Edexcel International A Level Chemistry: Transition metals: electronic structure, complexes and colour (YCH11)

This guide teaches outcomes 17.1 to 17.17 of Topic 17: Transition Metals and their Chemistry, from the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. The content is International A2 (IA2) material and sits in Unit 5: Transition Metals and Organic Nitrogen Chemistry. The later outcomes of Topic 17 (vanadium and chromium redox, reactions with hydroxide and ammonia, catalysis) are in the redox and catalysis guide.

Keep these open as you work: the Edexcel course hub for chemistry, the chemistry checklist to print, the revision notes for this unit, the practice questions and the free diagnostics. For background, revisit atomic structure revision notes, ionic and covalent bonding (dative bonds) and shapes of molecules (electron-pair repulsion).

What this unit covers

Spec ref	What you must be able to do
17.1	Define a transition metal
17.2	Write configurations of Sc to Zn atoms and ions
17.3	Explain variable oxidation number
17.4–17.6	Define ligand and complex ion; describe dative bonding in complexes
17.7–17.9	Explain why most aqueous ions are coloured and why some are not
17.10	Use the term coordination number
17.11	Explain colour changes from oxidation number, ligand or coordination number
17.12–17.14	Monodentate ligands; octahedral and tetrahedral shapes
17.15	Square planar complexes and cis-platin
17.16	Bidentate and hexadentate ligands: NH₂CH₂CH₂NH₂ and EDTA⁴⁻
17.17	Haemoglobin and carbon monoxide

17.1–17.2 Electronic structure

A transition metal is a d-block element with at least one stable ion whose d-orbitals are incompletely filled. Being in the d-block is not enough. Scandium forms only Sc³⁺, which is [Ar] (no 3d electrons). Zinc’s single ion, Zn²⁺, has the configuration [Ar] 3d¹⁰, a complete 3d sub-shell. Neither fits the definition, so neither is a transition metal.

Rules for Period 4 configurations:

• Fill 4s before 3d in the atom, but write 3d before 4s in the answer: Fe is [Ar] 3d⁶ 4s².
• Two exceptions: Cr is [Ar] 3d⁵ 4s¹ and Cu is [Ar] 3d¹⁰ 4s¹.
• When an atom forms a positive ion, 4s electrons are lost first, then 3d.

Worked example 1: titanium

Is titanium (Z = 22) a transition metal? Write the configurations of Ti, Ti³⁺ and Ti⁴⁺.

1. Ti has 22 electrons: [Ar] plus 4 more. Two go into 4s, two into 3d: [Ar] 3d² 4s².
2. Ti³⁺ loses three electrons. Both 4s electrons go first, then one 3d: [Ar] 3d¹.
3. Ti⁴⁺ loses all four: [Ar].
4. Ti³⁺ has a partly filled 3d sub-shell, and it is stable in compounds such as TiCl₃. So titanium is a transition metal. One qualifying ion is enough; it does not matter that Ti⁴⁺ is d⁰.

Worked example 2: cobalt ions

Cobalt is Z = 27. The atom is [Ar] 3d⁷ 4s². Co²⁺ loses the two 4s electrons: [Ar] 3d⁷. Co³⁺ then loses one 3d electron: [Ar] 3d⁶. A common slip is writing Co²⁺ as [Ar] 3d⁵ 4s², which removes the wrong electrons.

17.3 Variable oxidation number

Transition metals show more than one oxidation number because the 4s and 3d sub-shells are close in energy. Electrons can be removed from both, and successive ionisation energies rise fairly steadily rather than in one huge jump. The extra energy needed to form a more highly charged ion can be repaid by stronger lattice or hydration energies, or by forming covalent bonds (as in oxo-anions). So several oxidation numbers can be stable.

Manganese shows this well. Use oxygen = −2 and the overall charge:

Species	Working	Mn oxidation number
Mn₂O₃	2x + 3(−2) = 0	+3
MnO₂	x + 2(−2) = 0	+4
MnO₄²⁻	x + 4(−2) = −2	+6
Mn₂O₇	2x + 7(−2) = 0	+7

Compare this with a Group 2 metal such as calcium. Removing a third electron from calcium would break into a full inner shell, which costs far more energy than any bonding can return. So calcium is always +2.

17.4–17.6 and 17.10 Ligands, dative bonds and complex ions

• A ligand is a molecule or ion with a lone pair of electrons that it donates to a central metal ion.
• The bond formed is a dative (coordinate) covalent bond: both electrons in the shared pair come from the ligand.
• A complex ion has a metal ion at its centre with ligands bonded around it.
• The coordination number counts the dative bonds that the central metal ion receives. It equals the number of ligands only when every ligand is monodentate.

Worked example 3: reading a complex

Analyse [Co(NH₃)₅Cl]²⁺.

1. Ligands: five NH₃ (lone pair on N) and one Cl⁻ (lone pair on Cl).
2. Coordination number: 5 + 1 = 6.
3. Oxidation number of cobalt: x + 5(0) + (−1) = +2, so x = +3.
4. Configuration of the metal ion: Co³⁺ is [Ar] 3d⁶, an incomplete d sub-shell.
5. Shape: six-fold coordination, so octahedral (see 17.13).

17.12–17.14 Monodentate ligands and shapes

H₂O, OH⁻ and NH₃ are monodentate ligands: each forms one dative bond, through one lone pair (on O, O and N respectively). These ligands are small, so six fit round a Period 4 metal ion.

Why six-fold complexes are octahedral (17.13). The six bonding pairs around the metal ion repel each other equally. They move as far apart as possible, which puts them at the corners of an octahedron with 90° between adjacent bonds. Examples: [Fe(H₂O)₆]²⁺, [Ni(NH₃)₆]²⁺, and mixed complexes such as [Fe(H₂O)₅(OH)]²⁺.

Tetrahedral complexes (17.14). Larger ligands such as Cl⁻ cannot fit six round the ion, so four are bonded instead. Four bonding pairs repel to a tetrahedral arrangement with angles of 109.5°. Examples: [CuCl₄]²⁻, [CoCl₄]²⁻ and [NiCl₄]²⁻.

Coordination number	Shape	Bond angle	Example
6	octahedral	90°	[Cr(NH₃)₆]³⁺
4	tetrahedral	109.5°	[NiCl₄]²⁻
4	square planar	90°	[Pt(NH₃)₂Cl₂]

17.15 Square planar complexes and cis-platin

Some four-coordinate complexes are square planar: the metal and all four donor atoms lie in one plane at 90° to each other. Platinum(II) forms such complexes. Cis-platin is [Pt(NH₃)₂Cl₂]; platinum is +2 because x + 2(0) + 2(−1) = 0.

With two pairs of identical ligands on a square there are two arrangements:

• cis: the two Cl⁻ ligands are next to each other, Cl–Pt–Cl = 90°. This is cis-platin.
• trans: the two Cl⁻ ligands are opposite each other, Cl–Pt–Cl = 180°.

Cis-platin is used in cancer treatment. It binds to DNA in cancer cells and stops them dividing. The specification stresses that it is supplied as a single isomer, not mixed with the trans form; the two isomers are different compounds with different biological effects, so only the cis form is wanted in the drug.

17.16 Bidentate and hexadentate ligands

• A bidentate ligand forms two dative bonds to the same metal ion. Ethane-1,2-diamine, NH₂CH₂CH₂NH₂, donates the lone pair on each nitrogen.
• A hexadentate ligand forms six dative bonds. EDTA⁴⁻ uses two nitrogen atoms and four oxygen atoms (one from each carboxylate group).

Worked example 4: counting bonds, not ligands

(a) [Cr(NH₂CH₂CH₂NH₂)₃]³⁺. Three bidentate ligands give 3 × 2 = 6 dative bonds, so the complex is octahedral. The ligand is neutral, so the charge equals that of Cr³⁺: 3+.

(b) Mn²⁺ reacts with EDTA⁴⁻. One EDTA⁴⁻ supplies all six bonds, so the coordination number is 6. Charge: +2 + (−4) = −2, so the complex is [Mn(EDTA)]²⁻.

17.7–17.9 Colour

Most aqueous transition metal ions are coloured: for example [Cu(H₂O)₆]²⁺ is blue, [Ni(H₂O)₆]²⁺ is green and [Ti(H₂O)₆]³⁺ is purple.

How the colour arises (17.8):

1. In an isolated ion the five 3d orbitals have the same energy. When ligands bond, their electrons repel the d electrons, and the d-orbitals split into two groups at different energies. The gap is ΔE.
2. A d electron in the lower group can absorb a photon and move to the upper group. Only light with energy matching ΔE (where ΔE = hf) is absorbed, so certain frequencies of visible light are removed.
3. The light that is transmitted or reflected has the remaining frequencies. You see this complementary colour.

[Ti(H₂O)₆]³⁺ absorbs mainly yellow-green light and transmits red and blue, so it looks purple.

Lack of colour (17.9). An electron can only jump between split d-orbitals if there is an electron to move and a space to move into.

• d⁰ ions (Sc³⁺, Ti⁴⁺) have no d electrons. [Sc(H₂O)₆]³⁺ is colourless; TiO₂ is white.
• d¹⁰ ions (Zn²⁺, Cu⁺) have a full 3d sub-shell, so no transition is possible. [Zn(H₂O)₆]²⁺ is colourless; copper(I) chloride is white.
• Even when a d–d transition is possible, if ΔE is large enough that the absorbed light is ultraviolet, the complex looks colourless.

17.11 Why colours change

ΔE depends on the metal, its oxidation number, the ligands and the coordination number (and so the shape). Change any of these and a different frequency is absorbed.

Cause	Example	Colour change
Oxidation number	[Fe(H₂O)₆]²⁺ oxidised in air to iron(III)	pale green → yellow-brown
Ligand	excess NH₃ added to [Cu(H₂O)₆]²⁺ gives [Cu(NH₃)₄(H₂O)₂]²⁺	pale blue to deep blue
Coordination number	[Co(H₂O)₆]²⁺ + concentrated HCl → [CoCl₄]²⁻	pink → blue

In the cobalt example the ligand changes too, and the shape changes from octahedral (6) to tetrahedral (4). When asked, name every factor that changes.

17.17 Haemoglobin

Haemoglobin, which carries oxygen in red blood cells, is an iron(II) complex. Each iron(II) ion is held by a polydentate ligand (the haem group, which bonds through several nitrogen atoms) and by the protein. One coordination site is left for an O₂ molecule, which binds by a dative bond and is released in the tissues.

Carbon monoxide poisoning is ligand exchange. CO replaces the O₂ molecule on iron(II). CO binds much more strongly than O₂, so the exchange is not easily reversed. Those sites can no longer carry oxygen, and tissues are starved of it. You are not expected to know the structure of the haem group.

Worked example 5: oxygen capacity of blood

Use these values: a blood sample contains 15.0 g of haemoglobin per 100 cm³; Mr of haemoglobin = 64 500; each haemoglobin molecule has four iron(II) centres and each binds one O₂; molar gas volume = 24.0 dm³ mol⁻¹. Find the largest volume of O₂ that 100 cm³ of this blood can carry.

1. Amount of haemoglobin = 15.0 ÷ 64 500 = 2.33 × 10⁻⁴ mol.
2. Amount of O₂ = 4 × 2.33 × 10⁻⁴ = 9.30 × 10⁻⁴ mol.
3. Volume = 9.30 × 10⁻⁴ × 24 000 = 22.3 cm³.

If some sites hold CO instead, this capacity falls in proportion.

Common errors

• Taking 3d electrons out ahead of 4s (Fe²⁺ is [Ar] 3d⁶, not [Ar] 3d⁴ 4s²).
• Writing Cr or Cu with a 4s² configuration.
• Calling zinc a transition metal because it is in the d-block.
• Saying a complex’s colour is the colour it absorbs. You see the light that is not absorbed.
• Saying electrons “drop back and emit” the colour you see. Colour in complexes comes from absorption.
• Quoting how many ligands there are, instead of how many dative bonds, for bidentate or hexadentate ligands.
• Giving 90° for a tetrahedral complex, or 109.5° for a square planar one.
• Describing CO as “blocking” haemoglobin without saying it replaces O₂ by ligand exchange and binds more strongly.

Next steps

Condense with the revision notes, then try the practice questions. See the exam preparation guide for planning.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, published by Pearson Education Limited. Topic 17: Transition Metals and their Chemistry, outcomes 17.1 to 17.17.