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.You need to login to perform this action.
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