A) \[F{{e}^{3+}}\]
B) \[C{{o}^{3+}}\]
C) \[F{{e}^{2+}}\]
D) \[C{{o}^{2+}}\]
Correct Answer: A
Solution :
\[F{{e}^{2+}}\to [Ar]3{{d}^{6}}4{{s}^{0}}\] Number of unpaired electrons = 4 Hence, the correct option is [a].
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