How many unpaired electrons are present in the Brown Ring complex $[F e (H_{2} O)_{5} (N O)] S O_{4}$ ?

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Solution

The correct option is C 3
The oxidation state of Fe in the given complex is +1. Therefore, its configuration becomes $3 d^{6} 4 s^{1}$ . When the weak feild ligand like water and strong field ligand like NO attacks the configuration changes to $3 d^{7} 4 s^{0}$ . Hence , the number of unpaired electrons in the 3d subshell is 3. Therefore option B is correct.

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