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.
Identify the stable pair of ions:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Mn2+, Fe3+
An ion is especially stable when its d subshell is empty (d0), half-filled (d5) or full (d10). Working out the configurations:
Mn (Z = 25) [Ar] 3d54s2: \( \ce{Mn^{2+}} \) = 3d5, \( \ce{Mn^{3+}} \) = 3d4
Fe (Z = 26) [Ar] 3d64s2: \( \ce{Fe^{3+}} \) = 3d5
Ti (Z = 22) [Ar] 3d24s2: \( \ce{Ti^{3+}} \) = 3d1, \( \ce{Ti^{4+}} \) = 3d0
Sc (Z = 21) [Ar] 3d14s2: \( \ce{Sc^{2+}} \) = 3d1, \( \ce{Sc^{3+}} \) = 3d0
\( \ce{Mn^{2+}} \) and \( \ce{Fe^{3+}} \) are both 3d5, so both ions of the pair have the stable half-filled configuration.
Why the other options are incorrect
Option (1) Mn3+, Fe3+ — Incorrect. \( \ce{Fe^{3+}} \) is stable, but \( \ce{Mn^{3+}} \) (3d4) is unstable and strongly oxidising.
Option (3) Ti3+, Sc3+ — Incorrect. \( \ce{Sc^{3+}} \) (d0) is stable, but \( \ce{Ti^{3+}} \) (d1) is easily oxidised; Ti(IV) is the more stable state of titanium.
Option (4) Ti4+, Sc2+ — Incorrect. \( \ce{Ti^{4+}} \) (d0) is stable, but scandium(II) is virtually unknown.
NCERT (2026-27 edition):
Ionisation Enthalpies – “… A general trend of increasing values of second ionisation enthalpy is expected as the effective nuclear charge increases because one d electron does not shield another electron from the influence of nuclear charge because d-orbitals differ in direction. However, the trend of steady increase in second and third ionisation enthalpy breaks for the formation of Mn2+ and Fe3+ respectively. In both the cases, ions have d5 configuration.”