Part i): For atomic number 24 (chromium) the neat prediction \([Ar]3d^4 4s^2\) is not the real ground state. Nature prefers configurations where a subshell is exactly half filled, because that arrangement lowers the energy through maximum exchange stabilisation. Promoting one \(4s\) electron to \(3d\) makes both \(3d\) and \(4s\) half filled, so the actual configuration is \([Ar]3d^5 4s^1\).
Part ii): Start from neutral cobalt, \(Z = 27\), which is \([Ar]3d^7 4s^2\). Ionisation always strips the outermost \(4s\) electrons before the \(3d\), so removing two electrons for \(Co^{2+}\) empties the \(4s\) and leaves \([Ar]3d^7\). Distributing seven electrons over the five d orbitals singly first gives an arrangement \(\uparrow\downarrow\ \uparrow\downarrow\ \uparrow\ \uparrow\ \uparrow\), i.e. two paired orbitals and three singly occupied ones. So \(\boxed{3\ \text{unpaired electrons}}\).
Part iii): Compare the inner shells. In lanthanoids the \(4f\) orbitals are shielded and their electrons are held so firmly that they seldom join in bonding, so the metal is stuck near the +3 state. In actinoids the \(5f\) orbitals reach further out and lie almost level with the \(6d\) and \(7s\) orbitals; this small energy gap lets a variable number of \(5f, 6d, 7s\) electrons be used, producing a broad spread of oxidation states (up to +7). Therefore the actinoid range is far wider than the lanthanoid range.