How many unpaired electrons are present in the Brown Ring complex $\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\left(\mathrm{NO}\right)\right]{\mathrm{SO}}_4$?

Arrange the following in order of decreasing number of unpaired electrons.

(I) ${\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$    (II) ${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{3-}$    (III) ${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{4-}$   (IV) ${\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}$