Question 20.9: Draw a crystal field orbital energy-level diagram, and predi......
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_{2}O)_{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 d^{1} – d^{3} 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] 3d^{3}) has three unpaired electrons. In the complex, they occupy the lower-energy 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_{2}O 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_{2}O < en < CN^{-}):
