Read the Spectrochemical Series Before Ranking Splitting
Crystal field splitting energy depends on three things: the ligand's position in the spectrochemical series, the geometry (octahedral vs tetrahedral), and the metal's row (3d/4d/5d) — learn to rank all three independently.
Chemistry questions first require the material, particles or reaction condition to be identified before a formula or rule is selected.
The spectrochemical series, memorised once, used everywhere
Weak to strong field, roughly: I⁻ < Br⁻ < Cl⁻ < F⁻ < H₂O < NH₃ < en < NO₂⁻ (N-bonded) < CN⁻ < CO. This single order answers 'which ligand splits d-orbitals more', 'which absorbs shorter-wavelength light' (bigger Δo means higher-energy absorption), and 'which is more likely to be low spin.'
Octahedral splitting is always bigger than tetrahedral, by a fixed ratio
tetrahedral splitting is always smaller because only 4 ligands approach (not 6) and none point directly at the d-orbital lobes. This is why tetrahedral complexes are essentially always high-spin regardless of ligand strength: Δt rarely exceeds the pairing energy. Choose B.
High spin vs low spin only matters for d4–d₇
The high-spin/low-spin distinction only creates a real choice for d₄ through d₇ configurations, where t₂g and eg can be filled in more than one way depending on whether Δo beats the pairing energy P. Rh³⁺, Pt²⁺ and Ru²⁺ — second/third-row metals — favour low spin because Δo is intrinsically larger for 4d/5d metals than 3d metals with the same ligand.
3d < 4d < 5d for the same ligand and geometry
Crystal field splitting energy increases down a group of transition metals: 3d < 4d < 5d, because larger d-orbitals overlap more effectively with ligand orbitals. This is a separate trend from the spectrochemical series (which ranks ligands, not metals) — both apply simultaneously.
From the series to a wavelength ranking: Ni(II) with NO₂⁻, NH₃ and H₂O
A complex absorbs light whose energy matches its splitting Δo, so a stronger-field ligand means a larger Δo, higher-energy absorption and a SHORTER wavelength. For the same metal ion (Ni²⁺, d₈) and the same octahedral geometry only the ligand changes, so the spectrochemical series NO₂⁻ > NH₃ > H₂O orders the splitting as [Ni(NO₂)₆]4- > [Ni(NH₃)₆]2+ > [Ni(H₂O)₆]2+. The absorbed wavelength runs the opposite way: [Ni(NO₂)₆]4- < [Ni(NH₃)₆]2+ < [Ni(H₂O)₆]2+. Write the series, rank Δo, then flip the order to get wavelength. The commonest slip is to rank wavelength in the same direction as ligand field strength.
OlympIQ exam lens
- ✓Memorise one spectrochemical series and apply it to splitting magnitude, colour, and spin-state questions alike.
- ✓Δt = 4/9 Δo for the same metal and ligand — tetrahedral splitting is always smaller.
- ✓Tetrahedral complexes are essentially always high spin, because Δt rarely exceeds pairing energy.
- ✓High spin vs low spin is only a live question for d4–d₇; d1–d₃ and d8–d₁₀ have only one possible arrangement.
- ✓Splitting energy rises 3d < 4d < 5d for the same ligand — heavier metals favour low spin more easily.
- ✓For the same metal and geometry, a stronger-field ligand gives a larger Δo and a shorter absorbed wavelength (for Ni(II): wavelength NO₂⁻ complex < NH₃ complex < H₂O complex)
Don't apply the spectrochemical series alone when geometry or metal row also differs between the options being compared — Δt is always smaller than Δo for the identical ligand, and 5d metals split more than 3d metals even with a weaker-field ligand.
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