A compound with empirical formula Fe(H₂O)₄(CN)₂ has a magnetic moment corresponding to $2\frac{2}{3}$ unpaired electrons per iron. The molecular formula of the complex may be

Total number of unpaired electrons should be 8 and they should present in 3 iron atoms. So formula must be ${\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}_2\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]$ Each [Fe(H₂O)₆]²⁺ contain 4 unpaired electrons whole [Fe(CN)₆]^4– contain zero unpaired electrons. So total unpaired electrons are 8 which give 8/3 or $2\frac{2}{3}$