(a) (I) Account for the following : (3 + 2) (i) E^° value for Mn³⁺/Mn²⁺ couple is much more positive than that

Question

Q. (a) (I) Account for the following : (3 + 2)

(i)

E^\circ

value for

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

couple is much more positive than that for

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

(ii)

Sc^{3+}

is colourless whereas

Ti^{3+}

is coloured in an aqueous solution.

(iii) Actinoids show wide range of oxidation states. (II) Write the chemical equations for the preparation of

KMnO_4

from

MnO_2

(OR)

(b) (I) Account for the following : (2 + 2 + 1)

(i) Transition metals form alloys.

(ii)

Ce^{4+}

is a strong oxidising agent. (II) Write one similarity and one difference between chemistry of Lanthanoids and Actinoids. (III) Complete the following ionic equation :

Cr_2O_7^{2-} + 2OH^- \rightarrow

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Concept understanding — Stability of Oxidation States

Stability of Oxidation States – From Intuition to Precision

Imagine you're holding a ball on a hill. If you place it exactly at the top, it's balanced — but the slightest push sends it rolling down. That's an unstable position. If you place it in a small dip on the hillside, it stays put even if nudged — that's stable. Oxidation states work the same way: some are like the hilltop (easily changed), others like the dip (hard to change).

The Core Intuition

An oxidation state is just a number we assign to an atom to track how many electrons it has gained or lost compared to its neutral state. But atoms don't "want" to stay in arbitrary oxidation states — they want to reach a configuration that minimises their energy.

Stability here means: how reluctant is that oxidation state to change under normal conditions? A stable oxidation state resists being oxidised further or reduced further. An unstable one readily changes into something else.

The Precise Statement

Stability of an oxidation state refers to the tendency of an element to maintain that particular oxidation state under given conditions (temperature, pH, presence of other reagents). A stable oxidation state is one that does not easily undergo redox reactions — it is neither easily oxidised nor easily reduced.

This depends on three key factors:

Electronic configuration – Half-filled and fully-filled d or f subshells confer extra stability (e.g., $\text{Fe}^{3+}$ with $d^5$ is more stable than $\text{Fe}^{2+}$ with $d^6$ in some contexts).
Inert pair effect – Heavier p-block elements (like Tl, Pb, Bi) show lower oxidation states (e.g., +1 for Tl) as more stable than higher ones (+3 for Tl), because the s-electrons become reluctant to participate.
Disproportionation tendency – Some oxidation states are unstable because they spontaneously convert into two other states (e.g., $\text{Cu}^+$ in aqueous solution gives $\text{Cu}^{2+}$ and $\text{Cu}$ ).

Important

Stability is relative — it depends on the environment. $\text{Mn}^{2+}$ is stable in acidic solution but easily oxidised in alkaline medium. Always specify conditions when discussing stability.

Examples That Make It Concrete

Transition metals – $\text{Cr}^{3+}$ ( $d^3$ ) and $\text{Mn}^{2+}$ ( $d^5$ ) are exceptionally stable because half-filled/half-filled-like configurations have low energy. $\text{Cr}^{2+}$ ( $d^4$ ) is easily oxidised to $\text{Cr}^{3+}$ — it's unstable.

p-block elements – $\text{Pb}^{2+}$ is stable, $\text{Pb}^{4+}$ is a strong oxidising agent (unstable). $\text{Sn}^{2+}$ is a reducing agent (easily oxidised to $\text{Sn}^{4+}$ ), so $\text{Sn}^{4+}$ is more stable for tin.

Common pattern – For most elements, the most common oxidation state is the most stable one under standard conditions. But "most common" isn't always "most stable" — e.g., $\text{Fe}^{3+}$ is common but $\text{Fe}^{2+}$ is more stable in acidic solution.

Watch out

Do not confuse "stability" with "occurrence". $\text{Mn}^{7+}$ (as $\text{MnO}_4^-$ ) is common in the lab but is a powerful oxidising agent — it is not stable in the sense of resisting change. It readily accepts electrons.

How to Think About It in Exams …

Why this formula?

Stability of Oxidation States: Why It Works

This concept explains why certain oxidation states of an element are more stable than others — and why some states are never observed at all.

The Core Idea: Energy Minimisation

An oxidation state is stable when the total energy of the system is at a minimum. This depends on three competing factors:

Ionisation energy (energy needed to remove electrons)
Lattice energy (for ionic compounds) or bond energy (for covalent compounds)
Electronic configuration (half-filled / fully-filled subshells)

There is no single formula for stability — instead, we use trends and principles that act as "formulae" for prediction.

Key Principle 1: Inert Pair Effect (for p-block elements)

Why it holds:

For heavier elements (e.g., Tl, Pb, Bi), the 6s² electrons are held very tightly due to poor shielding and relativistic effects. They resist removal.

Result: Lower oxidation state (e.g., +1 for Tl, +2 for Pb) becomes more stable than the higher state (+3, +4).
Example: $\ce{Tl^3+}$ is a strong oxidising agent — it readily gains two electrons to become $\ce{Tl^+}$ .

Derivation logic:

The energy cost to remove the 6s² electrons is greater than the energy gained by forming additional bonds or lattice. So the system stays in the lower state.

Key Principle 2: Half-Filled / Fully-Filled Subshell Stability

Why it holds:

A half-filled ( $d^5$ , $f^7$ ) or fully-filled ( $d^{10}$ , $f^{14}$ ) subshell has extra exchange energy and symmetry — making it unusually stable.

Example: $\ce{Mn^{2+}}$ ( $d^5$ ) is more stable than $\ce{Mn^{3+}}$ ( $d^4$ ). $\ce{Fe^{3+}}$ ( $d^5$ ) is more stable than $\ce{Fe^{2+}}$ ( $d^6$ ).

Derivation logic:

The exchange energy (Hund's rule) is maximum for half-filled configurations. Removing an electron from a half-filled shell costs extra energy — so the half-filled state is favoured.

Key Principle 3: Lattice Energy / Hydration Energy Compensation

For transition metals, stability of a particular oxidation state in aqueous solution depends on:

\Delta G^\circ = \Delta H^\circ_{\text{hydration}} - \Delta H^\circ_{\text{ionisation}}

Why it holds:

Higher oxidation states have higher ionisation energy (harder to remove electrons).
But they also have higher hydration energy (smaller, more charged ions attract water more strongly).
The balance determines which state is stable.

Example:

$\ce{Cu^+}$ is unstable in water because its hydration energy is too low to compensate for the loss of the second electron.
$\ce{Cu^{2+}}$ is stable in water.

Key Principle 4: Disproportionation

Some oxidation states are unstable and spontaneously convert to two other states:

\ce{2Cu^+ -> Cu + Cu^{2+}}

Why it holds: …

(a) $Mn^{3+}/Mn^{2+}$ is strongly positive (stable $d^5$ $Mn^{2+}$ ); $Sc^{3+}$ ( $d^0$ ) colourless, $Ti^{3+}$ ( $d^1$ ) coloured; actinoids show many oxidation states; $KMnO_4$ made by fusing $MnO_2$ with KOH/ $O_2$ then oxidising $K_2MnO_4$ .

(b) Transition metals form alloys (similar radii); $Ce^{4+}$ is a strong oxidiser (→ $Ce^{3+}$ ); Ln & An both show +3 but An has more oxidation states; $Cr_2O_7^{2-}+2OH^-\to 2CrO_4^{2-}+H_2O$ .

(I)(i) $E^\circ_{Mn^{3+}/Mn^{2+}}$ vs $E^\circ_{Cr^{3+}/Cr^{2+}}$

A more positive reduction potential means the higher ion is reduced more easily. $Mn^{3+}$ ( $3d^4$ ) is reduced to $Mn^{2+}$ ( $3d^5$ , an exceptionally stable half-filled shell), so the reduction is very favourable ( $E^\circ = +1.57$ V). $Cr^{3+}$ ( $3d^3$ , half-filled $t_{2g}$ ) is itself stable and resists reduction to $Cr^{2+}$ ( $3d^4$ ), so $E^\circ_{Cr^{3+}/Cr^{2+}} = -0.41$ V.

(I)(ii) Colour of $Sc^{3+}$ vs $Ti^{3+}$

Colour arises from d–d transitions, which need a partially filled d-subshell. $Sc^{3+}$ is $3d^0$ (no d-electron) → no transition → colourless. $Ti^{3+}$ is $3d^1$ ; its single electron in the split $t_{2g}$ level absorbs visible light and jumps to $e_g$ → the ion appears violet.

(I)(iii) Wide range of oxidation states in actinoids

Actinoids: $[Rn]5f^{1-14}6d^{0-1}7s^2$ . The 5f, 6d and 7s orbitals lie very close in energy and the 5f orbitals are poorly shielded (more available for bonding than the buried 4f of lanthanoids), so electrons are removed from all these levels, giving oxidation states from +3 up to +7.

(II) Preparation of $KMnO_4$ from $MnO_2$

2MnO_2 + 4KOH + O_2 \xrightarrow{\Delta} 2K_2MnO_4 + 2H_2O

The green manganate is then oxidised to purple permanganate, e.g. by $CO_2$ (a disproportionation) or electrolytically:

3K_2MnO_4 + 2CO_2 \rightarrow 2KMnO_4 + MnO_2 + 2K_2CO_3

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Concept understanding — Actinoids

Actinoids: The Heavy, Radioactive Inner Transition Metals

Imagine you are walking through the periodic table, row by row. You know the lanthanoids — the 14 elements from cerium to lutetium that sit below the main table, where the 4f subshell is being filled. Now go one row lower. The elements that follow actinium (atomic number 89) are the actinoids: thorium (90) through lawrencium (103). Here, the 5f subshell is being filled.

The name comes from actinium, the first element of the series. But the real story begins with uranium and plutonium — the elements that made nuclear energy and atomic bombs possible. These are not just academic curiosities; they are the workhorses of nuclear physics.

The Precise Statement

Z=90

Why They Are Special

1. Radioactivity is the rule, not the exception. Every actinoid is radioactive. Some, like uranium-238, have half-lives comparable to the age of the Earth. Others, like lawrencium, exist for only seconds. This radioactivity is not a side effect — it is the defining property. It means their chemistry is often studied with trace amounts, using radiation detectors rather than test tubes.

2. Multiple oxidation states — more than you expect. Unlike the lanthanoids, which mostly stick to +3, the actinoids show a rich variety. Uranium, for example, exists as U $^{3+}$ , U $^{4+}$ , UO $_2^+$ (U $^{5+}$ ), and UO $_2^{2+}$ (U $^{6+}$ ). The early actinoids (Th, Pa, U, Np, Pu) are especially versatile because the 5f, 6d, and 7s orbitals are close in energy. As you move right, the +3 state becomes more stable — curium and beyond behave more like lanthanoids.

3. The actinoid contraction. As you go from Th to Lr, the nuclear charge increases by 14 protons. The 5f electrons are poor at shielding each other, so the effective nuclear charge felt by the outer electrons rises steadily. The result: ionic radii shrink smoothly across the series. This is exactly analogous to the lanthanoid contraction, and it has the same consequence — the chemistry of later actinoids becomes very similar to their lanthanoid counterparts, making separation difficult.

Tip

The actinoid contraction is why americium and curium are so hard to separate from the lanthanoids in nuclear waste. Their ionic radii are nearly identical.

A Quick Comparison with Lanthanoids

Property	Lanthanoids (4f)	Actinoids (5f)
Radioactivity	Mostly stable (except Pm)	All radioactive
Common oxidation state	+3 (strongly dominant)	+3, +4, +5, +6 (early ones)
5f vs 4f character	4f deeply buried, little bonding	5f more exposed, can participate in bonding

(I)(i) Transition metals form alloys

Their atomic and metallic radii are very similar, so atoms of one transition metal replace those of another in the crystal lattice with little distortion, forming solid solutions (alloys) such as brass and steel.

(I)(ii) $Ce^{4+}$ is a strong oxidising agent …

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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 27 on this concept.

CBSE 2026Set 56/3/11 markMCQ
$Cr_2O_7^{2-}$
›Reveal solutionSolution
In alkaline medium, dichromate ( $Cr_2O_7^{2-}$ ) converts to chromate ( $CrO_4^{2-}$ ) without any change in oxidation state — it’s a simple acid-base equilibrium, not a redox reaction. The correct option is (B).

The key to this question lies in understanding that the conversion of dichromate to chromate is not a redox reaction — the oxidation state of chromium remains +6 throughout. Many students instinctively think of reduction to $Cr^{3+}$ because they associate dichromate with strong oxidizing behaviour, but that only happens in acidic medium. In alkaline conditions, the chemistry is entirely different.

Let’s walk through the reasoning step by step.

Recall the oxidation state of chromium in dichromate.

In $Cr_2O_7^{2-}$ , each oxygen is -2, so total from seven oxygens is -14. The ion has a -2 charge, so the sum of oxidation states of the two chromium atoms must be +12. Hence each Cr is in the +6 state.

Now consider the alkaline medium.

When you add a base (like NaOH) to a solution of $K_2Cr_2O_7$ , the dichromate ion reacts with hydroxide ions. The reaction is:

$Cr_2O_7^{2-} + 2OH^- \rightarrow 2CrO_4^{2-} + H_2O$
Notice that the oxidation state of Cr in $CrO_4^{2-}$ is also +6 (four oxygens at -2 give -8, charge -2, so Cr = +6). No electrons are transferred — this is an acid-base equilibrium, not a redox change.

Why does this happen? Dichromate exists in equilibrium with chromate, and the position depends on pH. In acidic solution, the equilibrium shifts toward dichromate; in alkaline solution, it shifts toward chromate. The reaction is:

$2CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} + H_2O$
Adding $OH^-$ removes $H^+$ , pulling the equilibrium to the left — producing chromate. …
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CBSE 2026Set 56/1/11 markMCQ
$Q. Assertion (A) : Actinoids show wide range of oxidation states. Reason (R) : Actinoids are radioactive in nature. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.$
›Reveal solutionSolution
The assertion that actinoids show a wide range of oxidation states is true, but the reason given — that they are radioactive — is not the correct explanation. The correct explanation lies in the small energy gap between 5f, 6d, and 7s orbitals, which allows many electrons to participate in bonding. So the answer is option (B).

The question tests your understanding of why actinoids (elements 90–103, from thorium to lawrencium) exhibit so many different oxidation states. Many students memorise that “actinoids show variable oxidation states” and also know they are radioactive, so they assume the second explains the first. That’s a trap.

Let’s break it down properly.

Is Assertion (A) true?

Yes. Actinoids display a remarkably wide range of oxidation states. For example, uranium shows +3, +4, +5, and +6; neptunium and plutonium go from +3 to +7. This is far more varied than most d-block elements. The reason is that the 5f, 6d, and 7s orbitals are very close in energy. Electrons from all three can be lost with relatively little energy cost, so many different oxidation numbers become accessible.

Is Reason (R) true?

Yes, actinoids are indeed radioactive. All actinoid nuclei are unstable and decay over time. So the reason statement is factually correct.

Does the radioactivity explain the wide range of oxidation states?

No. Radioactivity is a nuclear property — it depends on the instability of the nucleus (proton/neutron ratio, nuclear binding energy). Oxidation states are an electronic property — they depend on how easily electrons are lost from the outer orbitals. These two phenomena are completely independent.

Watch out

A common mistake is to think that because both statements are true, the reason must be the explanation. But correlation is not causation. Radioactivity does not cause variable oxidation states; the orbital energy structure does.

What actually causes the wide range of oxidation states in actinoids? …
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CBSE 2026Set V11 markMCQ
$Q. Match the following given in List I with List II. List I (Elements) List II (Their maximum oxidation states) i) Thorium (Th) a) +7 ii) Protactinium (Pa) b) +6 iii) Lawrencium (Lr) c) +3 iv) Uranium (U) d) +5 v) Plutonium (Pu) e) +4 Choose the correct option :$
(a) i – e, ii – d, iii – a, iv – b, v – c
(b) i – e, ii – d, iii – c, iv – b, v – a
(c) i – b, ii – a, iii – c, iv – e, v – d
(d) i – b, ii – e, iii – c, iv – d, v – a
›Reveal solutionSolution
Matching each actinoid to its maximum oxidation state gives Th– $+4$ , Pa– $+5$ , Lr– $+3$ , U– $+6$ , Pu– $+7$ , i.e. option (b).

Early actinoids show a rise in maximum oxidation state as the number of available $5f/6d$ electrons increases, peaking around U–Np–Pu, then falling again toward the heavier members (which behave much like lanthanoids with a stable $+3$ state).

i) Thorium (Th): maximum oxidation state $= +4$ → (e)

ii) Protactinium (Pa): maximum oxidation state $= +5$ → (d) …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following oxidation states is common for all lanthanoids?$
(a) +2
(b) +3
(c) +4
(d) +5
›Reveal solutionSolution

All lanthanoids show a characteristic +3 oxidation state because it corresponds to a stable, similar electronic configuration achieved after losing the two 6s and one 4f (or 5d) electron.

Lanthanoids have the general electronic configuration [Xe] 4f^(1-14) 5d^(0-1) 6s2. Removal of the two 6s electrons and one more electron (from 4f or 5d) gives the Ln3+ ion, which is the most stable and commonly observed oxidation state across the entire series, from Ce to Lu.

…

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CBSE 2025Set 56/4/11 markMCQ
$I^{-}$
›Reveal solutionSolution
In alkaline medium, permanganate ( $MnO_4^-$ ) oxidises iodide ( $I^-$ ) to iodate ( $IO_3^-$ ), not to iodine or periodate. The balanced reaction shows $I^-$ loses 6 electrons to form $IO_3^-$ , while $MnO_4^-$ gains 3 electrons to form $MnO_2$ . The correct product is $IO_3^-$ , option (D).

Why the medium matters

The oxidation state of iodine in its products depends heavily on the pH of the solution. Permanganate is a powerful oxidising agent, but its reduction product changes with medium:

In acidic medium: $MnO_4^- \rightarrow Mn^{2+}$ (gains 5 electrons)

In neutral/alkaline medium: $MnO_4^- \rightarrow MnO_2$ (gains 3 electrons)

This difference in electron gain per mole of permanganate directly affects how far it can oxidise iodide. In alkaline medium, permanganate is a milder oxidising agent (gains only 3 electrons) compared to acidic medium (gains 5 electrons). Yet it still oxidises $I^-$ all the way to $IO_3^-$ , not stopping at $I_2$ .

Step-by-step reasoning

Identify the half-reactions

Iodide ( $I^-$ ) has oxidation state $-1$ . The possible products given are:

$IO_4^-$ : iodine in $+7$ state

$I_2$ : iodine in $0$ state

$IO^-$ : iodine in $+1$ state (hypoiodite)

$IO_3^-$ : iodine in $+5$ state (iodate)

In alkaline medium, permanganate reduces to $MnO_2$ (manganese in $+4$ state, from $+7$ in $MnO_4^-$ ).

Balance the oxidation half-reaction

Iodide going to iodate:

$I^- \rightarrow IO_3^-$
Balance oxygen with water (alkaline medium):
$I^- + 3H_2O \rightarrow IO_3^- + 6H^+$
Balance charge: left side has $-1$ , right side has $-1 + 6 = +5$ . Add 6 electrons to right:
$I^- + 3H_2O \rightarrow IO_3^- + 6H^+ + 6e^-$
In alkaline medium, add $OH^-$ to neutralise $H^+$ :
$I^- + 6OH^- \rightarrow IO_3^- + 3H_2O + 6e^-$
So each $I^-$ loses 6 electrons to become $IO_3^-$ .

Balance the reduction half-reaction

Permanganate to manganese dioxide in alkaline medium:

$MnO_4^- \rightarrow MnO_2$
Balance oxygen with water:
$MnO_4^- + 2H_2O \rightarrow MnO_2 + 4OH^-$
Balance charge: left $-1$ , right $-4$ . Add 3 electrons to left:
$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$
So each $MnO_4^-$ gains 3 electrons.

Combine the half-reactions

To equalise electrons: multiply reduction half by 2 (gives 6 electrons gained) and oxidation half by 1 (gives 6 electrons lost):

$2MnO_4^- + 4H_2O + 6e^- \rightarrow 2MnO_2 + 8OH^-$ $I^- + 6OH^- \rightarrow IO_3^- + 3H_2O + 6e^-$
Adding:
$2MnO_4^- + I^- + 4H_2O + 6OH^- \rightarrow 2MnO_2 + IO_3^- + 3H_2O + 8OH^-$
Cancel $3H_2O$ from both sides and $6OH^-$ from both sides: …
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CBSE 2025Set 56/6/11 markMCQ
$Q. For the following question, two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A) : Actinoids show wide range of oxidation states. Reason (R) : Actinoids are radioactive in nature.$
›Reveal solutionSolution
The assertion that actinoids show a wide range of oxidation states is true, but the reason given — that they are radioactive — does not explain this property. The correct answer is (B).

The question tests your understanding of why actinoids exhibit variable oxidation states. The key is to separate two distinct facts: actinoids are radioactive, and they do show many oxidation states — but the radioactivity is not the cause of the oxidation state variability.

Let’s break this down.

Why do actinoids show a wide range of oxidation states?

The 5f, 6d, and 7s orbitals in actinoids are very close in energy. This means electrons can be removed from any of these orbitals with relatively little energy cost. As you move across the actinoid series, the 5f orbitals gradually become more stable, but early actinoids (like Th, Pa, U, Np, Pu) can lose anywhere from 3 to 7 electrons. For example, uranium shows +3, +4, +5, and +6; plutonium shows +3, +4, +5, +6, and +7. This is the real reason for the wide range — it’s an electronic structure effect, not a nuclear one.

What about radioactivity?

Yes, all actinoids are radioactive — their nuclei are unstable and decay over time. But radioactivity is a nuclear property, while oxidation states depend on electron configuration. A nucleus decaying does not directly change how many electrons an atom can lose or gain in a chemical reaction. So while both statements are factually true, the reason does not explain the assertion. …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Element of actinoid series is -$
(a) Thorium
(b) Cerium
(c) Cadmium
(d) Chromium
›Reveal solutionSolution
Thorium (Th, Z = 90) belongs to the actinoid series; the other three options are not actinoids.

Actinoids are the 14 elements from Th (Z=90) to Lr (Z=103) that follow actinium, in which the 5f orbitals are progressively filled.

Checking each option:

Thorium (Z=90): an actinoid.

Cerium (Z=58): a lanthanoid (4f block), not an actinoid. …
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CBSE 2025Set ANNUAL1 mark
$Q. What is the common oxidation state of Lanthanoids?$
›Reveal solutionSolution

All lanthanoids overwhelmingly favour the +3 oxidation state, since their poorly-bonding 4f electrons are not readily involved, leaving the same outer 5d/6s electrons available across the series.

Across the entire lanthanide series, the +3 oxidation state is by far the most common and stable one, shown by essentially every lanthanoid. This is because the 4f electrons are deeply buried and well-shielded, taking little part in bonding, while the outer $5d^{0-1}6s^2$ electrons are readily lost to give the stable $Ln^{3+}$ ion. Occasional +2 or +4 states occur only for a few elements whe …

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CBSE 2024Set A11 markMCQ
$Q. Which of the following pair of metal oxides are amphoteric?$
(a) $\text{V}_2\text{O}_5$ , $\text{Cr}_2\text{O}_3$
(b) $\text{Mn}_2\text{O}_7$ , $\text{CrO}_3$
(c) $\text{V}_2\text{O}_5$ , $\text{V}_2\text{O}_4$
(d) $\text{CrO}$ , $\text{V}_2\text{O}_5$
›Reveal solutionSolution

$\text{V}_2\text{O}_5$ and $\text{Cr}_2\text{O}_3$ are the amphoteric pair — option (a).

For transition-metal oxides, the character changes from basic (low oxidation state) through amphoteric to acidic (high oxidation state). $\text{Cr}_2\text{O}_3$ (Cr in +3) is amphoteric — it dissolves in acids to give $\text{Cr}^{3+}$ salts and in alkali to give chromite. $\text{V}_2\text{O}_5$ (V in +5) is chiefly acidic but is genuinely amphoteric, dissolving in both acids and alka …
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CBSE 2024Set B1 mark
$Q. Write True or False: Lanthanoids is more reactive than actinoids.$
›Reveal solutionSolution

Because of their larger size and lower ionisation enthalpies (and comparable energies of 5f, 6d, 7s orbitals), actinoids are more reactive than lanthanoids, especially in the finely divided state.

Lanthanoids (4f series) are comparatively less reactive, tarnishing slowly in air, while actinoids (5f series) — especially the early actinoids like thorium, uranium, plutonium — are much more reactive: they combine readily with most non-metals, tarnish rapidly, and some (like plutonium) are pyrophoric in powdered form. This higher reactivity of actinoids is attributed to their larger atomic/ionic radii and t …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Which of the following oxidation state is common for all lanthanoids ?$
(a) +2
(b) +3
(c) +4
(d) +5
›Reveal solutionSolution

Every lanthanoid shows the +3 oxidation state as its characteristic and most stable state, even though a few also show +2 or +4 in special cases.

Lanthanoids (Ce to Lu) have the general electronic configuration $[Xe]4f^{1-14}5d^{0-1}6s^2$ . Losing the two 6s electrons and one 4f/5d electron gives the stable, half-filled/fully-filled-favouring $Ln^{3+}$ ion, which is why +3 is the predominant and universally shown oxidation state across the whole series. A handful of lanthanoids additi …

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CBSE 2023Set 56/1/11 markMCQ
$Q. The most common and stable oxidation state of a Lanthanoid is : (A) + 2 (B) + 3 (C) + 4 (D) + 6$
›Reveal solutionSolution
Lanthanoids overwhelmingly prefer the +3 oxidation state due to the stability gained from losing the two 6s and one 5d/4f electron, achieving a configuration analogous to noble gases or half-filled/filled f-subshells. The answer is (B) +3.

Why Lanthanoids Love +3: Electronic Configuration and Stability

The lanthanoid series (elements 57–71: La through Lu) sits in the f-block, where the 4f orbitals are being progressively filled. To understand their oxidation state preference, we need to look at what electrons are available and what configurations become stable upon ionization.

A typical lanthanoid has the general electronic configuration:
$[\text{Xe}] \, 4f^{0-14} \, 5d^{0-1} \, 6s^2$
The 6s electrons are outermost and easiest to remove. The 5d and 4f orbitals are close in energy, so sometimes one electron occupies 5d instead of 4f. When a lanthanoid forms a cation, it loses electrons in a specific order: 6s electrons go first, then 5d, then 4f (because 4f is more tightly held, being an inner orbital).

Step-by-Step Reasoning

First ionization removes 6s electrons

All lanthanoids have two 6s electrons. Removing both gives a +2 state, but this is rarely the stopping point because the resulting ion still has relatively accessible 5d or 4f electrons.

Third electron removal: the key to +3 stability

After losing the two 6s electrons, removing one more electron (from 5d if occupied, otherwise from 4f) produces the +3 oxidation state. This configuration turns out to be remarkably stable across the entire series.

Why? The resulting $\text{Ln}^{3+}$ ion achieves one of several favorable electronic arrangements:

For La ( $4f^0$ ): $[\text{Xe}]$ — a noble gas configuration.

For Gd ( $4f^7$ ): half-filled f-subshell with all spins parallel (exchange energy stabilization).

For Lu ( $4f^{14}$ ): completely filled f-subshell.

For others: partially filled 4f with reasonable stability.

Why not +2?

The +2 state does exist for a few lanthanoids (Eu, Yb) where it leads to half-filled or filled f-subshells ( $4f^7$ for Eu²⁺, $4f^{14}$ for Yb²⁺), but these are exceptions, not the rule. Most lanthanoids find +2 too reducing and unstable in aqueous solution.

Why not +4 or higher?

Removing a fourth electron means breaking into the tightly held 4f subshell (which is shielded and contracted). The ionization energy jumps dramatically. Only Ce commonly shows +4 (because Ce⁴⁺ achieves $4f^0 = [\text{Xe}]$ ), and even that is a strong oxidizing agent. Higher states like +6 are virtually unknown in lanthanoids—the 4f electrons are too stable to remove. …
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	List I (Elements)		List II (Their maximum oxidation states)
i)	Thorium (Th)	a)	+7
ii)	Protactinium (Pa)	b)	+6
iii)	Lawrencium (Lr)	c)	+3
iv)	Uranium (U)	d)	+5
v)	Plutonium (Pu)	e)	+4

Q.(a) (I) Account for the following : (3 + 2)

Concept understanding — Stability of Oxidation States

Stability of Oxidation States – From Intuition to Precision

The Core Intuition

The Precise Statement

Examples That Make It Concrete

How to Think About It in Exams …

Why this formula?

Stability of Oxidation States: Why It Works

The Core Idea: Energy Minimisation

Key Principle 1: Inert Pair Effect (for p-block elements)

Key Principle 2: Half-Filled / Fully-Filled Subshell Stability

Key Principle 3: Lattice Energy / Hydration Energy Compensation

Key Principle 4: Disproportionation

(I)(i) EMn3+/Mn2+∘E^\circ_{Mn^{3+}/Mn^{2+}}EMn3+/Mn2+∘​ vs ECr3+/Cr2+∘E^\circ_{Cr^{3+}/Cr^{2+}}ECr3+/Cr2+∘​

(I)(ii) Colour of Sc3+Sc^{3+}Sc3+ vs Ti3+Ti^{3+}Ti3+

(I)(iii) Wide range of oxidation states in actinoids

(II) Preparation of KMnO4KMnO_4KMnO4​ from MnO2MnO_2MnO2​

Concept understanding — Actinoids

Actinoids: The Heavy, Radioactive Inner Transition Metals

The Precise Statement

Why They Are Special

A Quick Comparison with Lanthanoids

(I)(i) Transition metals form alloys

(I)(ii) Ce4+Ce^{4+}Ce4+ is a strong oxidising agent …

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(I)(i) $E^\circ_{Mn^{3+}/Mn^{2+}}$ vs $E^\circ_{Cr^{3+}/Cr^{2+}}$

(I)(ii) Colour of $Sc^{3+}$ vs $Ti^{3+}$

(II) Preparation of $KMnO_4$ from $MnO_2$

(I)(ii) $Ce^{4+}$ is a strong oxidising agent …

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