In nitroprusside ion, the iron and $$NO$$  exist as $$F{e^{2 + }}$$  and $$N{O^ + }$$  rather than $$F{e^{3 + }}$$  and $$NO.$$  These forms can be differentiated by

Solution :
The nitroprusside ion is $${\left[ {Fe{{\left( {CN} \right)}_6}N{O^ + }} \right]^{2 - }}.$$
The magnetic moment measurements reveal the presence of 4 unpaired electrons in $$Fe$$  which must be then in $$F{e^{2 + }}\left( {3{d^6}} \right)$$   and not $$F{e^{3 + }}\left( {3{d^5}} \right)$$