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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$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$(Yellow coloured compound)
Consider the above reaction, the Product $"P"$ is:
$\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{\,\,\,\,C{H_2} - O - C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}} \\
{\,\,CH - O - C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,C{H_2} - O - C{H_3}}
\end{array}\xrightarrow{{HI(excess)}}$
(Critical micelle concentration ($CMC$) is marked with an arrow in the figures.)