Question

Give the number of unpaired electrons in octahedral complexes with strong- field ligands for (a) \(\mathrm{Rh}^{3+}\) (b) \(\mathrm{Mn}^{3+}\) (c) \(\mathrm{Ag}^{+}\) (d) \(\mathrm{Pt}^{4+}\) (e) \(\mathrm{Au}^{3+}\)

Short answer

Answer: (a) 0 unpaired electrons, (b) 0 unpaired electrons, (c) 1 unpaired electron, (d) 0 unpaired electrons, and (e) 1 unpaired electron.

Step-by-step solution

(1. Determine electron configurations)

(First, we need to find the electron configurations for each metal ion. To do this, we will use their oxidation states given in the exercise and take away that many electrons from the appropriate d and s orbitals.) (a) Rh^3+: Rhodium has an atomic number of 45. Its electron configuration is [Kr] 4d^8 5s^1. Removing 3 electrons from Rh gives [Kr] 4d^6. (b) Mn^3+: Manganese has an atomic number of 25. Its electron configuration is [Ar] 3d^5 4s^2. Removing 3 electrons from Mn gives [Ar] 3d^4. (c) Ag^+: Silver has an atomic number of 47. Its electron configuration is [Kr] 4d^10 5s^1. Removing 1 electron from Ag gives [Kr] 4d^9. (d) Pt^4+: Platinum has an atomic number of 78. Its electron configuration is [Xe] 4f^14 5d^9 6s^1. Removing 4 electrons from Pt gives [Xe] 4f^14 5d^6. (e) Au^3+: Gold has an atomic number of 79. Its electron configuration is [Xe] 4f^14 5d^10 6s^1. Removing 3 electrons from Au gives [Xe] 4f^14 5d^7.

(2. Determine unpaired electrons in octahedral complexes with strong-field ligands)

(To count the number of unpaired electrons in these octahedral complexes, we need to understand how the d-orbitals split in an octahedral crystal field and the order in which they fill for strong-field ligands. The d-orbitals split into two sets: lower-energy t2g and higher-energy eg. For strong-field ligands, we will fill the t2g orbitals first and then the eg orbitals.) (a) Rh^3+: [Kr] 4d^6. There are enough electrons (6) to fill the t2g set completely, leaving 0 unpaired electrons. (b) Mn^3+: [Ar] 3d^4. There are 4 electrons, which can all go into the t2g set, leaving 0 unpaired electrons. (c) Ag^+: [Kr] 4d^9. There are enough electrons (9) to fill the t2g set completely and then place one in each of the eg orbitals, leaving 1 unpaired electron. (d) Pt^4+: [Xe] 4f^14 5d^6. There are enough electrons (6) to fill the t2g set completely, leaving 0 unpaired electrons. (e) Au^3+: [Xe] 4f^14 5d^7. There are enough electrons (7) to fill the t2g set completely and then place one in one of the eg orbitals, leaving 1 unpaired electron.

(3. Summarize the results)

(Finally, let's summarize the number of unpaired electrons for each of the metal ion complexes.) (a) Rh^3+: 0 unpaired electrons (b) Mn^3+: 0 unpaired electrons (c) Ag^+: 1 unpaired electron (d) Pt^4+: 0 unpaired electrons (e) Au^3+: 1 unpaired electron