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Nickel (Z= 28) combines with a uninegative monodentate ligand to form a diamagnetic complex ${[{NiL}_{4}]}^{2 -}$ . The hybridisation involved and the number of unpaired electrons present in the complex are respectively:

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{sp}^{3}

, two

{dsp}^{2}

, zero

{dsp}^{2}

, one

{sp}^{3}

, zero

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detailed solution

Correct option is B

The electronic configuration of free ${Ni}^{2 +}$ is $3 d^{8}$ , i.e.

Since the complex is diamagnetic, there will no unpaired electrons and thus the configuration of ${Ni}^{2 +}$ in the complex will be

The hybridization involved in the complex is ${dsp}^{2}$ .

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Nickel (Z= 28) combines with a uninegative monodentate ligand to form a diamagnetic complex NiL42−. The hybridisation involved and the number of unpaired electrons present in the complex are respectively: