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 most stable oxidation state of Mn is;
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → +2
Transition metals show several oxidation states because both the ns and (n−1)d electrons can take part in bonding. An oxidation state is especially stable when the ion gets a half-filled (d5) or completely filled (d10) d subshell, since such arrangements have extra exchange energy.
Mn (Z = 25) is [Ar]3d54s2. Losing the two 4s electrons gives Mn2+: [Ar]3d5, a half-filled d subshell. This extra stability is what the passage refers to: the E°(\( \ce{Mn^{2+}} \)/Mn) value is more negative than the trend, because \( \ce{Mn^{2+}} \) forms so readily.
Manganese shows oxidation states from +2 to +7, but in the higher states it is an oxidising agent and tends to return to +2. So the most stable oxidation state of Mn is +2.
Why the other options are incorrect
Option (1) +3 — Incorrect. \( \ce{Mn^{3+}} \) (3d4) is a strong oxidising agent in water; it readily gains an electron to become the stable 3d5 \( \ce{Mn^{2+}} \).
Option (2) +1 — Incorrect. +1 is not a common oxidation state of manganese.
Option (4) +4 — Incorrect. Mn(IV), as in \( \ce{MnO2} \), is also an oxidising agent: \( \ce{MnO2} \) oxidises HCl to \( \ce{Cl2} \) and is itself reduced to \( \ce{Mn^{2+}} \).