Calculate the spin only magnetic momentum µ of K3 [ Mn(CN)6] compound .

Calculate the spin only magnetic momentum  µ of K₃[ Mn(CN)₆]  compound  .

In K₃ [ Mn(CN)₆]  the metal  manganese in Mn ⁺³ oxidation state . Mn ⁺³  ion is a d ⁴ system . Cyanide ion is a very strong field ligand . 

According to crystal field theory the given complex will be of low spin type  and its electron distribution will be t_2g⁴ e_g⁰ which is shown below.

There is two unpaired electron in the complex , it will be paramagnetic and magnetic moment will correspond to two unpaired electron which is given by µ .

Why tetrahedral metal complexes are usually not of low spin ?

According to crystal field theory splitting of d-orbitals takes place during complexation and they split into two different sets of orbitals in both octahedral and tetrahedral complexes. 

The value of splitting energy depends on the nature of the metal ion ligand and the nature of the geometry of the complexes. 

Depending of the value of this splitting energy Δo ( for octahedral ) and Δt ( for tetrahedral ) high spin and low spin complexes may result. 

As for example , for an octahedral complex of d4 system the electron distribution may either be  t_2g3 e_g1 ( high spin) when Δo < p or t_2g4 e_g0 (low spin ) when Δo > p . 

Spin pairing energies of different d4-system of a given d-series of transition element are almost the same.

Therefore it is necessary to obtain high spin low spin complexes the value of  Δo should be greater than pairing energy, p ( low spin ) in some cases and also it should be less than ,p ( high spin ) in some cases. Actually these are possible in octahedral complexes.

It has been found that Δt is always very small and is about 45% of Δo in all cases and never exceeds pairing energy.

Therefore these is no possibility of formation of low spin complexes . So it can be said that all tetrahedral metal complexes are usually not of low spin.

Summary.

Calculate the spin only magnetic momentum  µ of  K₃ [ Mn(CN)₆]  compound  .

 Why tetrahedral metal complexes are usually not of low spin ?

CLICK HERE:What is sodide ion ? How sodide ion stabilized in solution ? 

Which have more π-acceptor strength among the CO,NO+ and CN ligand ?

Which have more π-acceptor strength among the CO, NO ⁺ and CN ^– ligand ?

The three given species  CO, NO ⁺ and CN ^– are isoelectronic and isostructural . 
Now , which π-acceptor ligands have more strength , can be explained with the help of molecular orbital theory .

According to molecular orbital theory the electron distribution in molecular orbital in each case is ,

Now,bond order = ( bonding electron – anti bonding electron ) /2 = ( 10 – 4 ) /2 =3. So the bond order is 3.

 Each of them can behave as ligands . Since in each case there is a suitable lone pair of electrons for donation to metal.

 Again each possess vacant anti bonding  π* orbital and each can receive back-donated electrons from metal atom or ion in its vacant such orbital.

So they can behave as a Lewis base , that is, as σ-donors as well as they can behave as Lewis acid , that is, as π-acceptors. Bonding of such complexes can be represented as shown below,

Donor and acceptor  ability of these species depends on the virtual change on them. Donor ability increases with decrease in oxidation number. Hence the donating ability order is NO ⁺ <  CO < CN ^- .

Again π-acceptor ability increases with increase in the oxidation number and hence such ability , that is π-acceptor ability order is CN ^- < CO < NO ⁺ .

Practice problem:

Which have more π-acceptor strength among the CO, NO + and CN-ligand ?

Why CO molecule acts as a π-acceptor ligand ? 

What is uric acid ?  Click here 

What is sodide ion ? How sodide ion stabilized in solution ?

What is sodide ion ? How sodide ion stabilized in solution ?

The sodide ion is Na ^-1. It is an anion like non metallic anions such as Cl ^- , Br  ^-, I ^– etc . 
It has the electronic configuration  [ Ne ] 3s ². Group  (I) metals ( Na , K , Rb ) are stable in liquid NH₃. 

The solution is blue when dilute. The solution conduct electricity and are paramagnetic.

These measurement of transport number suggest that the main carrier is solvated electrons by which is implied electrons which are free from their parents Na-atoms occupy cavities in the ligand.

Electron spin resonance studies have shown the presence of free electrons , but the decrease of para magnetism with increasing 

concentration suggest that ammoniated electrons can associated to far diamagnetic electron pairs.

How the electrons are associated with the ammonia molecule ?

A  suggestion has been made that in addition to solvated cations negative ions stabilized by solution may exist in dilute solution . 

This can be involved in an oxidation-reduction equilibrium involving the solvent S’ with S ^-1  ( solvated  electron )

⇆      
  M ⁺ + 2 S ^-       M  ^-     +  2 S

In  addition ion pairing equilibrium are involved .

      ⇋  M  ⁺ + S ^-               M ⁺ S^-  

 M ⁺ + M^-         ⇋    M⁺ M^-

Where  M= Na , LI, K ….. etc.

Fairly stable other such solution of group  I ( A ) metals  Containing  M ^–  ions can be obtained  with solvents like  tetrahydrofuran ( THF) ethylene diethyl ether and orther methyl polyethers.

Summary

What is Sodide ion ? 
How sodide ion stabilized in solution ? 

 How do you explained the bonding of the carbonyl groups in the structure of  Fe2(CO)9 throughIR-spectra ? 

How do you explained the bonding of the carbonyl groups in the structure of Fe2(CO)9 through IR-spectra ?

How do you explained the bonding of the carbonyl groups in the structure of  Fe₂(CO)₉ through IR-spectra ?

The sixth member of 3d transition series is iron (Fe). Few properties of elemental iron are shown below pictorially,

Iron formed few carbonyl compounds like Fe(CO)₅ , Fe₂(CO)₉  etc. The most important carbonyl compound of iron is diiron nona carbonyl ,Fe₂(CO)₉ .

Infrared and  x-ray studies have shown that in this organometallic compound each Fe atom is directly linked with the other Fe atom by a sigma bond  ( bond distance 2.46 Å ) to three bridging carbonyl ( CO ) groups  by a sigma bond and for three terminal carbonyl groups by a coordinated bond. 

Presence of both type of carbonyl group ,that is,bridging and terminal can be easily proved with the help of I.R-spectra of the molecule. 

In bridging or ketonic  carbonyl groups , there is no possibility or necessity of back bonding , C – O bond order is purely 2.  The stretching frequency  of such group is in the order of 1700 – 1850 cm ^-1 .

 In terminal carbonyl  groups , there is a back bonding due to which the bond order of terminal carbonyl group is reduced to some extent from the value of three , which is the bond order of  free CO molecule.

Due to this back bonding stretching frequency of free CO molecule ( 2143 cm ^-1 ) is reduced, but obviously remains  above 1850 cm ^-1. 
The presence of two absorbance peaks one at around 1830 cm -1 for bridging CO group and other at 2020 cm -1 for terminal  CO group . 

This two value confirms the above structure of di-iron nona carbonyl ,Fe₂(CO)₉ .

The absorbance curve of stretching frequency of bridging and terminal carbonyl groups  in the structure of Fe₂(CO)₉  are given below.

Practice problem:

Discuss the structure and bonding of Fe₂(CO)₉ . 

Write down the structure of iron pentacarbonyl .

Click here: What is stereo specific and stereo selective reaction ? 

How -1 and 0 oxidation states of Ti are stabilized and What is Ziegler-Natta catalyst ? Give it’s use with example .

How -1 and 0 oxidation states of Ti are stabilized –explain

The element'Ti' is a second member of 3d-series of transition metal element. It’s electronic configuration and few properties are given below,

Being a member of transition element it can show variable oxidation states.The most common oxidation states  of Ti are  -2 , -1 ,  0 +2 , +3 , +4 . 

Among  this most stable and most common oxidation states is  +4.  The given oxidation states  -1  and  0 are unusually low oxidation states .  

In  low oxidation states each and every metal atom or ion are strongly  reducing and they can reduce any species which are  present around them. 

Ions or atoms which can be easily reduced can’t  remain in contact with such system .
  
Therefore   those molecules or ions which are themselves strongly reducing  and can’t  be  further reduced can stabilize such oxidation states of any metal . 

π-acid ligands like CO , NO , PX₃ , PR₃, AsR₃, 2,2- dipyridyl  etc  are very suitable for stabilization of such unusually low oxidation  states of metals . 

They are primarily electron  pair donor , that is , they act as a Lewis  base and form ligand  L-M σ-bond .  

Secondly they can receive back donated electrons from the metal  in their π- type suitable vacant orbitals. 

As a consequence the electron density around the metal  atom is considerably reduced through  such back bonding . 

In this way such low oxidation states are stabilized.Oxidation states of  -1 and  0 for Ti are known  in 2,2-dipyridyl  complex , Li [ Ti ( dipy)₃ ] and  [ Ti ( dipy ) ] respectively.  

These are prepared as back plates or purple needless respectively by  Li-reduction of TiCl₄ in presence of dipyridyl  in tetra  hydro  furan . 

These  compounds are stabilized by delocalization of electron density from the metal atom  or ion to the  aromatic ring .

What  is Ziegler-Natta  catalyst ? Give it’s use  with example .

When TiCl₄ is added to Al ( C₂H₅ )₃ in hexane , it yields a brown solid known as Ziegler-Natta  catalyst which is employed in the conversion ethene to straight chain polymer polyethene.

The probable  mechanism is as follows, 

Practice problems:

How -1 and 0 oxidation states of Ti are stabilized-explain ?

What is Ziegler-Natta  catalyst ?show one application .
What is most stable oxidation state of Ti metal and why ?

Why stable state of Cu is +2 while that of Au is +3 ?