(a) (i) When pyrolusite ore is fused with KOH, in presence of air, a dark green coloured product 'A' is obtain

Question

Q. (a)

(i) When pyrolusite ore is fused with KOH, in presence of air, a dark green coloured product 'A' is obtained which changes to purple coloured compound 'B' in acidic medium. (I) Write the formulae of 'A' and 'B'. (II) Write the ionic equation for the reaction when compound 'B' reacts with

Fe^{2+}

in acidic medium.

(ii) Give reasons : (I)

Ce^{4+}

in aqueous solution is a good oxidising agent. (II) The actinoid contraction is greater from element to element than lanthanoid contraction. (III)

E^{\circ}_{Zn^{2+}/Zn}

value is more negative than expected, whereas

E^{\circ}_{Cu^{2+}/Cu}

is positive.

(OR)

(b)

(i) While studying the periodic properties, Arti came across an abnormal behaviour in the atomic size of Hf. She found that, even though Hf is placed below Zr in the same group, both have almost similar atomic sizes. (I) Which phenomenon is responsible for the above behaviour ? Define it. (II) Mention any other consequence of the above phenomenon.

(ii) Give reasons for the following : (I) Transition metals exhibit catalytic properties. (II) Transition metals have high enthalpy of atomisation. (III) Sc is a transition element, while Zn is not.

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 — Lanthanoid Contraction

Lanthanoid Contraction – From Intuition to Precision

Imagine you are walking across a row of the periodic table — from lanthanum (atomic number 57) to lutetium (71). Your first instinct might be: as we add more protons and more electrons, the atom should get bigger. But the opposite happens. The atoms actually shrink, steadily and stubbornly, across these 15 elements. That is the lanthanoid contraction.

Why? The answer lies in the 4f orbitals.

The core intuition: a bad shield

Every new electron you add across the lanthanoid series goes into a 4f orbital. These 4f orbitals are shaped like cloverleaves, but they are tucked deep inside the atom — very close to the nucleus. They are also notoriously poor at shielding the outer electrons from the pull of the nucleus.

Here is the key: each step adds one proton to the nucleus. That proton yanks harder on all the electrons. Normally, the new electron you add would partly cancel that pull (shielding). But 4f electrons are so diffuse and so poorly penetrating that they do a terrible job of shielding. So the net effect is that the effective nuclear charge felt by the outer electrons increases steadily across the series. The outer electrons get pulled inward, and the whole atom shrinks.

Note

The 4f orbitals are "inside" the atom — they lie closer to the nucleus than the 5d and 6s orbitals. So adding electrons there does not push the outer shell outward; instead, the increasing nuclear charge dominates.

The precise statement

Lanthanoid contraction is the progressive and regular decrease in atomic and ionic radii of the lanthanoid elements (Ce to Lu) as atomic number increases. The contraction is about 1 pm per element, totalling roughly 15 pm from La to Lu.

This is not a small effect. It is so consistent that the radii of the later lanthanoids are almost identical to those of the 4d transition metals directly above them in the periodic table. For example, zirconium (Zr) and hafnium (Hf) have nearly the same atomic radius — a direct consequence of the lanthanoid contraction.

\text{Atomic radius (pm)} \approx 187 - 0.9 \times (Z - 57) \quad \text{(rough linear fit for trivalent ions)}

Why it matters

The lanthanoid contraction explains several important patterns in chemistry:

Similarity of 4d and 5d transition metals: Elements like Zr and Hf, Nb and Ta, Mo and W are nearly identical in size and chemical behaviour. Without the lanthanoid contraction, the 5d metals would be much larger. This is why separating Hf from Zr is famously difficult — they are chemical twins.
Trend in basicity: Across the lanthanoid series, the ionic radius decreases. This increases the charge density on the ion, making it more polarising. As a result, the basicity of the hydroxides decreases from La(OH)₃ (strong base) to Lu(OH)₃ (weak base). …

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Part (a): A = $K_2MnO_4$ (green), B = $KMnO_4$ (purple); $MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+}+5Fe^{3+}+4H_2O$ ; $Ce^{4+}$ oxidises (reverts to +3), actinoid contraction > lanthanoid (poor 5f shielding), Zn/Cu $E^\circ$ explained.

Part (b): The phenomenon is lanthanoid contraction (Hf $\approx$ Zr); consequences include similar 4d/5d properties and falling $Ln(OH)_3$ basicity; transition metals catalyse via variable states, have high atomisation enthalpy, Sc is transition, Zn is not.

Part (a)

(i)(I) Pyrolusite ( $MnO_2$ ) fused with KOH in air is oxidised to the green potassium manganate:

2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O

So A = $K_2MnO_4$ (Mn in +6, green). In acidic medium manganate disproportionates to the purple permanganate, so B = $KMnO_4$ (Mn in +7):

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

(i)(II) Permanganate is a strong oxidiser; with $Fe^{2+}$ in acidic medium:

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

(ii)(I) For the lanthanoids the +3 state is the most stable. $Ce^{4+}$ therefore has a strong tendency to gain an electron and return to $Ce^{3+}$ , which makes $Ce^{4+}$ a good oxidising agent in aqueous solution.

(ii)(II) Contraction in size along a series arises from imperfect shielding of the nuclear charge by the inner $f$ electrons. 5f electrons shield even more poorly than 4f electrons, so the effective nuclear charge felt by outer electrons increases more steeply — the actinoid contraction from element to element is greater than the lanthanoid contraction. …

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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 mark
$Q. Fill in the blank: Due to ............... Zirconium (Zr) and Hafnium (Hf) have identical radii.$
›Reveal solutionSolution

Zirconium (period 5) and hafnium (period 6) have almost identical atomic/ionic radii because of lanthanide contraction, which offsets the expected increase in size on descending a group.

Normally, atomic/ionic radius increases on going down a group as a new shell is added. But between Zr (Z=40) and Hf (Z=72), the 4f orbitals are filled across the lanthanides (Z=58–71). The 4f electrons shield the nuclear charge poorly, so the effective nuclear charge felt by the outer electrons increases steadily across the lanthanide series, causing a steady contraction in atomic/ionic size — the lanthanide contraction.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The reason of lanthanoid contraction is$
(a) negligible screening effect of f-orbital
(b) increasing nuclear charge
(c) decreasing nuclear charge
(d) decreasing screening effect
›Reveal solutionSolution

The lanthanoid contraction is caused by the poor ability of the diffuse, deeply-buried 4f orbitals to shield one another from the increasing nuclear charge, so effective nuclear charge rises steadily and atomic/ionic radii shrink across the series.

Mechanism: Going from $Ce$ to $Lu$ , the atomic number (and hence nuclear charge $Z$ ) increases by one unit at each step as an electron is added to the inner $4f$ subshell. The $4f$ orbitals are radially compact and have poor overlap/penetration properties, so one $4f$ electron shields another $4f$ electron (or the outer $5d^16s^2$ electrons) from the nucleus very inefficiently — far less efficiently than, say, $s$ or $p$ electrons shield each other. Because this poor screening does not keep pace with the steadily rising nuclear charge, the effective nuclear charge ( $Z_{eff}=Z-S$ ) experienced by the outer electrons increases continuously across the series, pulling the electron cloud in and steadily shrinking the atomic and ionic radii. The cumulative shrinkage across all 14 lanthanoids is the lanthanoid contraction, and it is large enough that the third-row transition elements following the lanthanoids (e.g. $Zr$ and $Hf$ , $Nb$ and $Ta$ ) end up with almost identical atomic radii to their second-row counterparts.

Why the other options are wrong/incomplete: …

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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 markMCQ
$Q. Lanthanoid contraction is due to increase in$
(a) atomic number
(b) effective nuclear charge
(c) atomic radius
(d) valence electrons
›Reveal solutionSolution

Lanthanoid contraction is the steady decrease in atomic/ionic radii across the lanthanoid series, caused by an increase in effective nuclear charge that is imperfectly screened by the diffuse 4f electrons.

Across the lanthanoid series (Ce to Lu), each successive element adds one proton to the nucleus and one electron to the inner 4f subshell. The 4f orbitals have poor shielding ability (diffuse, non-directional shapes), so they do not effectively screen the outer electrons from the increasing nuclear charge.

…

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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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Q.(a)

Part (a)

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