ionization energy of sulfur is lower than that of phosphorus.

(a) Draw orbital diagrams (boxes or lines) for the 3s and 3p electrons of a phosphorus atom and a sulfur atom. [2] (b) Use your diagrams to explain the lower value for sulfur, and say why this is evidence for sublevels. [2]

9. HL only. Transition elements.

(a) State four characteristic properties of transition elements. [2] (b) Zinc sits in the d-block. Explain why it does not count as a transition element. [2] (c) State what is meant by a ligand. [1]

10. HL only. The first four ionization energies of chromium are, in kJ mol⁻¹ (use these values): 653, 1591, 2987, 4743.

(a) Use the data to explain why chromium forms ions with several different charges. [3] (b) Deduce the electron configurations of Cr²⁺ and Mn⁴⁺. [2] (c) State the oxidation state of manganese in MnO₂ and name the compound. [1]

11. HL only. Use this colour wheel (approximate ranges): violet 400–450 nm, blue 450–490 nm, green 490–560 nm, yellow 560–590 nm, orange 590–630 nm, red 630–700 nm. Complementary pairs: red and green, orange and blue, yellow and violet. Take c = 3.00 × 10⁸ m s⁻¹.

(a) Write the electron configuration of Fe²⁺. [1] (b) Explain why complexes of Fe²⁺ in solution are coloured. [3] (c) A solution of a different complex absorbs light most strongly at a frequency of 4.80 × 10¹⁴ s⁻¹. Calculate the wavelength absorbed, in nm. [2] (d) Deduce the colour of the solution in (c). [1] (e) Explain why solutions of zinc compounds are colourless. [2]

Answers

1. (a) p-block (the last electron enters 3p) [1] (b) 1s² 2s² 2p⁶ 3s² 3p² (14 electrons; this is silicon) [1] (c) Metalloid [1] [3] Examiner insight: When a full configuration is asked for, write every sublevel; [Ne] 3s² 3p² may not earn the mark.

2. (a) Vanadium (Z = 18 + 5 = 23) [1]; period 4, group 5 (two 4s plus three 3d electrons) [1] (b) Transition element [1] [3] Examiner insight: Count both the 4s and the 3d electrons to find the group: 3d³ 4s² puts the element in group 5, not group 3.

3. (a) Nuclear charge increases across the period [1]; electrons enter the same main level with similar shielding, so they are pulled closer to the nucleus [1] (b) Mg²⁺ has lost its two 3s electrons, so it has one fewer occupied main level [1]; the remaining electrons are held closer by the same nuclear charge (12 protons, now 10 electrons) [1] (c) Down the group the atom has more occupied main levels, so the bonding pair is further from the nucleus [1]; more shielding by inner electrons means the shared pair is attracted less strongly [1] [6] Examiner insight: “More protons” alone rarely earns full credit in (a); pair it with the same-level, similar-shielding point.

4. (a) 2Rb(s) + 2H₂O(l) → 2RbOH(aq) + H₂(g): correct formulae [1]; balanced with state symbols [1] (b) Rubidium has an extra occupied main level, so its outer electron is further from the nucleus [1]; it is more shielded by inner electrons [1]; less energy is needed to remove it, so Rb forms Rb⁺ more readily [1] (c) Blue/purple, showing an alkaline solution (pH above 7) because hydroxide ions form [1] [6] Examiner insight: In (b), saying rubidium “has more electrons” without linking distance and shielding to ease of electron loss is likely to score little.

5. (a) Potassium iodide: the solution turns brown (iodine forms) [1]; potassium chloride: no change, it stays the yellow-orange colour of the added bromine [1] (b) Br₂(aq) + 2I⁻(aq) → 2Br⁻(aq) + I₂(aq) [1] (c) A bromine atom is smaller than an iodine atom, so it attracts an extra electron more strongly and can oxidize I⁻ [1]; a chlorine atom is smaller still and attracts electrons more strongly than bromine, so bromine cannot remove electrons from Cl⁻ [1] [5] Examiner insight: “Brown” must describe the iodine that forms; describing only the colour of the reagent added does not show a reaction.

6. (a) Cs₂O(s) + H₂O(l) → 2CsOH(aq) [1]; SO₂(g) + H₂O(l) → H₂SO₃(aq) [1] (b) Amphoteric [1] (c) More CO₂ dissolves: CO₂(g) + H₂O(l) ⇌ H₂CO₃(aq) [1]; H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq), so [H⁺] rises and pH falls [1]; H⁺(aq) + CO₃²⁻(aq) → HCO₃⁻(aq) [1]; less carbonate is left for organisms to make calcium carbonate shells [1] (d) Oxides change from basic to amphoteric to acidic as metallic character decreases across the period [1] [8] Examiner insight: In (c), equilibrium arrows are expected for the CO₂ and carbonic acid steps; a one-way arrow may cost the mark.

7. (a) −1 (magnesium is +2; a metal hydride) [1] (b) −1 (magnesium is +2; a peroxide) [1] (c) x + 3(−2) = −1, so +5 [1] (d) 2x + 7(−2) = −2, so +6 [1] (e) O₃ is an element: each oxygen bonds only to other oxygen atoms, so no electrons are transferred [1] (f) Potassium nitrate(III) (or potassium nitrite) [1]; sodium sulfate(IV) (or sodium sulfite) [1] [7] Examiner insight: Write oxidation states as −1 and +5, sign first; “1−” and “5+” read as ion charges and may not be accepted.

8. (a) P: 3s [↑↓], 3p [↑][↑][↑] [1]; S: 3s [↑↓], 3p [↑↓][↑][↑], with opposite spins in the paired orbital [1] (b) The electron removed from sulfur is one of a pair in a 3p orbital; repulsion between the paired electrons raises its energy, so less energy is needed to remove it [1]; electrons in the same main level therefore differ in energy, which shows that sublevels and orbitals exist [1] [4] Examiner insight: Answers that rely on “half-filled stability” of phosphorus, without the energy of the electron removed, are unlikely to gain the explanation mark.

9. (a) Any two of: variable oxidation states, high melting points, magnetic properties, catalytic properties, coloured compounds, complex ion formation [1]; two more from the list [1] (b) A transition element has an incomplete d-sublevel [1]; zinc has 3d¹⁰ in the atom and in its only ion, Zn²⁺, so the d-sublevel is always full [1] (c) A molecule or ion with a lone pair that it donates to a central metal ion, forming a coordination bond [1] [5] Examiner insight: In (b), mention the ion as well as the atom; the atom alone does not rule zinc out.

10. (a) The increases are 938, 1396 and 1756 kJ mol⁻¹: each value rises steadily with no very large jump [1]; this is because the 4s and 3d electrons are close in energy [1]; so the energy cost of losing two, three or more electrons is similar in scale, allowing several stable oxidation states [1] (b) Cr²⁺: [Ar] 3d⁴ [1]; Mn⁴⁺: [Ar] 3d³ [1] (c) +4; manganese(IV) oxide [1] [6] Examiner insight: Quote figures from the data in (a); a general statement with no reference to the values given is unlikely to earn the first mark.

11. (a) [Ar] 3d⁶ [1] (b) Ligands split the d-sublevel into two sets of orbitals of different energy [1]; a d electron absorbs a photon of visible light whose energy matches the gap and is promoted [1]; the colour seen is complementary to the colour absorbed [1] (c) λ = c / f = (3.00 × 10⁸) / (4.80 × 10¹⁴) [1] = 6.25 × 10⁻⁷ m = 625 nm [1] (d) 625 nm is orange, so the solution looks blue [1] (e) Zn²⁺ is [Ar] 3d¹⁰: the d-sublevel is full [1], so no electron can be promoted between split d orbitals and no visible light is absorbed [1] [9] Examiner insight: In (d), naming the absorbed colour (orange) as the colour seen is a common way to lose the mark.

Where marks are usually lost

• Writing a condensed configuration when a full one is asked for.
• Explaining radius or electronegativity trends with nuclear charge alone, without shielding or main levels.
• Writing CO₂ + H₂O → H₂CO₃ as one-way, or leaving out the carbonate step in ocean acidification.
• Using −2 for oxygen in a peroxide and +1 for hydrogen in a metal hydride.
• (HL) Removing 3d electrons before 4s in ion configurations.
• (HL) Giving the absorbed colour as the observed colour.
• (HL) Leaving nm in c = λf, or giving no unit for the final wavelength.

Next steps

• All free 10-minute diagnostics
• Structure 3.1 revision notes
• Structure 3.1 study guide
• Hub page for the course
• Topic checklist
• Book a free trial class

Official syllabus

Content here maps to the International Baccalaureate Organization’s Diploma Programme Chemistry guide, published in February 2023 for first assessment in 2025. Sub-topic covered: Structure 3.1, The periodic table: classification of elements. Understandings 3.1.1 to 3.1.6 are SL and HL; 3.1.7 to 3.1.10 are additional higher level.