The correct order about the number of unpaired electrons present in the following complexes is $$ \left…
The correct order about the number of unpaired electrons present in the following complexes is
$$
\left.\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-} \underset{\text { II }}{\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}} \underset{\text { III }}{\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6}\right]^{2+}
$$
II $>$ III $>$ I
II $>$ I $>$ III
I $>$ II $>$ III
III $>$ II $>$ I
Solution
I. $\left[\mathrm{Fe}(\mathrm{CN})_6 \mathrm{I}^{4-}\right.$
Atomic number of $\mathrm{Fe}=26$
Ground state configuration of $\mathrm{Fe}=[\mathrm{Ar}] 3 d^6 4 s^2$
$
\mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 d^6
$
Since, cyanide is a powerful ligand, therefore, all the electrons will be paired.
Number of unpaired electrons in $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}=0$
II. $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
Atomic number of $\mathrm{Fe}=26$
Ground state configuration of $\mathrm{Fe}=[\mathrm{Ar}] 3 d^6 4 s^2$
$
\mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 d^6
$
Number of unpaired electrons in $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}=4$
III. $\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
Atomic number of $\mathrm{Co}=27$
Ground state configuration of $\mathrm{Fe}=[\mathrm{Ar}] 3 d^7 4 s^2$
$
\mathrm{Co}^{2+}=[\mathrm{Ar}] 3 d^7
$
Number of unpaired electrons in $\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}=3$
So, the correct order is II $>$ III $>$ I