Identify each of the following coordination complexes as either diamagnetic or paramagnetic: (a) \(\left[\operatorname{CoBr}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+}\) (b) \(\left[\mathrm{W}(\mathrm{CN})_{6}\right]^{3-}\) (c) \(\left[\mathrm{VF}_{6}\right]^{3-}\) (d) \(\left[\mathrm{Rh}(\mathrm{o}-\mathrm{phen})_{3}\right]^{3+}\)

To determine the electronic structure of the metal ions in each complex, we can use the coordination number and metal ion's oxidation state, as well as the crystal field theory. The crystal field theory helps us predict whether the d orbital electrons will be paired or unpaired. We can then use this information to determine whether the complex is diamagnetic or paramagnetic.

For this complex, we see that the metal center is Co (cobalt). The coordination number is 6, with one bromide ligand and five ammonia ligands. In order to determine its oxidation state, we also need to consider the charge of the complex. The overall charge is +2, and the charge of the ligands is -1 (Br-) + 0 (NH3) = -1. Thus, the oxidation state of Co is +3. Cobalt has an atomic number of 27, so its electron configuration is \([Ar] 3d^7 4s^2\). In the +3 oxidation state, its configuration will now be \([Ar] 3d^6\). In a high-spin octahedral complex, the electrons are expected to fill the \(t_{2g}\) and \(e_g\) orbitals as follows: \(\uparrow \downarrow\) in \(t_{2g}\) and then one electron each for the remaining \(e_g\) orbitals, leading to four unpaired electrons. Thus, this complex will be paramagnetic.

In this complex, the metal center is W (tungsten), and the coordination number is 6. The overall charge of the complex is -3. Since the ligands all have a charge of -1 (CN-), the sum of the ligand charges is -6, so the oxidation state of W is +3. Tungsten has an atomic number of 74, so its electron configuration is \([Xe] 4f^{14} 5d^4 6s^2\). In the +3 oxidation state, its configuration will now be \([Xe] 4f^{14} 5d^3\). The strong-field ligand (CN-) causes the electrons to pair in the \(t_{2g}\) orbitals, leaving no unpaired electrons. Thus, this complex is diamagnetic.

In this complex, the metal center is V (vanadium), and the coordination number is 6. The overall charge of the complex is -3. Since the ligands all have a charge of -1 (F-), the sum of the ligand charges is -6, so the oxidation state of V is +3. Vanadium has an atomic number of 23, so its electron configuration is \([Ar] 3d^3 4s^2\). In the +3 oxidation state, its configuration will now be \([Ar] 3d^2\). The weak-field ligand (F-) allows electrons to fill the \(t_{2g}\) and \(e_g\) orbitals in a high-spin configuration, as follows: two unpaired electrons in the \(t_{2g}\) orbitals. Thus, this complex is paramagnetic.

In this complex, the metal center is Rh (rhodium), and the coordination number is 6. The overall charge of the complex is +3. Since the ligands are neutral (o-phen), the oxidation state of Rh is also +3. Rhodium has an atomic number of 45, so its electron configuration is \([Kr] 4d^8 5s^1\). In the +3 oxidation state, its configuration will now be \([Kr] 4d^6\). The strong-field ligand (o-phen) causes the electrons to pair in the \(t_{2g}\) orbitals, leaving no unpaired electrons. Thus, this complex is diamagnetic.

We have determined the following: (a) \(\left[\operatorname{CoBr}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+}\): Paramagnetic (b) \(\left[\mathrm{W}(\mathrm{CN})_{6}\right]^{3-}\): Diamagnetic (c) \(\left[\mathrm{VF}_{6}\right]^{3-}\): Paramagnetic (d) \(\left[\mathrm{Rh}(\mathrm{o}-\mathrm{phen})_{3}\right]^{3+}\): Diamagnetic