Q.List the oxidation states shown by vanadium and state which is the most stable, giving a reason.

Sc has only one stable oxidation state, +3, because its electron configuration leaves no other energetically accessible partial-loss options.

Sc (Z = 21) has configuration [Ar] 3d$^1$ 4s$^2$. It loses all three of these electrons (3d$^1$ + 4s$^2$) to attain the stable, empty 3d$^0$ configuration of Sc$^{3+}$ — isoelectronic with Ar. There is no other oxidation state that gives comparable stability, so Sc shows only +3.

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Mn, sitting at the middle of the 3d series with the maximum number of unpaired d and s electrons available for bonding, shows oxidation states from +2 right up to +7.

Across the 3d transition series, the number of accessible oxidation states rises from Sc to Mn and then falls again toward Zn, because Mn ([Ar]3d5 4s2) has the maximum number of electrons (5 d + 2 s = 7) that can be progressively involved in bonding/oxidation. This gives Mn the widest oxidation-state range of any 3d e …

Transition (d-block) elements characteristically show variable oxidation states because their (n-1)d and ns electrons are close in energy and can both participate in bonding.

Reasoning: Transition elements have partially filled d-orbitals, and the energy difference between the (n-1)d and ns subshells is small. This allows a variable number of electrons from both subshells to be lost/involved in bonding, giving transition metals multiple stable oxidation states (e.g. Fe: +2, +3; Mn: +2 up to +7; Cu: +1, +2).

• Alkali metals (Group 1) show only +1.
• Alkaline earth metals (Group 2) show only +2. …

Manganese reaches +7 in permanganate.

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