Transition elements

In a nutshell

The transition elements are the d-block metals from titanium to copper. They behave differently from s-block metals because their ions can have a partly filled d-sub-shell.

That one feature brings a set of properties you must be able to illustrate: more than one oxidation state, coloured ions, catalytic activity and the formation of complex ions with ligands.

This subtopic covers how complexes are built and shaped, their stereoisomers, and the ligand substitution, precipitation and redox reactions (with their colour changes) that you use to identify the metal ions.

Assumed knowledge: Electron structure: filling 4s before 3d; Bonding and structure: dative covalent bonds and shapes; Redox: oxidation numbers.

Core content

Properties: electron configurations of Sc to Zn

You need to write the electron configuration of any atom or ion from Sc to Zn, given its atomic number and charge, in sub-shell notation.

Two rules do almost all the work:

1. Filling: 4s is lower in energy than 3d, so 4s fills first. OCR writes the finished configuration in shell order, with 3d before 4s: iron is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s².
2. Forming ions: the 4s electrons are lost before the 3d electrons. Fe²⁺ is ...3d⁶ (not ...3d⁴ 4s²), and Fe³⁺ is ...3d⁵.

Chromium and copper are the two atoms that break the filling pattern: one 4s electron moves into 3d to give a half-full or full d-sub-shell.

[Ar] stands for 1s² 2s² 2p⁶ 3s² 3p⁶. If a question asks for the full electron configuration, write every sub-shell out from 1s.

Properties: what makes an element a transition element

A transition element is a d-block element that forms an ion with an incomplete d-sub-shell.

The ten d-block elements of Period 4 run from Sc to Zn, but only Ti to Cu are transition elements:

• Scandium forms only Sc³⁺, which has an empty d-sub-shell (3d⁰).
• Zinc forms only Zn²⁺, which has a full d-sub-shell (3d¹⁰).
• Copper counts, even though Cu⁺ is 3d¹⁰, because Cu²⁺ is 3d⁹. The definition needs only one ion with an incomplete d-sub-shell.

Still don't get it? · why Sc and Zn are not transition elements

Think of the d-sub-shell as a bus with ten seats. The test is not "is the bus part-full while parked at the depot?" (the atom). The test is "is the bus ever part-full while it is out on a route?" (an ion). A bus that only ever runs completely empty, or completely full, fails the test.

Now rebuild it step by step:

1. Being in the d-block just means the atom's highest-energy electron is in a d-sub-shell. Sc to Zn all qualify.
2. The transition-element test is about the ions the element forms, not the atom.
3. When an ion forms, the 4s electrons go first, then 3d electrons if more are needed.
4. Scandium (3d¹ 4s²) only ever forms Sc³⁺: it loses both 4s electrons and its one 3d electron, so the 3d sub-shell is empty.
5. Zinc (3d¹⁰ 4s²) only ever forms Zn²⁺: it loses just the two 4s electrons, so the 3d sub-shell stays full.
6. Titanium to copper each form at least one ion with between 1 and 9 d-electrons, so they pass.

In exam wording: Sc and Zn are d-block elements but not transition elements, because Sc³⁺ has an empty d-sub-shell and Zn²⁺ has a full d-sub-shell, so neither forms an ion with an incomplete d-sub-shell.

Properties: variable oxidation states, coloured ions and catalysts

You must be able to illustrate each property with at least two transition elements. These are the examples the rest of this note uses.

1. More than one oxidation state in their compounds.

2. Coloured ions. Most transition metal ions are coloured in solution, and the colour changes when the oxidation state or the ligands change (iron(II) green vs iron(III) orange-brown; [Cu(H₂O)₆]²⁺ pale blue vs [CuCl₄]^2- yellow). You do not need to explain how the colour arises.

3. Catalytic behaviour of the elements and their compounds:

No detail of how these catalysts work is required. What you should be able to discuss is why they matter to industry, and the downside:

• Benefit: a catalyst lets a process run at a lower temperature, so less energy is used, less fuel is burned and less CO₂ is released.
• Risk: many transition metals and their compounds are toxic (chromium(VI) compounds, for example), so they must be handled, recovered and disposed of carefully.

Ligands and complex ions: ligands and coordinate bonds

A ligand is a molecule or ion that donates a pair of electrons to a central metal ion (or atom) to form a coordinate bond.

A coordinate bond (dative covalent bond) is a covalent bond in which both electrons of the shared pair come from one atom, here the ligand. Every ligand therefore needs at least one lone pair.

Ligands are classed by how many coordinate bonds each one forms:

Exam questions can introduce other ligands. The test is always the same: find the atoms with lone pairs that can reach the metal ion. One bond per donor atom.

Ligands and complex ions: complex ions, coordination number and shapes

A complex ion is a central metal ion surrounded by ligands, bonded to it by coordinate bonds. It is written in square brackets with the overall charge outside, for example [Cu(H₂O)₆]²⁺.

The coordination number is the total number of coordinate bonds formed between the central metal ion and its ligands.

Count bonds, not ligands:

• [Cu(H₂O)₆]²⁺: 6 monodentate ligands, coordination number 6.
• [Ni(en)₃]²⁺: 3 bidentate ligands, 3 × 2 = coordination number 6.

The overall charge of a complex is the metal's oxidation number plus the charges on the ligands. For [CuCl₄]^2-: (+2) + 4(-1) = 2-.

The coordination number sets the shape:

A coordination number of 4 can give either four-fold shape, so you must learn which complexes are which: the tetrachloro complexes above are tetrahedral, the platinum complex is square planar.

Ligands and complex ions: stereoisomerism in complexes

Stereoisomers have the same structural formula but a different arrangement (of atoms) in space. Complexes show two kinds.

1. cis-trans isomerism. This needs two identical ligands that can sit either side by side or opposite each other:

• cis: the two identical ligands are next to each other, at 90°.
• trans: the two identical ligands are opposite each other, at 180°.

It appears in square planar complexes such as Pt(NH₃)₂Cl₂, and in octahedral complexes with four of one ligand and two of another, such as [Co(NH₃)₄Cl₂]⁺.

A tetrahedral complex cannot show cis-trans isomerism, because every position in a tetrahedron is next to every other one.

2. Optical isomerism. Optical isomers are non-superimposable mirror images of each other.

The spec example is [Ni(en)₃]²⁺, an octahedral complex with three bidentate ligands. The three en ligands wrap round the nickel like the blades of a propeller, and the propeller can twist one way or the other.

A complex can show both types. In an octahedral complex with two bidentate and two monodentate ligands, such as [Co(en)₂Cl₂]⁺, the trans isomer has a mirror plane, but the cis isomer does not, so the cis form exists as a pair of optical isomers (see Worked example 2).

Still don't get it? · optical isomers of [Ni(en)3]2+

Hold your two hands up, palms facing you. They are mirror images of each other. Now try to lay your right hand exactly on top of your left, palms both facing down: the thumbs end up on opposite sides. However you turn them, they never match up. That is "non-superimposable".

Now build the complex one step at a time:

1. Ni²⁺ sits at the centre of an octahedron: six positions, each at 90° to its neighbours.
2. Each en ligand is short, so its two N atoms can only reach two neighbouring positions (90° apart), never two opposite ones.
3. Three en ligands use all six positions, so they lie round the nickel like the three blades of a propeller.
4. A propeller can be built to twist clockwise or anticlockwise. Those two versions are mirror images.
5. Rotate either version any way you like: the twist direction never changes, so one can never be placed exactly on the other.
6. Two structures that are mirror images but cannot be superimposed are optical isomers.

In exam wording: [Ni(en)₃]²⁺ shows optical isomerism because it exists as two non-superimposable mirror images. To earn the drawing mark, draw both isomers in 3-D (wedges and dashed wedges) as reflections of each other, with each en spanning two adjacent positions.

Ligands and complex ions: cis-platin as an anti-cancer drug

cis-platin, the cis isomer of Pt(NH₃)₂Cl₂, is used as an anti-cancer drug. It works by binding to DNA, which prevents cell division.

How it acts, step by step:

1. Inside the cell, the Cl^- ligands are replaced (water molecules take their places first).
2. The platinum then forms coordinate bonds to nitrogen atoms in the DNA bases, usually to two guanine bases next to each other on the same strand.
3. This distorts the DNA, so it cannot be copied (replicated).
4. The cancer cell cannot divide, and it dies.

Only the cis isomer is used as a drug. In cis-platin the two Cl positions are next to each other, so the Pt can bind two neighbouring bases; in trans-platin they are on opposite sides.

Benefits and risks:

• Benefit: it is an effective chemotherapy drug that stops tumours growing.
• Risk: it also binds to the DNA of healthy cells, especially fast-dividing ones, causing unpleasant side effects such as hair loss, nausea and kidney damage. It is given in small, carefully controlled doses.

Ligand substitution: copper(II) and chromium(III)

In ligand substitution, one ligand in a complex ion is replaced by another. The metal's oxidation state does not change, but the colour usually does. Always use complexed formulae (square brackets) in these equations.

1. Copper(II) with excess ammonia. Four of the six water ligands are replaced. The shape stays octahedral and the coordination number stays 6, because NH₃ and H₂O are similar in size and both uncharged.

[Cu(H₂O)₆]²⁺(aq) + 4NH₃(aq) → [Cu(NH₃)₄(H₂O)₂]²⁺(aq) + 4H₂O(l)

pale blue solution → dark (deep) blue solution

(If you add the ammonia dropwise, a pale blue precipitate forms first and then dissolves in excess: see the precipitation section.)

2. Copper(II) with concentrated hydrochloric acid. All six water ligands are replaced by four chloride ions. The coordination number falls from 6 to 4 and the shape changes from octahedral to tetrahedral, because Cl^- ligands are larger than H₂O, so fewer fit around the Cu²⁺ ion.

[Cu(H₂O)₆]²⁺(aq) + 4Cl^-(aq) ⇌ [CuCl₄]^2-(aq) + 6H₂O(l)

pale blue solution → yellow solution (it looks green part-way, while both complexes are present). Adding water pushes the equilibrium back to the left and the blue colour returns.

3. Chromium(III) with excess ammonia. All six water ligands are replaced. Coordination number 6 and the octahedral shape are kept.

[Cr(H₂O)₆]³⁺(aq) + 6NH₃(aq) → [Cr(NH₃)₆]³⁺(aq) + 6H₂O(l)

green (or violet) solution → purple solution

Pure [Cr(H₂O)₆]³⁺ is violet, but most chromium(III) solutions you meet look green, because some water ligands have already been swapped for other ions in the solution.

Ligand substitution: iron in haemoglobin

Haemoglobin carries oxygen in red blood cells. Each haemoglobin molecule contains four haem groups, and each haem has an iron(II) ion, Fe²⁺, at its centre in an octahedral complex:

• four coordinate bonds from nitrogen atoms in the flat haem ring (a multidentate ligand);
• one coordinate bond from a nitrogen atom in the globin protein;
• the sixth position is where O₂ binds, by a coordinate bond to Fe²⁺.

In the lungs, where O₂ concentration is high, O₂ binds to the Fe²⁺. In respiring tissues, where it is low, the O₂ is released. This easy, reversible binding is what makes haemoglobin a carrier.

Carbon monoxide is toxic because of ligand substitution:

1. CO is also a ligand, and it binds to the Fe²⁺ at the same site as O₂.
2. CO binds more strongly than O₂, so it replaces O₂ (ligand substitution).
3. This substitution is not easily reversed, so that haem can no longer carry O₂.
4. Less oxygen reaches the tissues and cells.

Precipitation reactions: metal ions with sodium hydroxide and ammonia

Adding a small amount of aqueous sodium hydroxide or aqueous ammonia to these ions gives a coloured metal hydroxide precipitate. Ammonia gives the same precipitates as NaOH, because aqueous ammonia contains OH^- ions (NH₃ + H₂O ⇌ NH₄⁺ + OH^-).

The difference shows up when you add the reagent in excess. Learn this table: it is also how you identify the ions in qualitative analysis.

The two rules in the excess columns:

• Excess NaOH: only Cr(OH)₃ reacts further.
• Excess NH₃: only Cr(OH)₃ and Cu(OH)₂ react further.

The Fe(OH)₂ and Mn(OH)₂ precipitates change colour on standing because oxygen in the air oxidises them (iron(II) to iron(III), for example).

Ionic equations for precipitation. For precipitation, simple formulae are fine (complexed formulae such as [Cu(H₂O)₆]²⁺ and Cu(OH)₂(H₂O)₄ are also accepted). The number of OH^- ions equals the charge on the metal ion:

• Cu²⁺(aq) + 2OH^-(aq) → Cu(OH)₂(s)
• Fe²⁺(aq) + 2OH^-(aq) → Fe(OH)₂(s)
• Fe³⁺(aq) + 3OH^-(aq) → Fe(OH)₃(s)
• Mn²⁺(aq) + 2OH^-(aq) → Mn(OH)₂(s)
• Cr³⁺(aq) + 3OH^-(aq) → Cr(OH)₃(s)

Complex formation in excess:

• Excess NaOH: Cr(OH)₃(s) + 3OH^-(aq) → [Cr(OH)₆]^3-(aq)
• Excess NH₃: Cr(OH)₃(s) + 6NH₃(aq) → [Cr(NH₃)₆]³⁺(aq) + 3OH^-(aq)
• Excess NH₃: Cu(OH)₂(s) + 4NH₃(aq) + 2H₂O(l) → [Cu(NH₃)₄(H₂O)₂]²⁺(aq) + 2OH^-(aq)

Redox reactions: iron(II) and iron(III)

Transition metal ions change oxidation state in redox reactions, and the colour changes with it. You will not be asked to recall these equations, but you must be able to build them from half-equations and interpret them using oxidation numbers. So learn the reagents and the colours, and practise the method (see Redox and electrode potentials).

Fe²⁺ oxidised to Fe³⁺ by acidified manganate(VII), H⁺/MnO₄^-:

• MnO₄^-(aq) + 8H⁺(aq) + 5e^- → Mn²⁺(aq) + 4H₂O(l)
• Fe²⁺(aq) → Fe³⁺(aq) + e^- (× 5)
• Overall: MnO₄^-(aq) + 8H⁺(aq) + 5Fe²⁺(aq) → Mn²⁺(aq) + 4H₂O(l) + 5Fe³⁺(aq)

Colour: the purple manganate(VII) is decolourised as it is added; iron goes from pale green (+2) to yellow/orange-brown (+3).

Fe³⁺ reduced to Fe²⁺ by iodide ions, I^-:

• Fe³⁺(aq) + e^- → Fe²⁺(aq) (× 2)
• 2I^-(aq) → I₂(aq) + 2e^-
• Overall: 2Fe³⁺(aq) + 2I^-(aq) → 2Fe²⁺(aq) + I₂(aq)

Colour: orange-brown to pale green for the iron, but the brown iodine formed can mask the change, so the mixture looks brown.

Redox reactions: chromium(III) and dichromate(VI)

Chromium switches between +3 (Cr³⁺, green) and +6 (Cr₂O₇^2-, orange; CrO₄^2-, yellow).

Cr₂O₇^2- reduced to Cr³⁺ by zinc in acid, Zn/H⁺:

• Cr₂O₇^2-(aq) + 14H⁺(aq) + 6e^- → 2Cr³⁺(aq) + 7H₂O(l)
• Zn(s) → Zn²⁺(aq) + 2e^- (× 3)
• Overall: Cr₂O₇^2-(aq) + 14H⁺(aq) + 3Zn(s) → 2Cr³⁺(aq) + 7H₂O(l) + 3Zn²⁺(aq)

Colour: orange → green.

Cr³⁺ oxidised by hydrogen peroxide in alkali, H₂O₂/OH^-:

In alkali, chromium(III) is first present as the dark green [Cr(OH)₆]^3- (from excess NaOH). Warming with H₂O₂ oxidises it to yellow chromate(VI), CrO₄^2-. In alkaline solution the +6 product is chromate, not dichromate.

• 2Cr³⁺(aq) + 3H₂O₂(aq) + 10OH^-(aq) → 2CrO₄^2-(aq) + 8H₂O(l) (built step by step in Worked example 4)

Colour: dark green → yellow.

Chromate(VI) to dichromate(VI). Adding acid turns the yellow chromate into orange dichromate. This is not redox: chromium is +6 on both sides.

2CrO₄^2-(aq) + 2H⁺(aq) ⇌ Cr₂O₇^2-(aq) + H₂O(l)

Colour: yellow → orange.

Redox reactions: copper(II), copper(I) and disproportionation

Cu²⁺ reduced to Cu⁺ by iodide ions. Copper(I) iodide, CuI, is an insoluble white solid, and iodine is formed.

2Cu²⁺(aq) + 4I^-(aq) → 2CuI(s) + I₂(aq)

Colour: pale blue solution → white precipitate in a brown solution (the iodine colours the mixture brown, so the precipitate can look off-white or brown).

Disproportionation of Cu⁺. Disproportionation is the oxidation and reduction of the same element in one reaction. In aqueous conditions, Cu⁺ readily disproportionates into Cu²⁺ and copper metal:

2Cu⁺(aq) → Cu²⁺(aq) + Cu(s)

• One Cu⁺ is oxidised: +1 → +2 (Cu²⁺).
• The other is reduced: +1 → 0 (Cu).

A common example is copper(I) oxide with warm dilute sulfuric acid:

Cu₂O(s) + H₂SO₄(aq) → CuSO₄(aq) + Cu(s) + H₂O(l)

Ionic form: Cu₂O(s) + 2H⁺(aq) → Cu²⁺(aq) + Cu(s) + H₂O(l)

Colour: red-brown solid → blue solution and a brown (pinkish) solid of copper.

Interpreting and predicting unfamiliar reactions

Exam questions often give you a metal you have not studied (cobalt, nickel, vanadium) or a new ligand, and expect you to apply the same patterns. Decide first which type of reaction it is:

Then use these checks on any answer:

• Charge of a complex = oxidation number of the metal + total charge on the ligands.
• Coordination number = number of coordinate bonds (2 per bidentate ligand).
• 6 bonds → octahedral; 4 bonds → tetrahedral (or square planar, for Pt).
• Every equation balances for atoms and charge.

Worked examples

1. Electron configurations and the transition-element test

Task: give the full electron configurations of a Cr atom (Z = 24) and a Cr³⁺ ion, and decide whether chromium is a transition element.

1. Cr atom, 24 electrons. Fill to argon (18 electrons), then 4s and 3d. Chromium is an exception: one 4s electron moves to 3d, giving a half-full 3d sub-shell. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
2. Cr³⁺, 21 electrons. Remove the 4s electron first, then two 3d electrons. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³
3. Decide. Cr³⁺ has 3 electrons in a d-sub-shell that holds 10, so it is incomplete. Chromium forms an ion with an incomplete d-sub-shell, so it is a transition element.

Where marks go: the common wrong answers are 3d⁴ 4s² for the atom and 3d¹ 4s² for the ion (taking the 3d electrons first). Both lose the mark.

2. Charge, oxidation number, coordination number and isomers of a complex

Task: a cobalt(III) ion bonds to two en ligands and two chloride ligands. Give the formula and charge of the complex, its coordination number and shape, and the number of stereoisomers it has.

1. Charge. Add the metal's oxidation number to the ligand charges (en is neutral, each Cl^- is 1-):

$(+3) + 2(0) + 2(-1) = +1$

Formula: [Co(en)₂Cl₂]⁺

2. Coordination number. Count bonds, not ligands:

$2 \times 2 \text{ (en)} + 2 \times 1 \text{ (Cl)} = 6$

Coordination number 6, so the shape is octahedral.

3. Stereoisomers.
   • The two Cl^- can be opposite (trans) or next to each other (cis): cis-trans isomerism.
   • The trans isomer has a mirror plane, so it has no optical isomer.
   • The cis isomer has no mirror plane: it exists as two non-superimposable mirror images.
   • Total: 3 stereoisomers (one trans, two cis optical isomers).

Where marks go: students who give the coordination number as 4 (counting four ligands) lose the mark and then usually get the shape wrong too.

3. Building a redox equation: dichromate(VI) reduced by zinc in acid

Task: construct the overall equation for the reduction of Cr₂O₇^2- to Cr³⁺ by zinc in acid.

1. Oxidation numbers. Cr in Cr₂O₇^2-:

$2x + 7(-2) = -2 \quad\Rightarrow\quad x = +6$

Each Cr goes from +6 to +3, so it gains 3 electrons, and two Cr gain 6 electrons.

2. Reduction half-equation. Balance Cr, then O with H₂O, then H with H⁺, then charge with electrons: Cr₂O₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O Charge check: (-2) + (+14) + (-6) = +6 on the left; 2(+3) = +6 on the right.

3. Oxidation half-equation. Zn → Zn²⁺ + 2e^-

4. Match electrons. 6 electrons are needed, so multiply the zinc half-equation by 3:

$\frac{6 \text{ e}^- \text{ gained}}{2 \text{ e}^- \text{ lost per Zn}} = 3 \text{ Zn}$

5. Add and cancel the electrons: Cr₂O₇^2-(aq) + 14H⁺(aq) + 3Zn(s) → 2Cr³⁺(aq) + 7H₂O(l) + 3Zn²⁺(aq) Final check: charge +12 on each side; Cr 2, O 7, H 14, Zn 3 on each side.

Where marks go: writing 7H⁺ (balancing H with the wrong number) or leaving electrons in the overall equation.

4. Building a redox equation in alkali: chromium(III) oxidised by hydrogen peroxide

Task: construct the equation for the oxidation of Cr³⁺ to CrO₄^2- by H₂O₂ in alkaline solution.

1. Oxidation half-equation. Cr goes from +3 to +6, losing 3 electrons. In alkali, balance O using OH^- and H₂O (there is no H⁺): Cr³⁺ + 8OH^- → CrO₄^2- + 4H₂O + 3e^- Charge check: (+3) + (-8) = -5 on the left; (-2) + (-3) = -5 on the right.

2. Reduction half-equation. O in H₂O₂ goes from -1 to -2: H₂O₂ + 2e^- → 2OH^-

3. Match electrons. The lowest common multiple of 3 and 2 is 6:

$2 \times (3\text{e}^-) = 3 \times (2\text{e}^-) = 6\text{e}^-$

So use 2 × the chromium half-equation and 3 × the peroxide half-equation.

4. Add: 2Cr³⁺ + 16OH^- + 3H₂O₂ → 2CrO₄^2- + 8H₂O + 6OH^-

5. Cancel the OH^- that appears on both sides (16 - 6 = 10 left on the left): 2Cr³⁺(aq) + 3H₂O₂(aq) + 10OH^-(aq) → 2CrO₄^2-(aq) + 8H₂O(l) Final check: charge -4 on each side; Cr 2, O 16, H 16 on each side.

Where marks go: using H⁺ to balance a reaction done in alkali, or giving Cr₂O₇^2- as the product. In alkaline solution the product is chromate(VI).

5. Interpreting an unfamiliar reaction: cobalt(II) and concentrated HCl

Task: a pink solution containing [Co(H₂O)₆]²⁺ turns blue when concentrated hydrochloric acid is added. The blue species is a tetrahedral complex. Identify the reaction type, give the equation, and explain the change in shape.

1. Type. Cobalt stays +2 and only the ligands change, so this is ligand substitution.
2. Formula of the product. Tetrahedral means coordination number 4, so four Cl^- ligands. Charge:

$(+2) + 4(-1) = -2$

Product: [CoCl₄]^2- 3. Equation, with complexed formulae: [Co(H₂O)₆]²⁺(aq) + 4Cl^-(aq) ⇌ [CoCl₄]^2-(aq) + 6H₂O(l) Check: charge (+2) + (-4) = -2 on each side. 4. Explanation. Cl^- ligands are larger than H₂O ligands, so only four fit round the Co²⁺. The coordination number falls from 6 to 4 and the shape changes from octahedral to tetrahedral.

Where marks go: writing Co²⁺ instead of [Co(H₂O)₆]²⁺ (ligand substitution equations need complexed formulae), or giving the product charge as 2+.

Common exam mistakes

• Taking 3d electrons before 4s when forming an ion. Fe²⁺ is ...3d⁶, not ...3d⁴ 4s². The 4s electrons are always lost first.
• Getting the Cr and Cu atoms wrong. They are 3d⁵ 4s¹ and 3d¹⁰ 4s¹, not 3d⁴ 4s² and 3d⁹ 4s².
• Defining a transition element by the atom. The definition is about an ion: "a d-block element that forms an ion with an incomplete d-sub-shell". "Has a partly filled d-sub-shell" with no mention of an ion is not enough: the Cu atom (3d¹⁰ 4s¹) has a full d-sub-shell, yet copper is a transition element because of Cu²⁺.
• Explaining Sc or Zn without naming the ion. Say Sc³⁺ has an empty d-sub-shell and Zn²⁺ has a full d-sub-shell, not just "Sc has 1 d electron" or "Zn's d-shell is full".
• Defining a ligand as something that "donates electrons". It donates a pair of electrons (a lone pair) to form a coordinate (dative covalent) bond. Missing "pair" or the bond type costs the mark.
• Drawing the dative bond arrow the wrong way. The arrow points from the ligand's donor atom to the metal.
• Giving the number of ligands as the coordination number. [Ni(en)₃]²⁺ has coordination number 6, not 3: each en forms two coordinate bonds.
• Wrong overall charge on a complex. Forgetting the ligand charges: [CuCl₄] is 2-, not 2+; [Cr(OH)₆] is 3-.
• Wrong angles. Square planar is 90°, not 109.5°. Octahedral is 90°. Only tetrahedral is 109.5°.
• Flat drawings for stereoisomers. Octahedral and tetrahedral isomers must be drawn in 3-D with wedges and dashed wedges, and optical isomers must be drawn as a mirror-image pair.
• Describing cis-platin as "killing cancer cells" with no mechanism. The marks are for binding to DNA and preventing cell division (replication).
• Saying CO "binds to haemoglobin" and stopping. You need CO binds more strongly than O₂, it replaces O₂ (ligand substitution) and the change is not easily reversed, so less O₂ is transported.
• Saying the iron in haemoglobin is Fe³⁺. It is Fe²⁺ (iron(II)).
• Writing [Cu(NH₃)₆]²⁺ for copper with excess ammonia. Only four waters are replaced: [Cu(NH₃)₄(H₂O)₂]²⁺.
• Using Cu²⁺ in a ligand substitution equation. These equations need complexed formulae in square brackets; simple ions are only acceptable for precipitation equations.
• Saying the Fe(OH)₂, Fe(OH)₃ or Mn(OH)₂ precipitates dissolve in excess. They don't. Only Cr(OH)₃ dissolves in excess NaOH; only Cr(OH)₃ and Cu(OH)₂ dissolve in excess NH₃.
• Sloppy observations. Write "precipitate" for a solid and "colourless" (not "clear") for a solution with no colour. Give the colour of the precipitate and what happens in excess.
• Wrong product for Cr³⁺ + H₂O₂/OH^-. In alkali it gives yellow CrO₄^2-, not orange Cr₂O₇^2-; acid then converts chromate to dichromate.
• Calling the chromate to dichromate change redox. Chromium is +6 in both; it is an acid-base equilibrium.
• Disproportionation with only one change named. State both: Cu (+1) is oxidised to Cu²⁺ (+2) and reduced to Cu (0).

Key definitions

• Transition element: a d-block element that forms an ion with an incomplete d-sub-shell.
• Ligand: a molecule or ion that donates a pair of electrons to a central metal ion (or atom) to form a coordinate bond.
• Coordinate bond (dative covalent bond): a covalent bond in which the shared pair of electrons is supplied by one of the bonding atoms only.
• Monodentate ligand: a ligand that forms one coordinate bond to the central metal ion (e.g. H₂O, NH₃, Cl^-).
• Bidentate ligand: a ligand that donates two lone pairs, forming two coordinate bonds to the central metal ion (e.g. NH₂CH₂CH₂NH₂, 'en').
• Complex ion: a central metal ion surrounded by ligands, bonded to it by coordinate bonds.
• Coordination number: the total number of coordinate bonds formed between the central metal ion and its ligands.
• Stereoisomers: compounds with the same structural formula but a different arrangement (of atoms) in space.
• cis isomer (complexes): the two identical ligands are next to each other (at 90°); trans isomer: they are opposite each other (at 180°).
• Optical isomers: non-superimposable mirror images.
• Ligand substitution: a reaction in which one ligand in a complex ion is replaced by another ligand.
• Disproportionation: the oxidation and reduction of the same element in a reaction.

Specification

• I can write the electron configuration of any atom or ion of Sc to Zn in sub-shell notation, given the atomic number and charge, including Cr and Cu and losing 4s before 3d. (5.3.1(a))
• I can define a transition element and explain why Ti to Cu are transition elements but Sc and Zn are not. (5.3.1(b))
• I can illustrate, using at least two transition elements, the existence of more than one oxidation state in their compounds. (5.3.1(c)(i))
• I can illustrate, using at least two transition elements, the formation of coloured ions. (5.3.1(c)(ii))
• I can give examples of transition elements and their compounds as catalysts (Cu²⁺ with Zn and acid, MnO₂ with H₂O₂) and explain their importance to industry, including the benefits and the toxicity risks. (5.3.1(c)(iii))
• I can explain and use the term ligand in terms of coordinate (dative covalent) bonding, including monodentate (H₂O, Cl^-, NH₃) and bidentate (en) ligands. (5.3.1(d))
• I can use the terms complex ion and coordination number, and give examples of octahedral (six-fold) and tetrahedral or square planar (four-fold) complexes, drawn in 3-D. (5.3.1(e))
• I can explain and draw cis-trans isomerism in complexes such as Pt(NH₃)₂Cl₂. (5.3.1(f)(i))
• I can explain and draw optical isomerism in complexes such as [Ni(en)₃]²⁺. (5.3.1(f)(ii))
• I can describe how cis-platin acts as an anti-cancer drug by binding to DNA and preventing cell division, with its benefits and risks. (5.3.1(g))
• I can write equations and give colour changes for the formation of [Cu(NH₃)₄(H₂O)₂]²⁺ and [CuCl₄]^2- from [Cu(H₂O)₆]²⁺, and [Cr(NH₃)₆]³⁺ from [Cr(H₂O)₆]³⁺. (5.3.1(h))
• I can explain the importance of iron in haemoglobin, including ligand substitution involving O₂ and CO. (5.3.1(i))
• I can write ionic equations and give colour changes for Cu²⁺, Fe²⁺, Fe³⁺, Mn²⁺ and Cr³⁺ with NaOH(aq) and NH₃(aq), including complex formation in excess. (5.3.1(j))
• I can construct and interpret redox equations, with colour changes, for Fe²⁺/Fe³⁺, Cr³⁺/Cr₂O₇^2-, the reduction of Cu²⁺ to Cu⁺ and the disproportionation of Cu⁺. (5.3.1(k))
• I can interpret and predict unfamiliar ligand substitution, precipitation and redox reactions. (5.3.1(l))

Related notes

• Electron structure
• Bonding and structure
• Redox
• Reaction rates
• Alkenes
• Redox and electrode potentials
• Qualitative analysis (transition elements)
• Amino acids, amides and chirality