During the successive ionization of first-row (3d) transition metals to form stable cations, which s — d and f Block Elements Chemistry Question

💡 Solution & Explanation

While $4s$ fills before $3d$ (Aufbau), once the electrons are in place, the effective nuclear charge lowers the energy of the $3d$ orbitals below the $4s$. Thus, the outermost $4s$ electrons are always removed first during ionization.