Answer:
\[N{{i}_{28}}\to 1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{8}},4{{s}^{2}}\]
\[N{{i}^{2+}}\to
1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{8}},4{{s}^{0}}\]
Nickel will lose electrons from
4s orbitals.
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