Although fluorine is more electronegative than oxygen, but the ability of oxygen to stabilise higher oxidation

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Q. Although fluorine is more electronegative than oxygen, but the ability of oxygen to stabilise higher oxidation states exceeds that of fluorine. Why?

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Concept understanding — Lanthanide Contraction

Lanthanide Contraction: The Intuition

Imagine you are walking through a dense forest. With every step forward, you push through thick undergrowth. The deeper you go, the more tired you become — each step feels a little harder, and you find yourself hunching forward, your shoulders pulling inward. That inward pull is exactly what happens inside the lanthanide atoms.

The lanthanides are the 14 elements from cerium ( $Ce$ , atomic number 58) to lutetium ( $Lu$ , atomic number 71). As you move from one element to the next, you add one proton to the nucleus and one electron to the atom. The new electron goes into a 4f orbital — a set of orbitals that are shaped like clover leaves and sit deep inside the atom, close to the nucleus.

Here is the key: 4f orbitals are poorly shielded. They do not spread out far from the nucleus, and they do not block the nuclear charge from pulling on the outer electrons. So when you add a proton, the nucleus gets stronger, and the 4f electrons do almost nothing to stop that extra pull. The result? The entire electron cloud — especially the outermost electrons — gets pulled inward. The atom shrinks.

Note

Shielding is the ability of inner electrons to "block" the outer electrons from feeling the full positive charge of the nucleus. Electrons in s and p orbitals shield well; 4f electrons shield very poorly.

The Precise Statement

Lanthanide contraction is the steady and significant decrease in the atomic and ionic radii of the lanthanide elements as atomic number increases from 58 (Ce) to 71 (Lu).

\text{Atomic radius} \propto \frac{1}{Z_{\text{eff}}}

The total contraction across the entire series is about 15–20 picometers — roughly 10–15% of the initial radius. That is a substantial shrinkage for a single row of the periodic table.

Why It Matters

This contraction has two enormous consequences in chemistry:

Similarity of post-lanthanide elements: After lutetium, the next elements are hafnium ( $Hf$ , 72), tantalum ( $Ta$ , 73), and tungsten ( $W$ , 74). Because the lanthanide contraction has made the atoms so small, these elements have almost identical atomic and ionic radii to their counterparts directly above them in the periodic table — zirconium ( $Zr$ ), niobium ( $Nb$ ), and molybdenum ( $Mo$ ). This is why zirconium and hafnium are chemically almost inseparable — they are the same size.
Difficulty in separating lanthanides: All lanthanide ions ( $Ln^{3+}$ ) have nearly identical chemical properties because their radii change so gradually. Separating them requires hundreds of repeated steps (ion-exchange chromatography, solvent extraction) — a painstaking process that was a major challenge in early nuclear chemistry.

Watch out

A common mistake is to think lanthanide contraction means the atoms get smaller because the 4f orbitals are "full" or because of some repulsion effect. It is purely due to poor shielding of the 4f electrons, which lets the nuclear charge pull everything inward.

The Numbers (for reference)

Element	Atomic Number	Ionic Radius ( $Ln^{3+}$ , pm)
Ce	58	103.4
Pr	59	101.3
Nd	60	99.5

Why this formula?

Lanthanide Contraction: Why It Happens

The Lanthanide Contraction is the steady decrease in atomic and ionic radii of the lanthanide elements (Ce to Lu) as atomic number increases. The key observation: the radii shrink by about 1–2 pm per element, despite adding electrons to the 4f subshell.

The Core Question

Why does adding electrons not increase the size, but instead decrease it?

The Formula That Governs It

The effective nuclear charge ( $Z_{\text{eff}}$ ) experienced by an electron is:

Z_{\text{eff}} = Z - S

Where:

$Z$ = atomic number (protons in nucleus)
$S$ = shielding constant (screening by inner electrons)

The key formula for the trend in ionic radii ( $r$ ) across the lanthanides is:

r \propto \frac{n^2}{Z_{\text{eff}}}

Where $n$ is the principal quantum number of the outermost electron (here, $n=6$ for the 6s orbital).

The Derivation: Step by Step

1. What happens when you add a proton and an electron?

Each lanthanide adds:

+1 proton to the nucleus (increases $Z$ by 1)
+1 electron to the 4f subshell

2. The 4f orbital is "penetrating" but poorly shielding

The 4f orbital has a radial distribution that peaks close to the nucleus (inside the 5s and 5p shells).
However, 4f electrons are very poor at shielding the outer 6s electrons from the nuclear charge.

Why?

The 4f orbital is diffuse and deeply buried — it does not effectively screen the outer electrons because:

Its shape (complex, multi-lobed) means it doesn't occupy the space between the nucleus and the 6s electrons efficiently.
The 4f electrons are inside the 5s/5p shells, so they don't block the nuclear pull on the 6s electrons.

3. The net effect on $Z_{\text{eff}}$

When you add one proton ( $\Delta Z = +1$ ) and one 4f electron ( $\Delta S \approx 0.85$ to 0.95), the change in effective nuclear charge is:

\Delta Z_{\text{eff}} \approx +1 - 0.85 = +0.15 \text{ to } +0.05

Result: $Z_{\text{eff}}$ increases slightly with each element.

4. How this shrinks the radius

From the formula $r \propto \frac{n^2}{Z_{\text{eff}}}$ :

$n$ (the principal quantum number of the 6s orbital) stays constant at 6.
$Z_{\text{eff}}$ increases.
Therefore, $r$ decreases. …

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The key is that oxygen can form strong $\pi$ bonds (via its lone pairs) with high-oxidation-state metal centres, while fluorine cannot — this $\pi$ -donation stabilises the metal's high charge, making oxygen better at stabilising higher oxidation states despite its lower electronegativity.

This is a classic question that trips up many students because it seems to contradict the basic trend: if fluorine is the most electronegative element, shouldn't it be the best at pulling electrons away and stabilising a metal in a high oxidation state? The answer lies in a deeper factor — bond multiplicity and orbital availability.

Electronegativity measures an atom's ability to attract electrons in a covalent bond. But stabilising a high oxidation state isn't just about pulling electron density towards the ligand — it's also about the ligand donating electron density to the metal to relieve its high positive charge. A metal in a +6 or +7 state is extremely electron-deficient; it desperately needs electron density from its surroundings. The ligand that can donate more effectively will better stabilise that state.

Here, oxygen has a decisive advantage: it has two lone pairs available for $\pi$ back-donation into empty $d$ orbitals on the metal. Fluorine, with only one lone pair and a much higher tendency to hold onto its electrons (due to its extreme electronegativity), is a poor $\pi$ donor. Let's break this down step by step.

The problem with fluorine's approach.

Fluorine is small, highly electronegative, and forms strong $\sigma$ bonds. In a compound like $\text{OF}_6$ (which doesn't exist stably), fluorine would pull electron density away from oxygen via $\sigma$ bonds. But for a metal in a high oxidation state — say, $\text{MnO}_4^-$ (Mn in +7) — the metal needs more than just $\sigma$ withdrawal. It needs the ligands to share some of their own electron density to reduce the effective positive charge. Fluorine's lone pairs are held too tightly; they are not available for $\pi$ bonding. So fluorine can only form single $\sigma$ bonds, leaving the metal's high charge largely unscreened.
Oxygen's $\pi$ donation — the game changer.

Oxygen, in contrast, has two lone pairs in $p$ orbitals. When bonded to a transition metal in a high oxidation state, these lone pairs can overlap with empty $d$ orbitals on the metal (like $d_{xz}$ , $d_{yz}$ , $d_{xy}$ ) to form $\pi$ bonds. This $\pi$ donation pushes electron density onto the metal, stabilising the high charge. The classic example is the permanganate ion, $\text{MnO}_4^-$ : each Mn–O bond has significant $\pi$ character, making the bond order close to 2. This $\pi$ bonding is what makes $\text{Mn(VII)}$ stable in $\text{MnO}_4^-$ , whereas $\text{MnF}_7$ is unknown.
The $\pi$ bond strength argument.

The stabilisation energy from $\pi$ bonding is substantial. For a metal in a high oxidation state, the $d$ orbitals are contracted and low in energy, making them good acceptors for $\pi$ donation from oxygen's filled $p$ orbitals. Fluorine's $p$ orbitals are also low in energy (due to high electronegativity), but the key difference is that fluorine's lone pairs are non-bonding in the $\sigma$ framework and are not easily polarised towards the metal. Oxygen's lone pairs are more polarisable and form stronger $\pi$ overlaps.

Tip

A quick way to remember this: Oxygen is a better $\pi$ donor; fluorine is a better $\sigma$ withdrawer. For high oxidation states, $\pi$ donation matters more.

Experimental evidence — the oxyanions. …

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Here are the common mistakes students make when answering this question, along with the conceptual corrections to avoid them.

Mistake 1: Confusing Electronegativity with Oxidising Power

The Mistake: Students argue that because fluorine is the most electronegative element, it must be the best at stabilising high oxidation states. They assume electronegativity directly translates to "stabilising power."
Why it’s wrong: Electronegativity measures an atom's ability to attract shared electrons in a covalent bond. However, stabilising a high oxidation state (e.g., +7 in $\text{Mn}_2\text{O}_7$ ) requires the ability to accept and hold a large number of electrons from the central metal atom. This is about lattice energy (in ionic compounds) or multiple bond formation (in covalent compounds), not just pulling power.
How to avoid: Remember the key difference:
- Fluorine is a strong oxidising agent (it takes electrons away).
- Oxygen is better at sharing electrons to form strong $\pi$ -bonds (like $\text{M=O}$ ), which delocalises the charge and stabilises the high oxidation state.

Mistake 2: Ignoring the Role of Lanthanide Contraction

The Mistake: Students answer the question without mentioning lanthanide contraction at all, or they mention it but only as a fact about atomic size, failing to connect it to stabilisation.
Why it’s wrong: The question is specifically about higher oxidation states in transition metals (like $\text{CrO}_3$ , $\text{Mn}_2\text{O}_7$ , $\text{OsO}_4$ ). Lanthanide contraction causes the atomic radii of 4d and 5d series elements to be very similar. This similarity makes the 5d orbitals more accessible and better at overlapping with oxygen’s 2p orbitals to form strong $\pi$ -bonds.
How to avoid: Always link lanthanide contraction to orbital overlap. Write:

"Due to lanthanide contraction, the 5d orbitals of heavier transition metals (like Os, Re) are not much larger than the 4d orbitals. This allows for effective overlap with the small 2p orbitals of oxygen, forming strong $\text{M=O}$ double bonds that stabilise high oxidation states."

Mistake 3: Forgetting the "Size Mismatch" with Fluorine

The Mistake: Students say "fluorine is too small" but don't explain why that is a problem for stabilising high oxidation states.
Why it’s wrong: A small atom like fluorine cannot effectively accommodate the large number of electrons from a highly charged metal ion. The electron-electron repulsion in the small fluorine atom becomes too high, making the compound unstable.
How to avoid: Use the concept of steric hindrance and electron repulsion:

"Fluorine is the smallest halogen. To stabilise a +7 oxidation state (e.g., $\text{MF}_7$ ), seven large fluorine atoms would crowd around a small metal centre, causing severe steric repulsion and making the compound unstable. Oxygen, being smaller and able to form double bonds, avoids this crowding."

Mistake 4: Not Mentioning the "Double Bond" Advantage

The Mistake: Students only talk about ionic bonding (lattice energy) and forget the covalent aspect.
Why it’s wrong: High oxidation states in transition metals are almost always stabilised by covalent bonding, especially $\pi$ -bonding. Oxygen can form $\text{M=O}$ double bonds, which are very strong and delocalise the positive charge on the metal. Fluorine can only form single $\text{M-F}$ bonds.
How to avoid: Explicitly state: …

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

COMEDK 2025Set 2025-A1 markMCQ
$\mathrm{Ce}^{4+}$
›Reveal solutionSolution
The question asks which statements about lanthanoids are incorrect. After checking each statement, the incorrect ones are C and D, so the correct option is (A).

Concept & Intuition

Lanthanoids are the 14 elements from Ce (58) to Lu (71) where the 4f subshell is progressively filled. Their chemistry is dominated by the +3 oxidation state, but some elements show +4 or +2 states due to stability of empty, half-filled, or fully-filled f-subshells. Magnetic properties depend on unpaired electrons. Their atomic sizes show the “lanthanoid contraction” — a steady decrease across the series — which makes post-lanthanoid transition metals (like Hf, Ta, W) have nearly identical atomic radii to their 4d counterparts above them. Also, lanthanoid hydroxides are not water-soluble, and the general electronic configuration is often written with a possible 5d¹ electron for some elements (like La, Gd, Lu). Let’s examine each statement.

Step-by-step analysis

Statement A: $\mathrm{Ce}^{4+}$ is diamagnetic while $\mathrm{Sm}^{3+}$ is paramagnetic.

Ce (atomic number 58) has configuration $[\mathrm{Xe}]\,4f^1 5d^1 6s^2$ . Ce⁴⁺ loses all 4f, 5d, and 6s electrons → $[\mathrm{Xe}]$ (no unpaired electrons) → diamagnetic.

Sm (atomic number 62) has $[\mathrm{Xe}]\,4f^6 6s^2$ . Sm³⁺ loses 6s² and one 4f → $4f^5$ . With 5 unpaired electrons (Hund’s rule), it is paramagnetic.

So statement A is correct.

Statement B: The atomic size of the transition metals having atomic number greater than 71 are very close to that of the elements above them.

Elements with Z > 71 are the 5d transition metals (Hf, Ta, W, etc.). Their 4f counterparts (Zr, Nb, Mo, etc.) are directly above them in the periodic table.

Due to lanthanoid contraction (poor shielding by 4f electrons), the atomic radii of 5d metals are nearly equal to those of the 4d metals above them.

This is a well-known fact. So statement B is correct.

Statement C: Lanthanoids react with hot water forming water soluble $\mathrm{Ln}(\mathrm{OH})_3$ with the liberation of $\mathrm{O}_2$ . …
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COMEDK 2025Set 2025-E1 markMCQ
$(\mathrm{Zn}, \mathrm{Ca}, \mathrm{Fe}, \mathrm{Cu})$
›Reveal solutionSolution
The question tests knowledge of lanthanoid chemistry: chemical similarity to early lanthanoids, the stability of half-filled 4f⁷, colourless ions, and diamagnetic +2 states. The correct matching is A: Ca, B: Eu²⁺, C: Lu³⁺, D: Yb — option (D).

Concept & Intuition

Lanthanoids (elements 58–71) have similar chemistry due to the gradual filling of 4f orbitals, but subtle differences arise from electronic configurations, oxidation states, and magnetic properties.

Part A: The first few lanthanoids (La–Nd) are highly electropositive and reactive, resembling the alkaline earth metal Ca more than transition metals like Zn, Fe, or Cu.

Part B: A half-filled 4f⁷ subshell is exceptionally stable. Eu²⁺ has [Xe]4f⁷, but it readily loses one electron to become Eu³⁺ (still 4f⁷? No — Eu³⁺ is 4f⁶, but the driving force is the stability of the +3 state common to lanthanoids; Eu²⁺ is a strong reducing agent because it wants to reach +3).

Part C: Colour in lanthanoid ions arises from f–f transitions. Ions with empty (4f⁰), half-filled (4f⁷), or fully filled (4f¹⁴) subshells have no such transitions and are colourless. Lu³⁺ is 4f¹⁴ — colourless.

Part D: A diamagnetic +2 ion must have all electrons paired. Yb²⁺ has [Xe]4f¹⁴ — completely filled, hence diamagnetic and stable in +2 state.

Step-by-step reasoning

Part A: Chemical reactivity similar to early lanthanoids

Early lanthanoids (La, Ce, Pr, Nd) are highly electropositive, react readily with water and acids, and typically exhibit +3 oxidation state. Among the options, Ca (an alkaline earth metal) shares this high reactivity and electropositivity. Zn, Fe, and Cu are less reactive and have different chemical behaviour.

→ So A should be Ca.

Part B: Stable 4f⁷ configuration but acts as a strong reducing agent to M³⁺

Eu²⁺ has the configuration [Xe]4f⁷ — half-filled, stable. However, the standard reduction potential for Eu³⁺/Eu²⁺ is about –0.35 V, meaning Eu²⁺ is easily oxidised to Eu³⁺ (strong reducing agent). Ce²⁺, Pr²⁺, Dy²⁺ are less common and do not have the 4f⁷ stability.

→ So B should be Eu²⁺.

Part C: Colourless ion

Colour in lanthanoid ions is due to f–f transitions, which require partially filled 4f orbitals.

Tm³⁺: 4f¹² — coloured.

Lu³⁺: 4f¹⁴ — fully filled, no f–f transitions, colourless.

Gd³⁺: 4f⁷ — half-filled, also colourless in theory, but Lu³⁺ is more reliably colourless and is the classic example.

Sm³⁺: 4f⁵ — coloured. → So C should be Lu³⁺. …
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KCET 2024Set B-21 markMCQ
$+2$
›Reveal solutionSolution
The incorrect statement is the one about the colour origin: lanthanoid ion colours arise from f–f transitions, not d–d transitions. So option (D) is wrong.

The question tests your understanding of the lanthanoid series — their physical nature, variable oxidation states, common uses, and the origin of their colours. Each option touches a distinct property, so we need to check them one by one against known facts.

Option (A): Lanthanoids are silvery white soft metals

This is correct. All lanthanoids (elements 57–71, except perhaps promethium which is radioactive and less studied) are silvery-white, relatively soft metals. They tarnish quickly in air, but their fresh surfaces have that characteristic appearance. Softness increases across the series — they can be cut with a knife, like sodium, though they are harder than alkali metals.

Option (B): Samarium shows $+2$ oxidation state

This is correct. Samarium (Sm, atomic number 62) has the electronic configuration $[\ce{Xe}]\,4f^6\,6s^2$ . By losing the two 6s electrons, it reaches $+2$ (Sm $^{2+}$ ). The $4f^6$ configuration in Sm $^{2+}$ is half-filled (since f orbitals can hold 14 electrons, 7 is half-filled; 6 is one short, but still relatively stable). More importantly, the $+2$ state is stabilised by the proximity to the half-filled $4f^7$ configuration of Eu $^{2+}$ . In practice, Sm $^{2+}$ is known in compounds like SmI $_2$ and SmCl $_2$ , though it is less stable than the common $+3$ state.

Option (C): $\ce{Ce^{4+}}$ solutions are widely used as oxidising agents in titrimetric analysis

This is correct. Cerium(IV) (Ce $^{4+}$ ) is a strong oxidising agent — it gets reduced to Ce $^{3+}$ (with a standard reduction potential of about +1.72 V in acidic medium). Ce $^{4+}$ solutions are stable, have a sharp colour change (yellow to colourless), and are used in redox titrations, especially for determining iron(II), oxalates, and other reducing agents. This is a standard application in analytical chemistry.

Option (D): Colour of Lanthanoid ion in solution is due to d–d transition …
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COMEDK 2024Set 2024-A1 markMCQ
$Q. Consider the following statements in respect of lanthanides, which of the statements are incorrect?$
(i) $\mathrm{La}(\mathrm{OH})_3$ is least basic among the hydroxides of lanthanides
(ii) The lanthanide ions $\mathrm{Yb}^{2+}, \mathrm{Lu}^{3+}$ and $\mathrm{Ce}^{4+}$ are diamagnetic in nature.
(iii) $\mathrm{Ce}^{4+}$ can act as an oxidising agent
(iv) $\operatorname{Ln}$ (III) compounds are generally colourless
(v) Ionic radii of $\mathrm{Ce}^{3+}$ is greater than $\mathrm{Yb}^{3+}$ (A) (i),
(ii) and
(iii) (B)
(i) and
(iv) (C) (iii),
(iv) and
(v) (D)
(iii) and (iv)
›Reveal solutionSolution
The key idea is to evaluate each statement about lanthanide properties (basicity trends, magnetism, redox behavior, color, and ionic radii) against known periodic trends. The incorrect statements are (i) and (iv), so the correct option is (B).

Concept and Intuition

Lanthanides are the 4f-block elements (La to Lu). Their chemistry is dominated by the +3 oxidation state, but some ions show +2 or +4 states due to stability of empty, half-filled, or fully filled 4f subshells. Basicity of hydroxides decreases across the series as ionic radius decreases (lanthanide contraction). Magnetic behavior depends on unpaired 4f electrons; diamagnetic means all electrons paired. Color arises from f–f transitions; Ln(III) ions with no unpaired f-electrons (like La³⁺, Lu³⁺) are colorless, but most are colored. Ionic radii decrease from Ce³⁺ to Yb³⁺ due to lanthanide contraction.

Step-by-step reasoning

Statement (i): "La(OH)₃ is least basic among the hydroxides of lanthanides"

Basicity of lanthanide hydroxides decreases as the ionic radius decreases (smaller cation polarizes the OH bond more, making it more acidic).

La³⁺ has the largest ionic radius among Ln³⁺ ions, so La(OH)₃ is the most basic, not the least.

Therefore, (i) is incorrect.

Statement (ii): "Yb²⁺, Lu³⁺, and Ce⁴⁺ are diamagnetic"

Yb²⁺: Yb (atomic number 70) has electron configuration [Xe]4f¹⁴. Yb²⁺ loses two electrons, still 4f¹⁴ — all f-orbitals are fully filled, so no unpaired electrons → diamagnetic.

Lu³⁺: Lu (71) is [Xe]4f¹⁴5d¹6s²; Lu³⁺ loses three electrons → [Xe]4f¹⁴, fully filled → diamagnetic.

Ce⁴⁺: Ce (58) is [Xe]4f¹5d¹6s²; Ce⁴⁺ loses four electrons → [Xe] (no f-electrons) → diamagnetic.

Thus, (ii) is correct.

Statement (iii): "Ce⁴⁺ can act as an oxidizing agent"

Ce⁴⁺ has a strong tendency to gain an electron to become Ce³⁺ (which has a stable half-filled 4f¹ configuration? Actually Ce³⁺ is 4f¹, but the reduction potential Ce⁴⁺/Ce³⁺ is about +1.72 V in acidic medium, making Ce⁴⁺ a strong oxidizing agent).

Therefore, (iii) is correct.

Statement (iv): "Ln(III) compounds are generally colorless" …
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COMEDK 2024Set 2024-M1 markMCQ
$\mathrm{La}(\mathrm{OH})_3$
›Reveal solutionSolution
The key idea is that basicity of lanthanide hydroxides decreases across the series (La(OH)₃ most basic, Lu(OH)₃ least basic), Mn₂O₇ is acidic, and Fe₃H is an interstitial compound. The correct matching is A→S, B→P, C→Q, D→R, which corresponds to option (A).

Concept & Intuition

This question tests two separate ideas: (1) the trend in basicity of lanthanide hydroxides, and (2) the classification of oxides and hydrides. For the lanthanides, as atomic number increases, the ionic radius decreases (lanthanide contraction), making the M–OH bond stronger and harder to break — so basicity decreases. La³⁺ is the largest, so La(OH)₃ is the most basic; Lu³⁺ is the smallest, so Lu(OH)₃ is the least basic. Mn₂O₇ is a well-known acidic oxide (it’s the anhydride of permanganic acid). Fe₃H is a metallic hydride where hydrogen occupies interstitial sites in the iron lattice — hence an interstitial compound.

Step-by-step reasoning

Identify the nature of La(OH)₃ and Lu(OH)₃

Both are hydroxides of lanthanides. Basicity of lanthanide hydroxides decreases from La to Lu due to lanthanide contraction. La³⁺ has the largest ionic radius, so La–OH bond is weakest → most basic. Lu³⁺ has the smallest radius → least basic.

→ La(OH)₃ = Most basic (S)

→ Lu(OH)₃ = Least basic (Q)

Identify the nature of Mn₂O₇

Mn in +7 oxidation state forms an oxide that is strongly acidic. Mn₂O₇ reacts with water to give HMnO₄ (permanganic acid), a strong acid.

→ Mn₂O₇ = Acidic in nature (P)

Identify the nature of Fe₃H …
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COMEDK 2021Set 2021-B1 markMCQ
$Q. Lanthanides are a group of 14 elements which are metals. Identify the correct statement from among the 4 statements given below: (A) Shielding power of 4f electrons is quite strong. (B) As a result of Lanthanide contraction, the elements of the second and third transition series resemble each other in their chemical properties. (C) Due to Lanthanide contraction, the size of Lanthanoid ions increases regularly with increase in atomic number. (D) It is very easy to separate the Lanthanide elements from each other and obtain them in the pure state.$
›Reveal solutionSolution
[!TLDR]

Lanthanide contraction makes the 2nd and 3rd transition series resemble each other, so statement (B) is the correct one.

Concept

The 4f electrons shield the nuclear charge very poorly, so as atomic number rises across the lanthanoids the effective nuclear charge felt by outer electrons grows and the size of the atoms/ions shrinks steadily. This is the lanthanide contraction (CBSE/NCERT Class 12, d- and f-block elements).

Solution

Check each statement:

(A) 4f electrons have poor (diffuse) shielding power, not strong — false.

(B) The contraction cancels the expected size increase down a group, so pairs like Zr/Hf and Nb/Ta have nearly identical sizes and therefore very similar chemistry — true. …
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KCET 2020Set A-11 markMCQ
$\text{K}_2\text{O}_3$
›Reveal solutionSolution
Potassium forms oxides in which it exists as $\text{K}^+$ ions, and the only stable oxidation states of oxygen in these compounds are $-2$ (oxide), $-1$ (peroxide), and $-\frac12$ (superoxide). The formula $\text{K}_2\text{O}_3$ would require oxygen in an oxidation state of $-\frac43$ , which is not possible — so it does not exist.

The key to this question lies in understanding the oxidation states that oxygen can take in its compounds with alkali metals. Potassium, being a highly electropositive metal, always forms $\text{K}^+$ ions. The oxygen species present in the solid then determines the formula.

Recall the common oxides of potassium.

Potassium reacts with oxygen to form three well-known compounds:

Normal oxide: $\text{K}_2\text{O}$ — contains $\text{O}^{2-}$ (oxide ion, oxidation state $-2$ ).

Peroxide: $\text{K}_2\text{O}_2$ — contains $\text{O}_2^{2-}$ (peroxide ion, oxidation state $-1$ per oxygen).

Superoxide: $\text{KO}_2$ — contains $\text{O}_2^-$ (superoxide ion, oxidation state $-\frac12$ per oxygen). These are all stable and well-characterised.

Check the oxidation state of oxygen in $\text{K}_2\text{O}_3$ .

Let the oxidation state of oxygen be $x$ . Since each K is $+1$ , the total positive charge is $2 \times (+1) = +2$ . For a neutral compound:

$2(+1) + 3x = 0 \quad \Rightarrow \quad 3x = -2 \quad \Rightarrow \quad x = -\frac{2}{3}.$
This would mean oxygen exists in an average oxidation state of $-\frac{2}{3}$ , which is not a known stable oxygen species. Oxygen in ionic compounds only appears as $\text{O}^{2-}$ , $\text{O}_2^{2-}$ , $\text{O}_2^-$ , or (rarely) $\text{O}^-$ — never as a fractional or $-\frac{2}{3}$ state.

Why the other options are valid.

$\text{K}_2\text{O}$ : oxygen is $-2$ , perfectly normal. …
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Q.Although fluorine is more electronegative than oxygen, but the ability of oxygen to stabilise higher oxidation states exceeds that of fluorine. Why?

Concept understanding — Lanthanide Contraction

Lanthanide Contraction: The Intuition

The Precise Statement

Why It Matters

The Numbers (for reference)

Why this formula?

Lanthanide Contraction: Why It Happens

The Core Question

The Formula That Governs It

The Derivation: Step by Step

1. What happens when you add a proton and an electron?

2. The 4f orbital is "penetrating" but poorly shielding

3. The net effect on $Z_{\text{eff}}$

4. How this shrinks the radius

Method: Electronic Configuration & Lattice Energy Analysis

Step 1: Recall the Lanthanide Contraction

Step 2: Compare the nature of O²⁻ and F⁻ ions

Step 3: Apply Lattice Energy concept

Step 4: Compare stabilization of high oxidation states

Mistake 1: Confusing Electronegativity with Oxidising Power

Mistake 2: Ignoring the Role of Lanthanide Contraction

Mistake 3: Forgetting the "Size Mismatch" with Fluorine

Mistake 4: Not Mentioning the "Double Bond" Advantage

Concept

Solution

Q.Although fluorine is more electronegative than oxygen, but the ability of oxygen to stabilise higher oxidation states exceeds that of fluorine. Why?

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