d-Block Elements
The metals of the first series with the exception of copper are relatively more reactive and are oxidised by 1M H+, though the actual rate at which these metals react with oxidising agents like hydrogen ion (H+) is sometimes slow. For example, titanium and vanadium, in practice, are passive to dilute non oxidising acids at room temperature. The E° values for M2+/M indicate a decreasing tendency to form divalent cations across the series. It is interesting to note that the E° values for Mn, Ni and Zn are more negative than expected from the general trend. Whereas the stabilities of half-filled d subshell (d5) in Mn2+ and completely filled d subshell (d10) in zinc are related to their E° values; for nickel, E° value is related to the highest negative enthalpy of hydration.
The increasing oxidising power in the series VO2+ < Cr2O72− < MnO4− could be due to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → The increasing stability of the lower species to which they are reduced.
An oxidising agent works by being reduced. The more stable the product of that reduction (the lower oxidation state it forms), the stronger the driving force, and the stronger the oxidising agent.
Working out the oxidation states from the charge balance:
\( \ce{VO2^{+}} \): x + 2(−2) = +1, so V is +5; it is reduced to \( \ce{VO^{2+}} \) (V +4).
\( \ce{Cr2O7^{2-}} \): 2x + 7(−2) = −2, so Cr is +6; it is reduced to Cr3+ (Cr, Z = 24, is [Ar]3d54s1; losing three electrons gives 3d3, a stable half-filled t2g level).
\( \ce{MnO4^{-}} \): x + 4(−2) = −1, so Mn is +7; it is reduced to Mn2+ (Mn, Z = 25, is [Ar]3d54s2; losing the two 4s electrons gives 3d5, a very stable half-filled d subshell).
\( \displaystyle \ce{VO2^+ -> VO^2+},\quad \ce{Cr2O7^2- -> Cr^3+},\quad \ce{MnO4^- -> Mn^2+} \)
Why the other options are incorrect
Option (1) The increasing stability of the lower species to which they are oxidised — Incorrect. An oxidising agent is reduced, not oxidised, and oxidation cannot lead to a lower oxidation state.
Option (2) The increasing stability of the higher species to which they are reduced — Incorrect. Reduction always gives a lower oxidation state, not a higher one.
Option (3) Decreasing stability of the lower species to which they are reduced — Incorrect. A less stable product would make the oxidising agent weaker, the opposite of the trend.
NCERT (2026-27 edition):
Trends in Stability of Higher Oxidation States – …
