question_answer

Nickel\[(Z=28)\]combines with a uninegative monodentate ligand\[{{X}^{-}}\]to form a paramagnetic complex\[{{[Ni{{X}_{4}}]}^{2-}}\]. The number of unpaired electron (s) in the nickel and geometry of this complex ion are, respectively     AIEEE  Solved  Paper-2006

A) one, tetrahedral

B)        two, tetrahedral

C)        one, square planar         

D) two, square planar

Correct Answer: B

Solution :

\[_{28}Ni=[Ar]3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{8}}\] \[N{{i}^{2+}}=[Ar]3{{s}^{2}}3{{p}^{6}}3{{d}^{6}}\] Nickel has two unpaired electrons and geometry is tetrahedral due to\[s{{p}^{3}}\]hybridisation.