For each of the following metals, write the electronic configuration of the atom and its \(2+\) ion: $(\mathbf{a}) \mathrm{Mn},(\mathbf{b}) \mathrm{Ru},(\mathbf{c}) \mathrm{Rh} .$ Draw the crystal-field energy-level diagram for the \(d\) orbitals of an octahedral complex, and show the placement of the \(d\) electrons for each \(2+\) ion, assuming a strong-field complex. How many unpaired electrons are there in each case?

First, let's write the electronic configuration of each metal atom: a. Mn : \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{5}\) b. Ru : \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}5s^{2}4d^{7}\) c. Rh : \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}5s^{2}4d^{8}\) Now, we'll write the electronic configuration of the respective 2+ ions: a. Mn2+ : \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{0}3d^{5}\) b. Ru2+ : \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{0}4p^{6}5s^{0}4d^{6}\) c. Rh2+ : \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{0}4p^{6}5s^{0}4d^{7}\)

For an octahedral complex, the d orbitals split into two energy levels: \(e_g\) (the \(d_{z^2}\) and \(d_{x^2-y^2}\) orbitals) and \(t_{2g}\) (the \(d_{xy}\), \(d_{xz}\), and \(d_{yz}\) orbitals). The splitting between these levels is called the crystal-field splitting energy (\(\Delta\)).

In a strong-field complex, the crystal-field splitting energy is higher than the pairing energy, leading to the pairing of electrons in the lower energy orbitals before occupying the higher energy orbitals. a. Mn2+ : For Mn2+ ion, there are five d-electrons, they will fill the lower energy orbitals (\(t_{2g}\)) pairwise with all containing one unpaired electron. Unpaired electrons: 5 b. Ru2+ : For Ru2+ ion, there are six d-electrons. The first five will fill the \(t_{2g}\) orbitals, and the sixth electron will pair up in one of the \(t_{2g}\) orbitals. Unpaired electrons: 4 c. Rh2+ : For Rh2+ ion, there are seven d-electrons. The first five will fill the \(t_{2g}\) orbitals, and the two remaining electrons will fill up the \(e_g\) orbitals with one unpaired electron in each. Unpaired electrons: 5 To summarize, the number of unpaired electrons for Mn2+, Ru2+, and Rh2+ are 5, 4, and 5, respectively.