IB Chemistry · Unit 6 · Structure

A table that
explains
everything.

118 elements. One periodic table. A grid that predicts behaviour from position — once you understand why.

Lessons Key terms
8Lessons
0HL extensions
42Key terms
SL+HLLevel
PERIODIC TABLE
Unit 6 · Standard & Higher Level

8 lessons to work through.

The required syllabus content for Unit 6, in order. Each card is one lesson-sized checkpoint.

Lesson 1

The Periodic Table

The arrangement of elements in the periodic table helps to predict their electron configuration.

Lesson 2

+3- Periodicity: Trends in the periodic table

Lesson 2 of Unit 6.

Lesson 4

Chemical Trends

Members of a group often have very similar reactivity. Why do you think this is?

Lesson 5

Oxides

There is a gradual transition from basic to acidic character, reflecting a gradual transition from metallic to non-metallic nature

Lesson 6

Oxidation States

The sum of the oxidation states of all the atoms or ions in a neutral compound is zero.

Lesson 7

Properties of transition metals

Lesson 7 of Unit 6.

Lesson 8

Complex ions

(HL ONLY) Structure 3.1.8 – Transition elements have incomplete d-sublevels that give them characteristic properties.

Lesson 9

Colours of complex ions

Lesson 9 of Unit 6.

Lessons in detail

The unit, lesson by lesson.

Each lesson card below mirrors the original teacher deck — syllabus refs, content, worked examples and practice questions in order.

Lesson 1

The periodic table

Structure 3.1.1Structure 3.1.2

Lesson outcomes

Mendeleev (1869) arranged elements by atomic mass into columns of similar reactivity, leaving gaps for elements not yet discovered. Moseley (1913) showed that organising by atomic number (Z) rather than mass resolved the anomalies — the modern periodic table.

The four blocks

Period number = highest occupied principal quantum number. Group number (1, 2, 13–18) gives valence electron count.

Metals, non-metals, metalloids

The "staircase" running down from B/Si/Ge/As/Sb/Te splits metals (left) from non-metals (right). Elements touching the staircase are metalloids — intermediate properties; many are semiconductors.

Try these

  1. Who was Dmitri Mendeleev and what role did he play in the development of the periodic table?
    Show answer
    Russian chemist. In 1869 he produced the first widely-accepted periodic table by arranging elements in order of increasing atomic mass and into columns (groups) sharing similar reactivity. He left gaps for elements not yet discovered (Ga, Ge, Sc) and predicted their properties — predictions later confirmed when those elements were found.
  2. Who was Henry Moseley? What was his contribution?
    Show answer
    British physicist. In 1913 he proposed arranging elements by atomic number (Z = number of protons) rather than atomic mass. This resolved anomalies in Mendeleev's table (e.g. Ar/K order) and is the basis of the modern table.
  3. What does the period number tell you about an element? What does the group number tell you?
    Show answer
    Period number = highest occupied principal energy level (n). Group number indicates the column. For groups 1, 2 and 13–18, the group number tells you the number of valence electrons (with conversion: group 13 has 3 valence; group 17 has 7 valence; etc.).
  4. Name the element in period 2 group 16. Give its full electron configuration and number of valence electrons.
    Show answer
    Oxygen. 1s² 2s² 2p⁴. 6 valence electrons.
  5. Name the element in period 3 group 2. Full configuration and valence count.
    Show answer
    Magnesium. 1s² 2s² 2p⁶ 3s². 2 valence electrons.
  6. Name the element in period 4 group 8 (transition metal). Full configuration.
    Show answer
    Iron, Fe. [Ar] 4s² 3d⁶.
  7. Name the element in period 7 group 7 (transition metal). Full configuration.
    Show answer
    Bohrium, Bh. [Rn] 7s² 5f¹⁴ 6d⁵.
  8. Identify the families containing: Na, Ca, Cl, He, Ti.
    Show answer
    Na = alkali metal (group 1). Ca = alkaline earth metal (group 2). Cl = halogen (group 17). He = noble gas (group 18). Ti = transition metal (d-block, group 4).
  9. Indicate on the table the position of metals, non-metals and metalloids.
    Show answer
    Metals: left and centre of the table (groups 1–2 and the d/f-blocks). Non-metals: top-right region (most of the p-block, plus H). Metalloids: along the diagonal 'staircase' from B → Si → Ge → As → Sb → Te → Po.
  10. Where are the four blocks (s, p, d, f) of the periodic table? What does the block of an element tell you about its valence electron?
    Show answer
    s-block: groups 1–2 (left) — valence ns. p-block: groups 13–18 (right) — valence np. d-block: transition metals (centre) — valence (n−1)d. f-block: lanthanides + actinides (bottom) — valence (n−2)f. The block tells you which sub-shell the highest-energy valence electron occupies.
  11. Why are the f-block elements normally drawn separately at the bottom of the table?
    Show answer
    Purely a space-saving convention. The full ('wide-format') table places them between groups 2 and 3 of periods 6 and 7. The detached layout makes printed tables narrower.
Lesson 2-3

Periodicity · trends across and down

Structure 3.1.3

Lesson outcomes

Atomic radius

Across a period: decreases. Same shell, more protons → stronger pull on outer electrons → smaller radius. Down a group: increases. More shells → outer electrons further from nucleus and more shielded.

Ionic radius

Cations (lost electrons) are smaller than parent atom — fewer electrons, often a whole shell gone, less e-e repulsion. Anions (gained electrons) are larger — more e-e repulsion in the outer shell.

Isoelectronic species (same number of electrons) have radius decreasing as Z increases: O²⁻ > F⁻ > Ne > Na⁺ > Mg²⁺.

First ionisation energy

Across a period: increases (with characteristic dips at s→p and at the half-filled p — see Unit 1). Down a group: decreases. Outer electrons further from nucleus, more shielding, easier to remove.

Electronegativity

Tendency of a bonded atom to attract bonding electrons. Pauling scale: F = 4.0 (highest), Cs = 0.79 (lowest among metals). Across a period: increases. Down a group: decreases.

Worked example

Comparing radii

Problem. Arrange in order of increasing radius: Na, Mg, K.
Solution. Mg < Na (across period 3, Mg has more protons → smaller). Na < K (down group 1, K has more shells → larger). Order: Mg < Na < K.

Try these

  1. State the trend in atomic radius across a period and down a group. Explain each.
    Show answer
    Across a period: decreases. Same shell, more protons → stronger pull on electrons → smaller radius. Down a group: increases. New shells added → outer electrons further from nucleus.
  2. State the trend in ionic radius across a period for cations and anions.
    Show answer
    Cations are smaller than parent atoms (lost electrons; sometimes a whole shell). Anions are larger (more e-e repulsion). Across period 3 cations (Na⁺ > Mg²⁺ > Al³⁺) decrease in size. Across period 3 anions (P³⁻ > S²⁻ > Cl⁻) decrease in size.
  3. Arrange in order of increasing radius: F⁻, Na⁺, Mg²⁺, O²⁻, Ne.
    Show answer
    All isoelectronic (10 electrons). Radius decreases as nuclear charge Z increases (more protons pulling on same electron count). Z values: O = 8, F = 9, Ne = 10, Na = 11, Mg = 12. So Mg²⁺ < Na⁺ < Ne < F⁻ < O²⁻.
  4. Why does the first IE of B (Z=5) drop below that of Be (Z=4) despite B having more protons?
    Show answer
    Be is [He] 2s². B is [He] 2s² 2p¹. Removing a 2p¹ electron is easier than removing a 2s² electron because 2p is higher in energy. This dip is direct evidence for the s/p sub-shell split.
  5. Why does the first IE of O (Z=8) drop below that of N (Z=7)?
    Show answer
    N is 2p³ (three unpaired electrons in three orbitals, Hund-stable). O is 2p⁴ — must have one pair, which experiences e-e repulsion → easier to remove. Evidence for the orbital occupancy / Hund's rule.
  6. Predict the trend in electron affinity down group 17. Justify. Note any anomalies.
    Show answer
    Generally exothermic; magnitude decreases down the group (Cl > Br > I) because smaller atoms attract added electrons more strongly. Anomaly: F is less exothermic than Cl. The 2p shell of F is so small that the incoming electron experiences significant e-e repulsion → less energy released than expected.
  7. Explain the trend in electronegativity across a period and down a group.
    Show answer
    Across a period: increases. Greater nuclear charge with similar shielding pulls bonding electrons more strongly. Down a group: decreases. More distance and shielding between nucleus and bonding pair.
  8. Why does first IE decrease down group 1 (Li to Cs)?
    Show answer
    Outer s¹ electron is in a higher principal energy level → further from nucleus + more shielding by inner electrons → weaker attraction → easier to remove.
Lesson 4

Chemical trends · groups 1 and 17

Structure 3.1.3Reactivity 3.1

Lesson outcomes

Group 1 · alkali metals

Reactivity increases down the group: Li < Na < K < Rb < Cs. Reason: the outer s¹ electron is further from the nucleus and more shielded going down → easier to lose.

All react vigorously with water: 2 M(s) + 2 H₂O(l) → 2 MOH(aq) + H₂(g). Sodium fizzes; potassium ignites; rubidium/caesium explode on contact with water.

Group 17 · halogens

Reactivity decreases down the group: F > Cl > Br > I. The smaller atoms attract the gained electron more strongly. Halogens are oxidising agents; a more reactive halogen will displace a less reactive halide ion from solution: Cl₂ + 2 KBr → 2 KCl + Br₂.

Trend in physical state: F₂(g) → Cl₂(g) → Br₂(l) → I₂(s) — bigger molecules have stronger London forces.

Try these

  1. Why does reactivity increase down Group 1 (Li to Cs)?
    Show answer
    Reactivity in Group 1 means tendency to lose the outer s¹ electron. Down the group, outer electron is further from nucleus and more shielded → easier to lose → more reactive.
  2. Write balanced equations for: (a) Na + H₂O, (b) K + H₂O.
    Show answer
    (a) 2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g). (b) 2 K(s) + 2 H₂O(l) → 2 KOH(aq) + H₂(g). Both vigorous; K is faster and the H₂ produced often ignites.
  3. Why does reactivity decrease down Group 17 (F to I)?
    Show answer
    Reactivity in Group 17 means tendency to gain an electron (oxidising power). Smaller atoms (F, Cl) attract added electrons more strongly. Down the group, atoms get larger, the added electron experiences less effective pull, and reactivity falls.
  4. You are given solutions of Cl₂, Br₂ and I₂. Deduce a procedure to confirm the reactivity order using only solutions of KBr, KCl and KI.
    Show answer
    Mix each halogen with each halide solution. A reaction (colour change indicating displaced free halogen) confirms the halogen is more reactive than the displaced halide. Expected: Cl₂ displaces Br⁻ and I⁻; Br₂ displaces I⁻ but not Cl⁻; I₂ displaces neither. Confirms Cl > Br > I.
  5. Why is the trend in physical state F₂(g), Cl₂(g), Br₂(l), I₂(s)?
    Show answer
    More electrons in the molecule down the group → larger and more polarisable electron clouds → stronger London dispersion forces between molecules → higher b.p. and m.p. F₂ and Cl₂ are gases; Br₂ a liquid; I₂ a solid at room T.
  6. Why is sodium stored under oil, but lithium is generally less of a concern?
    Show answer
    Both react with O₂ and water vapour in air. Sodium reacts much more rapidly (lower IE), so storing under oil prevents air contact. Lithium reacts more slowly, but is still stored carefully.
Lesson 5

Oxides · acid-base trends

Structure 3.1.4

Lesson outcomes

Across period 3, the character of element oxides shifts from basic (metal oxides) through amphoteric (Al₂O₃) to acidic (non-metal oxides).

OxideBondingIn waterCharacter
Na₂O, MgOIonicForms strong/weak base (NaOH, Mg(OH)₂)Basic
Al₂O₃Ionic/covalentInsoluble; reacts with acids and basesAmphoteric
SiO₂Covalent networkInsoluble; reacts with hot conc. NaOHWeakly acidic
P₄O₁₀, SO₃, Cl₂O₇Covalent molecularForm strong acids (H₃PO₄, H₂SO₄, HClO₄)Acidic
Worked example

Amphoteric behaviour of Al₂O₃

Problem. Write balanced equations showing Al₂O₃ acting as a base (with HCl) and as an acid (with NaOH).
Solution. As a base: Al₂O₃(s) + 6 HCl(aq) → 2 AlCl₃(aq) + 3 H₂O(l).
As an acid: Al₂O₃(s) + 2 NaOH(aq) + 3 H₂O(l) → 2 Na[Al(OH)₄](aq) — forms the tetrahydroxoaluminate(III) complex.

Try these

  1. Predict the products when SO₃ is added to water. Classify the resulting solution.
    Show answer
    SO₃ + H₂O → H₂SO₄. Strongly acidic solution. SO₃ is a non-metal oxide, predicted acidic.
  2. Write the equation for Na₂O reacting with water. Classify the oxide.
    Show answer
    Na₂O(s) + H₂O(l) → 2 NaOH(aq). Strongly basic — sodium oxide is a strong basic oxide (metal oxide).
  3. Write the two equations showing that Al₂O₃ is amphoteric.
    Show answer
    As base: Al₂O₃ + 6 HCl → 2 AlCl₃ + 3 H₂O. As acid: Al₂O₃ + 2 NaOH + 3 H₂O → 2 Na[Al(OH)₄] (sodium tetrahydroxoaluminate(III)).
  4. Why is Al₂O₃ amphoteric but Na₂O is purely basic?
    Show answer
    Al³⁺ has a high charge density and polarises the oxide ion strongly, giving the Al–O bond significant covalent character. This lets Al₂O₃ behave as a covalent acidic oxide as well as as an ionic basic oxide. Na⁺ has too low a charge density for this dual behaviour.
  5. Predict whether MgO would be more strongly basic or less strongly basic than Na₂O. Why?
    Show answer
    Less strongly basic. Mg²⁺ has higher charge density than Na⁺, so MgO has somewhat more covalent character. Less ionic = less effectively basic. Mg(OH)₂ is a weak base; NaOH is strong.
  6. Why does the pH of oxide-water solutions change predictably across a period?
    Show answer
    Metal oxides (left of period) are ionic and strongly basic — give high pH. Towards the right, oxides become more covalent. Al₂O₃ is amphoteric (in the middle). Non-metal oxides (right) are acidic and give low pH. The shift mirrors the metal → non-metal trend across the period.
Lesson 6

Oxidation states

Reactivity 3.2.1

Lesson outcomes

The oxidation state (oxidation number) of an atom in a compound is the hypothetical charge it would have if all bonds were ionic. A useful bookkeeping tool for tracking electrons in redox.

Rules (apply in order of priority)

  1. Free elements have OS = 0 (Na, O₂, Cl₂, S₈).
  2. Monatomic ions: OS = charge (Na⁺ = +1; Cl⁻ = −1).
  3. Group 1: always +1. Group 2: always +2. F: always −1.
  4. H: +1 (except in metal hydrides like NaH, where H is −1).
  5. O: −2 (except in peroxides H₂O₂ where O = −1, and OF₂ where O = +2).
  6. Sum of all OS in a neutral compound = 0; in an ion = the ion's charge.

IUPAC names with oxidation state

For elements with variable oxidation state, the value is written in Roman numerals: iron(III) chloride = FeCl₃; manganese(VII) oxide = Mn₂O₇.

Worked example

Assigning oxidation states in KMnO₄

Problem. Find the oxidation state of Mn in KMnO₄.
Solution. K = +1 (group 1). O = −2 × 4 = −8. Sum = 0 (neutral compound). So Mn = 0 − (+1) − (−8) = +7.
Worked example

Assigning oxidation states in Cr₂O₇²⁻

Problem. Find the oxidation state of Cr in the dichromate ion, Cr₂O₇²⁻.
Solution. O = −2 × 7 = −14. Sum = −2 (ion charge). 2 × Cr + (−14) = −2 → 2 × Cr = +12 → Cr = +6.

Try these

  1. State the rules for assigning oxidation states in priority order.
    Show answer
    (1) Free elements = 0. (2) Monatomic ions = charge. (3) F always −1; group 1 always +1; group 2 always +2. (4) H = +1 (except in metal hydrides: NaH, MgH₂, H = −1). (5) O = −2 (except in peroxides H₂O₂: −1; in OF₂: +2). (6) Sum of OS = 0 (compound) or = charge (ion).
  2. Assign oxidation states to all atoms in: (a) KMnO₄, (b) Cr₂O₇²⁻, (c) NH₄⁺, (d) ClO₃⁻.
    Show answer
    (a) K = +1, O = −2, Mn = +7. (b) O = −2 × 7 = −14. Sum = −2 → Cr = +6. (c) H = +1 × 4 = +4. Sum = +1 → N = −3. (d) O = −2 × 3 = −6. Sum = −1 → Cl = +5.
  3. Identify the species oxidised and reduced in: 2 KI(aq) + Cl₂(g) → 2 KCl(aq) + I₂(s). Give the OS changes.
    Show answer
    I: −1 → 0 (oxidised, lost e⁻). Cl: 0 → −1 (reduced, gained e⁻). KI is the reducing agent; Cl₂ is the oxidising agent.
  4. Name CuSO₄ using IUPAC oxidation-state notation. Why must we use Roman numerals here?
    Show answer
    Copper(II) sulfate. Roman numerals are needed because Cu has more than one common oxidation state (+1 and +2). Without the (II), CuSO₄ could in principle be ambiguous.
  5. Name: (a) Fe₂O₃, (b) Mn₂O₇, (c) NO, (d) PCl₅.
    Show answer
    (a) iron(III) oxide. (b) manganese(VII) oxide. (c) nitrogen(II) oxide (or nitric oxide). (d) phosphorus(V) chloride.
Lesson 7

Properties of transition metals

Structure 3.1.5

Lesson outcomes

Transition metals are d-block elements with at least one stable ion having a partially-filled d sub-shell. (By this definition, Sc and Zn are not strictly transition metals in IB — Sc³⁺ has empty d, Zn²⁺ has full d.)

Six characteristic properties

  1. Variable oxidation states. 4s and 3d are very close in energy — losing 4s and various numbers of 3d gives a range of stable ions.
  2. Coloured compounds. d-d transitions in complexes absorb visible light.
  3. Catalytic activity. Variable OS lets the metal accept and donate electrons in cycles.
  4. Form complex ions. Empty d-orbitals accept lone pairs from ligands.
  5. Magnetic properties. Unpaired d electrons give paramagnetism.
  6. High melting and boiling points + density. Strong metallic bonding from 4s and 3d delocalised electrons.

Why +2 is the most common

The 4s electrons are removed first (4s is higher in energy than 3d once 3d is occupied). So every transition metal forms an M²⁺ ion by losing its two 4s electrons. Higher OS removes 3d electrons too.

Examples of catalysts

Try these

  1. List the six characteristic properties of transition metals.
    Show answer
    (1) Variable oxidation states. (2) Coloured compounds. (3) Catalytic activity. (4) Form complex ions. (5) Magnetic properties (paramagnetism from unpaired d electrons). (6) High m.p./b.p./density (strong metallic bonding).
  2. Why is Sc not considered a 'true' transition metal in some classifications?
    Show answer
    Sc³⁺ has no d electrons ([Ar] only). It can't show variable OS, form coloured complexes, or display d-electron magnetism. Some sources define a transition metal as one whose most stable ion has a partially-filled d sub-shell — by which Sc fails.
  3. Why does Zn also fail the 'partially-filled d' test for transition metals?
    Show answer
    Zn²⁺ has [Ar] 3d¹⁰ — fully filled. No d-d transitions, no coloured compounds, no paramagnetism. Zn doesn't show variable OS (essentially always +2).
  4. Write the electron configuration of (a) Fe, (b) Fe²⁺, (c) Fe³⁺. Explain why Fe³⁺ is particularly stable.
    Show answer
    (a) Fe: [Ar] 4s² 3d⁶. (b) Fe²⁺: [Ar] 3d⁶ (lose 4s² first). (c) Fe³⁺: [Ar] 3d⁵ (lose 4s², then one 3d). Fe³⁺ is half-filled d, which is unusually stable (Hund).
  5. Give four examples of industrial reactions catalysed by transition metals.
    Show answer
    (1) Fe in the Haber process (N₂ + H₂ → NH₃). (2) V₂O₅ in the Contact process (SO₂ → SO₃). (3) Ni in hydrogenation of vegetable oils. (4) Pt/Pd/Rh in catalytic converters (CO + NOx + hydrocarbons → CO₂ + N₂ + H₂O). Also: MnO₂ for H₂O₂ decomposition.
  6. Why are zinc compounds not coloured, unlike most transition-metal compounds?
    Show answer
    Zn²⁺ has full d¹⁰ — no possibility of d-d electron transitions. With no available energy gap in the visible region, the compound absorbs no visible light and appears colourless / white.
Lesson 8

Complex ions

Structure 3.1.6Reactivity 3.4 (HL)

Lesson outcomes

A complex ion consists of a central metal ion bonded to one or more ligands through coordinate (dative) covalent bonds — the ligand donates a lone pair into an empty d-orbital of the metal.

Common ligands

Coordination number and shape

CNShapeExample
2Linear[Ag(NH₃)₂]⁺ — Tollens'
4Tetrahedral or square planar[CuCl₄]²⁻ tetrahedral; [Pt(NH₃)₂Cl₂] sq. planar
6Octahedral[Fe(H₂O)₆]³⁺, [Co(NH₃)₆]³⁺
Worked example

Naming and shape of [Cu(H₂O)₆]²⁺

Problem. State the coordination number, shape, ligand denticity and charge on the central metal in [Cu(H₂O)₆]²⁺.
Solution. Coordination number = 6 (six water ligands). Shape = octahedral. H₂O is monodentate. Charge on Cu = +2 (water is neutral, complex has +2 overall).

Try these

  1. Define complex ion, ligand and coordination number.
    Show answer
    Complex ion: a central metal ion bonded to surrounding ligands via coordinate (dative) covalent bonds. Ligand: a molecule or ion that donates a lone pair into an empty orbital of the metal. Coordination number: the number of bonds the metal makes to ligand donor atoms (most commonly 4 or 6).
  2. Classify each as monodentate, bidentate, or hexadentate: H₂O, NH₃, Cl⁻, ethanedioate (C₂O₄²⁻), 1,2-diaminoethane (en), EDTA⁴⁻.
    Show answer
    Monodentate: H₂O, NH₃, Cl⁻. Bidentate: ethanedioate, 1,2-diaminoethane. Hexadentate: EDTA⁴⁻.
  3. What is the coordination number, shape, and charge on Fe in [Fe(H₂O)₆]³⁺?
    Show answer
    CN = 6 (six water ligands). Shape = octahedral. H₂O is neutral, complex charge +3, so Fe³⁺.
  4. Why is EDTA called a 'hexadentate chelating' ligand?
    Show answer
    Hexadentate because it has 6 donor atoms (four carboxylate oxygens and two amine nitrogens) which all bond to the same metal centre. Chelating because it 'grabs' the metal in a multi-point grip, like a claw (Greek chele = claw), making the complex extra stable.
  5. Write the formula of: (a) the tetraamminecopper(II) ion, (b) the hexaaquairon(III) ion, (c) the diamminesilver(I) ion (used in Tollens' test).
    Show answer
    (a) [Cu(NH₃)₄]²⁺ (often [Cu(NH₃)₄(H₂O)₂]²⁺ in solution). (b) [Fe(H₂O)₆]³⁺. (c) [Ag(NH₃)₂]⁺.
Lesson 9

Colours of complex ions

Structure 3.1.6Reactivity 3.4 (HL)

Lesson outcomes

In a free transition-metal ion, the five d-orbitals are degenerate. In a complex, the surrounding ligands split them into two energy levels (in an octahedral complex: lower t2g and higher eg). The energy gap ΔE typically corresponds to a visible-light photon.

When a photon of energy ΔE is absorbed, an electron jumps from the lower set to the upper set — a d-d transition. The colour we see is the complementary colour of the wavelength absorbed.

Complementary colours (rough guide)

Absorbs red→ looks green/blue
Absorbs yellow→ looks blue/violet
Absorbs green→ looks red/purple
Absorbs blue→ looks orange/yellow

Four factors affecting ΔE (and so colour)

  1. Metal identity — Fe vs Cu vs Co all give different gaps.
  2. Oxidation state — Fe²⁺ (pale green) vs Fe³⁺ (yellow-brown). Higher OS pulls ligands closer → larger ΔE.
  3. Ligand — the "spectrochemical series" CN⁻ > NH₃ > H₂O > OH⁻ > F⁻ > Cl⁻ > I⁻ orders ligands by ΔE produced.
  4. Geometry — octahedral, tetrahedral and square planar give different splitting patterns and gap sizes.

Try these

  1. Why are most transition-metal complexes coloured?
    Show answer
    Ligand field splits the 5 d-orbitals into two energy levels with gap ΔE. ΔE typically corresponds to a visible photon. Electron absorbs the photon, jumps from lower to higher d-orbital — a d-d transition. The complementary colour of the wavelength absorbed is what we see.
  2. List the four factors that affect the colour of a transition-metal complex.
    Show answer
    (1) Identity of the metal. (2) Oxidation state of the metal (higher OS pulls ligands closer → larger ΔE). (3) Identity of the ligand (spectrochemical series). (4) Coordination number / geometry (octahedral vs tetrahedral split d differently).
  3. Compare Fe²⁺ and Fe³⁺ colour. Why are they different?
    Show answer
    Fe²⁺ (e.g. [Fe(H₂O)₆]²⁺) is pale green. Fe³⁺ (e.g. [Fe(H₂O)₆]³⁺) is yellow-brown. Different OS → different effective nuclear charge → different ΔE → different colour.
  4. [Cu(H₂O)₆]²⁺ is pale blue; [Cu(NH₃)₄(H₂O)₂]²⁺ is deep blue. Explain.
    Show answer
    Replacing 4 H₂O with 4 NH₃ ligands (NH₃ is higher in the spectrochemical series than H₂O) produces a larger ΔE. The absorbed wavelength shifts to higher energy (lower λ — yellow rather than red), and the complementary colour seen shifts to a more intense blue-violet.
  5. Why is Zn²⁺ colourless in complexes despite being a d-block element?
    Show answer
    Zn²⁺ is d¹⁰ — fully filled. There is no empty d-orbital for an electron to jump into, so no d-d transition is possible. No absorption in the visible → colourless.
  6. A complex absorbs at 600 nm (orange). What colour does it appear?
    Show answer
    Complementary to orange = blue. (Colour wheel: orange ↔ blue, red ↔ green, yellow ↔ violet/purple.)
Vocabulary

42 terms to own.

If you can't define one of these in a sentence, that's where to revise next. Click any term for its definition.

elementcompoundheterogeneousdepositionatomprotonelectronatomic numberenergy levelorbitalsublevelelectron configurationionisation energysuccessive ionisation energiesmoleconcentrationsolutiongraphcatalysttemperatureheterogeneous catalystendothermicperiodic tableperiodgroupalkali metalsalkaline earth metalshalogensnoble gasestransition metalsmetalloidmetallic characteratomic radiusionic radiuselectronegativityshieldingeffective nuclear chargeoxideamphotericcomplex ionligandvariable oxidation state