Coordination Complexes and Molecular Formulas

Question 1

a. η⁵-Cp₂Ni

b. fac-CuF₃(H₂O)₃^–

c. cis-NiCl₂(PMe₃)₂

d. CoCl₂Br₂^–

e. μ-Cl₂-(Cu(dppe))₂

d Orbital Splitting Diagrams

Question 1

a. Octahedral

b. Octahedral with π-acceptor ligands

c. Octahedral with π-donor ligands

d. Tetrahedral

e. Square Planar

f. Trigonal Bipyramidal

g. Square-based Pyramid with axial-base angle = 90 °

h. Square-based Pyramid with axial-base angle > 90 °

Ligand Field Stabilisation Energy (LFSE)

Question 1

a. d³

b. d⁵ (high-spin)

c. d⁵ (low-spin)

d. d⁹

Question 2

a. d²

b. d⁴

c. d⁷

d. d¹⁰

Coordination Complexes and Colour

Question 1

CrO₃ is d⁰, therefore the colour is not due to d→d transitions. Since Cr is in a high oxidation state (+6) and attached to electron-rich oxygen atoms, the colour is caused by ligand-to-metal charge transfer (LMCT). The colour is dark because LMCT is not restricted by any selection rules.

Question 2

PCy₃ is a stronger σ donor and π acceptor than PPh₃, which results in the PCy₃ complex to be square planar and the PPh₃ complex to be tetrahedral. The d→d transtion is higher in energy for a square planar complex compared to a tetrahedral because the d_x2-_y2 orbital used in the square planar geometry lies along the bond axis, resulting in the anti-bonding b_1g to be high in energy compared to the antibonding orbitals in a tetrahedron that result from poor orbital overlap (and are therefore lower) in energy. If red is observed (square planar), then green light is absorbed amd if green is observed (tetrahedral) then red light is absorbed. Green light is higher energy than red light, corresponding to the energy differences between Δ_sp and Δ_t.

Question 3

Cl and H₂O are weak field ligands, which will result in the Mn complex to be high-spin d⁵. All d orbitals will be occupied by 1 unpaired electron, therefore the transitions are forbidden because ΔS = 0 (there can be no spin flips). With six of the same ligands and all d orbitals equally occupied, the molecular geometry will be a perfect octahedron, so transiitons will only occur during vibrational distortions. Both of these factors will result in pale colour.

CO is a strong field ligand, which will result in the Mn complex to be low-spin d⁶. The t_2g orbitals will each be occupied by 2 electrons, so electrons can jump to e_g* without flipping their spins. Since there are 5 CO ligands and 1 Br ligand, the complex will be permanently distorted from O_h symmetry, and so there will be a permanent mixing of p and d orbitals. Both of these factors will result in an intense colour.