JEE Main (Engineering) · Chemistry (JEE & NEET)
Coordination Compounds
Werner theory, ligands, nomenclature, isomerism, valence bond and crystal field theory and applications of coordination compounds.
Eight concepts from Werner's two valences to crystal-field splitting and spin-only magnetism. Almost every JEE mark here is a classification — ligand denticity, isomer type, high vs low spin — with one arithmetic check attached (CFSE, μ, EAN or an oxidation state inside a name).
- JEE Main (Engineering)
- Medium level
- 8 concepts
- 5 practice questions
1Werner's primary and secondary valence
Werner separated two jobs a metal ion does in a complex. The primary valence is the oxidation state: it is satisfied by anions that neutralise the metal charge — some ionisable outside the brackets, others already bound inside the coordination sphere. The secondary valence is the coordination number: it is satisfied by ligands bound directly to the metal and fixes their spatial arrangement.
The classic CoCl₃·nNH₃ series makes the split measurable. Six ammine ligands give [Co(NH₃)₆]Cl₃ with three free Cl⁻; replace one NH₃ by Cl inside the sphere and only two Cl⁻ remain ionisable. Conductivity and AgNO₃ precipitation count the outer ions; they never count ligands already inside.
Figure. Ionisable Cl⁻ falling as NH₃ is replaced by Cl inside the coordination sphere. Bar length is the outer-ion count AgNO₃ actually sees; primary valence stays +3 for every member while secondary valence stays 6. Lengths follow the integers by construction.
How it works
- Write the spherePut every ligand that stays bound in square brackets with the metal; leave only ionisable counter-ions outside.
- Read primary valenceOxidation state of the metal equals the charge needed once ligand charges and the complex charge are set. In the CoCl₃·nNH₃ series that OS stays +3 even while the ionisable outer Cl⁻ count falls from 3 to 0 — the missing anions have moved inside the sphere.
- Read secondary valenceCoordination number is the number of ligand donor atoms attached to the metal (six in every member of the CoCl₃·nNH₃ series).
Ionisable chlorides in CoCl₃·5NH₃
CoCl₃·5NH₃ behaves as [Co(NH₃)₅Cl]Cl₂ in water. How many moles of AgCl precipitate per mole of complex with excess AgNO₃, and what are the primary and secondary valences of cobalt?
- Species written[Co(NH₃)₅Cl]Cl₂
- Outer Cl⁻ (ionisable)2 → 2 mol AgCl
- Primary valence (OS of Co)+3
- Secondary valence (CN)6 (5 NH₃ + 1 Cl)
Pro tip. AgNO₃ only sees counter-ions outside the brackets. The Cl already inside [Co(NH₃)₅Cl]²⁺ does not precipitate, which is the whole experimental point of Werner's split.
In [Co(NH₃)₆]Cl₃, the secondary valence of cobalt is
- 3, equal to the number of ionisable Cl⁻
- 6, equal to the number of NH₃ ligands bound to Co
- 9, the sum of ligands and outer ions
Secondary valence is the coordination number — six ammine ligands in the sphere. The three Cl⁻ are the primary valence (oxidation state +3) expressed as outer ions.
2Ligands, denticity and the chelate effect
A ligand is classified by how many donor atoms it uses on one metal. Monodentate ligands (Cl⁻, NH₃, H₂O, CN⁻) donate through one atom. Bidentate ligands such as ethane-1,2-diamine (en) and oxalate (ox²⁻) bite with two. EDTA⁴⁻ is hexadentate — six donors wrapping one metal.
When a polydentate ligand closes a ring with the metal, the complex is a chelate. Five- and six-membered chelate rings are the most stable. Replacing two monodentate ligands by one bidentate ligand raises the formation constant — the chelate effect — mainly because the entropy of freeing two separate ligands beats freeing one linked pair.
Figure. Denticity as bar length: one donor, two donors, six donors. Lengths are the integer donor counts, not formation constants — the chelate effect (why en outbinds two NH₃) is the entropy argument in the prose, not a second quantity on this chart.
How it works
- Count donor atomsDenticity is the number of metal–ligand bonds one ligand molecule forms, not the number of atoms in the ligand.
- Add to coordination numberCN is the total donor-atom count around the metal: three en ligands give CN = 6.
- Prefer the chelateAt equal donor count, the polydentate ligand usually wins on K_f — especially with five- or six-membered rings.
| Ligand | Donors | Denticity | Notes |
|---|---|---|---|
| Cl⁻, NH₃, H₂O, CN⁻, CO | 1 | Monodentate | CN⁻ and CO are strong-field |
| en, ox²⁻ | 2 | Bidentate | Forms five-membered chelate rings |
| dien | 3 | Tridentate | Diethylenetriamine |
| EDTA⁴⁻ | 6 | Hexadentate | Fills an octahedral sphere alone |
EDTA⁴⁻ bound to a single metal centre is best described as
- Monodentate, because it is one ion
- Bidentate, like en
- Hexadentate — six donor atoms on one metal
EDTA⁴⁻ offers two N and four carboxylate O donors. One EDTA fills an octahedral coordination sphere (CN = 6), which is why it is the textbook hexadentate ligand.
3IUPAC naming of coordination compounds
Name the ligands first, in alphabetical order of their ligand names, then the metal. Neutral ligands keep ordinary names (ammine for NH₃, aqua for H₂O, carbonyl for CO); anionic ligands end in -o (chlorido, cyanido, oxalato). Multiplicity prefixes are di-, tri-, tetra- — or bis-, tris- when the ligand name already contains a prefix (bis(ethane-1,2-diamine)).
An anionic complex takes the metal stem with -ate (ferrate, cuprate, argentate). The oxidation state of the metal is written in Roman numerals in parentheses at the end. Cations outside the complex are named first as ordinary salts.
Figure. Name left to right: ligands alphabetical (prefixes ignored for order), metal stem with -ate if the complex is an anion, then the oxidation-state Roman numeral from the charge balance — as in potassium hexacyanidoferrate(II).
How it works
- Split the saltName outer cations (or anions) first; the complex comes next.
- List ligands A–ZAlphabetical by ligand name, ignoring multiplicative prefixes. Then the metal.
- Deposit the OSCompute the metal oxidation state from the complex charge and ligand charges; write it as (II), (III), … For an anion, use the -ate metal name.
Naming K₄[Fe(CN)₆]
Give the IUPAC name of K₄[Fe(CN)₆].
- Outer cation + complex anionK⁺ + [Fe(CN)₆]⁴⁻
- Ligandssix cyanido → hexacyanido
- Fe OS: x + 6(−1) = −4x = +2
- Anionic complex → ferrate(II)potassium hexacyanidoferrate(II)
Pro tip. Anionic complexes always take the Latin metal stem with -ate: ferrate, cuprate, argentate, plumbate. 'Ironate' is never the name.
The IUPAC name of [Co(NH₃)₄Cl₂]Cl is
- cobalt tetraammine dichloride chloride
- tetraamminedichloridocobalt(III) chloride
- tetraamminedichloridocobalt(II) chloride
Ligands alphabetical: ammine before chlorido → tetraamminedichlorido, then cobalt. Charge of complex is +1 with two Cl⁻ and four NH₃, so Co is +3 → (III). Outer ion is chloride.
4Structural isomerism in complexes
Structural isomers share a formula but differ in which atoms are bonded to which. Four JEE types recur. Ionisation isomers exchange a ligand with a counter-ion ([Co(NH₃)₅Br]SO₄ vs [Co(NH₃)₅SO₄]Br). Linkage isomers differ at the donor atom of an ambidentate ligand (M–NO₂ vs M–ONO; M–SCN vs M–NCS).
Coordination isomers swap ligands between the two metals of a double-salt complex ([Co(NH₃)₆][Cr(CN)₆] vs [Cr(NH₃)₆][Co(CN)₆]). Hydrate (solvate) isomers exchange water between the sphere and the lattice ([Cr(H₂O)₆]Cl₃ vs [Cr(H₂O)₅Cl]Cl₂·H₂O).
Figure. Four structural classes share a formula but differ in connectivity or which ions sit inside the coordination sphere. Linkage isomerism (M–NO₂ vs M–ONO) is the bond-level case; the others swap ligands with counter-ions or solvent.
Picking the type
- Same sphere, different outer ion?Ionisation isomerism — the ligand and the counter-ion have swapped roles.
- Same ligand, different donor atom?Linkage isomerism — NO₂⁻/ONO⁻ or SCN⁻/NCS⁻.
- Two metals swapping ligands, or water moving in/out?Coordination isomerism for the first; hydrate isomerism for the second.
| Type | What changes | Stock example |
|---|---|---|
| Ionisation | Ligand ↔ counter-ion | [Co(NH₃)₅Br]SO₄ / [Co(NH₃)₅SO₄]Br |
| Linkage | Donor atom of ambidentate ligand | [Co(NH₃)₅NO₂]²⁺ / [Co(NH₃)₅ONO]²⁺ |
| Coordination | Ligand distribution between two metals | [Co(NH₃)₆][Cr(CN)₆] / swap |
| Hydrate | H₂O in sphere vs lattice | [Cr(H₂O)₆]Cl₃ / [Cr(H₂O)₅Cl]Cl₂·H₂O |
[Co(NH₃)₅SCN]Cl₂ and [Co(NH₃)₅NCS]Cl₂ are
- Ionisation isomers
- Linkage isomers
- Coordination isomers
SCN⁻ is ambidentate: S-bound thiocyanato versus N-bound isothiocyanato. The outer ions and the rest of the sphere are unchanged, so the pair is linkage isomerism.
5Geometrical and optical isomerism
Stereoisomers keep the same bonds and rearrange them in space. Geometrical (cis/trans) isomerism appears in square-planar Ma₂b₂ and octahedral Ma₄b₂ or Ma₃b₃: cis has identical ligands adjacent; trans has them opposite. Octahedral [Co(NH₃)₄Cl₂]⁺ is the textbook cis (violet) / trans (green) pair.
Optical isomerism needs a non-superimposable mirror image. Octahedral [Co(en)₃]³⁺ and cis-[Co(en)₂Cl₂]⁺ are chiral and resolve into Δ and Λ enantiomers; the trans-[Co(en)₂Cl₂]⁺ isomer has a mirror plane and is optically inactive.
Figure. Geometrical isomerism for [Ma₄b₂] is adjacent (cis) versus opposite (trans) identical ligands. Optical isomerism needs a non-superposable mirror pair — [Co(en)₃]³⁺ is the classic propeller. Flat boxes name the relation; they are not octahedra.
How it works
- Ask geometrical firstFor Ma₄b₂ octahedral or Ma₂b₂ square planar, count whether the two b ligands share an edge (cis) or sit at 180° (trans).
- Then ask opticalIf the complex has no improper axis (no mirror plane / inversion), it has enantiomers. Tris(chelate) and cis-bis(chelate) octahedral complexes usually do.
- Kill optical with a planetrans-bis(chelate) octahedral complexes are typically meso — geometrical isomers that are not optical.
Which statement about [Co(en)₃]³⁺ is correct?
- It shows cis–trans isomerism but is optically inactive
- It is optically active (Δ and Λ) with no cis–trans pair
- It is square planar, so only cis–trans applies
Three bidentate en ligands give a tris-chelate octahedron. There is no cis/trans pair of that formula; the two stereoisomers are mirror-image Δ and Λ enantiomers.
6Octahedral crystal-field splitting
Crystal Field Theory treats ligands as point charges that split the metal d-orbitals. In an octahedral field the d set divides into a lower t_{2g} trio (d_{xy}, d_{xz}, d_{yz}) and an upper e_g pair (d_{x^2-y^2}, d_{z^2}), separated by the octahedral splitting \Delta_o (or 10\,Dq).
Relative to the unsplit barycentre, each t_{2g} electron sits at -0.4\,\Delta_o and each e_g electron at +0.6\,\Delta_o. Tetrahedral fields invert the order and shrink the gap: \Delta_t = \frac{4}{9}\Delta_o, which is why tetrahedral complexes almost never pay the pairing energy.
Figure. Octahedral CFT ladder drawn to the −0.4 / +0.6 barycentre rule. The dashed free-ion level sits at the weighted mean; e_g is three-fifths of \Delta_o above it and t_{2g} is two-fifths below, so the bracket from t_{2g} to e_g is exactly \Delta_o. Orbital shapes are not drawn.
How it works
- Place the barycentreThe weighted average of the split levels stays at the free-ion d energy: 3(-0.4)+2(+0.6)=0.
- Read the gape_g lies \Delta_o above t_{2g}. Strong-field ligands enlarge \Delta_o; weak-field ligands shrink it.
- Compare tetrahedral\Delta_t\approx 4/9\,\Delta_o and the e set lies below t_2. Pairing almost never wins.
Tetrahedral splitting from Δₒ
If a metal–ligand pair gives \Delta_o = 18000\,\mathrm{cm^{-1}} in an octahedral complex, estimate \Delta_t for the tetrahedral complex of the same pair.
- RelationΔ_t = (4/9) Δ_o
- Δ_t = (4/9) × 180008000 cm⁻¹
- Compare to typical pairing energy (~15000 cm⁻¹)Δ_t ≪ P
- Consequence for spintetrahedral → high spin
Pro tip. The 4/9 rule is why textbooks say tetrahedral complexes are always high-spin: \Delta_t almost never exceeds the pairing energy. Do not apply the octahedral spectrochemical cut-offs to a tetrahedral complex without rescaling.
In an octahedral field, the energy of an e_g orbital relative to the barycentre is
- −0.4 Δₒ
- +0.6 Δₒ
- +0.4 Δₒ
Octahedral CFT puts e_g at +0.6\,\Delta_o and t_{2g} at -0.4\,\Delta_o. The weights 2\times 0.6 + 3\times(-0.4) cancel, preserving the barycentre.
7Spectrochemical series, high/low spin and CFSE
Whether a d^4–d^7 octahedral ion goes high-spin or low-spin is a fight between \Delta_o and the pairing energy P. The spectrochemical series orders ligands by the \Delta_o they produce: I^- < Br^- < Cl^- < F^- < H_2O < NH_3 < en < CN^- < CO. Strong-field ligands (CN^-, CO, en) make \Delta_o > P and force pairing (low spin); weak-field ligands (halides, often H_2O) leave high spin.
Crystal-field stabilisation energy for an octahedral configuration is \mathrm{CFSE}=(-0.4\,n_{t_{2g}}+0.6\,n_{e_g})\Delta_o, quoted before pairing-energy corrections unless the question asks for net energy. Valence-bond language still appears in older stems: strong-field d^2sp^3 (inner orbital) versus weak-field sp^3d^2 (outer orbital) — the same pairing decision under another name.
Figure. Unpaired electrons on the two Fe(III) complexes from the worked example. Bar length is the unpaired count (5 vs 1), which is what enters \mu=\sqrt{n(n+2)} — not CFSE and not a colour scale. Both are d^5; only the ligand field changed.
How it works
- Find dⁿ and the ligandOxidation state → d count; place the ligand on the spectrochemical series.
- Decide pairingStrong field → fill t_{2g} first (low spin). Weak field → put one electron in each of the five d orbitals before pairing (high spin).
- Compute CFSE and μCFSE from the -0.4/+0.6 weights; spin-only \mu=\sqrt{n(n+2)} from the unpaired count.
Comparing [Fe(CN)₆]³⁻ and [FeF₆]³⁻
Predict hybridisation, spin and magnetic moment of [Fe(CN)_6]^{3-} and [FeF_6]^{3-}.
- Both: Fe is +3 → 3d⁵d⁵
- CN⁻ strong-field → t₂g⁵ e_g⁰ (1 unpaired)d²sp³, low spin, μ = √3 = 1.73 BM
- F⁻ weak-field → t₂g³ e_g² (5 unpaired)sp³d², high spin, μ = √35 = 5.92 BM
- CFSE: LS −2.0 Δₒ; HS (−0.4×3+0.6×2)Δₒ0 (HS) vs −2.0 Δₒ (LS)
Pro tip. Same metal ion, opposite spin: the ligand's place in the spectrochemical series decides pairing. Quoting hybridisation (d^2sp^3 vs sp^3d^2) without the unpaired count loses the magnetism mark.
For octahedral d^6, which ligand set is most likely to give a low-spin complex?
- 6 F⁻
- 6 H₂O
- 6 CN⁻
CN⁻ sits at the strong-field end of the spectrochemical series, so \Delta_o > P and the six electrons pair into t_{2g}^6. F⁻ and usually H₂O leave high-spin t_{2g}^4 e_g^2.
8Spin-only magnetic moment and EAN
The spin-only magnetic moment of a complex is \mu = \sqrt{n(n+2)} Bohr magneton, where n is the number of unpaired electrons. It is the quantity JEE asks when a stem says "magnetic moment" and gives no orbital contribution. Reading n still needs the high/low-spin decision from the spectrochemical / CFSE concept.
The effective atomic number (EAN) rule is Sidgwick's electron count: \mathrm{EAN} = Z - (\text{oxidation state}) + 2\times(\text{coordination number}). Many stable complexes land on 36, 54 or 86 — the next noble-gas count — but the rule is a guide, not a law: known exceptions exist, and CFSE or the 18-electron count usually explains stability more cleanly.
Figure. Spin-only \mu=\sqrt{n(n+2)} for n=0–5 unpaired electrons. Bar height is the moment in BM — \sqrt{1\cdot3}=1.73, \sqrt{2\cdot4}=2.83, \sqrt{3\cdot5}=3.87, \sqrt{4\cdot6}=4.90, \sqrt{5\cdot7}=5.92. EAN is the electron-count rule in the ledger, not a second series on this chart.
How it works
- Get n for μOxidation state → dⁿ → high or low spin → count unpaired electrons → \sqrt{n(n+2)}.
- Get EANSubtract the metal oxidation state from the atomic number, then add two electrons per coordinate bond (each ligand donor atom).
- Compare to noble gas36 (Kr), 54 (Xe), 86 (Rn) are the common targets; treat a miss as a prompt to check OS and CN, not as an automatic instability verdict.
EAN of [Fe(CN)₆]⁴⁻
Calculate the EAN of iron in [Fe(CN)_6]^{4-} and state the noble-gas count it matches.
- Z(Fe) = 26; complex charge −4 with 6 CN⁻OS of Fe = +2
- Coordination number6
- EAN = 26 − 2 + 2×636
- Noble-gas matchKr (Z = 36)
Pro tip. EAN counts electrons the metal "owns" after bonding, not the dⁿ of CFT. For the same [Fe(CN)_6]^{4-} ion, CFT says low-spin d^6 (t_{2g}^6) while EAN reports 36 — both can be right because they answer different questions.
A complex has two unpaired electrons. Its spin-only magnetic moment is closest to
- 1.73 BM
- 2.83 BM
- 3.87 BM
\mu=\sqrt{2\times 4}=\sqrt{8}=2.83 BM. 1.73 BM is one unpaired electron; 3.87 BM is three.
Notes
- Werner's theory: a metal shows primary valence (its oxidation state, ionisable) and secondary valence (its coordination number, fixing the spatial arrangement of ligands).
- Ligands are classified by donor atoms: monodentate (Cl^-, NH_3), bidentate (en, oxalate), and polydentate (EDTA^{4-} is hexadentate); chelate complexes are extra stable.
- IUPAC naming: ligands are named alphabetically before the metal; an anionic complex takes the suffix -ate, and the metal oxidation state is given in Roman numerals.
- Isomerism: structural (ionisation, linkage, coordination, hydrate) and stereoisomerism (geometrical cis/trans and optical), e.g. [Co(en)_3]^{3+} is optically active.
- Crystal Field Theory: in an octahedral field d-orbitals split into t_{2g} and e_g separated by \Delta_o; strong-field ligands cause pairing (low spin), weak-field give high spin.
Formulas
- EAN =Z-(\text{oxidation state})+2\times(\text{coordination number})
- CFSE (octahedral) =(-0.4\,n_{t_{2g}}+0.6\,n_{e_g})\Delta_o
- Spin-only moment \mu=\sqrt{n(n+2)} BM
- \Delta_t=\frac{4}{9}\Delta_o (tetrahedral splitting is smaller)
- Spectrochemical series: I^-<Br^-<Cl^-<F^-<H_2O<NH_3<en<CN^-<CO
Exam traps & shortcuts
- Strong-field ligands (CN⁻, CO, en) give low-spin inner-orbital complexes (d^2sp^3); weak-field ligands (H₂O, halides) give high-spin outer-orbital (sp^3d^2).
- Tetrahedral complexes are always high spin because \Delta_t is too small to force pairing.
- The chelate effect: five- and six-membered rings give maximum complex stability.
Reference tables
Five relations this topic actually computes. Isomer type and ligand denticity stay as lookups in their concepts.
| Quantity | Relation | Watch |
|---|---|---|
| EAN | Z − OS + 2×CN | 36 / 54 / 86 are common targets, not laws |
| CFSE (octahedral) | (−0.4 n_t₂g + 0.6 n_eg) Δₒ | Quote pairing energy only if asked |
| Spin-only μ | √[n(n+2)] BM | n = unpaired electrons after spin choice |
| Tetrahedral gap | Δ_t = (4/9) Δₒ | Tetrahedral ≈ always high spin |
| Spectrochemical | I⁻ < … < H₂O < NH₃ < en < CN⁻ < CO | Strong field → low spin (octahedral d⁴–d⁷) |
Ligands ordered by increasing Δₒ. The cut between high and low spin also depends on the metal and on P; the series only ranks the ligand contribution.
| Ligand | Field | Typical spin (oct. Fe³⁺ / Co³⁺) |
|---|---|---|
| I⁻, Br⁻, Cl⁻, F⁻ | Weak | High spin |
| H₂O | Weak–intermediate | Usually high spin for Fe³⁺ |
| NH₃, en | Intermediate–strong | en often low spin for Co³⁺ |
| CN⁻, CO | Strong | Low spin |
Recap
Read only this the night before.
- Werner
- Primary = OS (anions inside or outside may satisfy it). Secondary = CN (ligands in the brackets). AgNO₃ counts only the outer ions — which need not equal the OS.
- Denticity
- Mono (NH₃, Cl⁻), bi (en, ox), hexa (EDTA⁴⁻). Chelates with 5–6 membered rings win on K_f.
- Naming
- Ligands A–Z, then metal; anionic complex → -ate; OS in Roman numerals. K₄[Fe(CN)₆] = potassium hexacyanidoferrate(II).
- Structural isomers
- Ionisation / linkage / coordination / hydrate. Ambidentate NO₂⁻ and SCN⁻ → linkage.
- Stereo
- cis = adjacent, trans = opposite. [Co(en)₃]³⁺ is optical (Δ/Λ); trans-[Co(en)₂Cl₂]⁺ is not.
- CFT
- Octahedral: t₂g at −0.4 Δₒ, e_g at +0.6 Δₒ. Δ_t = (4/9) Δₒ → tetrahedral almost always high spin.
- Spin
- Spectrochemical: CN⁻, CO strong → low spin; halides weak → high spin. Same Fe³⁺: [Fe(CN)₆]³⁻ μ=1.73, [FeF₆]³⁻ μ=5.92.
- μ and EAN
- μ = √[n(n+2)] BM. EAN = Z − OS + 2×CN; [Fe(CN)₆]⁴⁻ → 36.
Practise Coordination Compounds
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