Chapter XVI. Clusters

We are now going to look at much larger molecules. My objective is to get you to look at the great diversity and beauty of structures in transition metal and main group chemistry. Furthermore, there is a way to count the electrons and actually predict the structure of the cluster. Let's first take a look at some typical clusters and carry out electron counting with an eye on what is the M-M bond order in them.

Here are some examples where the cluster has an encapsulated atom:

Now check out these examples. The geometry has changed!

Here is an interesting pair! They are obviously isoelectronic.

Now check out this set of compounds!

Obviuosly we are not going to be able to rationalize every structure. But the simple ones and compounds that we would not call clusters can have their shapes predicted by using Wade's rules. In this formulation all structures are viewed to be deltahedra. Memorize those shown below.

The number N in each of these structures gives the number of vertices present in the structure. Notice that the most simple deltahedron, the tetrahedron, is not included. Each of the vertices will correspond to the position of an atom that makes up the cluster. There can be one or more atoms (groups) connected to this atom. In fact this vertex may be occupied by an atom or not. There are three classes of clusters then that we will worry about (there are a few more).

Number of skeletal electrons to make the structure stable:

2n+2 where n = the number of surface atoms present in the cluster, i.e. n = N

2n+4 Now N = n+1(so the number of electrons = 2n'+2 where n' = n+1)

2n+6 Now N = n+2 (so the number of electrons is still 2n"+2 whwere n" = n+2)

So what are skeletal electrons? Let us first consider that the cluster is made up of only main group atoms. Then there are basically three possibilities:

It is the number of electrons in the radial plus tangential orbitals that are counted as the skeletal electrons of the cluster.

Hydrogen atoms frequently migrate from bridging to terminal positions in these clusters. Notice that one of the A-H s orbitals is used as a tangential orbital. From the shape of the deltahedra (always using equilateral triangles) we are enforcing tetrahedral coordination at the surface atoms in all of these cases. Let's take some examples, starting with boron containing groups:

therefore, BH contributes 2 skeletal electrons

CH = 3 skeletal electrons

NH = 4 skeletal electrons

OH = 5 skeletal electrons

therefore, B contributes 1 skeletal electrons

C = 2 skeletal electrons

N = 3 skeletal electrons

O = 4 skeletal electrons

therefore, BH₂ contributes 3 skeletal electrons

CH₂ = 4 skeletal electrons

NH₂ = 5 skeletal electrons

OH₂ = 6 skeletal electrons

A complete table of these is given in your book. For transition metal complexes the coordination for each surface atom is based on the octahedron. Therefore, we can use the isolobal analogy for ML₃ and MCp cases.

For ML₂ we say that there is an additional lone pair pointed away from the cluster (like the :A case). So for Ni(CO)₂ it contributes 2 skeletal electrons:

For an ML₄ unit (like the AH₂ case) one M-L bond with its two electrons is counted:

Notice that there is an important difference in the way electrons are counted using Wade's rules and the isolobal analogy:

CH₂ Fe(CO)₄ Ni(CO)₂

# frontier e-'s = 2 2 2

# skeletal e-'s = 4 4 2

Why do Wade's rules work? Really, why do we have 2n+2 skeletal electrons for the closo molecules which occupy all bonding + nonbonding MO's. Alternatively, why are there n+1 bonding + nonbonding MOs? This is extremely difficult to show in a generalized manner. One can work through several examples (as the book does). A hint is offered by:

So the number of bonding + nonbonding MOs is in fact n+1.

Lets now use Wade's rules to predict the structure given a molecular formula for the compound. One needs to count the skeletal electrons and

# e-'s = 2n + x

where n = the number of surface atoms present

x = 2 closo structure

then n = N (the number of vertices in the deltahedron)

x = 4 nido structure

then n +1 = N

x = 6 arachno structure

then n + 2 = N.

a. B₆H₆^2-: b. C₂B₇H₉:

c. C₃B₄GaH₉ = C₂B₄H₆GaMe

Now lets take some transition metal examples:

a. Rh₆(CO)₁₅[P(OMe)₃] b. H₂Ru₆(CO)₁₈

count as: count as: Ru₆(CO)₁₈^2-

[Rh(CO)₃]₄[Rh(CO)(PR₃ )][Rh(CO)₂]

c. Cp₃Ni₃(CO)+

We handle encapsulated atoms by donating all of the electrons from the encapsulated atom for cluster bonding, i.e.

a.Ru₆(CO)₁₇C = b. Os₃(CO)₁₅C

[Ru(CO)₃]₅[Ru(CO)₂ ]C

Here are just a couple of examples for molecules that might not seem like clusters:

a. Fe₂(CO)₆S₂ b. CpMn(CO)₃