Why this chapter is so predictable
Give a student the formula [Fe(CN)₆]³⁻ and the questions that follow are almost fixed: what is the oxidation state of iron, how many unpaired electrons does it have, what is the magnetic moment, is it inner or outer orbital, and what is its hybridisation. Every one of those follows from the same two inputs - the metal's d-electron count and whether the ligand is strong or weak field.
That regularity makes the chapter unusually efficient to prepare. The memorisation is small - the spectrochemical series and a handful of ligand names - and the reasoning is repetitive, so practice converts directly into speed.
How JEE actually asks Coordination Compounds
NTA does not publish chapter-wise weightage, so figures online are coaching estimates from past papers; check the official NTA information bulletin for your session's syllabus and pattern.
The recurring types are: determine the oxidation state and coordination number from a formula, count the geometrical or optical isomers, decide high or low spin from the ligand and compute the magnetic moment, and name a complex or deduce a formula from its name. Metal carbonyls and biological coordination compounds supply short recall questions.
Key concepts, compressed
- ✓Werner distinguished primary valency, satisfied by anions outside the coordination sphere and ionisable, from secondary valency, satisfied by ligands inside it and non-ionisable.
- ✓A ligand donates at least one electron pair; the number of donor atoms it provides makes it mono-, bi- or polydentate.
- ✓Crystal field theory treats the ligands as point charges that split the metal's d orbitals into two sets of different energy.
- ✓Whether electrons pair up depends on the competition between the splitting energy and the pairing energy.
- ✓Strong-field ligands produce large splitting and low-spin complexes; weak-field ligands produce small splitting and high-spin complexes.
- ✓Colour and magnetism both follow directly from the resulting d-electron arrangement.
Results you need before attempting the questions
| Item | Result | Note |
|---|---|---|
| Spin-only moment | μ = √(n(n + 2)) BM | n unpaired electrons |
| μ for n = 1, 2, 3 | 1.73, 2.83, 3.87 BM | |
| μ for n = 4, 5 | 4.90, 5.92 BM | |
| Octahedral splitting | t₂g lower, e_g higher | 3 and 2 orbitals |
| Tetrahedral splitting | Δt ≈ (4/9)Δo | always high spin |
| CFSE (octahedral) | (−0.4x + 0.6y)Δo | for t₂g^x e_g^y |
| Strong-field ligands | CN⁻, CO, NO₂⁻, en, NH₃ | large Δ, low spin |
| Weak-field ligands | I⁻, Br⁻, Cl⁻, F⁻, H₂O | small Δ, high spin |
| EAN | Z − oxidation state + 2 × coordination number | |
| [Ma₄b₂] octahedral | 2 geometrical isomers | cis and trans |
| Colour | d-d transition | d⁰ and d¹⁰ colourless |
The five mistakes that cost the most marks
- ✓Ignoring ligand charges when computing the metal's oxidation state. Ammonia and water contribute zero; chloride, cyanide and hydroxide each contribute −1.
- ✓Predicting a low-spin tetrahedral complex. Δt is too small to force pairing, so tetrahedral complexes are effectively always high spin regardless of the ligand.
- ✓Equating coordination number with the number of ligand molecules. Three bidentate ligands give a coordination number of six, not three.
- ✓Forgetting that the spin-only formula ignores orbital contribution. It is an approximation, though a good one for first-row transition metals and the one JEE expects.
- ✓Confusing ionisation isomerism with linkage isomerism. The first exchanges a ligand with the counter-ion; the second uses a different donor atom of the same ligand.
Practice set 1: fundamentals
1. What did Werner's theory distinguish?
Primary valency, satisfied by anions outside the coordination sphere and therefore ionisable, from secondary valency, satisfied by ligands directly bonded to the metal and not ionisable. In [Co(NH₃)₆]Cl₃ the three chlorides are primary and precipitate with silver nitrate, while the six ammonia ligands are secondary and do not.
2. What are monodentate, bidentate and polydentate ligands?
A monodentate ligand donates one electron pair through one donor atom, such as ammonia or chloride. A bidentate ligand donates through two, such as ethylenediamine or oxalate. Polydentate ligands donate through several - EDTA is hexadentate. The count of donor atoms, not molecules, is what matters.
3. What is the coordination number of a complex?
The number of donor atoms directly bonded to the central metal, not the number of ligand molecules. So [Co(en)₃]³⁺ has coordination number 6, because each of the three ethylenediamine ligands occupies two sites. Miscounting this invalidates the geometry and everything that follows.
4. What is the oxidation state of cobalt in [Co(NH₃)₆]Cl₃?
+3. The three chlorides outside the sphere carry −3 in total, so the complex ion must be +3. Ammonia is neutral and contributes nothing, so cobalt itself is +3. Always account for ligand charges first - this single step underpins the whole question.
5. What is the chelate effect?
The extra stability of a complex containing a polydentate ligand compared with one containing an equivalent number of monodentate ligands. It is largely an entropy effect: replacing six water molecules with three bidentate ligands increases the number of free particles, which favours the chelated product.
6. What is an ambidentate ligand?
One that can bind through either of two different donor atoms - the nitrite ion through nitrogen or oxygen, the thiocyanate ion through sulphur or nitrogen. This is what makes linkage isomerism possible, and it is the distinguishing feature that questions test.
7. What are the main IUPAC rules for naming a complex?
Name the ligands alphabetically before the metal, with prefixes di, tri and so on for simple ligands and bis, tris for those already containing a numerical prefix. Anionic ligands end in -o. The metal's oxidation state follows in Roman numerals in brackets. If the complex ion is anionic, the metal name takes the -ate suffix.
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Practice set 2: isomerism
8. What types of structural isomerism occur in coordination compounds?
Ionisation, hydrate, linkage and coordination isomerism. Ionisation isomers exchange a ligand with the counter-ion; hydrate isomers differ in how many water molecules are coordinated rather than free; linkage isomers use different donor atoms of an ambidentate ligand; coordination isomers redistribute ligands between two complex ions.
9. Give an example of ionisation isomerism.
[Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br. The bromide and sulphate swap places between the coordination sphere and the counter-ion position. They are distinguished experimentally by precipitation tests - the first gives a precipitate with barium chloride, the second with silver nitrate.
10. Give an example of linkage isomerism.
The nitrite ion bound through nitrogen (nitrito-N, written as NO₂) or through oxygen (nitrito-O, written as ONO). Both isomers have identical formulas and differ only in which atom bonds to the metal. Linkage isomerism requires an ambidentate ligand, which is why it is far less common than the other types.
11. When does geometrical isomerism occur in a complex?
When ligands can occupy different relative positions around the metal. Square planar [Ma₂b₂] complexes give cis and trans forms, and octahedral [Ma₄b₂] complexes do the same. Tetrahedral complexes show no geometrical isomerism, because all four positions are equivalent - a point questions test directly.
12. When does optical isomerism occur?
When the complex is non-superimposable on its mirror image, which typically requires chelating ligands. Octahedral complexes of the type [M(en)₃] are the standard example. The presence of a plane of symmetry rules it out, so trans isomers are usually optically inactive while cis isomers may not be.
13. How many geometrical isomers does an octahedral [Ma₄b₂] complex have?
Two - cis, where the two b ligands are adjacent at 90°, and trans, where they are opposite at 180°. Counting isomers is a standard question type, and the reliable method is to fix one ligand's position and enumerate the distinct arrangements of the rest rather than guessing.
Practice set 3: bonding and crystal field theory
14. What distinguishes inner and outer orbital complexes in valence bond theory?
An inner orbital complex uses (n−1)d orbitals in hybridisation, giving d²sp³ and a low-spin arrangement. An outer orbital complex uses nd orbitals, giving sp³d² and a high-spin arrangement. Strong-field ligands force pairing and produce inner orbital complexes; weak-field ligands leave electrons unpaired and give outer orbital ones.
15. How do d orbitals split in an octahedral field?
Into a lower-energy t₂g set of three orbitals and a higher-energy e_g set of two. The e_g orbitals point directly at the approaching ligands and are therefore destabilised more by the electrostatic repulsion. The energy gap between the two sets is Δo, the crystal field splitting energy.
16. What determines whether a complex is high or low spin?
The competition between the splitting energy Δ and the pairing energy P. If Δ exceeds P, electrons pair up in the lower set rather than occupy the higher one, giving a low-spin complex. If Δ is smaller than P, they spread out and the complex is high spin. The ligand determines Δ, which is why the ligand determines the spin state.
17. What are strong-field and weak-field ligands?
Strong-field ligands such as cyanide, carbon monoxide and ethylenediamine produce large splitting and favour low-spin complexes. Weak-field ligands such as the halides and water produce small splitting and give high-spin complexes. The distinction is not about charge - the neutral CO is far stronger field than the anionic fluoride.
18. State the spectrochemical series in outline.
Roughly I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO, from weakest to strongest field. The order is experimental rather than derived from simple electrostatics, which is one of crystal field theory's acknowledged limitations - a point-charge model cannot explain why neutral CO outranks anionic fluoride.
19. What is the difference between high-spin and low-spin arrangements?
A high-spin complex has the maximum number of unpaired electrons, with electrons occupying the higher orbitals rather than pairing. A low-spin complex has the minimum, with pairing occurring in the lower set first. The distinction only arises for d⁴ to d⁷ configurations - other counts give the same arrangement either way.
20. Why are tetrahedral complexes almost always high spin?
Because the tetrahedral splitting Δt is only about four-ninths of Δo for the same metal and ligands. That is smaller than typical pairing energies, so electrons occupy the higher set rather than pairing, regardless of where the ligand sits in the spectrochemical series. Predicting a low-spin tetrahedral complex is essentially always wrong.
21. How is crystal field stabilisation energy calculated for an octahedral complex?
CFSE = (−0.4x + 0.6y)Δo for a t₂g^x e_g^y configuration, since each t₂g electron is stabilised by 0.4Δo and each e_g electron destabilised by 0.6Δo. For a low-spin d⁶ complex with all six in t₂g, that gives −2.4Δo - the largest possible value, which is why such complexes are exceptionally stable.
22. Why are most coordination compounds coloured?
Because the crystal field splitting is typically the energy of visible light, so an electron can absorb a photon and jump from the lower set to the higher one. The observed colour is the complement of the absorbed wavelength. Complexes with d⁰ or d¹⁰ configurations have no such transition available and are usually colourless.
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Practice set 4: magnetic behaviour
23. How is the spin-only magnetic moment calculated?
μ = √(n(n + 2)) Bohr magnetons, where n is the number of unpaired electrons. It ignores any orbital contribution, which is a good approximation for first-row transition metals and is what JEE expects. The formula works in both directions - a measured moment can be used to deduce the unpaired count.
24. What are the moments for one to five unpaired electrons?
1.73, 2.83, 3.87, 4.90 and 5.92 Bohr magnetons respectively. These five values are worth memorising, because questions frequently supply a measured moment and expect you to identify the number of unpaired electrons - which then fixes the spin state and the ligand field strength.
25. What is the magnetic moment of [Fe(CN)₆]³⁻?
1.73 BM. Iron is +3, giving a d⁵ configuration. Cyanide is a strong-field ligand, so the complex is low spin with all five electrons in t₂g - four paired and one unpaired. With n = 1 the moment is √3 = 1.73 BM. The complex is therefore weakly paramagnetic despite iron having five d electrons.
26. What is the magnetic moment of [FeF₆]³⁻?
5.92 BM. Iron is again +3 with a d⁵ configuration, but fluoride is a weak-field ligand, so the complex is high spin with all five electrons unpaired across t₂g and e_g. With n = 5 the moment is √35 = 5.92 BM. Comparing with question 25 shows the ligand alone changes the moment from 1.73 to 5.92 with the same metal ion.
Practice set 5: applications
27. What is the effective atomic number rule?
EAN = Z − oxidation state + 2 × coordination number, and complexes whose EAN equals the atomic number of the next noble gas tend to be especially stable. It works well for metal carbonyls but has many exceptions elsewhere, so it is a guideline rather than a law - which is itself an examinable point.
28. How does bonding work in metal carbonyls?
By synergic bonding. Carbon monoxide donates a lone pair to the metal in a sigma bond, and the metal donates electron density back from its filled d orbitals into CO's empty pi antibonding orbitals. The two effects reinforce each other, which makes the bond unusually strong and lengthens the C-O bond measurably.
29. Give three biologically important coordination compounds.
Haemoglobin contains iron in a porphyrin ring and transports oxygen; chlorophyll contains magnesium in a similar ring and drives photosynthesis; vitamin B12 contains cobalt in a corrin ring. The recurring pattern of a metal held in a large chelating ring is worth noting, along with which metal belongs to which.
30. What does a stability constant measure?
The equilibrium constant for the formation of a complex from the metal ion and its ligands. A larger value means a more stable complex. Stability generally increases with higher metal charge, smaller ionic radius, and chelating ligands - which is the quantitative expression of the chelate effect from question 5.
How to study this chapter efficiently
- ✓Always compute the oxidation state first, accounting for every ligand charge. Everything else depends on it.
- ✓Memorise the spectrochemical series in outline. It is short, and it decides spin state, magnetic moment and colour.
- ✓Learn the five spin-only magnetic moment values so you can work in either direction.
- ✓Note that tetrahedral means high spin, without exception in practice.
- ✓For isomer counting, fix one ligand and enumerate the rest systematically rather than guessing.
- ✓Practise the standard chain: formula to oxidation state to d-count to spin state to unpaired electrons to magnetic moment. That single chain answers a large share of the chapter.
Turn this into active practice
Because the question types repeat so closely, this chapter rewards drilling the standard chain until it runs without conscious effort. A student who can go from formula to magnetic moment in twenty seconds has effectively mastered the chapter.
The JEE Coordination Compounds quiz on QUFF generates fresh questions across nomenclature, oxidation states, isomerism, crystal field theory and magnetic behaviour, marks them instantly and explains each answer. Do timed sets and check where in the chain your errors occur - it is usually the oxidation state or the spin state, and both are quick to fix.
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