Read the passage given and answer the question.
The transition 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 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). At the other end, the only oxidation state of zinc is +2 (no d electrons are involved). The maximum oxidation states of reasonable stability correspond in value to the sum of s- and d-electrons upto manganese \((Ti^{IV}O_2, V^VO_2^+, Cr^{VI}O_4^{2-}, Mn^{VII}O_4^- )\) followed by a rather abrupt decrease in stability of higher oxidation states, so that the typical species to follow are \(Fe^{II, III}\), \(Co^{II, III}\), \(Cu^{I, II}\), \(Zn^{II}\).
The variability of oxidation states, a characteristic of transition elements, arises out of incomplete filling of d-orbitals in such a way that their oxidation states differ from each other by unity, e.g., \(V^{II}\), \(V^{III}\), \(V^{IV}\), \(V^V\). This is in contrast with the variability of oxidation states of non transition elements where oxidation states normally differ by a unit or two.
An interesting feature in the variability of oxidation states of the d-block elements is noticed among the groups (groups 4 through 10). Although in the p-block, the lower oxidation states are favoured by the heavier members (due to inert pair effect), the opposite is true in the groups of d-block. For example, in group 6, \(Mo^{(VI)}\) and \(W^{(VI)}\) are found to be more stable than \(Cr^{(VI)}\). The \(Cr^{(VI)}\) in the form of dichromate in acidic medium is a strong oxidising agent, whereas \(MoO_3\) and \(WO_3\) are not.
Low oxidation states are found when a complex compound has ligands capable of \(\pi \)-acceptor character in addition to the \(\sigma \)-bonding. For example, in \(Ni(CO)_4\) and \(Fe(CO)_5\), the oxidation state of nickel and iron is zero.
The elements giving minimum number of oxidation states are in extreme left and right in 3d series. This is because:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → their elements have least number of unpaired electrons
The oxidation states of a transition element come from its 4s electrons and the unpaired 3d electrons that can be lost or shared. Working out the configurations:
Sc (Z = 21): [Ar] 3d14s2, one unpaired d electron
Ti (Z = 22): [Ar] 3d24s2, two unpaired d electrons
Mn (Z = 25): [Ar] 3d54s2, five unpaired d electrons
Zn (Z = 30): [Ar] 3d104s2, no unpaired d electron
So the elements at the two ends of the series have the fewest unpaired electrons and show the fewest oxidation states (Sc only +3, Zn only +2), while manganese in the middle has the most.
Why the other options are incorrect
Option (2) their last electrons enter in ‘f’ orbitals — Incorrect. In the 3d series the differentiating electron enters the 3d orbitals; f orbitals are filled only in the lanthanoids and actinoids.
Option (3) these have high enthalpy of atomization — Incorrect. The enthalpy of atomisation is highest near the middle of the series; zinc at the right end has the lowest value, so this does not explain fewer oxidation states.
Option (4) these elements are inert — Incorrect. Sc and Zn are reactive metals; they readily form \( \ce{Sc^{3+}} \) and \( \ce{Zn^{2+}} \) compounds.
NCERT (2026-27 edition):
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).”