Latest
  • Admissions openClass 11 Science, 2027‑28: JEE, NEET and MHT‑CET with junior college and hostel under one roofApply now
  • IMGSAT 2027Free scholarship and admission test for Class 10 students, every Saturday and Sunday at our Nashik campusRegister
  • Free Foundation 2026‑27Evening classes for Class 10 in Physics, Chemistry, Maths and Biology, taught by our IITian and doctor facultyJoin free

+91 70303 00666

Chemistry · Class 12 · Chapter 5

Coordination Compounds

Haemoglobin, chlorophyll, vitamin B12 and the anticancer drug cisplatin are all coordination compounds. The chapter asks you to name them, count their isomers and explain their colour and magnetism, and each of those skills follows a fixed routine that you can practise.

In this chapter: Werner's theory, double salts and complexes, key terms (ligand, denticity, chelate, coordination number, oxidation number), IUPAC nomenclature, structural and stereoisomerism, valence bond theory, crystal field theory in octahedral and tetrahedral fields, colour, bonding in metal carbonyls, and uses of coordination compounds.

Werner's theory

Werner prepared a series of cobalt(III) chloride-ammonia compounds and counted how many chloride ions each gave as AgCl with silver nitrate in the cold:

CompoundMol AgCl per molFormula
CoCl3·6NH3 (yellow)3[Co(NH3)6]Cl3
CoCl3·5NH3 (purple)2[CoCl(NH3)5]Cl2
CoCl3·4NH3 (green and violet forms)1[CoCl2(NH3)4]Cl (trans is green, cis is violet)
  • A metal shows two kinds of valence. Primary valence is ionisable and corresponds to the oxidation state; it is satisfied by negative ions. Secondary valence is non-ionisable and equals the coordination number; it is satisfied by neutral molecules or negative ions.
  • Groups satisfying the secondary valence are arranged in definite directions in space, giving geometries such as octahedral, square planar and tetrahedral.

A double salt such as Mohr's salt, FeSO4·(NH4)2SO4·6H2O, or potash alum dissociates completely into simple ions in water. A complex such as K4[Fe(CN)6] does not: its solution gives no test for Fe2+ or CN−.

Key terms

  • Coordination entity: a central metal atom or ion bonded to a fixed number of ions or molecules, e.g. [CoCl3(NH3)3] or [Fe(CN)6]4−. The part in square brackets is the coordination sphere; ions outside it are counter ions.
  • Ligands: ions or molecules bonded to the central atom through donor atoms. Unidentate: Cl−, H2O, NH3. Didentate: ethane-1,2-diamine (en, H2NCH2CH2NH2) and oxalate (C2O42−). Hexadentate: ethylenediaminetetraacetate (EDTA4−), which binds through two N and four O atoms.
  • Chelate: a di- or polydentate ligand that binds through two or more donor atoms to the same metal ion, forming a ring. Chelate complexes are more stable than similar complexes of unidentate ligands.
  • Ambidentate ligand: can bind through two different atoms, e.g. NO2− (through N as nitrito-N, or through O as nitrito-O) and SCN− (through S as thiocyanato-S, or N as thiocyanato-N).
  • Coordination number (CN): the number of ligand donor atoms directly bonded to the metal. In [Co(en)3]3+ it is 6, not 3.
  • Oxidation number of the central atom: the charge it would carry if all ligands were removed along with their electron pairs.
  • Homoleptic complexes have only one kind of ligand ([Co(NH3)6]3+); heteroleptic ones have more than one ([Co(NH3)4Cl2]+).
Worked example: Find the coordination number and oxidation state of cobalt in [Co(en)2Cl2]Cl.
Solution: each en is didentate, so the donor atoms are 2 × 2 + 2 Cl = 6. The complex ion carries +1 (one Cl− outside). en is neutral and each Cl is −1, so x + 0 − 2 = +1, giving x = +3.

IUPAC nomenclature

  1. Name the cation first, then the anion, whether or not the complex is the cation.
  2. Inside the coordination entity, name ligands in alphabetical order before the metal. Multiplying prefixes (di, tri) are ignored for alphabetical order.
  3. Anionic ligands end in -o: chlorido, cyanido, hydroxido, oxido, oxalato, nitrito-N. Neutral ligands keep their names except aqua (H2O), ammine (NH3), carbonyl (CO) and nitrosyl (NO).
  4. Use di, tri, tetra for simple ligands; bis, tris, tetrakis for ligands whose names already contain a number, with the ligand name in brackets, e.g. tris(ethane-1,2-diamine).
  5. Give the oxidation state of the metal in Roman numerals in brackets after it.
  6. If the complex ion is an anion, the metal name ends in -ate: cobaltate, ferrate (Fe), cuprate (Cu), argentate (Ag), aurate (Au), plumbate (Pb), stannate (Sn).
FormulaIUPAC name
[Co(NH3)6]Cl3Hexaamminecobalt(III) chloride
[CoCl(NH3)5]Cl2Pentaamminechloridocobalt(III) chloride
K3[Fe(CN)6]Potassium hexacyanidoferrate(III)
K4[Fe(CN)6]Potassium hexacyanidoferrate(II)
[Pt(NH3)2Cl(NO2)]Diamminechloridonitrito-N-platinum(II)
[Co(en)3]2(SO4)3Tris(ethane-1,2-diamine)cobalt(III) sulphate
[Ni(CO)4]Tetracarbonylnickel(0)

In the formula, the central atom is written first, followed by the ligands in alphabetical order of their symbols, whatever their charge. So the formula is written [CoCl(NH3)5]Cl2, with Cl before NH3.

Isomerism

Structural isomerism

TypeWhat differsExample
LinkageDonor atom of an ambidentate ligand[Co(NH3)5(NO2)]Cl2: red form (O-bonded) and yellow form (N-bonded)
CoordinationLigands exchanged between cationic and anionic complexes[Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6]
IonisationWhich ion is inside and which is the counter ion[Co(NH3)5(SO4)]Br and [Co(NH3)5Br]SO4
Solvate (hydrate)Water as ligand or as free water of crystallisation[Cr(H2O)6]Cl3 (violet) and [Cr(H2O)5Cl]Cl2·H2O (grey-green)

Geometrical isomerism

Geometrical isomers differ in the relative position of ligands. It occurs in square planar complexes of type [MX2L2] and in octahedral complexes of types [MX2L4] and [MX2(L-L)2] (cis and trans), and [MA3B3] (fac, with the three A on one face, and mer, with them around a meridian). Tetrahedral complexes do not show geometrical isomerism, because all four positions are adjacent to each other.

cis and trans isomers of [Pt(NH3)2Cl2]www.iitmedicoguide.comPtClClH₃NNH₃90°PtClClH₃NNH₃cis isomerCl-Pt-Cl angle 90° (polar)trans isomerCl-Pt-Cl angle 180° (non-polar)www.iitmedicoguide.com
Square planar [Pt(NH₃)₂Cl₂] exists as two geometrical isomers. The cis form, with the chlorides side by side, is the anticancer drug cisplatin.

Optical isomerism

Optical isomers are non-superimposable mirror images (enantiomers), called dextro (d) and laevo (l) forms. They are common in octahedral complexes with didentate ligands: [Co(en)3]3+ is optically active, and so is cis-[PtCl2(en)2]2+. The trans isomer of [PtCl2(en)2]2+ has a mirror plane and is optically inactive.

Valence bond theory

The metal uses hybridised orbitals to accept lone pairs from the ligands. The hybridisation fixes the shape, and the number of unpaired electrons fixes the magnetic behaviour.

ComplexMetal ionHybridisation, shapeMagnetism
[Co(NH3)6]3+Co3+, 3d6d2sp3 (inner orbital, low spin), octahedralDiamagnetic
[CoF6]3−Co3+, 3d6sp3d2 (outer orbital, high spin), octahedralParamagnetic, 4 unpaired
[NiCl4]2−Ni2+, 3d8sp3, tetrahedralParamagnetic, 2 unpaired
[Ni(CN)4]2−Ni2+, 3d8dsp2, square planarDiamagnetic
[Ni(CO)4]Ni(0), 3d84s2sp3, tetrahedralDiamagnetic

VBT has clear limits: it involves several assumptions, does not explain colour, gives no quantitative account of magnetic data, cannot distinguish weak and strong field ligands, and cannot say exactly why some 4-coordinate complexes are tetrahedral and others square planar.

Crystal field theory

CFT treats the metal-ligand bond as purely ionic, with ligands acting as point charges (or dipoles). In a free ion the five d orbitals have the same energy. The approaching ligands raise their energy and, because the ligands come from particular directions, split them into groups. This is crystal field splitting.

Crystal field splitting in octahedral and tetrahedral fieldswww.iitmedicoguide.comegt2gΔo+0.6Δo−0.4Δofive d orbitalst2eΔtOctahedral fieldeg = dx²−y², dz²t2g = dxy, dyz, dxzBarycentre(average energy)Tetrahedral fieldΔt = (4/9) Δowww.iitmedicoguide.com
In an octahedral field the two eg orbitals, which point at the ligands, rise and the three t2g orbitals fall. In a tetrahedral field the order is reversed and the splitting is smaller, Δt = (4/9)Δo.

Octahedral complexes

Six ligands approach along the x, y and z axes. The dx²−y² and dz² orbitals point straight at them and are raised (eg set, +0.6Δo); dxy, dyz and dxz point between the axes and are lowered (t2g set, −0.4Δo). The size of Δo depends on the ligand. The spectrochemical series, found from experiment, lists ligands in increasing order of field strength:

I− < Br− < SCN− < Cl− < S2− < F− < OH− < C2O42− < H2O < NCS− < edta4− < NH3 < en < CN− < CO

For d1 to d3 the electrons go singly into t2g. From d4 onwards there is a choice. If Δo is smaller than the pairing energy P (weak field ligands), the fourth electron enters eg: t2g3eg1, a high spin complex. If Δo > P (strong field ligands), it pairs in t2g: t2g4eg0, a low spin complex.

Tetrahedral complexes

The splitting is inverted (e below, t2 above) and smaller, Δt = (4/9)Δo. The energy needed for pairing is generally larger than Δt, so low spin tetrahedral complexes are rarely observed. The subscript g is not used because a tetrahedron has no centre of symmetry.

Worked example: Predict the d electron arrangement and magnetic behaviour of [Fe(CN)6]4− and [CoF6]3−.
Solution: in [Fe(CN)6]4−, Fe is +2 (3d6) and CN− is a strong field ligand, so the arrangement is t2g6eg0: no unpaired electrons, diamagnetic, CFSE = 6 × (−0.4Δo) = −2.4Δo (plus pairing energy). In [CoF6]3−, Co is +3 (3d6) and F− is a weak field ligand, so it is t2g4eg2: 4 unpaired electrons, μ = √24 = 4.90 BM.

Colour

When white light falls on a complex, an electron absorbs light of energy Δ and moves from t2g to eg (a d-d transition). The colour seen is complementary to the colour absorbed. [Ti(H2O)6]3+ (d1) absorbs in the blue-green region and looks violet. Without ligands there is no splitting, so anhydrous CuSO4 is white while CuSO4·5H2O is blue. Changing the ligand changes Δ and therefore the colour.

CFT also has limits: treating ligands as point charges means anionic ligands should give the largest splitting, yet they sit at the weak end of the series; and it ignores the covalent character of metal-ligand bonds.

Bonding in metal carbonyls

Homoleptic carbonyls: [Ni(CO)4] is tetrahedral, [Fe(CO)5] trigonal bipyramidal and [Cr(CO)6] octahedral. The M-C bond has both σ and π character. CO donates its carbon lone pair into a vacant metal orbital (σ bond), and the metal donates electrons from a filled d orbital into the empty antibonding π* orbital of CO (π back bonding). This synergic bonding strengthens the M-C bond and weakens the C-O bond.

Importance and applications

  • Analysis: hardness of water is estimated by titration with Na2EDTA, which forms stable complexes with Ca2+ and Mg2+. Ni2+ is detected with dimethylglyoxime.
  • Biology: chlorophyll is a Mg complex, haemoglobin an Fe complex and vitamin B12 (cyanocobalamin) a Co complex.
  • Metallurgy: gold and silver are extracted as cyanide complexes such as [Au(CN)2]−; nickel is purified through [Ni(CO)4].
  • Catalysis: Wilkinson's catalyst, [(Ph3P)3RhCl], is used for hydrogenation of alkenes.
  • Medicine: EDTA removes excess lead from the body in lead poisoning; cis-platin inhibits the growth of tumours.
  • Photography: undecomposed AgBr on a developed film dissolves in hypo as [Ag(S2O3)2]3−.
Common mistakes: (1) Counting ligands instead of donor atoms for the coordination number. (2) Alphabetising by prefix ("dichlorido" under d). Use the ligand name, so ammine comes before chlorido. (3) Writing "ammine" for an amine or "amine" for NH3. (4) Forgetting that a trans isomer with didentate ligands is usually optically inactive. (5) Assuming every Co3+ complex is diamagnetic. With weak field F− it is high spin.

JEE and NEET focus

  • Werner's theory problems: number of ions in solution, moles of AgCl precipitated, conductivity comparisons.
  • IUPAC names both ways, including anionic complexes (-ate names) and ambidentate ligands.
  • Counting geometrical and optical isomers for [MX2L2], [MX2L4], [MA3B3] and [M(AA)3] types.
  • Hybridisation, geometry and spin-only magnetic moment from VBT and from CFT with the spectrochemical series.
  • Octahedral and tetrahedral splitting diagrams, CFSE, the reason for colour, and synergic bonding in carbonyls.

Practice questions

One mole of CoCl3·5NH3 is treated with excess AgNO3. The number of moles of AgCl precipitated is:

  1. 1
  2. 2
  3. 3
  4. 0
Show answer
B. The compound is [CoCl(NH3)5]Cl2; only the two Cl− outside the sphere precipitate.

The IUPAC name of K3[Fe(CN)6] is:

  1. Potassium hexacyanidoiron(III)
  2. Potassium hexacyanidoferrate(II)
  3. Potassium hexacyanidoferrate(III)
  4. Tripotassium hexacyanoiron(II)
Show answer
C. The complex is an anion, so -ate (ferrate) is used; 3(+1) + x − 6 = 0 gives x = +3.

Which of the following is optically active?

  1. trans-[CoCl2(en)2]+
  2. [Co(NH3)6]3+
  3. [Co(en)3]3+
  4. [Pt(NH3)2Cl2]
Show answer
C. The three en rings make a propeller-like shape with no mirror plane.

[Ni(CN)4]2− is:

  1. sp3, tetrahedral, paramagnetic
  2. dsp2, square planar, diamagnetic
  3. sp3, tetrahedral, diamagnetic
  4. dsp2, square planar, paramagnetic
Show answer
B. Strong field CN− pairs the eight 3d electrons, leaving one 3d orbital free for dsp2.

Which ligand produces the largest crystal field splitting?

  1. Cl−
  2. H2O
  3. NH3
  4. CO
Show answer
D. CO is at the strong field end of the spectrochemical series.

[Co(NH3)5(NO2)]Cl2 and [Co(NH3)5(ONO)]Cl2 are:

  1. Ionisation isomers
  2. Coordination isomers
  3. Linkage isomers
  4. Geometrical isomers
Show answer
C. NO2− binds through N in one and through O in the other.

Tetrahedral complexes are almost always high spin because:

  1. Δt is larger than the pairing energy
  2. Δt is small, only about 4/9 of Δo
  3. They have no d electrons
  4. Ligands in them are always weak field
Show answer
B. Pairing energy is usually greater than Δt, so electrons occupy the upper t2 set before pairing.

The number of unpaired electrons in [CoF6]3− is:

  1. 0
  2. 2
  3. 3
  4. 4
Show answer
D. Co3+ is d6; weak field F− gives t2g4eg2.
Call WhatsApp Apply
Chat with us on WhatsApp