Question
Why do transition metals (d-block elements) show variable oxidation states? Explain with examples. Why is Sc(III) and Zn(II) the only stable oxidation states for scandium and zinc respectively?
Solution — Step by Step
Transition metals are elements in which the d subshell is partially filled — either in the element itself or in one of its stable ions. They belong to groups 3–11 of the periodic table (excluding Zn which has a full d-subshell).
General electronic configuration:
For the 3d transition series (Period 4): Sc (3d¹4s²) through Zn (3d¹⁰4s²).
Transition metals show variable oxidation states for two key reasons:
Reason 1: Small energy difference between (n-1)d and ns orbitals
In transition metals, the energy difference between the and orbitals is very small (unlike main-group elements where the energy gap between valence levels is large). This means electrons from BOTH the and orbitals can be involved in bonding/ionisation.
For example, iron (Fe: [Ar] 3d⁶4s²):
- Remove 2 electrons (both 4s): Fe²⁺ (3d⁶) — iron(II)
- Remove 3 electrons (2 from 4s + 1 from 3d): Fe³⁺ (3d⁵) — iron(III)
Both Fe²⁺ and Fe³⁺ are stable. The energy required to remove the third electron (from 3d) is not much larger than removing the first two (from 4s), so both oxidation states are accessible under normal chemical conditions.
Reason 2: Stability through d orbital filling patterns
Extra stability arises with completely empty d orbitals (d⁰), half-filled d orbitals (d⁵), and completely filled d orbitals (d¹⁰). This stability influences which oxidation states are preferred.
Manganese (Mn: [Ar] 3d⁵4s²) shows +2 (d⁵, half-filled — especially stable), +3, +4, +5, +6, +7 (permanganate, d⁰ for Mn in +7).
Chromium (Cr: [Ar] 3d⁵4s¹) shows +2, +3 (most stable, d³), +6 (chromate/dichromate, d⁰ — also stable).
The table shows the variety:
| Element | Config | Common Oxidation States | Highest O.S. | Most Stable |
|---|---|---|---|---|
| Sc | 3d¹4s² | +3 only | +3 | +3 |
| Ti | 3d²4s² | +2, +3, +4 | +4 | +4 |
| V | 3d³4s² | +2, +3, +4, +5 | +5 | +5 |
| Cr | 3d⁵4s¹ | +2, +3, +6 | +6 | +3 |
| Mn | 3d⁵4s² | +2, +3, +4, +6, +7 | +7 | +2 |
| Fe | 3d⁶4s² | +2, +3 | +3 (common) | +2, +3 |
| Co | 3d⁷4s² | +2, +3 | +3 | +2 |
| Ni | 3d⁸4s² | +2 (mainly) | +4 (rare) | +2 |
| Cu | 3d¹⁰4s¹ | +1, +2 | +2 | +2 |
| Zn | 3d¹⁰4s² | +2 only | +2 | +2 |
The maximum oxidation state increases from Sc(+3) to Mn(+7), then decreases. The maximum oxidation state roughly equals the number of (n-1)d + ns electrons available for bonding, until the d orbitals become too stabilised to participate easily.
Scandium (Sc: [Ar] 3d¹4s²): Removal of all three valence electrons (3d¹ + 4s²) gives Sc³⁺ with the [Ar] noble gas configuration — extremely stable. Adding back to Sc²⁺ (3d¹4s²) would require more energy than is gained. All three electrons are easily available (small ionisation energies for a 4s²3d¹ system relative to the stability of the noble gas core). Lower oxidation states (Sc⁺, Sc²⁺) are very unstable. So Sc shows only +3.
Zinc (Zn: [Ar] 3d¹⁰4s²): Zinc has a completely filled d orbital (3d¹⁰). Removing the 4s² electrons gives Zn²⁺ with a 3d¹⁰ configuration — completely filled d subshell, which is very stable. To form Zn³⁺ or Zn⁴⁺, electrons would have to be removed from the stable, filled 3d subshell — this requires very high energy and does not occur under normal conditions. Zn¹⁺ (3d¹⁰4s¹) is unstable because of disproportionation. So Zn shows only +2.
This is also why zinc is NOT considered a true transition metal by strict IUPAC definition — it has no partially filled d orbital in either the metal or its common ion.
Why This Works
The variable oxidation states of transition metals arise from the unique energetic situation of having partially filled d orbitals with energies close to the s orbitals. This is a quantum mechanical consequence of effective nuclear charge and orbital shielding — the d electrons shield each other poorly, but collectively they stabilise the outer s electrons to different extents.
The stability of particular oxidation states is determined by which electron configuration gives the lowest total energy, considering both the ionisation energy cost and the lattice/solvation energy gain.
Alternative Method
A simpler version for quick revision: “Transition metals lose both ns and (n-1)d electrons (not just ns electrons like typical metals), giving multiple possible positive oxidation states.” This is the essence, without the deeper stability arguments.
Common Mistake
A frequent error is writing that all transition metals show ALL oxidation states from +1 to their maximum. This is wrong — each metal shows a specific set of stable states. For example, iron shows +2 and +3 commonly, but iron +7 is essentially non-existent. The stable oxidation states are determined by the specific electron configuration and stability factors (half-filled, fully-filled d), not just by “how many electrons you can remove.”
For JEE, the most tested aspects of this topic are: (1) the colour of transition metal ions (due to d-d transitions), (2) the highest oxidation state shown by different metals, (3) why some states are more stable than others (d⁵ half-filled, d⁰, d¹⁰), (4) why Zn²⁺ solutions are colourless (d¹⁰ → no d-d transition possible).