For each of the following metals, write the electronic configu- ration of the atom and its \(2+\) ion: \((\) a) \(M n,\) (b) \(R u,\) (c) \(R h\). 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, we must determine the electron configurations of the given metals and their 2+ ions. We can find the atomic numbers of these metal atoms from the periodic table: Mn (25), Ru (44), and Rh (45). For Mn, the electron configuration is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5\] For the Mn2+ ion, we remove two electrons, which results in the following electron configuration: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^0\] For Ru, the electron configuration is: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 5s^2 4d^6\] For the Ru2+ ion, we remove two electrons, which results in the following electron configuration: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^0 4p^6 5s^0 4d^6\] For Rh, the electron configuration is: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 5s^2 4d^7\] For the Rh2+ ion, we remove two electrons, which results in the following electron configuration: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^0 4p^6 5s^0 4d^7\]

Now, we will draw the crystal-field energy-level diagram for the d orbitals of an octahedral complex, which consists of two sets of d orbitals: the \(t_{2g}\) (lower energy) and the \(e_g\) (higher energy). The splitting between the two sets of orbitals is called the crystal-field splitting energy \(\Delta\).

Next, we will show the placement of the d electrons for each 2+ ion, assuming a strong-field complex (where the electrons fill the lower-energy orbitals before occupying the higher-energy orbitals). For Mn2+ (3d5), the electron configuration in an octahedral complex is: \[t_{2g}^3 e_g^2\] For Ru2+ (4d6), the electron configuration in an octahedral complex is: \[t_{2g}^4 e_g^2\] For Rh2+ (4d7), the electron configuration in an octahedral complex is: \[t_{2g}^5 e_g^2\]

Finally, we will calculate the number of unpaired electrons for each metal ion in the octahedral complex. Mn2+ has 3 unpaired electrons, as the t2g orbitals are not completely filled: \[t_{2g}^3 (↑, ↑, ↑)\] Ru2+ has 2 unpaired electrons, as one of the t2g orbitals is completely filled while the other two have unpaired electrons: \[t_{2g}^4 (↑↓, ↑, ↑)\] Rh2+ has 1 unpaired electron, as two of the t2g orbitals are completely filled and one has an unpaired electron: \[t_{2g}^5 (↑↓, ↑↓, ↑)\] In conclusion, Mn2+ has 3 unpaired electrons, Ru2+ has 2 unpaired electrons, and Rh2+ has 1 unpaired electron in their octahedral complexes.