Chemistry - Complete d-block Question with Solution | TestHub

Key Idea For the absorption of visible light, presence of unpaired d-electrons is the necessity. (a) In ${\left[\mathrm{Ni}(\mathrm{CN}{)}_4\right]}^{2-},\mathrm{Ni}$ is present as ${\mathrm{Ni}}^{2+}$. $${\mathrm{Ni}}^{2+}=[\mathrm{Ar}]3{\mathrm{d}}^{8}4{\mathrm{s}}^0$$
(Pairing occurs because ${\mathrm{CN}}^{-}$is a strong field ligand). Since, in ${\left[\mathrm{Ni}(\mathrm{CN}{)}_4\right]}^{2-}$, no unpaired electron is present in d-orbitals, it does not absorb visible light. (b) In ${\left[\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_6\right]}^{3+},\mathrm{Cr}$ is present as ${\mathrm{Cr}}^{3+}$. ${\mathrm{Cr}}^{3+}=[\mathrm{Ar}]3{\mathrm{d}}^{3}4{\mathrm{s}}^0$ (Three unpaired electrons) (c) In ${\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+},\mathrm{Fe}$ is present as ${\mathrm{Fe}}^{2+}$. ${\mathrm{Fe}}^{2+}=[\mathrm{Ar}]3{\mathrm{d}}^{6}4{\mathrm{s}}^0$ (Four unpaired electrons) (d) In ${\left[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$, $\mathrm{Ni}$ is present as ${\mathrm{Ni}}^{2+}$. ${\mathrm{Ni}}^{2+}=[\mathrm{Ar}]3{\mathrm{d}}^{8}4{\mathrm{s}}^0$ (Two unpaired electrons) The complexes given in option (b), (c), (d) have unpaired electrons, thus absorb visible light.