NEET UG 2010 — Chemistry Inorganic Chemistry

Key Idea For the absorption of visible light, presence of unpaired d-electrons is the necessity. (a) In $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}, \mathrm{Ni}$[Ni(CN)4​]2−,Ni is present as $\mathrm{Ni}^{2+}$Ni2+. $$\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^0$$Ni2+=[Ar]3 d84 s0

(Pairing occurs because $\mathrm{CN}^{-}$CN−is a strong field ligand). Since, in $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$[Ni(CN)4​]2−, no unpaired electron is present in d-orbitals, it does not absorb visible light. (b) In $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}, \mathrm{Cr}$[Cr(NH3​)6​]3+,Cr is present as $\mathrm{Cr}^{3+}$Cr3+. $\mathrm{Cr}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^3 4 \mathrm{~s}^0$Cr3+=[Ar]3 d34 s0 (Three unpaired electrons) (c) In $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}, \mathrm{Fe}$[Fe(H2​O)6​]2+,Fe is present as $\mathrm{Fe}^{2+}$Fe2+. $\mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^6 4 \mathrm{~s}^0$Fe2+=[Ar]3 d64 s0 (Four unpaired electrons) (d) In $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$[Ni(H2​O)6​]2+, $\mathrm{Ni}$Ni is present as $\mathrm{Ni}^{2+}$Ni2+. $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^0$Ni2+=[Ar]3 d84 s0 (Two unpaired electrons) The complexes given in option (b), (c), (d) have unpaired electrons, thus absorb visible light.