Which of the first transition series elements shows the maximum number of oxidation states and why?

Question

Accepted Answer

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 $\text{Fe}^{2+}$ or three to become $\text{Fe}^{3+}$ . Manganese can show +2, +3, +4, +6, and +7. This is not random — it follows a clear pattern rooted in energy.

The Precise Statement

Important

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 ( $[\text{Ar}]\,3d^6 4s^2$ ), 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 $\text{Fe}^{2+}$ ). The next electron lost comes from the 3d orbital (giving $\text{Fe}^{3+}$ ). 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.

Note

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 $3d^5 4s^2$ , can lose all seven — giving $\text{MnO}_4^-$ 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: $\text{Cr}^{3+}$ is stable in acidic solution, but $\text{Cr}^{6+}$ (as chromate) is stable in alkaline medium.
The ligand: Some oxidation states are stabilised by certain ligands (this is where coordination chemistry meets redox).
Half-filled and fully-filled stability: $\text{Mn}^{2+}$ ( $d^5$ ) and $\text{Zn}^{2+}$ ( $d^{10}$ ) are particularly stable because of extra exchange energy and spherical symmetry.

Watch out

Do not memorise the list of oxidation states blindly. Understand that the range comes from the energy closeness of $d$ and $s$ orbitals, and the limits come from how many electrons can be removed before the ion becomes too unstable.

The Big Picture

Variable oxidation states are what make transition metals so useful in catalysis, batteries, and biological systems. An iron atom in haemoglobin can cycle between +2 and +3 to carry oxygen. Manganese in the photosynthesis enzyme cycles through +2, +3, and +4 to split water. The ability to change oxidation state with a small energy cost is the fundamental reason transition metals are the workhorses of redox chemistry.

The core idea: when the $d$ and $s$ orbitals are close in energy, the atom can lose different numbers of electrons from both — giving a range of stable oxidation states.

Because variable Oxidation States trends are easy to frame as one-mark or assertion-reason questions, Variable Oxidation States is a genuinely high-yield topic across CBSE Class 12 board exams, JEE Main, and NEET — searches for "Variable Oxidation States important questions" and "Variable Oxidation States class 12 chemistry notes" consistently point back to this exact d- and f-Block Elements concept from the NCERT curriculum.

Manganese (Z = 25, [Ar]3d⁵4s²) shows the maximum number of oxidation states in the first transition series: every state from +2 (loss of the two 4s electrons) up to +7 (all five 3d plus both 4s electrons involved, as in $\mathrm{MnO_4^-}$ ). The half-filled 3d⁵ arrangement means each of the five d electrons is singly occupied and available for bonding, so the accessible states run without a gap from +2 to +7.

✓Final answer

Mn — all states from +2 to +7, because its 3d⁵4s² configuration makes seven electrons usable

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 22 on this concept.

CBSE 2026Set V11 markMCQ
$Q. The common oxidation state shown by the element with atomic number 21 is$
(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 $[\text{Ar}]\,3d^1\,4s^2$ .

Scandium loses its two $4s$ electrons and its single $3d$ electron to attain the stable, noble-gas [Ar] configuration:
$\text{Sc} \rightarrow \text{Sc}^{3+} + 3e^-$
Having no further $d$ electrons to lose easily, +3 is the characteristic and stable oxidation state of scandium (higher states such as +4, +5 are not shown). Options (b), (c) and (d) are therefore incorrect.

✓Final answer
$(a) +3$
CBSE 2026Set ANNUAL1 markMCQ
$Q. What is the maximum oxidation state of Mn in its compounds?$
(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).

Thus Mn readily shows +7 in compounds such as potassium permanganate ( $KMnO_4$ ) and manganese heptoxide ( $Mn_2O_7$ ), where all 7 valence electrons are formally lost/shared. Common lower oxidation states like +2, +4, +6 are also seen, but +7 is the observed maximum.

✓Final answer
$KMnO_4$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following transition metals does not show variable oxidation state?$
(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 variability. In contrast: titanium shows +2, +3, +4; chromium shows +2, +3, +4, +6 (most stable +3); copper shows +1 and +2. So only Sc fails to show variable oxidation states.

✓Final answer
$(d) Sc.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Which element does not show variable oxidation state ?$
(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.

Scandium ([Ar]3d¹4s²): loses all 3 outer electrons to give Sc³⁺ ([Ar], empty 3d), which is exceptionally stable (empty d-subshell, noble gas core). Scandium essentially shows only the +3 state; no other oxidation state is significant/stable for it.

✓Final answer
$Scandium (option iv) does not show a variable oxidation state — it exists almost exclusively as +3.$
CBSE 2024Set 56/3/11 markMCQ
$Q. Which of the following does not show variable oxidation states ? (A) Fe (B) Cu (C) Mn (D) Sc$
›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⁺.

Could it show +1 or +2? In principle, losing only the $4s$ electrons would give +2, but this is extremely unstable and essentially never observed in normal chemistry. The +3 state is so overwhelmingly favorable (achieving a noble-gas configuration) that scandium effectively shows only one oxidation state.

Watch out

Don't confuse "variable oxidation states" with "multiple oxidation states that might theoretically exist." For an element to show variable oxidation states in practice, those states must be reasonably stable and observable in common compounds. Scandium's +1 and +2 states are so unstable they're not considered part of its normal chemistry.

The pattern

Notice that elements with more $d$ -electrons (Fe with 6, Mn with 5, Cu with 10) have multiple stable ways to lose electrons. Scandium, with just one $d$ -electron, has essentially one path: lose everything to get to [Ar].

✓Final answer
$The correct option is (D) Sc, as scandium shows essentially only the +3 oxidation state in its chemistry.$
CBSE 2024Set FZ1 markMCQ
$not$
(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.

Why Sc. Sc $=[Ar]3d^1 4s^2$ . On losing its two $4s$ and one $3d$ electrons it reaches the stable $[Ar]$ core as $Sc^{3+}$ ( $3d^0$ ); no other stable state exists, so its oxidation state is fixed at $+3$ .

The others. Ti ( $+2,+3,+4$ ), V ( $+2,+3,+4,+5$ ) and Cr ( $+2,+3,+6$ ) each show several states.

✓Final answer
$+3$
CBSE 2024Set D1 markMCQ
$Q. The maximum oxidation state of chromium is$
(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 (Cr2O7^2-) ions and in CrO3.

While +3 (e.g. Cr2O3, Cr3+) is the most stable state, the maximum oxidation state chromium exhibits is +6.

✓Final answer
$(D) +6.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. Element showing the highest number of oxidation states is -$
(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.

Mn (Z = 25, [Ar]3d5 4s2) exhibits oxidation states of +2, +3, +4, +5, +6 and +7 - all seven of its valence electrons (5 from 3d + 2 from 4s) can be involved in bonding, giving the maximum +7 state (as in KMnO4).

Compare: Fe mainly shows +2, +3 (rarely +4, +6); Co shows +2, +3; Ni mainly +2 (rarely +3, +4); Cr shows +2 to +6 (a wide range but not as many as Mn).

So Mn shows the greatest number of oxidation states among the options.

✓Final answer
$(a) Mn.$
CBSE 2024Set ANNUAL1 mark
$Q. Which element of the 3d series of the transition metals exhibits the largest number of oxidation states?$
›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).

Manganese sits at this peak: its electron configuration [Ar]3d5 4s2 allows it to display the widest range of oxidation states, most notably +2 (Mn2+), +4 (MnO2), +6 (MnO4^2-, manganate), and +7 (MnO4-, permanganate - its highest and most well-known oxidation state).

✓Final answer
$Manganese (Mn) exhibits the largest number of oxidation states (+2 to +7) among the 3d transition elements.$
CBSE 2023Set A1 mark
$Q. Match the following. Column A item: 'Mn'. Choose its correct match from Column B:$
(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 electrons (3d⁵4s²) — is +7, as seen in KMnO4 (potassium permanganate) and Mn2O7.

✓Final answer
$Mn \to (h) +7.$
CBSE 2023Set ANNUAL1 mark
$Q. Name a transition element which does not exhibit variable oxidation states.$
›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 $[Ar]\,3d^1 4s^2$ . 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. $Sc^{3+}$ ). 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 several $3d$ electrons giving rise to multiple stable ions), scandium exhibits only the $+3$ oxidation state in its compounds — it does not show the variable oxidation states characteristic of the rest of the series.

✓Final answer
$+3$
CBSE 2022Set E1 markMCQ
$Q. In which of the following is the oxidation state of Mn lowest ?$
(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.

Mn3O4: 4 oxygens give -8, so 3 Mn = +8, average = +8/3 ≈ +2.67 (a mixed +2 and +4 oxide).

Mn2O7: 7 oxygens give -14, so 2 Mn = +14, Mn = +7.

The lowest oxidation state, +2, occurs in MnSO4.

✓Final answer
$(a) MnSO4.$

> [!ANSWER]
> Mn — all states from +2 to +7, because its 3d⁵4s² configuration makes seven electrons usable

Q.Which of the first transition series elements shows the maximum number of oxidation states and why?

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