JEE Main 2024 — Coordination Compounds Question with Solution

In ${\left[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$, ${\mathrm{H}}_2\mathrm{O}$ is a weak field ligand. Therefore, there are unpaired electrons in ${\mathrm{Ni}}^{2+}$. In this complex, the d electrons from the lower energy level can be excited to the higher energy level i.e., the possibility of $\mathrm{d}-\mathrm{d}$transition is present. Hence, ${\left[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$ is coloured.

In ${\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}$, the electrons are all paired as ${\mathrm{CN}}^{-}$ is a strong field ligand. Therefore, $\mathrm{d}-\mathrm{d}$ transition is not possible in ${\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}$. Hence, it is colourless.

${\left[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{+2}\to$Green colour solution due to $\mathrm{d}-\mathrm{d}$ transition.

${\left[\mathrm{Ni}(\mathrm{CN}{)}_4\right]}^{-2}\to$is diamagnetic and it is colourless.