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.
Which amongst the following is a better oxidizing agent than the other?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Mn3+ and Mn2+
Note: The given answer is as per the official NTA Answer Key. However, there is more than one correct option for this question, as explained below.
Each option names two ions, and the first is claimed to be the better oxidising agent. An ion is a strong oxidising agent when it is easily reduced, that is, when its reduced form is more stable. The higher the standard reduction potential (E°) of the ion, the stronger it is as an oxidising agent.
Option (1) Mn2+ and Mn3+ — Incorrect. \( \ce{Mn^{2+}} \) is 3d5, a half-filled and stable configuration, so it has no tendency to gain an electron. Of the two, \( \ce{Mn^{3+}} \) is the better oxidising agent, so the order is reversed.
Option (2) Mn3+ and Mn2+ — Correct. \( \ce{Mn^{3+}} \) is 3d4. By taking one electron it becomes \( \ce{Mn^{2+}} \), which is 3d5, half-filled and extra stable. So \( \ce{Mn^{3+}} \) is strongly oxidising and is a better oxidising agent than \( \ce{Mn^{2+}} \).
\( \ce{Mn^{3+} + e^- -> Mn^{2+}} \) E° = +1.57 V
Option (3) Cu+ and Cu2+ — Is also correct. \( \ce{Cu^{+}} \) is reduced more easily than \( \ce{Cu^{2+}} \): E°(\( \ce{Cu^{+}} \)/Cu) = +0.52 V is higher than E°(\( \ce{Cu^{2+}} \)/Cu) = +0.34 V. \( \ce{Cu^{2+}} \)(aq) is the more stable ion because of its much more negative hydration enthalpy. So copper(I) compounds are unstable in water and disproportionate; one \( \ce{Cu^{+}} \) ion oxidises another:
\( \ce{2Cu^{+} -> Cu^{2+} + Cu} \)
So \( \ce{Cu^{+}} \) is a better oxidising agent than \( \ce{Cu^{2+}} \), and this option is also correct.
Option (4) Ni2+ and Ni3+ — Incorrect. +2 is the stable oxidation state of nickel (the typical species after manganese include \( \ce{Ni^{II}} \)). \( \ce{Ni^{3+}} \) would readily take an electron to become \( \ce{Ni^{2+}} \), so \( \ce{Ni^{3+}} \) is the better oxidising agent and the order is reversed.
NCERT (2026-27 edition):
Chemical Reactivity and E° Values – “… An examination of the Eo values for the redox couple M3+/M2+ (Table 4.2) shows that Mn3+ and Co3+ ions are the strongest oxidising agents in aqueous solutions.”
Trends in Stability of Higher Oxidation States – “… The stability of Cu2+ (aq) rather than Cu+(aq) is due to the much more negative ΔhydHo of Cu2+ (aq) than Cu+, which more than compensates for the second ionisation enthalpy of Cu.”