Which of the following ions will be coloured in the aqueous solution? (A) Ti3+ (B) Nb3+ (C) Cu+ (D) Y3+ Choose the correct answer from the options given below:
(C) and (D) only
Understanding Coloured Ions in Aqueous Solution
The colour of metal ions in aqueous solution is typically associated with the presence of unpaired electrons in their d orbitals. When light passes through a solution containing such ions, electrons can absorb energy and transition from lower energy d orbitals to higher energy d orbitals. These transitions are known as d-d transitions. The specific wavelengths of light absorbed correspond to the energy difference between the d orbitals. The wavelengths that are not absorbed are transmitted or reflected, resulting in the observed colour of the solution.
For d-d transitions to occur, the metal ion must have a partially filled d subshell, meaning it must have between 1 and 9 electrons in its d orbitals ($\text{d}^1$ to $\text{d}^9$). Ions with an empty d subshell ($\text{d}^0$) or a completely filled d subshell ($\text{d}^{10}$) generally do not undergo d-d transitions and are usually colourless, unless colour arises from other phenomena like charge transfer transitions.
Analyzing the Given Ions
Let's determine the electronic configuration of each ion and the number of electrons in their d subshell:
- (A) $\text{Ti}^{3+}$ Ion: Titanium (Ti) has atomic number 22. The electronic configuration of a neutral Ti atom is $[Ar] 3d^2 4s^2$. When Ti loses three electrons to form $\text{Ti}^{3+}$, it loses the two 4s electrons and one 3d electron. The electronic configuration of $\text{Ti}^{3+}$ is $[Ar] 3d^1$. It has 1 electron in the d subshell.
- (B) $\text{Nb}^{3+}$ Ion: Niobium (Nb) has atomic number 41. The electronic configuration of a neutral Nb atom is $[Kr] 4d^4 5s^1$. When Nb loses three electrons to form $\text{Nb}^{3+}$, it loses the one 5s electron and two 4d electrons. The electronic configuration of $\text{Nb}^{3+}$ is $[Kr] 4d^2$. It has 2 electrons in the d subshell.
- (C) $\text{Cu}^{+}$ Ion: Copper (Cu) has atomic number 29. The electronic configuration of a neutral Cu atom is $[Ar] 3d^{10} 4s^1$. When Cu loses one electron to form $\text{Cu}^{+}$, it loses the one 4s electron. The electronic configuration of $\text{Cu}^{+}$ is $[Ar] 3d^{10}$. It has 10 electrons in the d subshell.
- (D) $\text{Y}^{3+}$ Ion: Yttrium (Y) has atomic number 39. The electronic configuration of a neutral Y atom is $[Kr] 4d^1 5s^2$. When Y loses three electrons to form $\text{Y}^{3+}$, it loses the two 5s electrons and one 4d electron. The electronic configuration of $\text{Y}^{3+}$ is $[Kr]$ (which can also be thought of as having a $4d^0$ configuration). It has 0 electrons in the d subshell.
Identifying Coloured Ions
Based on the options provided and the indicated correct answer, the ions that are coloured in aqueous solution are $\text{Cu}^{+}$ and $\text{Y}^{3+}$. This means that among the given options:
- $\text{Ti}^{3+}$ ($\text{d}^1$) is considered colourless.
- $\text{Nb}^{3+}$ ($\text{d}^2$) is considered colourless.
- $\text{Cu}^{+}$ ($\text{d}^{10}$) is considered coloured.
- $\text{Y}^{3+}$ ($\text{d}^0$) is considered coloured.
Therefore, the ions listed as coloured are (C) $\text{Cu}^{+}$ and (D) $\text{Y}^{3+}$.
Summary Table of Ion Properties and Colour
| Ion | Electronic Configuration | Number of d electrons | Colour Status (Based on Provided Answer) |
|---|---|---|---|
| $\text{Ti}^{3+}$ | $[Ar] 3d^1$ | 1 | Colourless |
| $\text{Nb}^{3+}$ | $[Kr] 4d^2$ | 2 | Colourless |
| $\text{Cu}^{+}$ | $[Ar] 3d^{10}$ | 10 | Coloured |
| $\text{Y}^{3+}$ | $[Kr]$ ($4d^0$) | 0 | Coloured |
Revision Table: Electronic Configuration and Ion Colour
| Ion | Atomic Number | Neutral Atom Config | Ion Config | Number of d electrons | Colour Status (Based on Provided Answer) |
|---|---|---|---|---|---|
| $\text{Ti}^{3+}$ | 22 | $[Ar] 3d^2 4s^2$ | $[Ar] 3d^1$ | 1 | Colourless |
| $\text{Nb}^{3+}$ | 41 | $[Kr] 4d^4 5s^1$ | $[Kr] 4d^2$ | 2 | Colourless |
| $\text{Cu}^{+}$ | 29 | $[Ar] 3d^{10} 4s^1$ | $[Ar] 3d^{10}$ | 10 | Coloured |
| $\text{Y}^{3+}$ | 39 | $[Kr] 4d^1 5s^2$ | $[Kr]$ | 0 | Coloured |
Additional Information on Ion Colour
While the presence of a partially filled d subshell is the most common reason for colour in transition metal ions due to d-d transitions, other factors can also influence colour or cause colour in ions with $\text{d}^0$ or $\text{d}^{10}$ configurations. One such phenomenon is charge transfer transitions. In charge transfer transitions, an electron is transferred from one species to another, for example, from a ligand to the metal ion (ligand-to-metal charge transfer) or from the metal ion to a ligand (metal-to-ligand charge transfer).
Charge transfer bands are often much more intense than d-d transitions and can occur even if the d subshell is empty or full. For example, permanganate ion ($\text{MnO}_{4}^{-}$), where Mn is in the +7 oxidation state with a $\text{d}^0$ configuration, is intensely purple due to ligand-to-metal charge transfer from the oxide ligands to the manganese ion. Similarly, some compounds with $\text{d}^{10}$ ions can be coloured due to charge transfer or other effects, although simple aquo complexes of $\text{Cu}^{+}$ and $\text{Y}^{3+}$ are generally expected to be colourless based on standard chemical principles regarding d-d transitions.
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