Here, oxidation state of $F e=+3$
Electronic configuration of $F e$ (ground state) $=[A r] 3 d^{6} 4 s^{2}$
Electronic configuration of $F e$ (excited state) $=[A r] 3 d^{5} 4 s^{0}$
Number of unpaired electrons $=5$
(ii) $\left[F e(C N)_{6}\right]^{4-}$
Here, oxidation state of $F e=+2$
Electronic configuration of $F e$ (ground state) $=[A r] 3 d^{6} 4 s^{2}$
Electronic configuration of $F e$ (excited state) $=[A r] 3 d^{6} 4 s^{0}$
Number of unpaired electrons $=4$
(iii) $\left[\operatorname{Cr}\left(H_{2} O\right)_{6}\right]^{3+}$
Here, oxidation state of $C r=+3$
Electronic configuration of $C r^{3+}=[A r] 3 d^{3} 4 s^{0}$
Number of unpaired electrons $=3$
(iv) $\left[ Cu \left( H _{2} O \right)_{6}\right]^{2+}$
Here, oxidation state of $C u=+2$
Electronic configuration of $C u^{2+}=[A r] 3 d^{9} 4 s^{0}$
Number of unpaired electrons $=1$
A number of unpaired electrons present is directly proportional to a paramagnetic character.
So, $\left[ Fe ( CN )_{6}\right]^{3-}$ has maximum paramagnetic character.
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$C{H_3} - CH = CH - COOH\xrightarrow{{B{r_2}}}$