WAEC Chemistry 2014 Paper

Concept:For positive ions, electrons are removed from the outermost shells, starting with the highest energy level. For iron, electrons are lost first from the 4s orbital, then from the 3d orbital.Explanation:The neutral iron atom, $_{26}\mathrm{Fe}$, has the electron configuration:$1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^6$This is equivalent to $[\mathrm{Ar}]\,4s^2\,3d^6$.For $\mathrm{Fe}^{3+}$, three electrons must be removed.The first two electrons removed come from the 4s orbital, leaving $4s^0$.The third electron is removed from the 3d orbital, leaving $3d^5$.Thus, the electron configuration becomes:$1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^0\,3d^5$Which is written as:$[\mathrm{Ar}]\,4s^0\,3d^5$Answer:D. $[\mathrm{Ar}]\,4s^0\,3d^5$