Question

Which among the following cations produces colourless aqueous solution in their respective oxidation state?

• A. $Ti^{3+}$
• B. $V^{3+}$
• C. $Sc^{3+}$
• D. $Cu^{2+}$

Hint:

Chemistry Tip: Always check for $d^0$ (like $Sc^{3+}, Ti^{4+}$) or $d^{10}$ (like $Zn^{2+}, Cu^+$) configurations when asked to identify colourless or diamagnetic transition metal ions.

Answer 1

The Correct Option is C

Concept: Chemistry (d-Block Elements) - Colour of Transition Metal Ions.

Step 1: Recall the condition for colour in transition metal ions. Transition metal ions typically exhibit colour in aqueous solutions due to d-d electronic transitions. This requires the presence of at least one unpaired electron in the d-orbitals (i.e., a partially filled d-subshell, $d^1$ to $d^9$). If the d-subshell is completely empty ($d^0$) or completely full ($d^{10}$), no d-d transition is possible, and the ion will be colourless.

Step 2: Determine the electronic configuration of $Ti^{3+}$. Titanium (Ti) has an atomic number of 22, with the ground state configuration $[Ar] 4s^2 3d^2$. Losing three electrons to form $Ti^{3+}$ results in the configuration $[Ar] 3d^1$. Since it has one unpaired electron, it undergoes d-d transitions and produces a coloured (violet) solution.

Step 3: Determine the electronic configuration of $V^{3+}$. Vanadium (V) has an atomic number of 23, with the ground state configuration $[Ar] 4s^2 3d^3$. Losing three electrons to form $V^{3+}$ yields $[Ar] 3d^2$. It has two unpaired electrons, so it is coloured (green/blue).

Step 4: Determine the electronic configuration of $Cu^{2+}$. Copper (Cu) has an atomic number of 29, with the anomalous ground state configuration $[Ar] 4s^1 3d^{10}$. Losing two electrons to form $Cu^{2+}$ gives $[Ar] 3d^9$. It has one unpaired electron, making its aqueous solution coloured (blue).

Step 5: Determine the electronic configuration of $Sc^{3+}$ and conclude. Scandium (Sc) has an atomic number of 21, with the ground state configuration $[Ar] 4s^2 3d^1$. To form the $Sc^{3+}$ ion, it loses all three of its valence electrons, resulting in the stable noble gas configuration $[Ar] 3d^0$. With absolutely zero d-electrons available for transitions, it cannot absorb visible light. Therefore, its aqueous solution is entirely colourless. $$ \therefore \text{The cation that produces a colourless solution is } Sc^{3+}. $$