Chapter 4 · Chemistry
Fluorine Fails, Oxygen Wins: Highest Oxidation States in Oxides and Fluorides
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A metal reaches its highest oxidation state only with oxygen or fluorine. Fluorine is the more electronegative, yet manganese's highest fluoride is MnF₄ (+4) while its oxide reaches +7 (Mn₂O₇, MnO₄⁻): one oxygen forms a multiple bond and counts two electrons, one fluorine counts only one. From Sc₂O₃ to Mn₂O₇ the highest oxide's oxidation state equals the group number.
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Key points
- The highest oxidation state of a metal appears only in its oxides and fluorides (small, highly electronegative partners).
- Fluorine: single bond, 1 electron per atom; Mn's highest fluoride is MnF₄ (+4), MnF₇ has never been made.
- Oxygen: multiple bond, 2 electrons per atom; only four O are needed: MnO₄⁻ and Mn₂O₇ (+7). Oxygen does not "pull harder".
- Highest oxide = group number: Sc₂O₃ +3, TiO₂ +4, V₂O₅ +5, CrO₃ +6, Mn₂O₇ +7. A state this high is a strong oxidising agent (KMnO₄).
Why oxygen goes further than fluorine
| Partner | Bond to Mn | Electrons per atom | Atoms for +7 | Known |
|---|---|---|---|---|
| Fluorine | single | 1 | 7 | MnF₄ (+4) only |
| Oxygen | multiple | 2 | 4 | MnO₄⁻, Mn₂O₇ (+7) |
In MnO₄⁻, 4 × 2 = 8 electrons: 7 from Mn and 1 from potassium, hence the 1− charge (x + 4(−2) = −1, x = +7). The oxidation state is a counting rule; there is no free Mn⁷⁺ ion.
Highest oxides of the 3d metals (NCERT Table 4.6)
| Group | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|
| Oxide | Sc₂O₃ | TiO₂ | V₂O₅ | CrO₃ | Mn₂O₇ |
| State | +3 | +4 | +5 | +6 | +7 |
Beyond Mn, no iron oxide above Fe₂O₃ is known. Mn₂O₇ is covalent: two MnO₄ tetrahedra sharing one oxygen.
Highest fluorides and the Cu⁺ case
- Highest halides: TiX₄ (+4), VF₅ (+5), CrF₆ (+6), MnF₄ (+4), FeX₃ and CoF₃ (+3). Mn reaches +7 with fluorine only in the oxofluoride MnO₃F.
- Fluorine stabilises high states by high lattice energy (CoF₃) or high bond enthalpy (VF₅, CrF₆).
- Cu⁺(aq) disproportionates: 2Cu⁺ → Cu²⁺ + Cu, because Cu²⁺'s much more negative hydration enthalpy outweighs the second ionisation enthalpy.
- Cu²⁺ oxidises iodide: 2Cu²⁺ + 4I⁻ → Cu₂I₂ + I₂, so CuI₂ is not known.
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