Question

a. Predict the structure of [Cr(OH₂)₆]²⁺

- I wrote that the complex will undergo Jahn-Teller distortion, making the octahedral complex become tetragonal (don't know if that's correct)

b. Account for the variation in the lattice enthalpy of Mn (2780 kJ/mol), Fe (2926 kJ/mol), Co (2976 kJ/mol), Ni (3060 kJ/mol), and Zn (2985 kJ/mol) flouride given that the metal center in all of the flourides is surrounded by an octahedral array of F-ions.

c. Explain the effect on the d-orbital energies when an octahedral ceomplex is: a) compressed along the z-axis; and b) stretched along the z-axis.

d. Explain why square planar geometry occurs most commonly with d⁸ complexes containing strongly interacting ligands.

e. A d⁸ nickel complex containing a neutral macrocyclic ligand with four N-donor atoms as well as a weakly coordinating perchlorate ligand is red, diamagnetic, and low spin. When the perchlorate ligand is replaced by two thiocyanate ions (SCN-), the complex turns violet and is high spin with two electrons. Interpret the chnage in terms of the structure of the complexes.

- I wrote down that it's due crystal field splitting, the square planar complex becomes octahedral and therefore paramagnetic (don't know if that's correct)

Answer 1

a) [Cr(H₂O]₆²⁺ is a d⁴, high-spin complex with three t₂g electrons and one eg electron.So there is an unequal distribution of electrons in eg.Since the d _x²-y² and d_z²orbitals are directed towards the aqua ligands, there will be repulsion between the orbital containing the electron and the electrons of water.As a result, the structure will undergo distortion to reduce the repulsion and increase the stability.Even though it can undergo z-in and z-out distortion, the current structure will prefer Z-out because it involves the elaboration of only two ligands.Hence the structure of the [Cr(H₂O]₆²⁺ complex is tetragonal.

b)Lattice enthalpy or energy is the energy needed to break an ionic compound into its component ions in the solid state.

• Fluorine is a weak field ligand.So all the complexes are octahedral and high spin.
• Since lattice energy is proportional to (1/r⁺+r^-), as the radius of cation decreases, lattice energy increases.
• But for the metals with d⁰,d⁵ and d¹⁰ electrons, the lattice energy is found to be less compared to the expected value.
• lattice energy follows the same trend of crystal field stabilization energy(CFSE).As the CFSE increases, lattice energy also increases.
• CFSE= [−0.4 x n(t₂g) + 0.6 x n(e_g)] ∆o.Since CFSE of Mn and Zn are zero,the lattice energy is also a smaller value compared to predicted value.

c)The octahedral complex has three degenerate lower energy orbitals called t₂g set and two degenerate higher energy orbitals called e_g set.t2g set consists of the orbitals d_xz,d_yz and d_xy.eg set consists of the orbitals d _x²-y² and d_z² .

(a)When the complex is undergoing Z-in distortion, all the orbitals with z component will increase in energy and all the other orbitals will decrease in energy compared to the octahedral energy diagram.

(b)When the complex is undergoing Z-out distortion, all the orbitals with z component will decrease in energy and all the other orbitals will increase in energy compared to the octahedral energy diagram

d)The d8 metals with strong ligands will prefer square planar complex since, in the energy level diagram, the electrons are filling all the low energy orbitals.

e)In the first complex, there are five ligands and in the second complex, there are six ligands.

• The first complex is square pyramidal and the second complex is octahedral.
• In the square pyramidal complex, the eight electrons filled all the four lower energy orbital hence is low spin.
• In the octahedral complex, the eight electrons are arranged in the t2g and the eg set hence is high spin.