Introduction
Why are coordination compounds colored? Why is [Ti(H₂O)₆]³⁺ purple while [Cu(H₂O)₆]²⁺ is blue? Why can VBT predict geometry but not color? The answer lies in Crystal Field Theory (CFT), developed by Hans Bethe and John Hasbroek van Vleck in the 1930s.
The Fundamental Idea
VBT vs CFT: Key Difference
| Theory | Treats Ligands As | Focus |
|---|---|---|
| VBT | Electron pair donors (covalent bonding) | Geometry, magnetism |
| CFT | Point charges or dipoles (electrostatic) | Color, magnetic properties quantitatively |
Core Concept of CFT
- Ligands are negative point charges (or dipoles)
- They create an electrostatic field around the metal
- This field splits the d orbitals into different energy levels
- The splitting pattern depends on geometry
- d-d electron transitions between split levels cause color
d Orbitals in Free Ion vs Ligand Field
Free Metal Ion (Spherical Field)
All five d orbitals have the same energy (degenerate):
d_xy d_yz d_xz d_x²-y² d_z²
↑ ↑ ↑ ↑ ↑ ← All same energy
In Ligand Field (Octahedral)
The d orbitals split into two sets:
graph TB
A[Free ion: 5 d orbitals
all same energy] --> B[Octahedral field]
B --> C[e_g set
Higher energy
d_x²-y², d_z²]
B --> D[t_2g set
Lower energy
d_xy, d_yz, d_xz]
style C fill:#e74c3c
style D fill:#3498dbWhy this splitting?
- t₂g orbitals (d_xy, d_yz, d_xz): Point between the axes → less repulsion with ligands
- e_g orbitals (d_x²-y², d_z²): Point along the axes → more repulsion with ligands
Crystal Field Splitting in Octahedral Complexes
Energy Level Diagram
e_g ────┬──── (d_x²-y², d_z²)
│
│ Δ₀
Barycenter ───────┼───── (average energy)
│
t₂g ────┴──── (d_xy, d_yz, d_xz)
Free ion Spherical Octahedral
field field
Key Parameters
Δ₀ (or 10 Dq) = Crystal Field Splitting Energy (octahedral)
- Energy difference between e_g and t₂g levels
- Determines whether complex is high spin or low spin
- Depends on metal and ligand
Energy changes:
- e_g orbitals: increased by +0.6Δ₀ (or +6 Dq)
- t₂g orbitals: decreased by -0.4Δ₀ (or -4 Dq)
Factors Affecting Δ₀
1. Nature of Ligand (Most Important!)
Spectrochemical Series:
$$\boxed{I^- < Br^- < SCN^- < Cl^- < F^- < OH^- < H_2O < NCS^- < NH_3 < en < NO_2^- < CN^- < CO}$$Weak field ←─────────────────────────→ Strong field (Small Δ₀) ←─────────────────────────→ (Large Δ₀)
Examples:
- Δ₀[CoF₆]³⁻ ≈ 13,000 cm⁻¹ (weak field)
- Δ₀[Co(NH₃)₆]³⁺ ≈ 23,000 cm⁻¹ (moderate)
- Δ₀[Co(CN)₆]³⁻ ≈ 33,000 cm⁻¹ (strong field)
2. Oxidation State of Metal
Higher oxidation state → Larger Δ₀
Example:
- Δ₀[Fe(H₂O)₆]²⁺ < Δ₀[Fe(H₂O)₆]³⁺
Why? Higher charge → stronger electrostatic interaction
3. Principal Quantum Number of d Orbitals
Same group, lower period → Larger Δ₀
Example:
- Δ₀(3d) < Δ₀(4d) < Δ₀(5d)
- Δ₀[Co(NH₃)₆]³⁺ < Δ₀[Rh(NH₃)₆]³⁺ < Δ₀[Ir(NH₃)₆]³⁺
Why? Larger orbitals → better overlap with ligands
High Spin vs Low Spin Complexes
The Critical Question
When electrons fill the split d orbitals, do they:
- Spread out to minimize pairing energy?
- Pair up to occupy lower energy t₂g orbitals?
Answer depends on: Is Δ₀ greater or less than pairing energy (P)?
Case 1: Weak Field (Δ₀ < P)
Electrons prefer to remain unpaired → HIGH SPIN
Example: [Fe(H₂O)₆]²⁺ (Fe²⁺ = d⁶)
e_g: ↑ ↑
t₂g: ↑ ↑ ↑ ↑
- 4 unpaired electrons
- Paramagnetic
- μ = √[4(4+2)] = 4.90 BM
Case 2: Strong Field (Δ₀ > P)
Energy saved by occupying lower t₂g > pairing energy → LOW SPIN
Example: [Fe(CN)₆]⁴⁻ (Fe²⁺ = d⁶)
e_g: __ __
t₂g: ↑↓ ↑↓ ↑↓
- 0 unpaired electrons
- Diamagnetic
- μ = 0 BM
High spin vs low spin only matters for:
- d⁴, d⁵, d⁶, d⁷ configurations in octahedral complexes
For other configurations (d¹, d², d³, d⁸, d⁹, d¹⁰), there’s only one possible arrangement!
Electron Distribution Examples
d⁴ Configuration: [Mn(H₂O)₆]³⁺ vs [Mn(CN)₆]³⁻
High Spin (weak field):
e_g: ↑ __
t₂g: ↑ ↑ ↑
4 unpaired electrons
Low Spin (strong field):
e_g: __ __
t₂g: ↑↓ ↑ ↑
2 unpaired electrons
d⁵ Configuration: [Fe(H₂O)₆]³⁺ vs [Fe(CN)₆]³⁻
High Spin:
e_g: ↑ ↑
t₂g: ↑ ↑ ↑
5 unpaired electrons, μ = 5.92 BM
Low Spin:
e_g: __ __
t₂g: ↑↓ ↑ ↑
1 unpaired electron, μ = 1.73 BM
d⁷ Configuration: [Co(H₂O)₆]²⁺ vs [Co(CN)₆]⁴⁻
High Spin:
e_g: ↑ ↑
t₂g: ↑↓ ↑↓ ↑
3 unpaired electrons
Low Spin:
e_g: ↑ __
t₂g: ↑↓ ↑↓ ↑↓
1 unpaired electron
Crystal Field Stabilization Energy (CFSE)
Definition
CFSE = Energy gained by distributing electrons in split d orbitals compared to a spherical field
Calculation Formula
$$\text{CFSE} = [(-0.4) \times n_{t_{2g}} + (+0.6) \times n_{e_g}] \times \Delta_0$$If pairing occurs, subtract pairing energy (P)
Examples
1. [Ti(H₂O)₆]³⁺ (d¹)
e_g: __ __
t₂g: ↑ __ __
CFSE = (-0.4)(1) × Δ₀ = -0.4Δ₀
2. [Fe(CN)₆]⁴⁻ (d⁶, low spin)
e_g: __ __
t₂g: ↑↓ ↑↓ ↑↓
CFSE = (-0.4)(6) × Δ₀ = -2.4Δ₀ But 2 pairings occurred: CFSE = -2.4Δ₀ + 2P
3. [Fe(H₂O)₆]²⁺ (d⁶, high spin)
e_g: ↑ ↑
t₂g: ↑ ↑ ↑ ↑
CFSE = (-0.4)(4) + (0.6)(2) × Δ₀ = -0.4Δ₀
CFSE Table for Octahedral Complexes
| d electrons | High Spin CFSE | Low Spin CFSE |
|---|---|---|
| d⁰ | 0 | 0 |
| d¹ | -0.4Δ₀ | -0.4Δ₀ |
| d² | -0.8Δ₀ | -0.8Δ₀ |
| d³ | -1.2Δ₀ | -1.2Δ₀ |
| d⁴ | -0.6Δ₀ | -1.6Δ₀ + P |
| d⁵ | 0 | -2.0Δ₀ + 2P |
| d⁶ | -0.4Δ₀ | -2.4Δ₀ + 2P |
| d⁷ | -0.8Δ₀ | -1.8Δ₀ + P |
| d⁸ | -1.2Δ₀ | -1.2Δ₀ |
| d⁹ | -0.6Δ₀ | -0.6Δ₀ |
| d¹⁰ | 0 | 0 |
Tetrahedral Complexes
d Orbital Splitting in Tetrahedral Field
In tetrahedral geometry, ligands approach between the axes, so the splitting is reversed!
t₂ ────┬──── (d_xy, d_yz, d_xz)
│
│ Δ_t
Barycenter ──────┼─────
│
e ─────┴──── (d_x²-y², d_z²)
Tetrahedral field
Key differences from octahedral:
- e orbitals are now lower in energy
- t₂ orbitals are now higher in energy
- Δₜ ≈ (4/9)Δ₀ (tetrahedral splitting is smaller!)
Why Tetrahedral Complexes Are Always High Spin
- Δₜ is already small (4/9 of Δ₀)
- Only 4 ligands vs 6 in octahedral
- Δₜ is ALWAYS less than pairing energy
- Therefore: All tetrahedral complexes are high spin!
CFSE in Tetrahedral Complexes
$$\text{CFSE}_{tetrahedral} = [(-0.6) \times n_e + (+0.4) \times n_{t_2}] \times \Delta_t$$Example: [CoCl₄]²⁻ (Co²⁺ = d⁷)
t₂: ↑ ↑ ↑
e: ↑↓ ↑↓
CFSE = (-0.6)(4) + (0.4)(3) × Δₜ = -1.2Δₜ
Square Planar Complexes
d Orbital Splitting
d_x²-y² ─────── (highest)
d_xy ──────────
d_z² ──────────
d_xz, d_yz ──── (lowest, degenerate)
Square planar field
Common for:
- d⁸ configuration (Ni²⁺, Pd²⁺, Pt²⁺, Au³⁺)
- Strong field ligands
Example: [Ni(CN)₄]²⁻ All 8 electrons pair up in lower orbitals → Diamagnetic
Color in Coordination Compounds
Why Are Complexes Colored?
Reason: d-d electron transitions absorb specific wavelengths of visible light.
The Process
- White light hits the complex
- Energy matching Δ₀ is absorbed
- Electron jumps from t₂g → e_g
- Complementary color is observed
Example: [Ti(H₂O)₆]³⁺ (d¹)
Before absorption: After absorption:
e_g: __ __ e_g: ↑ __
t₂g: ↑ __ __ t₂g: __ __ __
- Absorbs light at λ ≈ 500 nm (green-yellow)
- Appears purple (complementary color)
Energy-Wavelength Relationship
$$\Delta_0 = h\nu = \frac{hc}{\lambda}$$For [Ti(H₂O)₆]³⁺:
- Δ₀ = 20,300 cm⁻¹
- λ = 493 nm (green region)
- Observed color = Purple
Color Wheel
graph TD
A[Absorbed Color] --> B[Observed Color]
C[Red] --> D[Green]
E[Orange] --> F[Blue]
G[Yellow] --> H[Violet]
I[Green] --> J[Purple/Red]
style C fill:#ff0000
style E fill:#ffa500
style G fill:#ffff00
style I fill:#00ff00Why Some Complexes Are Colorless
d⁰ and d¹⁰ Configurations
Examples:
- [Sc(H₂O)₆]³⁺ (d⁰) → Colorless
- [Zn(H₂O)₆]²⁺ (d¹⁰) → Colorless
Reason:
- d⁰: No electrons to excite
- d¹⁰: All d orbitals full, no empty orbitals for transition
No d-d transition possible → No visible light absorbed → Colorless
Comparison: Octahedral vs Tetrahedral
| Property | Octahedral | Tetrahedral |
|---|---|---|
| Splitting | Δ₀ | Δₜ ≈ (4/9)Δ₀ |
| Lower energy orbitals | t₂g (3 orbitals) | e (2 orbitals) |
| Higher energy orbitals | e_g (2 orbitals) | t₂ (3 orbitals) |
| High/Low spin | Both possible (d⁴-d⁷) | Always high spin |
| Color intensity | More intense | Less intense |
| CFSE | Generally larger | Smaller |
Memory Tricks
Spectrochemical Series
“I Broke Several Chairs, Felt Obliged; Had Witnessed Ninjas Entering, Noting Crazy Combat”
I⁻ < Br⁻ < SCN⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO
Strong Field = Low Spin
“STRONG ligands FORCE pairing → LOW spin → FEW unpaired”
Splitting Pattern
“Oct-t₂g-Below” → Octahedral has t₂g below “Tet-t₂-Top” → Tetrahedral has t₂ on top
Common Mistakes
Wrong: In octahedral, e_g is lower energy Right: In octahedral, t₂g is lower energy (points between ligands)
In tetrahedral, this reverses!
Wrong: d³ configuration can be high or low spin Right: Only d⁴, d⁵, d⁶, d⁷ show this variation
d¹, d², d³, d⁸, d⁹ have only ONE arrangement!
Wrong: [Ti(H₂O)₆]³⁺ absorbs purple, so it’s purple Right: It absorbs green-yellow, so appears purple (complementary!)
Observed color = Complementary to absorbed color
Practice Problems
Level 1: Basic Concepts
Q1. Arrange the following in increasing order of Δ₀: [CrCl₆]³⁻, [Cr(H₂O)₆]³⁺, [Cr(NH₃)₆]³⁺, [Cr(CN)₆]³⁻
Q2. Which of the following will be colored? a) [Sc(H₂O)₆]³⁺ (d⁰) b) [Ti(H₂O)₆]³⁺ (d¹) c) [Zn(NH₃)₄]²⁺ (d¹⁰) d) [Cu(H₂O)₆]²⁺ (d⁹)
Q3. Calculate CFSE for d³ configuration in octahedral field.
Level 2: Application
Q4. [Fe(H₂O)₆]²⁺ is pale green while [Fe(CN)₆]⁴⁻ is pale yellow. Explain: a) The difference in magnetic moments b) The difference in colors
Q5. Why are tetrahedral complexes generally less intensely colored than octahedral complexes?
Q6. A complex of Co³⁺ (d⁶) absorbs light at 600 nm. Calculate Δ₀ in: a) cm⁻¹ b) kJ/mol (Given: h = 6.626 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s, N_A = 6.022 × 10²³)
Q7. Predict which will have larger Δ₀: a) [Fe(H₂O)₆]²⁺ or [Fe(H₂O)₆]³⁺ b) [Co(NH₃)₆]³⁺ or [Rh(NH₃)₆]³⁺ c) [CoF₆]³⁻ or [Co(CN)₆]³⁻
Level 3: JEE Advanced
Q8. The magnetic moment of [Mn(H₂O)₆]²⁺ is 5.92 BM. When treated with excess CN⁻, the moment drops to 1.73 BM. Explain using: a) Electron distribution diagrams b) CFSE calculations for both complexes
Q9. [Ti(H₂O)₆]³⁺ absorbs light at 20,300 cm⁻¹ (purple color observed). a) Calculate the wavelength of absorbed light b) Explain using CFT why this complex is colored c) Predict the color if Ti³⁺ is replaced by Sc³⁺
Q10. Explain the following observations: a) [Ni(CN)₄]²⁻ is square planar and diamagnetic b) [NiCl₄]²⁻ is tetrahedral and paramagnetic c) Both have same metal and coordination number
Q11. Calculate the CFSE for: a) [Co(H₂O)₆]³⁺ (high spin d⁶) b) [Co(NH₃)₆]³⁺ (low spin d⁶) c) Which is more stable based on CFSE alone?
Q12. A student observes:
- [Fe(H₂O)₆]²⁺ is pale green
- [Fe(H₂O)₆]³⁺ is pale violet
- [Fe(CN)₆]⁴⁻ is pale yellow
- [Fe(CN)₆]³⁻ is red
Explain the trend in colors using CFT.
Solutions to Selected Problems
Q1. [CrCl₆]³⁻ < [Cr(H₂O)₆]³⁺ < [Cr(NH₃)₆]³⁺ < [Cr(CN)₆]³⁻ (Based on spectrochemical series)
Q2. b) and d) will be colored (d¹ and d⁹ allow d-d transitions)
Q3. CFSE = (-0.4)(3) Δ₀ = -1.2Δ₀
Q7. a) [Fe(H₂O)₆]³⁺ (higher oxidation state) b) [Rh(NH₃)₆]³⁺ (4d orbitals, larger splitting) c) [Co(CN)₆]³⁻ (CN⁻ stronger field than F⁻)
Q9. a) λ = 1/ν̃ = 1/20,300 = 493 nm b) d¹ configuration: electron transitions from t₂g to e_g c) Sc³⁺ is d⁰ → colorless (no d electrons for transition)
Q11. a) CFSE = (-0.4)(4) + (0.6)(2) Δ₀ = -0.4Δ₀ b) CFSE = (-0.4)(6) Δ₀ + 2P = -2.4Δ₀ + 2P c) Low spin is more stable if -2.4Δ₀ + 2P < -0.4Δ₀, i.e., if Δ₀ > P
Advanced Topic: Jahn-Teller Effect
The Problem
For d⁹ Cu²⁺ octahedral complexes, CFT predicts perfect octahedral geometry. But experiments show distortion!
Jahn-Teller Theorem
“Any non-linear molecular system in a degenerate electronic state will undergo geometric distortion to remove the degeneracy.”
Example: [Cu(H₂O)₆]²⁺
Cu²⁺ = d⁹:
e_g: ↑↓ ↑ ← Asymmetric filling causes instability
t₂g: ↑↓ ↑↓ ↑↓
Result: Complex distorts - usually elongation along z-axis
- 4 short Cu-O bonds (in xy plane)
- 2 long Cu-O bonds (along z)
This is beyond JEE syllabus but explains why Cu²⁺ complexes often appear square planar!
Related Topics
Within Coordination Compounds
- Bonding Theories — VBT vs CFT comparison
- Isomerism — Geometry affects isomerism
- Stability Constants — CFSE affects stability
- Applications — Color in biological systems
Cross-Chapter Connections
- Atomic Structure — d orbitals and electron configurations
- d-Block Elements — Colored compounds
- Chemical Bonding — Molecular orbital theory
- Light and Electromagnetic Radiation — Absorption and color
Interactive Demo: Crystal Field Theory Visualizer
Explore how d orbitals split in different crystal field environments. Change the metal ion, geometry, and ligand to see real-time changes in electron configuration, CFSE, and predicted color!