Difficulty: Intermediate to Advanced | Prerequisites: Electron configurations, Lewis structures, VSEPR, basic knowledge of transition metals
Coordination chemistry is the study of metal complexes: a central metal ion surrounded by molecules or ions (ligands) that donate electron pairs to it. Crystal field theory (CFT) explains why these complexes have particular colours, magnetic properties, and stabilities by modelling how the ligands split the metal's d orbitals. This topic ties together electron configurations, geometry, colour, and magnetism. If you can assign an oxidation state and count d electrons, you can predict most of the properties the exam will test.
Metal complexes have a coordination number (how many bonds the metal makes to ligands) and an oxidation state (the charge on the metal after removing all ligands with their electron pairs). The spectrochemical series ranks ligands by how strongly they split d orbitals, which determines whether a complex is high spin or low spin, its colour, and its magnetic behaviour. Diamagnetic means all electrons paired; paramagnetic means unpaired electrons present.
Coordination number
The number of ligand donor atoms directly bonded to the central metal ion. For octahedral complexes it is 6; for tetrahedral, 4; for square planar, 4. In simple terms, count the bonds from ligands to the metal.
Oxidation state (oxidation number)
The charge left on the metal ion after all ligands are removed with their bonding electrons. Determined by balancing the overall charge of the complex against the known charges of the ligands.
Ligand
A molecule or ion that donates one or more electron pairs to a metal ion. Monodentate ligands (NH3, Cl-, H2O, CN-) donate through one atom. Bidentate ligands (en = ethylenediamine) donate through two atoms.
Spectrochemical series
An experimentally determined ranking of ligands by the size of the crystal field splitting (Δ) they produce: Cl- < F- < H2O < CH2CN < NH3 < en < NO2- < CN-. Weak-field ligands (left) produce small splitting; strong-field ligands (right) produce large splitting.
Crystal field splitting (Δ)
The energy gap between the lower-energy (t2g) and higher-energy (eg) sets of d orbitals in an octahedral complex. The size of Δ determines the colour of the complex and whether it is high spin or low spin.
High spin vs low spin
In a high-spin complex, Δ is small (weak-field ligands), so electrons spread across all five d orbitals before pairing. In a low-spin complex, Δ is large (strong-field ligands), so electrons fill the lower set first and pair up before occupying the upper set. The distinction matters for d4 through d7 configurations in octahedral geometry.
Diamagnetic
A substance with all electrons paired. It is weakly repelled by a magnetic field. Think of it as: no unpaired electrons, no attraction to magnets.
Paramagnetic
A substance with one or more unpaired electrons. It is attracted to a magnetic field. The more unpaired electrons, the stronger the paramagnetism.
Complementary colour
The colour a complex appears is the complement of the colour it absorbs. A colour wheel relates absorbed wavelength to observed colour. If a complex absorbs green light (around 520 nm), it appears red.
Cis-trans isomerism (geometric isomerism)
In octahedral or square planar complexes with two identical ligands, cis places them adjacent (90°) and trans places them opposite (180°).
Optical isomerism (enantiomers)
Non-superimposable mirror images of a complex. A complex is optically active if it lacks an internal mirror plane. In octahedral complexes, the cis isomer of [M(en)2X2] type complexes often has optical isomers, while the trans does not.
Count the number of donor atoms bonded to the metal. Each monodentate ligand contributes 1; each bidentate ligand contributes 2.
Example: Na[Cr(NH3)3(CN)3]
The complex ion is [Cr(NH3)3(CN)3]-.
Three NH3 (monodentate) + three CN- (monodentate) = 6 donor atoms.
Coordination number = 6.
Set up a charge balance: (charge on metal) + (sum of ligand charges) = (overall charge on complex ion).
Example: Na[Cr(NH3)3(CN)3]
Na+ is the counter ion, so the complex ion has charge -1.
NH3 is neutral (charge 0). CN- has charge -1.
Metal charge + 3(0) + 3(-1) = -1
Metal charge - 3 = -1
Metal charge = +2
Example: FeCl3 · 3NH3 (written incorrectly, the proper formula is [Fe(NH3)3Cl3])
Overall complex is neutral (no counter ion mentioned in the correct formula).
3 Cl- at -1 each = -3. 3 NH3 at 0 each = 0.
Metal charge + 0 + (-3) = 0
Metal charge = +3
Neutral: NH3, H2O, en (ethylenediamine), CO
Charge -1: Cl-, F-, Br-, CN-, NO2-, OH-
Charge -2: O2-, S2-
Cl- < F- < H2O < CH2CN < NH3 < en < NO2- < CN-
Ligands on the left are weak-field (small Δ, absorb low-energy / long-wavelength light). Ligands on the right are strong-field (large Δ, absorb high-energy / short-wavelength light).
When a complex absorbs a certain colour of light, it appears as the complementary colour (the colour opposite on the wheel).
Absorbs at 520 nm (green) → appears red
Absorbs at 460 nm (blue) → appears orange
A larger Δ means the complex absorbs shorter-wavelength (higher-energy) light.
Consider three Co(III) complexes with different ligands:
[Co(NH3)6]3+: six NH3 (strong-field). Large Δ, absorbs short wavelength (toward violet). Appears yellow/orange. This is the one that looks orange.
[CoCl(NH3)5]2+: five NH3 + one Cl-. Slightly weaker field than pure NH3. Intermediate Δ. This is the violet complex.
[CoCl2(NH3)4]+: four NH3 + two Cl-. Even weaker average field. Smallest Δ. Absorbs longest wavelength. This is the green complex.
Ordering violet, orange, green: II, I, III (the correct answer is c).
A complex is colourless when there is no d-d transition possible. This happens when:
The metal has a d0 configuration (no d electrons to excite)
The metal has a d10 configuration (no empty d orbitals to receive an excited electron)
Sc3+ is d0, so [Sc(H2O)6]3+ is colourless. The other complexes listed (Cr3+, Ni2+, Co2+, Au3+) all have partially filled d orbitals and are coloured.
First find the oxidation state, then write the electron configuration of the metal ion. Remove electrons from the 4s orbital first, then from 3d.
Example: Fe in [Fe(H2O)3Cl3]
Fe with Cl- and H2O ligands. Overall neutral complex: Fe charge + 3(0) + 3(-1) = 0, so Fe is +3.
Fe is [Ar] 3d6 4s2. Fe3+ loses 2 from 4s and 1 from 3d → d5.
For octahedral complexes, draw the d orbital splitting diagram (three t2g orbitals below, two eg orbitals above) and fill according to high spin or low spin.
[Fe(H2O)3Cl3] (high spin)
Fe3+ is d5. Both H2O and Cl- are weak-field ligands.
High spin d5: all five d orbitals get one electron each before any pairing.
Unpaired electrons: 5
Na4[NiCl6]
Na4 means the complex ion is [NiCl6]4-. Ni charge + 6(-1) = -4, so Ni is +2.
Ni2+ is d8. Cl- is a weak-field ligand → high spin.
d8 in octahedral: t2g has 6 electrons (3 pairs), eg has 2 electrons (1 in each).
Unpaired electrons: 2
K3[Ru(CN)6] (low spin)
K3 means [Ru(CN)6]3-. Ru charge + 6(-1) = -3, so Ru is +3.
Ru3+ is d5. CN- is a strong-field ligand → low spin.
Low spin d5: all five electrons go into t2g (3 orbitals). Two orbitals get pairs, one gets a single electron.
Unpaired electrons: 1
All electrons paired → diamagnetic (weakly repelled by a magnetic field).
Any unpaired electrons → paramagnetic (attracted to a magnetic field).
A gaseous substance with only paired electrons is diamagnetic. This is a definition question that appears frequently.
[Pt(H2O)2Cl2] is square planar. With two pairs of identical ligands, there are two geometric isomers:
Cis: both Cl- ligands adjacent (90°), both H2O adjacent.
Trans: Cl- ligands opposite (180°), H2O ligands opposite.
Total: 2 isomers (cis and trans).
[Mn(en)2(NH3)2]: this is an octahedral complex with two bidentate en ligands and two monodentate NH3 ligands.
The cis isomer (NH3 ligands on the same side) has a non-superimposable mirror image, so cis has optical isomers.
The trans isomer (NH3 ligands opposite) has a mirror plane, so it does not have optical isomers.
Answer: cis only has optical isomers.
[Fe(NH3)(CN)BrF] is tetrahedral with four different ligands. A tetrahedral complex with four different substituents is chiral (no mirror plane), so it exists as a pair of enantiomers.
Total: 2 isomers (the molecule and its enantiomer).
C5H12 has 3 structural isomers: pentane, isopentane (2-methylbutane), and neopentane (2,2-dimethylpropane). None of these are chiral.
Students often forget to remove 4s electrons before 3d electrons when finding the d electron count of a transition metal ion. Iron is [Ar]3d6 4s2, but Fe3+ is [Ar]3d5, not [Ar]3d3 4s2.
Confusing coordination number with oxidation state. Coordination number is the number of bonds to ligands. Oxidation state is the effective charge on the metal.
Thinking that all octahedral complexes with the same metal have the same number of unpaired electrons. The ligand field strength matters: [Fe(H2O)6]3+ (high spin, 5 unpaired) vs [Fe(CN)6]3- (low spin, 1 unpaired).
Forgetting that bidentate ligands like en contribute 2 to the coordination number, not 1.
⚠️ The spectrochemical series will likely be given on the exam. Memorise the relative positions; you need to apply them, not reproduce the full list from memory.
⚠️ Colour prediction using the colour wheel is very commonly tested. Know how to go from absorbed wavelength to observed colour.
⚠️ Determining unpaired electrons for high-spin vs low-spin complexes is a multi-step problem that integrates oxidation state, d electron count, and ligand field strength. Practise the full workflow.
True or false: A complex that absorbs red light appears red. (False, it appears the complementary colour, which is green.)
Fill in the blank: Ethylenediamine (en) is a ___ ligand that contributes ___ to the coordination number. (Bidentate, 2.)
True or false: A d10 metal complex is always colourless (assuming no charge transfer). (True.)
The oxidation state of Cr in [Cr(NH3)4Cl2]+ is ___. (+3.)
True or false: A paramagnetic substance has all electrons paired. (False, it has unpaired electrons.)
Q: What is the coordination number and oxidation state of Cr in Na[Cr(NH3)3(CN)3]?
A: Coordination number = 6 (three NH3 + three CN-). Oxidation state = +2 (complex charge is -1; Na+ is the counter ion; +x + 0 + 3(-1) = -1, so x = +2).
Q: A metal complex absorbs light at 520 nm. What colour does it appear?
A: Red. 520 nm is in the green region; the complementary colour is red.
Q: How many unpaired electrons does high-spin [Fe(H2O)3Cl3] have?
A: 5. Fe3+ is d5. Both H2O and Cl- are weak-field, so high spin: one electron in each of the five d orbitals.
Q: Does the cis or trans isomer of [Mn(en)2(NH3)2] have optical isomers?
A: Only the cis isomer. The cis arrangement lacks a mirror plane and produces non-superimposable mirror images. The trans isomer has a mirror plane.
Q: Which is most likely colourless: [Sc(H2O)6]3+, [Cr(NH3)5Cl]2+, [Ni(NH3)6]2+, or [Co(NH3)6]2+?
A: [Sc(H2O)6]3+. Sc3+ has a d0 configuration, so there are no d electrons to undergo d-d transitions. The others all have partially filled d orbitals and absorb visible light.
Q: A substance with only paired electrons is described as what?
A: Diamagnetic.
Q: How many unpaired electrons are in K3[Ru(CN)6] (low spin)?
A: 1. Ru3+ is d5. CN- is strong-field, so low spin: five electrons fill the three t2g orbitals (two pairs + one single).
Crystal field theory connects to molecular orbital theory (Doc 3), which provides a more complete picture of bonding in metal complexes. The magnetic properties tested here reappear when you study MO diagrams for diatomic molecules. Coordination chemistry is also the basis for understanding catalysis, bioinorganic chemistry (haemoglobin, chlorophyll), and materials science (pigments, dyes).
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