Q. 47

Using valence bond theory, explain the following in relation to the complexes given below:

[Mn(CN)6]3– , [Co(NH3)6]3+, [Cr(H2O)6]3+ , [FeCl6]4–

(i) Type of hybridisation.

(ii) Inner or outer orbital complex.

(iii) Magnetic behaviour.

(iv Spin only magnetic moment value.

Answer :

• [Mn(CN)6]3–: The central metal atom is Mn and oxidation state is, x=-6(-1)-3=+3.


Therefore , having a d4 configuration (1s2 2s2 2p6 3s2 3p6 3d4).


(i)Mn has +3 oxidation state with d4 configuration and left with 2 empty d orbitals and 1 empty 4s orbital which in interaction with 3,4p orbitals and attain the hybridisation d2sp3 and the 6 CN- electrons will occupy these 6 hybridised orbitals. And the hybridisation scheme will be :1.PNG


(ii) Since, in the formation of the complex [Mn(CN)6]3– inner d orbital (3d) is used, it will be an inner orbital complex.


(iii) The orbital energy diagram of [Mn(CN)6]3– is :



Hence, there will be 2 unpaired electrons and therefore, it will be paramagnetic.


(iv) The Spin only magnetic moment value for [Mn(CN)6]3– will be:


= 2.87 BM.


• [Co(NH3)6]3 +: The central metal atom is Co and oxidation state is x=+3;as NH3 is a neutral molecule.


Therefore, having a d6 configuration (1s2 2s2 2p6 3s2 3p6 3d6) .


(i)Co has one +3 oxidation state with d6 configuration and left with 2 empty d orbitals and 1 empty 4s orbital which in interaction with 3,4p orbitals and attain the hybridisation d2sp3 and the 6 NH3 electrons will occupy these 6 hybridised orbitals. And the hybridisation scheme will be :1.PNG


(ii) Since in the formation of the complex [Co(NH3)6], 3+ inner d orbital (3d) is used, it will be an inner orbital complex.


(iii) The orbital energy diagram of [Co(NH3)6]3+ is :


2.PNG


Hence, there will be no unpaired electrons as ammonia is a strong field ligand and therefore, it will be diamagnetic.


(iv)The Spin only magnetic moment value for will be equal to zero because there are no unpaired electrons.


• [Cr(H2O)6]3 +: The central metal atom is Cr and oxidation state is x=+3 as water molecules are neutral.


Therefore , having a d3 configuration (1s2 2s2 2p6 3s2 3p6 3d3).


(i)Cr has one +3 oxidation state with d3 configuration and left with 2 empty d orbitals and 1 empty 4s orbital which in interaction with 3,4p orbitals and attain the hybridisation d2sp3 and the 6 H2O electrons will occupy these 6 hybridised orbitals. And the hybridisation scheme will be :1.PNG


(ii) Since in the formation of the complex [Cr(H2O)6], 3+inner d orbital (3d) is used, it will be an inner orbital complex.


(iii) The orbital energy diagram of [Cr(H2O)6]3+is :



Hence, there will be 3 unpaired electrons and therefore, it will be paramagnetic.


(iv) The Spin only magnetic moment value for [Mn(CN)6]3– will be:


= 3.87 BM


• [FeCl6]4–: The central metal atom is Fe and oxidation state is-


x=-6 (-1)-4=+2.


Therefore, having a d6 configuration (1s2 2s2 2p6 3s2 3p6 3d6).


(i)He has a +2 oxidation state with d6 configuration and left with 2 empty outer d orbitals (4d not 3d as all the 3d will be occupied) and 1 empty 4s orbital which in interaction with 3,4p orbitals and attain the hybridisation sp3d2 and the 6 Cl- electrons will occupy these 6 hybridised orbitals. And the hybridisation scheme will be :



(ii) Since, in the formation of the complex [FeCl6]4–the outer d orbital (4d) is used, it will be an outer orbital complex.


(iii) The orbital energy diagram of [FeCl6]4–is:


image019.png


Hence, there will be 4 unpaired electrons and therefore, it will be paramagnetic.


(iv)The Spin only magnetic moment value for [FeCl6]4–will be:


= 4.9 BM.


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