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

(P) ${\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$       (Q) ${\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{3-}$

(R) ${\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{4-}$         (S) ${\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}$

${\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+},{\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+},{\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{3-},{\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{4-}$ have 5,4,1 and 0 respectively.
The order of decreasing number of unpaired electrons is,
${\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}>{\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}>{\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{3-}>{\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{4-}$.  
So, the correct option is A.