Transition Metals: Color & Magnetism — JEE Main

The color and magnetic properties of transition metal complexes are among the most conceptually elegant topics in JEE Main Chemistry. The same framework — Crystal Field Theory — explains both phenomena, and JEE tests them in two to four questions per session. Once you understand why the d-orbital degeneracy breaks in a crystal field and how electrons fill the split levels, you can answer any color/magnetism question systematically.

Why Do Transition Metal Ions Have Color?

Transition metal ions with partially filled d-orbitals (d¹ to d⁹ configurations) can absorb visible light by exciting an electron from a lower d-level to a higher one. The energy of absorbed light = ΔE = hν = hc/λ. The colour we see is the complementary colour of the absorbed wavelength (colour wheel: red-green are complementary; blue-orange are complementary). Ti³⁺ ([Ti(H₂O)₆]³⁺, d¹) absorbs green-yellow light (λ ~500 nm) and appears violet-purple. Cu²⁺ ([Cu(H₂O)₆]²⁺, d⁹) absorbs red light and appears blue-green.

Conditions for colour: (1) partially filled d-orbitals (d¹–d⁹); (2) a ligand that causes d-orbital splitting. Exceptions: d⁰ (Sc³⁺, Ti⁴⁺) and d¹⁰ (Cu⁺, Zn²⁺) complexes are colourless because no d-d transition is possible. Charge transfer spectra (not d-d) give very intense colours in MnO₄⁻ (Mn is d⁰ but intensely purple — the colour arises from O→Mn charge transfer, not d-d). JEE tests: which ions are coloured (check d-count), and why MnO₄⁻ is coloured despite Mn(VII) being d⁰. Take a free coordination chemistry mock. See our coordination chemistry guide.

Crystal Field Theory: Splitting the d-Orbitals

In an octahedral field (6 ligands), the five d-orbitals split into two sets: t₂g (d_xy, d_xz, d_yz — pointing between ligands, lower energy) and e_g (d_z², d_x²-y² — pointing toward ligands, higher energy, destabilised). Energy gap = Δ_o (crystal field splitting energy, also called 10 Dq). Crystal Field Stabilisation Energy (CFSE) = (number of electrons in t₂g × (−0.4Δ_o)) + (number of electrons in e_g × (+0.6Δ_o)). For high-spin d⁶: t₂g⁴ e_g² → CFSE = 4(−0.4) + 2(0.6) = −0.4Δ_o. For low-spin d⁶: t₂g⁶ e_g⁰ → CFSE = 6(−0.4) = −2.4Δ_o (but add pairing energy penalty). JEE asks you to calculate CFSE for a given d configuration in octahedral or tetrahedral field.

Magnetic Properties: Spin-Only Magnetic Moment

The spin-only magnetic moment formula: μ = √(n(n+2)) BM where n = number of unpaired electrons, BM = Bohr magneton. This is the most tested formula in the entire Coordination Chemistry chapter. Steps: (1) determine the metal ion's oxidation state; (2) find the d-electron count; (3) draw the high-spin or low-spin d-orbital filling; (4) count unpaired electrons; (5) apply μ = √(n(n+2)). For high-spin Fe³⁺ (d⁵): n = 5 → μ = √35 ≈ 5.92 BM. For low-spin Fe³⁺ (d⁵ with strong field ligand): n = 1 → μ = √3 ≈ 1.73 BM. Paramagnetic (n > 0) vs diamagnetic (n = 0).

Strong vs weak field ligands: spectrochemical series (increasing Δ_o): I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en (ethylenediamine) < CN⁻ < CO. Strong field ligands (CO, CN⁻, en) cause large Δ_o → low-spin complexes (electrons pair up in t₂g). Weak field ligands (halides, H₂O) cause small Δ_o → high-spin complexes. JEE question: [Fe(CN)₆]⁴⁻ has μ ≈ 0 BM (low spin, all paired); [Fe(H₂O)₆]²⁺ has μ ≈ 4.9 BM (high spin, 4 unpaired). For the full d-block and f-block coverage see our d and f block guide and for coordination isomers see our coordination isomers guide.