Oxidation states and the stability of half-filled d
When a first-series atom becomes an ion, the two 4s electrons leave first, so +2 is found in every element except scandium, which settles at +3. Higher states also take 3d electrons. The highest state rises from +3 at Sc to +7 at Mn, where it equals the group number, and then falls: iron is usually +2 or +3, cobalt and nickel +2, copper +1 or +2, zinc only +2. Why so many states? The 4s and 3d electrons have similar energies, so the next electron costs only a little more than the last and the successive ionisation energies climb gradually. In sodium the 2nd electron comes from an inner shell and the energy jumps from about 500 to about 4560 kJ/mol, so sodium has only +1. Half-filled 3d matters too. Fe²⁺ is 3d⁶; losing one more electron gives Fe³⁺, 3d⁵, half-filled, so iron(II) is oxidised readily (in air iron(II) compounds slowly change to iron(III)). Mn²⁺ is already 3d⁵; making Mn³⁺ (3d⁴) would destroy the half-filled arrangement, so manganese(II) resists oxidation. Ions are formed by removing 4s electrons before 3d.
| Element | Common oxidation states | Ion that forms with +2 or +3 |
|---|---|---|
| Sc | +3 | Sc³⁺: no d electrons |
| Ti | +2, +3, +4 | Ti⁴⁺: no d electrons |
| V | +2 to +5 | V³⁺: 3d² |
| Cr | +2, +3, +6 | Cr³⁺: 3d³ |
| Mn | +2 to +7 | Mn²⁺: 3d⁵ |
| Fe | +2, +3 | Fe²⁺: 3d⁶, Fe³⁺: 3d⁵ |
| Co | +2, +3 | Co²⁺: 3d⁷ |
| Ni | +2 | Ni²⁺: 3d⁸ |
| Cu | +1, +2 | Cu⁺: 3d¹⁰, Cu²⁺: 3d⁹ |
| Zn | +2 | Zn²⁺: 3d¹⁰ |
| Change | d electrons before → after | Half-filled result? |
|---|---|---|
| Fe²⁺ → Fe³⁺ | 6 → 5 | yes |
| Co²⁺ → Co³⁺ | 7 → 6 | no |
Boxes and coins again. Draw five 3d boxes. For Fe²⁺ put 6 coins in them, one in each box and the sixth on top of one. Take one coin away from a doubled box: you now have the Fe³⁺ picture. What do you see? For Mn²⁺ put 5 coins, one in each box, and try to take one away: what is left? Write in a sentence which change keeps a tidy arrangement and which spoils it.