Which of the following complex ions is not expected to absorb...

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 $<img src="/api/proxy-image?url=https$ \\ (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.