The salt of $Sc^{3+}$ ion is colourless and the salt of $Mn^{3+}$ ion is coloured. Explain. [Z of Sc = 21 and Z of Mn = 25]

Question

Statement-I : Consider the following pairs of ions $(\text{Sc}^{3+}, \text{Ti}^{3+}), (\text{Ti}^{4+}, \text{Ni}^{2+}), (\text{Cu}^{2+}, \text{Zn}^{2+})$ and $(\text{Cr}^{3+}, \text{Mn}^{3+})$. Out of these pairs three pairs consist of ions that are both coloured :
Statement-II : Among the lanthanide ions $\text{Eu}^{2+}$, $\text{Gd}^{3+}$, $\text{Ce}^{4+}$ and $\text{Tb}^{4+}$, the ion $\text{Tb}^{4+}$ is the strongest reducing agent.
Choose the correct option.

Answer 1

Step 1: Understanding the Concept:
The color of transition metal ions is generally due to $d-d$ electronic transitions. For such transitions, the metal ion must have a partially filled $d$-orbital ($d^1$ to $d^9$).
Step 3: Detailed Explanation:
1. $Sc^{3+}$ ion:
Configuration of $Sc$ ($Z=21$): $[Ar] 3d^1 4s^2$.
Configuration of $Sc^{3+}$: $[Ar] 3d^0$.
Since there are no electrons in the $d$-orbital, $d-d$ transitions are impossible. Hence, its salts are colourless.
2. $Mn^{3+}$ ion:
Configuration of $Mn$ ($Z=25$): $[Ar] 3d^5 4s^2$.
Configuration of $Mn^{3+}$: $[Ar] 3d^4$.
It has 4 unpaired electrons in the $3d$ subshell. When light falls on the salt, electrons can jump between split $d$-orbitals ($d-d$ transition), absorbing specific wavelengths and reflecting the rest. Thus, it is coloured (typically violet/purple).
Step 4: Final Answer:
$Sc^{3+}$ is $3d^0$ (colourless) and $Mn^{3+}$ is $3d^4$ (coloured).

The salt of \(Sc^{3+}\) ion is colourless and the salt of \(Mn^{3+}\) ion is coloured. Explain. [Z of Sc = 21 and Z of Mn = 25]

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Solution and Explanation

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