ChemistryCoordination compoundsClass 12

Coordination chemistry & crystal field: interactive simulation

Crystal Field Theory (CFT) treats ligands as point negative charges that split the five degenerate d-orbitals of a transition metal ion into groups of different energies depending on coordination geometry.

Useful for: JEE Main, JEE Advanced, NEET, CBSE Boards · 3 modes in one lab

What you can explore

  1. 1

    Octahedral & Tetrahedral Splitting

    Interactive energy level diagram showing t₂g, e_g and barycentre offsets.
  2. 2

    Spectrochemical Series & Spin

    Electron filling across d¹ to d⁸ with high/low spin magnetic moment calculation.
  3. 3

    d-d Absorption & Magnetism

    Absorbed wavelength vs observed complementary solution colors and BM magnetic moments.

Try this

Explore octahedral t₂g / e_g splitting, see how strong-field ligands force electron pairing to create low-spin diamagnetic complexes, and discover why [Ti(H₂O)₆]³⁺ appears purple while [Cu(H₂O)₆]²⁺ appears sky blue.

  1. 1

    Set d-electrons to 5 (d⁵ Fe³⁺) and switch from Weak to Strong ligand.

    What you should see

    In weak field (High spin), n=5 unpaired electrons give μ = 5.92 BM. In strong field (Low spin), 4 electrons pair in t₂g, leaving only n=1 unpaired electron with μ = 1.73 BM!

  2. 2

    Compare Octahedral vs Tetrahedral splitting.

    What you should see

    In octahedral, e_g is higher (+0.6 Δ_o); in tetrahedral, t₂ is higher (+0.4 Δ_t), and tetrahedral Δ_t is always too small to cause pairing (always high spin).

  3. 3

    In d-d Absorption, select [Ti(H₂O)₆]³⁺.

    What you should see

    It absorbs yellow-green light (λ ≈ 498 nm) to promote its single d-electron, transmitting complementary purple/violet light.

Formulas behind it

Octahedral Crystal Field Stabilization Energy (CFSE)

CFSE=(−0.4nt2g+0.6neg)Δo+mP\text{CFSE} = \left(-0.4 n_{t_{2g}} + 0.6 n_{e_g}\right) \Delta_o + m P

Tetrahedral vs Octahedral splitting relation

Δt=49Δo\Delta_t = \frac{4}{9} \Delta_o

Spin-only magnetic moment formula

μs=n(n+2) BM(n=unpaired electrons)\mu_s = \sqrt{n(n + 2)} \text{ BM} \quad (n = \text{unpaired electrons})

Absorption wavelength of d-d transition

Δo=hν=hcλ\Delta_o = h \nu = \frac{h c}{\lambda}

Common mistakes

  • Thinking tetrahedral complexes can form low-spin configurations — Δ_t is only 4/9 of Δ_o and virtually never exceeds pairing energy P.
  • Calculating CFSE without taking into account the barycentre: t₂g is stabilized by −0.4 Δ_o while e_g is destabilized by +0.6 Δ_o.
  • Confusing absorbed color with observed color — a complex absorbing red light appears cyan/blue.
Stuck on a coordination chemistry & crystal field question?Snap it and get a whiteboard video solution.

FAQ

What is the spectrochemical series?

An experimentally determined series ranking ligands in order of increasing field strength: I⁻ < Br⁻ < S²⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < CN⁻ < CO.

Why are d⁰ and d¹⁰ complexes colorless?

Because d-d transitions require an electron to jump between partially filled d-orbitals; d⁰ has no electrons, and d¹⁰ has no vacant d-orbitals.

What is a bidentate chelating ligand?

A ligand like ethylenediamine (en) or oxalate (ox²⁻) that binds through two donor atoms to form a stable ring structure (chelate effect).

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