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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.

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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

For the same metal and ligand,
Δt=4/9Δo\Delta t = 4/9 \Delta o

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)
OlympIQ trap

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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