JEE Main 2022 — Coordination Compounds Question with Solution

${\mathrm{K}}_3\left[\mathrm{Cu}{\left(\mathrm{CN}\right)}_4\right]$:

${\left[\mathrm{Cu}{\left(\mathrm{CN}\right)}_4\right]}^{-3}$

Oxidation number of of copper is ${\mathrm{Cu}}^{+1}$

Electronic configuration of ${\mathrm{Cu}}^{+1}=\left[\mathrm{Ar}\right]3 {\mathrm{d}}^{10}4{\mathrm{s}}^0$

Due to the absence of unpaired electrons, their spin only magnetic moment is zero, they are diamagnetic in nature

${\mathrm{K}}_2\left[\mathrm{Cu}{\left(\mathrm{CN}\right)}_4\right]$:

${\left[\mathrm{Cu}{\left(\mathrm{CN}\right)}_4\right]}^{2-}, \left({\mathrm{Cu}}^{2+}=\left[\mathrm{Ar}\right]3{\mathrm{d}}^9\right)$ one unpaired electron is present.Hence, paramagnetic in nature.

${\mathrm{K}}_4\left[{\mathrm{FeCl}}_6\right]$:

${\left[{\mathrm{FeCl}}_6\right]}^{-4}$, ${\mathrm{Fe}}^{+2}: \left[\mathrm{Ar}\right] 3{\mathrm{d}}^6 4{\mathrm{s}}^0,\mathrm{one} \mathrm{unpaired} \mathrm{electron}$.Hence, paramagnetic in nature.