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$$3{d^5}$$  has maximum number of unpaired electrons.

IUPAC name of sodium nitroprusside $$N{a_2}\left[ {Fe{{\left( {CN} \right)}_5}NO} \right]$$    is sodium pentacyanoni trosyl ferrate (III) because in it $$NO$$  is neutral ligand and the oxidation number of $$Fe$$  is + 3. Which is calculated as
$$\eqalign{ & N{a_2}\left[ {Fe{{\left( {CN} \right)}_5}NO} \right] \cr & 2 \times \left( { + 1} \right) + x + 5 \times \left( { - 1} \right) + 1 \times 0 = 0 \cr & \,\,\,\,\,\,{\text{[where }}x{\text{ }} = {\text{ oxidation state of }}Fe{\text{]}} \cr & 2 + x - 5 = 0 \cr & \,\,\,\,\,\,\,\,\,x - 3 = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = + 3 \cr} $$