Question

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

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• A. $\left[Cr\right(N{H}_3{)}_6{]}^{3+}$
• B. $\left[Ni\right(CN{)}_4{]}^{2-}$
• C. $\left[Fe\right(H_{2}O{)}_6{]}^{2+}$
• D. $\left[Ni\right(H_{2}O{)}_6{]}^{2+}$

Answer 1

The correct option is B $\left[Ni\right(CN{)}_4{]}^{2-}$

$\left[Ni\right(CN{)}_4{]}^{2-}$
For the absorption of visible light, the presence of unpaired d-electrons is the necessity.

a) In $\left[Ni\right(CN{)}_4{]}^{2-},Ni\text{is present as}N{i}^{2+}$
$N{i}^{2+}=\left[Ar\right]3d^{8}4s^0$
Therefore,


(pairing occurs because $C{N}^{-}$ is a strong field ligand).
Since in $\left[Ni\right(CN{)}_4{]}^{2-}$, no unpaired electron is present in $d-$ orbitals, it does not absorb visible light.


$\left[Cr\right(N{H}_3{)}_6{]}^{3+}\to C{r}^{3+}$ = Have three unpaired electrons
$\left[Fe\right(H_{2}O{)}_6{]}^{2+}\to F{e}^{2+}$ = Have four unpaired electrons
$\left[Ni\right(H_{2}O{)}_6{]}^{2+}\to N{i}^{2+}$ = Have two unpaired electrons

Hence, option (a) is correct.