Oxidation states and the stability of higher oxidation states
Transition metals show many oxidation states differing by one, because d and s electrons are close in energy. Fluorine and oxygen stabilise the highest states.
Why transition metals show variable oxidation states
In a transition metal, the (n−1)d and ns orbitals have very similar energies. So the atom can lose its ns electrons and then one or more d electrons without a big jump in energy. This gives a range of oxidation states that usually differ by one unit, such as Fe²⁺ and Fe³⁺, or Cu⁺ and Cu²⁺. In the p-block, oxidation states usually differ by two units (for example, Sn²⁺ and Sn⁴⁺).
Oxidation states of the 3d elements (common ones in bold)
| Element | Oxidation states |
|---|---|
| Sc | +3 |
| Ti | +2, +3, +4 |
| V | +2, +3, +4, +5 |
| Cr | +2, +3, +4, +5, +6 |
| Mn | +2, +3, +4, +5, +6, +7 |
| Fe | +2, +3, +4, +6 |
| Co | +2, +3, +4 |
| Ni | +2, +3, +4 |
| Cu | +1, +2 |
| Zn | +2 |
Patterns in the table, with reasons
- Up to manganese, the highest state equals the total number of 3d and 4s electrons: Sc +3, Ti +4, V +5, Cr +6, Mn +7.
- The middle of the series (Mn) shows the most states. At the start, there are too few electrons to lose or share (Sc shows only +3). At the end, there are too many d electrons, so fewer empty orbitals are left for bonding (Zn shows only +2).
- After Mn, the higher states quickly become less stable: Fe shows mainly +2 and +3, Co +2 and +3, Ni +2.
- Going down a group, higher states become more stable: Mo(VI) and W(VI) are more stable than Cr(VI). So dichromate, Cr(VI), is a strong oxidising agent, but MoO₃ and WO₃ are not.
- Some complexes show zero oxidation state, such as Ni(CO)₄ and Fe(CO)₅.
Higher oxidation states in halides
Fluorine stabilises the highest states in halides, as in TiX₄, VF₅ and CrF₆. It does this either through a higher lattice energy (as in CoF₃) or through higher bond enthalpies (as in VF₅ and CrF₆). V(V) appears only as VF₅; the other V(V) halides are hydrolysed to oxohalides, VOX₃. Copper(II) forms all halides except the iodide, because Cu²⁺ oxidises iodide: 2Cu²⁺ + 4I⁻ → Cu₂I₂(s) + I₂. Many copper(I) compounds are unstable in water and disproportionate: 2Cu⁺(aq) → Cu²⁺(aq) + Cu(s). Cu²⁺(aq) is more stable than Cu⁺(aq) because the hydration enthalpy of Cu²⁺ is much more negative, more than making up for the second ionisation enthalpy of copper.
Oxygen vs fluorine in stabilising high oxidation states
Tetrahedral oxoanions such as VO₄³⁻, CrO₄²⁻ and MnO₄⁻ also hold the metal in its highest state.
Iron switches easily between Fe²⁺ and Fe³⁺. This one-step change lets iron in cytochromes pass electrons along during respiration, and is why iron supplements are usually given in the Fe²⁺ form.
2 marks: “Why do transition elements show variable oxidation states? Which 3d element shows the largest number, and why?” (n−1)d and ns electrons have similar energies, so different numbers can be lost. Mn (3d⁵ 4s²) shows +2 to +7 because it can use all seven electrons.
2 marks: “Cu⁺ is not stable in aqueous solution. Explain with an equation.” 2Cu⁺(aq) → Cu²⁺(aq) + Cu(s). The hydration enthalpy of Cu²⁺ is much more negative than that of Cu⁺, which more than makes up for the second ionisation enthalpy.
3 marks: “Give reasons: (i) The highest oxide of Mn is Mn₂O₇ but its highest fluoride is MnF₄. (ii) CuI₂ does not exist. (iii) Scandium shows only the +3 state.” (i) Oxygen forms multiple bonds with Mn. (ii) Cu²⁺ oxidises I⁻: 2Cu²⁺ + 4I⁻ → Cu₂I₂ + I₂. (iii) Losing all three (3d¹ 4s²) gives the stable noble-gas configuration Sc³⁺.
Middle metals do the most (Mn +2 to +7). Oxygen beats fluorine because it double-bonds: Mn₂O₇ but only MnF₄.