Studying The Cluster Compounds: The LNCC |ChemFam #39|

Greetings to everyone! We have already studied the 18 electron rule and hapticity. Today, we shall be studying about cluster compounds, an important topic in the organometallic chemistry. Compounds in which several metal atoms are bound together directly through M-M bonds are called metal cluster compounds. the metal toms in cluster compounds tend to agglomerate to from the maximum number of M-M bonds in such a way that ML_n fragments should be sufficiently sterically unhindered to approach to within M-M bonding distance of each other.

The structures of the clusters resemble the close packed structures of the elemental metals themselves. This is due to the reason that the cluster contain unsaturated L_nM fragments. The unsaturated 16 electron fragment Os(CO)₄ is unstable and form a stable triangular cluster Os₃(CO)₁₂ which obeys 18 electron rule. Similarly the 15 electron fragment Rh(CO)₃ forms tetrahedral cluster Rh₄(CO)₁₂ which also obeys 18 electron rule. The tendency to form M-M bond increases on descending the group in the periodic table.

The stable transition metal clusters are divided into (i) organometallic clusters and (ii) inorganic clusters.

The most extensive class of organometallic clusters belong to metal carbonyls because M(CO)_n fragments are sufficiently unhindered to approach to within metal-metal bonding distance of each other. In these clusters the metals are in low oxidation states i.e., -1, 0, +1, +2.

In inorganic clusters such as Re₂Cl₈^2-, Mo₂Cl₈^4-, the metals have higher oxidation states. there are also some naked metal clusters of the post transition elements such as Sn₉^2-.

Metal Carbonyl Clusters

The metal carbonyl clusters have CO as ligands in the metal clusters involving M-M bonds. The metal carbonyl clusters are classified into two categories :

(1) Low Nuclearity Carbonyl Clusters

(2) High Nuclearity Carbonyl Clusters

Low Nuclearity Carbonyl Clusters (LNCC)

The first metal carbonyl clusters having M-M bond is Mn₂(CO)₁₀. The low nuclearity carbonyl clusters contain metal atoms ≤4.

Calculation of number of M-M bonds using 18/16 electron rule in LNCC

The procedure for calculation of M-M bonds in a compound involves the following steps :

Step I :

Calculate total valence electron (TVE) on central metal atom in the compound say A, which is equal to the number of valence electrons of the metal plus the number of electrons from each ligand and the charge.

Step II :

Substract A from n×18 where n is the number of metals atoms in the compound. and say it is B, i.e.,

B = n×18 - A

Step III :

The value of B/2 gives the total number of M-M bonds. The value of A/n gives the number of electrons per metal.

If A/n = 18, then there will be no M-M bond

If A/n = 17, there will be one M-M bond per metal

If A/n = 16, then there will be two M-M bond

If A/n = 15, there will be three M-M bond per metal

If A/n = 14, there will be four M-M bonds per metal

This procedure can not be applied for HNCC.

Calculations of M-M bonds in some multinuclear complex is illustrated below:

(1) Mn₂(CO)₁₀

Here, A = 2 × 7 + 2 × 10 = 34
B = 2 × 18 - 34 = 2
B/2 = 2/2 = 1
∴ Total number of M-M or Mn-Mn bonds = 1
A/n = 34/2 = 17
∴ There is one Mn-Mn bond per metal.

Drawn via ChemDraw by @splash-of-angs63

(2) Co₂(CO)₈

Here, A = 2 × 9 + 2 × 8 = 34
B = 2 × 18 - 34 = 2
B/2 = 2/2 = 1
∴ Total number of M-M or Co-Co bonds = 1
A/n = 34/2 = 17
∴ There is one Co-Co bond per metal.

(3) Fe₂(CO)₉

Here, A = 8 × 2 + 2 × 9 = 34
B = 2 × 18 - 34 = 2
B/2 = 2/2 = 1
∴ Total number of M-M or Fe-Fe bonds = 1
A/n = 34/2 = 17
∴ There is one Fe-Fe bond per metal.

Drawn via ChemDraw by @splash-of-angs63

(4) Fe₃(CO)₁₂

Here, A = 3 × 8 + 12 × 2 = 48
B = 3 × 18 - 48 = 2
B/2 = 6/2 = 3
∴ Total number of M-M or Fe-Fe bonds = 3
A/n = 48/3 = 16
∴ There is o two Fe-Fe bond per metal.

(5) Rh₄(CO)₁₂

Here, A = 4 × 9 + 12 × 2 = 60
B = 4 × 18 - 60 = 2
B/2 = 12/2 = 6
∴ Total number of M-M or Rh-Rh bonds = 6
A/n = 60/2 = 15
∴ There is three Rh-Rh bond per metal.

Drawn via ChemDraw by @splash-of-angs63

(6) Os₂(CO)₉

Here, A = 2 × 8 + 9 × 2 = 34
B = 2 × 18 - 34 = 2
B/2 = 2/2 = 1
∴ Total number of M-M or Os-Os bonds = 1
A/n = 34/2 = 17
∴ There is one Os-Os bond per metal.

(7) Ru₃(CO)₁₂

Here, A = 3 × 8 + 12 × 2 = 48
B = 3 × 18 - 48 = 6
B/2 = 6/2 = 3
∴ Total number of M-M or Ru-Ru bonds = 3
A/n = 48/3 = 16
∴ There is two Ru-Ru bond per metal.

Drawn via ChemDraw by @splash-of-angs63

(8) Ir₄(CO)₁₂

Here, A = 4 × 9 + 12 × 2 = 60
B = 4 × 18 - 60 = 2
B/2 = 12/2 = 6
∴ Total number of M-M or Ir-Ir bonds = 6
A/n = 60/2 = 15
∴ There is three Ir-Ir bond per metal.

The 18 electron rule is of great importance for predicting the number of M-M bonds, but it does not play any role to distinguish the bridging and terminal ligands. The factors causing the CO ligands to bridge or not bridge are the steric factors. V(CO)₆ does not dimerise as it would become seven coordinate and increase in steric hinderance. In Mn₂(CO)₁₀, if bridging occurs, there will be increase in coordination number beyond six and therefore it will be destabilized.

So, this is how one could predict the number of M-M bonds in a carbonyl cluster compound, which are crucial for studying the organometallic compounds. I shall discuss the HNCC later. Till then happy learning!

B I B L I O G R A P H Y

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Organometallic Chemistry (Evans)

Organometallic compounds- An overview

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