A) Mn
B) Fe
C) Co
D) Ni
Correct Answer: B
Solution :
[b] The electronic configurations of the given metals and in their +3 state are: \[Mn:3{{d}^{5}}4{{s}^{2}}\text{ }\,\,M{{n}^{3+}}:3{{d}^{4}}\] \[Fe:3{{d}^{6}}4{{s}^{2}}\text{ }\,\,\,F{{e}^{3+}}:\,\,3{{d}^{5}}\] \[Co:\,\,3{{d}^{7}}\,4{{s}^{2}}\text{ }\,\,\,\,C{{o}^{3+}}:\,\,3{{d}^{6}}\] \[Ni:\,\,3{{d}^{8}}4{{s}^{2}}\text{ }\,\,\,\,N{{i}^{3+}}:3{{d}^{7}}\] Since \[F{{e}^{3+}}\]has stable configuration of\[3{{d}^{5}}\], the Third ionization energy of Fe is minimum.
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