Colour and Magnetic Properties of Complexes
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Direct answer
Colour and magnetism in complexes flow from one source: crystal-field splitting of the d orbitals. In an octahedral field the t2g set drops and eg rises by Δo; electrons pair only if Δo exceeds the pairing energy P. A photon spanning the t2g-to-eg gap paints the colour — the transmitted complementary hue — while unpaired electrons set the spin-only moment μ = √(n(n+2)) BM. Same metal, different ligand, different answers: [Fe(CN)6]3− keeps one unpaired electron (1.73 BM) while [FeF6]3− keeps five (5.92 BM), and [Ti(H2O)6]3+ wears violet from a single d electron's jump.
What you must remember
- Octahedral arithmetic: each t2g electron contributes −0.4 Δo and each eg electron +0.6 Δo; low-spin d6 (t2g6) collects −2.4 Δo, high-spin d6 (t2g4 eg2) only −0.4 Δo before pairing terms.
- Spectrochemical series to recite: I− < Br− < Cl− < F− < OH− < H2O < NH3 < en < NO2− < CN− < CO — iodide weakest, carbon monoxide strongest; the series' ends carry the marks.
- Spin-only moments: μ = √(n(n+2)) BM — one unpaired 1.73, two 2.83, three 3.87, four 4.90, five 5.92; pair them with [Fe(CN)6]3−, tetrahedral [NiCl4]2− (2.83), and [Fe(H2O)6]3+ (5.92).
- Colour logic: absorb a d-d photon, transmit the complement — [Ti(H2O)6]3+ absorbs green-yellow and shows violet, [Cu(H2O)6]2+ absorbs red and shows blue; a stronger field shifts absorption to shorter wavelength.
- Intensity caveat: high-spin d5 (Mn2+ aqua) transitions are spin-forbidden and pale; charge-transfer bands need no d electrons at all — MnO4− is intensely purple with d0 manganese(VII).
- Tetrahedral amendment: Δt = (4/9)Δo, far below pairing energy — tetrahedral complexes are essentially always high spin and more intensely coloured.
- Geometry switch example: square-planar d8 [Ni(CN)4]2− is diamagnetic while tetrahedral d8 [NiCl4]2− carries two unpaired electrons — same formula count, opposite magnetism.
- Pairing boundary: d4-d7 octahedral complexes exist in both spin states; d1-d3 and d8-d10 are spin-unambiguous whatever the ligand.
One ligand swap, two magnetic destinies
Take cobalt(III), d6. In [CoF6]3−, fluoride sits low on the spectrochemical series, so Δo is small; electrons spread as t2g4 eg2 with four unpaired — μ = √24 ≈ 4.90 BM and CFSE of (−0.4 × 4 + 0.6 × 2)Δo = −0.4 Δo with one electron pair. Replace fluoride with ammonia in [Co(NH3)6]3+ and Δo grows past the pairing energy: electrons fill t2g6 completely — zero unpaired, diamagnetic, CFSE −2.4 Δo bought at the cost of two extra pairs. The balance sheet explains the switch: low spin wins whenever −2.4 Δo outweighs the additional 2P. The same arithmetic answers colour: larger Δo for the ammine means the absorbed photon is more energetic, so the hue shifts between the two complexes even though the metal never changes. Run the parallel with iron(III), d5: cyanide's strong field gives t2g5 (one unpaired, 1.73 BM) while water's weak field gives all five unpaired (5.92 BM) — the exam's most repeated pairing.
Calculation slips in magnetism
JEE Main calculates: given the complex, find n and μ — the universal slip is counting unpaired electrons from the free-ion d count without consulting ligand field strength, which turns d5-with-cyanide into a five-unpaired error. JEE Advanced layers pairing energy into CFSE sums, asks why tetrahedral complexes shun low spin (Δt = 4/9 Δo never beats P), and separates d-d from charge-transfer colour using permanganate as the exhibit. Two more traps close the list: quoting a nonzero moment for a diamagnetic complex (the answer is zero, full stop), and reading geometry from the formula — the nickel pair above shows the same d8 count going both magnetic ways depending on shape. Coordination chemistry remains fully on both syllabi, and these two-property questions are its most reliable scoring segment.
Frequently asked questions
Calculate the spin-only moment of [Fe(CN)6]3−.
Iron(III) is d5; cyanide's strong field forces low-spin t2g5 with one unpaired electron, so μ = √(1 × 3) = 1.73 BM.
Why is permanganate coloured though Mn(VII) has no d electrons?
Its intense purple comes from ligand-to-metal charge-transfer transitions from oxygen 2p to manganese orbitals — a mechanism that needs no d electrons, unlike d-d colour.
Which order do ligands follow in the spectrochemical series?
I− < Br− < Cl− < F− < OH− < H2O < NH3 < en < NO2− < CN− < CO, from the weakest field (smallest Δ) to the strongest.
Why do tetrahedral complexes almost always stay high spin?
The tetrahedral splitting is only 4/9 of the octahedral value, too small to repay the pairing energy, so electrons stay unpaired.
What is the CFSE of a low-spin d6 octahedral complex?
t2g6 eg0 contributes −0.4 × 6 = −2.4 Δo, the largest octahedral stabilisation available — one reason low-spin Co(III) ammines are so robust and kinetically inert.