VBT, Crystal Field Theory and Colour

• Apply VBT to predict hybridisation and geometry of complexes
• Distinguish inner orbital (low spin) from outer orbital (high spin) using d-orbital availability
• Understand crystal field splitting of d-orbitals in octahedral and tetrahedral fields
• Apply the spectrochemical series to predict spin state from ligand type
• Calculate Crystal Field Stabilisation Energy (CFSE)
• Explain colour of complexes as result of d-d transitions

Level 1 — Foundations

Valence Bond Theory (VBT)
Central metal provides empty hybridised orbitals; ligands donate electron pairs into them.

Crystal Field Theory (CFT) — Key Idea
Ligands are treated as point charges/dipoles. Their approach breaks the degeneracy of the five d-orbitals. In an octahedral field, d-orbitals split into two sets:

• t₂g (d_xy, d_xz, d_yz) — lower energy, each lowers energy by 0.4Δ_o
• e_g (d_z², d_x²-y²) — higher energy, each raises energy by 0.6Δ_o

Level 2 — JEE Depth

VBT — Inner vs Outer Orbital Complexes

Inner orbital (low spin): Uses inner (n−1)d orbitals for hybridisation → d²sp³

• Requires pairing of d-electrons to free up inner d-orbitals
• Formed with strong field ligands (CN⁻, CO, NO₂⁻, en, NH₃)
• Fewer unpaired electrons → lower magnetic moment
• Example: [Co(NH₃)₆]³⁺ — Co³⁺ is d⁶, strong field forces all 6 into t₂g with pairing → d²sp³

Outer orbital (high spin): Uses outer (n)d orbitals → sp³d²

• No need to pair electrons; uses 4s, 4p, 4d
• Formed with weak field ligands (F⁻, Cl⁻, Br⁻, I⁻, H₂O)
• More unpaired electrons → higher magnetic moment
• Example: [CoF₆]³⁻ — Co³⁺ d⁶ with weak field F⁻ → sp³d² → 4 unpaired electrons

Limitation of VBT: Cannot explain colour, magnetic properties quantitatively, or why some ligands are strong field. CFT does this better.

Crystal Field Splitting in Octahedral Field

Energy stabilisation for each t₂g electron: −0.4Δ_o
Energy destabilisation for each e_g electron: +0.6Δ_o

CFSE = (number of t₂g electrons × −0.4Δ_o) + (number of e_g electrons × +0.6Δ_o)

For d⁶ high spin (weak field): t₂g⁴ e_g²
CFSE = 4(−0.4) + 2(+0.6) = −1.6 + 1.2 = −0.4Δ_o

For d⁶ low spin (strong field): t₂g⁶ e_g⁰
CFSE = 6(−0.4) + 0 = −2.4Δ_o (more stable)
But also add pairing energy P: net stabilisation = −2.4Δ_o + P × (extra pairs)

Spectrochemical Series (weak to strong field ligands)
I⁻ < Br⁻ < S²⁻ < SCN⁻ < Cl⁻ < NO₃⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < NH₃ < en < bipy < phen < NO₂⁻ < CN⁻ < CO

Strong field (right side) → large Δ_o → electrons pair up (low spin)
Weak field (left side) → small Δ_o → electrons spread out (high spin)

Colour in Coordination Compounds
d-d transitions: electron absorbs a photon of energy = Δ_o and jumps from t₂g to e_g
The colour we see = complementary colour to what is absorbed

d⁰ and d¹⁰ complexes have no possible d-d transition → usually colourless
([TiCl₄] is colourless: Ti⁴⁺ is d⁰; [Zn(NH₃)₄]²⁺ is colourless: Zn²⁺ is d¹⁰)

Magnetic Moment Formula
μ = √(n(n+2)) BM, where n = number of unpaired electrons

• n=0: μ=0 (diamagnetic)
• n=1: μ=1.73 BM
• n=2: μ=2.83 BM
• n=3: μ=3.87 BM
• n=4: μ=4.90 BM
• n=5: μ=5.92 BM

JEE Traps

• [Fe(CN)₆]⁴⁻: Fe²⁺ is d⁶, CN⁻ is strong field → low spin, t₂g⁶ e_g⁰, 0 unpaired, diamagnetic
• [FeF₆]³⁻: Fe³⁺ is d⁵, F⁻ is weak field → high spin, t₂g³ e_g², 5 unpaired
• Δ_t (tetrahedral) ≈ 4/9 Δ_o → nearly all tetrahedral complexes are high spin
• Colour requires BOTH an absorbing electron AND a vacancy to accept it → d⁰ and d¹⁰ are colourless

Example 1: Spin State and Magnetic Moment

Compare [Fe(CN)₆]⁴⁻ and [FeF₆]³⁻

[Fe(CN)₆]⁴⁻:
  Fe oxidation state: let x + 6(−1) = −4 → x = +2 → Fe²⁺
  Fe²⁺: electron configuration [Ar]3d⁶ → 6 d-electrons
  CN⁻: strong field ligand (right side of spectrochemical series)
  → Large Δ_o → electrons pair up (low spin)
  → d-electron filling: t₂g⁶ e_g⁰
  → n (unpaired) = 0
  → μ = √(0×2) = 0 BM (diamagnetic)
  → Hybridisation: d²sp³ (inner orbital, uses 3d orbitals)

[FeF₆]³⁻:
  Fe oxidation state: let x + 6(−1) = −3 → x = +3 → Fe³⁺
  Fe³⁺: [Ar]3d⁵ → 5 d-electrons
  F⁻: weak field ligand (left side of spectrochemical series)
  → Small Δ_o → electrons remain unpaired (high spin)
  → d-electron filling: t₂g³ e_g² (maximum spreading, Hund's rule)
  → n (unpaired) = 5
  → μ = √(5×7) = √35 = 5.92 BM (paramagnetic)
  → Hybridisation: sp³d² (outer orbital)

Summary: CN⁻ (strong field) → low spin, diamagnetic Fe²⁺;  
         F⁻ (weak field) → high spin, paramagnetic Fe³⁺

Example 2: Colour of [Ti(H₂O)₆]³⁺ — Wavelength Absorbed

Given: [Ti(H₂O)₆]³⁺ is violet, Δ_o = 20,300 cm⁻¹
Find: wavelength of light absorbed

Step 1: Ti in [Ti(H₂O)₆]³⁺
  Ti³⁺: [Ar]3d¹ → 1 d-electron in t₂g level

Step 2: Energy of d-d transition
  Electron absorbs photon to jump from t₂g to e_g
  Energy = Δ_o = 20,300 cm⁻¹

Step 3: Convert to wavelength
  E (in cm⁻¹) = 1/λ (in cm), so λ = 1/20300 cm = 4.926×10⁻⁵ cm = 493 nm

Step 4: Identify colour absorbed and observed
  λ_absorbed = 493 nm → blue-green light absorbed
  Complementary colour to blue-green = violet/red-purple
  → Complex appears violet ✓

Answer: Wavelength absorbed = 493 nm (blue-green); complex appears violet
(This is why [Ti(H₂O)₆]³⁺ is violet — a d¹ system with exactly one possible d-d transition)

• Q1: [Ni(CN)₄]²⁻ is square planar. What is the hybridisation and number of unpaired electrons? (Ni²⁺ is d⁸)
• Q2: Calculate CFSE for a d³ ion in an octahedral field with Δ_o = 18,000 cm⁻¹.
• Q3: Which absorbs light of longer wavelength — a complex with large Δ_o or small Δ_o?
• Q4: Why is [Zn(NH₃)₄]²⁺ colourless despite NH₃ being a strong field ligand?
• Stretch: Q5: Two complexes [Co(NH₃)₆]³⁺ and [CoF₆]³⁻ have different colours. Using CFT, predict which absorbs higher energy photons. Calculate the ratio of wavelengths if Δ_o(NH₃) = 22,900 cm⁻¹ and Δ_o(F⁻) = 13,100 cm⁻¹.

NCERT Chapter 9 link: Sections 9.6 (VBT) and 9.7 (CFT) cover hybridisation of coordination compounds, crystal field splitting, spin states, CFSE, colour, and magnetic properties. NCERT examples include [Ti(H₂O)₆]³⁺ colour and magnetic moment calculations.

Exam connections: JEE Mains: predict hybridisation from geometry, calculate μ from unpaired electrons, identify colour absorbed vs observed. JEE Advanced: CFSE calculation for specific d^n configurations, comparison of complexes with same metal but different ligands (spin states), tetrahedral vs octahedral splitting ratio (4/9).

Study strategy: Build a d-electron filling table for d¹ through d¹⁰ under both strong and weak fields — this is the engine of all CFT MCQs. Practise reading the spectrochemical series left-to-right. For colour: "absorbed + complementary = white light" is the mental model.

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This material emphasises experiential "learning by doing", competency (apply/create/analyse), vocational exposure, critical thinking, and multidisciplinary connections. Designed to feed live worlds, AI Mentor (with memory), gamification, robotics, parent analytics, and future skills — not just exam prep.

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Open the Practice tab for aligned questions (easy/medium/hard + case-based) with full AI scaffolding.

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