Which of the following complex ions is expected to absorb visible light? (At. no. $\text{Zn} = 30$, $\text{Sc} = 21$, $\text{Ti} = 22$, $\text{Cr} = 24$)

Absorption of visible light by a transition metal complex is generally due to $d-d$ transitions. This requires the central metal ion to have an incompletely filled $d$-subshell ($d^1$ to $d^9$ configuration). Let's check the electronic configurations of the central metal ions in the given complexes: (A) In $[\text{Sc}(\text{H}_2\text{O})_3(\text{NH}_3)_3]^{3+}$, the oxidation state of Sc is $+3$. The electronic configuration of $\text{Sc}^{3+}$ ($Z=21$) is $[\text{Ar}] 3d^0$. Since it has no $d$-electrons, no $d-d$ transition can occur. (B) In $[\text{Ti}(\text{en})_2(\text{NH}_3)_2]^{4+}$, the oxidation state of Ti is $+4$. The electronic configuration of $\text{Ti}^{4+}$ ($Z=22$) is $[\text{Ar}] 3d^0$. It also has no $d$-electrons, so no $d-d$ transition is possible. (C) In [\text{Cr(\text{NH}_3)}_6]^{3+}, the oxidation state of Cr is $+3$. The electronic configuration of $\text{Cr}^{3+}$ ($Z=24$) is $[\text{Ar}] 3d^3$. It has three unpaired $d$-electrons ($t_{2g}^3 e_g^0$), making it capable of undergoing $d-d$ transitions and absorbing visible light. (D) In $[\text{Zn}(\text{NH}_3)_6]^{2+}$, the oxidation state of Zn is $+2$. The electronic configuration of $\text{Zn}^{2+}$ ($Z=30$) is $[\text{Ar}] 3d^{10}$. Since the $d$-subshell is completely filled, there are no empty $d$-orbitals available for electrons to transition into, meaning $d-d$ transitions are not possible. Therefore, [\text{Cr(\text{NH}_3)}_6]^{3+} is the only complex expected to absorb visible light.