Syllabus topics (RTMNU)
- Limitations of Valency bond theory
- Crystal field theory: Splitting of d-orbital in octahedral, tetrahedral and square planar complexes
- Factors affecting the Magnitude of 10Dq
- Crystal Field Stabilisation Energy of Octahedral and Tetrahedral complexes, Numericals
- Jahn Teller Effect
- Selection Rules (Laporte and Spin selection Rules)
- Hole formalism principle
- Electronic spectrum of [Ti(H2O)6]3+ and [Cu(H2O)6]2+ complex ions
Limitations of Valence Bond Theory
Valence bond theory (Pauling) explains the shape and magnetic nature of complexes through hybridisation (inner-orbital d2sp3 and outer-orbital sp3d2), but it has many limitations:
- It cannot explain the colour and electronic spectra of complexes.
- It does not explain why some ligands are strong (cause pairing) and others weak.
- It cannot explain the orbital contribution to magnetic moment or its variation with temperature.
- It gives no quantitative idea of the energies or stability of complexes.
- It cannot explain distortions such as the Jahn–Teller effect, or why Cu(II) square planar complexes use a 4p orbital (promotion of an electron).
- The distinction between inner- and outer-orbital complexes is often arbitrary.
Crystal Field Theory: Splitting of d-Orbitals
Crystal field theory (CFT), given by Bethe and Van Vleck, treats ligands as point negative charges (or dipoles) and the metal–ligand bond as purely electrostatic. In a free metal ion the five d-orbitals are degenerate (equal energy). When ligands approach, electrons in d-orbitals pointing towards the ligands are repelled more, so the d-orbitals split into groups of different energy. The energy gap is the crystal field splitting energy, Δ or 10Dq.
Octahedral complexes: six ligands lie along ±x, ±y, ±z axes. The eg set (dx²−y², dz²), pointing at ligands, is raised by +0.6Δo (+6Dq); the t2g set (dxy, dyz, dxz), lying between axes, is lowered by −0.4Δo (−4Dq). The barycentre (average energy) stays the same.
Tetrahedral complexes: four ligands approach between the axes, so the t2 set (dxy, dyz, dxz) is raised by +0.4Δt and the e set (dx²−y², dz²) is lowered by −0.6Δt — the reverse of octahedral. Δt = (4/9)Δo, so tetrahedral complexes are almost always high-spin.
Square planar complexes: obtained by removing the two z-axis ligands from an octahedron. Orbitals with a z-component fall in energy. Order: dx²−y² (highest) > dxy > dz² > dxz = dyz (lowest). The large gap below dx²−y² makes d8 ions (Ni2+ with strong ligands, Pd2+, Pt2+) square planar and diamagnetic, e.g. [Ni(CN)4]2−.
Factors Affecting the Magnitude of 10Dq
- Nature of the ligand: the spectrochemical series (increasing Δ): I− < Br− < S2− < SCN− < Cl− < F− < OH− < C2O42− < H2O < NCS− < NH3 < en < bipy < phen < NO2− < CN− < CO. Ligands at the left are weak-field; at the right, strong-field.
- Oxidation state of the metal: higher charge pulls ligands closer → larger Δ; e.g. Δo for [Co(H2O)6]3+ is much larger than for [Co(H2O)6]2+.
- Nature of the metal (row): Δ increases by about 30–50% from 3d to 4d and again to 5d, so 4d/5d complexes are usually low-spin.
- Geometry: Δt ≈ 4/9 Δo; square planar Δ is largest.
- Number of ligands and metal–ligand distance (shorter distance → larger Δ).
Crystal Field Stabilisation Energy (CFSE) and Numericals
CFSE is the net lowering of energy of the d-electrons in a crystal field compared with the hypothetical spherical field. Whether electrons pair up depends on Δ compared with the pairing energy P: if Δo < P → high spin (weak field); if Δo > P → low spin (strong field). This choice exists only for d4–d7 octahedral ions.
Tetrahedral: CFSE = (−0.6 × ne + 0.4 × nt2) Δt
| Complex | Configuration | CFSE |
|---|---|---|
| [Ti(H2O)6]3+ (d1) | t2g1 | −0.4Δo |
| [Cr(H2O)6]3+ (d3) | t2g3 | −1.2Δo |
| [Fe(H2O)6]2+ (d6, high spin) | t2g4eg2 | 4(−0.4) + 2(+0.6) = −0.4Δo |
| [Fe(CN)6]4− (d6, low spin) | t2g6 | 6(−0.4) = −2.4Δo (+ 2P extra pairing) |
| [Mn(H2O)6]2+ (d5, high spin) | t2g3eg2 | 3(−0.4) + 2(0.6) = 0 |
| [CoCl4]2− (d7, tetrahedral) | e4t23 | 4(−0.6) + 3(0.4) = −1.2Δt |
Numerical: [Ti(H2O)6]3+ absorbs at 20,300 cm−1. This equals Δo. Using 1 cm−1 = 11.96 J mol−1: Δo = 20,300 × 11.96 = 242.8 kJ mol−1. CFSE = −0.4 × 242.8 = −97.1 kJ mol−1 (or −8,120 cm−1).
Jahn–Teller Effect
Jahn–Teller theorem (1937): "Any non-linear molecule in a degenerate electronic state will undergo a distortion to remove the degeneracy and lower its energy." In octahedral complexes, unequal filling of the eg orbitals causes strong distortion — usually tetragonal elongation (two axial bonds longer, four equatorial shorter).
Strong Jahn–Teller distortion occurs for d9 (Cu2+), high-spin d4 (Cr2+, Mn3+) and low-spin d7. Weak distortion occurs with unequal t2g filling (d1, d2). No distortion for d3, high-spin d5, low-spin d6, d8, d10. Example: in [Cu(H2O)6]2+ (t2g6eg3), the two axial Cu–O bonds are longer than the four equatorial ones.
Selection Rules: Laporte and Spin
- Laporte (orbital) rule: allowed transitions involve a change in parity (Δl = ±1, g ↔ u). d–d transitions (g → g) are Laporte-forbidden in centrosymmetric (octahedral) complexes. They still appear weakly (ε ≈ 1–10 L mol−1 cm−1) because molecular vibrations momentarily destroy the centre of symmetry (vibronic coupling). Tetrahedral complexes have no centre of symmetry, so their bands are more intense (ε ≈ 100–1000) — e.g. deep blue [CoCl4]2− versus pale pink [Co(H2O)6]2+.
- Spin selection rule: ΔS = 0 — the number of unpaired electrons must not change. Spin-forbidden transitions are very weak; high-spin Mn2+ (d5) complexes have only spin-forbidden d–d bands and are almost colourless (pale pink).
Hole formalism: a dn configuration can be treated as (10 − n) "holes" (positive charges) in a filled d-shell. The holes behave like electrons but with the energy order reversed. So d9 (one hole) behaves like d1, d8 like d2; both d1 and d9 octahedral ions give a single d–d band (2T2g ↔ 2Eg, inverted for d9).
Electronic Spectra of [Ti(H2O)6]3+ and [Cu(H2O)6]2+
[Ti(H2O)6]3+ (d1): ground state t2g1 (2T2g). One electron is promoted to eg (2T2g → 2Eg), giving one broad band at about 20,300 cm−1 (≈ 493 nm). It absorbs yellow-green light, so the solution looks purple/violet. The band is broad and has a shoulder because the excited state (eg1) undergoes Jahn–Teller distortion, splitting the eg level.
[Cu(H2O)6]2+ (d9): t2g6eg3 — one hole. By hole formalism it is like d1 inverted: the single transition 2Eg → 2T2g appears as one broad, asymmetric band around 12,500 cm−1 (≈ 800 nm) in the red region, so the solution looks blue. The band is broad because the ground state is Jahn–Teller distorted (tetragonal), which splits the levels into several close transitions.
Likely exam questions
- (Long, 8 marks) Explain the splitting of d-orbitals in octahedral and tetrahedral complexes according to crystal field theory. Calculate the CFSE of [Fe(H2O)6]2+ and [Fe(CN)6]4−.
- (Long, 8 marks) State the Jahn–Teller theorem and the Laporte and spin selection rules. Discuss the electronic spectra of [Ti(H2O)6]3+ and [Cu(H2O)6]2+.
- (Short, 4 marks) Give any four limitations of valence bond theory.
- (Short, 4 marks) What are the factors affecting the magnitude of 10Dq?
- (Short, 4 marks) Explain the hole formalism principle.