A) 2
B) 4
C) 5
D) 0
Correct Answer: A
Solution :
\[Ni\to 1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{8}}4{{s}^{2}}\] \[N{{i}^{2+}}\to 1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{8}}\] \[3{{d}^{8}}=10-2=2\] unpaired electron.
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