Question

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]

Hint:

Any ion with \(d^0\) or \(d^{10}\) configuration is colourless. Ions with \(d^1\) to \(d^9\) configurations are typically coloured.

Answer 1

Step 1: Understanding the Concept:
The color of transition metal ions is generally due to \(d-d\) transitions of electrons. This requires the presence of partially filled \(d\)-orbitals.
Step 2: Detailed Explanation:
1. For \(Sc^{3+}\):
Atomic number of \(Sc\) is 21. Configuration: \( [Ar] \, 3d^1 \, 4s^2 \).
For \(Sc^{3+}\), all three valence electrons are removed. Configuration: \( [Ar] \, 3d^0 \).
Since the \(3d\) subshell is empty, there are no electrons to undergo \(d-d\) transitions. Thus, it is colourless.

2. For \(Mn^{3+}\):
Atomic number of \(Mn\) is 25. Configuration: \( [Ar] \, 3d^5 \, 4s^2 \).
For \(Mn^{3+}\), three electrons are removed. Configuration: \( [Ar] \, 3d^4 \).
It contains unpaired electrons in the \(d\) subshell. These electrons can absorb specific frequencies of visible light to jump between split \(d\)-orbital energy levels (\(d-d\) transition). The reflected/transmitted light gives the salt its color.
Step 3: Final Answer:
\(Sc^{3+}\) is colourless due to empty \(d\)-orbitals (\(d^0\)), while \(Mn^{3+}\) is coloured due to partially filled \(d\)-orbitals (\(d^4\)).