A Level Chemistry: Colour, Stereoisomerism and Kstab — Revision Notes

Condensed for revision. For the full explanation, use the Colour, Stereoisomerism and Stability Constants study guide, then test yourself with the practice questions. For oxidation states, ligands and redox, see the Transition Elements revision notes.

Syllabus: Cambridge International AS & A Level Chemistry 9701, 2025–2027, A Level content: subtopics 28.3 Colour of complexes, 28.4 Stereoisomerism in transition element complexes and 28.5 Stability constants, Kstab.

Colour of complexes (28.3)

Degenerate and non-degenerate d orbitals

• Degenerate orbitals have the same energy. In an isolated transition-metal ion, all five 3d orbitals are degenerate.
• When ligands bond to the ion, their lone pairs repel the d electrons unequally, so the d orbitals split into two non-degenerate sets (sets of different energy). The energy gap is ΔE.

Shape of complex	Higher-energy set	Lower-energy set
Octahedral (6 ligands)	two d orbitals	three d orbitals
Tetrahedral (4 ligands)	three d orbitals	two d orbitals

For comparable ligands, ΔE is smaller in a tetrahedral complex than in an octahedral one.

Why complexes are coloured: the chain of reasoning

1. The ligands split the d orbitals into two non-degenerate sets, separated by ΔE.
2. An electron in the lower set absorbs a photon of visible light and is promoted to the higher set.
3. The frequency absorbed is given by ΔE = hf.
4. That frequency is removed from white light; the colour we see is the complementary colour of the light absorbed.

This needs a partially filled d sub-shell. A d⁰ ion (for example Sc³⁺) has no electron to promote, and a d¹⁰ ion (for example Zn²⁺) has no vacancy, so no d–d transition is possible: scandium(III) and zinc compounds are white or colourless.

Ligands change ΔE. Different ligands split the d orbitals by different amounts. A different ΔE means a different frequency absorbed, so a different complementary colour is seen. That is why ligand exchange changes the colour. A change of shape (octahedral to tetrahedral) also changes the splitting.

Copper(II) and cobalt(II) examples

Reagent added to the aqua ion	Copper(II)	Cobalt(II)
none: aqua ion	[Cu(H₂O)₆]²⁺, pale blue solution	[Co(H₂O)₆]²⁺, pink solution
OH⁻(aq), or a little NH₃(aq)	[Cu(OH)₂(H₂O)₄], pale blue precipitate	[Co(OH)₂(H₂O)₄], blue precipitate
excess NH₃(aq)	[Cu(NH₃)₄(H₂O)₂]²⁺, deep blue solution	[Co(NH₃)₆]²⁺, pale brown (straw) solution, which darkens in air
concentrated HCl (excess Cl⁻)	[CuCl₄]²⁻, yellow-green solution (tetrahedral)	[CoCl₄]²⁻, blue solution (tetrahedral)

Equations:

[Cu(H₂O)₆]²⁺ + 2OH⁻  → [Cu(OH)₂(H₂O)₄] + 2H₂O
[Cu(H₂O)₆]²⁺ + 4NH₃  ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O
[Cu(H₂O)₆]²⁺ + 4Cl⁻  ⇌ [CuCl₄]²⁻ + 6H₂O
[Co(H₂O)₆]²⁺ + 2OH⁻  → [Co(OH)₂(H₂O)₄] + 2H₂O
[Co(H₂O)₆]²⁺ + 6NH₃  ⇌ [Co(NH₃)₆]²⁺ + 6H₂O
[Co(H₂O)₆]²⁺ + 4Cl⁻  ⇌ [CoCl₄]²⁻ + 6H₂O

With chloride, the coordination number falls from 6 to 4 because Cl⁻ is a larger ligand than H₂O. The complex changes from octahedral to tetrahedral. The overall charge also changes (2+ to 2−) because each Cl⁻ carries a negative charge.

Stereoisomerism in complexes (28.4)

en = ethane-1,2-diamine, H₂NCH₂CH₂NH₂, a bidentate ligand (it bonds through both N atoms).

Geometrical (cis/trans) isomerism

Complex	Shape	cis	trans
[Pt(NH₃)₂Cl₂]	square planar	two Cl⁻ at 90° (cisplatin, an anti-cancer drug)	two Cl⁻ at 180°
[Co(NH₃)₄(H₂O)₂]²⁺	octahedral	two H₂O at 90°	two H₂O at 180°
[Ni(en)₂(H₂O)₂]²⁺	octahedral	two H₂O at 90°	two H₂O at 180°

Optical isomerism

A complex shows optical isomerism when it is not superimposable on its mirror image (at this level: it has no plane of symmetry).

• [Ni(en)₃]²⁺: three bidentate ligands wrap around the Ni²⁺ like a propeller; the two mirror images cannot be superimposed, so there is a pair of optical isomers.
• [Ni(en)₂(H₂O)₂]²⁺: the cis form is chiral and has a pair of optical isomers. The trans form has a plane of symmetry, so it is not optically active. This complex therefore has three stereoisomers in total: trans, and the two optical isomers of cis.

Polarity

Decide from the symmetry of the whole complex:

• trans isomers of the complexes above: the bond dipoles are opposite each other and cancel, so the complex is non-polar.
• cis isomers: the bond dipoles do not cancel, so the complex is polar.
• [Ni(en)₃]²⁺ has identical ligands arranged symmetrically, so it has no overall dipole.

Stability constants, Kstab (28.5)

Definition: Kstab is the equilibrium constant for the formation of the complex ion in a solvent from its constituent ions or molecules.

Writing the expression: products over reactants, as for Kc, but [H₂O] is not included.

[Cu(H₂O)₆]²⁺ + 4NH₃ ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O

Kstab = [Cu(NH₃)₄(H₂O)₂²⁺] / ( [Cu(H₂O)₆²⁺] [NH₃]⁴ )      units: mol⁻⁴ dm¹²

For [Co(NH₃)₆]²⁺, with six ligands exchanged, the concentration of NH₃ is raised to the power 6 and the units are mol⁻⁶ dm¹⁸.

Worked example. (The Kstab value here is illustrative, for practice only; use the value given in your question.) Kstab for the copper–ammonia complex above is 1.0 × 10¹³ mol⁻⁴ dm¹². At equilibrium, [NH₃] = 0.20 mol dm⁻³. Calculate the ratio [Cu(NH₃)₄(H₂O)₂²⁺] : [Cu(H₂O)₆²⁺].

ratio = Kstab × [NH₃]⁴ = 1.0 × 10¹³ × (0.20)⁴
      = 1.0 × 10¹³ × 1.6 × 10⁻³ = 1.6 × 10¹⁰

So almost all the copper(II) is present as the ammine complex.

Ligand exchange and Kstab:

• A large Kstab means the equilibrium lies far to the right: a stable complex ion is formed.
• When two ligands compete for a metal ion, the exchange favours the complex with the larger Kstab.
• A large excess of a ligand can still shift the position of equilibrium (for example, concentrated HCl forming [CuCl₄]²⁻), and the change can be reversed by dilution.

Exam traps

• Colour comes from light absorbed as an electron is promoted between d orbitals. It is not light emitted as an electron falls back, which is the flame-test explanation.
• The colour seen is the complementary colour of the light absorbed, not the colour absorbed.
• Octahedral: two up, three down. Tetrahedral: three up, two down. Do not swap them.
• Leave [H₂O] out of every Kstab expression, but keep the aqua ion [Cu(H₂O)₆]²⁺ in it.
• The power on the ligand concentration equals the number of ligands exchanged (4 for CuCl₄²⁻, 6 for Co(NH₃)₆²⁺).
• trans-[Ni(en)₂(H₂O)₂]²⁺ is not optically active; only the cis form is.

Self-test

1. Define degenerate orbitals.
2. In an octahedral complex, how many d orbitals are in the higher-energy set?
3. Explain why [Cu(H₂O)₆]²⁺ is coloured but [Zn(H₂O)₆]²⁺ is not.
4. State the colour change when excess concentrated hydrochloric acid is added to aqueous cobalt(II) ions, and give the formula of the new complex.
5. Why does the coordination number fall when Cl⁻ replaces H₂O around Cu²⁺?
6. Which isomer of [Pt(NH₃)₂Cl₂] is polar? Explain.
7. How many stereoisomers does [Ni(en)₂(H₂O)₂]²⁺ have?
8. Write the Kstab expression, with units, for [Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O.
9. (Illustrative values.) Kstab = 2.0 × 10⁵ mol⁻⁴ dm¹² for the reaction in question 8. At equilibrium, [Cl⁻] = 1.0 mol dm⁻³ and [Cu(H₂O)₆²⁺] = 1.0 × 10⁻⁶ mol dm⁻³. Calculate [CuCl₄²⁻].

Answers:

1. Orbitals of the same energy.
2. Two.
3. Cu²⁺ is 3d⁹: the ligands split the d orbitals, and an electron absorbs visible light as it is promoted from the lower set to the higher set; the complementary colour is seen. Zn²⁺ is 3d¹⁰: the d sub-shell is full, so no electron can be promoted and no visible light is absorbed.
4. Pink to blue; [CoCl₄]²⁻.
5. Cl⁻ ligands are larger than H₂O molecules, so only four fit around the metal ion.
6. The cis isomer: its bond dipoles do not cancel. In the trans isomer they are opposite each other and cancel.
7. Three: the trans isomer and the two optical isomers of the cis form.
8. Kstab = [CuCl₄²⁻] / ( [Cu(H₂O)₆²⁺] [Cl⁻]⁴ ); units mol⁻⁴ dm¹².
9. [CuCl₄²⁻] = Kstab × [Cu(H₂O)₆²⁺] × [Cl⁻]⁴ = 2.0 × 10⁵ × 1.0 × 10⁻⁶ × 1.0⁴ = 0.20 mol dm⁻³.

These are original notes written for revision. Kstab values marked illustrative are not data-book values. Check the full syllabus wording in the official 9701 syllabus.