Question

What are different possible oxidation states exhibited by scandium?

• A. +4
• B. +5
• C. +4, +5
• D. +2, +3

Hint:

Logic Tip: While +3 is overwhelmingly the most common and stable oxidation state for Scandium in its compounds, the +2 state is theoretically possible and observed in some rare compounds. It can never exceed +3 due to a lack of valence electrons.

Answer 1

The Correct Option is D

Concept:
Scandium (Sc) is the first transition metal in the 3d series, with an atomic number of 21. Its electronic configuration is $[Ar] 3d^1 4s^2$. The possible oxidation states depend on the removal of valence electrons from the $4s$ and $3d$ subshells.
Step 1: Analyze the electron loss for oxidation states.
To form ions, scandium first loses electrons from the outermost $4s$ orbital. Losing the two $4s$ electrons results in a $+2$ oxidation state: $$Sc \rightarrow Sc^{2+} + 2e^- \quad \text{(Configuration: } [Ar] 3d^1)$$
Step 2: Identify the most stable oxidation state.
After losing the two $4s$ electrons, scandium can easily lose the single remaining electron in the $3d$ orbital. Losing all three valence electrons results in a $+3$ oxidation state: $$Sc \rightarrow Sc^{3+} + 3e^- \quad \text{(Configuration: } [Ar])$$ The $+3$ state is highly stable because it achieves the stable noble gas configuration of Argon.
Step 3: Rule out incorrect options.
Scandium only has 3 valence electrons ($two~4s$ and $one~3d$). Therefore, it cannot lose 4 or 5 electrons under normal chemical conditions. Thus, oxidation states of +4 and +5 are impossible. The possible states are +2 and +3.