(a) (i) Account for the following : (I) Orange colour of Cr₂O₇²⁻ ion changes to yellow when treated with an al

Question

Q. (a)

(i) Account for the following : (I) Orange colour of

Cr_2O_7^{2-}

ion changes to yellow when treated with an alkali. (II) Zn, Cd and Hg are non-transition elements. (III)

E°

value for

Mn^{3+}/Mn^{2+}

couple is highly positive (+1·57 V) as compared to

Cr^{3+}/Cr^{2+}

(ii) What happens when : (I) Manganate ion undergoes disproportionation reaction in acidic medium ? (II)

KMnO_4

is heated ?

(OR)

(b) Answer the following questions :

(i) What is 'Misch metal' ? Give its one use.

(ii) Write the formula of an oxoanion of chromium in which it shows the oxidation state equal to its group number.

(iii) Why does Vanadium pentoxide (

V_2O_5

) act as a catalyst ?

(iv) Why do transition elements have high enthalpies of atomisation ?

(v) How do you prepare

Na_2Cr_2O_7

from

Na_2CrO_4

CBSECBSE Class XII Board 2025Subjective· 5mImportance★★★★★

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Part (a)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). …

Part (b)Concept understanding — Variable Oxidation States

Variable Oxidation States – The Intuition

The Precise Statement

Important

Why This Happens – The Energy Picture

Note

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

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). …

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Part (a): $Cr_2O_7^{2-}+2OH^-\rightarrow2CrO_4^{2-}+H_2O$ ; Zn/Cd/Hg are $d^{10}$ non-transition; high $Mn^{3+}/Mn^{2+}$ from stable $d^5$ ; $3MnO_4^{2-}+4H^+\rightarrow2MnO_4^-+MnO_2+2H_2O$ ; $2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2$ .

Part (b): Misch metal = lanthanoid alloy (flints); $CrO_4^{2-}$ ; $V_2O_5$ catalyses via variable states; high atomisation from strong metallic bonding; acidify $Na_2CrO_4 \rightarrow Na_2Cr_2O_7$ .

Part (a)

(i)(I) Dichromate and chromate are in a pH-dependent equilibrium. In alkali the orange dichromate is converted to the yellow chromate:

Cr_2O_7^{2-}\ (\text{orange}) + 2OH^- \rightarrow 2CrO_4^{2-}\ (\text{yellow}) + H_2O

(i)(II) A transition element must have a partially filled $d$ -subshell in its atom or a common ion. Zn, Cd and Hg have a completely filled $d^{10}$ configuration in both their atoms and their usual $+2$ ions, so they are classified as non-transition (d-block but not typical transition) elements.

(i)(III) After losing electrons, $Mn^{2+}$ attains the especially stable half-filled $3d^5$ configuration; hence $Mn^{3+}$ is strongly reduced to $Mn^{2+}$ , giving a large positive $E^\circ$ (+1.57 V). In contrast $Cr^{3+}$ has the stable $3d^3$ ( $t_{2g}^3$ ) configuration, so it resists reduction and $E^\circ_{Cr^{3+}/Cr^{2+}}$ is small (negative).

(ii)(I) In acidic medium manganate ( $Mn^{+6}$ ) disproportionates into permanganate ( $Mn^{+7}$ ) and $MnO_2$ ( $Mn^{+4}$ ):

3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O

(ii)(II) On strong heating potassium permanganate decomposes: …

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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:

$\text{Sc} \rightarrow \text{Sc}^{3+} + 3e^-$ …
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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).

…

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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 v …

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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. …
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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⁺. …
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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.

…

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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 (C …

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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. …

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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).

…

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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 electro …

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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 sev …

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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. …
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Q.(a)

Part (a)

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