The complex ion \(\operatorname{Ru}(\text { phen })_{3}^{2+}\) has been used as a probe for the structure of DNA. (Phen is a bidentate ligand.) a. What type of isomerism is found in $\operatorname{Ru}(\text { phen })_{3}^{2+} ?$ b. \(\operatorname{Ru}(\text { phen })_{3}^{2+}\) is diamagnetic (as are all complex ions of \(\mathrm{Ru}^{2+} \)). Draw the crystal field diagram for the \(d\) orbitals in this complex ion.

The given complex ion is \(\operatorname{Ru}(\text { phen })_{3}^{2+}\). The central metal ion is Ruthenium (Ru) and the complex ion has a 2+ charge. In order to determine the oxidation state of Ruthenium, we need to deduce the charges on the phen ligands. Since phen is a neutral ligand (with no charge), the oxidation state of Ru is +2.

The coordination sphere includes the metal ion (Ru) and the phen ligands. Each phen ligand is a bidentate ligand, meaning it can form two coordinate bonds with the central metal ion (Ru). Since there are three phen ligands and each can form two coordinate bonds, there are six coordination sites occupied in this complex. All six positions are equivalent, so we have an octahedral coordination complex. In an octahedral complex, the possible types of isomerism are geometrical (cis-trans) and optical isomerism. However, in this case, since the three ligands are all phen, there is no difference in the positions of the ligands. Thus, geometrical isomerism is not possible. Optical isomerism occurs when a complex has non-superimposable mirror images. However, all three phen ligands are equivalent and arranged symmetrically in this complex. Therefore, optical isomerism is not possible either. Thus, no type of isomerism is found in \(\operatorname{Ru}(\text { phen })_{3}^{2+}\).

In an octahedral crystal field, the d orbitals split into two groups: the \(\mathrm{t_{2g}}\) set (lower energy) containing the \(d_{xy}\), \(d_{xz}\), and \(d_{yz}\) orbitals, and the \(\mathrm{e_{g}}\) set (higher energy) containing the \(d_{x^{2}-y^{2}}\) and \(d_{z^{2}}\) orbitals. Since the \(\mathrm{Ru}^{2+}\) ion is diamagnetic and has a +2 oxidation state, it has four d electrons (Ru has atomic number 44 and has 4d5 5s1 electron configuration in the ground state). These four d electrons will fill the lower-energy \(\mathrm{t_{2g}}\) set completely, with no unpaired electrons. The crystal field diagram for the d orbitals in the \(\operatorname{Ru}(\text { phen })_{3}^{2+}\) complex ion should show a fully filled \(\mathrm{t_{2g}}\) set ( \(d_{xy}\), \(d_{xz}\), and \(d_{yz}\) orbitals with two electrons each) and an empty \(\mathrm{e_{g}}\) set (\(d_{x^{2}-y^{2}}\) and \(d_{z^{2}}\) orbitals with no electrons).