A) \[N{{i}^{2+}}>C{{r}^{2+}}>F{{e}^{2+}}>M{{n}^{2+}}\]
B) \[Sc>Ti>Cr>Mn\]
C) \[M{{n}^{2+}}>N{{i}^{2+}}<C{{o}^{2+}}<F{{e}^{2+}}\] (unpaired electron)
D) \[F{{e}^{2+}}>C{{o}^{2+}}>N{{i}^{2+}}>C{{u}^{2+}}\] (unpaired electron)
Correct Answer: A
Solution :
[a] In a period on moving from left to right ionic radii decreases. [a] So order of cationic radii is \[C{{r}^{2+}}>M{{n}^{2+}}>F{{e}^{2+}}>N{{i}^{2+}}\]and [b] \[Sc>Ti>Cr>Mn\] (correct order of atomic radii) [c] For unpaired electrons \[{{\underset{\text{(Five)}}{\mathop{\text{Mn}}}\,}^{\text{2+}}}\text{}\underset{\text{(Two)}}{\mathop{\text{N}{{\text{i}}^{\text{2+}}}}}\,\text{}\underset{\text{(Three)}}{\mathop{\text{C}{{\text{o}}^{\text{2+}}}}}\,\text{}\underset{\text{(Four)}}{\mathop{\text{F}{{\text{e}}^{\text{2+}}}}}\,\] [d] For unpaired electrons \[{{\underset{\text{(Four)}}{\mathop{Fe}}\,}^{2+}}>{{\underset{\text{(Three)}}{\mathop{Co}}\,}^{2+}}<{{\underset{\text{(Two)}}{\mathop{Ni}}\,}^{2+}}<{{\underset{\text{(One)}}{\mathop{Cu}}\,}^{2+}}\]
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