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  • Please,please help me - p- d- and f-block elements - BITSAT

Amongst Ni(CO)_{4},  [Ni(CN)_{4}]^{-2}     and  [NiCl_{4}]^{-2}

  • Option 1)

    Ni(CO)_{4}, and [NiCl_{4}]^{-2} are diamagnetic and [Ni(CN)_{4}]^{-2} is paramagnetic.

  • Option 2)

    [NiCl_{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  [NiCl_{4}]^{-2} is paramagnetic.

     

     

  • Option 4)

    Ni(CO)_{4}, is diamagnetic and [NiCl_{4}]^{-2}  and  [Ni(CN)_{4}]^{-2}  are paramagnetic.

 

Answers (1)

best_answer

As we learnt in 

Diamagnetism -

When substance unaffected by a magnetic field or the central metal atom doesn't have unpaired e- called Diamagnetic substance or character.

- wherein

eg\ \;Sc^{+++}\Rightarrow 3d^{\circ }9s^{2}

No unpaired e-

 

Paramagnetism -

When a substance is attracted by a magnetic field or have unpaired  e-  called paramagnetic substance.

- wherein

eg\Rightarrow \:\:Cr^{+}-3d^{5}

5 unpaired e- 

 

 Lets take each species individually

Ni (CO)4 : Ni config: 3d10 and 4s0

CO is a strong ligand

Ni(CN)^{2-}_{4}:Ni^{2+}\: \: config: 3d^{8}\: \: 4s^{0}

CN- is a strong Ligand

Ni(Cl)^{2-}_{4}:Ni^{2+}\: \: config: 3d^{8}\: \: 4s^{0}

Cl- is a weak Ligand 

As we can see, only Ni(Cl)^{2-}_{4} has unpaired electrons. 

 

 

 

 


Option 1)

Ni(CO)_{4}, and [NiCl_{4}]^{-2} are diamagnetic and [Ni(CN)_{4}]^{-2} is paramagnetic.

This answer is incorrect

Option 2)

[NiCl_{4}]^{-2}  and  [Ni(CN)_{4}]^{-2}  are diamagnetic and Ni(CO)_{4}, is paramagnetic.

This answer is incorrect

Option 3)

Ni(CO)_{4},  and  [Ni(CN)_{4}]^{-2}  are diamagnetic and  [NiCl_{4}]^{-2} is paramagnetic.

 

 

This answer is correct

Option 4)

Ni(CO)_{4}, is diamagnetic and [NiCl_{4}]^{-2}  and  [Ni(CN)_{4}]^{-2}  are paramagnetic.

This answer is incorrect

Posted by

prateek

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