What can be inferred from the magnetic moment of the following complex species?

To calculate magnetic moment of the complex species, we use the spin formula:

μ =√n(n+2) BM

When n= 1

⇒ μ = √1(1+2) 

⇒ u= √3 

⇒ u=1.73 BM

When n=2

⇒ μ = √2(2+2)  

⇒ μ = √8 

⇒ μ = 2.83 BM

 When n= 3

⇒ μ = √3(3+2) 

⇒  μ = 15

⇒ μ = 3.87 BM

When n = 4

⇒ μ = √ 4(4+2)

⇒ μ = √24

⇒ μ = 4.899 BM

When n= 5

⇒ μ = √5(5+2)

⇒ μ = √35 

⇒ μ = 5.92 BM

1.[K4 [Mn(CN)6]

⇒μ = 2.2 BM (given)

We can see from the above calculation that the given value(2.2) is close to n=1. It means that it has only one unpaired electron Also in this complex Mn is in +2 oxidation state,i.e., as Mn2+. Thus when CN-  ligands approach Mn2+ ion, The electrons in 3d do not pair up. 

The atomic number of Manganese (Mn) is Z = 25

The electronic configuration of 25Mn= [Ar] 3d5 4s2

And, the electronic configuration of Mn2+=[Ar] 3d5  

Thus CN- is a strong ligand.

The hybridization involved is d2sp3 forming inner orbital octahedral complex

2.[Fe(H2O)6]2+.      

⇒ μ = 5.3 BM (given)

We can see from the above calculation that the given value(5.3) is close to n=4. It means that it has four unpaired electrons. In this complex, Fe is in +2 state, i.e., as Fe2+. This means that the electrons in 3d do not pair up when the ligands, H2O molecules approach.

The atomic number of Iron(Fe)is Z = 26

The electronic configuration of 26Fe= [Ar] 3d6 4s2

And, the electronic configuration of Fe2+=[Ar] 3d6  

 Thus H2O is a weak ligand. To accommodate the electrons donated by six H20 molecules, the hybridization will be sp3d2. Hence it will be an outer orbital octahedral complex.

3. K2 [MnCl4]

⇒μ = 5.9 BM (given)

We can see from the above calculation that the given value(5.9) is close to n=5. It means that it has five unpaired electrons

 In this complex, Mn is in +2 state, i.e., as Mn2+. Hence, we can say that Cl- is a weak ligand and does not cause the pairing of electrons.

The atomic number of Manganese (Mn) is Z = 25

The electronic configuration of 25Mn= [Ar] 3d5 4s2

And, the electronic configuration of Mn2+=[Ar] 3d5  

Thus, the hybridization involved will be sp3 and the complex will be tetrahedral complex

Note: Being Lewis bases (those who donate electrons) the ligands with less electronegativity will be stronger. Therefore, in general halogen or oxygen donors (F-,Cl-,Br-, H2O) are weak field ligands and the ones in which carbon or nitrogen atom is the donor (eg-CN-,CO,NH3) are strong field ligands.