Coordination Compounds · Lesson 2 of 12
Ligands and the Language of Complexes
“Read a complex at the donor-atom level: identity, denticity, charge, geometry and oxidation state.”
• Identify metal centres, ligands, donor atoms and counter ions. • Calculate oxidation state and coordination number independently. • Distinguish chelation from ambidentate binding. • Relate coordination numbers to possible geometries.
The electron pair that builds a coordination bond
Ammonia contains a nitrogen lone pair, and water contains oxygen lone pairs. An available electron pair can interact with an accepting metal centre to make a metal–ligand bond. In Lewis language the ligand is a base and the metal centre is an acid. The pair is assigned to the ligand in the donor–acceptor description, but the completed bond is not a permanently labelled special species of electron pair. Metal–ligand bonding may contain both covalent and electrostatic contributions.
A central atom or ion together with its directly attached ligands, described as a single unit such as [CoCl₃(NH₃)₃], [Ni(CO)₄] or [Fe(CN)₆]⁴⁻. An entity may be neutral, positive or negative.
The coordination sphere is enclosed in square brackets. Counter ions stand outside the bracket. Thus K₄[Fe(CN)₆] contains one iron centre, six cyanido ligands and four potassium counter ions. A ligand molecule can contribute more than one donor atom, so counting molecules and counting coordination sites need not give the same result.
Denticity measures simultaneous attachment
| Ligand | Charge | Usual donor atoms to one metal | Description |
|---|---|---|---|
| NH₃ | 0 | one N | Monodentate (unidentate) |
| H₂O | 0 | one O | Monodentate |
| Cl⁻ | −1 | one Cl | Monodentate |
| en: H₂NCH₂CH₂NH₂ | 0 | two N | Bidentate (didentate) |
| oxalate: C₂O₄²⁻ | −2 | two O | Bidentate in the examples here |
| N(CH₂CH₂NH₂)₃ | 0 | four N | Potentially tetradentate |
| EDTA⁴⁻ | −4 | two N + four O | Commonly hexadentate |
Ethane-1,2-diamine, abbreviated en, binds through both nitrogen atoms to one metal and closes a ring that includes the metal. Oxalate similarly supplies two oxygen donors. This ring-forming attachment is chelation; the ligand behaves like a claw. EDTA has six potential donors, but the denticity actually adopted depends on the metal and conditions. A formula by itself may not establish an unusual binding mode.
A ligand with alternative donor atoms that can attach through one or another site in the binding mode under discussion. Nitrite can bind through N or O; thiocyanate through N or S.
Ambidentate does not mean bidentate. In a monodentate nitrite complex, one nitrite occupies one coordination site whether the donor is N or O. In an en complex the two N donors act simultaneously and occupy two sites. Nitrite has more than one possible attachment mode, while en has more than one simultaneous attachment in the examples being studied. The distinction predicts linkage isomerism for nitrite and chelate-ring geometry for en.
Three separate calculations for a single formula
First infer the entity charge from the superscript or counter-ion balance. Then assign each ligand its conventional charge and solve for the metal oxidation state. Finally count directly attached donor atoms to obtain coordination number. For the mononuclear examples, count one attachment per monodentate ligand, two per coordinated en or oxalate, and six for hexadentate EDTA. Do not count metal–ligand π bonding a second time.
Problem
Analyse [Co(H₂O)(CN)(en)₂]²⁺.
- 1.Water and both en molecules are neutral; cyanido has charge −1. Solve x − 1 = +2, giving Co +3.
- 2.Water contributes one O donor, cyanido one C donor and two en molecules four N donors. CN = 1 + 1 + 4 = 6, although there are only four ligand molecules/ions.
- 3.The entity is heteroleptic because more than one ligand type is present. Co³⁺ is d⁶: neutral Co is [Ar]3d⁷4s²; removing 4s electrons first and one 3d electron gives [Ar]3d⁶. Six-coordinate geometry is commonly octahedral here; magnetic behaviour requires additional bonding information.
Problem
Find metal oxidation states in [CoBr₂(en)₂]⁺, [PtCl₄]²⁻, K₃[Fe(CN)₆] and [Cr(NH₃)₃Cl₃].
- 1.For cobalt, x − 2 = +1 gives +3; its CN is 2 + 4 = 6. For platinum, x − 4 = −2 gives +2; four Cl donors give CN 4.
- 2.Three K⁺ require [Fe(CN)₆]³⁻. Solve x − 6 = −3 to obtain Fe +3; CN is 6.
- 3.The chromium entity is neutral. x − 3 = 0 gives Cr +3, and three N plus three Cl donors give CN 6. Entity charge, oxidation state and CN are therefore three independent descriptions.
Geometry describes donor positions
An octahedron has six vertices, with four donors in an equatorial plane and two axial donors; adjacent ideal angles are 90° and opposite angles are 180°. Four-coordinate square-planar complexes have all four donors in one plane, also with adjacent 90° angles. Tetrahedral complexes have four donors toward tetrahedral corners, with ideal angles about 109.5°. Five-coordinate complexes may be trigonal bipyramidal or square pyramidal. CN alone cannot uniquely choose between these alternatives.
| Example | Metal state and d count | CN | Established geometry |
|---|---|---|---|
| [Co(NH₃)₆]³⁺ | Co(III), d⁶ | 6 | Octahedral |
| [Ni(CO)₄] | Ni(0), formal d¹⁰ | 4 | Tetrahedral |
| [PtCl₄]²⁻ | Pt(II), d⁸ | 4 | Square planar |
| [Fe(CO)₅] | Fe(0), formal d⁸ | 5 | Trigonal bipyramidal |
One ligand type or several
Homoleptic means all ligands are of one kind, as in [Co(NH₃)₆]³⁺ or [Cr(C₂O₄)₃]³⁻. Heteroleptic means more than one kind, as in [CoCl₂(NH₃)₄]⁺. A homoleptic complex may contain several donor atoms per ligand; three identical oxalates remain one ligand type. Conversely, two kinds of ligand may share an element as donor, so donor-element identity alone does not define homoleptic.
Use a fixed reading order whenever a new complex appears: locate brackets; identify ligands and their charges; balance the entity charge; count donor atoms; determine d count; then investigate geometry, magnetism and colour. This avoids the common error of deducing all properties from the sign of the complex ion.
Quick Quiz
Quiz
What is the coordination number in [Co(en)₃]³⁺?
Three bidentate ligands supply six N donors.
Which pair illustrates alternative rather than simultaneous donor attachment?
Nitrite changes its donor atom while occupying one site in these forms.
What is the oxidation state of Fe in [Fe(CN)₆]⁴⁻?
Each cyanido ligand carries −1, so x − 6 = −4 gives Fe in oxidation state +2.
Which statement about CN 4 is correct?
Four donors can adopt more than one geometry and may come from fewer than four ligands.
Which entity is homoleptic?
All three ligands are oxalate even though each uses two donor atoms.
Practice Problems
- Analyse [Fe(C₂O₄)₃]³⁻ for ligand count, CN, oxidation state and d count.
- Compare en, nitrite and EDTA⁴⁻ by donor behaviour.
- A complex has one EDTA⁴⁻ ligand surrounding a metal of oxidation state +3. Give the usual entity charge and CN.
- Explain why π back-bonding to a carbonyl does not double its contribution to CN.
- Give two examples each of coordination entities, ligands and coordination polyhedra.
- What is wrong with equating oxidation number, entity charge and CN in [Cu(CN)₄]³⁻?
1. Three oxalates carry total charge −6. x − 6 = −3 gives Fe +3. 2. Three bidentate ligands give CN 6. 3. Fe³⁺ is d⁵ after removing two 4s and one 3d electrons. 4. The six-coordinate entity is octahedral and homoleptic; spin cannot be decided from charge balance alone.
Key Takeaways
• Ligands donate electron density through donor atoms. • Denticity counts simultaneous donor attachments to one metal. • Chelation forms a ring involving the metal. • Ambidentate ligands offer alternative donor atoms. • Entity charge is the sum of metal oxidation state and ligand charges. • CN counts donor atoms rather than ligand molecules or π bonds. • Four-coordinate geometry cannot be inferred from CN alone.