Q.Which of the 3d transition metals exhibits the largest number of oxidation states?
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Variable Oxidation States
Variable Oxidation States – The Intuition
Think of an atom as having a wallet with two compartments. In most elements, one compartment is much easier to open than the other — you can only take money from the shallow one, so the amount you can spend (the oxidation state) is fixed. For transition metals, both compartments are at nearly the same depth. You can reach into either, and you can take different combinations of notes from each. That is variable oxidation states in a nutshell.
Iron, for example, can lose two electrons to become Fe2+ or three to become Fe3+. Manganese can show +2, +3, +4, +6, and +7. This is not random — it follows a clear pattern rooted in energy.
The Precise Statement
Transition metals exhibit variable oxidation states because the (n−1)d and ns subshells have similar energies. Electrons can be removed from both subshells in different numbers, producing a range of stable positive oxidation states.
The key is similar energies. In main-group elements (like sodium or chlorine), the outermost ns and np electrons are far higher in energy than the inner core — you lose only the valence electrons, and the oxidation state is fixed. In transition metals, the (n−1)d orbital is not much lower than the ns orbital. Both are close enough that losing a few d electrons along with the s electrons costs comparable energy.
Why This Happens – The Energy Picture
For a transition metal like iron ([Ar]3d64s2), the 4s orbital is actually slightly lower in energy than the 3d when the atom is neutral. But once you start removing electrons, the energy ordering shifts. The first two electrons lost are from the 4s orbital (giving Fe2+). The next electron lost comes from the 3d orbital (giving Fe3+). Because the 3d and 4s are so close in energy, removing that third electron does not require a huge jump in energy — it is feasible.
The actual order of filling is 4s before 3d, but the order of removal is also 4s first. This is not a contradiction — it is a consequence of how orbital energies change as the nuclear charge increases.
The Pattern Across the Series
For the first transition series (Sc to Zn), the common oxidation states are:
| Element | Common oxidation states |
|---|---|
| Sc | +3 |
| Ti | +3, +4 |
| V | +2, +3, +4, +5 |
| Cr | +2, +3, +6 |
| Mn | +2, +3, +4, +6, +7 |
| Fe | +2, +3 |
| Co | +2, +3 |
| Ni | +2 |
| Cu | +1, +2 |
| Zn | +2 |
Notice the trend: the maximum oxidation state increases from Sc (+3) to Mn (+7), then decreases. The maximum possible oxidation state equals the total number of electrons in the (n−1)d and ns orbitals (the "group number" for many). Manganese, with 3d54s2, can lose all seven — giving MnO4− where Mn is +7. After manganese, the d orbitals become more stable (higher effective nuclear charge), and it becomes harder to remove all of them.
Stability and the Environment
Not all oxidation states are equally stable. The stability depends on:
- The medium: Cr3+ is stable in acidic solution, but Cr6+ (as chromate) is stable in alkaline medium.
- The ligand: Some oxidation states are stabilised by certain ligands (this is where coordination chemistry meets redox). …
The number of oxidation states a 3d metal can show generally tracks how many d and s electrons are available for bonding, which peaks partway through the series. …
Mn shows oxidation states from +2 to +7 (all consecutive states are known), the widest range in the 3d series, because its half-filled 3d⁵4s² configuration allows a variable number of electrons to be lost/shared.
Among the 3d series (Sc to Zn), the number of oxidation states shown by an element generally increases up to Mn (group 7) and then decreases towards Zn, because:
- Up to Mn, the number of unpaired d and s electrons available for bonding increases (Mn: [Ar]3d⁵4s²), giving a maximum possible spread of oxidation states.
- The exactly half-filled 3d⁵ configuration is exceptionally stable, so Mn can lose anywhere from 2 electrons (Mn²⁺) up to all 7 valence electrons (Mn⁷⁺, e.g. in KMnO4, MnO4⁻) with reasonable stability. …
Showing the 12 most recent of 22 on this concept.
- CBSE 2026Set V11 markMCQQ.The common oxidation state shown by the element with atomic number 21 is(a) +3(b) +4(c) +5(d) Both +3 and +5
›Reveal solutionSolution
The element with Z = 21 is scandium, whose common (and essentially only stable) oxidation state is +3.
Atomic number 21 corresponds to scandium (Sc) with electronic configuration [Ar]3d14s2.
Scandium loses its two 4s electrons and its single 3d electron to attain the stable, noble-gas [Ar] configuration:
Sc→Sc3++3e− …
- CBSE 2026Set ANNUAL1 markMCQQ.What is the maximum oxidation state of Mn in its compounds?(a) +4(b) +5(c) +6(d) +7
›Reveal solutionSolution
Manganese shows a maximum oxidation state of +7, equal to the sum of its 4s and 3d valence electrons.
Manganese has the ground-state electronic configuration [Ar]3d^5 4s^2, giving it 7 electrons in its outermost (4s + 3d) shells. For the early-to-middle members of the 3d transition series, the maximum oxidation state shown is equal to the total number of 4s and 3d electrons, since all of them can, in principle, take part in bonding (this trend peaks around Mn and then declines as d-electrons become increasingly core-like towards the end of the series).
…
- CBSE 2025Set ANNUAL1 markMCQQ.Which of the following transition metals does not show variable oxidation state?(a) Ti(b) Cr(c) Cu(d) Sc
›Reveal solutionSolution
Sc has only one electron beyond the noble-gas+d0 core to lose (3d1 4s2 -> Sc3+ is d0), so there is no intermediate oxidation state available; Ti, Cr and Cu all show at least two.
Scandium's configuration is [Ar]3d¹4s²; losing all three of these electrons gives the very stable, empty-d-subshell Sc³⁺ (3d⁰) ion, which is the only oxidation state scandium is practically found in — it has no partly-filled-d intermediate oxidation state to show v …
- CBSE 2025Set ANNUAL1 markMCQQ.Which element does not show variable oxidation state ?(i) Vanadium(ii) Iron(iii) Mercury(iv) Scandium
›Reveal solutionSolution
Scandium has only one stable, common oxidation state (+3) because losing all three electrons outside its noble-gas-like [Ar] core empties the 3d subshell completely — there is no other accessible, stable configuration.
Most transition metals show variable oxidation states because both the (n-1)d and ns electrons are close in energy and can be lost in different numbers.
- Vanadium: shows +2, +3, +4, +5 — clearly variable.
- Iron: shows +2 and +3 (and rarely +6) — variable.
- Mercury: shows +1 (as Hg₂²⁺) and +2 — variable. …
- CBSE 2024Set 56/3/11 markMCQQ.Which of the following does not show variable oxidation states ? (A) Fe (B) Cu (C) Mn (D) Sc
›Reveal solutionSolution
Transition metals show variable oxidation states when they can lose different numbers of d-electrons along with their s-electrons. Scandium has only one d-electron, giving it essentially one stable oxidation state (+3), while Fe, Cu, and Mn have multiple d-electrons that can be removed in different combinations. The answer is (D) Sc.
Why transition metals show variable oxidation states
Transition metals are famous for their ability to exist in multiple oxidation states. This happens because their (n−1)d and ns orbitals are close in energy, so electrons from both can participate in bonding. The more d-electrons available, the more combinations of electron loss are possible, leading to a richer variety of oxidation states.
The key is to look at the electronic configuration and see how many electrons can realistically be removed to form stable ions.
Analyzing each element
Let's examine the electronic configurations and common oxidation states:
1. Iron (Fe): [Ar] 3d⁶ 4s²
Iron can lose its two 4s electrons to give Fe²⁺ ([Ar] 3d⁶). It can also lose one more 3d electron to give Fe³⁺ ([Ar] 3d⁵), which is particularly stable due to the half-filled d-subshell. Higher oxidation states like +4, +5, and +6 exist in certain compounds, though they're less common.
Common oxidation states: +2, +3 (and higher in special cases)
2. Copper (Cu): [Ar] 3d¹⁰ 4s¹
Copper readily loses its single 4s electron to form Cu⁺ ([Ar] 3d¹⁰), which has a stable filled d-subshell. It can also lose one 3d electron to give Cu²⁺ ([Ar] 3d⁹), which is actually more common in aqueous chemistry due to higher hydration energy.
Common oxidation states: +1, +2
3. Manganese (Mn): [Ar] 3d⁵ 4s²
Manganese is the champion of variable oxidation states among first-row transition metals. With five d-electrons and two s-electrons, it can lose anywhere from two to all seven electrons, giving oxidation states from +2 all the way to +7 (as in permanganate, MnO₄⁻).
Common oxidation states: +2, +3, +4, +6, +7
4. Scandium (Sc): [Ar] 3d¹ 4s²
Here's the critical case. Scandium has only one d-electron. When it forms compounds, it loses both 4s electrons and its single 3d electron to achieve the stable [Ar] configuration, giving Sc³⁺. …
- CBSE 2024Set FZ1 markMCQQ.The transition element in which variable oxidation state is not found, is:(a) Sc(b) Ti(c) V(d) Cr
›Reveal solutionSolution
Scandium exhibits only the +3 state, so variable oxidation state is not found in Sc → option (a).
Concept. Transition metals normally show variable oxidation states because their (n−1)d and ns electrons have similar energies, so a variable number can be involved in bonding. The exception is an element that has just one accessible state.
…
- CBSE 2024Set D1 markMCQQ.The maximum oxidation state of chromium is(a) +2(b) +3(c) +4(d) +6
›Reveal solutionSolution
Cr has the configuration [Ar]3d5 4s1, i.e. six electrons (5 in 3d + 1 in 4s) available for bonding, so its highest oxidation state is +6.
Chromium (Z = 24) has electronic configuration [Ar]3d5 4s1. All six electrons in the 3d and 4s subshells can participate in bonding, so chromium can reach the +6 oxidation state, seen in chromate (CrO4^2-) and dichromate (C …
- CBSE 2024Set ANNUAL1 markMCQQ.Element showing the highest number of oxidation states is -(a) Mn(b) Ni(c) Fe(d) Cr
›Reveal solutionSolution
Manganese, with the electronic configuration [Ar]3d5 4s2, shows oxidation states ranging from +2 to +7, the widest range of any 3d transition element.
Among the first transition series, the number of oxidation states shown by an element is generally maximum near the middle of the series, where the largest number of unpaired d and s electrons are available for bonding. …
- CBSE 2024Set ANNUAL1 markQ.Which element of the 3d series of the transition metals exhibits the largest number of oxidation states?
›Reveal solutionSolution
Manganese, with its half-filled 3d5 4s2 configuration, can lose varying numbers of electrons to give the widest spread of oxidation states among the first-row (3d) transition metals: +2, +3, +4, +5, +6, and +7.
Across the 3d transition series, the number of accessible oxidation states generally increases from Sc to Mn (as more d and s electrons become available for bonding/removal) and then decreases again from Fe to Zn (as the increasing nuclear charge makes it harder to remove d electrons and pairing energy effects set in).
…
- CBSE 2023Set A1 markQ.Match the following. Column A item: 'Mn'. Choose its correct match from Column B:(a) Ether(b) Primary amine(c) Lactose(d) C12H22O11(e) Glucose(f) Negative ions(g) C6H5SO2Cl(h) +7
›Reveal solutionSolution
Manganese exhibits a maximum oxidation state of +7, seen in the permanganate ion.
Among the given items, 'Mn' pairs with option (h) +7, because manganese's highest possible oxidation state — using all its 3d and 4s electro …
- CBSE 2023Set ANNUAL1 markQ.Name a transition element which does not exhibit variable oxidation states.
›Reveal solutionSolution
Most transition elements show variable oxidation states because both 4s and partially-filled 3d electrons can be lost with comparable ease; scandium is the exception among the first-row transition series, showing only the +3 state.
Scandium has the electronic configuration [Ar]3d14s2. On forming compounds, it loses all three of these electrons (the two 4s and the single 3d) to attain the very stable, empty-d, noble-gas-like configuration [Ar] (i.e. Sc3+). Because this +3 ion is exceptionally stable and there is no comparably accessible alternative oxidation state (there is only one 3d electron to begin with, unlike elements such as Mn or Fe that have sev …
- CBSE 2022Set E1 markMCQQ.In which of the following is the oxidation state of Mn lowest ?(a) MnSO4(b) MnO2(c) Mn3O4(d) Mn2O7
›Reveal solutionSolution
Mn is +2 in MnSO4, +4 in MnO2, +8/3 in Mn3O4, and +7 in Mn2O7 — so the lowest is +2 in MnSO4.
Compute the oxidation state of Mn in each:
- MnSO4: SO4 is -2, so Mn = +2.
- MnO2: 2 oxygens give -4, so Mn = +4. …
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