Q.Which of the 3d series transition elements exhibits the largest number of oxidation states? What is the highest oxidation state of that element? OR Write the general electronic configuration of lanthanides.

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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