The brown ring complex is $\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\right(\mathrm{NO}\left)\right]{\mathrm{SO}}_4.$ The oxidation state of Fe in the brown ring complex is +1 and its outer electronic configuration becomes $3{\mathrm{d}}^{6}4{\mathrm{s}}^1.$ NO ligand is a strong ligand, therefore when NO approaching to Fe, its electronic configuration becomes $3{\mathrm{d}}^{7}4{\mathrm{s}}^0.$ Therefore, the number of unpaired electrons of Fe becomes 3.