This mini-lesson covers the whole of Edexcel Topic 15: what makes a d-block element a transition metal, variable oxidation states, complex ions and ligands, coordination number and shape, ligand substitution and the chelate effect, colour and d-orbital splitting, catalysis, and redox titrations with manganate(VII).
Work through each screen, answer the questions as you go (several are calculations) and collect ⭐ stars. Every calculation is worked through for you first. Press Start when you're ready.
Definitions
d-block element vs transition metal
A d-block element is one whose highest-energy electron is in a d sub-shell. That is a position on the Periodic Table. A transition metal is defined by its chemistry:
A transition metal forms at least one stable ion with a partially filled d sub-shelli.e. a d¹ to d⁹ ion — not d⁰ and not d¹⁰
Scandium is not one. Sc = [Ar] 3d¹ 4s². Its only stable ion is Sc³⁺ = [Ar] 3d⁰ — empty d sub-shell.
Zinc is not one. Zn = [Ar] 3d¹⁰ 4s². Its only stable ion is Zn²⁺ = [Ar] 3d¹⁰ — full d sub-shell.
Both are d-block elements. Neither shows the hallmark properties that need a partially filled d sub-shell: variable oxidation states, coloured compounds and catalytic activity. (They do still form complex ions — [Zn(H₂O)₆]²⁺ and [Zn(NH₃)₄]²⁺ exist — but those complexes are colourless.)
Remember: 4s fills before 3d, but the 4s electrons are lost first when the ion forms. Fe = [Ar] 3d⁶ 4s² → Fe²⁺ = [Ar] 3d⁶ → Fe³⁺ = [Ar] 3d⁵.
Quick check
Why isn't zinc a transition metal?
1Zinc sits in the d-block but is not classed as a transition metal. Why?
Oxidation states
Variable oxidation states
In a transition metal the 4s and 3d sub-shells are very close in energy, so a variable number of electrons can be removed with similar amounts of energy. That gives a range of stable oxidation states — and every one of them has its own colour and its own chemistry.
Species
Oxidation state of the metal
Colour
[Fe(H₂O)₆]²⁺
+2
pale green
[Fe(H₂O)₆]³⁺
+3
pale violet / yellow-brown
[Cu(H₂O)₆]²⁺
+2
pale blue
MnO₄⁻
+7
deep purple
Cr₂O₇²⁻
+6
orange
How to work out an oxidation state: oxygen is −2 and the oxidation states must add up to the overall charge on the ion. Water is a neutral ligand (0); Cl⁻ and CN⁻ contribute −1 each.
Calculate
Your turn — oxidation state of Mn
2Work out the oxidation state of manganese in the manganate(VII) ion, MnO₄⁻. Type the number only (e.g. type 3 for +3).
(as +n)
Hint: let the oxidation state of Mn be n. Each O is −2, and there are four of them. So n + 4(−2) = −1, the overall charge.
Calculate
Your turn — oxidation state of Cr
3Work out the oxidation state of chromium in the dichromate(VI) ion, Cr₂O₇²⁻. Type the number only.
(as +n)
Hint: 2n + 7(−2) = −2, so 2n − 14 = −2 and 2n = 12. Careful — there are two chromium atoms.
Complexes
Complex ions and ligands
A complex ion is a central metal ion surrounded by ligands joined to it by dative (coordinate) covalent bonds.
A ligand is a species with a lone pair of electrons that it donates to the metal ionthe metal ion accepts the lone pair — it is a Lewis acid; the ligand is a Lewis base
Monodentate: donates one lone pair, forming one dative bond — H₂O, NH₃, Cl⁻, CN⁻, OH⁻.
Bidentate: donates two lone pairs from the same molecule — 1,2-diaminoethane (H₂NCH₂CH₂NH₂, "en") and the ethanedioate ion, C₂O₄²⁻.
Multidentate:EDTA⁴⁻ is hexadentate — it donates six lone pairs (two from N, four from O) and wraps a single metal ion completely. Haem in haemoglobin is another multidentate example.
The coordination number is the number of dative bonds from ligands to the central metal ion — not the number of ligand molecules. One EDTA⁴⁻ ligand gives a coordination number of 6.
Quick check
What makes something a ligand?
4Which statement best defines a ligand?
Shapes
Coordination number and shape
Octahedral, 6 ligands, 90° bond angles — the commonest. Small ligands such as H₂O and NH₃ pack six around the metal: [Cu(H₂O)₆]²⁺, [Fe(H₂O)₆]³⁺, [Co(NH₃)₆]²⁺.
Tetrahedral, 4 ligands, 109.5° — with larger ligands such as Cl⁻ only four will fit: [CuCl₄]²⁻, [CoCl₄]²⁻.
Square planar, 4 ligands, 90° — for d⁸ metal ions such as Pt(II) and Ni(II). Cisplatin, cis-[Pt(NH₃)₂Cl₂], is square planar; only the cis isomer is an anticancer drug, because its two Cl⁻ ligands are adjacent, so it can bond to nitrogen atoms on two neighbouring guanine bases in DNA.
Linear, 2 ligands, 180° — [Ag(NH₃)₂]⁺, the active species in Tollens' reagent.
Purple/green/orange = the central metal ion; pale circles = ligands. Square planar (4 ligands, 90°, all in one plane) is the fourth shape you must know.Sort it
What shape is this complex?
Tap a complex ion, then tap its shape. Count the ligands — and remember that Cl⁻ is a big ligand.
🟦 Octahedral
🟩 Tetrahedral
🟧 Linear
Match it
Match the term to its meaning
Tap a term on the left, then its definition on the right.
Term
Meaning
Calculate
Your turn — coordination number
51,2-diaminoethane ("en") is a bidentate ligand. What is the coordination number of the nickel ion in [Ni(en)₃]²⁺?
dative bonds
Hint: coordination number = number of dative bonds, not number of ligand molecules. Each bidentate ligand makes two bonds, and there are three of them.
Ligand substitution
Ligand substitution reactions
One ligand can replace another. If the incoming and outgoing ligands are similar in size (H₂O and NH₃ are almost identical), the coordination number and shape do not change. If they differ in size (H₂O → Cl⁻), both can change.
[Cu(H₂O)₆]²⁺ + 4NH₃ ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂Opale blue → deep blue · still octahedral · only 4 of the 6 waters swap
[Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂Opale blue → yellow-green · octahedral → tetrahedral, because Cl⁻ is a much larger ligand
Check the charges. NH₃ and H₂O are neutral, so the 2+ charge is unchanged in the first reaction. Four Cl⁻ ligands add 4− to a Cu²⁺, giving 2 − 4 = 2− — hence [CuCl₄]²⁻.
Quick check
Copper(II) with excess ammonia
6Excess concentrated ammonia is added to aqueous copper(II) sulfate. A deep blue solution forms. Which is the product complex?
Chelate effect
Why multidentate ligands win: entropy
A multidentate ligand will displace monodentate ligands, even though the bonds formed and broken are of similar strength (so ΔH ≈ 0). The driving force is entropy.
[Cu(H₂O)₆]²⁺ + 3en ⇌ [Cu(en)₃]²⁺ + 6H₂O4 particles on the left → 7 particles on the right
Every bidentate ligand that binds releases two water molecules. The number of free particles goes up, so ΔS⦵ is positive and large.
ΔG⦵ = ΔH⦵ − TΔS⦵. With ΔH⦵ ≈ 0 and ΔS⦵ positive, ΔG⦵ is negative — the substitution is spontaneous, and the equilibrium lies far to the right.
This is the chelate effect. It is why EDTA⁴⁻ (hexadentate, releasing six waters for one ligand) forms such extremely stable complexes, and why it is used to treat heavy-metal poisoning and to soften water.
Exam trap: do not say the chelate complex is more stable "because the bonds are stronger". Say the number of moles of particles increases, so the entropy change is positive, making ΔG negative.
Calculate
Your turn — the entropy driver
7For [Cu(H₂O)₆]²⁺ + 3en ⇌ [Cu(en)₃]²⁺ + 6H₂O, calculate the increase in the total number of moles of free particles (right-hand side minus left-hand side).
mol
Hint: left = 1 complex + 3 en = 4 particles. Right = 1 complex + 6 H₂O = 7 particles. Now subtract.
Colour
Colour and d-orbital splitting
In a free metal ion the five d orbitals are degenerate (equal in energy). When ligands approach, they repel the d electrons unequally — the orbitals pointing straight at the ligands are pushed higher. The d sub-shell splits into two levels separated by an energy gap ΔE.
ΔE = hν = hc ÷ λh = 6.63 × 10⁻³⁴ J s · c = 3.00 × 10⁸ m s⁻¹ · ν = frequency in Hz · λ = wavelength in m
An electron absorbs a photon of exactly energy ΔE and jumps from the lower to the upper level — a d–d transition. Only visible light of that frequency is absorbed.
The colour you see is the complementary colour of the light absorbed. [Cu(H₂O)₆]²⁺ absorbs orange-red light, so it looks pale blue.
No partially filled d sub-shell → no d–d transition → no colour. That is why Sc³⁺ (d⁰) and Zn²⁺ (d¹⁰) solutions are colourless.
ΔE — and so the colour — changes with the ligand, the oxidation state and the coordination number/shape. Swapping H₂O for NH₃ in the copper complex increases ΔE, which is why the blue deepens.
Measuring it: a colorimeter (with a filter of the complementary colour) measures absorbance. A calibration curve of absorbance against known concentration then lets you find an unknown concentration.
Quick check
Why are Zn²⁺ solutions colourless?
8Aqueous solutions of Zn²⁺ and Sc³⁺ are colourless, whereas Cu²⁺ and Fe³⁺ are coloured. Why?
Calculate
Your turn — frequency of the absorbed light
9A copper complex absorbs light of wavelength 600 nm. Calculate the frequency of that light, giving your answer in units of 10¹⁴ Hz to 3 significant figures. (c = 3.00 × 10⁸ m s⁻¹.)
× 10¹⁴ Hz
Hint: convert first — 600 nm = 600 × 10⁻⁹ m = 6.00 × 10⁻⁷ m. Then ν = c ÷ λ = 3.00 × 10⁸ ÷ 6.00 × 10⁻⁷.
Calculate
Your turn — the splitting energy ΔE
10Using your frequency of 5.00 × 10¹⁴ Hz, calculate ΔE for that d–d transition, in units of 10⁻¹⁹ J, to 3 significant figures. (h = 6.63 × 10⁻³⁴ J s.)
× 10⁻¹⁹ J
Hint: ΔE = hν = (6.63 × 10⁻³⁴) × (5.00 × 10¹⁴). Multiply the numbers, add the powers: 6.63 × 5.00 = 33.15, and 10⁻³⁴ × 10¹⁴ = 10⁻²⁰.
Catalysis
Why transition metals catalyse
Transition metals and their compounds are excellent catalysts because they can change oxidation state (accepting and donating electrons) and because they can adsorb reactant molecules onto their surface using partially filled d orbitals.
Heterogeneous — catalyst in a different phase from the reactants (usually a solid with gases). Reactants adsorb onto active sites, bonds weaken, the reaction occurs, products desorb. Examples: Fe in the Haber process, V₂O₅ in the Contact process, Ni in hydrogenation of alkenes.
Homogeneous — catalyst in the same phase (usually all aqueous). It forms an intermediate which then reacts on to release the catalyst unchanged.
Multiply the iron half-equation by 5 so the electrons cancel, and add:
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂Ocharge check: (−1) + (+10) + (+8) = +17 on the left; (+2) + (+15) = +17 on the right ✔
The reacting ratio is 1 MnO₄⁻ : 5 Fe²⁺ — this is the number you use in every calculation.
No indicator is needed — MnO₄⁻ is self-indicating. The end point is the first permanent pale pink colour, when one drop of MnO₄⁻ is left unreacted.
Acidify with dilute sulfuric acid. Not hydrochloric (the Cl⁻ would itself be oxidised to Cl₂, giving a false high titre) and not nitric (it is an oxidising agent and would oxidise the Fe²⁺ itself).
Calculate
Your turn — moles of MnO₄⁻
12A 25.0 cm³ sample of an acidified Fe²⁺ solution needs 22.40 cm³ of 0.0200 mol dm⁻³ KMnO₄ to reach the end point. Calculate the moles of MnO₄⁻ used, in units of 10⁻⁴ mol, to 3 significant figures.
× 10⁻⁴ mol
Hint: n = c × V, with V in dm³. 22.40 cm³ = 0.02240 dm³. So n = 0.0200 × 0.02240.
Calculate
Your turn — concentration of the Fe²⁺
13You found 4.48 × 10⁻⁴ mol of MnO₄⁻ reacted with the 25.0 cm³ Fe²⁺ sample. Using the 1 : 5 ratio, calculate the concentration of the Fe²⁺ in mol dm⁻³, to 3 significant figures.
mol dm⁻³
Hint: n(Fe²⁺) = 5 × 4.48 × 10⁻⁴ = 2.24 × 10⁻³ mol. Then c = n ÷ V = 2.24 × 10⁻³ ÷ 0.0250 dm³.
Recap
The big ideas to know
Transition metal: forms at least one stable ion with a partially filled d sub-shell — so Sc (d⁰) and Zn (d¹⁰) are excluded
Variable oxidation states: 4s and 3d are close in energy; the 4s electrons are lost first
Ligand: donates a lone pair to the metal ion → dative covalent bond. Mono- (H₂O, NH₃, Cl⁻), bi- (en, C₂O₄²⁻), hexadentate (EDTA⁴⁻)
Coordination number: number of dative bonds. 6 → octahedral (90°), 4 → tetrahedral (109.5°) or square planar (90°, d⁸), 2 → linear (180°)
Ligand substitution: [Cu(H₂O)₆]²⁺ + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O; with Cl⁻ the shape changes to tetrahedral [CuCl₄]²⁻
Chelate effect: multidentate ligands displace monodentate ones because the number of particles rises → ΔS positive → ΔG negative
Colour: ligands split the d orbitals; an electron absorbs a photon of ΔE = hν; d⁰ and d¹⁰ ions are colourless
Catalysis: heterogeneous (Fe, V₂O₅, Ni — adsorption on active sites) vs homogeneous (Fe²⁺/Fe³⁺ — via an intermediate)