For the four successive transition elements (Cr, Mn, Fe and Co), the stability of +2 oxidation state will be there in which of the following order? (At. No Cr= 24, Mn= 25, Fe=26, Co= 27)

The stability of the +2 oxidation state in aqueous solution is determined by the standard electrode potential ($E^{\ominus}$) for the reaction $M^{2+} + 2e^- \rightarrow M$. A more negative $E^{\ominus}$ value indicates a higher tendency to form the divalent ion ($M^{2+}$) from the metal, implying greater stability.

1. Values from NCERT (Table 4.4): $Mn$: $-1.18\text{ V}$ $Cr$: $-0.90\text{ V}$ $Fe$: $-0.44\text{ V}$ $Co$: $-0.28\text{ V}$

2. Analysis:

• Manganese (Mn): Has the most negative $E^{\ominus}$ due to the stability of the half-filled $d^5$ configuration in $Mn^{2+}$. It is the most stable +2 species.
• Chromium (Cr) vs Iron (Fe): Although $Cr$ has a more negative $E^{\ominus}$ for formation than $Fe$, $Cr^{2+}$ ($d^4$) is a strong reducing agent and easily oxidizes to the stable $t_{2g}^3$ configuration of $Cr^{3+}$ . In contrast, $Fe^{2+}$ ($d^6$) is less prone to oxidation than $Cr^{2+}$ in acidic solution (though it can be oxidized to $Fe^{3+}$). Therefore, in terms of resistance to chemical change (redox stability), $Fe^{2+}$ is often considered more stable than the reducing $Cr^{2+}$.
• Cobalt (Co): Has the least negative $E^{\ominus}$, indicating the least tendency to form the +2 ion from the metal among the given elements.

3. Conclusion: The order considering both formation tendency and resistance to further oxidation aligns with Mn > Fe > Cr > Co.