The most stable oxidation state of Sc is:
A+2
B+3 (Sc³⁺ has noble-gas Ar configuration)
C+4
D+1
Answer & Solution
Correct answer: B. +3 (Sc³⁺ has noble-gas Ar configuration)
**Sc: [Ar] 3d¹ 4s²** → losing all 3 valence electrons gives **Sc³⁺ ([Ar])** = stable noble-gas configuration. Sc(II) and Sc(I) are virtually unknown. Only +3 state is meaningful for Sc.
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