EAN:Effective Atomic number (By-Sidgwick)
The metal ion accepts electron pairs from the ligands until it achieve the next noble gas configuration. this is called EAN rule.
At.no. of Fe=26
No. Of electrons in Fe2+ = 24
Electrons from 6CN– ions= 12
EAN of [Fe(CN)6]4- = 36
•this Rule fails in [Ni(CN)4]2- and other Complexes , EAN of [Ni(CN)4]2- =26 + 8 =34
Limitations of Sedgwick Theory
•there are number of complexes which are stable but do not follow EAN rule.
•this theory Does not predict the geometry of the Complexes & their magnetic behavior.
•This theory cannot explain as to why certain metal ions show more than one Coordination number.
|Bonding in Coordination Compounds|
In order to explain the nature of bonding , structure and their charateristic properties such as colour ,geometry and magnetic properties , three theories are proposed.
1.Valance bond theory (VBT)
2.Crystal field theory (CFT)
3.Molecular orbital Theory (MOT)
|VBT : Valence Bond theory|
The bonding in metal Complexes Arises when a filled ligand orbital containing a pair of electrons overlaps with a vacant hybrid orbital on the metal atom or ion forming a Coordination covalent bond.
•the appropriate atomic orbitals of metal hybridise to give a set of new orbitals of equivalent energy called hybrid orbitals.
•the d-orbitals involved in the hybridisation may be either inner (n-1)d orbital or outer(nd) orbitals.
•each ligand has at least one orbital of donar atom containing a lone pair of electrons.
•The empty hybrid orbitals of metal ion overlap with filled orbitals of ligands to form M-L coordinate covalent bonds.
•in addition to sigma bond ,there is also a possibolity of π-bond provided suitable orbitals are available on the metal atom and the ligands.
Note-VBT assumes the bonding between the metal and ligands to be purely covalent.
|Application of VBT|
VBT is used to explain the geometry & magnetic behavior of the complex compounds.
Examples of Complexes of C.N.-6
Cr has electronic configuration 3d54s1 and it is in +3 oxidation state.here inner d-orbitals are vacant and two vacant 3d ,one 4s and three 4p orbitals ,hybridised to form d2sp3 hybrid orbitals.six pairs of electrons one from each NH3 molecule, occupy the six hybrid orbitals.thus molecule has Octahedral geometry.
Since complex contains three unpaired electrons ,it is paramagnetic.
•[Co(NH3)6]3+ complex ion
Electronic configuration of Co = 3d74s2
Oxidation state of Co= +3
This complex is found to be Diamagnetic and to account this ,two electrons in 3d orbitals are paired up leaving two 3d orbitals empty.then these Six vacant orbitals( two 3d ,one 4s, and three 4p orbitals) hybridise to form six vacant d2sp3 hybrid orbitals.
Thus complex has octahedral geometry and is diamagnetic.
Here Co is in +3 oxidation & has electronic configuration 3d6.this complex has found to be paramagnetic due to presence of 4 unpaired electrons.
In this complex , electrons in 3d orbitals are not distributed and the outer 4d orbitals are used for hybridisation. the six orbitals (one 4s ,three 4p and two 4d ) are hybridised to form six sp3d2 hybrid orbitals.Six pairs of electrons,one from each F- ion are donated to the vacant hybrid irbitals forming Co-F bonds.
Thus ,the complex has octahedral geometry and is paramagnetic.
Inner orbital complex
If the complex is formed by use of inner d-orbitals for hybridisation (written as d2sp3) ,it us called inner orbital complex.in the formation of inner orbital complex , the electrons of the metal are forced to pair up and hence the complex will be either diamagnetic or will have lesser number of unpaired electrons. Such complexes are Low Spin Complexes.
For example, [Fe(CN)6]3- is inner orbital complex.
Outer orbital complex
If the complex is formed by use of outer d-orbitals for hybridisation (written as sp3d2) ,it is called as outer orbital complex.outer orbital complex will have larger number of unpaired electrons since the configuration of the metal ion remains undisturbed. Such complex is also called high spin complexes.
For example, [Fe(H2O)6]3+ is outer orbital complex.
•ligands such as H2O and halides have high spin and there is no overlapping of electrons in orbitals of metal.
|Outer Orbital complex(uses outer nd orbitals) ~ High spin complex|
|Inner orbital Complexes(uses inner (n-1)d orbitals) ~ Low spin Complex|
Examples of Complexes of C.N.- 4
Ni(ii) ,Pt(ii) and Pd(ii) form mostly ,4-coordinated Complexes and there geometry will be tetrahedral or square planar depending upon whether sp3 or dsp3 hybridisation is involved.
•Ni2+ complexes with C.N.-4
Ni has ground state electronic configuration as 3d84s2 ,and here it is in +2 oxidation state with electronic configuration 3d8.
Depending upon the type of hybridisation, Ni with C.N.-4 has two possibilities. If complex is sp3 hybridised then it would have tetrahedral structure. if on the other hand,complex with dsp2 hybridisation, has square planar geometry.
For formation of tetrahedral structure ,3d orbitals remains unaffected , therefore two unpaired d-electrons remain as such.Consequently ,the complex would be Paramagnetic like Ni2+ ion itself.
For example:- [NiCl4]2- ion
On the other hand ,for the formation of square planar structure through dsp2 hybridisation, One of the 3d-orbitals should be Empty and available for hybridisation. This is possible ,if the two unpaired d-electrons are paired up thereby making one of the 3d-orbitals empty.thus complex whould be diamagnetic.
For example [Ni(CN)4]2-
Ni(0) has 3d84s2 as outer electronic configuration .for complexes with C.N.-4 ,the central atom may involve sp3 or dsp2 type hybridisation ,for each of which the 4s orbital must be empty.
The electrons of 4s orbitals are forced into 3d-orbitals to pair up with 2 unpaired d-electrons. Therefore complex is diamagnetic. this results in sp3 hybridisation and complex has tetrahedral structure.
[Cu(NH3)4]2+ ion Exceptional
Oxidation state of Cu is +2 with outer electronic configuration 3d9.thus only one d-orbital will be singly filled.we know that there are two possible geometries and if complex is tetrahedral it involve sp3 hybridisation and comPlex will be paramagnetic due to presence of one unpaired electron. thus ,tetrahedral geometry can explain the magnetic behavior of the complex ion.
However X-ray studies shows that complex ion has square planar geometry.therefore , the metal ion must involve dsp2 hybridisation and one of the 3d orbital must be vacated.this can be achieved by permoting one electron from one of the 3d orbitals into higher energy vacant 4p orbitals. but if electron occupies higher energy level ,it becomes unstable and will be easily lost.this means that complex could be easily oxidised to Cu3+ . However this is not true because Cu3+ ions are rare.
The spectroscopic studies also shown that electron is not present in 4p orbital. To solve this anomaly,it has been suggested that the electrons in 3d orbitals remain undisturbed and it involves the use of outer 4d orbitals for hybridisation. Thus ,this complex is Square planar involving sp2d hybrid orbitals.
Example of Complexes of C.N.-5
Oxidation state of Fe in this complex is Zero and it has outer electronic configuration 3d64s2.Now there are 4 unpaired electrons in 3d orbital and complex with C.N.-5 ,central atom may involve dsp3 hybridisation ,thus 4s must be empty. Two electrons of 4s orbital and one electron of 3d orbital are pushed into 3d orbitals to pair up with the three unpaired 3d electrons. since the complex has no unpaired electron, it will be diamagnetic and has trigonal bipyramidal geometry.
Limitations of VBT
•it provides only qualitative explainations for complexes.
•it does not explain the Colour and Electronic Spectra of Complexes.
•this theory fails to explain why some metal complexes in a particular oxidation state are low spin(inner orbital) complexes while some other complexes of same metal ion in the same oxidation state are high spin (outer orbital) complexes.
•It does not explain the variations of magnetic moment values with temperature.
•It does not take into account the splitting of d-energy levels.
• it does not predict the relative stabilities of different structures.it does not explain the thermodynamic and kinetic stabilities of different coordination compounds.
•it does not distinguish between weak and strong ligands. ie does not tell us why water and halide ions commonly form high spin complexes while cyanide ion form liw spin complexes.