question_answer

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

A) \[s{{p}^{3}}\], two                  

B) \[ds{{p}^{2}}\], zero

C) \[ds{{p}^{2}}\], one                

D) \[s{{p}^{3}}\], zero

Correct Answer: A

Solution :

[a] \[{{[Ni{{L}_{4}}]}^{2-}}\] i.e, no. of unpaired electron = 2 hybridization -\[s{{p}^{3}}\].