Eduqas A-level Chemistry (A410QS) · PI2 — More complex patterns of the Periodic Table
Mini-Lesson
PI2 · Complex patterns of the Periodic Table
Eduqas PI2 takes the Periodic Table beyond simple trends: the p-block (amphoteric aluminium, the oxides and chlorides of Period 3) and then the d-block transition metals — variable oxidation states, complex ions and ligands, the origin of colour, ligand substitution, and catalysis.
Work through each screen, answer every question as you go — the multiple-choice checks, the calculations and the sorting games — and collect ⭐ stars. Press Start when you are ready.
PI2 · the p-block
Aluminium and amphoteric behaviour
Aluminium sits on the metal / non-metal borderline, so Al₂O₃ and Al(OH)₃ are amphoteric — they react with both acids and alkalis.
Al(OH)₃ + 3HCl → AlCl₃ + 3H₂Obehaving as a BASE · Al 1 = 1 · Cl 3 = 3 · H 6 = 6 · O 3 = 3
Al(OH)₃ + NaOH → NaAl(OH)₄behaving as an ACID, forming the aluminate ion [Al(OH)₄]⁻
Aluminium metal resists corrosion because a thin, tough, impermeable layer of Al₂O₃ forms instantly on the surface — anodising thickens it deliberately.
Group 4 and the inert pair effect: going down C → Si → Ge → Sn → Pb, the +2 oxidation state becomes progressively more stable relative to +4. So Sn²⁺ is a reducing agent (it wants to become Sn⁴⁺), whereas Pb⁴⁺ is an oxidising agent (it wants to become Pb²⁺).
Quick check
What does amphoteric mean?
?Aluminium hydroxide dissolves both in dilute hydrochloric acid and in aqueous sodium hydroxide. What term describes this?
PI2 · Period 3
Oxides of Period 3
Across Period 3 the oxides change from ionic and basic to covalent and acidic:
Na₂O and MgO — ionic, basic. Na₂O dissolves to give a strongly alkaline solution (pH ≈ 13); MgO is only sparingly soluble (pH ≈ 9).
Al₂O₃ — amphoteric, and insoluble in water.
SiO₂ — giant covalent, weakly acidic, insoluble in water but attacked by hot concentrated alkali.
P₄O₁₀ and SO₃ — simple covalent, strongly acidic; they react vigorously with water to give phosphoric(V) acid and sulfuric acid (pH ≈ 0–2).
The chlorides follow the same story: NaCl and MgCl₂ simply dissolve, but AlCl₃, SiCl₄ and PCl₅ are hydrolysed, fuming in moist air and producing acidic solutions of HCl.
Sort it
Acidic, basic or amphoteric?
Tap an oxide, then the box that describes its reaction with water, acid and alkali.
🔵 Basic oxide
🟣 Amphoteric oxide
🔴 Acidic oxide
PI2 · transition metals
What makes a transition metal?
A transition element is a d-block element that forms at least one stable ion with a partially filled d subshell.
That single definition explains all four characteristic properties:
Variable oxidation states — the 4s and 3d orbitals are very close in energy, so a variable number of electrons can be lost.
Coloured ions — d–d transitions are only possible if the d subshell is partly filled.
Catalytic activity — variable oxidation states let them accept and donate electrons; and their surfaces adsorb reactants.
Complex ion formation — the empty d orbitals accept lone pairs from ligands.
The two exceptions:Sc forms only Sc³⁺, which is 3d⁰ (empty), and Zn forms only Zn²⁺, which is 3d¹⁰ (full). Neither has a partly filled d subshell in its ion, so neither is a transition metal — and their compounds are white.
Quick check
Why is zinc not a transition metal?
?Zinc is in the d-block but is not classed as a transition metal. Why?
PI2 · complexes
Complex ions, ligands and shapes
A ligand is a species with a lone pair that forms a dative covalent (coordinate) bond to a central metal ion. The coordination number is the number of coordinate bonds.
Monodentate — one donor atom: H₂O, NH₃, Cl⁻, CN⁻, OH⁻.
Bidentate — two donor atoms: ethane-1,2-diamine ("en"), the ethanedioate ion.
Multidentate — EDTA⁴⁻ is hexadentate; haem is tetradentate.
Shapes: 6 coordinate bonds → octahedral (90°), the commonest, with small ligands like H₂O and NH₃. 4 coordinate bonds → usually tetrahedral (109.5°) with the bulkier Cl⁻; but Pt(II) complexes such as cisplatin are square planar. 2 → linear, e.g. [Ag(NH₃)₂]⁺ in Tollens' reagent.
The chelate effect: a multidentate ligand displaces monodentate ones because the reaction produces more particles — a large positive entropy change makes ΔG negative.
Quick check
Count the bonds
?Which is the shape of the complex ion [Cr(H₂O)₆]³⁺ ?
Calculate
Your turn — oxidation state in a complex
1What is the oxidation state of iron in the complex ion [Fe(CN)₆]³⁻ ? Each CN ligand carries a charge of −1.
(sign + number)
Hint: x + 6(−1) = −3.
PI2 · colour
Why transition metal complexes are coloured
Ligands split the five d orbitals into two groups of slightly different energy. An electron can absorb a photon of visible light and jump the gap (a d–d transition) — but only if the d subshell is partly filled, so there is both an electron to promote and a space to promote it into.
ΔE = hν = hc ÷ λh = 6.63 × 10⁻³⁴ J s · c = 3.00 × 10⁸ m s⁻¹ · λ in metres
The colour you see is the complementary colour to the light absorbed. Change the ligand, the oxidation state or the coordination number and you change ΔE — and therefore the colour.
Colorimetry: because the absorbance is proportional to concentration, a colorimeter (with a filter of the complementary colour) can measure the concentration of a coloured complex.
Calculate
Your turn — energy of a d-d transition
2A complex absorbs light of wavelength 600 nm. Calculate ΔE, giving your answer as a multiple of 10⁻¹⁹ J (so if ΔE = 4.50 × 10⁻¹⁹ J, type 4.50). Use h = 6.63 × 10⁻³⁴ J s and c = 3.00 × 10⁸ m s⁻¹.
One ligand replaces another. If the new ligand is a similar size (H₂O and NH₃), the coordination number is unchanged; if it is much larger (Cl⁻), the coordination number usually drops from 6 to 4.
[Cu(H₂O)₆]²⁺ + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂Opale blue → deep royal blue · still octahedral (only 4 of the 6 are swapped)
[Cu(H₂O)₆]²⁺ + 4Cl⁻ → [CuCl₄]²⁻ + 6H₂Opale blue → yellow-green · octahedral → TETRAHEDRAL, because Cl⁻ is bulky
Carbon monoxide and haemoglobin: the Fe(II) in haem normally binds O₂ reversibly. CO binds to the same site far more strongly, so it is not readily displaced — the blood can no longer carry oxygen. This is a ligand substitution reaction.
Quick check
Add concentrated HCl
?Concentrated hydrochloric acid is added to a pale blue solution of [Cu(H₂O)₆]²⁺. What happens?
PI2 · catalysis
Transition metals as catalysts
Heterogeneous — the catalyst is in a different phase from the reactants. Reactants adsorb onto active sites on the surface, bonds weaken, they react, and the products desorb. Examples: Fe in the Haber process, V₂O₅ in the Contact process, Ni in hydrogenation, Pt/Rh/Pd in a catalytic converter. Increase the surface area to increase activity — and beware poisoning (sulfur poisons the Haber iron; lead poisons a catalytic converter).
Homogeneous — the catalyst is in the same phase, and works by forming an intermediate. Example: Fe²⁺ catalysing the reaction between S₂O₈²⁻ and I⁻ (two low-activation-energy steps replace one high-Ea collision between two negative ions).
Autocatalysis — a product catalyses the reaction. Mn²⁺ autocatalyses the MnO₄⁻ / ethanedioate reaction, so the rate speeds up as the reaction proceeds.
All of this depends on variable oxidation states — the metal can accept electrons and then give them back, returning unchanged at the end.
Quick check
Name the catalyst
?Which catalyst is used in the Contact process, 2SO₂ + O₂ ⇌ 2SO₃ ?
PI2 · redox titrations
Redox chemistry of the transition metals
Variable oxidation states make transition metals ideal for redox titrations.
Dichromate titrations need an indicator (e.g. barium diphenylamine sulfonate) because green Cr³⁺ masks the colour change.
Calculate
Your turn — dichromate titration
325.0 cm³ of 0.100 mol dm⁻³ Fe²⁺ is titrated with 0.0200 mol dm⁻³ potassium dichromate(VI). Using the 1 : 6 ratio, calculate the volume of dichromate needed, in cm³.
cm³
Hint: n(Fe²⁺) = 0.0250 × 0.100 = 2.50 × 10⁻³ mol. n(Cr₂O₇²⁻) = that ÷ 6 = 4.167 × 10⁻⁴ mol. V = n ÷ c, then × 1000.
Calculate
Your turn — percentage of iron
4A 0.500 g iron tablet is dissolved and titrated; it needs 20.0 cm³ of 0.0200 mol dm⁻³ KMnO₄ (ratio 1 MnO₄⁻ : 5 Fe²⁺). Calculate the percentage by mass of iron in the tablet. (Ar(Fe) = 55.8)