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(H2O)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 d8 – d10 complexes, the electrons occupy the orbitals in accord with Hund’s rule so as to give the maximum number of unpaired electrons. For d4 – d7 complexes, the orbital occupancy and number of unpaired electrons depend on the position of the ligand in the spectrochemical series.
(a) Cr3+ ([Ar] 3d3) has three unpaired electrons. In the complex, they occupy the lower-energy set of d orbitals as shown below.
(b) Mn3+ ([Ar] 3d4) 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) Co2+ ([Ar] 3d7) has a high-spin configuration with three unpaired electrons because H2O 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 (H2O < en < CN−):