Question

Which of the following complex ion is expected to absorb light in the visible region ?

• A. $\left[Ti\right(en{)}_3{]}^{4+}$
• B. $\left[Sc\right(N{H}_3{)}_4(H_{2}O{)}_2{]}^{3+}$
• C. $\left[Cr\right(N{H}_3{)}_6{]}^{3+}$
• D. $\left[Zn\right(CN{)}_2{]}^{2-}$

Answer 1

The correct option is C $\left[Cr\right(N{H}_3{)}_6{]}^{3+}$

Only those transition metal complexes are expected to absorb visible light, in which d-subshell is incomplete and excitation of an electron from a lower energy orbital to higher energy orbital is possible.

(a) In $\left[Sc\right(H_{2}O\left)\right(N{H}_3{)}_3{]}^{3+}Sc$ is present in $S{c}^{3+}=\left[Ar\right]3d^0,4s^0$
Since in this complex excitation of the electron is not possible as it does not have d electron, it will not absorb visible light.

(b) $\left[Ti\right(en{)}_2(N{H}_3{)}_2{]}^{4+},Ti$ is present as $T{i}^{4+}$
$T{i}^{4+}=\left[Ar\right]3d^0,4s^0$
it does not have d electron, Hence it will not absorb visible light

(c) In $\left[Cr\right(N{H}_3{)}_6{]}^{3+}Cr$ is present as $C{r}^{3+}$
$C{r}^{3+}=\left[Ar\right]3d^3,4s^0$
Since this complex has three unpaired electrons excitation of electrons is possible and thus, it is expected that this complex will absorb visible light.

(d) In $\left[Zn\right(N{H}_3{)}_6{]}^{2+}Zn$ is present as $Z{n}^{2+}.$
$Z{n}^{2+}=\left[Ar\right]3d^{10},4s^0$
it does not have d electron, Hence, this complex will not absorb visible light. Thus the correct option is $c$.