The brown ring complex compound is formulated as `[Fe(H_(2)O)_(5)(NO)]SO_(4)`.The oxidation state of iron is :

Answer

1

Explanation

The complex is `[Fe(H_2O)_5NO^(+)]^(2+)`. Here iron is in +1 oxidation state. Water is weak field ligand so following rearrangement of electrons occurs , then

The complex is outer orbital complex having three unpaired electrons and the principal quantum number of d-orbital participating in hybridisation is 4.
`NO^(+)` is positive ligand and it is isoelectronic with nitrogen, so its bond order is 3.