Why is a ruby red, an emerald green, and a beaker of copper sulfate sky-blue? Each one is a metal atom wearing a "cage" of neighbouring atoms (called ligands), and that cage is what tints the colour. Picture five identical chairs around a table — all equally comfy. Now shove the table against a wall: some chairs are now squashed against it, others still have legroom. The chairs haven't changed, but their comfort has split into "good" and "cramped." A metal's five $d$-orbitals do exactly this when ligands crowd in around them.
The squashing isn't random. Orbitals whose lobes point straight at the incoming ligands get repelled hardest and rise in energy; orbitals that point between the ligands stay lower. The gap between the lowered set and the raised set is the crystal-field splitting $\Delta$. Its size sets the colour: the complex swallows a photon whose energy matches $\Delta$, and you see whatever light is left over. For violet $[\mathrm{Ti(H_2O)_6}]^{3+}$, $\Delta_o \approx 20{,}300\ \mathrm{cm^{-1}}$, so it absorbs at $\lambda = 10^7/20{,}300 \approx 493\ \mathrm{nm}$ (blue-green) — and the eye reads the leftover light as purple.
In an octahedron the two "point-at-ligand" orbitals ($d_{z^2},\ d_{x^2-y^2}$) form the upper $e_g$ set, sitting $+0.6\,\Delta_o$ above centre; the three "point-between" orbitals ($d_{xy},\ d_{xz},\ d_{yz}$) form the lower $t_{2g}$ set at $-0.4\,\Delta_o$. How the electrons arrange themselves is a tug-of-war: spreading out into the upper set costs $\Delta$, while doubling up two electrons in one orbital costs the pairing energy $P$. When $\Delta < P$ the electrons stay spread out — high-spin, many unpaired electrons; when $\Delta > P$ they pair up low down first — low-spin. The "$d$-electron count," "$\Delta_o$," and "$P$" sliders are exactly these three numbers, and the readouts turn them into spin state, CFSE, and the magnetic moment $\mu_{\text{eff}} = \sqrt{n_u(n_u+2)}$.
① Load the preset $[\mathrm{Fe(CN)_6}]^{4-}$, then $[\mathrm{Fe(H_2O)_6}]^{2+}$. Both are $d^6$, but the strong-field $\mathrm{CN^-}$ pushes $\Delta_o > P$ (low-spin, 0 unpaired) while weak-field water leaves it high-spin (4 unpaired) — same metal, different colour and magnetism.
② In Octahedral mode set $d^5$ and slowly drag $\Delta_o$ up past $P$: watch the spin state flip from High to Low and $\mu_{\text{eff}}$ collapse from $\approx 5.9$ to $\approx 1.7\ \mu_B$.
③ Switch to the Tetrahedral tab. The gap shrinks to about $\tfrac{4}{9}\Delta_o$ — too small to beat $P$ — which is why almost every tetrahedral complex is high-spin.
| Symbol | Meaning | Unit |
|---|---|---|
| Δ_o | Octahedral splitting between t2g and eg | cm⁻¹ (or kJ/mol) |
| Δ_t | Tetrahedral splitting (e ↔ t2); $\Delta_t \approx \tfrac{4}{9}\Delta_o$ | cm⁻¹ |
| nt2g, neg | Number of d-electrons in the lower / upper set | — |
| P | Mean spin-pairing energy (Coulomb + exchange) | cm⁻¹ |
| m | Number of extra electron pairs vs the free ion | — |
| μeff | Effective magnetic moment (spin-only) | μB (Bohr magnetons) |
| S | Total electron spin = (unpaired e⁻)/2 | — |
| 10 Dq | Equivalent symbol for Δ_o in older literature | cm⁻¹ |
| Simulation | Equation |
|---|---|
| "d-electron count" slider | $n = n_{t_{2g}} + n_{e_g}$ |
| "Δ_o ligand-field strength" slider | $\Delta_o$ in cm⁻¹ |
| "Pairing energy P" slider | $P$ in cm⁻¹ |
| "Spin state" readout | Compare $\Delta_o$ vs $P$: HS if $\Delta_o |
| "CFSE" readout | Direct evaluation of the boxed equation |
| "μ_eff" readout | $\mu_{\text{eff}} = \sqrt{n_u(n_u+2)}$ where $n_u$ = unpaired e⁻ |
| 3D orbital lobes (Three.js) | Real $|d_{z^2}|^2$, $|d_{x^2-y^2}|^2$, $|d_{xy}|^2$ angular shapes |
| "d-d absorption" graph | $\lambda = hc/\Delta_o$ → photon wavelength absorbed |
Ti³⁺ has electron configuration [Ar]3d¹. With six H₂O ligands in Oh geometry:
• Measured $\Delta_o = 20\,300\;\text{cm}^{-1}$ (from absorption spectrum at 493 nm).
• d¹ configuration: one electron in t2g, $n_{t_{2g}}=1$, $n_{e_g}=0$, no pairing change.
$$ \text{CFSE} = (-0.4)(1)(20\,300\;\text{cm}^{-1}) = -8120\;\text{cm}^{-1} = -97.1\;\text{kJ/mol} $$
Conversion: 1 cm⁻¹ = 11.96 J/mol = 0.01196 kJ/mol. So $-8120 \times 0.01196 \approx -97.1$ kJ/mol of stabilization.
The complex absorbs at $\lambda = 10^7/20\,300 \approx 493\;\text{nm}$ (blue-green) → transmitted light is purple. Hence Ti(H₂O)6³⁺ solutions are violet. ✔ confirmed experimentally.