Why do transition metals have more oxidation states than other elements? Give the highest oxidation states for scandium to copper.

Transition metals have electrons in their d orbitals, in addition to their outermost s orbitals. Because both the \(4s\) and \(3d\) orbitals are close in energy, electrons can be removed from either when forming ions, leading to multiple oxidation states. For example, losing one electron, typically from the \(4s\) orbital first, creates a +1 oxidation state. Further loss of electrons from the \(3d\) orbitals gives higher oxidation states.

Scandium (Sc) has the electronic configuration \([Ar] 3d^1 4s^2\), and can form Sc^3+ by losing all three of its valence electrons, so its highest oxidation state is +3. Titanium (Ti) has the configuration \([Ar] 3d^2 4s^2\), and can form Ti^4+ by losing all four valence electrons, so its highest oxidation state is +4. Vanadium (V) has \([Ar] 3d^3 4s^2\), and can form V^5+ by losing all five valence electrons, giving it a highest oxidation state of +5. Chromium (Cr) has \([Ar] 3d^5 4s^1\), and can lose all six valence electrons to form Cr^6+, so its highest oxidation state is +6. Manganese (Mn) has \([Ar] 3d^5 4s^2\), and can form Mn^7+ by losing all seven valence electrons, so its highest oxidation state is +7. From iron (Fe) to copper (Cu), the number of valence electrons decreases again, so iron can form Fe^3+, cobalt (Co) also can form Co^3+, nickel (Ni) can form Ni^2+, and copper can form Cu^2+ so their highest oxidation states are +3, +3, +2, +2 respectively.