Coordination Compounds

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Chemistry Class XII, Chapter 5.

One drop of ammonia, a completely new compound

Pour a little ammonia into pale blue copper sulphate solution. A cloudy precipitate appears first, and then the mixture clears into a deep, inky blue. Nothing exotic was added, yet the ion in solution is no longer the same one:

[Cu(H₂O)₄]²⁺ + 4NH₃ → [Cu(NH₃)₄]²⁺ + 4H₂O

Four ammonia molecules have each pushed out a water molecule and attached to Cu²⁺ through a lone pair on nitrogen. The result is a coordination entity, a metal ion surrounded by a fixed number of molecules or ions. It has its own shape, colour, magnetism and reactivity. Haemoglobin, chlorophyll, vitamin B₁₂ and the anticancer drug cisplatin are all of this kind. This chapter teaches you to name them, draw them, count their isomers and predict their colour and magnetism.

Werner's puzzle: why do some chlorides precipitate and others not?

In the 1890s Alfred Werner studied compounds with the formula CoCl₃ with ammonia and made a simple test. He added excess silver nitrate and counted how much AgCl precipitated. Only chloride that is a free ion precipitates.

In every case the cobalt held on to exactly six groups, but only some of its chlorides could be removed. Werner explained this with two kinds of valence.

The groups attached by secondary valence form the coordination polyhedron: octahedral for 6, square planar or tetrahedral for 4. In [Co(NH₃)₆]Cl₃ the primary valence of 3 is satisfied by three free Cl⁻ ions, while the secondary valence of 6 is satisfied by six NH₃. In [CoCl(NH₃)₅]Cl₂ one Cl⁻ does double duty, as a primary and a secondary valence, so it does not precipitate.

A complex is not a double salt. A double salt such as carnallite (KCl·MgCl₂·6H₂O), Mohr's salt or potash alum breaks up completely in water into all its simple ions. A complex ion such as [Fe(CN)₆]⁴⁻ stays intact and does not release free Fe²⁺ or CN⁻.

The language of coordination compounds

Coordination number counts donor atoms, not ligand molecules. Only σ bonds count, π bonds do not. Ethane-1,2-diamine (en) has two donor atoms, so [Co(en)₃]³⁺ has coordination number 6, not 3.

Types of ligand. A ligand is named by its denticity, the number of donor atoms it uses.

A ligand that grips the metal through two or more donor atoms at once forms a ring and is called a chelate ligand. Chelate complexes are more stable than comparable complexes with unidentate ligands (the chelate effect).

Beyond the textbook. For the exam: chelates are more stable than analogues with unidentate ligands. In reality: the main reason is entropy. When one en replaces two NH₃ molecules, three particles become two free ones, so disorder in the system increases and the exchange is favoured. A single hexadentate EDTA⁴⁻ replaces six water molecules and releases even more. Why it matters: that is why EDTA removes lead and hard water ions so effectively.

Worked example: oxidation number and coordination number. Find both for the metal in [Fe(C₂O₄)₃]³⁻ and in [Co(en)₂Cl₂]⁺.

The common wrong step is to count ligand molecules (3 for the oxalate complex) instead of donor atoms.

Naming a complex: the routine

Follow these steps in order.

Name of K₃[Fe(C₂O₄)₃]: potassium trioxalatoferrate(III). There is no space between the ligand names and the metal, and no space before the bracket.

Worked example: formula from a name. Write the formula of tetraamminedichloridocobalt(III) chloride.

Pause and predict: how many ions does K₄[Fe(CN)₆] produce in water, and is the compound a double salt?

Check your answer: five ions, four K⁺ and one [Fe(CN)₆]⁴⁻. It is not a double salt, because the complex ion does not split into Fe²⁺ and CN⁻.

Isomerism: same formula, different compounds

Isomers share a formula but differ in structure. Coordination compounds show two big families.

Geometrical isomers have two or three ligands of the same kind next to each other (cis) or opposite each other (trans).

Optical isomers are non-superimposable mirror images (enantiomers), named d and l by the direction in which they rotate plane polarised light. They need a complex with no mirror plane. Among the common cases are octahedral complexes with didentate ligands: [Co(en)₃]³⁺ gives two enantiomers, and cis-[CoCl₂(en)₂]⁺ gives two, while trans-[CoCl₂(en)₂]⁺ has a mirror plane and is not optically active. So [CoCl₂(en)₂]⁺ has three stereoisomers in all.

Structural isomers, with a clue for each.

Beyond the textbook. For the exam: cis-[Pt(NH₃)₂Cl₂] and trans-[Pt(NH₃)₂Cl₂] are geometrical isomers. In reality: only the cis isomer is the anticancer drug cisplatin. It binds to two neighbouring bases on a DNA strand, which needs the two Cl positions to be next to each other, and stops cancer cells from dividing. The trans isomer cannot do this. Why it matters: it is the clearest example that geometry, not just formula, decides what a compound does.

Worked example: counting isomers. How many stereoisomers does [Co(NH₃)₄Cl₂]⁺ have, and does any show optical activity?

Total: 2 stereoisomers, none optical. The common wrong step is to say the cis form is optical. Optical activity in octahedral complexes usually needs didentate ligands, as in [CoCl₂(en)₂]⁺.

Pause and predict: can a tetrahedral complex of type [MA₂B₂] show cis and trans forms?

Check your answer: no. In a tetrahedron every vertex is next to every other, so there is only one way to place A and B.

Valence bond theory: hybrid orbitals and spin

VBT treats each ligand as donating a lone pair into an empty hybrid orbital on the metal. The hybridisation fixes the shape, and the number of unpaired electrons fixes the magnetism.

The deciding factor is the ligand. A strong field ligand (CN⁻, CO, NH₃) forces the metal's d electrons to pair up, which frees inner d orbitals for d²sp³ bonding. A weak field ligand (F⁻, Cl⁻, H₂O) leaves them unpaired, so the metal must use outer d orbitals and sp³d² hybrids.

Worked example: hybridisation and magnetic moment. Work out the hybridisation, shape and spin-only magnetic moment of (a) [Fe(CN)₆]³⁻, (b) [CoF₆]³⁻ and (c) [Ni(CN)₄]²⁻.

The common wrong step is to decide the hybridisation from the number of ligands alone. Six ligands can give either d²sp³ or sp³d², and four ligands can give either sp³ or dsp². You must look at the ligand strength and the d count.

More cases worth knowing:

Notice [NiCl₄]²⁻ and [Ni(CO)₄]: both tetrahedral, but one is paramagnetic and the other is diamagnetic. [Ni(CN)₄]²⁻ is the square planar one.

Limits of VBT. It tells you the shape and the number of unpaired electrons, but it cannot explain the colour of complexes, cannot give magnetic data quantitatively, says nothing about stability, cannot predict tetrahedral versus square planar geometry, and cannot rank ligands by field strength. Crystal field theory handles all of these.

Crystal field theory: the d orbitals split

CFT is an electrostatic model. The ligands are treated as negative point charges (or the negative end of a dipole). In a free ion the five d orbitals have the same energy. Once ligands approach, the electrons in the d orbitals are repelled by them, and orbitals pointing at the ligands are raised in energy more than those pointing between them.

Octahedral field. Six ligands sit on the x, y and z axes. The two orbitals that point at them, d(x²−y²) and d(z²), form the e_g set, and are raised by +0.6Δₒ. The three that point between the axes, d(xy), d(yz) and d(xz), form the t₂g set, and are lowered by −0.4Δₒ. The gap is the crystal field splitting Δₒ. The average energy stays the same, because $2(0.6) + 3(-0.4) = 0$.

Spin state depends on Δₒ versus P. An electron entering the upper set must pay the energy cost Δₒ. An electron pairing in a lower orbital must pay the pairing energy P. The cheaper option wins.

This choice exists only for d⁴, d⁵, d⁶ and d⁷. For d¹ to d³ and d⁸ to d¹⁰ there is only one way to fill the orbitals.

The spectrochemical series ranks ligands by how strongly they split the d orbitals. This order is fixed by experiment, and you must learn it:

I⁻ < Br⁻ < SCN⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < EDTA⁴⁻ < NH₃ < en < CN⁻ < CO

I⁻ is the weakest and CO the strongest. Ligands to the left give high spin complexes and ligands to the right give low spin complexes.

Beyond the textbook. For the exam: learn the series as given, with halides at the weak end and CN⁻ and CO at the strong end. In reality: the series cannot be explained by the point charge picture of CFT, which would predict that charged ligands such as F⁻ split strongly and neutral NH₃ weakly. It is explained by ligand field and molecular orbital theory, where CN⁻ and CO are π acceptors that take electron density back from the metal. Why it matters: this is a limitation of CFT, and CUET asks about it directly.

Worked example: spin and magnetism from CFT. Compare [Fe(H₂O)₆]²⁺ and [Fe(CN)₆]⁴⁻. Give the t₂g and e_g occupation, the unpaired electrons and μ.

The common wrong step is to treat the two complexes as the same because both contain Fe²⁺. The ligand sets the spin state.

Tetrahedral field. With four ligands the splitting is the reverse of the octahedral case. The two-orbital set (e) lies lower and the three-orbital set (t₂) lies higher, and the gap is only

$$\Delta_t = \dfrac{4}{9}\Delta_o$$

Because Δₜ is smaller than the pairing energy, electrons spread out, so tetrahedral complexes are almost always high spin. The subscript g is dropped because a tetrahedron has no centre of symmetry.

Colour from d–d transitions. An electron in the lower set absorbs light whose energy equals the gap and jumps up. The complex shows the colour that is left, the complementary colour.

[Ti(H₂O)₆]³⁺ is d¹. Its one electron jumps from t₂g to e_g by absorbing blue green light at 498 nm, so the solution looks violet. A stronger ligand gives a larger gap, which shifts the absorption to shorter wavelengths and changes the colour. This is seen when ethane-1,2-diamine replaces water on nickel one step at a time in [Ni(H₂O)₆]²⁺, which moves from green to pale blue to blue to violet. Without ligands there is no splitting, which is why anhydrous CuSO₄ is white while CuSO₄·5H₂O is blue. A d⁰ or d¹⁰ metal ion has no d–d transition and its complexes are colourless.

Limits of CFT. Treating ligands as point charges cannot explain why anionic ligands sit at the weak end of the series. It also ignores covalent character in the metal–ligand bond, which ligand field theory and molecular orbital theory add.

Metal carbonyls

Carbonyls are compounds of metals with CO. In every one, the CO donates through carbon. The bond has two parts, and the second explains why CO is so strong in the series.

The two flows strengthen each other, so the effect is called synergic bonding. Back donation strengthens the M–C bond and weakens the C–O bond.

Why coordination compounds matter

Gold is extracted from its ore by dissolving it in cyanide solution and then displacing it with zinc: 4Au + 8CN⁻ + 2H₂O + O₂ → 4[Au(CN)₂]⁻ + 4OH⁻, then 2[Au(CN)₂]⁻ + Zn → 2Au + [Zn(CN)₄]²⁻.

Beyond the textbook. For the exam: haemoglobin contains iron and carries O₂. In reality: carbon monoxide is poisonous because it binds the iron in haemoglobin more than 200 times as strongly as O₂ does. CO is a strong field, π acceptor ligand, the same property you saw at the top of the spectrochemical series. Why it matters: a ligand's position in the series is not abstract, since it decides which ligand wins a competition for the metal.

Exam traps

Formula and fact sheet

60-second revision

Where this chapter connects

• Before this: The d- and f-Block Elements introduces the partly filled d orbitals, the colour and the magnetism that this chapter explains.
• Class XI link: Chemical Bonding and Molecular Structure is where hybridisation and shapes come from, and Structure of Atom gives the d orbitals themselves.
• Next: Haloalkanes and Haloarenes begins organic chemistry for Class XII, where the same ideas of Lewis acids, bases and electron pair donors reappear as nucleophiles.