Explain in detail the Crystal Field Theory. Using
examples where applicable, how does the theory
explain the following for a transition metal complex;
1.The structure.
2.Colours behavior
3.Compound formation
4.Magnetic properties
Prepare and submit a one-page rep
For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
Given: complexes
Asked for:Â structure, high spin versus low spin, and the number of unpaired electrons
Strategy:
Solution
B The fluoride ion is a small anion with a concentrated negative charge, but compared with ligands with localized lone pairs of electrons, it is weak field. The charge on the metal ion is +3, giving a d6Â electron configuration.
C Because of the weak-field ligands, we expect a relatively small Δo, making the compound high spin.
D In a high-spin octahedral d6 complex, the first five electrons are placed individually in each of the d orbitals with their spins parallel, and the sixth electron is paired in one of the t2g orbitals, giving four unpaired electrons.
B C Because rhodium is a second-row transition metal ion with a d8 electron configuration and CO is a strong-field ligand, the complex is likely to be square planar with a large Δo, making it low spin. Because the strongest d-orbital interactions are along the x and y axes, the orbital energies increase in the order dz2dyz, and dxz (these are degenerate); dxy; and dx2−y2.
D The eight electrons occupy the first four of these orbitals, leaving the dx2−y2. orbital empty. Thus there are no unpaired electrons.
Comments
Leave a comment