Magnetic properties, coloured ions and complex formation
Unpaired d electrons make ions paramagnetic: μ = √(n(n+2)) BM. d-d transitions absorb visible light, so most transition metal ions are coloured.
Magnetic properties
A substance placed in a magnetic field is either diamagnetic (weakly repelled) or paramagnetic (attracted). Paramagnetism comes from unpaired electrons: each one spins and moves around the nucleus, so it acts like a tiny magnet. Most transition metal ions have unpaired d electrons, so they are paramagnetic. Ferromagnetism, as in iron, is an extreme form of paramagnetism. For the 3d metals, the magnetic moment can be worked out from the spin alone: μ = √(n(n+2)) BM. The more unpaired electrons, the larger μ.
Spin-only magnetic moments
| Unpaired electrons (n) | μ = √(n(n+2)) BM | Example ions |
|---|---|---|
| 0 | 0 (diamagnetic) | Sc³⁺, Ti⁴⁺ (d⁰); Zn²⁺, Cu⁺ (d¹⁰) |
| 1 | √3 = 1.73 | Ti³⁺ (d¹), Cu²⁺ (d⁹) |
| 2 | √8 = 2.83 (the textbook table prints 2.84) | V³⁺ (d²), Ni²⁺ (d⁸) |
| 3 | √15 = 3.87 | Cr³⁺ (d³), Co²⁺ (d⁷) |
| 4 | √24 = 4.90 | Mn³⁺ (d⁴), Fe²⁺ (d⁶) |
| 5 | √35 = 5.92 | Mn²⁺, Fe³⁺ (d⁵) |
Worked example: spin-only magnetic moment
(a) Calculate the spin-only magnetic moment of Fe²⁺ (Z = 26). (b) A 3d ion has μ = 2.83 BM. How many unpaired electrons does it have?
- (a) Fe is [Ar] 3d⁶ 4s². Fe²⁺ loses the two 4s electrons: [Ar] 3d⁶.
- In five d orbitals, 6 electrons give one pair and 4 unpaired electrons: n = 4.
- μ = √(n(n+2)) = √(4 × 6) = √24 = 4.90 BM (√24 lies between 4.8² = 23.04 and 4.9² = 24.01).
- (b) √(n(n+2)) = 2.83, so n(n+2) = 2.83² ≈ 8.
- Try n = 2: 2 × 4 = 8. So n = 2.
(a) 4.90 BM; (b) 2 unpaired electrons
Why transition metal ions are coloured
In a transition metal ion surrounded by ligands (water molecules, in solution), the five d orbitals no longer have the same energy. An electron can absorb light and jump from a lower-energy d orbital to a higher-energy d orbital. This is a d-d transition. The energy needed matches light in the visible region. The ion absorbs some colours, and we see the complementary colour of the light absorbed. The exact colour depends on the ligand. Ions with d⁰ (Sc³⁺, Ti⁴⁺) or d¹⁰ (Zn²⁺, Cu⁺) cannot have a d-d transition, so they are colourless.
Colours of some 3d ions in water
| Ion | Configuration | Colour |
|---|---|---|
| Sc³⁺ | 3d⁰ | Colourless |
| Ti³⁺ | 3d¹ | Purple |
| V³⁺ | 3d² | Green |
| Cr³⁺ | 3d³ | Violet |
| Mn²⁺ | 3d⁵ | Pink |
| Fe²⁺ | 3d⁶ | Green |
| Fe³⁺ | 3d⁵ | Yellow |
| Co²⁺ | 3d⁷ | Pink |
| Ni²⁺ | 3d⁸ | Green |
| Cu²⁺ | 3d⁹ | Blue |
| Zn²⁺ | 3d¹⁰ | Colourless |
Formation of complex compounds
- In a complex compound, a central metal ion is bonded to several anions or neutral molecules called ligands, and the whole unit behaves as one species with its own properties.
- Examples: [Fe(CN)₆]³⁻, [Fe(CN)₆]⁴⁻, [Cu(NH₃)₄]²⁺ and [PtCl₄]²⁻.
- Transition metals form many complexes because their ions are small, carry a high charge, and have vacant d orbitals of suitable energy to accept lone pairs from ligands.
- The deep blue colour seen when ammonia is added to copper sulphate solution is the complex [Cu(NH₃)₄]²⁺.
The green of emeralds comes from traces of Cr³⁺, and the blue of copper sulphate (neela thotha), sprayed on grapevines as Bordeaux mixture, comes from Cu²⁺. Both colours come from d-d transitions.
2 marks: “Calculate the spin-only magnetic moment of Cr³⁺ (Z = 24).” Cr³⁺: [Ar] 3d³, n = 3. μ = √(3 × 5) = √15 = 3.87 BM.
2 marks: “Why are most transition metal ions coloured? Why is Ti⁴⁺ colourless?” They have partly filled d orbitals; d-d transitions absorb visible light and the complementary colour is seen. Ti⁴⁺ is 3d⁰, so no d-d transition is possible.
2 marks: “Give reasons why transition metals form a large number of complex compounds.” Small size of the ions, high ionic charge, and availability of vacant d orbitals to accept lone pairs from ligands.
No unpaired, no pull; no d-d, no colour. d⁰ and d¹⁰ are white and diamagnetic.