One of the notable features of the transition element is the great variety of oxidation states it may show in its compounds. The elements which give the greatest number of oxidation states occur in or near the middle of the series. Manganese, for example exhibits all the oxidation states from +2 to +7. The lesser number of oxidation states at the extreme ends is either due to the loss or sharing of a few electrons (Sc, Ti) or too many d electrons of higher valence. Early in the first series scandium (II) is virtually unknown and titanium (IV) which is more stable than Ti (III) or Ti (II). On the other end, the only oxidation state of zinc is +2.
Oxidation state of scandium is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → +3
Scandium (Z = 21) has the configuration [Ar] 3d1 4s2. It has only three electrons beyond the argon core, and it loses all of them to form \( \ce{Sc^{3+}} \), [Ar] 3d0, a stable noble-gas configuration. Scandium therefore shows only the +3 oxidation state; Sc(II) is virtually unknown. Like zinc (+2 only), scandium shows a single oxidation state.
NCERT (2026-27 edition):
Oxidation States – “The lesser number of oxidation states at the extreme ends stems from either too few electrons to lose or share (Sc, Ti) or too many d electrons (hence fewer orbitals available in which to share electrons with others) for higher valence (Cu, Zn). Thus, early in the series scandium(II) is virtually unknown and titanium (IV) is more stable than Ti(III) or Ti(II).”