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) on too many d-electrons (hence fewer orbitals available in which to share electrons with others) for higher valence (Cu, Zn). Thus early in the early 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 as 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 noticedamong 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 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 other?

\(Mn^{3+}\) and \(Mn^{2+}\)

The correct answer is option 2. \(Mn^{3+}\) and \(Mn^{2+}\).

Let us go through the concepts step by step to understand which among the given options is the better oxidizing agent.

Oxidizing Agent: A substance that oxidizes another substance by accepting electrons. In the process, the oxidizing agent itself gets reduced.

Reduction Potential: The tendency of a chemical species to gain electrons and be reduced. The higher (more positive) the reduction potential, the stronger the oxidizing agent.

Oxidation States and Stability: The stability of different oxidation states depends on the element and its electron configuration. A higher oxidation state generally corresponds to a species that is more easily reduced, making it a stronger oxidizing agent.

Analysis of Each Pair

Option 1: \(Mn^{2+}\) and \(Mn^{3+}\)

Manganese (Mn):

Electronic Configuration: \( [Ar] 3d^5 4s^2 \)

\(Mn^{2+}\): \( [Ar] 3d^5 \) (stable half-filled \(d\)-orbital)

\(Mn^{3+}\): \( [Ar] 3d^4 \) (less stable)

Reduction Potential:

\(Mn^{3+} + e^- \rightarrow Mn^{2+}\)

The reduction of \(Mn^{3+}\) to \(Mn^{2+}\) has a high positive reduction potential, indicating that \(Mn^{3+}\) is a strong oxidizing agent because it readily gains an electron to become the more stable \(Mn^{2+}\).

Conclusion: Between \(Mn^{2+}\) and \(Mn^{3+}\), **\(Mn^{3+}\)** is the better oxidizing agent.

Option 2: \(Mn^{3+}\) and \(Mn^{2+}\)

This is just the reverse order of Option 1, but the conclusion remains the same. \(Mn^{3+}\) is the better oxidizing agent compared to \(Mn^{2+}\).

Option 3: \(Cu^{+}\) and \(Cu^{2+}\)

Copper (Cu):

Electronic Configuration: \( [Ar] 3d^{10} 4s^1 \)

\(Cu^{+}\): \( [Ar] 3d^{10} \)

\(Cu^{2+}\): \( [Ar] 3d^9 \)

Reduction Potential:

\(Cu^{2+} + e^- \rightarrow Cu^{+}\)

The reduction of \(Cu^{2+}\) to \(Cu^{+}\) has a positive reduction potential, indicating that \(Cu^{2+}\) is a better oxidizing agent than \(Cu^{+}\).

Conclusion: Between \(Cu^{+}\) and \(Cu^{2+}\), \(Cu^{2+}\) is the better oxidizing agent.

Option 4: \(Ni^{2+}\) and \(Ni^{3+}\)

Nickel (Ni):

Electronic Configuration: \( [Ar] 3d^8 4s^2 \)

\(Ni^{2+}\): \( [Ar] 3d^8 \)

\(Ni^{3+}\): \( [Ar] 3d^7 \)

Reduction Potential:

\(Ni^{3+} + e^- \rightarrow Ni^{2+}\)

The reduction of \(Ni^{3+}\) to \(Ni^{2+}\) makes \(Ni^{3+}\) a stronger oxidizing agent, though it’s less common.

Conclusion: Between \(Ni^{2+}\) and \(Ni^{3+}\), \(Ni^{3+}\) is the better oxidizing agent.

Comparison and Final Conclusion

Now, let’s compare the strongest oxidizing agents from the pairs:

\(Mn^{3+}\) (from Option 1 and 2)

\(Cu^{2+}\) (from Option 3)

\(Ni^{3+}\) (from Option 4)

Among these, \(Mn^{3+}\) is a very strong oxidizing agent, more so than \(Cu^{2+}\) or \(Ni^{3+}\) due to its position in the periodic table and the instability of \(Mn^{3+}\) in comparison to its other oxidation states.

Final Answer: The correct option is Option 2: \(Mn^{3+}\) and \(Mn^{2+}\), indicating that \(Mn^{3+}\) is a better oxidizing agent than \(Mn^{2+}\).