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  •  Nickel (Z=28)   combines with a uni-negative monodentate ligand to form a diamagnetic complex [NiL4]2-. The   hybridisation inv

 Nickel (Z=28)   combines with a uni-negative monodentate ligand to form a diamagnetic complex [NiL4]2-. The  
hybridisation involved and the number of unpaired electrons present in the complex are respectively :            

 

Option: 1

sp3, two


Option: 2

dsp2, zero


Option: 3

dsp2,one


Option: 4

sp3, zero


Answers (1)

best_answer

 

Strong Field Ligand -

These are the ligands which pair the electron in inner d-orbital and makes required d-orbitals empty/available. 

- wherein

For example: CO, CN etc

 

 

Hybridisation -

sp3d2 - square bipyramidal or octahedral 

d2sp3 - octahedral 

sp3 - tetradedral 

dsp2 - square planar

- wherein

sp3d2 - outer complex

d2sp3 - inner complex

sp3 - [Ni(Cl)_{4}]^{2-}

dsp2 - [Pt(CN)_{4}]^{2-}

 

 In  \left [ NL_{4} \right ]^{2-}  the oxidation state of N_{i}\: 2t

N_{i}^{2t}\Rightarrow 3d^{^{8}}

dsp^{2}

Ans= 2

number of unpaired election

Posted by

Sanket Gandhi

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