Which of the following statements is not correct? (i) Copper liberates hydrogen from acids. (ii) In its higher

NCERT Exemplar · Q17

Q. Which of the following statements is not correct?

(i) Copper liberates hydrogen from acids.

(ii) In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine.

(iii)

Mn^{3+}

and

Co^{3+}

are oxidising agents in aqueous solution.

(iv)

Ti^{2+}

and

Cr^{2+}

are reducing agents in aqueous solution.

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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: …

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The key idea is to check each statement against known trends in transition-metal chemistry — particularly the stability of oxidation states and their redox behaviour in aqueous solution. The incorrect statement is (i), because copper does not liberate hydrogen from acids (it lies below hydrogen in the electrochemical series).

Statement (i): Copper liberates hydrogen from acids.

This is a classic point. The electrochemical series places hydrogen above copper — meaning copper has a lower tendency to get oxidised than hydrogen does. For a metal to displace hydrogen from an acid, its standard reduction potential must be more negative than $0\ \text{V}$ (or at least less positive than hydrogen’s). Copper’s standard reduction potential for $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ is $+0.34\ \text{V}$ , which is positive. So copper cannot reduce $\text{H}^+$ to $\text{H}_2$ under normal conditions.

Watch out

A common mistake is to think that because copper reacts with concentrated nitric acid (producing NO₂, not H₂), it reacts with all acids. That’s wrong — the reaction with nitric acid is due to the oxidising nature of the nitrate ion, not the acidity. Copper does not liberate hydrogen from any acid.
Statement (ii): In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine.

Manganese in its +7 oxidation state (as in $\text{MnO}_4^-$ ) is very stable with oxygen. Similarly, $\text{MnF}_4$ and $\text{MnF}_3$ exist, though $\text{Mn(VII)}$ fluoride is not stable (fluorine cannot stabilise the +7 state as well as oxygen does). The statement says “with oxygen and fluorine” — it does not claim all higher states are stable with both, but that stable compounds exist. This is correct: $\text{Mn}_2\text{O}_7$ (with oxygen) and $\text{MnF}_4$ (with fluorine) are known.
Statement (iii): $\text{Mn}^{3+}$ and $\text{Co}^{3+}$ are oxidising agents in aqueous solution.

Both ions have a strong tendency to gain an electron and get reduced to a lower oxidation state.
- $\text{Mn}^{3+}$ is unstable in water and readily disproportionates: $2\text{Mn}^{3+} + 2\text{H}_2\text{O} \rightarrow \text{Mn}^{2+} + \text{MnO}_2 + 4\text{H}^+$ . It acts as an oxidising agent. …

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Here are the common mistakes students make on this question, broken down by each option, along with the conceptual fix.

Mistake 1: Assuming Copper behaves like other transition metals

The Trap: Students see "Copper" and think it will react with acids like Zinc or Magnesium to give $H_2$ gas.

The Reality: Copper is below hydrogen in the electrochemical series. It cannot displace $H^+$ from acids.

Why? $E^\circ_{Cu^{2+}/Cu} = +0.34V$ (positive), meaning the reduction of $Cu^{2+}$ is easier than the reduction of $H^+$ .
The correct fact: Copper reacts only with oxidising acids (like conc. $HNO_3$ or hot conc. $H_2SO_4$ ), but these produce $NO_2$ or $SO_2$ , not $H_2$ .
How to avoid: Memorise the reactivity series and the position of Cu. If a metal has a positive standard reduction potential, it will not liberate $H_2$ from dilute acids.

Mistake 2: Forgetting the "Fluorine Exception" for Manganese

The Trap: Students recall that Mn in +7 state (like $MnO_4^-$ ) is a strong oxidising agent, so they assume it cannot form stable compounds with any non-metal.

The Reality: The statement says "with oxygen and fluorine." This is a specific exception.

Why? Oxygen and fluorine are the most electronegative elements. They can stabilise very high oxidation states.
- Oxygen: $Mn_2O_7$ (Mn in +7) is stable (though explosive).
- Fluorine: $MnF_4$ (Mn in +4) and $K_2MnF_6$ (Mn in +4) are stable. Mn in +7 with F is not stable, but the statement says "higher oxidation states" (not necessarily the highest), so +4 qualifies.
How to avoid: Remember the "Electronegativity Rule": High oxidation states are stabilised by highly electronegative atoms (O, F). Do not confuse "oxidising power" with "stability of the compound."

Mistake 3: Confusing Oxidising vs. Reducing Agents

The Trap: Students mix up which species gets oxidised/reduced.

The Logic:

Oxidising agent: Gets reduced itself (its oxidation number decreases). It takes electrons from others.
Reducing agent: Gets oxidised itself (its oxidation number increases). It gives electrons to others.

For Option (iii): $Mn^{3+}$ and $Co^{3+}$

Why they are oxidising agents: They want to gain an electron to become $Mn^{2+}$ and $Co^{2+}$ (more stable $d^5$ and $d^6$ configurations).
Common error: Thinking $Mn^{3+}$ is stable because it is a common ion. It is actually unstable in water and readily reduces.
How to avoid: Check the electronic configuration of the product. If the product has a half-filled ( $d^5$ ) or fully-filled ( $d^{10}$ ) d-subshell, the reactant is likely an oxidising agent.

For Option (iv): $Ti^{2+}$ and $Cr^{2+}$ …

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

KCET 2025Set D-41 markMCQ
$[\text{NiCl}_4]^{2-}$
(a) The complex has tetrahedral geometry
(b) Co-ordination number of Ni is 2 and oxidation state is +4
(c) The complex is $sp^3$ hybridised
(d) It is a high spin complex
(e) The complex is paramagnetic (A) a, c, d and e (B) a, b, d and e (C) b, c, d and e (D) a, b, c and d
›Reveal solutionSolution
Find the oxidation state (+2, so $d^8$ ), note Cl⁻ is a weak-field ligand so no pairing occurs, giving a high-spin $sp^3$ tetrahedral paramagnetic complex — every statement is true except (b), whose CN and oxidation state are both wrong.

Step 1 — Oxidation state and d-configuration of nickel

Let the oxidation state of Ni be $x$ . Chloride carries $-1$ each, and the overall charge is $-2$ :
$x + 4(-1) = -2 \quad\Longrightarrow\quad x = -2 + 4 = +2$
So the metal is $\ce{Ni^{2+}}$ .

Nickel is $Z = 28$ : $\;\ce{Ni} = [\text{Ar}]\,3d^{8}\,4s^{2}$ . Removing the two $4s$ electrons:
$\ce{Ni^{2+}} = [\text{Ar}]\,3d^{8} \qquad (\text{a } d^8 \text{ ion})$
Step 2 — Coordination number

Chloride is a unidentate ligand (one donor atom, Cl). With four of them:
$\text{Coordination number} = 4$
Step 3 — This already settles statement (b)

(b) "Co-ordination number of Ni is 2 and oxidation state is +4"

Both halves are wrong — the CN is 4 (Step 2) and the oxidation state is +2 (Step 1). (b) is FALSE.

This is decisive: every option containing (b) — namely (B), (C) and (D) — is eliminated at once. Only (A) a, c, d and e remains. Let us verify that all four of those statements are indeed true.

Step 4 — Ligand field strength decides everything else

In the spectrochemical series, chloride sits near the weak-field end:
$\ce{I- < Br- < \underline{Cl-} < F- < H2O < NH3 < en < CN- \approx CO}$
$\ce{Cl-}$ is a weak-field ligand, so the splitting energy $\Delta$ it produces is small — smaller than the electron pairing energy $P$ :
$\Delta < P$
When $\Delta < P$ , it costs less energy for an electron to occupy a higher orbital than to pair up in a lower one. Therefore no pairing occurs — the $3d^8$ configuration is left untouched, with its two unpaired electrons intact.

Since the $3d$ orbitals are not vacated, no inner $3d$ orbital is available for hybridisation.

Step 5 — Hybridisation and geometry ⇒ statements (a) and (c)

With the $3d$ set unavailable, Ni²⁺ must use its outer orbitals: one $4s$ and three $4p$ .
$4s + 4p_x + 4p_y + 4p_z \;\longrightarrow\; \textbf{four } sp^3 \textbf{ hybrid orbitals}$
Four $sp^3$ orbitals point to the corners of a tetrahedron ( $109.5^\circ$ ).

(a) "The complex has tetrahedral geometry" — TRUE ✓

(c) "The complex is $sp^3$ hybridised" — TRUE ✓ …
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COMEDK 2025Set 2025-A1 markMCQ
$Q. Larger number of oxidation states are exhibited by the actinoids than those of lanthanoids. The reason is: (A) Lesser energy difference between 5 f and 6 d than between 4 f and 5 d orbitals (B) More energy difference between 5 f and 6 d than between 4 f and 5 d orbitals (C) 4 f orbitals are more diffused than 5 f orbitals (D) Highly reactive nature of the actinoids$
›Reveal solutionSolution
The key idea is that actinoids show more oxidation states because their 5f and 6d orbitals are closer in energy than the 4f and 5d orbitals of lanthanoids, making it easier to involve f-electrons in bonding. The correct option is (A).

The question asks why actinoids exhibit a larger number of oxidation states than lanthanoids. This is a classic comparison in f-block chemistry, rooted in the electronic structure of the two series.

Concept and Intuition

Oxidation states arise when an atom loses electrons. For f-block elements, the electrons lost can come from both the f and d orbitals. The ease of removing f-electrons depends on how tightly they are held, which is related to the energy gap between the f and d orbitals. A smaller gap means f-electrons can be promoted to d orbitals more readily, allowing a wider range of oxidation states. Actinoids (5f series) have a smaller 5f–6d energy difference than lanthanoids (4f–5d), so they can access more oxidation states.

Let’s break it down step by step.

Understand the orbital energy trends

In lanthanoids, the 4f orbitals are deeply buried inside the atom, shielded by outer electrons. The 5d orbitals are at a significantly higher energy. This large 4f–5d energy gap makes it difficult to remove or promote 4f electrons, so lanthanoids typically show only +3 (and occasionally +2 or +4) oxidation states.

In actinoids, the 5f orbitals are less shielded and more extended (diffuse). The 5f and 6d orbitals are much closer in energy. This small energy difference allows 5f electrons to be easily promoted to 6d orbitals or directly lost, enabling a variety of oxidation states (e.g., +3, +4, +5, +6, and even +7 in some cases like neptunium and plutonium).

Evaluate the options

(A) Lesser energy difference between 5f and 6d than between 4f and 5d orbitals — This matches the explanation above.

(B) More energy difference — This would make it harder to involve f-electrons, reducing oxidation states, so incorrect.

(C) 4f orbitals are more diffused than 5f orbitals — Actually, 5f orbitals are more diffused (less tightly held) due to poorer shielding, so this is false. …
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COMEDK 2024Set 2024-E1 markMCQ
$\mathrm{Cr}(\mathrm{VI})$
›Reveal solutionSolution
The question tests knowledge of transition metal chemistry: magnetic properties of manganate vs. permanganate, properties of interstitial compounds, oxidation states of cerium, and stability of Cr(VI) vs. W(VI). Only statement (C) is correct.

Let’s examine each statement carefully, using chemical principles to decide which one is accurate.

Statement (A): “The green manganate ion shows diamagnetic nature but the permanganate ion exhibits paramagnetic nature.”

The manganate ion is $\text{MnO}_4^{2-}$ , where manganese is in the +6 oxidation state. Electronic configuration of Mn in +6: $[\text{Ar}]\,3d^1$ . That’s one unpaired electron → paramagnetic, not diamagnetic.

The permanganate ion is $\text{MnO}_4^{-}$ , with Mn in +7: $[\text{Ar}]\,3d^0$ . No unpaired electrons → diamagnetic.

So the statement gets both magnetic natures backwards. False.

Statement (B): “Interstitial compounds of transition metals have lower melting points than that of pure transition metals and their compounds are chemically reactive.”

Interstitial compounds (e.g., carbides, nitrides, hydrides) form when small atoms like C, N, or H occupy holes in the metal lattice. This usually increases hardness and melting point (often very high), and they are chemically inert (not reactive).

Both claims here are opposite to reality. False.

Statement (C): “Cerium is a lanthanoid metal which exists in a stable oxidation state of +4, besides exhibiting an oxidation state of +3.”

Cerium (Ce, atomic number 58) has the electron configuration $[Xe]\,4f^1\,5d^1\,6s^2$ . The common +3 state arises from losing the 5d and 6s electrons.

The +4 state is also stable because losing one more electron gives a $4f^0$ configuration (empty f-subshell), which is especially stable. Ce(IV) is well-known in compounds like $\text{CeO}_2$ and ceric ammonium nitrate.

This is a textbook fact. True. …
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COMEDK 2024Set 2024-M1 markMCQ
$\mathrm{Cu}$
›Reveal solutionSolution
The key idea is to evaluate each statement about transition-metal chemistry using periodic trends and redox stability; the incorrect statement is (A) because fluorine cannot exceed oxygen in stabilising high oxidation states.

Let’s go through each option carefully, building the reasoning step by step.

Option (A): “The ability of Fluorine to stabilise the higher oxidation states of transition metals exceeds that of Oxygen”

In transition-metal oxyanions (like $\mathrm{CrO_4^{2-}}$ , $\mathrm{MnO_4^-}$ ), oxygen stabilises high oxidation states via strong $\pi$ -bonding (O donates electron density to the metal, reducing its effective charge).

Fluorine is more electronegative than oxygen, but it is a poor $\pi$ -donor (it has no available d-orbitals for back-bonding) and forms weaker multiple bonds.

For example, $\mathrm{Mn_2O_7}$ (Mn in +7) is stable, but $\mathrm{MnF_7}$ does not exist; the highest fluoride of Mn is $\mathrm{MnF_4}$ (+4).

Conclusion: Oxygen stabilises high oxidation states better than fluorine. So statement (A) is false.

Option (B): “Cu(I) compounds in aqueous medium undergo disproportionation reaction”

Disproportionation: $2\mathrm{Cu^+} \rightarrow \mathrm{Cu} + \mathrm{Cu^{2+}}$ .

In water, the standard reduction potentials: $\mathrm{Cu^+ + e^- \rightarrow Cu}$ ( $E^\circ = +0.52\,\mathrm{V}$ ) and $\mathrm{Cu^{2+} + e^- \rightarrow Cu^+}$ ( $E^\circ = +0.16\,\mathrm{V}$ ).

The net cell potential for disproportionation is $E^\circ_{\text{cell}} = 0.52 - 0.16 = +0.36\,\mathrm{V} > 0$ , so it is spontaneous.

Conclusion: Statement (B) is true.

Option (C): “ $\mathrm{Cr^{2+}}$ is a stronger reducing agent than $\mathrm{Fe^{2+}}$ ”

Standard reduction potentials: $\mathrm{Cr^{3+} + e^- \rightarrow Cr^{2+}}$ ( $E^\circ = -0.41\,\mathrm{V}$ ) $\mathrm{Fe^{3+} + e^- \rightarrow Fe^{2+}}$ ( $E^\circ = +0.77\,\mathrm{V}$ )

A more negative reduction potential means the reduced form (here $\mathrm{Cr^{2+}}$ ) is more easily oxidised — i.e., a stronger reducing agent. …
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KCET 2023Set D-21 markMCQ
$(n-1)d^{10}ns^2$
›Reveal solutionSolution
The configuration $(n-1)d^{10}ns^2$ is the ground-state pattern of group-12 elements. Copper is the exception — it is $(n-1)d^{10}ns^1$ , not $ns^2$ — so the pair in which the configuration fails is (D) Cu, Zn.

The pattern $(n-1)d^{10}ns^2$ means a filled $(n-1)d$ subshell together with a filled $ns$ subshell. This is the hallmark of the group-12 elements — Zn, Cd, Hg and Cn — in their ground state. To find the pair that breaks the pattern, we check each element's configuration.

Option (A): Zn, Cd.

Zn ( $Z=30$ ) is $[\text{Ar}]3d^{10}4s^2$ ; Cd ( $Z=48$ ) is $[\text{Kr}]4d^{10}5s^2$ . Both group 12 — both fit the pattern.

Option (B): Cd, Hg.

Cd fits; Hg ( $Z=80$ ) is $[\text{Xe}]4f^{14}5d^{10}6s^2$ . Both group 12 — both fit.

Option (C): Hg, Cn.

Hg fits; Cn ( $Z=112$ ) is the group-12 element $[\text{Rn}]5f^{14}6d^{10}7s^2$ . Both fit.

Option (D): Cu, Zn.

Zn is $3d^{10}4s^2$ (fits), but Cu ( $Z=29$ ) is the classic exception: its ground state is $[\text{Ar}]3d^{10}4s^1$ , not $3d^9 4s^2$ , because a completely filled $d^{10}$ subshell is especially stable and pulls one electron out of $4s$ . So Cu is $(n-1)d^{10}ns^1$ and does not show the $ns^2$ configuration. …
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COMEDK 2023Set 2023-E1 markMCQ
$\mathrm{Mn}^{2+}$
›Reveal solutionSolution

The incorrect statement is (A): fluorine stabilises higher oxidation states because of the high lattice (and bond) enthalpy of its compounds, not the low lattice enthalpy. B, C and D are correct NCERT statements.

Option (A) — incorrect. Fluorine, being small and highly electronegative, forms fluorides with high lattice enthalpy (and high M–F bond enthalpy). It is this high lattice/bond enthalpy that lets fluorine stabilise the highest oxidation states of transition metals. The statement wrongly attributes it to "low lattice enthalpy."

Option (B) — correct. The irregular ionisation enthalpies in this series arise from the stability of the half-filled $d^5$ configurations produced ( $\mathrm{Mn^{2+}}$ , $\mathrm{Fe^{3+}}$ ), as noted in NCERT.

Option (C) — correct. Transition metals form alloys readily because their metallic radii are similar (within ~15%). …
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KCET 2019Set A-11 markMCQ
$Ce (Z = 58)$
›Reveal solutionSolution
The question tests your understanding of lanthanide chemistry, specifically the stability and redox behaviour of cerium. The incorrect statement is (A): $Ce^{4+}$ is an oxidising agent, not a reducing agent.

Concept & Intuition

Cerium is the first element in the lanthanide series (Z=58). Its ground-state electronic configuration is $[Xe]\,4f^1\,5d^1\,6s^2$ . The key to its chemistry lies in the stability of the empty $4f$ subshell ( $4f^0$ ) and the half-filled $4f$ subshell ( $4f^7$ ). For cerium, losing four electrons gives the $Ce^{4+}$ ion with a $[Xe]$ configuration — a noble gas core, which is exceptionally stable. This stability makes $Ce^{4+}$ a strong oxidising agent (it readily accepts electrons to go back to the more common $Ce^{3+}$ state). In contrast, $Ce^{3+}$ has a $[Xe]\,4f^1$ configuration and is the most stable oxidation state for cerium in aqueous solution.

Now let’s examine each statement.

Statement (A): $Ce^{4+}$ is a reducing agent.

A reducing agent is a substance that donates electrons (gets oxidised itself). $Ce^{4+}$ has a strong tendency to gain one electron and become $Ce^{3+}$ (the $4f^1$ configuration is more stable than $4f^0$ in most chemical environments). This means $Ce^{4+}$ is an oxidising agent, not a reducing agent. In fact, $Ce^{4+}$ is a well-known oxidising agent in analytical chemistry (e.g., in cerimetric titrations). So this statement is false.

Statement (B): Atomic size of Ce is more than that of Lu.

This is true. Across the lanthanide series (from Ce, Z=58, to Lu, Z=71), there is a steady decrease in atomic and ionic radii — the lanthanide contraction. The 4f electrons are poorly shielding, so as nuclear charge increases, the electron cloud is pulled inward. Ce is near the beginning of the series, Lu at the end, so Ce has a larger atomic radius than Lu.

Statement (C): Ce in +3 oxidation state is more stable than in +4. …
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Q.Which of the following statements is not correct?

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

Method: Electronic Configuration & Stability Analysis

Steps:

Mistake 1: Assuming Copper behaves like other transition metals

Mistake 2: Forgetting the "Fluorine Exception" for Manganese

Mistake 3: Confusing Oxidising vs. Reducing Agents

Step 1 — Oxidation state and d-configuration of nickel

Step 2 — Coordination number

Step 3 — This already settles statement (b)

Step 4 — Ligand field strength decides everything else

Step 5 — Hybridisation and geometry ⇒ statements (a) and (c)

Q.Which of the following statements is not correct?

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