Manganese shows all the oxidation states from +2 to +7, but scandium and titanium, at the beginning of the first transition series, show fewer oxidation states. What is the reason?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → They have too few electrons to lose or share.
The oxidation states of a transition element come from the 4s and 3d electrons it can lose or share. The elements in or near the middle of the series have the most such electrons and show the greatest number of oxidation states.
Sc (Z = 21): 3d1 4s2, only 3 such electrons; Sc shows +3, and scandium(II) is virtually unknown.
Ti (Z = 22): 3d2 4s2, only 4 such electrons; titanium(IV) is more stable than Ti(III) or Ti(II).
Mn (Z = 25): 3d5 4s2, 7 such electrons; Mn shows every state from +2 to +7.
Option (1) They have too many d electrons, so fewer orbitals are available to share electrons with others. — Incorrect. This is the reason at the other end of the series, for copper and zinc. Scandium and titanium have only one and two d electrons.
Option (2) They have too few electrons to lose or share. — Correct. With only 3 (Sc) or 4 (Ti) outer electrons, they cannot reach many states, unlike manganese with 7.
Option (3) They have completely filled d orbitals in the ground state. — Incorrect. A completely filled d10 set belongs to zinc, whose only oxidation state is +2. Sc (3d1) and Ti (3d2) are far from full.
Option (4) Their d subshells are half-filled, which makes them very stable. — Incorrect. A half-filled 3d5 set is found in manganese, the element with the most oxidation states. Sc and Ti have only one or two d electrons.
NCERT (2026-27 edition):
4.3.4 Oxidation States – “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 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).”