The halides of transition elements become more covalent with increasing oxidation state of the metal. Why?

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Q. The halides of transition elements become more covalent with increasing oxidation state of the metal. 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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Higher oxidation states in transition metals increase the charge-to-size ratio (ionic potential), which strongly polarises the halide anion’s electron cloud. This increased polarising power shifts the bonding from ionic toward covalent character — the Fajans’ rule explanation.

The question touches on a beautiful pattern in transition metal chemistry: as you oxidise the metal to a higher state, its halides behave less like salts and more like molecular compounds. For example, $\ce{TiCl4}$ is a fuming liquid at room temperature, while $\ce{TiCl3}$ is a solid. The reason lies in how the metal ion’s charge and size change.

1. The core idea: ionic potential

Covalent character in an ionic bond arises when the cation distorts the electron cloud of the anion — a process called polarisation. The ability of a cation to polarise an anion depends on its ionic potential, defined as:

\text{Ionic potential} = \frac{\text{charge}}{\text{radius}} = \frac{Z}{r}

A higher charge and a smaller radius both increase the polarising power. When a transition metal is in a higher oxidation state, two things happen simultaneously:

The positive charge $Z$ increases.
The ionic radius $r$ decreases (because removing electrons reduces electron-electron repulsion and the nuclear pull becomes more effective).

Both changes push the ionic potential sharply upward.

Fajans’ rule: Covalent character in an ionic bond increases with: Higher charge on the cation Smaller size of the cation Larger size of the anion (more easily polarised)

2. Step-by-step reasoning

Step 1: Compare the same metal in two oxidation states.

Take manganese as an example. $\ce{Mn^{2+}}$ has a radius of about $83\ \text{pm}$ , while $\ce{Mn^{7+}}$ has a radius of roughly $46\ \text{pm}$ (these are approximate crystal radii). The charge jumps from $+2$ to $+7$ , and the radius shrinks. The ionic potential goes from:

\frac{2}{83} \approx 0.024 \quad \text{to} \quad \frac{7}{46} \approx 0.152

That’s a six-fold increase. The $\ce{Mn^{7+}}$ ion is an extremely powerful polariser.

Step 2: What happens to the halide anion?

The halide ion (say $\ce{Cl-}$ ) has a diffuse electron cloud. When a highly charged, small cation approaches, it pulls the anion’s electron density toward itself. This distorts the spherical symmetry of the anion, creating a dipole. The bond is no longer purely electrostatic — it acquires a covalent component because electron density is now shared to some extent.

Step 3: The trend across oxidation states.

For any given transition metal, as you go from the lowest to the highest stable oxidation state, the halides show a clear progression:

Metal	Low oxidation state halide	Nature	High oxidation state halide	Nature
Fe	$\ce{FeCl2}$	Ionic solid	$\ce{FeCl3}$	More covalent (low melting solid, sublimes)
Cr	$\ce{CrCl2}$	Ionic solid	$\ce{CrCl3}$	Covalent character (insoluble in water)
Mn	$\ce{MnCl2}$	Ionic, pink solid	$\ce{MnO3Cl}$ (permanganyl chloride)	Covalent, explosive liquid

The higher oxidation state halides are often volatile, soluble in organic solvents, and have lower melting points — all hallmarks of covalent compounds.

Watch out

A common mistake is to think that the metal itself becomes more electronegative in higher oxidation states. That’s not quite right — electronegativity is a property of the element, not the ion. What changes is the polarising power of the cation, which is a function of its charge and size.

3. Why this is especially pronounced for transition metals …

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Common Mistakes: Covalent Character in Transition Metal Halides

Mistake 1: Confusing Covalent Character with Ionic Character

The error: Students often think higher oxidation state means more ionic character, because the metal has a higher positive charge.

Why it's wrong: Higher charge density (charge/size ratio) actually increases polarising power of the cation. A small, highly charged cation distorts the electron cloud of the anion more strongly, pulling electron density toward itself — this is covalent character, not ionic.

How to avoid: Remember Fajan's Rules:

Small cation + large anion → more covalent
High charge on cation → more covalent
High charge on anion → more covalent

Key insight: Higher oxidation state → smaller ionic radius (due to greater nuclear pull on fewer electrons) → higher charge density → more covalent halides.

Mistake 2: Forgetting Lanthanide Contraction Applies Here

The error: Students treat transition metal halides in isolation, not connecting to the lanthanide contraction concept.

Why it matters: For 4d and 5d series (e.g., $W^{6+}$ , $Mo^{6+}$ ), the lanthanide contraction makes 5d elements have nearly identical radii to 4d elements. This means:

$W^{6+}$ has very high charge density
$WF_6$ is highly covalent (low melting point, volatile)
$WCl_6$ is even more covalent

How to avoid: Always connect:

\text{Higher oxidation state} \rightarrow \text{Smaller radius} \rightarrow \text{Higher polarising power} \rightarrow \text{More covalent}

Mistake 3: Thinking All Halides Behave Similarly

The error: Assuming the trend is identical for fluoride, chloride, bromide, and iodide.

The correction: Covalent character increases as:

\text{F}^- < \text{Cl}^- < \text{Br}^- < \text{I}^-

Because larger anions are more easily polarised (softer). So:

$TiCl_4$ is covalent (liquid at room temp)
$TiF_4$ is more ionic (polymerises, higher melting point)

How to avoid: Always specify which halide — and remember: fluoride is the most ionic, iodide is the most covalent for the same metal oxidation state.

Mistake 4: Ignoring the "Why" — Electron Cloud Distortion

The error: Memorising "higher oxidation state = more covalent" without understanding the mechanism. …

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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.The halides of transition elements become more covalent with increasing oxidation state of the metal. 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

1. The core idea: ionic potential

2. Step-by-step reasoning

3. Why this is especially pronounced for transition metals …

Method: Fajan’s Rules for Covalent Character in Ionic Bonds

Steps

Common Mistakes: Covalent Character in Transition Metal Halides

Mistake 1: Confusing Covalent Character with Ionic Character

Mistake 2: Forgetting Lanthanide Contraction Applies Here

Mistake 3: Thinking All Halides Behave Similarly

Mistake 4: Ignoring the "Why" — Electron Cloud Distortion

Concept

Solution

Q.The halides of transition elements become more covalent with increasing oxidation state of the metal. Why?

1. The core idea: ionic potential

2. Step-by-step reasoning

3. Why this is especially pronounced for transition metals …

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