Question

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

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

Answer 1

Answer: (A)

For the absorption of visible light, presence of unpaired d-electrons is the necessity.
(a) In ${\left[Ni(CN{)}_4\right]}^{2-},Ni$ is present as $N{i}^{2+}$
$N{i}^{2+}=[Ar]3d^{8}4s^0$

Since, in ${\left[Ni(CN{)}_4\right]}^{2-}$, no unpaired electron is present in $d$-orbitals, no unpaired electron is presen $t$ in $d$-orbitals it does not absorb visible light.
(b) In ${\left[Cr{\left(N{H}_3\right)}_6\right]}^{3+},Cr$ is present as $C{r}^{3+}$
$C{r}^{3+}=[Ar]3d^{3}4s^0$ (Three unpaired electrons)
(c) In ${\left[Fe{\left(H_{2}O\right)}_6\right]}^{3+},Fe$ is present as $F{e}^{2+}$.
[AI] $3d^{6}4s^0$ (Four unpaired electrons)
(d) In ${\left[Ni{\left(H_{2}O\right)}_6\right]}^{2+},Ni$ is present as $N{i}^{2+}$.
$N{i}^{2+}=[Ar]3d^{8}4s^0$ (Two paired elections)
The complex given in option (b), (c), (d) have unpaired electrons, thus absorb visible light.