Question 20.11: Drawing Crystal Field Energy-Level Diagrams for Octahedral C......
Drawing Crystal Field Energy-Level Diagrams for Octahedral Complexes
Draw a crystal field orbital energy-level diagram, and predict the number of unpaired electrons for each of the following complexes:
(a) [Cr(en)_3]^{3+}
(b) [Mn(CN)_6]^{3-}
(c) [Co(H_2O)_6]^{2+}
STRATEGY
All three complexes are octahedral, so the energy-level diagrams will show three lower-energy and two higher-energy d orbitals. For d¹ -d³ and d^8 -d^{10} complexes, the electrons occupy the orbitals in accord with Hund’s rule so as to give the maximum number of unpaired electrons. For d^4 -d^7 complexes, the orbital occupancy and number of unpaired electrons depend on the position of the ligand in the spectrochemical series.
(a) Cr^{3+} ([Ar] 3d³ ) has three unpaired electrons. In the complex, they occupy the lowerenergy set of d orbitals as shown below.
(b) Mn^{3+} ([Ar] 3d^4 ) can have a high-spin or a low-spin configuration. Because CN^- is a strong-field ligand, all four d electrons go into the lower-energy d orbitals. The complex is low-spin, with two unpaired electrons.
(c) Co^{2+} ([Ar] 3d^7 ) has a high-spin configuration with three unpaired electrons because H_2O is a weak-field ligand.
In the following orbital energy-level diagrams, the relative values of the crystal field splitting Δ agree with the positions of the ligands in the spectrochemical series
(H_2O < en < CN^-):
