Friday, August 19

Noorul Huda

18 Electron Rule: Application and Problems

The rule states that thermodynamically stable transition metal organometalliccompounds  are formed when the sum of the metal d electrons and the electrons  conventionally considered as being supplied by the surrounding ligands  equals 18.

In general, the conditions favoring adherence to the 18 electron rule are, an electron rich  metal  (one that is in a low oxidation state) and ligands that are good π‐acceptors

The hapto symbol, η,with a numerical superscript, provides a topological description  by indicating the connectivity between the ligand and the central atom. For example,  if all the five carbon atoms of a cyclopentadienyl moiety are equidistant from a metal  atom, we term it as η5‐cyclopentadienyl





Examples:  η1‐R,  η1‐Ar,  η2‐C2R4,  η1‐allyl,   η3‐allyl, η4‐Cb,  η5‐Cp,   η6‐C6H6, η8‐C8H8, η2‐C60, η5‐ R5C60.

The symbolμindicates bridging;  normally we have   μ1 and rarely  μ3bridging

Examples:  μ2‐CO, μ3‐CO, μ2‐CH3, μ2‐H, μ2‐Cl, μ3‐Cl,  μ2‐OR, μ2‐PR2, μ2‐NR2


Methods Of Counting:

Neutral Atom Method  &  Oxidation State Method











Neutral atom method: Metal is taken as in zero oxidation state for counting purpose

Oxidation state method: We first arrive at the oxidation state of the metal by considering the  number of anionic ligandspresent and overall charge of the complex


[Suggestion: Focus on one counting method till you are confident]







Easy Way to Remember ligand electron contribution for NEUTRAL ATOM COUNTING method:


APPLICATIONS OF 18 Electron Rule

How to determine the total number of metal ‐metal bonds


Determine the total valence electrons (TVE) in the entire molecule (that is, the number of valence  electrons of the metal plus the number of electrons from each ligand and the charge); say, it is A.
Subtract this number from n×18  where n is the number of metals in the complex, that is, (n×18) –A;   say, it is B.


(a) B divided by 2 gives the total number of M–M bonds in the complex.
 

(b) A divided by n gives the number of electrons per metal. If the number of electrons is 18, it indicates  that there is no M–M bond;
 if it is 17 electrons, it indicates that there is 1 M–M bond; 
if it is 16  electrons, it indicates that there are 2 M–M bonds and so on.

 



 Few More Examples:









Exceptions to the 18 electron rule

• Square planar organometallic complexes of the late transition  metals (16e).  


• Some organometallic complexes of the early transition metals (e.g.Cp2 TiCl2, WMe 6, Me 2NbCl3 ,CpWOCl3) [A possible reason for the same is that some of the orbitals of these complexes are too high in energy for effective utilization in bonding or the ligands are mostly σ donors.]
 


• Some high valent d0 complexes have a lower electron count  than 18.
 

• Sterically demanding bulky ligands force complexes to have  less than 18 electrons.
 

• The 18 electron rule fails when bonding of organometallic clusters of  moderate to big sizes (6 Metal atoms and above) are considered.
 

• The rule is not applicable to organometallic compounds of main    group metals as well as to those of lanthanide and actinide metals. 








 

Why We Study Metal Carbonyls????

Simplest of organometallic compounds where M‐C σbonding is well understood. CO is one  of the strongest  π acceptor ligands.  Back bonding(πbonding) and variation in electronic  properties of CO can be monitored very efficiently by Infrared spectroscopy

A range of metal carbonyls are used as catalysts in Chemical Industry





:BONUS:


Molecular Orbital diagram of CO

Why does  CO bind a metal through its less  electronegative carbon atom  than its more  electronegative oxygen ? What makes it a  good  π acceptor 


The highest occupied molecular orbital (HOMO)  of CO is weakly antibonding (compared with the O atomic orbitals)  and is an MO which is carbon based. Secondly, the π* antibonding orbital  which is the lowest unoccupied molecular orbital (LUMO) is also of  comparatively lower energy which makes it possible to interact with metal t2g orbitals for πbonding. There exists a strong back bonding of metal electrons to the π* anti bonding orbitals of CO





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Problems Asked in CSIR and Other Examinations:



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Noorul Huda

About Noorul Huda -

A chemist, a teacher and a passionate blogger. Currently pursuing his PhD from School of Chemistry, University of Hyderabad is creative head of this blog and lives with a motto of teaching what he knows and exploring what he don't.

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