question_answer

The number of unpaired electrons in \[F{{e}^{3+}}(Z=26)\] are [KCET 2000]

A)                 5             

B)                 6

C)                 3             

D)                 4

Correct Answer: A

Solution :

           \[F{{e}^{3+}}\]\[(z=26)\]            \[F{{e}^{3+}}=[Ar]\,3{{d}^{5}}4{{s}^{0}}\]                                                                Total no. of unpaired electron=5