Main Group Systems for Activation and Catalysis

What is the TVEC of this complex?

d0 + 3(4) + 6 = 18e

What is the TVEC of this comeplex?

d2 + 5(2) + 4 = 16e in this form.

O can form a triple bond to make it an 18e complex.

Draw the structure formed by group 6 metals in the presence of CO at high pressure.

Draw the structure formed by group 7 metals in the presence of CO at high pressure.

Staggered conformation to reduce steric clash.

Draw the structure formed by group 8 metals in the presence of CO at high pressure.

Draw the structure formed by group 9 metals in the presence of CO at high pressure.

Draw the structure formed by group 10 metals in the presence of CO at high pressure.

Draw an MO diagram for the bonding between two metals.

Which orbitals have the worst/best overlap?

The d_z² orbitals overlap the best. If this is the only occupied MO, then full rotation around the M-M bond is possible to reduce steric clash of ligands:

Explain why a d7 complex has one less ligand than a d6.

The extra electron occupies the d_z² orbital. This prevents one of the z direction ligands from binding, hence giving a square pyramidal geometry.

Draw the MO diagram for M-M bonding of a d7 metal and give the bond order.

Bond order = 1/2(8 - 6) = 1

Hence metals form a single bond with d_z² orbital, allowing full rotation to produce the lower energy staggered conformation.

How does M-M bond strength change down the d7 group? Why?

The M-M bond gets stronger down the group because the σ - σ* energy gap becomes larger, hence the bonding MO is the more preferred form

The σ - σ* energy gap gets larger down the group due to improved overlap of the larger d orbitals:

Explain M-M quadruple bonds with an MO diagram.

Theres 8 electrons in bonding orbitals, so bond order of 4.

Use a MO diagram to explain how a bond order of 6 can be achieved.

The (n+1)s orbital overlap can be so stabilised that it reaches the bonding manifold and therefore increases the bond order.

Give the oxidation state and therefore d electron count of the metals in this complex

Re(III) so d4

Give the oxidation state and therefore d electron count of the metals in this complex

Mo(II) so d4

Give the oxidation state and therefore d electron count of the metals in this complex

W(II) so d4

Explain why d4 metals form eclipsed dimers using this MO diagram

d_xy MO is occupied.

Four overlaps causing fixed orientation of metals and therefore ligands.

What is the difference between these two complexes: Re₂Cl₈^2- (group 7) and Os₂Cl₈^2- (group 8)

In the Re (d4) complex, the δ dxy MO is occupied, causing fixed eclipsed comformation of the dimer. Bond order is 8 so M-M quadruple bond.

In the Os (d5) complex, the δ* dxy MO is also occupied, which cancels out the δ MO and allows rotation. Bond order is now 6 so M-M triple bond so staggered and eclipsed can both form.

Draw molecular orbital diagrams to explain why d³ (M≡M) complexes can be eclipsed or staggered.

d³ complexes fill the bottom 3 MOs.

The d_xy and d_yz orbitals hybridise with d_x²-y² and d_xy respectively, causing a tilt. They then overlap as above (with some δ character), giving the eclipsed conformation as the most stable.

Aligning these orbitals in a staggered conformation gives weaker π orbital overlap

However, large steric bulk of ligands can cause the staggered conformation to be preferred anyway.

Explain the trend shown in this table

Cr → Mo → W is going down group 6 (d³)

This means that the metals are getting larger (3d → 4d → 5d).

Therefore, there is greater orbital overlap between the metals and thus the M≡M bond is shorter.

This allows the unpaired electron of each metal to pair together, hence making the dimer more diamagnetic.

Mo is variable because the M-M distance depends on the cation in the crystal