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  Amongst Ni(CO)_4, [Ni(CN)_4]^{2-} and Ni(Cl)_4^{2-}

  • Option 1)

    Ni(CO)_4, and Ni(Cl)_4^{2-}are diamagnetic and  [Ni(CN)_4]^{2-} is paramagnetic

  • Option 2)

    Ni(Cl)_4^{2-} and [Ni(CN)_4]^{2-} are diamagnetic and Ni(CO)_4,  is paramagnetic.

  • Option 3)

    Ni(CO)_4, and [Ni(CN)_4]^{2-}– are diamagnetic and Ni(Cl)_4^{2-}  is paramagnetic.

  • Option 4)

    Ni(CO)_4,    is diamagnetic and,Ni(Cl)_4^{2-}  and  [Ni(CN)_4]^{2-}    are paramagnetic

 

Answers (1)

best_answer

As we have learned

Delta o and P (Pairing energy) -

\Delta \circ < P No pairing, weak field ligand

\Delta \circ > P Pairing, strong field ligand 

-

 

  In complex, [Ni(CO)4] nickel is in zero oxidation state. The CO is strong field ligand and, therefore compels for the pairing of electrons. The hybridisation scheme is as shown in figure.

Ni^0[Ar]3d^84s^2  

[Ni(CO)_4]   Four pairs of electrons from four CO.

 

It is tetrahedral and as all electrons are paired so diamagnetic.

            In the diamagnetic  and square planar complex [Ni(CN)4]2–, the nickel is in +2 oxidation state and the ion has the electronic configuration 3d8. The hybridisation scheme is as shown in figure.

Ni^2+, [Ar]3d^8 

[Ni(CN)_4]^2-                 

 

In the paramagnetic  and tetrahedral complex [NiCl4]2–, the nickel is in +2 oxidation state and the ion has the electronic configuration 3d8. The hybridisation scheme is as shown in figure.

 

Ni^2+, [Ar]3d^8               

    [NiCl_4]^2-                                            

 


 

 

 

 

 

 


Option 1)

Ni(CO)_4, and Ni(Cl)_4^{2-}are diamagnetic and  [Ni(CN)_4]^{2-} is paramagnetic

Option 2)

Ni(Cl)_4^{2-} and [Ni(CN)_4]^{2-} are diamagnetic and Ni(CO)_4,  is paramagnetic.

Option 3)

Ni(CO)_4, and [Ni(CN)_4]^{2-}– are diamagnetic and Ni(Cl)_4^{2-}  is paramagnetic.

Option 4)

Ni(CO)_4,    is diamagnetic and,Ni(Cl)_4^{2-}  and  [Ni(CN)_4]^{2-}    are paramagnetic

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