Q.Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?

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

1. 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.

2. 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.

3. Statement C: Lanthanoids react with hot water forming water soluble $\mathrm{Ln}(\mathrm{OH})_3$ with the liberation of $\mathrm{O}_2$.

   • Lanthanoids do react with hot water to give $\mathrm{Ln}(\mathrm{OH})_3$ and hydrogen gas ($\mathrm{H}_2$), not oxygen.
   • Example: $2\mathrm{Eu} + 6\mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{Eu}(\mathrm{OH})_3 + 3\mathrm{H}_2$
   • Also, $\mathrm{Ln}(\mathrm{OH})_3$ are insoluble in water (like most metal hydroxides).
   • So statement C is incorrect on two counts: wrong gas and wrong solubility.

4. Statement D: The general electronic configuration of Lanthanoids is $(n-2) \mathrm{f}^{1-14} 5 \mathrm{~d}^0 6 \mathrm{~s}^2$, where $n=6$.

   • For lanthanoids, $n=6$ (6th period), so $(n-2)=4$. The configuration is $4f^{1-14} 5d^{0-1} 6s^2$.
   • The statement says $5d^0$ always, but exceptions exist:
     • La (57): $5d^1 6s^2$ (no 4f electron)
     • Gd (64): $4f^7 5d^1 6s^2$ (half-filled f-shell stability)
     • Lu (71): $4f^{14} 5d^1 6s^2$
   • So the $5d^0$ part is not universally true. Hence statement D is incorrect.

Conclusion: Incorrect statements are C and 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

1. 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.

2. 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⁺.

3. 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⁺.

4. Part D: Stable +2 oxidation state and diamagnetic

   A diamagnetic ion has no unpaired electrons.

   • Ce: common +3, +4; +2 is rare and paramagnetic (4f²).
   • Sm: +2 exists but Sm²⁺ is 4f⁶ — paramagnetic (6 unpaired).
   • Ho: +2 is unstable, Ho²⁺ is 4f¹¹ — paramagnetic.
   • Yb: Yb²⁺ is [Xe]4f¹⁴ — fully filled, diamagnetic, and stable in +2 state (e.g., YbCl₂). → So D should be Yb.

5. Match with options

   • (A) A: Cu ✗, B: Dy²⁺ ✗, C: Sm³⁺ ✗, D: Ho ✗
   • (B) A: Zn ✗, B: Ce²⁺ ✗, C: Gd³⁺ ✗, D: Sm ✗
   • (C) A: Fe ✗, B: Pr²⁺ ✗, C: Tm³⁺ ✗, D: Ce ✗
   • (D) A: Ca ✓, B: Eu²⁺ ✓, C: Lu³⁺ ✓, D: Yb ✓

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.

1. 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.

2. 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.

3. 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.

4. Option (D): Colour of Lanthanoid ion in solution is due to d–d transition

   This is incorrect. Lanthanoid ions have partially filled $4f$ orbitals. The colours arise from f–f transitions — electrons in the $4f$ subshell absorb visible light and get excited to higher $4f$ levels. These transitions are Laporte-forbidden (f→f is parity-forbidden) but become weakly allowed due to vibronic coupling, giving pale colours. In contrast, d–d transitions occur in transition metal ions (d-block elements), where the d-orbitals split in the ligand field. Lanthanoids are f-block elements; their f-orbitals are deeply buried and do not split significantly by ligand fields, so the colour mechanism is fundamentally different.

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

1. 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.

2. 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.

3. 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.

4. Statement (iv): "Ln(III) compounds are generally colorless"

   • Most Ln³⁺ ions have unpaired 4f electrons (e.g., Pr³⁺, Nd³⁺, Sm³⁺, Eu³⁺, etc.) and exhibit characteristic colors due to f–f transitions. Only La³⁺ (4f⁰) and Lu³⁺ (4f¹⁴) are colorless.
   • So "generally colorless" is false; they are generally colored.
   • Thus, (iv) is incorrect.

5. Statement (v): "Ionic radii of Ce³⁺ is greater than Yb³⁺"

   • Lanthanide contraction: ionic radii decrease steadily from La³⁺ to Lu³⁺. Ce (58) comes before Yb (70), so Ce³⁺ is larger than Yb³⁺.
   • Hence, (v) is correct.

Summary of correctness:

• Incorrect: (i) and (iv)
• Correct: (ii), (iii), (v)

This matches option (B).

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

1. 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)

2. 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)

3. Identify the nature of Fe₃H

   Fe₃H is a non-stoichiometric hydride where hydrogen atoms occupy interstitial sites in the iron metal lattice. It is not a salt or covalent hydride; it’s a classic example of an interstitial compound.

   → Fe₃H = Interstitial compound (R)

4. Match the pairs

   • A (La(OH)₃) → S
   • B (Mn₂O₇) → P
   • C (Lu(OH)₃) → Q
   • D (Fe₃H) → R

   This matches option (A).

[!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.
• (C) Lanthanoid ion size decreases regularly with atomic number, not increases — false.
• (D) Because all Ln$^{3+}$ ions are so similar in size, separating them is notoriously difficult — false.

Only (B) is correct.

[!ANSWER]

(B) As a result of lanthanide contraction, the elements of the second and third transition series resemble each other in their chemical properties.

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.

1. 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.

2. 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:

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.

3. Why the other options are valid.
   • $\text{K}_2\text{O}$: oxygen is $-2$, perfectly normal.
   • $\text{KO}_2$: oxygen is $-\frac12$ (superoxide), stable for large alkali metals like K, Rb, Cs.
   • $\text{K}_2\text{O}_2$: oxygen is $-1$ (peroxide), also stable.