CBSE Class 12 Chemistry Coordination Compounds Notes Set 10

These Class 12 CBSE Chemistry notes for Unit 5 Coordination Compounds (Set 10) explain bonding in complexes by the valence bond theory and the crystal field theory, the magnetic behaviour and colour of complexes, the structural and stereoisomerism of coordination compounds, and the main applications. Six solved examples work through hybridisation, spin-only magnetic moments, d-orbital occupancy and counting of isomers.

Important Notes and Formulas

Valence Bond Theory (VBT)

1. The metal ion provides empty orbitals ((n−1)d, ns, np or ns, np, nd). They hybridise to give equivalent orbitals of definite geometry.
2. Each ligand donates a lone pair into a hybrid orbital, forming a coordinate (sigma) bond.
3. Inner orbital (low spin) complex: uses (n−1)d orbitals, as in d²sp³. Strong field ligands force the electrons of the metal to pair up, so there are few unpaired electrons.
4. Outer orbital (high spin) complex: uses nd orbitals, as in sp³d². The d electrons keep their unpaired arrangement.
5. The geometry follows the hybridisation: sp³ is tetrahedral, dsp² is square planar and d²sp³ or sp³d² is octahedral.

In the diagrams the shaded boxes are the orbitals that hybridise, and red arrows are the electron pairs donated by the ligands.

The spin-only magnetic moment is given by μ = \( \sqrt{n(n+2)} \) BM, where n is the number of unpaired electrons.

Limitations of VBT: it does not explain the colour of complexes, gives no quantitative treatment of magnetic data, cannot explain why some d⁴ and d⁵ complexes behave as they do, does not tell strong from weak field ligands, and gives no information about thermodynamic or kinetic stability.

Crystal Field Theory (CFT)

1. CFT treats the metal-ligand bond as purely electrostatic. Anionic ligands are point negative charges, and neutral ligands such as NH₃ and H₂O are point dipoles with the negative end towards the metal.
2. In the free ion the five d orbitals have the same energy. In the field of the ligands the degeneracy is lost, and the orbitals split. This is crystal field splitting.
3. The pattern of splitting depends on the arrangement of the ligands.

Splitting in an Octahedral Field

1. The ligands approach along the x, y and z axes. The orbitals d(x²−y²) and d(z²) point towards the ligands and are repelled strongly: they rise in energy and form the e_g set.
2. The orbitals dxy, dyz and dzx point between the ligands, so they are repelled less. They are lowered and form the t_2g set.
3. The energy gap between the two sets is Δ₀. Relative to the barycentre, e_g is raised by 0.6 Δ₀ and t_2g is lowered by 0.4 Δ₀.
4. For d⁴ to d⁷ ions two arrangements are possible. If Δ₀ < P (pairing energy), the next electron goes into e_g: a high spin complex (weak field ligand). If Δ₀ > P, the electron pairs in t_2g: a low spin complex (strong field ligand).

For d¹, d², d³, d⁸, d⁹ and d¹⁰ ions only one arrangement is possible in an octahedral field, so there is no high spin or low spin choice.

Spectrochemical Series

The order of ligands by the size of the splitting they produce (weak to strong field), found by experiment:

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

Ligands up to H₂O are generally weak field and those from NH₃ onwards strong field. For the same ligand, Δ₀ is larger for a higher oxidation state of the metal and for a metal from the second or third transition series.

Splitting in a Tetrahedral Field

1. The splitting is inverted: the e set (d(x²−y²) and d(z²)) is lower and the t₂ set (dxy, dyz, dzx) is higher.
2. Only four ligands are present and none points directly at the orbitals, so Δt = 4/9 Δ₀ for the same metal and ligands.
3. Because Δt is small, it is almost always less than the pairing energy, so tetrahedral complexes are high spin. Low spin tetrahedral complexes are rare.

Colour of Coordination Compounds

1. A complex with unpaired d electrons and a partly filled d set absorbs visible light to excite an electron from a lower d level to a higher one (a d-d transition). The colour seen is the complement of the colour absorbed.
2. [Ti(H₂O)₆]³⁺ has one electron in t₂g. It absorbs blue-green light (about 498 nm) and shows the electron jump t₂g¹ e_g⁰ → t₂g⁰ e_g¹, so it appears purple (violet).
3. The colour changes with the ligand because Δ₀ changes. A larger Δ₀ means light of shorter wavelength is absorbed.
4. Without ligands there is no splitting, so the substance is colourless: anhydrous CuSO₄ is white, but CuSO₄.5H₂O is blue. Ions with d⁰ or d¹⁰ configuration such as Sc³⁺, Zn²⁺ and Cu⁺ give colourless complexes.

Limitations of CFT: it treats ligands as point charges, so it cannot explain why anionic ligands such as Cl⁻ and OH⁻ lie low in the spectrochemical series, and it does not account for the covalent character of the metal-ligand bond.

Isomerism in Coordination Compounds

Isomers have the same molecular formula but different arrangement of atoms. They fall into two families: structural isomers and stereoisomers.

1. Structural Isomerism

2. Stereoisomerism

Stereoisomers have the same bonds but a different arrangement of the ligands in space.

Geometrical isomerism occurs in heteroleptic complexes with coordination number 4 (square planar) and 6 (octahedral). In the cis form two identical ligands are adjacent (90°); in the trans form they are opposite (180°).

1. A tetrahedral complex shows no geometrical isomerism, because all four positions are equivalent.
2. A square planar complex [MA₂B₂] has two isomers, cis and trans, e.g. [Pt(NH₃)₂Cl₂]. [MABCD] has three.
3. An octahedral [MA₄B₂] has two isomers, cis and trans.
4. An octahedral [MA₃B₃] has two isomers: facial (fac), where the three identical ligands occupy one triangular face, and meridional (mer), where they lie in a plane through the metal.
5. A square planar M(AA)₂ with a symmetrical didentate ligand, such as [Pt(en)₂]²⁺, has no geometrical isomers. Complexes of the type [MA₄], [MA₃B] and [MA₅B] also have none.

Cis and trans isomers of the square planar complex [Pt(NH₃)₂Cl₂]

Cis and trans isomers of the octahedral complex [Co(NH₃)₄Cl₂]⁺

Facial and meridional isomers of [Co(NH₃)₃Cl₃]

Optical isomerism occurs when the complex and its mirror image cannot be superimposed (the complex is chiral). The two forms, dextro (d) and laevo (l), rotate plane-polarised light in opposite directions. Such a pair is called an enantiomeric pair.

1. Octahedral complexes with didentate ligands often show optical isomerism, e.g. [Co(en)₃]³⁺ and cis-[CoCl₂(en)₂]⁺.
2. The trans form of [CoCl₂(en)₂]⁺ has a plane of symmetry, so it is optically inactive. cis-[Co(NH₃)₄Cl₂]⁺ also has a plane of symmetry and is inactive.
3. So [CoCl₂(en)₂]⁺ has three stereoisomers in all: trans (inactive) and a pair of cis enantiomers.
4. Square planar complexes are not chiral, because the plane of the molecule acts as a mirror plane.

The two enantiomers of cis-[CoCl₂(en)₂]⁺ (each curved line joining two N atoms is one en ligand; the two structures are mirror images that cannot be superimposed)

Importance and Applications of Coordination Compounds

1. Biological systems: chlorophyll is a complex of magnesium, haemoglobin (carries oxygen in blood) is a complex of iron, and vitamin B₁₂ (cyanocobalamin) is a complex of cobalt.
2. Analytical chemistry: many metal ions are detected or estimated through coloured complexes, e.g. dimethylglyoxime gives a red precipitate with Ni²⁺ in ammoniacal solution, and EDTA is used to estimate the hardness of water (Ca²⁺ and Mg²⁺).
3. Extraction of metals: silver and gold dissolve in aqueous NaCN in the presence of air to form [Ag(CN)₂]⁻ and [Au(CN)₂]⁻. The metal is then displaced by zinc.
4. Purification of metals: impure nickel is converted into volatile [Ni(CO)₄] (Mond process), which decomposes on stronger heating to give pure nickel.
5. Electroplating: the metals deposit more evenly from solutions of [Ag(CN)₂]⁻ and [Au(CN)₂]⁻.
6. Photography: in the fixing bath, hypo (sodium thiosulphate) dissolves the unexposed AgBr as the complex [Ag(S₂O₃)₂]³⁻.
7. Medicine: EDTA is used in the treatment of lead poisoning through chelation, and cis-platin [Pt(NH₃)₂Cl₂] is used against certain cancers.
8. Catalysis: Wilkinson's catalyst [(Ph₃P)₃RhCl] is used for the hydrogenation of alkenes.

Solved Examples

Example 1. Explain, using VBT, why [Fe(CN)₆]³⁻ is weakly paramagnetic while [FeF₆]³⁻ is strongly paramagnetic. Give the spin-only magnetic moment of each.
Answer:
Fe³⁺ has the configuration 3d⁵ (five unpaired electrons).
CN⁻ is a strong field ligand. It pairs the electrons into t₂g (3d⁵ → t₂g⁵), leaving two 3d orbitals empty for d²sp³ hybridisation. Unpaired electrons n = 1, so μ = \( \sqrt{1(1+2)} \) = 1.73 BM (inner orbital, low spin).
F⁻ is a weak field ligand and cannot pair the electrons. The ion uses 4s, 4p and 4d orbitals (sp³d²). Unpaired electrons n = 5, so μ = \( \sqrt{5(5+2)} \) = 5.92 BM (outer orbital, high spin).

Example 2. Predict the hybridisation, shape and magnetic behaviour of (i) [Ni(CN)₄]²⁻, (ii) [NiCl₄]²⁻ and (iii) [Ni(CO)₄].
Answer:
(i) Ni²⁺ is 3d⁸. CN⁻ is a strong field ligand, so the 3d electrons pair up and one 3d orbital becomes free: dsp², square planar, diamagnetic.
(ii) Cl⁻ is a weak field ligand and cannot pair the 3d electrons: sp³, tetrahedral, paramagnetic (2 unpaired electrons, μ = 2.83 BM).
(iii) Ni in [Ni(CO)₄] is in the zero oxidation state (3d⁸4s²). CO pairs the electrons into 3d¹⁰: sp³, tetrahedral, diamagnetic.

Example 3. Write the d-electron configuration of the central ion in (i) [Fe(H₂O)₆]²⁺ (weak field) and (ii) [Fe(CN)₆]⁴⁻ (strong field). Give the number of unpaired electrons.
Answer:
Fe²⁺ is d⁶.
(i) H₂O is a weak field ligand, so Δ₀ < P: t₂g⁴ e_g², 4 unpaired electrons (high spin).
(ii) CN⁻ is a strong field ligand, so Δ₀ > P: t₂g⁶ e_g⁰, 0 unpaired electrons (low spin, diamagnetic).

Example 4. [Ti(H₂O)₆]³⁺ is purple but [Sc(H₂O)₆]³⁺ is colourless. Explain with CFT.
Answer:
Ti³⁺ is d¹. Its single electron sits in the t₂g level. The complex absorbs light of about 498 nm (blue-green) to move the electron to e_g, and the transmitted light is purple.
Sc³⁺ is d⁰. There is no d electron to excite, so no d-d transition is possible and no visible light is absorbed. The complex is colourless.

Example 5. How many stereoisomers are possible for (i) [Pt(NH₃)₂Cl₂] (square planar) and (ii) [Co(en)₂Cl₂]⁺? Which of them are optically active?
Answer:
(i) Two isomers, cis and trans. Neither is optically active, because square planar complexes have a mirror plane.
(ii) Three stereoisomers: trans (has a plane of symmetry, inactive) and cis, which exists as a pair of enantiomers (d and l). Only the cis form is optically active.

Example 6. Name the type of isomerism shown by each pair: (i) [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br, (ii) [Cr(H₂O)₆]Cl₃ and [CrCl₂(H₂O)₄]Cl.2H₂O, (iii) [Co(NH₃)₅(NO₂)]Cl₂ and [Co(NH₃)₅(ONO)]Cl₂.
Answer:
(i) Ionisation isomerism: Br⁻ and SO₄²⁻ have exchanged places between the sphere and the counter ion.
(ii) Solvate (hydrate) isomerism: the number of water molecules in the coordination sphere differs.
(iii) Linkage isomerism: NO₂⁻ is bound through N in one and through O in the other.