A) 0
B) 4
C) 8
D) 1
Correct Answer: B
Solution :
\[N{{i}^{3+}}=\left[ Ar \right]3{{d}^{7}}4{{s}^{0}}\Rightarrow 3\text{unpaired electrons}\] \[CuS{{O}_{4}}+2Kl\to Cu{{l}_{2}}+{{K}_{2}}S{{O}_{4}}\] \[2Cu{{l}_{2}}\to \underset{\text{Cuprous iodide}}{\mathop{2Cul}}\,+{{l}_{2}}\] \[S{{c}^{3+}}>C{{r}^{3+}}>F{{e}^{3+}}>M{{n}^{3+}}\] \[C{{r}^{3+}}>M{{n}^{3+}}>F{{e}^{3+}}>S{{c}^{3+}}.\]
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