The correct option is
A [Fe(CN)6]3−(i) [Fe(CN)6]3−
Here, oxidation state of Fe=+3
Electronic configuration of Fe (ground state) =[Ar]3d64s2
Electronic configuration of Fe (excited state) =[Ar]3d54s0
Number of unpaired electrons =5
(ii) [Fe(CN)6]4−
Here, oxidation state of Fe=+2
Electronic configuration of Fe (ground state) =[Ar]3d64s2
Electronic configuration of Fe (excited state) =[Ar]3d64s0
Number of unpaired electrons =4
(iii) [Cr(H2O)6]3+
Here, oxidation state of Cr=+3
Electronic configuration of Cr3+=[Ar]3d34s0
Number of unpaired electrons =3
(iv) [Cu(H2O)6]2+
Here, oxidation state of Cu=+2
Electronic configuration of Cu2+=[Ar]3d94s0
Number of unpaired electrons =1
A number of unpaired electrons present is directly proportional to a paramagnetic character.
So, [Fe(CN)6]3− has maximum paramagnetic character.