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Question

(A) transition metal complex absorbs visible light only if it has unpaired electrons,which allows for $d-d$ transitions.$1$. In $[Ni(CN)_4]^{2-}$,$Ni$

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$[Ni(CN)_4]^{2-}$

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(A) transition metal complex absorbs visible light only if it has unpaired electrons,which allows for $d-d$ transitions.$1$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in the $+2$ oxidation state $(3d^8)$. Since $CN^-$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals,resulting in no unpaired electrons ($d^8$ configuration: $t_{2g}^6 e_g^2$ is not applicable here as it is square planar,but all electrons are paired).$2$. $[Cr(NH_3)_6]^{3+}$ has $Cr^{3+}$ $(3d^3)$,which has $3$ unpaired electrons.$3$. $[Fe(H_2O)_6]^{2+}$ has $Fe^{2+}$ $(3d^6)$,which has $4$ unpaired electrons.$4$. $[Ni(H_2O)_6]^{2+}$ has $Ni^{2+}$ $(3d^8)$,which has $2$ unpaired electrons.Since $[Ni(CN)_4]^{2-}$ has no unpaired electrons,it is not expected to absorb visible light.

Which of the following complex ions is not expected to absorb visible light?

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