Why do transition elements show variable oxidation states? Explain the trend across the 3d series and how fluorine and oxygen stabilise higher oxidation states.

1. 1 · Give the reason. (n−1)d and ns electrons have similar energies, so states differ by one unit.
2. 2 · Trend across the series. Highest state = 3d + 4s electrons up to Mn (+7); most states in the middle; fewer at the ends (Sc +3 only, Zn +2 only).
3. 3 · Fluorides. F stabilises high states by lattice energy (CoF₃) or bond enthalpy (VF₅, CrF₆).
4. 4 · Oxides. Oxygen does even better because it forms multiple bonds: Mn₂O₇ (+7) but only MnF₄ (+4).
5. 5 · Copper as an example.
   • 2Cu²⁺ + 4I⁻ → Cu₂I₂ + I₂ (no CuI₂).
   • 2Cu⁺(aq) → Cu²⁺(aq) + Cu(s): Cu²⁺ is stabilised by its large hydration enthalpy.
6. 6 · Close in one line. So stability of an oxidation state depends on the electron configuration and on the atom bonded to the metal.

Scoring tip: Quote one pair of formulas (Mn₂O₇ and MnF₄); a named example earns more than a general statement.