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

Vedclass · Accepted Answer

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

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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?

Similar Questions

Arrange the following coordination compounds in the increasing order of magnetic moments. (Atomic numbers: $Mn = 25$; $Fe = 26$)
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$(D)$ $[Mn(CN)_{6}]^{3-}$

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