(a) (i) Describe giving reason which one of the followi — CBSE Class 12 Chemistry 2026 PYQ

Question

Q. (a)

(i) Describe giving reason which one of the following pairs has the property indicated : (I) Fe or Cu

-

higher melting point (II)

Ti^{3+}

Sc^{3+}

-

coloured in aqueous solution (III) Cr or Zn

-

higher third ionisation enthalpy

(ii) Write the ionic equations for the oxidizing action of

MnO_4^-

in acidic medium with : (I)

Fe^{2+}

ion (II)

I^-

ion

(OR)

(b)

(i) A black-brown coloured solid (A) when fused with KOH in the presence of air, produces a dark green coloured compound (B) which on electrolytic oxidation in alkaline medium gives a dark purple coloured compound (C). Identify (A), (B) and (C). Write the reactions involved.

(ii) What happens when an acidic solution of the green compound (B) is allowed to stand for some time ? Also write the equation involved. What is this type of reaction called ?

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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 — Disproportionation Reaction

Disproportionation Reactions: The Self-Oxidation-Reduction

The Intuition

Imagine you have a group of friends who are all equally wealthy — each has exactly ₹100. Now suppose one friend decides to give ₹50 to another. After this transaction, one friend has ₹50 (lost money), another has ₹150 (gained money), and the rest are unchanged. Notice something: the same action — transferring money — made one person poorer and another richer.

A disproportionation reaction works on a similar principle, but with electrons instead of money. One atom of an element simultaneously gets oxidised (loses electrons) and reduced (gains electrons). The same element ends up in two different oxidation states — one higher, one lower — starting from a single intermediate oxidation state.

Note

The word "disproportionation" literally means "breaking apart into unequal parts." The original state splits into two different states.

The Precise Definition

A disproportionation reaction is a redox reaction in which a single substance (element or compound) in an intermediate oxidation state is simultaneously oxidised and reduced, producing two different products — one with a higher oxidation state and one with a lower oxidation state.

The general form looks like this:

\text{Element in intermediate state} \longrightarrow \text{Higher oxidation state} + \text{Lower oxidation state}

The Key Condition

For disproportionation to occur, the element must be in an intermediate oxidation state — meaning it can both increase and decrease its oxidation number. If the element is already in its highest possible oxidation state, it can only be reduced. If it's in its lowest, it can only be oxidised. No disproportionation possible.

Important

Disproportionation requires the element to have at least three accessible oxidation states: one lower, one intermediate (the starting point), and one higher.

Classic Example: Hydrogen Peroxide

Hydrogen peroxide ( $H_2O_2$ ) is the textbook example. Oxygen in $H_2O_2$ has an oxidation state of -1. This is intermediate — oxygen can go to 0 (in $O_2$ ) or to -2 (in $H_2O$ ).

When $H_2O_2$ decomposes:

2H_2O_2 \longrightarrow 2H_2O + O_2

Let's track the oxygen:

In $H_2O_2$ : oxidation state = -1
In $H_2O$ : oxidation state = -2 (reduction — gained an electron)
In $O_2$ : oxidation state = 0 (oxidation — lost an electron)

The same oxygen atoms (from the same molecule) undergo both oxidation and reduction. That's disproportionation.

Another Common Example: Copper(I) in Solution

Copper(I) ion ( $Cu^+$ ) is unstable in aqueous solution and disproportionates:

2Cu^+ \longrightarrow Cu + Cu^{2+}

$Cu^+$ (oxidation state +1) is the intermediate
$Cu$ (oxidation state 0) is the reduced product
$Cu^{2+}$ (oxidation state +2) is the oxidised product

Watch out

A common mistake is to think that a single atom does both oxidation and reduction. In reality, two atoms of the same element are involved — one gets oxidised, the other gets reduced. The reaction requires at least two formula units of the starting substance.

How to Identify a Disproportionation Reaction

Look for a single reactant that contains an element in an intermediate oxidation state.
Check the products — the same element must appear in two different oxidation states (one higher, one lower than the starting state). …

Why this formula?

Disproportionation Reaction — Understanding the Why

A disproportionation reaction is a redox reaction where the same element in one oxidation state simultaneously undergoes oxidation (increase in oxidation number) and reduction (decrease in oxidation number).

The key formula that governs whether such a reaction is spontaneous is based on the standard electrode potentials ( $E^\circ$ ) of the two half-reactions.

The Core Idea: Why Does Disproportionation Happen?

For an element in an intermediate oxidation state, it can be both oxidised and reduced.

Whether this happens spontaneously depends on the relative ease of these two processes.

Consider an element $X$ in oxidation state $+n$ :

Oxidation half-reaction:

$X^{+n} \rightarrow X^{+(n+1)} + e^-$

(loss of electron, oxidation number increases)
Reduction half-reaction:

$X^{+n} + e^- \rightarrow X^{+(n-1)}$

(gain of electron, oxidation number decreases)

The overall disproportionation reaction is:

2X^{+n} \rightarrow X^{+(n+1)} + X^{+(n-1)}

The Key Formula: Spontaneity Condition

For a disproportionation reaction to be spontaneous (under standard conditions), the overall cell potential $E^\circ_{\text{cell}}$ must be positive.

Derivation:

Identify the two half-reactions and their standard reduction potentials ( $E^\circ$ ):
- Reduction half-reaction (the one that gains electrons):
  
  $X^{+n} + e^- \rightarrow X^{+(n-1)}$
  
  Let its standard reduction potential be $E^\circ_{\text{red}}$ .
- Oxidation half-reaction (the one that loses electrons):
  
  $X^{+n} \rightarrow X^{+(n+1)} + e^-$
  
  This is the reverse of a reduction. So its standard oxidation potential is $-E^\circ_{\text{ox}}$ , where $E^\circ_{\text{ox}}$ is the standard reduction potential for:
  
  $X^{+(n+1)} + e^- \rightarrow X^{+n}$
Overall cell potential is:

E^\circ_{\text{cell}} = E^\circ_{\text{reduction half-cell}} - E^\circ_{\text{oxidation half-cell}}

But careful: The oxidation half-cell is the reverse of a reduction. So we write:

E^\circ_{\text{cell}} = E^\circ_{\text{red}} - E^\circ_{\text{ox}}

where:

$E^\circ_{\text{red}}$ = standard reduction potential for $X^{+n} \rightarrow X^{+(n-1)}$
$E^\circ_{\text{ox}}$ = standard reduction potential for $X^{+(n+1)} \rightarrow X^{+n}$

Spontaneity condition:

\boxed{E^\circ_{\text{cell}} > 0 \quad \Rightarrow \quad E^\circ_{\text{red}} > E^\circ_{\text{ox}}}

In words: Disproportionation is spontaneous if the reduction potential for the lower oxidation state is greater than that for the higher oxidation state.

Why This Makes Sense — A Conceptual Explanation

$E^\circ_{\text{red}}$ tells you how easily $X^{+n}$ gets reduced to $X^{+(n-1)}$ . …

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Part (a): Fe > Cu in melting point (more unpaired d-electrons); $Ti^{3+}$ ( $d^1$ ) is coloured, $Sc^{3+}$ ( $d^0$ ) colourless; Zn has higher $IE_3$ (breaks $d^{10}$ ); $MnO_4^-$ oxidises $Fe^{2+}\to Fe^{3+}$ and $I^-\to I_2$ in acid.

Part (b): $MnO_2\to K_2MnO_4\to KMnO_4$ ; acidified manganate disproportionates: $3MnO_4^{2-}+4H^+\to 2MnO_4^-+MnO_2+2H_2O$ .

Part (a)

(i) Comparisons

(I) Fe or Cu – melting point: melting point rises with the number of unpaired d-electrons available for metallic bonding. Fe ( $3d^64s^2$ ) has 4 unpaired 3d electrons; Cu ( $3d^{10}4s^1$ ) has a filled 3d shell that hardly bonds. Fe has the higher melting point.
(II) $Ti^{3+}$ or $Sc^{3+}$ – colour: colour needs a $d$ – $d$ transition, i.e. at least one d-electron. $Ti^{3+}$ is $3d^1$ (coloured, violet); $Sc^{3+}$ is $3d^0$ (colourless). $Ti^{3+}$ is coloured.
(III) Cr or Zn – third ionisation enthalpy: After losing two electrons, $Cr^{2+}$ is $3d^4$ (3rd electron removed easily to give stable $3d^3$ ), whereas $Zn^{2+}$ is $3d^{10}$ (removing the 3rd electron destroys a stable filled shell). Zn has the higher $IE_3$ .

(ii) Oxidising action of $MnO_4^-$ in acidic medium

Half-reaction: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ .

With $Fe^{2+}$ ( $Fe^{2+}\to Fe^{3+}+e^-$ ):

MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}

With $I^-$ ( $2I^-\to I_2+2e^-$ ): …

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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 28 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 2026Set ANNUAL1 markMCQ
$Q. In the following reaction 4P + 3KOH + 3H2O -> 3KH2PO2 + PH3, which statement is correct?$
(a) 'P' is oxidised only
(b) 'P' is reduced only
(c) 'P' is oxidised as well as reduced
(d) 'P' is neither oxidised nor reduced
›Reveal solutionSolution

When the same element, starting from a single oxidation state, ends up in both a higher and a lower oxidation state among the products, that is a disproportionation reaction.

Reaction: 4P + 3KOH + 3H2O -> 3KH2PO2 + PH3

Oxidation state of P in elemental phosphorus (P4, written here as P): 0 (element in its standard state).

Oxidation state of P in KH2PO2 (potassium hypophosphite): Using K = +1, H = +1 (bonded to O, standard H), O = -2:

(+1) + 2(+1 for the two H bonded to O) ... more directly: for the hypophosphite ion H2PO2-, charge = -1. With 2 H at +1 and 2 O at -2: 2(+1) + x + 2(-2) = -1 => 2 + x - 4 = -1 => x = +1.

So P is +1 in KH2PO2 -- this is an INCREASE from 0, i.e. P is OXIDISED here.

Oxidation state of P in PH3: H bonded to P (a less electronegative element than H here, since P and H have similar/P slightly higher electronegativity by the modified scale used for this compound) is taken as -1 in this hydride convention: x + 3(-1) = 0 => x = +3?

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. In the disproportionation reaction 2 Cu⁺ (aq) ⇌ Cu (s) + Cu²⁺ (aq), the cuprous ion, Cu⁺$
(a) undergoes reduction only
(b) undergoes both reduction and oxidation
(c) undergoes oxidation only
(d) does not undergo redox
›Reveal solutionSolution
Disproportionation means the same species is simultaneously oxidised and reduced; here half the Cu⁺ goes to Cu (reduction) and half to Cu²⁺ (oxidation) — option (B).

Disproportionation is a redox reaction in which an element in one intermediate oxidation state is simultaneously oxidised and reduced.

Track the oxidation number of copper in $2\,Cu^{+} \rightarrow Cu + Cu^{2+}$ :

One $Cu^{+}$ (oxidation state $+1$ ) → $Cu$ (oxidation state $0$ ): gain of electron = reduction. …
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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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Q.(a)

Part (a)

(i) Comparisons

(ii) Oxidising action of MnO4−MnO_4^-MnO4−​ in acidic medium

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(ii) Oxidising action of $MnO_4^-$ in acidic medium

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